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CONTENTS

CHAPTER SKELETONS WITH EXAMS iI

a.apter 1 MATHEMATICAL INTRODUCTION I 1.1 PI.an.or VCC1Qf1, Scientific NoIa!ion .• rod Unit. I 1.2 Thrtt- Dimcnsional Vec!o,,; Dot and Croso ProductS

a..pltt' 1 EQUILIBRIUM OF CONCURRENT FORCES 1.1 Ropu, KI>Q!<, and Fri<tionlel6 Pulleys I 2.2 Fri<1ion and ]ncUne<l P1ant$ f 2.3 G.aphic:al and OdIe. Problems

CIaapIcr 3 KINEMATICS IN ONE DIMENSION 3. 1 Dim.miens and Units; ConOlant·Aca:leration Problems

Chpter 4 NEWfON'S LAWS OF MOTION 4.1 Force. Mass. ~nd Acceleration I 4.2 Friction; Inclined Plane.; Vector Notation I 4.3 Tw(»()bjea and Other Prob/tlll$

CIuoptcr 5 MOTION IN A PLANE I 5.1 Projectile Motioo I S.2 Relati~e Mohl)d

Cupter Ii MOnON IN A PLANE II 6.1 Circula. Motion; Centripetal f orce I 6.2 La .. of Vo;v.",,) Gravitation ; Satellite Motioft I 6.3 General Motion in I Pl ....

a..pttt'1 WORK AND ENERGY 7.1 Work I)Qnc b~ . Fo"", f 7.2 Work, Kine!;" Energy. aDd POIentil1 E.lcrgy I 7.3 Cons.c,valioo of Mechanical Enerv I 1.4 Additional Pmbkllll

Cbapler I POWER AND SIMPLE MACHINES 8.1 Power I 8.2 Simple Machine>

CIIaptn 9 IMPULSE AND MOMENTUM 9.1 Elemenllry Problerm I 9.2 Elallic CoIlisionl I 9.3 1""1,,,'ic Colli ....... and Balli.lie PenJulu .... I 9.-4 CoJli$iQns in Two Dimen ....... I 9.5 Recoil and Reac\ion I 9.6 Cenle' of MOM (""" abo Chap. 10)

OUopter 10 STATICS OF RIGID BODIES 10.1 Equilibrium of Rigid Bodi<1 f 10.2 ~n'er of MOM (Cent .. of Gravity)

ClaapI:H 11 ROTATIONAL MOTION I: KINEMATICS AND DYNAMICS 11.1 Anguli, MOIion and To' que I 11.2 ROIliional Kinematics I I U TOIq"" and ROlatiQn f t 1.4 Mo .... nl of Inertia I 11 .5 T",,,,lal;o,w_R<>to,ional Rclalion>llipo I 11 .6 Problem. Involving Cordi. Around Cylin<len. RollinS Ot>jCClS. ele.

CIuoptH 12 ROTATIONAL MOTION II : KINETIC ENERGY. ANGULAR IMPULSE.

"

"

" .. 1Il

IJ6

'"

17'

'"

ANGULAR MOMENTUM 128 12.\ Energy and Pow" , \2.2 Angular Impulse ; ,he Phy>ical hndulum f 12.3 Angular Momcmum

Claapter l3 MATTER IN BULK '" 1J.1 DcMity and Spccifi< (l",vi'~ I 13.2 Eiallic Properties

Iii

Iv a CONTENTS

Chapler 14 SIMPLE HARMON IC MOTION ZS6 14.1 o.all11ion. of a M~ on a Spring I 14.2 SliM of Pendulum, and Other Sy>te ....

CIuoplH 15 HYDROSTATlCS \5. \ Pr • ..., ... ond Demit)" I 15.2 r~_I·, ond A",him'&'· r rincipla; Surlau T''''''''n

CUpl"" 16 HYDRODYNAMICS 16.1 Ilqualiooo of Cootinuity. Bcmoul!i·, Equation. Torri:etu·, "Theor.m I 16.2 Vi~ty . Sto« ... • Law. Po;"uille·, Law. Turbulence . R.yoolds Number

CUpler 17 TEMPERATURE AND THERMAL EXPANSION 11.1 Teml"'r.lIUre &ales; Linear E..pansion I 17.2 Are. and Volume E..pall$ion

CUpln 18 HEAT AND CALORIMETRY 18.1 Heal and Energy; Mechanical Equivalent of Heal I 18.2 Calorimelry. Specific Heats. Heats of Fu~ and Vap>riLlilioo .

Chaptn L9 HEAT TR.ANSFER 19. \ c<;,.,d.."..,., I 19.20>"",,,,000, I \9. 3 Rlodi.,"""

CUpltr 20 GAS LAWS AND KINETIC THEORY 20.1 The Mol< Concept; ,be Ideal 00$ Law f 20.2 Kinetic Th<Or)" I 20.3 Atmoopbcric Pn'penies; Specific 11001$ of Solids

Chapin 11 THE FIRST LAW OF THERMODYNAMICS 21.1 Buic Thermodynamic Concept! I 21 .2 The F,rst Law of ThermodynamQ. Inrem.1 EnerlY. p-VDia"am •. Cydical System.

Chapler 22 THE SECOND LAW OF THERMODYNAMI CS 22.1 Ileal En&ir>e>~ Kelvi n - Planck and tlausiu. St.le ..... o .. of tbe Second Law I 22.: Entropy

CUpler 13 WAVE MO T ION 23 .1 Cha. a:teristic PTOperties I 23.2 Standing w. ,-es ar.d Rewnantt

Cluopler z.a SOUND 24.1 SOUOO Velocity; !kat>; D<>pple. Shift I 24 .2 ru"·er. Inl.n.ity. Reverb ... tion TI ..... . Sbock Wa~

CUptft" 2S COULOMS·S LAW AND ELECTRIC FIELDS :<!i. I C"DI~mb·, La .. of EleotrO$tJl, ic F""'e I 25.2 TIM: Electric Focld; CootinUOllO Charge Oi5tributioo~~ Motion of Char&«! Panicles in an Electric Field I 25.1 Elc:otric Flux and Oau,,·, La· ..

Cluopter 26 ELECTRIC POTENTIA L AND CAPACITANCE 26.1 Potential 0 ... to Point Charge. or Cha'l!< Distributions I 26.2 T1>c Potentilt Fun •• ;"" and ,ho ~a'w Et«1ri< Rold f 26.1 E""'JO'i<" Proble .... "";,h Moving CharI!<' f 26.4 Copa<itaooo and Fock! E ... ,gy I 26.~ CopacitOfS in CombinaTi""

Cluoplfr 1:1 SIMPLE ELECTRIC CIRCUITS 27.\ Ohm·s Law, Cu....,nT. R~si'tal\CC I 27.2 Resist"'"' '' Combination I 27.3 EMf and ElenrocMmica\ Systems I 27.4 Elenrie Me~ur~ .... nl J n.s Electric P .... er I 27.6 More Complex Circuil!. Kirchhoff. O=;T Rules . eimoil$ ,,;th Clpa<itar>a:

Cltapler 1II THE MAGNETlC FI ELD !!J. I Fortt()fl a M",·ing Charg~ I UU fortt on an EI«Iric Currcnt I 28.3 Torque and Magr><{ic Dipole Momc:"' I 28A SoUKeS of the Malnetc field ; La .. of BicIt aoo Savan I U.S Mor. Compln Owmc:tri .. ; Ampt .. ·, La ...

Z7I

"" ". ". J4S

'57

'"

J87

'"

<6,

maJ:lf -

CONTENTS a v

Chapter 29 MAGNETIC PROPERTIES Of MATTER 29 .1 n.. H aDd M f;'Id.: Su~pI;bilily: Re\ali'~ Permeabilily f 29.2 Magr.et.: Polo SI~nglh

ChapluJO INDUCED EMF: GENERATORS AND MOTORS J(},I aw.,. in Magnelic Au ... faraday" La ..... ..... n ... \..0 .... f 30.2 MOliOfla\ EMf; IndP<ed CurnnlS and f orces I 3O,1 Time·Varying Magr>el~ aDd lndu<ed Electric F .. \d, I 30 .• Electric Generalors nd MOl""

Chapter 31 INDUCTANCE I I. I Self·\ndllttar.co f 11 .2 MUlual IOOu"a"",,: The Ideal T .... noformtr

Chapler 32 ELECTRIC C IRCUITS 32.1 R·C. R· L L-C.nd R·L-C Circuits: TLme RespoMO I 32.2 AC Circuiu in It.: S,.ady S",te I 32.1 Tim< Behavior of AC Ci"'ui"

Chapin 33 ELECTROMAGNETIC WAVES H.I Oispl..,.menl Curnn1. Mu"" ell'. &jualions. lhe Speed of Lighl f 33.2 MalhemalicaIDescription of W ...... inO...alld Th .... [); mc",ion. I 3) ,] n.. Component Field. of an Electromagrocl i< W.""; IndP<ed EMF I 33 .• Enom ar.d Momentum Au .....

Chapin 34 llGHT AND OPTICAL PHENOM ENA 3-'. 1 R.tloction alld Refraction I 34.2 Di,,,,, ... io" alld Color I 34.3 Pbolomelry and Illumination

Chapin 35 MIRRORS, lENSES. AND OPTICAL INSTR UMENTS 35.1 MiTT<) .... I 3S.2 Thin ].<,n$<1 f ]5.1 L<-... mak .... &juI.ion; Compooite L<-II$ SY' .. m. I 3S .4 Optical [",.rumcnlS: PrO;"C'o .... , Camer ... ,I>< Eye I l' .S Oplicall notrumen,,: Microoc.,..,.. alld Teleocopes

'"~

52.

'"~

607

Chapltr 36 INTERFERE NCE, Di ffRACTION . AN D POLARIZA nON 668 36.1 lnterf.~n"" of Ught f 36.2 Diffracl ion and ,I>< Diffroction Gratina I J6.J Pol,riution of l ist"

Chapter 37 SPECIAL RELATIVITY n.1 l orenu Tran.Iormation, L<-"8Ih Contraction, Time Dilation, and Velocily Transformalion I 37,2 M .. l- E ... rIY Relat;on; Relativisl ic Dynamics

Clu.pltr)3 PARTICLES OF llGHT AND WAVES Of MAlTER 38. 1 Photo ... aDd tbe Pbotoclectri< Effm I )8,2 ComP'OB Scat .. rina; X·royt; Pair P,oductiott alld Ann ihilatiott I 38.1 de Broglie Wa ..... and lhe Ur.conainty Principle

... 10.

Chlple,39 MODERN PHYSICS; ATOMS. NUCLEI, SOllO--ST ATE ELECTRONICS no 39.[ Aloms and Mol"",ulcs I J9.2 Nudei aDd Rod ioacli"ity I 19,] Solid-Stale Electronics

INDEX

C :lPY nghted m na

TO THE STUDENT

This book is intended for use by students of general physi(':S, eit her in ca!cutulo or noncalculus' based coursel, Proble ms TC'(juiring real calculus (no( me":ly calculus notation) are marked with a sman superscript f .

The onLy .... ay 10 maSler general ph)Sics is 10 gain abili ty and sophistication in problcm-sol~inl. This book is meant 10 ma ke you a m.Sler of the on - and should do .0 if used properly. As a rule. a prob lem Can be solved once you h.,". learned Ihe ideas behind il; sometimes these very ideas are brought inlo sharper focus by lookil1@ at sample problems and llIei, 50lUlions. If you ha."c difficulty with a 10pic, you un selcct a few problems in Iha1 are a, u amine the rolmions carefully, and IMn try 10 ro[,·. rda!...:! p.obkms before looking at t he prinlN! solut iOM_

TM,e aro nurn .. out ways of posing a problem and, frequ~ntly. numerOuS ways of $olying One. You should try !O gain und~r1tand i n8 of ho .... to approach various classes of probl~ms. rather than memorizing partkular wlutions. Understanding is better Ihan memory for su~s in phys;c,; .

The problems in Ihi" book CO,'er eYery imponanl topic in a typical N 'O- Or thm:-scmeOier general phy.iCf sequen«. Ranginl from the simple IQ the cQm pltx. Ihe y will prQYide you .... ilh plenty Qf practic~ and food fQr thQughl.

Tile Chaprtr ShlrwfIJ With Exams. t>cginning Qn lhe nUl page . .... as devised tQ hel p studenl5 with limited lime gain maximum bend,t from Ihis book. It is hop«! Ihal II\( uSC Qf Ih is feature is sclf­eyidenl; still, Ihe fo\1o" ; ng remarks may help:

• The Chaplt.f Skfkw" , divide lhe pmblems in t his book imo Ihru calegories: SCAN, HOM EWORK and EXAMS. (Turn 10 page ix 10 s«an uample.)

• To ga in a quid o,·ervi~ .... of 1M basic ideas in a chaplOr, r.vie .... Ih. SCAN problems and >("oJy Ih.i, I' l inl~ .>Olu,ioll • .

•

•

•

HOM EWORK problems Kre for prael icing your problem~olving skill,,; (OW'r thf salu/i(m .... ith all I",J", rara as you ",ad. ar.-d Iry 10 w ive. the problem. Do /)()/h H/J if your course is cakulus bascU.

No problem from $CAl">' or HO!o1EWUR K i. dup liulc-d in EX AMS. aoo no two EUmli onrlap. Calculus-based students an: urged al$O 10 take the Hard Exam. hams run about 60 minutes. unless otl\(rwise in-dicated.

Stilt funMr prQblems conslilUt~ 1M IWO groups of Final Exams. Stay in )'ou r cat~gory(ics). and lood luck.

CHAPTER SKELETONS WITH EXAMS

o.pt .. I

5 1.1, ,., ... 1.4. I .~ . 'A (.I I. 1.13.

C I.'''' I. I $. 1.11, 1.60, U l, ,..,. , ... A '" I.n. , ... N

Everybody 1.8. 1. 11 . 1.1" 1.11. I .ll. ,."- 1.31,

H 1.J3, 1.35, Ul. 1.41. 1.48, I.SO. ,...,. 0 us, , .... I.'''' , ... 1.92

M , W

CaJc.-based 0 I.SI. 1.51, , .... 1.57. 1.59. 1.75. 1.76,

R ~'Y 1.77, ,.." 1.11. 1.117, 1.93

K

, '''Y 1.10. 1.21 . 1.3'. , .~

X Hard 1.40, 1.49, 1.67. , ... A M C3lc.-based , .... 1.71. , ... 1.91 5 ,"Oy

"

x I CHAPTER SKELETONS Wtni EXAMS

ChaptPl"l

S 1.1. 1.1. ,., 2.1 1. 1.17_ 1.U!_ ,." C U

A

N

H E .... erybody " ,." l.l l_ W , 1.31_ ," '"

0 M

E w 0 calc.-based 1.1 • . l.lO. "" "" ". R ,," , E Easy ,." 1.7_ ". ,~ X A Hard 1.15- w, "" '" M S Calc.-based 2.1 6, 1.11. l.l1_ W

001,

CUpter 3

s '., "" U. 3. 19. 3.13_ 3.29. ,-'" C ).37

A N

H e .... erybody ,., "" "''' 1.21_ ,,'- "", 1.41,

0 "" "", '" M E W

0 R Calc.-based "'- "'" "''- us. 1.5S. 1.59. 1.60. , 001, U ,

E Easy 1.10. "" "'. ""

X A

Hard '" "'. ,,"- '" M Calc.-based "'" "" l.62 S

00"

,.

CHAPTER SKELETONS WITH EXAMS J xl

S 4.1. <L ., •• <0, 4.1l. .,. C ,37, ." . ..... 4.73 • .. , A N

H Everybody ••• .,. ." '.u. 4.38, UO. • •• 50

0 ..... . . ." ."

M E W 0

Calc.-based R

,)1, .,. ." .. , • . 58, 4061,

K 00" • • 71. ."1, ." E .. ,' .... , u, • n . ., . X A H". 4.17. ..... .... . "'" .... M S Celc.-bllsod .. ~

00' ' .... ''< ." '" - 70 ";"'1

s ". .... " " . 5.19, 5.34. 5.37.

C 5.5!

A

N

H Ewryt>od, ••• S.9. '" ''' . U . ", 0 M E W 0 R Calc.-based 5..11 . . ~, "'. 5.41 • .... .." K 00'

E Euy 5.3. 5.21. 5.1<1. '" X Hard 5.7, ••• 50 11. ,u. ,~ 160·70 ..... ) A M

Calc.-based ". 5.11, W . ", ( .. _70 .... ) S

00'

xN I CHAPTER SKELETONS Willi EXAMS

o.pter6

s ". ". ... '-21. '." 6.27. '-" C A N

H ,.- 6.11 . ... .,. u" ." . .., 0 M , W 0

Calc.-based " 41. .. , ..... . ". 6.51. .." • K "". , Easy " .... u, .... , X A

Hard ... '.1" .,. ..... M Calc. -based ". '.44, "'. .... S only

ClIaptfr 7

S 7. 1, ,.., '., 7.1. 7.11, 7.lt, , .... C UL 7,41, , ..... 7.51, 7.'1 , 7.6l,

A 7.67, 7.79, ,., N

H Everybody 7.' . 7.1t. 7.1" 7.10. ,,,,, 7.4l, ' ....

0 , .... 1.51, 7.", 7.1!!. 7.81. ' ... 7.87.

M Ul, , ... ,

w 0

• Calc.-based 7.17, 7.U. ,,,,, 7.11. 7.76- 7.93. 1.1M.

K '". 1.107, 7.110. 7.11" 1.117

E Easy 7,1'. 7..17. "' .. 7.70. , ...

X A

Hard 1 .... , 1 .• S. 1.n. 7.17. ",.

M Calc.-baSfld 7.14. '-'I. 7.1"- 7. 113 S

"""

CHAPTER SKELETONS WITH EXAMS I xiII

s u. .-'. ... 1.19. 1.20. 8.U. U2. C ..... .J< A

N

H Everybody .. .. , 8.11. 8.11. ." .,. 8.33

0 M E W 0 R Calc.-based ... ... 1.27, '" K "'" , Easy '., 8.7. 1.23. .. " 8.37

X Hard 809, 8.15. 1.29 • ... A

M Calc.-based 8. 11. 1. 17 • ..... .'" s ooty

CluI pltr 9

s ••• U. '-II. U6, 9.17, 9.19, 9.32. C ..... ..... 9.87 A

N

H Everybody ... 9.21. ..... ...... .. " 9.!9, ' .70,

0 .... .. "

M , W 0

CaIc.-baS8d R

'-11. '-17. ' .• 7. 9.50. 9.79. '-'7

K ooty UG4

, Easy '. 18. ..... ' .71. .. " (60 _ 70 ...... j

X Hard ' .3', M7. 9.", 9.102 (6CI - 70 .... ' A

M Calc.-based 9.60. 9.15. 9.101. 9.106 (60 - 70 onl .. j S

ooty

righted IT £na

_Iv I CHAPTER SKELETONS WITH EXAMS

o..pw 10

S IU. 10.7. 10.14, 10.32, 11.37. 10.44,

C 10.4 (_Ioo 9.101) , N

H Everybody 10.2, ,.., ID.l5, 10.3}' 10.41. 11.51.

0 10.60 (_100 9.113)

M , W 0 R Galc.-based ID.lD. 10.31. 10.64, 10.45- 10.66,

K 00' 10.75, 10.79 ( .... 9.110)

, Easy 10.17. 10.20. , ..... 10.67

X Hard , 10.12. ''''' ' .... ID.l II

M S Calc.-based IO.lot.

only 10.52. 10.57. 10.77 (" -70"')

CluIpter 11

S 11.10. 11.l1, 11.1" 11 . .\4. 11.52.

C 11.61. 11.72. 11.11 , N

H Everybody 11.11. 11.21. 1I.Jo. 11.3S, II .S},

0 11.6' , 11.18, I U 3

M , W 0

Galc.-based R " .. 11.40, " .... II .... 11.62. 1I.7S, , 00' 11.79, " .. , Easy II.U 1!.l3. 11.5' . 11.73

X Hard , 1I.l7, 11-'0, 11.31, 11.17

M Galc.-based S

only 11.45, " ... 11.16, 11.13

C JPYnghted malenal

CHAPTER SKELETONS WITH EXAMS I xv

s ", ". ,U 12.11. 12.19. IUS. C IUO. ",. A N

H Everybody ", ,,< 12.11. n .1" n .lO. U.H.

0 IL5L n.'! M E W 0 R Calc.-basad IL l !. l U!. "" 12.37. 12.49.

K 00', 12.65. 12.69. 11.71

• Eo" 1l.9. 1L17. 11.31, ",. X

Hard A 11.11. ,,'" ",. 1%.59

M Calc.-basad S "" "" 1%.57. 11.71 (60 _ 70 001 .. )

00',

CbpltT 13

S 13.1. ", n .l6, 13. 17. D .18. 13. 19.

C 13.%0. 13.11

A N

H Everybody '" 1l.1I. 13.12. 13.31. 1l.41. ,,,.

0 M E W 0

Calc.-basad R "" ,,,. K 00', E

Easy 1). 7. IUS. 13.38, 11.45

X Hard 13.1(1. IJJL 13.44- 13.41 A M Ca!c.-basad n.ls, '''' 11.48. J3.!1 S 00',

opvr hIed IT lIa

xvi I CHAPTER SKeLETONS WITH e XAMS

CUpltr 14

S 101.1 , I .. ' , 14.11 . 14.1&. 14.27, " ..... C IU4, 14.S6

A

N

H Everybody 1 .. 3, 1 .. 9, IU!, l UI, 1 .. 21, 14.n.

0 .... ....

M , W 0

Calc.-based R

14.31, '0, lUI, 14.4?, 14.51, 14.S'

K 0"

, Easy 14.17. 1 .. 2l. ''', 1 .. 4S, ,,,. X A Hard 14.19. ,,, .. 14.5S, '''' M S Calc.-based 1'.3e, IUS, IUl, 1 .. 53

0''

o..pt .. I S

S I Sol. ISo?, 15.1'(' lSol!, I So39, '' '' C IU', ,,,. A

N

H evarybody 15.1, 15.1, 15.11, 15.101. 15.40, ,,,.

0 IS.so. lSoU

M , w 0

Calc.-based R

IS.S. 15.16, \S.Js, '''' IS.S'

K ~'Y

, Easy lSoI', l U I, ,,,. '''' X Hard lSo31 . lSo31, 15041, IS.S&. 15.66 A M Calc.-based lSo27, ,,,. 15061, IS.', (61 · 1' aiL) S

0"

.,

CHAP'reR SKELETONS WITH EXAMS I xvii

c..pler .6

S 16.1. ..~ ' '-' . U I U . 16.10. e 16.lS, 16.31. .. ~ A N

H Everybody 16.7, 16.10, 16.12, 16.1" 16.18, ......

0 16.41

M , W 0 Calc.-based R

..... 16.17. 16.32, . .,,, l U ... JUS

K 00"

, Ea" .. " l Ul. .. ,., 16.37

X Hard A

J6.11, 16.11, .. ,.. 16.19 • 16.43

M Calc.-ba.sed 16.17, 16.12, 16.1" 16.S1 S 00'

Ch.pters 17 . nd II

S 17.1. " .. 17.1"- 17.31, .. ~ 1&.16,

e .. ,., 18.16. "",

A N

H EYQf'ybody 17.2, 17. 11 . 17.10, 17.36. II .S. 11.19.

0 18.17. .. .... 18.46 .

M , W 0 Cale.-based 17.18. 17.~ l U9. 18.12, 18.45 R 00" K

, ,." 17.10. 17.37. 11.17. 18.33

X A Hard 17.1S, " ... IU, ...... M

Calc.·based S 17.11. 17 ...... .... I&'S I

00'

opvr h

xvlW f CHAPTER SKELETONS WITH EXAMS

CUpit. 19

5 1~. I. 19.1. 111.7. 111.30, 111.31, III.l%. C 111.38. 19.311. III.~

A

N

H Everybody 1'.3, 19.1D. 19.1S, ' 'I.n 19.14, 111.41

0 M , W 0 R Calc.-based 111.17. 19.10, 19.1%. 1'-111

K only

E Easy ,,~ 111.13, III.lS, >OM

X Hard 19.1 I, 19.14. 19.36- 1' .48 A

M Calc.-bllsed 19.2S, 19.19, 19.37, 19 . .51 5

'"' a._pit. 20

5 10.1. ". 10.7. " •. 1O.1D. "'" C lO.l1. ,. .... 1O.4S, 1O.7t A N

H Everybody ,.,

~ 20.10- ,,'. 10,31, >OJ.

0 "., ,..., 10.11, 10.73

M , W 0 R Calc.-based 10.16- ".,., 10.41 . ,.'" " .... 20.62,

K 00" N .... " .. " ... , , .. , 10.11, "" "" 10.74

X Ha.d 10.1<1, lO.17, ,." 10.71, ".n A M Calc.-based " .... "". "'". lo.a9 (60 -" .... j 5

'"'

"

CHAPTER SKElETONS WITH EXAMS { .1.

Chapin'S 1 1 . nd 11

S 11.1. 11.1. 21.3, " ... 11.6, 11.17.

C 11,.41. n . l. "L '" '" ,,.

A J1. I4, ' ''' N

H Everybody " ... 11 ... 11.11. 11.30. 11.31. 11.<16.

0 ,,. ,u, 2L1S. "-" n."

M , W 0 R Cl lc.-blSed 11.14, 11.l0, 11.l6, 11.39. " ... 11.9.

K 00' 21.1&. 22..19

, ElsV 11 .1" 11.29. 11.47. 12..11

X Hard

A 11 .13, 11.34, 12..13, l1.lS

M S CaIc.·oosed ILlS, 11.4S. 11.l1. ,,,. (60· 7e __ )

00' Finrol Exams (or Glpters 1·22 (160 • 180 min.)

t :AlyA U l. l .3<l, 5.13, 1.101. 10. 1a. 11.11. I~.

l U I. 15.21. 11.40, JUS

£Aly 8 2..lS, 4,90, • . s. 8.32.. 9.57, 11.51, 14,38, 16.13, ""2.. 11.35

Hani.. 1.11, U4, s,g, 7,99, 1"'1. 1l.6I , 13.31, 16.l I, 11.41. lO.18

Hani B 2..11. U I, 01, 7.111, UOI, II .S., 14.J9, Is .n 19.46, 21.1

Co.k. " 1.79. l.56. 6.10, 6.S6. 7.11" 11.79, n .70, 15.30, 17.40, 20.' 1

Co./c. B 1.41. 4,78, $..49, 6.Sl, 1.120, 9.61. 14,ll, IU 9. I ...... 19."', 21.27

,.

xx I CHAPTER SKELETONS WITH EXAMS

5 n,. "" 23.1. no. ll.1S, n,.. C ll." I. w,. "". :u.1I. ,.." A "' .. "''' N

H Everybody n, n. 13.1'. 23.11. 23.l3, ".,

n ... ,.., 2<4.1., ~. I S, 'U .. "" 0 M E W 0 Calc.-based lll'. ll.2!l. ll3<, 23.37. M. I$. R

~" 14.". """ .. ... .. "., .. ". K

E Easy " .. n", "''' "''' X A Hard llloo, ll."7. "''' "''' M

Cak:.-based S ~"

n,. n,.. 14.21, ,..'"

S 15.1 . ,,,. "'" "" 15.11. 15.1"

C 15.29. 2S .. 12, 2S.46- 15.S'. " .. A N

H Everybody ",. "''' "''' "" .. """ ,.,. 0 "... 15.60. =, "-" M E W 0 R Cale.-based '''It. lS.1'. " ... "''' " .... K ~" "'''. "' ... "''' E Easy 15.15- "''' 15.45- ISoU

X Hard

A "',.. lS.31. "',.. 15."

M 5 Calc.-based " .n

~" 15 ... 2, 15.65- 1S.71. 25.71

•

CHAPTER SKELETONS WITH EXAMS I ul

S 21.1. 26.1. 2 ..... ,.. 26.1. 2U5,

C 26.3l. ,." ,.." , .. ., ,.. ... A 26.". ,.." ,..~. 26.\11. ,. .. N

H Everybody ,... ,..,. ,.. .. 26.U. l6.19. ,.,..

0 "'" ,.. ... 26.61, 26.65, 26.76. 26.11),

M "'"' ,.." U ... , W 0

Calc.-based R "'" "" 26.21. 26.25. ,..,. 1'-27.

K ~I, "'''. 26.43, 26.47. 26.63, 26.66, ,.. ...

U .. 26.16. 26.106

, ,,,' 16.17. , .. " , .... 2Ul, ,.." X

Hard A

lUll. ' ...... ,.'" 26.101. l6.102

M S Calc.-based 2'-2., 26.1'. 26.37. 26.11)9 ('" _ 70 .IL)

001,

o.plH 27

S 27.1. 27.1, 27 .... 21.5, 27.11. 27.31.

C ,,-'" 27.41. 27.049. 27.50. 17.~ 27.64-

A 27.65, 27.66. 17.67. 17.611. 17." . 17.111.

N "" 17.11\1. 17.91, 27."" 27.95. 17.110

H Everybody " .. 17.13, 27.11. 27.25. """ 17.43,

0 17.47. 27~1 . "". 27.71. 17.74. 27.113,

M 27.9 .... 27."'. 27.1111, 27. 117. 27.121 ,

W 0

C81c.-based R

27.l. 17.17. 17.18. 27 . .l6, 27.117. 17.11l.

K ooly 27.114. 27.125.. 27.126. 27.131. 27.14 1.

27.142

E Easy 27.10. 2Hl, 27M. 27.104

X A

Hard 27.15, 27 ..... 27.60. 27.11)5

M C.lc.-based 27.37. 27.80. 27. 136- 27.145 S

ooly

C ;.pvrlghted malarial

uN I CHAPTER SKELETONS WITH EXAMS

Chapltt 1lI

s ,.." ~,' ,.., 21.11. ,..'" 1lI.ll.

C ,.." ~,'" 28.61. IS.I>4. IS.S7. 28.89.

A ,. ... ,.. .. , N

H Everybody '''' ~,' IS.". 2I.J./o. 28 .• S. ,.." 0 ~ .... ~, ... ,..,. ,.." ,..,"" M IS. I06. 18.i19. 28.I3<t. 13.1". 28. I~ , W

0 R Calc.·based IS.l6, IS .• 9. IS.50. 28.60. IS.70- ,..,. K 001, IS.n. u.n. lU8. 28.112. 1&.112.

1lI.1l1. IS.1311

, Easy "-'" 28 •• 7. 28.". ,...., 28. '36

X Hard A ,.." 28.57. "''', U.1I7

M S Calc. ·based 28.11. ""', 28.1 18. 28.1. ' (60·70 lllia-. ,,'

Chl.pl~r 29

S 19.1. ~"' n .... 2U. n,,. 211.11 . C I~. u. 19.17. n.'" 19.35. n ..... A 19 ...... n." 29 .• 7

N

H EVllrybody I'." 2'.11. 29.10- 29.16, 29.17. n-"" 0 " ..... 2'.J7. 2'.38. n .... 29.0. n ....

M 2MO , W 0 R Calc.·based 19.5. 29.27. N.'" 19.29. 29.32.

K ,,' 19.)9. N ,,"

, Easy 29.7. 29.8. 29.111. 29 .• 5. N."

X Hard 29.1'" A

2US. 29,49. 29.55

M Caic.-oased 19.12. 29.%3, lUI. 29.51 S

00'

CHAPTER SKEt.ETONS WITH EXAMS I uiii

S 30.1. "', JO.II. 3O.1S, JO.11. 30.3(1.

C 30.11. ,.,. ..... 30.67. J41.7'" ".n A ,.,. ..... J41.91. 3O. loe N

H Everybody .... .. .. .10.11. 30.13. .10.16- 30.21.

0 .10.39. ...... 3O.5CI • " ... 30.76- 30."'.

M .. " 30.93. 30. 1111

E W 0

• Calc.-based "A 3 ...... .. " 30.)7 • 30.441. ,. .... K 00' ",. 30.44. 30.71. .10.110

E Easy 30. 1' . ".,. " .. .IO. Ia8

X Hard

A 30.17. 311.18. ".,. " ... " ...

M Calc.-ba&ed S

00' JO.41. " .... JO.7S, ,.,~

S 31.1. 31.1. 31.10. 31.13. 31.16- 31.29. C 31.lO, 31.31. 31.3IL, 31.41. 31.59

A N

H Everybody 31 .•• 31.S, 31.1"- 31 .11. 31.17. 31 .31.

0 31.3l. " ... 31.6(1

M E W 0

CaIC. -based • 31.l. 31.1S, 31.17. 31.15- " .... 3 .... 3.

K 00' 3 ... 9. ",. " ... E Easy lU. lU!), 11.35- 3 ... 0

X A

Hard 31.11. 3 1.11. 31.37. 31.4<11

M S Calc.-oased lU. 31.:1;1, 3 1.S1, ,,-" (.e-,,".)

00'

opvr hIed IT ria

ulv ( CHAPnR SKELETONS WITH EXAMS

OIjKtr 32

s n .l, ", 31.9, 32-1G. ,,'" ,'-" C "." 31.37, "" ,'-". 31.40,

A 3Ul, ,,'" 11.S7, ",., ,," N

H Everybody ,,< 31.7, 31.13, 32.11. "" 'U. 0 3UI, lUl, ",. ,,'" '''0, M 31.61, " .. , E w 0 , Calc.-basad 31.6. 31.15, 31.11, 32.11, ,,,,, " ... K 001, 31.15. ,,..., 31.17, 3UI

E Ea., """ 31. 11, 31.71. 31.74

X Hard 31.11 , 32.1S. 32.l5. 32.66

A

M Ca!c.-based ,U. 3UI. 32.1-4. 31.~ S 001,

CluIpler JJ

s n ... m. ll17. un. "-". 33.37.

C ",... """ "". "",. ,,,. A H

H Everybody "',. lll'. llJJ. ",." "'''- n.'" 0 "',., "''' M E w 0 , Ca!c.-based "',. "', "'< "" 33.6, 33.14,

K 00' "',. "' ... "''' E Easy llll . ""'- """ """ "..,

X Hard A "'''' ""'- "' ... 33.71

M S Ca!c.-basad "',. ",., ",.,. '''''

00'

C JPYnghted matena

,00 s ('IIfIII OL -") (,l l '~ 'or" '?Nt 'L'Sf pes8q.':>!8:> ~

tn'if 'tll 'lif U'Sf ' IrK PJIIH v X

III'S( ""!if '6E"S( " I"if ~S83 ,

en'if 'unr 'n'S( '1L"S£ ,00 X

'S,'Sf ' ..,'Sf 'it'Sf 'fYi£ '\Tit '«WK pen:q-'OI I1:J • 0 ... " '"

'OTiC 'tn 'Sf '£1I 'i£ ' illWS( 'U'if ,

'''nf ·"W 'uw """Sf 'WK "!i","£ ~

'S1i'S( 'ti"K ' Int 'O>S£ 'LNf 'trW 0

' Int 'tI1'Sf 'L1"'O£ '1110£ "'if ""Sf ,i.poqAJe"3 H

... " 'St1'Sf 't Il 'S£ " "'S( '9OI 'i£ 'OOI 'S£ N

'9/O'S£ WSC '"w 'ti;"S( 'K'S( '00'0£ V

' !H"S£ " !>"Sf '9I>'S£ 'Srw 'tt"S£ '9'"0£ 0

'i I 'Sf 1010£ • .. ·Sf ,£'5( 't'S( "'Sf S

lif mdlnlJ

,,00 ton,: 'tL't{ 'S9't{ 'is'to( peSvq-'0I10 S

~

IOI 't£ "" 'o9't\" 'tl"tf PJIH V X

... " .,."C .,,'K ' 1I"t( ,(S1I3 ,

U '" , 00 X

, L'tt: ""'10'£ ',.10'£ ""10'£ trK '9"K p&$IIIQ-':)t Iil::> • 0

'" , K'K '1)6'10'£ " 1J'K ~

"." 'tt'l'( "S'I'( 'tn( ...... 1'£ ' IYK 0

', C'W ',t'W "t'K '9" K 'iTK ~K ,(poqAJe"3 H

" ' 10'£ 'WK N 'SS't( ' 11'1'£ '08"10'£ ' IL"K ...... V '~'K '9i'K "St"K 'rYK '££'t\" 0

'9nr "n"tf 'S I 't\" '01"t£ , ... ', ... S

K ulClWU

AlIII I SW'tX3 HlJM SNOJ313)1S I:13.1.dYHO

IUlvi I CHA.PTER SKELETONS WITH EXAMS

s ,., ,.. 36. 11 . ,.,. "'''. ~-" C ,." """. 36.S1. ,. .... "' .. A " .. "" N

H Everybody "". "" "''' 36.31. "'" .M.37.

0 36..1. "" ,." 36.". 36.63. 36065

M E W

0 Caic.-based R 36. 1. ",. "" ",. ,..,. J6.~ 2.

K ,," ,.. ... "' .. "',. "" E Easy "'. 36.29. "'''- ""- "'53 X Hard "-,. "'''' "'" "-" A M S

Caic.-based 36.19, ,," 36.27, 36..7, ,,"n """

S 37.1. n.l. 37.04. 37.11. n . l4- 37.21.

C 17.341, 37.~7. 37.50. 37.S1. 37.58 A

N

H Everybody 17.1- 17.S, n .7, n . l s, 37.16. 17.18-

0 37. 19, 37.104. 37.-48. 37.53. 37.61, 37.64

M E w 0 R CaIc.-based 37. 10. 37. 11 . 37.26- 37.32. 37.'w,

K ~'Y 37.36. " .... 37."

E Easy 37. 11, 37.21, 37.28- 37.49, 37.Sl. 37.51

X A HMd 37.6- ".., 37.31, "",

M Calc.-based S 37.27, 37.37. 37.59. 37.66 (61 - 78 .,i .. , ,,"

•

CHAPTER SKELETONS WITH EXAMS I uvii

S 31.1. ,., 38.18- 38.19. ,." ,.". C 311.37, 31U1. ,." 311.53. " ... A N

H Everybody ,., ,. .. 38.17. 38.18, ".,., " .... 0 ,.'" "". ,. ... ",. """ 38.67

M , W

0 R Calc.-based ,. .. ,.,., ,." ".,., 38.47. .\S.49.

K ~'Y ,.,. 38.6 1. 38.6l. .\S.6!. .\S.71

, Easy ,,~ ,." ,." "'" X A Hard .\S.1l. 38..11 • J8.!7. " ... M S Calc.-based .)11. 14, .\S. I!. "J' .\S .~ I • J8.~5 (100 _ 70 riLl

~'Y

Quo,I« 39

S 3U. 39.4, 39.11. 39.14. 39.1S, "". C 3Ul-. J,..lA. J9.39. )9.4l. " ... A 39.!-I. )9.57. " .... 39.61. 39.73. N 39.7S, 39.76, "" H Everybody "., 39.S, 3'. 16- J9.18, J9.18, ".,. 0 " ... 39.43. "." ".,., "."- 3MS,

M ".,. )9.61. " .... , .. , , W 0 R Calc.-baSed 39.10, " .n. "." 3'-27. 39 . .1l. ".,., K "0' }9.17. " .... " ... 3'.7l, 3'.77, 39.78. " ... , Easy ,,~ 19.41, 39.49. 19.61

X A Hard 39. 10, 39.1 1, 39.H . 3Ut

M S Calc.-based 3'. 1l. 39. 17 . .lU6, 39.52 (60 -" IlliL)

~'Y

D CHAPTER 1

Mathematical Introduction

• . 1 PLANAR VECTORS, SCIEI'il1f1 C NOTATION, AND UNITS

1.1 What is a scala. quantity?

, A ~aJ.r '1.nntity It&> only magnitude; il i. a pur. number. positiVi: 01 "'Iali~c, Scalars, be;n, $impl. numben. Ire added. rubtracttd. CIC .. in tM usuil way. 11 may have I unit aftcr il. e.,. mau . J k._

1.1 Whal is. ~OI quan.j.y?

, ... vector quantity ha. hoIh magnilude ilI>d dil~ion_ For • • ampic, • "", movin8 ><luth a1 40 km/ h It ••• we"" vrlocily of 40 tm / h "" .. thward .

A VUlor qIJantily tan be reprflt'n1ed by an arrow drawn 10 ..,ak. The lenllh of Ow arrow is proponional to lhe ma",itude ofth. vK'\Q' q .... ntity (40 km/ It in the above example) . The direction of 1M arrow '''P'' .... nt> the direction of 1be vc('lO, quan.j.y_

1.3 Wha' is II>< ·, ..... han!' ~or~

, The ..,,,,haM of . number of .imilar v.Clo .... !OHOe v«1l1rS, for ... am~. ;, th.t oinlie VCClOf " 'hick would have th. $lime effect '" all''''' original V«IOI'$ ,aten ,"Celhe •.

U De~ibc d .. gnph.ical addition of ,'.aon.

, The m"(bod for ~ndin, 1M resuhanl of <e"cral vectors comi". in beginnin, II any co,,,cnionl poinl and drawing (10 sule) each VttIor a,lOW in lurn . They may be takon in any order of sUOC<'$$ion, "The lail end of each armw is al!AChed 10 Ihe lip 000 of Ihe preceding OM.

"The ~$\Illanl is ~~nt.d by an an'OW wilh i .. lail end ailM 'lattin, poinl and ito tip ud al lhe tip of lhe last _tor added.

t .! Dc$(ri\>c the par_ltelogum melhod of addilion of lwo VttIors.

, "The ~",IUlnl of fW<> VttIOT> acting at any an.,. may be "'I""c ... nted by ,"" diagonal of a parallelogram. The tWO VttIon a~ dJ'll.\II"t\ as ,"" sides of the l"',aileIOll'llm and ,"" , •• Wlanl is ito diagonal, a. ,I>own in Fi • . I ·!, The diru1ion of the .. sultant is OWly from the origin of t"" two VttIon,

f1c. 1·1

1.6 How do you. .... bt""" ve<"lors1

, To ,ubt ... 1 a VttIOr 8 hom I '·"",or A . ... "" ... Ihe dircdion of B and add il v"",orill1y 10 '"«Ior A , Ihal ; •. A - 8 . "+ (- 8) ,

1.7 Describe the t rigooomctric fuJlClions.

, For lhe right 'riangle shown in fig. 1·2. by de~nilion

Lj"";;'" .. _"-, . •

~.O<lj"".,_ . ",: 1.1

opynghtoo IT £rla

2 a CHAPTER 1

U E>:p~ ~ach of the lollowing in scienl i1ic notation, (II) 627.4, (1I) O.OOO:W;, (t) 2000] , (4) 1.(1067, (t) 0.0067.

, (II) 6.274 x 10'. (It) 3.M x 10- ' . (t) 2.00] x 10'. (4) 1.0067 x let'. (t) 6.7 x 10- >.

U Elprns each of the follo"';n, as simple nu mbe .. x let': (II) 31.M x 10- ' (61 OAI5 x 10' Ie) 1 1(2.~ x 10-') (4) ]1(43 x 10').

, (II) 0.031&5. (b) 4 IS.000. ( t ) 488. Cd) 0.{Kr0Z33.

1. lt TII.e dia_t .. of lhe eanh is aboul 1.27 x ]0' m. Fi rxl it> diameter in (II) "..Ihmele ... (6) megamet.n. Ie) milco .

I (_) ( 1.l7 x 10' m)(1000m m/l m). 1.27 x 1O,0mm. (b) Multiply meten by I MmllO' m 10 obtain 12.7 Mm. (e) Tll.en use (I ~ml I OOO m)(1 mi/l .6] km) : 1M diameler is 7.69 x 10' mi.

1.11 A 100000m .ate is run on a .!OtJ..m-circumfe,c.x;e circular I_ t . ll\e run"" .. run eastward al lhe >Ian and berxl soulh . What iii the di!oplacemcnt of tile endpoinl of llIe rate from tile >laning poinl?

I ,.,.,. runners move ., .h"",n in Fig. 1-3. The ""'" i. halfway around lhe track 10 lhe di!oplaccmenl is one di.me'." - il9J..4 - R.l..I!! dO>< '''<l1h .

1.11 Whal .. a rontponmr of a ,·ec,or?

, A rompo ... nt of a ,·octor is ilS ··shadow·· (l"'~ndi""lar dmp) Qtl an.u:is in a g:i~n dirOCliocJ. For eX;impk. the !'-<ompo .... nl of I displacemenl is 1M diltance along Iht P ax\$ corr..,ponding to ' he gi"en displacement . !t i. a !otalar quanlity. being positive or negati"" IS it ;. positi""ly or ""gati>'ely directed alon, the ax .. in que" ion . In Fig. ].-4. A., is positive . (Ont $<Imetimes de~...., I "<"<lor comflOllt'nl as a !l<'aIW pointing alonJ 'he u is and having Ih .. size of lhe >C;ol..,- compo".nl. If Ihe IC;Ilor component is negali"e til<: v«1or '""""ponent poinlS in 'ht ""ga'i'~ di. «1ioo along the axis. ) II .. ""stomary. and u$Ceful, 10 ,esoI,~ a veeto. inlO romponenls olong mU'Ulllly ,n'F"ndicuulr dir«1ioo. (rtoCIang"l«t compontn<».

1.\3 What is the rom(lOflCnt method for adding vecto .. ?

, Each v«lor i. ,ciOh'c<! inlo its x. y. and: components. wilh negali~ly di.«led romponen" la ~en .. negali"" . TII.e x componenl of Ihe resultanl. R,. i. the algehraic ... m of all ,he z componenl •. The y arxl , rompo""n" ol the re""llanl arc fourxl in I similar way,

1.14 Dcfi"" Ihe m~hipli<aliort 01 a vectOr by I >C;olat .

, "The quanlity bF i, a v«1or having magnitude I'" F (Ille absolule val"" gf b lime. Ihe magnilude of f ): 'he direction 01 bF i. lilal 01 F 0. - f . depending on ",'hetht' b i. posilive o. negative,

1.15 U,ing Ibe graphical melhod. firxllbe resultant of the following lwo displ~menls: 2 m al 4(f arxl ~ m al 127". ,I>< angle. bei~g tak.n ,dative '0 tht +x axis.

I Choose z. y . xes as ~own in Fig. 1·5 arxl layout lhe di.placemenl. 10 scale lip 10 tail from the origin . Not. Ih., .n angles are mcasur~d from lhe +x axis. TIle resultant v.eto • . R. poinll f. om ,taning point 10 endpoint a •• t"",,'n. Me.,ure iI' len!lh "" tht $Celie di,,,am 10 find il' magllilude. 4.6m. U,ing I prot.actor. me .. ure ils argle 11 10 be 101 ' . The Tcoullanl di'placement i. therefore 4.6 II al 101' .

MATHEMATICAL INTROOUCTION a 3

, ,

• 21 oia )(!"

•

1.1' Find t~e x aDd y component. of a ZS·m di",laoemeot It In In&le of 210'.

, The vec10r di",'-mcnt IDd il$ comPOD¢ll1$ an iIbo..-n in Fig. 1-4. TIle romponon15 are

x component_ -25<0$ 30' - =lL1.m , component _ -n sin3O'_ - 12 ~ m

Note in particular that each romponent point. in tbt negltive coordinate di rection and ",ull lbe,e/orc b<' taken as M •• li .....

1.17 Solve !'rob. 1.15 by use of reetangulo, components.

, Re>oIve uch vector inlo realJlgul., componentS .,Ihooon in Fig. 1_7(~) Ind (1)). (Piace. crou-hat"" ')'IObol on 1M oripul ~~O< '0 """'" Iha. i, <"HI. "" ,..,pI-.:I by 1M ."'" of ill yo""" corn_oro.) n.. mull.", IIu tbt .".11, <:omponenu

R, _ 1.~3 - 2.40_ -O.87m R, _ 1.29 + 3.20 - 4.49 m

NOle that components pointi", io 1M nc",tivc dirtttion mUlt "" .!Signed. no",t;ve vaw.. . "The resultant is .00 .. 0 in Fil. 1-7te); we $« that

,

R .. V(O.87)' +(4.49)' _Ul.m.

•

• ~n SJ ' _ J ': O

."""S)· ·U(J

'" J"iI. 1_1

,

. ., ton. _ _

' "'

•

.. ~

"I

, •

• • '" •

1.11 Add 1~ followinl ,wo fon:e v""lon by use of II>< pataJleqtam "",.bod: 30 pOUnd. II JO" Ind 20 poundo •• 140". (A powuJ "'fora''' chos.cn welt thai • I·kg object weigh> 2.211b 00 earth . One !'<lund i. e<juivalentto I fora oi4.4SN.)

, "Tho fora: .""I0fJ arc shown in Fi • . 1-8. CoDltruct. parallelogram using tl><m as sidtl. I S .hown in Fig. 1·9. The resultant. R, .. then shown as 1M diap111. Meaourcmcnt shows thl' R is JO lb II 11" .

.... " rted malenal

4 D CHAPTER 1

1.19 Find In. <:<:>mP'll><: nts of the ~eCtM }' in Fig. 1·\0 alooJ! Ihe .. and J' axes.

I In Fig. 1- 10 the da.hed p"rpc:ndicula .. from P t .. X and Y delermil><:.he magm''''''' and diro«ion. of lhe _e~'t'" romponcnts f , and f , <>f _0«", f . The .i8"~d magnitlldts <>f lhe", \<ttl", rom""""n". which are lhe .... l.r wml"'I><:ni' Q( 1' . arc ... ·rilten a. f:. F, . It ;" .. en that f ; _ f ' cos 9. F, _ F . in 8.

" " ,

• u''''::''---'--_''"L-~x' AI. I.IO

' .:0 ,_) LeI f h.,.., a ma",ilude <>f 300 N and mate "ngle e - .lO' ... ·jth the positi"e" ditcClion, Find F. and F,. Ib) Su~ Ihat F .. .lOON .n<! 0 - I~S" (}' is here iG 11>0 second Guadranl). Find F. and F.. I i_ I F, - 3OO=.W' - illJ!1!. f: - JOO:sin 3()"-~. ibo) F. _ )00005 14S" _ (300)( - 0.6192) _ - 24' .75 N (in the negal i~c direction of Xl, f; .. 300 ,in 145" .. (300)( +0.5736) -.l12.lI1.l:!

1.11 A car 8"'" j.Okm ca.l. 3.0 km >Qu!h.2.0 km " CSt . al>d 1.0 km n<>f1h . (.) Dcte.mir>C how f.r nonh al>d 00 ... · fa, USI it Mos !lttn displattd . ( 110 ) fiOld the displattn><nt ,·ttt()T IXllh graphicall)' and algebraically.

I I.' 1I.«""inl t""t ,.""'",., can be added in any o.dor " 'e can immediately add lhe 3.O-km ,"",Ih and LO-km oon h displ""'m.m "cClnn I .. ~ct a ""t Z,O·km .I<lUlh displacement ,·ector . Simil .. ly tM S.O- km east and H Hm .. ..,,, .-ee")" odd 10 a 3-km east di .... attme nt ''''''Of . B«au .. the ~asl di5placem~nl oonlriootes no rompoocnt along !he ""nh-south linc al>d the sout h diSfJlattmcnt hOI) n" OOmponent . loog the . ast·l>'e$' li .... the ca . is -2,0 km oonh and 3.0 ~m east of its starting point. (6) Using the he.d·to--t" il method . we easily Can ronl"""t th~ resultant dill'lattm,nt D as sho .... n in Fig. 1· 11. AI~.br~ically "'.., noIe Ihat

[) - YO;+ 0;-0.: +3' - Mlm lin q,., -J or tan O _ i 8 - l!: SQuth oI,a>l.

-----·i;::l5~--"<'''---l- E " " ",

, "C-__ ~.,,,..J r-q:. HI

I .U Fil>d the x and , . roml"'"e tu, <>f " -IOO-N f<)lcc at an a"81e of 1 ZS' '0 Ih. x ",;' .

I Formal m.thod (uSt> . ngle "b<.",~ posi ti". x 3.isl :

Vi,ual method (u>cs only acute angles abo"e 0. btlo ... p<lSit;~e Of negati"~.r ax;'):

Sy in.~ction Q( Fig. 1·12 . F. - - 11: 1- -2.."'9 N: f: - II ;I - llll!.

I.ll Add the foll",,';n, t ..... ",plan. t f<>.o;s: 30 N •• 37' .I>d SO N at 180".

I Split eII<h into romponcnlS and fi nd lht f • • ullan\: R, .. H - SO . -16N. R. _ IS + 0 - 18 N. TlI.cn R .. 3U'N a l>d Ia n 0 .. IS/ -16. so 0 _~.

MATHEMATICAL INTROOUCTION D 5

F"II. I ·U

1.2<1 For tbe "ector. A .nd B ~J"'n in Fig, 1·13. find (pI A + B. (bl A - B. and k l I _ A ,

I The rompon~nu ... A. _ 6m. A , - 0. Ii, _ 12 eo> 60" _t. m, and 8, _ 12 sin 6C' .. lOA m, i., f A + I ), - 12 m 000 ( A + II), - lOA m. SO Ihat A + B '"l1,!m at~. fb ) (A - B). _ 0 and (A - B). _0 _ 10, 4 so,, - B _Io.4m al =..I!!!: «, (lI - A ), - 0 and (11 - Al. - IOA -O so B - A -l!M.m at ,!!".

I .ZS F()O" 'h~ ~ecton .ho" 'n In Fig. I'''. nrKI (a l A ... " ... C and (b) A.,. " - C.

I Tht ~ and y rompon~nt' of C are ~ . S m and -7.8 m. (. ) The x wmpo .... nl i. A. + B, + C. _ 16.5 000 for tbe y romponent .. '~ ~nd 2.~. SO Ilk: _ector i~ 16.7 m al 2&:. (6) A. + B. - C. _ 7, S and Ilk: y rompon.nt i$ 0+ lOA - (-7.8) - 11\.2; changing thi$ 10 a magnitoo. and angle . w~ find.l2.2..m al !i!!:.

1.26 For lbe ,.." IOJ"$ sh(ro,n in F"'I!. I·IJ. find ( a ) A - le. (b) B - (A + C). and (r) - A - B - C.

I i'" The ~ «Imp ..... n' i! A. - 2e, .. -3 and tbe y rompo",,"1 is - 2( -7 .8) _ IS,6. livi", ll.i.m '" .lQ.L. (b ) The .. romp,m.nt - 6 - (6 + 4. S) - - 4.5; tbe r wmpo .... n! - 10A - [0 "1- (-7.8)1 - 182; therefort (4 .5' + 18,2' ) '" '' ll1..!!! at !lli:. le i This is . be ""gali'·. of lhe '"«lor of Prob . LZS (.) . SO .ha. il _ ~ al !W ' + 18O" .. ~ .. - 171".

1.27 A di.placement of 20 m i. ",ade in the » ' plan. at In angl. 01 7f1' (i. • .• 7f1' counl<",I<)Cl ",;'" fr<>m tlk: +~ axis) . Find it< .. and Y COIIIponenl •. Repcol if Ihe angk i\ 110', if Ih. angl. is !5O".

I In •• ,h ca", s, - S «IS tland " - $ :<in 8. Tht ,,,,,, lIS are U . JL.lm: =--lM. 17} m; ~, .=..!.8.&.m,

1.23 It i. found Ih.1 an object .... 11 Ilang po-""",l~ if an x fMel' of 20 N and a Y IOfet of -3O N . '. applied 10 il. Find the wnlle lor« {m~gnilude and d;lectionl .. n"'h "'ould do the "'m~ job.

I Addoin, compon<"nll of ,be fOlet~ yi.ld. R. " 20 N and R, " - 30 N, R _ (400 + 9001 '·7 .. Jhl:!, Calling 11 the count.rclod",,* angle fro .. Ihe "1--< a;o;i>. tan 9 .. - 301lO and so 9 - mor- .=...l:fI..l:.

1.29 Find the magnilude and diroction of Ih. 10' '''' .. 'hien has an .. component of - -ION and a y component of -60N,

1.30 Find lhe magnilude and dir«1 ion of the ",'" of the 1"lIo"'inl tWO coplanar di.pla<emenl ... cton: Zf) m "' 0-and IOma112O',

I Splinin& each inlo romp;lncnIS. R. " 20 -, _ I'm and R. _ 0 + 8.7 .. 8.7 m, "Thtn R - l1l..!n "';110 Ian 8 _ S. 7{ IS giorinl 9 .. lIt.

1.31 FOUl coplano, fo,,,,,. act on a body al point 0 a. >I>o,,'n in Fig 1·I4(QI· Find lh.ir ,eSultanl graphi<'ally,

I Slaning I, om 0 , lhe four ~cctuo or<: plon.d in 'urn a$!hown in Fig . 1.14(1)), f'lact the t. il end of on. ~ectOf lllhe lip end of the ?",,,,,ding one . The ar""" from 0 to the lip of 11k: la" "«Ior ,epr=nl. lhe resultanl of Ih. ,·ecton.

6 D CHAPTER 1

l iON

" , . _·,,-LO'Q~-,-;;.....L.I"'--_. . . , ~ ,., ,.,

MeASII~ R hom tM ~ drawin. in Fi • . 1.14(b) and lind ;110 be 119 N. An&le a is mculI.ed by J>I"OCI"KtOI" I nd is found 10 be 37". He",," tIM: resultant makes an InJlc 8 _ ]80" - 37"" ]4J" with 1M """live :.: Di •. The re5IIltant is ll2.b: .t ~.

1.12 SoN. Prob. 1.31 by u~ of the .eaonpln eomponcnl method .

, The: VC<:Ion and [heir componcnu I rc IS foUows .

.......... , N

'" '00

'" '"

:.: ca ..... .t, N

'" ]00 _ 4$' _ 7] - 110 _lO" __ 9:'!

- 1(1)_ 2(1' - - ISO

NOIe the sign of earn component. To lind the resultant . we hi ••

R • .. flO . 7] - 9S - 1SO_ - 94 N

, U.,.I",I, N

" ]00 ';n45· .. 7] 110';n3O"' - ~,

- ](1) sin 20" _ - 55

The reoultont io >hown in Fig. I·U: .... ..., lhat R '"' .,1(94)'. (7]), - 1l8..I!!.. Funher. tan a " 71/901. frOom wlU(h a - )7", Thereto« lhe re.ul~nt is 118 N It 180 -)7 '"'~.

FI&. 1-15

1.33 Perform grlphica1ly the follo"":ns veaOf addition. and oubtraction •• ,,'her. A. B. and C ar. the ,'eaOl'$ !.hOW" in Fi • . 1-16: (I' A + B. (.) A . B . C, (e) A - B, (d) A+ B - C.

, See Fig. 1-16(", through (d). In (e ). A - B .. A + (- B); that is. to oubtr~ B from A. rnnw the direction of B and add il veaorilUy to A.. Simil.,!y. in (d ). A + B - C .. A + B + ( - C) . where - C is equal in magnitude but opposite in direaion 10 C.

- <

< , ,

/..1 • 0' • , , , ~, .' ., I)' .,

~ ., :-. "/ " ,

" <, , -,., {O' "~I '" FiC. l -Ui

do CO

MATHEMAT~AL INTAOOUGnON 0 7

nco 1·)1

1.)1 Filld Ihe rewltlnl ll: of III<: following forttS an acting on II>e.ame poinl in II>e &i.en direCliom: 30Ib 10 Ih' IIOrIMasl; 10 Ib 101M SO\llb, alld SO Ib 11!' north of we ...

, Choose ,asl UIM pooilive -" direaion (Figure 1· 17).

x co.p ....... lit

3Oros 4SO .. 21.2 - SOoo. 20' _ ~

Total _ -25.8Ib

R _ V(-2Hr + ( )1.7)' - v'66j.8+ 1004.9 - 40,9Ib

30 li n 4SO .. 2!.2 SOlin2O" " 17.1

- 7(1 .. =l!L Total . - JUlb

(j _ 39" we .. of S9!ub

1.35 Find the ansk belween IWO vectOf forttS of equal magniIOOo .• u.ch Iballhe . esultanl is "",·Ihird as much as aIM. of 1M onpnal forees.

, In 1M vector force dia"am (F',. 1-18). 1M diagonals of Ihe rhombus bi$ttl nob otMr. Thus.

F/6 I 00. '" -T- (; -0, 1667 "'"' 80.4' H '" 160.!r

'The anile between tM IWO forces is~.

- -FIe· 1·11

1.36 find the . ector sum of II>e follo"';n! four dis.placemenlS on a map: 60mm nonh; 30mm "''''1 ; 40mm al «1' ... e .. of IIOfth ; SO mm al )If .... esl of SQUlh . Solve (a) graphically alld (b ) 1I"brakally.

I (.) Wit!l ruler and protractor . ronSlrua 1M sum of v« lOr displacemUIS by lhe lail.I~Mad melhod a. shown in Fig. 1·19. 'The resultanl vector from t.il of fin! 10 Mad of 1 ... 1 i. 'M" lIso ""asured wilh ruler and protractor . AIlS, : 21.mm at 67.r W of N. (bo) LeI D .. resultant displacemenl .

D, .. -30 - 4O"n«1' - SOlin JO" .. - 89,6 mm D, _ 60 +40001 fir - SOrosJO"_ +36. 7 mm

8 D CHAPTER 1

, _.

,; "'''''' _--.ii:. ~4-----,

fiI. ).19

1.37 Two fOl<es.1!O N and lOON actinl al an angle 0( flY' wilh cK h orl>er, pull on an obje<!. Whal sinpe 10= ",..,..ld "pl''''' the rwo /o"",,? Whal ";ngIc for"" (called Ihe .quilib"",, ) would b>ola""" Ihe 1 .... 0 forces? Sol ve algeb<aitaUy,

, Ch<>oHe lhe or "";. alonl lhe 8(l.N for"" • ...:Ilhe y "";. 50 11111 11>1: Ii)).N force fH' a\>(we lhe posil;"".r am has. ~iti,'e y rompo""nl . Then rhe . ;npe lorce A Ihal ",pi""", lhe two forus i. lhe "eel'" sum 0( lhese ,-

H, .. 80 + UXlros6O' .. lJON R. " ]OO ,in 60''' 8? N

R _ VH! + H;" Uhl::l.

The fo,""" rhal bala""", A is - R wirh • magnilude of l~ N bu. poinlin, in the <>pp<»i •• direction to R 34' bd<7W ""prj"t x axis (0< 21"* .b<nI. positi" • .r ",,;') .

I.J& Two forces a<t 01\ • poinr objttt ItS follo-..$; 100 N ot 17f!' l i d 100 N al sao. FiDd Ih.ir ~Itant,

, t ", _ lOON . t 17f!' aoo..'t -" axi.; F, " 100 N II 5(1" 000... r axi.,

II _ F, + F, H, _ lOO<:os 170" + 100 <:OS 5(1" .. - 34.1 N R, .. 100";n )7(f + lOOsin sao _ 94.0N

H _ VH! + H;- ll!!U:!

h • • 1" -0 s"lul ions: 19O' and 110*. From a \oo~ at i" rompo .... nl . ... ' • ..., 11111 II I;'. in rh. oerond quadr. n< . 50 1M aM" er is.LLt.! (0. 2lr ab<nle n.pliv. x axi.). In . I",s fo,mal appr(>iloCh '. c. can find: 4> _ tan- ' IR,/R. I • ... 'hid! .. ilI give an IK'Ule anglc >OiuliOll. in .I\is ca .. of 2lr . .. hich .1 .... ]'1 TCp"..wnls .1>I: .nak wilh lhe posil;'-. Of nepl;'..,.r •• is. and either ab<nle or below Ih ••• ~i • . Since ..... b.ady know f, om .Ite componen" ".-hich quad,a.1 A Ii." in. "'. koowthe dirffiion pruisc ly , In our <3IC tile 70* is abo"" the ""pli"" or ""is,

1.39 A 10''''' of lOON mak., an anll. of 8 ... itll lhe x ""is aDd bas" y rompo"""t of 30 N. Find both Ih. x rompor;em of tl>l: fo,"" and the anglt 9.

I 1M dal. a", sk.rchc<l in Fig. 1·2(). w. "" isIl 10 find F. " lid 8. W •• ""'" Ihat

o ~ ";n9 - ;; - 100 · 0,3O

from which 9 - 1L£. 1Mn .• i""" Q " h COS 9, "" e b,o"c

F. _ IOO cos 17's* _~

'. • fiI. '.:0

1 . .0 A boal can Ifa'·.! al a sp«d of 8 kml h in . till " ·a.t, on a la ••. In lhe: lInwing .. 'al., of a ."cam, it can 11>0\'.

al 8 kmlll relalive 10 lhe: ... ·ale. in the mUm. If the: "'cam IJIe"d i. 3 km/ h . 00 ... fa" ~an lhe boal mo'·. pi" a tre. on lh. silo •• in tr&,..,li"8 (.) u~ream? (6) do .. n", • • m?

MATHEMATICAL INTRODUCTION a 9

, (_) If \1Ie: .. ate • ...,rt ... nding 1Iill. 1M boat'. opeed "" .. tile: tr.e would be 8 km/ h. Bu •• IM: stream is carrying;, in the OJIposil. dire<1ion .t ) km/h. lbcrefore tIM: boat'. speed relat;ve 10 the tree is 8 - 3 .. ~. (Ill In thi. ca .... lhe Strum is carrying the boa, in ,1M: $I. .... dir«tion tile: boat is uyina to mO'o"e. H.~ il$ 5p<ed pasllhe tree is 8 + 3 .. Ilkm /h .

1.41 A plane: i. 'u~lin& eosTWard .t an ainpced of SOO km/ h. Bu, a 90 kml h ... ·ind is bIowinl $Outlt ...... d. Wha, llC Ihe direction and speed of the pI.De •• Iotiv. to the ground?

, The plane', relUhant velocity is tile: sum of two velocili .... SOO km/h .astward .nd 90 km/h $OUthwud. lbe ... componenl velocit; .... re sbown in Fig. 1·21. Th. pl ..... ·' relUhant v.locity i, f""nd by u ... of

R _ 'As-XII' + (90)' - S08kmlh

The '''11. '" is given by

from .... hich .... ~. The planc', velocity r.'ativ. to the """nd is 508 km lh 0'10.2" $OUth of CI".

, • , •

I .U Wi,h the: .. me ain.pecd a. in P,ob. 1.41. in .. hat direction mus' the pia .... h .. d in order to mov. due .'"t rclati~ to tile: earth?

, lbe Sum of the plane's vt"locity thtOUJh the .1, and rhe velocity of ,he wind mllSl be ,he .... ultan' cast"',"d >'docl,y or ,he p'ane ,d.ti"" ,,, ,he earth . Thl.;, ono...·n in ,he ",,<to< di ."am of F" . '·U. II is IICCn tha, lin (J _ 901SOO. from which (J .. 10 ... •. lbe pll ncll>ould head.ll!£ nonh of .... , if it is In """'" .... Iw.rd nn Ih. canh.

If..., wir.h tn fi ..... ,he pia .... ·• castward 5p<ed. Fig. 1·U 1.lh us lbat R .. SOOCO$ (J _ m ~m!b .

1.4.1 A. child pull. a rope attached to l!okd with a for"" nl roN. The rope mak .. an anslc of ~ 10 the "ound. (0) Compul. the ctrcctil'C value of Ih. pull I.ndin, to move the ,led alonl 1M ground. (II) Comput. ,he 10<"" 'ending 10 tift the sled l'Crtically.

, '" sJr,own in Fil. ]-23 , the components of the 60 N fOftt are 19 N I nd #tN. ,_) The pull along t ... ground is the OOrilQnlaJ componenl . ~. (III The lifli". fOftt is the vertical <OmpnllCnl. U .

F _«hin 0\10" , _ _ WN

•

1.44 Find the resultanl of the copI ...... r fo."" . )'St.m.tlown in Fil. 1·24.

fIc. 1·13

, R _ Y,+ Y, + ' , R. - - 40+80cosJO"+O_29.)lb R,-O-80linJO"+60_20lb

10 D CHAPTER 1

-I. - "" ~

" __ ,' ,-,-,-:-" .~~ _L __ _ •

I ', .. "'''~

FiI. 1·}.4

1.45 Repeal Prob. 1. 44 1m- fig. l·lS.

I R .. F , -+ V, -+ V, -+ V. R, _ lS -+ 0 - 9000<> 50' -+0_ - 32.8Ib R, _ 0 - .0- 90 lLin W' -+ (,0 .. - 48.9 1b

R" VR:-+R! .. ~ #- " ... 'I;:I-&.!.: below - ..: u;'

I a _ a , -+ a , -+ a , -+ R. R, _ - llS <05 2.~. -+ 0 -+ ISO <05 23' -+ ISO """ 62" .. 122.8 Ib

R, _ - 12hin lS' - 130 - ISOwn 23' -+ ISOlLin 62" - -120.71b

R .. VR: -+ R: - J..ZZ,lJJ;z . 4> - Ian 'It. 1-~ below -+ .. a. is

r , _ I"" I~

r, _ I .~H~

I."" Compute algebraically lbe resuhanl (R) and «juilibranl ( IE ) ollhe f<)llowing ooplanar force" lOOkN 1130'. 141.' k N al 4S· . • nd 100 kN al UI:f ,

, a _ F, -+ F, -+ . ",

100 .in 240" - - 100 sin 60'.

R, .. 100 sin .lO'"'" 14 1.4 wn 45' -+ ](10 .in 2-10' _ 6JA kN

R- VR;-+R;-IW·6kN abo>'<: + .r •• i •.

1.48 Compule algebra,,:ally Ihe ,e.uilanl of Ih. foll""'inl di. placem.nts: 20 m .. JO". 40 m at 120'. lS m I. lilt)'. 42 m al l7f1'. and 12 m al l iS '

, I) .. "" +"" +"" -+ ... -+ "', _ ",,,,,I, anl displ"".rncn, - I ~.)m

NOlo: I!O" is al""8 - .. ni" 27(:f i. I lunS - , .... i$; <05 120" .. - <05 60'; ";n 120' .. ";,, 60'; . 0 0315' .. <:OS 4~";

sin lIS' .. - ,in 45' .

"

MATH EMATICAL INTRODUCTIO N a 11

D,- V D;+D;_m.z..m

1.49 Refe. to Fig . 1·Z7. In I.m» olV«ton A Ind II . exp . .... th. veclo .. P. R. S. an<! Q.

I W. h",·. he •• a pafal:.logram. SO Ii: .. B and p o. A + R _ A + II . a.arly S .. - A ond Q is lbe .... m 01 - II wilh A (lf O "A- II .

FIe. 1·27

1.5Ci Ref.r 10 Fig. 1·28. In tenns 01 V«tor.; A and B .• ~preM lbe V«tor.; E. D - C. and E ... D - C.

I Clearly - E _ A '" B or E o. -(A + 11) _ - A - II . D - C _ D + (_ q _ A. Then E + D - C _ [ + A _ - II .

1.51 A displatemenl D 01 100m from lhe origi n al an angle of 31' l bo'·.th . .... is i, th • • esull of Ih",. socct"loliv. displatemenu: d ., "'hieh i> 100 malon, Ihe negaliv. ~ axis: d, . "'hieh i. 200 m al an angle of I~ abo>-e lbe ~ a, i.; and I displace"",n, d,. Find d,.

I D .. d ,+ d, + d, D ... d ... +d ... +d,. or loooos 31' _ _ loo+200oos15O"+d .. d,, _ 353 m ( Nol.: oos 1.5(1' - -oos)lf) D," d., + dl> + d." Or 100 sin 31' .. 0 + 200 . in 150" + d"

ti.,, __ 40m (Nol. :sin I5O" _ .in)(l'). d,- Vti.t+ d~- illm ~ - tan "I~I_U below+ .. ui, 1.52 The '""sultanl IMCe d"" 'n lbe a<lion of 100' fOTttS is R. whieh is 100 N along Ihe ""ptiw y uis Thr.e of

Ihe fo. ces a •• ]00 N. fIJ" ab<:>w lbe ~ nis : 200 N. 14lr above Ihe ~ u is: !SO N. 320" aboy. the ~ lri • . Find lhe loonh loroe .

'M _ !!',+ !!',+ !!', + !!'. R.- ~ . + F,. + Iit.. .j. Fo. Of O_IOOeootn'+2OCIcoo 140'+ ZSOrol J2O"+ F". Fo. .. - 88 3 N R, " F", + F".j..o;, + F., '" - 100 - lOO si nflJ"'" 200sin 14(1' + 2SOsin 320" + F.,.

Fo, - - 15,1.5 N F. - \ IFt .j. F~ .. !l!!.1::! ~ .. tan·j I~ I - .6!.!: below - z axis

1.53 A ca. " 'I\osc wei,l" is ... is on a ramp whiell mat • • an an"" 910 lhe borironlal. How la.ge a perpendicula. loroe mUll lbe ramp wilMland if it is 00110 bttak uoo., 1M ca,', ..... ighl?

, A, . hown in Fig. 1·2'). Ihe car', " 'eight is . fora: .. Ihal pulls r.I .aighl down on lbe co •. W. 'a •• CQlllponenu of .. alonllbe irw:line and p<rpendicula. 10 it . "The .Imp must Nlane. lbe IMCe co"1"'f'Cnl l!:..M..!I if Ih. ear is 001 In crash Ihmugh lhe .amp.

12 a CHAPTER 1

, , , •

1.54 "The fiy~ """Ianar f"r~ M1""-n in Fi8- I·JO(A ) act on an objM. Find Ihe ,""IIBnl lor~ due 10 them.

"" •

,.1 FIt. I·JO

, ( I ) Fil\d the x· and y-rompo .... nlO 01 eaoch force.

lIPJIIinod~. N x CO"'",",UI. N

" 19.0

" 15 em 60" _

" " - 16 em 45' _ - 11.3

" - II em)Q" - -9.S

" 0

Note Ihe 'i8"5 10 indicate + al\d - directions . (2) "The .~l!ant R ha. <lOfnp<>ncnh

,

'"

, colnpOMnt. N

0 IS":n tI:/_ 13.0 16":.45" .. 11 .)

-lhinlO" _ _ 5.5

-22.0

R. - 1: t: _ 1\1 0 + 7.5- 11.3-9.5+0 - +51 N R. -1: F. -0 + 13.0+ 11.3 - 5.S - 22.0 .. -3.2 N

(3) Fil\d Ihe ma8"ilUdc 0 1 Ihe "SIIlla"1 from

R .. \IR~+ R!_ 6.5N

(4) SketCh 'he . ,,>ullant 3\ s/lO"'n in Fig. 1·19(b) and find iI, anile . We see Ihal

" unq.-n- O.S6

from ... ·hich q. .. Y . "Thcn ... ·e ha.~ 9 _ 360" _ 29"' _ 331". "Thc r •• ullanl is U.I:i al ur (or - 2'1").

•

alenal

MATHEMATICAL INTROOUCTION a 13

l.5S find algebfaically the .c.ul.am (It) an<! C<juilib .. nt (E) of the following ooplanar f=: )Q() N at (f', 400N at 30". aOO 400 N at I~.

, R_ F, + .·,+ f , E _ _ II. R,-300 +.\OO cos 3O"+ 400ro< lS(f'_300N NOIc th.t 400 0)0 1$(1" - - .\00 c<>s JO' 400 >in U(f' _ 400 >in 30"

R, _ 0+ 4O:)sin 30' + 4OO>in 15O"_-IOON R- VR: +R; - m1::l:

1.2 ntREE-DlMENSIONAL VECTORS: DOT AND CROSS PRODUcrs

1 • .56 Find the magnitooe of the ,·«tor A in Fig. ].) I. "h_ tail I~ ~t t.ht <>figin and whose head lies at the point (1.0m. -l.O m. ~.O mi.

, First. nolO th.tthe ,·ector A .nd iii vectOf component A. are the h~po"nu~ and one >ide of a right tri angle "'hOSt plane is ~rpcndio:\llar to th. ~). plln • . Hence the pythlgo ... n th"",cm gjVf> A' _ 8 ' + A:. BUI the ,·ector B ~ itself the hypotenuse of the right t. iangle in the xy plallC ",Ilo:l;c »<Its arc Ihe vcctm componentS A , and A,. Thus )"Ou can ..,. .hep)1hallo.ean theQrcm apin !O obtain B' . A! + ... ;. Combining the two C<juations •

... hich ;s the fom! taken b~ the p)·thago.un theo.em in th ... dimclUions. 1'0. the douo.

A - V(7.0mr + (4 .0m)' + (~.O ", )'- U.m

"'" ."

" •. I,. <, ' - " 0 ". <.0", . ) .0 .. '

- - ...,---,

:.. ) " " '"' "

" 'C;:;:*;:j==:.=:-=-:';-:-:-:_::';_~_~. , .' .. " . 0' _ ~ . ,'» ..... 0 ... 0 .. '

•

• ."0 .. ' U7 find the KIIla. oompollCnts of the thlce-dimensional vector f in f ig, 1-32.

FIJ. l·ll

, In Fig_ 1·32. f . , f •. f . are the rffIanJutar ' -ectOr comp''''''''t. or r : Ihe """Iar component< F,. F .. F, . r. siven by

F. - Fc<>s9, F. - Foo.e, Or fo, oonvenience .... ,i.ing cos 6 , _ I. cos 6, _ m, 00. 6,. n.

F,. Fm F, - Fn

I, "' . and n are It/erred!O as the dir«'ion <",inn of r . By.he three-dimensional pythagorean {hfflrem (Prob_ L.~) .I' +m ' +n ' .1.

14 a CHAPTER 1

,

, , I f 1// --'--------- - -"

I ', '

, , ,

flC. 1·)2

1..511 In Fig. I·n, lei F r~preKnI a fOf"" of 200 N, Lei fI, _ {(l". fI, " 40'. Find F" F, . and F, .

/II _ 0.766 " .. (l-t' - /II ' ) '11 .. 0,~

(assuming thai F, is positi~e; OIhe.-.ise. " - - 0.4()(), and the re<:rangular components of F are

F. - (200)((1.5) - lWl':! 1', - Wll::!.

A. a tl>e<k . (100' + I B.2' + 80.8')"''' 200. NoIe Ihal fl. _ 66.17".

1.59 find the ~ect"r",m (R l of Ihree .'ect<ll'J. F" F" ' •. dra ... n from the origin <II a Ihr~-<limen>ional ~inole ~Iem such as thai of Fig. 1.32.

I Follo";ng lhe oomponenl method.

R, - 1',, " F,, " F" R." F" .. F" + l'i.

R _ 1(1'" + fl," 1),)' .. (F" + F" .. F,, )' + (F" + F" + F,.,)'J'" .. here F" i, .he X componen. of F " ele. The di,c<'ioo «»inco of R arc ,i>e:! 1»

/II_ 1'" +1), +§. R

l1tc re",'"nt (m.",itude and dire<tion) of any number of ~ectOlS d.a-..n from 0 (an be oblained in ,he loa...., way.

I." Define a unit ""<tOf .

I Any nonzero ~ectOf f may he ...-rillen as r oo Fe . ... he.e F is the ma,nituo. of F and where e i,. unit 'W/D' (a >e<tOf whOO( ma.,ull,.1e is 1) in the dircctioo of f . That is. the "",,,,,tude of f is indicated by l' and i:. di,ectioo is llIal of t , If , tam" unili (c .g., N, ml,), F am ... the LI...., units ; e is • di...., .... onicss ve<:ror.

1-'1 bpr .... the VectOf , in Flj. I .J2 in tennl of unil ve<"tOfI olon, the OO()r<lina~ .... "

, In Fi,. 1·32. lei lIS introd"", unit Y«IOn I, J. " along X. y, Z. 'especti~.ly. l1tcn the ve<tor oompomnts of f can he ...-rinen a.

f , _ 1',1 F, " F,J

oooording to Prob . 1.60, I[f C>nc of the "",)ar component. is neptive---soy. F, _ -IF. I - -IF,I-then we ha~e r. _ -IF. I I -l r , I ( - II. wbich is still as prcscrihed by Prob. 1.611 1 Si""" F isthe resultant of ill ~ectt>r components. ~ oblain the vt'1')' important e~prn.s;on f - F, I .. F,J + F, .. . In this expression , F, _ l' <OS 9, _ f1. etc .. '" previow.ly"""""' ; and, as bofa.-c, ,he magnitude and direction (direction cosine •. that i,) are obtained '"

;tpVIIghted material

MATHEMATICAL INTROOUCTION a 15

1.62 A 10"", F has compootnll F. _ lOON. ,, _ 153.2 N. F, _ !KI8N. E~pr_ t· in leffil> of unit ""Clon.rId find ill ma&niIOOc Ind direction .

, "The ~ector f COn ~ written as F .. 1001 + IUlj + !KI.1lk wilh magnitooe F .. (l00' + 1~3. 2' + 81,) '7 .. 2OON ... d direction:

m _0.766 ~ - O.$

Strictly • ..., ,""""Id h."" ""rinen

f ... (100 N)I + (153.2 NlJ + (BO.B N")k " f - UlH 1~3.Zj + 1ll.8k N

1.63 A 1o"", A il added to a s..oond IOfce " 'hi<h has x and y oomp<mc:nls 3 Nand -S N. 1lIe ~",ltanl 01 Ihe two forca ill in Ihc -x di rection and has a malnilude of .IN. Find lbe x and y cumpo~nlS of A .

I leI A _A, ! + A,j . l1Icn A, "'" 3 _ - .\ and A , - 5 _0. So A, _ ~ and A , - U::!

1.64 Express A . II . Ind C of FiS . 1·]3 in term. of the unil vector< I. j . and k.

I A " II, ! "," AJ + A,k - !lLm. B _ til + lOA1 m. and C _ ~ .SI _ 7. 81 m.

1.65 Find I,," oomp<mc:nl> of a di spl.ceme nt "'hkh wben added 10. d;~.cemenl of 71 - ~j m will Ii"" a , .. ulli nt di""laoemenl of SI - 3j m.

I We have A , "," 7 _ ~ and A , - .\ _ - 3. 50 II , - -im and II, -l..!!!.

1.66 Find the magnitude and dir«lion of!he ~ectO, Sum of the lollowing th.ee V=OO: 21 - 3j . - 91 - 5j . .Ii + 8j .

I We have (2 - 9 + 4)1 + (-3 -, + 8lJ - - J(. Thus.;1 i. ~ unil> along Ihe -x di~C!ion .

1.67 What mu" be the componen1> of a ",,<10' "'hkh "'hen addcd 10 I,," follOwing two ~mors gi''''' rise 10 I veclor 61 : ](j1- 7J and 41 + 2j?

I CallI,," '"«'0. A. "Then II, + 10+ 4 ... 0 and A, -7 + 2 _ 6. So A, -::::.H and II, -11.

1.68 A ce,," in .oom has a noor whkh is 5 x 6m and I,," ceiling height;" 3 m. Wril. an exp • ..woo for Ihe ""<10' diltoncc from One ror ... , of lbe room 10 Ihc oor ... , diallon. lI~ opp"'il. il. Whal is 'h" magnitude "f the distance?

I Th. x, y. z di<placcment~ in IOing from one co,ne. \0 !he (111\(. a •• 5.6. and 3m. '~~ti~!y. Th .. ef=. D - 51 + 6j + 3l m. Also, D' - D; + D; + D;. """i<h give> D - 1i!.J:!J..

1.69 Find Ihe displaoement Yffio. from the poinl (O. 3. -I) m 10 the point (-2.6. 4) m. Oi~. YOU' ~ns"'~. in I. j . k nOlation . Also gi~e lhe magn itude of the displaC<'ment.

I "The oompo ... nl d;"placcmc n15 ~'" D, _ -2 - 0 _ -2. D, _ 6 -3_3. and D," '\ - (- t) _ 5: Ihu, 0 _ - 21 + 3J + 51; m and D-u.m

1.70 An objet! , originally al tl\( point (2 . 5. t) om. is gi"en a di""iaC<'me nt 81 - 2j + k ern . Find the coordinale. of it .... '" po$ilion.

I The ... ", coo.dinale> a~ x _2+8_ IO:y _ S + (-2) - 3; z - I + I .. 2 Th~ objm il at nO.l. 2)<m.

1,11 Find the . e>pltanl di5p1acement <IUIed by Ih~ fallOWing th.ee diljll"""m~nts: 21- 3l. 5J - :11. . 01>\1 -~ + J + Ilk . aU in millimete rs. Oi"e i15 magnituOc IS "'~II a. it! i . j . k "'p~n'a'ion

I R _ (2 - 6)1 + (5 + IU + ( -3 - 2 + 8). _ - .\I + §:! "'" 3l mm . and SO R .. 61 '~ -1..8.m.m.

I.n Oive tbe I. j . • ~p~..,nl.!ion and magnituOc of lhe force "'h;oh must be: .ddcd I" lhe followi ng two fOfces 10 gi" • • fot« 71- 61 - . : 21 - 711 and 3j + 2k. AU form 0'" in ... w,o"..

I Call the force F. "Then F, +2_7: F, +3 _ -6: F, -7+2 - -\. "The~fo'e F - 51- ?j + 4kN and F_~.

1.73 Vectors A ~nd II afC in Ih~ .o:y plaM. If A is ill N al 90" and B is 120 N al 210' . find (_) A - B. and (bl ",,<lor C ... dothat A - B + C _ O.

16 a CHAPTER 1

, " aoo II ,"'nncn in lemu <>f unit ¥ectoru,e " " 7Qj and II .. -120= JO'I - 120Jin 3O"j" - 1041 -6Oj . 1.1 A - I _ (0 + 1(4)1+ (10 + (0)J _1661'1 at 51", (h I C Oo - (A - I ) _ - 1041 - 1)0j _ 1661'1 4t 2.3 1".

1.74 If A - ;>.i - .lj + 51>. mm and 1 _ - I - :!j + 7k mm. find (in component fo,m) , . ) A _ I . (b ) B - A. (e) Vtt\OT C IUC~ th"t A ... B '" C _ 0

I (. 1 A - I .. [2 - (- 1))1 + [- 3 - ( - I) lJ + (5 - 7)k _ 31 - J 2k mm (6) B - A .. - (A - Il) _ - .l .... J + ,/.11. m m (~I C - - (A ... II ) - - I ... 5J - I lk mm.

1.705 Vtt\(>T A - 3i + 5J - 2. and ¥ ...... o. B Oo - 3J ... ill . Find a vtttOT C wch that 2A '" 7B + 4C _ O.

I A ~tt\OT is uro if and only if each of ill component. is zero . F", eumple. lA, + 7B, +4C, '" 0, "" C, - - 1 ~, Fo. C, . 'l<)1~< 2(5) ... 7( - .l) ... 4C, - 0 to obtain C, .. 2.75. Simila<ly, C, _ - 9.5. The.efore C Oo - 1,51.+ 2.75:1 - ',1.5. ,

1.76 A ~rlain '-ectOf is ~iv,," by 31 + 4j + 71. Find Ihe .ngle il makes wilh Ihe z u is

I Fi. >! find th. projmion of the ","",lOT in the .. y plan. , this is (3' ... 4' ) '" .. s, The magnilU<k of tbt ,-ector is (7' + 5')' ~ _ 8,6 and il makes an angle of Ian - ' (5/7) _ ~ to th. z axis, Oth" ..... ;se. from P,ob 1_61. ,

cos 8 , - VY + 4' + 7, - 0-814

l .n Wh"t mus1 be Ihe ,eI.tion bet ",'","n ,·ec:to,., A and I if th. following condition is 10 he lrue :

" - 2B - - 3("+ II )

If .,.,"!O. A - 61 - 2. m. ,"'hat is B~

I One has A - 28 _ - JA - .lB. "" 4A _ - 8 . Sub$lilutinS "" -1. 10' A gi>'es B _ - 201; + 8!c m,

1.71 What mu~t "" Ih. , elation ""tween tWO ".c:tOfi A and 8 if the magniludt of A + I «juab the map ilud< of A - I . Ihal is,

1" + 8 1'" IA - 8 1

I A and I mu>! h< mutually perpend;rulat. To "". 'hat . let P .nd Q he ,."" .'.ct.,... of the .. me magnitude, A. >i>own in Fig. 1·.l3. P and Q fOfm a rhomh;n wlsosc diagonab. P + Q and P - Q . arc necessarily pe'pendic" lar, N_ .. t P _ ! (A + I ) . Q _ j(A - Bl_

1.19 The ,·~cto. di\pIlIcemcnl~ of 1",'0 points 1'1 and B from tl>< origin "'e

'"' •• - - I - SJ +2. nn

Find Ihe magnilude and I. j . • "p •• ",nlalion of Ih. ' ·.Cl0. f. om point A 10 poinl B.

I The vtttor in q...,"ion hascomponen1\ D. _ - 1- 3 - - 4: D, - - ~ - ( - 2) _ - 3: D, - 2 - S - - 3 Thus D _ - 41 - 3J - 3k em and from D' _ 4' + 3' + 3' . we find D _ S,8cm.

l .tIO The r«1angular <:<om""nen" of an actt~,.! ion vector _ ar. a, - 6, ..,. _ 4. a, _9 m/. '. Find lhe "cClOr . xP"'S>iUtt for _ and;t> di.eclion COIiJltS_

MATHEMATICAL INTRODUCTION D 17

, In ' -«1or 101m. I " 6i + ~j + 911 m/~'. The maK"i tude of. i, Q " (6' + ~' .. 9')' ' - !Lll.m...'i. " Kl th~ di,tct:on cmine> of. ,If.

; ,---lUl

, '" ---llll

, .---lU1

1.81 Find . "CClOr e.pre>!;;"" lOt' • line !.egment.

, TIle , u aight Ii .... al>. ~18. 1·J.l. i, determi ... d i)y point. P, and P,. Re~.dinJ the line ~g""'n t lrom P, 10 P, as a >':e1'" s ...... con ",n t.

• - (~ , - ' ,)1'" (y, - >'dJ" Il , - :,lk ,.it h n agnitude s .. [(x , - ... , .' + {y, "I' " (:, - "j'l ' , and dirffiion

.... -" , ' --'--, L

e. ,

1" ,

" _ J " ",

t , - y, .--­, 0

• 1

1

I" I

I __ V

: , - : , .--­,

, ..-; ,

A. impo. unl special ca~ of ' his is ''''' 5O..,al ~.d •• diu, vffiOl •. 0 .. di,«ted ><:gmcnllrorn the o rigin 0 to a po:nt P ( .... y. :):

r r xl .. yJ + : k

with . .. (x' .. y' + : ' .' " and

, 1- -, '" .. r . -­, ,

1.1Il Conside r the ,-.Iocity ,-«tot .. _ Hi! + J()j .. 24. ",I •. wi,h ,. _ (16' + .lO' + 24' )' " .. ~ 1.1>2 ml .. di . tction gi'-.n byl - Hi/4[,62, .t.

No ... let U\ muttipty . I>~ 10: 10. _ 1601 + JCq + Ha. . Y, Find the magnitude aad di.«tioo of Y,.

, ", "' I( 160)' + 1)0))' .. (240)'1" .. ( 101(4 1.62) .. It) ••

and tl>< dirtction cmi...s of Y, are

I.. 160 .. ....!!.. .. I , (1O)(H 62) 4 1.62

m ,_ m

,,·hich!.bows tllat Y, has the direClion of Y,

1.I!13 D<:~"" th. """Ia. o. dot I'" .• .\",t of 1"''0 '-CClOf' .

, 'The dol pmdU<'1 of an~ t"-,, .«w~. as 1' , aM }', . Fig. 1·3S. i. written . , 1',' 1', aoo "dcIi .... d ., the proo:h:ct 01 their malnill>des and the COloi .... of the inchKlood angl • . Th.t is . }', . 1'," F,F, roo II. " 'hieh i •• """I", quantity . In Fig , 1·]5. F, " 15. F, " 100. I) .. tJ1'. Thill.. 1', • • ', " (751(1 00I(O.5) . 37:.0.

1 .~ Find ,ht dot prodll.:u of the unit vectors along X. Y. Z.

, Sit>«: L j. • are mutwllly pc:rpcooicular and <Ii unit masnit ..... _ the d~finit"'" Qf the dot pr<:>d .... 8;'-'"

I · I - J ·j - . ·~ - I l · j _ l · k _ j · . _ O

18 0 CHAPTER 1

o

I I

I

I I

,

I .IS Roo th~ dol p",dUCI of any t .... ~cctOT> in terms of Tedangulu components.

, Write Iny two veclon 3S

f ,- F" I"''';,J'''F" . f , " fi. 1 ... F" J '" 1'; ••

r, ' r,. (1'; .1 ... F,,J'" F,.k)· (F" I '" F"J ... F..k) Tbc: ri",,·hand side may be oimplified hy noting that tloe di"ribllti~ I . ... holds 000 employing tloe values of I· I. etc . . loul\d in Pr<>b. 1.114 .

f , · f , - F"fi. ... "; ,1';, ... F .. fi.

To cbed. that f , · F, i. juS! the qWlmity Fil';eo& 9. where 9 is the .",Ie he""""n f , and f , . multiply aDd di";& tho right >i"" rhrou'" hy F,I';. living

F, · F, - F,l';(i,' ~ ... ~~ ... ~, ~.) .. M(I,I,'" m,m,'" ~ ,~ ,) N ...... . tM lam ilia, addition lormula in ''''0 <li"",",ioll5 .

CO< 9"' .... (9, - 9,) _ .... 9, COlI 9, ... <in 9, <in 9, _1 ,1, ... ... ,m, ~xrends K> three di_mion< as "'" 6 - I ,I, ... "',"' , ... lI,n ,. IItnce, the ab<we becomes F, F, "'" 8. aDd this .... thod of multiplication is, ... np«ted, in aoc:o,d "';tll tM ""~n;tion of the dot prod""t.

1.116 ut f , _ 101- I~j - 201<. F,. 61 ... 8J - 12k. Find their dot pr<><h><1 and the angle bet .... en them, , F, • F, _ (10)(6) ... (- 15)(8) + ( - 20)( - 12) - llIQ

Now no:e that F, _ (10' ... 15' ... 20')'~ _ 26.93, l'i _ 15.62. Hence . the angle 9 between F, al\d f , is given hy

f , ·F, 180 "", 9 . F,I'; -(2t..93)(15.62)· O,4279

Of roune. lhe loa .... ~al"" un be o!)uil\ed f""" cos 6 - 1,1, + m, ... , + /I ,n,.

1.87 Find lhe projtaion of I ny "tao. I lonl a >I.ai"'t line .

, Tbc: projtaioo of vector A . (A, . A ,. A .) alon,lhe \i". ""t.nni".d by lhe radius "tao • • _ (~ . ,.. z ) is A, " A <011 6, ... hu . 9 is tM anile het...,en r .1\d A, F""" lhe de6nition 01 the dol prodUCI . A· r '"' (Ar COl 1/) _ ..t , •. lIence. writi", r _ (I I ' )' '"' (I, "', /I) _ unit veclo. llonl _e have A, ... A . , '"'

04.1 ... A,m'" 04 , 11_ This .x~ for A, rema;ns .. alid . ven ,"'hen the line does not pass throoYt the origin ,

I ." Roo 1M pmjection of A .. 101 ... aJ - 611 I lonl r _ 51 ... 6j ... 91<.

, Her. r _ (~' + 6' ... 9')' ~ .. lUT.! and P,oI> , 1.87 giv ..

A,'" .01,1 ... A,m'" .01 , 11 .. 1 ~1 1 ~'I2 )'" SC1692) -11(11992) - I ~-tul

•

MATHEMATICAL INTRODUCTION a 19

Ut De6nc the cross prod..a (01' vector prod..a) of two VCClOB.

, The ~r<>tf prodUd of two vecton, as F, and F, . FI,. ' ·36 . ..ntten .. F _ ' , )( F" is dt~ned IS. vCClor F hamg a magnitude F .. F,F; sin 9 and a di~ion which is the dir«tioo of ..:InnoI: of . ri&!n·h.nd SCJ¢'"

.... h.n tumc:d hom F, to ' , throuSh . ope 8, it beio! .....".,...:1 that the 1Xi. 01 the ocrow n nomlal to the pi .... dotcrmincd by F, and F, (!he ~Iu-luutd scrtW ,lilt). Or. il the curled ~ngOB of the right hand point ftvtn F, 10 F, . the utended Ihumb poinu in the dit<:Ction of F (riglu-Mm/ nJ~).

Note !hat in accord with lhe right·band :!CT.'" rule , F, X " .. -(f, X F,).

I." Fmd the eros. prod"," of the /XIOfdiMtc unit vCClon.

, Si~ I, J. IF. are mutuall~ ptrptndkular and of unit ma,nitu<k . it follows from lhe do~nition 0( the cro.s produCl thll

1 )( I .. J xJ .. " X" .. 0

.X\ " J JXI - - IF. .xJ - -1

I.'. Given two vCClon, a. in Fig. 1·37.

A -A, I-+- AJ -+-A, IF.

~nd their cross prodl><l in 'CClUlgulat axwdin.leJ.

FIt. 1037

II _ B, I + B,.j + B, .

, C " A X 11 _ (,01.1 -+- AJ + 1'1, . ) x (B, I + BJ + B, t )

Applying the di. lribu tive , . .. 1<;> the right-hand . ide and uwn, the val .... o f I X I. el •.. loond in Prob. 1_90. wI' -. C _ A X II .. (A, B, - A, B, )I + (A ,B. - A, B, )J -+- (A , B, - A, B,)II.

Equivalently, A X II may be uprnse<l a •• dotermi ... nl :

, J • C " AXB _ A,A, A ,

B, B; B,

,.. may he verified by Upl-nding Ihe dolerm;nanl wilh reopcct to the ~BI row. NOIelhat the X, Y. Z romponenu 0( C are

C,_ A, iJ, _ A, B, C, _(A ,B._ A, B, ) C, _ A ,B, _ A, B,

Hence the magnitudo of C is C .. (e; -+- C: -+- c;) '~ and it. direction DOli ... . are

/_ C. C

C m '~

C

VCClor C n , of counc. normal 10 the pl . ... of vCClon A and B.

c, , . -C

" ,.

2Q D CHAPTER 1

1.91 Assuming Ih.1 v.olOn A 000 B. Fig. ]-38 .• '" in 1M Xl' plane. delormi"" the magnilude aOO direction of C _ A)(IS . , C .. (2OO)(]OO) . in (55' - 15") .. 2000lhin 40" _ ll~S

and by Ihe right-haOO ru~ the dirC<."lion of C is Illat of +Z . V«Iorially...., ~an ""';te C _ 1288S11 .

, C IJ Z

•

FIt:. 1-311

l .t3 Rcierrin810 Fi, . ]-37. lei A .. n - ]OJ + nand B _ -61 + ISJ - 25 • . (.) fiOO the magniludes of A 100 B. Ib) fiOO the dircclion.,.,.; ..... of A . Ie) Find Ihc vectOr produ<1 C _ A)( 8 . (~ FInd the ma!nit""" aOO dircclion of C. (~) find anglc: 6 between A and 8 (fJ FiOO the value!! of the di.C<."Iion <Oline!!I,. m,. II, of B, and ~nd an!les a". a". an bel"'eon B and the X. Y. and Z axe!!. """"""ivcly.

, I .. ) A - {2O' +10'+3O')'~-~ B -Jl22!

(b) 20 -10 30 " " 37.42 m ' ·37,42 II, • .l1:4i

1<) Applying the delOfminanl fOfmula.

• C - 20-10 ](I

-6 1~-25

C . 111 -10)1 - lS) - I i51(](I) I - JI(20)( - 25) - (:lOj( -6)1 + . /(20)(15) - ( - ]0)(-6»)

- - 2001 + 320j + 240< .. 200(-1 + L6J + U k)

(d) "The magnitude of C is C - 200(1 ' + 1.6' + 1.2')'~ -~

The direction""';""" a.e - 200 320 2-40

I, - 447,21 m, " 447.21 n, " 447.21 Note Ihat C - ql,l + mJ + ",k) .

It) C· ... B)in6 447.21· 137.42)(29.nlsin6 )in 11 .. 0.401<45 8-l.l&r V) IS - - 61 + ISJ - 2SIr. .. B(I,I + m,J + N-,k )

Thus 81, .. - 6 Bm, _ 15 BII , - -25 B_(6' + iS' +25' )"' .. ~ I, ,, - 0.2016 m,_ o.SO:W II , " - Q8399

au " 101.6.J" "'" _ 147. 13"

C :rpynghted malenal

D CHA~ER2

Equilibrium of Concurrent Forces

1.I ROPES. KNOTS. AND FRICllONlESS PULLEYS

U TIle "bje<' in Fig. 2·1( .. ) "" ";&1" SO N and is support"" by a <""d . Find the «.won in Ih.t eo,d .

I Two forces;o;I ~P'>n 'he <!bjed. the up,,-ard pull of II>< co,d and .he do",,.,,-. ,d pull of ,,..vi11. Rep'"..,n, 1M pull of the cord by T. (he ICIlSion in .he ",:"d. "The pull of V"';Iy. 'M , .. ight of the "bjed. ~ ... .. 50 N. 'Thnc 1M) fo.tt1 are .ru,,,.,, in the free-body d'."am. Fig. 2- l(b J.

TM loren arc alrcldy in component form and so "'-e can \tIlite the ~rst coodition for equilibrium .t ~.

~ r, - O "'-" 0 - 0

L: F,"O berome$ T - 50N - O

fmm "hicb T .. ~

" , ,

• • ' - SO N

,., ,., .1J. 2·\

: .Z A, ,hown in fig. 2-2( .. 1. the ,c",ion in the horiZQnta) cord is JON. Find the .. eight of the objed.

,

f, ~. -10'

T,_ .'

,., ,.,

, M !lCen in f'Tob. 2.1. the tension in rord 1 ;1 equal 10 .hoc wei,hl of.tIe ohject hangin, from il . Therdore T, .. .... _rod ,,'t"""" 10 ~r>d T, 0 ' M',

1"01" (ha l {hot untnlY"'n fQl'tt. T •. and (he knMiA fQftt . )0 N. DoIh pull on thc ).; 0>01 ~l point p, II the rdorc makes ""nsc !O i>Olu. th. kn01 at P ti our obj..", 1M f, • • ·body diagram .howing the fOl""'" 00 the knot i. d,a"'n ti f ig_ 2·2( b). Th. 10":" compo"""t, are al§<> found the ",_

N • • t write the ft ... condition 1m equilibrium fOf tM knot, From the fr.., -body wovam.

I F, • (I become, JO N - T, co. JI;f • 0

I F.- o bccromei T, sin oI()" - .... 0

SolvinK tile first . ,,""tloo for T, Ki"es T," 39.2 N, SU~litulin • • hi. ~.lue in the ....""d . "u •• joo Ki"ei .. _ ~.,.n.. " 'cianI of tho objoCl,

"

22 a CHAPTER 2

,

to.l N

,., ,., , CoMidc. lhe kl'lOllO be in equilibrillm lIlHk. the action of th.ee: forces, IS $hown in Fi., 2-3(b).

L F. - 0 yields T,cosSO" - T,-O Of 0.643T,-T,

L F, - 0 yiflds 1; sin 50" - til)) N - 0 Of 0.766T,- 6OO N

This gi_ T, - Z8J1!. SubMilU!ing into the 1: F, eqWlUoo yields T, - mJ!!..

:.. The Iollowinl coplana. fOftfl pull on I ring: 200 N at )(1'. ~ N at I!O', 300 N at 240". and an unknown force, Find lite m .... itudt and direction of the unkn"",n force if the rilll is to be in equilibrium.

, ","ume that the (I' line is tbe .. uis and 9(1' opccifics tM J uis. The Ih.ee: known forca arc then IS shown in Fig. 2-4. If r. is the unk""",,, 10."". IMn r, + r, + r, " r. _0. Lei R _ r ," r, + F,. IM n R + r. _0:) F. _ - R. To 6nd r ...... e need only 6nd R.

• f, _ !OO N

f, _ ~N

'-,

R," f,, " f,. .. Ii." Ii, cosJO" .. I'; cos I!O'- F, cos6O' , R. - 2((1(0.866) +500(0. 17") - 300(0.~) .. liON R, " F, ";nJO" " F,sin 8tT - F, sinfR'. R, _ 200(0.500) .. ~0.9I!S) - 3CXl(O.Il66) " 333 N, R '" (R: + R:lc~ ..

R llil::!. tan 9. _::l => 9. _ 71-7". F. _ R and 9 ... .. 9 . .. 18(l' -~ R,

2.5 In Fi,_ 2'5(. ) the VlIluc of W is IIll N. Find 1M lensions in ropes A and B.

, Refe. to Fig . 2-5(b). Slimming Ioroes in lhe .. and ~ directions 10 ze.o yields A _ B cos33' and B sin ~J' _ w .. 180, F.om the I'tte., B '" ml:!. whicb wben i_ned into the form« givn A .. lJ11:!.

[f the identical ropes A and B in F"'1. 2·5(,.) can eacb suppan teMinns no IarJer lhan 2OON, .. bit is lbe muimum .,Iuc thai W can b,ovc' WIlli is the tension;n tbe otber rope..men W has lhis m.uimum vallie?

, From Prob. 2.5. B .. iIl ClpOriencc the '-'geu tension. 200N in this cue. Solve for fOfCCS , 1008 the vcnical: W - 201h,inS3" "~. and in the horizonlal dit«tion 6nd A .. 2OOoos~l" -il2l!:.

2.7 A rope Htends belween lwo poles. A 9I).N boy h ...... from it . IS IbowJI in Fl,. 2-6(.). Find the tensions in the two parts 01 the rope .

EOUILIBRIUM OF CONCURRENT FORCES a 23

, .. ,.,

,.

,., , Labclthe two tensions T, aOO 7;, aOO iso[a,,, 1I><.ope al tbtboy'. hands .. the ob;ea_ "The free-body diayam for Ih;,; object i •• ho...-n in fig. 2-6(b).

T, COS S" - T, CO$ 10"" 0

LF, .. O T,sio S" + T, sin 10" -90 N .. 0

E~al""ling IIot: sines and cosines, these e<jualions b«:ome

O.9116 T, - O.m T, .. O and O.081T, +0.174T,-90 .. 0

Solvin, Ihe fint fOf T, giYf"$ 1) _ O.II89T,. Sul>ltituling this in lhe oeroOO e<juatioo gives

O.086T, +O.I74 T, - 90 . 0

from .. hid! T, _ 346 N. "Then , b«ausc T, _O.II89T" .. ~ have T, _ 342 N.

1.8 "The tension in oordll in FiB_ ,,7 i. JO N. FiOO Ihe'ension in Band tllo ' 'al ... of W.

"&. 1-7

, DTa .. a free_body diagram fOl th. poin' on Ihe ro~ whe.e ,110 oordo meet: ,110 equilil>rium rei";"'" in the z and y directions are T~ COl SO" _ T.oos toO' aOO W 'O' T. sin SO" + T. sin6O", where T .. " JON . Solving: T. - .121! and W -~.

1.' In Fig. 2-7, how large are T. and T. if W - 80 Nt

, "The equilil>rium equation> have . lready been obtained in Prot>_ 2.8. "The y-equilibrium equation when W _ !IO N is 80 .. T.!sin SO" + (tao toO')(ros SO")I, where .... have substi,utcd for T. from tllo horizontal e<juation . Thu. the tensions are T .. -!Ul!. aOO T. 'O'.u..t::!.

24 a CHAPTER 2

2.1 0 A boy of ,..,i! hl W hanv hom lhe cen,., of. clothesline and <Ii.,on. Ih. Ii"" iii) Ihal it make. 2(r angles with lh. hornonla l al each end . Find lhe lens;.", in the dOlh ... li"" in term< of W.

, Frvrn Fig . HI. 2T sin 2(r .. w. The.efore T .. 1.46W.

...,.. 2.11 In ,hooling an • .,ow from a bow. an .",hor hold> I~ bow wrtic.1 and pulls back On Ihe anow "-;Ih • 101"CC

of !ION. The 1 .. -0 hal, ·., of lhe >!ring mab~' 'Rgles ... ilh II>e ~eniea!' Whal is lh. te"';on in tl>e >lrin,?

, Stning horizonlal forces 10 ler() al lhe point On lhe >Iring l>eld by II>e .. <he . .... obloin!lO _ 2T sin 25' , .. I>cnce T .. 21b:

1.U In Fig . 2·</(8 ). Iht pulleys al. fricti ()nks< and Ih. syslem hanV.' equilibrium_ If .', is . lOON .... ight , ... hat .,e the values of W, and .... 1

, , ,., ,.,

, The knot aoo.'. w, i, in equitib. ium under th. action of lh.ec fOfCCi. as ibo'A'n in Fi,. 2-9(1) ). Since Ihe pulleys .'. f.ic1ionlei.l. T, _ ,.., ; To " w,. Also T, .. w,. W. It< eiwn T, .. .., - :!OO N. From r F, _ 0, T, si n 15"' - T" in W" " O. (Note th • .<-compo"""' equation in>"OI>'". sine function. becau~ an"', ." "ith le!opeCl 10 Y .~is , ) Then 200(o. j 74) " T,(O_ 761i) ~ T, " U!U! " ",. From L F, " o. 1, «<3$"' ... T, 1m SO'- T," o or 200(0 ,819) + I~O.(3 ) - T, => T, " Z!i!.!J.'!. .. "',_

1.13 Suppose w, in Fig. 2.9(Q) "'eigh, SOON, Find the ~.lu .. of..., and ... , if Ihe syslem is 10 han, in e<juilibrium,

, NQW T, .. 500 N in Fig. 2-9(b). r F, .. O=> T, sin 35"' - 1i sin SO"' .. 0 =>0.574 T, .. 0, 7667;. 0. IOI>'ing for T, .

T," J.3J7;

1: F, .. 0 => T, (:OS 35' + T, (:OS SO' - T, _ O=> O. ~19T, ... 0.6431j .. 500 N. Subo,!1!u!in, 1.33 T, for T, we have

0.1I1</( 1.33 T, l + 0.643 T, .. 500 N => 1.737; a 500 N

From (1). ,..~ gel T, - lIl:I...t::!-T, " ml!

2.14 Find Ihe lensioo~ in Ihe fopes >h()WII in Fig. 2· 10 if Ih~ supported obj«1 ,"'eigh$ 600 N.

flI. 2-1.

U)

EQUILIBRIUM OF CONCURRENT FORCES a 25

• Lel~' ",lee!. a. ou, object Ihe tl>Ol ot A. beau", .... know one lora: actin, on it. The ,...,ilhl pull< down on it witb a fo',", of 600 N al>d $0 Ihe fr.e·body dia .... m for the knol i, as shooon in Fi,. 2-l!(d). Applyin, !lie first condition for equilibrium to Iblt diagram. ",', ha ...

"The: tint equation yield. T, " T, . Substitution of T, for T, in lh. S«O<Id equation gives r, - .J:iIil:!. al>d Ihis is also T, .

leI us DO .. iooIate kllOl B as ou, obj-cct. I" free.body doiagnlm ;, .I>o ... ·n ;n Fig 2-11{b). We hi .... already found Ihat T, _ 346 N .00 $0 Ihe equilibrium equations are

"The: last equation yields T,_ 8:ZZB. SUbsl;lut;n, Ihi. in lhe prior equal;on gi~ T, .. ~. , " .' , ,

,., '"

Wo can no:rw proceed 10 lhe knot al C 000 lbe I ... -body diagram of Fig 2-11(0). Recallin! Ihal 1;" 346 N,

Tbc Laner equation yields To _ m N.

T, -+- 346.in)(l' - To _20' .. 0

T.';n 20' - 346001 )(l' .. 0

[NOlO lbat from the symmetry of lhe sy:slem .... C<)Uld ha"e ded~d T, .. T, al>d T. _ T, .]

1.15 II II' - 40N in the equilibrium ';",ation oIoo ... n in Fi,. 2-12(d) fil>d T, and T,.

,., '" Fl&. z.,U

, Tbc knot i. in eqwlibrium under I"" action of Ih .... forces. al>d I"" free-body d'iagtam is a. shown in Fi,. 2.\2(b ). T, _ .... _ 40N.

Ir. - O!}T, sin M-T,00II6O'_ O or (0.940)7; _ (O.500)T" T, _I.SST,

~ F,. - O=> T, sin 60" - T,00Ii 70' - T, .. ~. (0.866)T, - (0.342)T,_ T, _40N

26 0 CHAPTER 2

Subsliluling 10. 7;.

(O.866)(I.88T,J - (O.}42)T, _ 4() N I.29T,- 4()N T, - llJlli and T, - (1.88}(31.0) - all!.

1.16 Rd~, 10> Fig. 2·12(a). The co.ds an: 'Iroog C"o>ugh to "';tI\sWld. maximlm t."'ion a l SON. What Is the large!! "lIIlue o>f .. that t~y an iUppon IS shown?

I F.om Pmb. 2.15 t~ eq uIlibrium cquatIom .,e.la, Iny "',.

T,_I.SST, (1)

0.8667; - 0.342T, _ ... (2)

From Eq. (I) it I. C~llf that T, > T, a1 .... ys. Therel"", T, win .nch the brUking po:>int first. We thus lei T, _ SO 1'1 10> ~nd lhe corresponding .... From (I).

I . SST,- 80N~ T, - W

from (2). .. - (O.866)(SO N) - (O. }42)(42.6N) _~

1.17 n.e ... Ighl W, In Fig. 2·13{Q) isJOON. Find T,. T" T, . and W"

'.

,., '" '" fl&.2- 13

I from Fig. 2·13(b): T, sin 37' - JOO 10 T, .. ~. AI50. T," T, coo37' - .!OO..t! . From Fig_ 2-13{c). T, coo 53" - T,. loa T,_~, BUI T,sin 53"" W" 10 W, _~. (N",e In.wers..., la lwo-pla.ce lIX\I.acy) .

1.18 [18, _II, in fig. 2·14. whll can be .... d lbout T" T" T" W,' ond W, pr<l'o'id«llhe pulley Is friction[ ... ?

FIc. 2-14

I Wh. r. Ille III ••• ropes join. T, loin II , - T, loin 11,. so 7; - T, . Also T, -T, Ind W, " T,. Furtber . lhe .qui[ibrium condition lor Ihe ~.rlical direction i, W, -2T, COlI II, . Therelore T, " T, " T," W, " W,{(2cos II,).

2.19 In reference to Fi._ 2·14. if II, ,, 53" Q~d 9, .. 31", how I.rle i. W,;., ".,.......ri..,n '0 W,?

I S;nce T, -T, - W,. the .qui[ibrium equllions ar. W, sin 37" _ T, si n 51' Ind T, COS 53" + W,cos31" _ W,. S<.>lv;ng 1m T, in the 1i"1 .quation ond pl.cing il into lhe M'COnd yields Wt _ I.2:l W,.

1.lQ Suppose Ihal W, - w, in R I . 2·1 4 and Ihll II, " 53' , Find 11,.

I The equilibrium equalio..s ... W, sin 9, .. 7; sin 53' Ind T, COS 53" + W; COS 9, .. W" Eliminating T, in the

EQUILIBRIUM OF CONCURRENT FORCES a 27

second equation by usc of the ~rst yield>

sin 9,ro(5)' + em 9, sin 53'· sin 53' or sin(9, + SJ"). sin ll'

Thu •.• itbe. 9 , _ 0 [the two .... i&hu a •• on tM .. me ",,"ieal line and T, - 01 Or ebe (8 , + ~)') + 53' - 18O"~

9, .14' [Ihe "normal" . ns .... r] .

1.21 Ref. , to F". 2-15(_). What " the 'ension in cord Air!

,., '"

, Fint ~nd Iht tension in the cords htlow the cord in q",,"ion by balancing """,ieal fortts .1 11>1: lowe, jU<>CIion in Fig. 2·1S(~): 2 T COS 30" • 70 .... T. 4{1.4 N. Equilibrium conditions fo r jun<tion B [fi, 2·15(b)1 are T' coo 4(r. TAO + T oin 30"; T' sin 40". T cos 30". Subslituting 10. T and elimina'ing T ' from Ihe ,_ relations yield. TAO . z.L.l./:!, From sym me'ry ,h . .. me equation. are found at ju<>CIion <1 .

2..22 The: weight w in Fig. 2·16(~);s 80 N and i. in equilibrium. Fir.d T,. Ji. T, . and T. ,

" " "

" " " , , ,

" •

so>s

" ,., '" '" t". 2.16

, UMling the lower knOT <1 and Ihe urope' knot B. IO'e hl", 1M fre. ·body dilgrams of Fig. 2·1 6(b ) and (e ). We address tIM: equilibrium of knol A /i"1 bccau,"" il invulv .. 11>e only known lorce ,

rl".·O~ Tl 'in 25· - T, .. 0 T, - 0.423T,

L F, " O~ T,ros 25" - 80N _ 0 0.9(I6T, _ 80 N T, .!!a,ll!

Then f.om 1:>0>10.0 . 1; - ll.!l::!. Turning now U} knOT B. and "'m.mboring Ihal IO~ know T,.

L 1", .0:::0 T.sin 55' - 1; - T,sin 25' .. 0

• 0819 7; - T, .0,423T, .17.4 N 0'

L F. • O~ T.""" S5' - T, coo2S' - O:::OO.573T, " 80 N:::O r. - HQ..IS.

From (i). ·r, - O.HI<n, - 37.4 .. Z11:!.

rghtoo matmal

28 a CHAPTER 2

l.l.l Th< pulleys """"n in Fig . 2-17(<1' ha"e I>C,ligible _illJl' and friction. Who, is , he value of .. if i . .. mains \.UfIpontd., ...... ' .. n by the 7G-N weigh'?

,., '" I Con~idr:. 'he I",,'e, pulley system ( just above weight wI IS ,lie ')'SIem in equilibrium. Sinct';' if, weichtl .... il if, in <'<!uilibrium uf\de. the action of five fOlta .... hown in Fig. 2-17(b). T, and T. and T, a.e due .o!be common co.d ..... ppoed abou! ,lie ceilinl pulley di.ec,ly .~. arQund IIIe ,)'S,em pulley. and around tile OIM' ceiling pulley. and fiully COIlI>CClcd .o ,he 7G-N weilh., SiDCC the pulle)'S •• e friClionleS$. T, _ T,_ T._ 70 N, For the body in equilibrium.

L F.. - 0::> T, CO!! 40" - T, _0::> T, - (O.7661(70N)-ll&l!

L 1';- - O=> T, + To + T, sin4O' - w _O=> W " 70N + 70N + (70N)(O.642) - ill.!::!.

2.14 How la"e is 1M force tha, ""'ohes tM patient 'slc, in Fi,. 2-]81 How la.S" an upward forO!' does 1M de"", exen on foot and Ie, togetM'? A.>ulllC fri<:l;.,nl~ Ind m ... l ... pulle)'S.

f1c.2--11

I 3 k& .... ei~ abo<ot JON. Sina: the pulleys lie fricrionles$ and witb tlClUgiblc m .... the tension T in the <:Qrd is the ....... vctyWb.c •• . T hold!. up 1M .... ei.h •• $0 T .. JO N , Th< folCeS on the Ie, and fOOl from the de"", OR' ""u~ by the ,ension. in ,he _d. The I>orizontal or .".,dUn. forO!' is T + T CO!! )0" _ ~. wbile the upward force i. T + T si n )t)' '' :!ll:!.

1.1S For ,he situation shown in Fig. 2·]9. wi.h wha' fom: mu!ll ,he IiOO-N man pull down ..... '" on ,he rope: 10 suppon him ... \f free from ,he floor? """me 'M pulle)'S have I>C&ligible friction and weicht,

I Call T the ,."""" in the top< the man if, holdin.; T is ,he ~me throughoollhe one pie<:e of rope. The 01 .... ven ..... l force on the man is.he .ension in the rope: IlIarlINI.O the pulley above the man', he.d ..... hid! nlll" he 2T fOl' the pulley in equilibrium. 'The .... , v.rt ..... 1 force is 3T. wbioh i. b.al.~ by hi> .... i.h' of 600 N. Therd"", the man e.eru . downward pull of 200 N.

loU In [he ~.up of RI. 2.20. the mobile pu]ley and tt.. ~.rd pu]ley. both fm"",Jess. a.e ....ociated wi,h equal ..... iJhI> ... Find II><: .ngIc 8.

I s;""" tM t.nsion in the cord is .... the condition f'" _ertkal equilibrium of the mobile pulley is 2 .. ';n 9 - .... or sin 9 .. \ . Of 9_JO',

atanal

EQUILIBRIUM OF CONCURRENT FORCES D 29

fIt· lol'

- - 6 -- -- ----

• .

2.2 FRlcnON AND INCLINED PL" NES

2.27 A 2(W).N w...,n is In I:M: pull<d up a 30" indine al ronslanllp(<<i . How lalle a force parall<llo the incliJle is needed if fria;"" e!fro! are negligible?

,., '" ",. loll

, The silu.alion i. shown in Fig. 2-21(Q). Ikca_ 1M wagon "")\ItS al constanl speed along. slraight liJle. ill .elocil)" .""'or ~ """Slant . Tbcrefore Ih. wa!oo is in uanolltional equilibrium. and Ihe tim condition for <quilibriu m a;JPIie. 10 il

We ;solaf. Ih. "'"!Oft '" ,be ob;'ct, Thrn nonnegligible form act on iI: (1 ) !be pulJ of pa'oi,y ... (ill weighl). dire<le<! iII.igh! down; (2) Tbe (0Ke P e~erTed Oft tbe "'agon parallel 10 lbe incline 10 pull il up lbe incline ; (3) tbe push Yof Thc ineli"" thl' OUpponl Thc WI..,., . Thtoe three force> . re 011""", in Ihe frce-body diagram. Fig. 2·21(b),

For siluations invol'oing i""lin«. it is conv. ni.n, 10 take Tbe ~ ... is pa.aUcllO lbe incline and lbe y ui!. pc1p<'ndi<ula, 10 iT. Afte. lakin, componcnn llonglhC'$t "C' . "'e can wriTe ,be firsT """dilion for equilibrium.

P - O.5Qw _ O

Solving The ti~ equaTion and recaUing that"' .. lOON, w. find that P _ 0.30 ... _ lQ2l:!, Tbc . equi.e<! pulling force i. lOON,

30 a CHAPTER 2

1.18 A box wciJ.hin, lOON i. at res! on a IIoriIOntal tloor. TIle coefficient of . t. tic frictioo between tile box and tbe l\ooT i1 0.4. What il the ...,.ll«t force F exetted eastwa.d and upward at an angle 01 JIr "';th tM horizontal that can $I.n , 1>0 bolt in motion?

,

• FIc· 1-22

I Fint d ..... a fOl"C<: Ilia""m. Fil . 2·Z2. Nu,. ""n,.;de. tho force. in.1>o x di' O<"Iion and apply tho conditions for "'Iuilibrium • ..otin! I "'Iual. ito maximum val"" to >I.n ""olion.

FroiO - I_O O.866F -I - ",N _O. ~N

NQw apply tho condition. rm equilibrium to tile forces in y direction.

N+FJinO-W _ O N+0.5F - l00 _0 N _ lOO _ 0.5F

Sub>ti'uting 'hi. equ,';oo for N in O.866F _ 0.4N ."", . .,.

O.866F - 0.4( 100 - O.SF) O.866F + 0.2F .-40 F -l1.1l!.

1.19 A bklck "" an i""lin«l plane jn!! btal'" to.tip if the in<!inat;oo 01 tbt plane is 50'. (.) Wha' is ,1>< oodl\cien. of lillie ftiction? (II) If tho bklck ha. a m ... of 2 kg. what i. the actual frictional force just befo.e il be,;ns to . Iip?

FIc· 1-U

I ,,~ , - "'t , in 5(1' " , .. I .,_ I N" tan 5(1' _ 1.19"2

III ) I. "' mg Jin 50' - 2(9.8)(0.766) .. UJ!l!

1." A SQ.N box i •• Iid .traight ..,ross ,he tloo. at «",5\onl Sf>Ced by. force of Z5 N. I •• hown in Fig. 2·24(a). II ...... la.,e a friction rme. imped .. the moI;on of the box? How I •• ~ ;, the normal force?

I

,., '" ",. H4

op nghted m

EQUILIBRIUM OF CONCURRENT FOfICES 0 31

, NQ!~ rh~ f",..,e< acrin. on rM box a •• hown in Fil. 2·201{s). ".., fri<1;"n i$ I and rl>c nonnal fo,.,.. Ihe . upporting fOT« u~n(d by 1M Hoor. i, Y. The fr«_body diogram and O)ttIponen .. are , """'n in Fit. 2·24(1)). Becouse Ihe box i. movin, wilh <OnSlanl >elo<ity . it i. in .Guilibrium. 1l>e firoc condition fOT eGu;l;brium tells UI that 25 COil .10" - I _ 0

w. can !oOI~ 10' 1.1 onct 10 find thaI I '" 19 N. 1l>e friction fOTce;' l.!!l!. To fir><! Y w. u'" Ih. flC! thaI

I,<;. - o or Y + 25 r.in.lO" - SO - 0

Solving give 1M m<>fm.1 i()Tte a, r _ }! N.

2_\1 Each oIlhe objects in Fi&. 2·25;' in equilibrium. Find the normal 10'00. Y. in cadi ca ...

(.,

, We apply I: F, _ 0 in each case.

(sl Y+2OJsin3O" -~- 0 frum .. hich (1)) Y - 200sin3O" - ISO _ 0 frumwhkfl (e) Y-200 cos8 _0 fromwhi<h

y • ..,.

y . """ Y - {200 COil 61) N

1.32 For the si.ual;"n. (l( Prot>. 2.31. find 1M e<xffici. nt (l( kineti< fric. ion if the objM i. mo';n! ";th con.u nt 'p«d.

, W. have already fouod Y 10 ' eadl case in Prob. 2.31. To find I. the slidin, tri<tiooo force. w. _ 1: F, _ 0,

(. ) 2OOcos)(l'-I - O and ," 1 - 173N

1l>en. 1'. - I IY - 173/400-!Ml,

(6) 2OOcos)(l'-1 - 0 and," I - IBN

"Thcn.I'. _IIY. I73.'250 .. ~, Ie) - 200r.in9 + 1 _ 0 ar><!w 1 - (200r.in9)N

1l>en, P . - I IY - (2OIlsin 9)1(200 _ 9) - .Il!il!

1.ll SUP\l'05C Ihal in FiS· 2-251~) tM bl""k i. al •• st . The a".te nf tM i""linc is slowly ;,,""ased. AI an anp. 8 _ 41', Ihe bto.:k begin. to slide . What;' 1M coefficient 01 "atic friction bet ..... en thc bl""k and tM incline? (The block and $U.rfac:e.", not rh ... me as in Probs. 2.3 \ ond 2.32.)

, Allh. instant the bkd begin. 10 slide, 1M lriction force has its criric.1.-al .... Then-for., p , - fl Y al rhat inSlant . Following the melhod of Probs. 2.31 ar><! 2.32. w. have

Y _wcos8

The.efore. WM" iliding jusl "an",

I wr.in 8 "' ______ I&n9

Y .. cos9

BUI fJ was found by uperimentlO be 41'. The refore . p, _ tan4? -~.

opynghtoo IT £na

32 0 CHAPTER 2

2.34 Two ..... tltt. are hung O\'er lWO frictionless pulleys. as shoo"'n in Fi,. 2-26(Q). What ... eight W";II faU5e iM )OO.lb block to juSt otan moving 101M nth!?

, ,

r. F. W .

• ~" Ih

'"' .. , flt. 2-1.6

, From 1M force diagram. Fig. 2·26(b).

1; - 40 cos.f5" - 28.3 Ib T, _40';n 45"" 28.3Ib

T,+ W,+ N -300lb-O 28.3Ib + W sin 30'+ N _ 300ib NO' JOO lb - 28.3Ib - O.SW

W 00030' - 28.3Ib -0.3N _ 0

We substitute NO' 271.7 -O.SW into lhe Il$f equat;';'" to get

W cos )l)' - 28.3 lb - 0.3(271.1 - O.SW) _ O.

Sol';n, for W &i~es W .. IOI! lb.

l.J5 A ... ume that W .. 60 Ib, 9 .. 43". and II, _0.3 in Fig. 2·27(Q )_ Whal pu.h directed up the plane ";11 move the block with coMlanl "",ed I.) up the plane at>d (II) oo...n tM plane?

w, w,

'"' .. , ,.,

, The w"ght componenu. an.:

w, .. (601b).in43"_~ w, _(601b)cos43"_~

N - W,-O N "USS lb

F. .. II,N .. O. 3(43.88) F, - 13_161b

~F. - o P-F,- W. _ O P - 13_16+40.92 P - M-llb

(II) The puoIl is now just enough to keep motion down plane to COMl 0.1 .p<e<I_ From Fig. 2·27(c),

P+F. - W, _ O P - 40.'12 lb -13.16 Ih P_Z1.8 1b

EOUIUBRIUM OF CONCURRENT FORCES a 33

2.36 Whal I\oriuldtal pw.b P is required 10 ju>l hold • 200-1" block on a &:!' inclined pi .... if " , _ 0, 41

,.

,., ,., , Figur( 2-28(~) is !he h_body di ..... m.

W. - 200 COl JO"-lll.l.t! w,. - 200 sin 30' - ll!!!l!. F, - ",N N .. P <0$ 30' .. W, " (). 866P .. 100 I"

P, +" ,N - W, -O " , .. 0. '

P sin JO" + 0.'(O.8Il6P" 100 N) - 173.2 N - () O.~f + O. J46f + 40 N - \73 .2 N - 0

1.17 In Prob, 2.36 . what hori:wntal push _Id just sran the blod: moving up the plane?

, No .... Fig. 2·28(1)) applies.

2F,-0 N_looN-+-O.866P Psin 30' - W, - " ,N _ ()

0.5P - 173.2 N - (0.4)(0,866P -+- 100 N) .. 0 O.S P - 173.2 N - O.146P - 40 Ib .. O

p _ mN

2.J1 In FiJ. 2-29(a). the ')">Icm i> ;n equilibrium. t.) What is the maximum value that II' can havc if the friction fort>: 1)(\ !he 4().N bIod cannot exceed 12.0 N? (t) What i>!be wcmccnt QI ~atic hictillll hct1ljttn!be bIod: and tabletop?

• • I I I f .101' t ",

,., ,.,

'. • ,.

, (a, The {.u-body diagrall\$ for the bl<:ld and the knot are <.bown in F". 2·29(1)) and (e),

/,,-,-12.0 1" implies T",-.. '" 12.0N

For tbe knoc:

7i - T, <os)/J'

Eliminating T" .. e gel II' _ T, tanJO" _ 0.!l17T, . Thus ""1_ " .. O.~nT "'_I - 6.92 N. (t) 1', - f ,,_JN _ 12/40 .. ~.

• ,,'

2.)9 The block 01 Fig. 2·29\a) i. JUSt on lbe verse (>{ slipping, If .. _ 8.0 N. " 'hal is 1M """,fici.cnl 01 ~alic friction between the block and lablclop1

34 D CHAPTER 2

, At the verge of dipping. / . " /"_,, and the ""rre_ding hanging .. ..,ight is ... _ KO N, Min I'rob_ 2.38 • ... _0,5177; and T," / .. Thu. 1. ,_. - ... /0.517 - 8.0/0.517 - 13.9 N ~ 1'. - 1),11/40 - \!....HI.

lAG Find the no", .. 1 fl>l"" ""ing QI'I the blcxk in e""h of the equilibrium <ituat;"ns $lIo ... " in Fig 2·30_

, In each ca .. lriction i. nuc,,,,J}' lor equilibrium_ BUI friction 00e. not ente, in wl"ing lor N

,

f . ,

Ih ) '" nc· 1-31

1_' Free·bod~ diagram is Fig _ 2·31(s ) (F - 2ON,'" - 50 N).

L F, - O:)N - ... + F sin 55'- 0 No. 50 N - (20 N)(0.819)_~

, II ) F,ee·bod~ di.pam i. Fi,. 2·31 (b ) ( ... _ 60 NJ.

N - {60N)(0.166) -~

1<) F"",.body diagt'am i. Fig 2·31«): (F - 10 N . ... - 60 N).

N - 60(0.166) + 10(0,643) - tl&./'!,

2.41 'The bloo;k shown in fig_ 2·30(~ ) d ido. "';tb romOlant .p«d u...se. tbe Klion of the lora: shown. I_I Ho ... large i. the retanling ftictio.m fon;e' III) WIIal is the ooeflXienl oi kinelM:!rict;"n bel"'een lhe block and the IIoor'

, U .. Fig. 2·31(s) and 1', 01> , 2, 4O(s ).

(. I LF,_O=> F COIoSS"_ /_O (N ... ... ,.O)

L l U N IIII !"-N- 34N-!lH

20(0.571) -I -!.lll:i

Z.4Z 'The block . hown in Fig. 2.3O(b) didos 01 oonstanl speed down the incline . I_I 110'" I*,II" i.lbe friction force that ~ ilS moIion? (II, What is the ooe~nt of didin, (kiMti<:) friction betw....,n the block and plane?

, U .. F~. z.]1(b) and I'rob. 2.-I0(b).

I_I}: F." 0=>1 - ...... n 010" - 0 / - 6ON(0 ,643)_~

1 38.6 N (II) 1'. - -----..,

N "N

z.e The block in Fig. 2·.10«) just begins to ,Iide up the inchne when the pushing fora: <ho-..'n is i""reased 10

EQUlLlBRlUM OF CONCURRENT FORCES D 35

,., '" '" Fi&.1-32

7ON. (or, What is the critical .. ati~ lriction 10<..., on it? (II) Whit is the volue of the ~ftkient of .. atic '-'- ' ? ""~ .

, U'" Fi8' l-31(c) and P,ob. 2,4il(c),

(or, L F. • (I::> F cos .so" - ... >in .\0' - / • (I. F. 70 N and block ju" " afl. to frlfWe:f? / ,. / .,_., .. 70(0,766) -

60(0,643). Jl..!!1:I. (II) " . 1.(_, _ 15.0 N.

W • N 91 N

2M The 'ystem in Fig. 2-32(Q) remaim at rest when the hanginl _iAllt ... i. 220 N. What a.e the maw"tOOe and direction of the f.iction fum: OIl the 200 N bloct ? The pull-cy is l!idionle<$,

, Sin~ the ptllley i. lri<tionle". the len.ion i$ ' he $a"", lhroDg./>out the rord. Th. I'n-body diasram. for the t ... o blocks II •• al in Fig, 2-32ib) and (c ), In principle the frictional fo= / oould be .ith • • do "11 lhe incline o. up the inclinc depe nding on the details of lhe Pfohlem, In this ca", "'e can quickly....crt tl'lol it is down the incline wilh the following re~"""ing: T .. 22O N . Opposing T along the indine: is tbe compone:nt of lhe 2(1).N " 'eighl do"'n th. ind ine:. This is sure!y Ie.. than lhe /u]1 200 N and thus is ill$l>8'.cient 10 bal""",, T, The.eIOf. the help 01 friction do"", lhe incline is nc:ccnar)'. To obtai n 1 "". solve

(along incline) T - 200 <i n 3S - / . 0::>/.220 - 200(0.574) .. ~

(Not. th31 fo. this proble m tile oormal fo=. N, doe. not enler the <.kulation.)

CHAPTER 3

Kinematics in One DimensionD

3.1 DIMENSIONS AND UNITS; CONSTANT.ACCELERATION PROBLEMS

U A car', odomet ... eads 226if1 km at the lIat1 of a trip alld 22791 km at tile end. llH: nip look 4 h. What was 1M "",'s .,·~rltge spu-d in kilometers per hour? In melerS per S«Ond1

, d di>l",,"lra~lcd (12791 - 22687)btI ~~k-'" Ave,agc'pce" limetaten" 4h -~

3.2 An au'" ,ra~cl •• r • r'''' ol25 krtl/ h for 4 min, then a, 5Okm/ h for 8 min. and finaUy a. 20 km/ h for 2 min. Find (OIl the 100al diSlance OO>'crcd in kilometen an<! (b, ,be ."",age >peed for tho compklr trip in meters pc. secoO>d .

, (.1 O;,;,a""" tray.kd .. d , + d, + d,. wl\to,...

<I," (25 k:)(4 mil!)(601~ ) -1\ km

d, .. (20 k;)(2 rhi(I)(601~) .. j km Thus d, +d, +d,-'l.i!!!.

( 'm),., " ) <1, _ SOT (Sm,n'\60 il\;q _6j km

d.+d,+ d, .:..ow (b) A"e • .,. opecd.. _ .. (9 k1 )( I((() m/-.)I( 141hi(o x 60., ,,,;,,,) '" 10.7 mi.

total tu'"

3.3 A runner Ir.""," I .S laps around . circular u .ck in I Ii"", of SOs. n... diameter of lhe !Tack is 40 m and ill artUmf.,...""" is 126m. Roo (<lI !~ a~trale IP«<! of 1M runMT al'ld (6') lhe magnill1<k of lhe runner', .vt"'~ nloeit~.

, (., Average >p«d .. dio .... _ .. (L.:,!:oe)(126 m~) .. UlmiJ. Ume ""I

(b) Average nlocily is a veclor . Ii i'lhe displacement VC<:Ior for the time lap'" ol inle,esl divided by the time I.pse--in lhis case, SO I . Sinot I! laps ha~e Uolen pI~, the displ_ment points from the $Iarting point on Ihe lrack to a poinl on lhe track ! lap ' '''y, ,.h ieh i, of COIl .... dire<:tly IOCfO!,S • diamete, ol til<, lrack. The mllInilude ollhe av.ra~ nloeity is lherefore lhe m.1&nilude of lhe displacement divided by the lime lap"' . Thus mllnimde of a~.nl8C' ~Iocily _ 40 m/.'iO, -!!..I!mlI.

3.4 U!>C dimensional anal)"lis 10 determine ,.hicll of 1M following equalions is ocnlinly .,.onl:

, ..

36

F _ ~ , ..tM:re A aDd 10 are length< . nd [F1 - [MLT-'l. 'The OIhe. symbols have IIIei. u ..... 1 meaning.

, [wl_[LT-'nT) - IL1 , bullA) - IL), SO equation.l. _ VI an he......-. [ "" ~1 - IMllT'L - '1 - [MT' L - 'I, bul IF) - IMLT-' 1; hen<e F - ",1. io ir>«JfnCI. [",,,11[ - [M][Lr-'l[T- '1- [MLr- ' [ aDd F_ "'~ II is di ..... Mionally corrcct. [u'I2&I- \(L'IT' )I(LIM[ '"' IL[; aDd >i n<e I~ I " [L[, h - ~'f2& is dimen>ionally correct . Sin<e [u l _ [L T-'I, [(2&10 )'''1 - I(L "'T- ') L '~I " [L T- 'I, ~ - (2&10 )'~ is olso dl ..... "';'Onally COIT«' . w. not. IMI pol'" numbers .... dimell$ionle ...

If. is diManoo and I is lime , ",hal midi be the dimcn>ions of e,. e" C" .nd e . in each of the folJowing tquatiom1

J "C,. J " je,f . .. e, sin C,I

(HiN: The .. gument of any triROfll>metric function mUll be dimeMionkss.)

ngntea IT na

KINEMATICS IN ONE DIMENSION a 37

, The dimensioa of sis [L]. 10 all exprenions on 1M ri&hl. haDd <ides of 1M eq uations muOl also ha"" Ih~ dimensions of lenglh. [C,]_ [LI T). since [C,I] a.e then I(L/T)T] - L IC,I- [Lr'J. rC,)- [L ) !>cause 1M ,ine i. dime ........ lem. and sinee 1M a.gument of. trigonolndfi( fundion has no uniu. [C.]_ [r'J.

1.' The speed v 0(. WI"" on a >lrin, dtpe:od> on rhe r~nsioa Fin lhe orring aDd lhe nil .. pe. unil length ",I to( tile mingo If ir is tooW1llhat [F) - IMLIlTr '. fiDd rhe conWnU a and b in ,he following equation for rhe speed of a wi"" on • sIring: v - (constant)F"(m/ f) ' .

, II is siYen Ibar Ivl -I Frl", lfr. Aft • • inserting inlO Ihis . xp .... ion Iv] and IF] . ..... have. _ ailing M" means no II\a$I units invol""d. [M"L'T" ) _ [ML T" "rIM L " 1' - IM-r''lLr- '1 r ' f - A fundamental dimell$ion mUSI appea. 10 lhe >MIe po..-~, On both sides of an equarion~ benee ... t h .. 0 ... - h .. I, and -:z.,_ -I. From lhis"_1 and h --I.

1.1 The frequency of vibralion / of a mOS$ .. al 1M end of a spring rh., has a orift'n.ess ronstanl k is •• Ialed 10 m and k by a relation of the form / .. (cor.nanl)",· k'. U .. dimensional anaJ~i$ 10 fiDd Q and b. II is known thar Ill-In ' and Ik[ .. [Mlln'· , A. in Prob. 3.6.

.. +h - O and -2/1 .. - 1 0. h_ - a. !

3.8 A body wilh initial velocilY 8 m/$ moves along .. lraig)1I line with constanl acct'leration and rr.",,1s MO m in 40 I. For 1M 40, inlerval. find (a) lhe overage velocity. (hi lhe final ""Io<:ity. and (e) rhe ..uierarion.

, A .. llrne :t .. 0 a, I • O. . displacemenl 640 m

(a) Average velocily" _ .. ~.]6 mI. nme dllpxd .... S

(b, c) v, .. v. + /II. We k""'" V. _ 8 mls and I .. 40 I. bUI we don 'l know a. lIowcv~ .. ,

640 m • (8 mJ.)(40 sj t IQ( 40 sj'

• _ 2(640- 3~jm _ nAOmi.' , ... Su~liluting inlo ou. velocil~ formuLo VI _ 8 mi. + (0.40 ml.')( 40 sj '" hlnlI.

l .t A u"d SUlrII from ,..$1 and move. wirb a oonStanr acct'leration of , mi.' . Find ito ,peed and rhe distance ","yeled after 4 . hao elapKd.

, v, .v . . .. I,V. _ 0 . ... Sm/.' ,1.4.

V,,,O t (S ml.'j(4 sj.lQJnlI

To JC1 distance , wbich in rhi. caK is lhe same as lhe I1I.Ignilude of lbe di.placement . .. .., ba""

%. ~.I + \ .. 1' - 0 + 1(5 mll')(4 .j'·b

3.10 A box slides down an incline wilh uni form a=leration. It otart. from rnt and ." ..... speed oIl.7 mi. in h. Find (a) !be IKCeIc",tioI\ and (b) lhe di~ moved in !be 6 .. 1 61.

, LeI s . v rep.-esenl rhe displacernenr Ind ""Iocily doW1l lbe: indine. respcrtiY~Jy. Then, (a) v, _ v. t at and " .., are gi""n v. " O. v, .27 mis, I • • lh; thull.7 m/$ _ 0 + ,, (lO s). o. Q _ 0,90 Jl I.'. Now wt' .,e jn,erelled in lirne .... 6.0 S. So (b ) a - ..... + 1<,,' _ 0 + IIn.90 m/.,)(6.0 I)' .. l!l.m.

l. n A ca. OTarts from rCSl and was" doW1l a hill wilb a comranl accelerarion . If it goes 90 m in 8 f. ~n.d (a) tbe: a=ler.tion and (b l Ibe yelocity . fter 81.

, ' (.) '''2'''' (sinccv. _ O)

(b) v - '" - 28 >( 8 - ZZ.4m/i

, 9O - -a(8)' ,

opvr hIed IT ria

38 0 CHAPTER 3

l.U A ear is a=leralins uniformly .. il JW"'OS lWO theckpoinll IILaI an: XI III apan. lbc lime laken beTWftn checkpoinlS i1 4.01 and lhe ear', speed at lhe liM checkpoinT is 5.0ml •. Find lbe ear'. a=leralioa and ilS speC<! II lhe >eeond checkpoinl.

, CaUinllhe fiRI <be<:kpoinl 1M initial posilioa and the 5eWnd lhe final pOOlioa. ~ ha~. ~. - 5.0 mI •. I _ 30m, , _4.0 •. To find. _ ~ lhe displact"mmt fijuation

x_V ... .. !.I' Of .lOm _ (5.0ml .)(4.0.) .. j.(4.0,)' 0'1 ._I.25ml. >

lbcn v, _ v . .. III yickb v, _ 5.0 mI." (1.25 m/.')(4.01) -1lmlJ.

l . ll An auto'. wlocity incru$« uni formly from 6.0 mI. 10 20 mi . ... bile <:(IV.rinl 70 m. Find .be ...,.,..\enlioa and lhe lime taken.

, We a.e Jiven v, - 20 m/. and ~.- 6.0m/1 and Ihe O$$OCialed displac(ment i. 70 m. Si""" lhe elllflKd lime , i, not &iven . _ find • uoinl vi .. v; + 2ax. o. (20 m/.)' _ (6.0 m/I)' + 201(70 mI. ThIlS • - il.mJ.i.. I\I{)w 1 folio"",, immedialely from

2Om/. _ 6.0m/.+ (2.6 m/I')I ... , -M>

l .14 A pia ... lIam from rw and .oocleralcs alon, lbe pound before lakeol'r. It moves 600m in I2s. Find (a) (he ...,.,..Iefllioa. (b) opced al lhe end 01121, (e, distance moved durinllhe twelfth """"n<!..

, Weall:&i~nvG -O, A -600m,' - I2I. (a) I " V ... "\111 ' . or 600m _ 0+\_(12 Ij'. So _ _ 8,33 mil'. (b) v, _ V."III. or v, _0 .. (8,33 ml l'XI2.j y<\d u, - Wmli, (t) Reme mberinllh.al lhe filll second is bel",«n I - 0 and I - I s. ~ ualiu Ihal 1M 1_lfih oecond ill bwl<ffn I - 11 and I _ 12 s. $i""" we aln:a.dy koow I (I .. 12 sj. _ solve for I (t .. I L . j :

I _ 0 + 1(&.33 m/.')(1 I .j' .. 504 m

Our .",we. is lhen w,(twetf! h 5) _ %(1 _ 12.j - %(1- II s) -'i!t!!!.

l.U A rrain ruM;n* .I.lOm/s i. slowed uniformlY '1) a .. op in 44 1. Find ,he .=Ierat;"n and tM lIoppinl dim ......

, Here v. -30m/. and v, -O al 1_441. v, _v. +./ yiekb O- .lOm/s+.(44 s). or d _ - 0 68 ml1' . I - v .. + jar' - (.lO m(.)( 44 0) + j( -0.68 ml.')(44 I j' - lm2..m..

3.1' An object movinl al 13 mI. slows uniformly al 1M rate of 2.0ml . eadl scrond for a lime of 6,. ~Iennine (a) ;11 Anal speed. (.) its "'enge speC<! durinS 1M 6 •. and (e) 1M dist.""" moved in lhe 6s.

, Slowinl "atIM rale 012.0 mI. ead! 5«OI"Id"' means a _ -2.0 m/I' . We arc &i""n .... . 13 mi. and 1_ 6.0. (.) lbcn v, . v.+ a' _ 13 mI . .. ( - 2.0 m/ .')(6.0 1) -ilmIJ. Because iMTantanwul speed is li>c ml",iludc 01 mllantlUlCOIH velocity. our ans"""r is l&.mJJ. (~, c) Aver.,. speed _ dill ...... /time. lbc disl''''''' "';lli>c . i>c same as lhe magnitude <>f Ihe displ.aoemenl as Iona'" Ihe object docs 1101 backtrack. Le . . reverw direction . In our ""'" the ""Iocily al li>c end <>f 6, is still positive. '" IMre ind«d has been nl) backtra(kin*. f Qf diOJll_m. I . v .. + ! III' .. (Il m(.X6.01) + 1(-2.0 m/.')(6.0.)' - 42 m (whid! incidenlally solves p&n «)). A.erage speC<! .. 42 m/6.0._ L2.mlJ.

l . n A rocket·propelled car lIam from rt"st al I - 0 and mo..-es in li>c + dircdioa 0( X wilh oonstanl a=le.ation Q _ S mi.' 1m II. until the Iud is exhausted. I llhen rontin"e' with ron.lant .docily. Whal di.I.""" does the Car ~T in Ih?

, The dista""" from x. allhe moment fuel is uhau>led;" x , _ (0)(8) + 1(5)(8)' -lli!l.m. Ind at 1M poinl v _ (2....-, )m _ 40 m/s. Hen<:e 1M clistana: rovered in 12~ is x , - x , + ... (Il - 8) - 160+ (40)(4) - m.w,.

l .11 lbc p&nic\e shown in FiS . }-I moves along X wilh • OOD$t""l ~Ieralion of _ 4 mls' . As il pasw. lhe ,,"gin. mavinS in lhe + di.caioa <>f X. ;1. ~Iocily is 2Om( •. In Ihis problem. timet i. meaSllred from lhe mOIIICnl The pani<1c: ;,. fio,l 31 the: origin. (.) AI ... hal diu.""" %. and lime I" doe. v _ O'! (. ) Al whalTime il Ihe p&nicle al I _ Ij m, and what is ill velocily &lthi. poinT? If ) WILaI is li>c . elocit) of lhe p&niclc al % .. +25 m? al I . -25 m1 Try Andi", Ihe ... ooty of 1M p&nidc II x _ ~S m.

KINEMATICS IN ONE DIMENSION 0 39

, . _1 AP9lyin,,, '"' " . +111, 0_ 20 + (-4)t ', or~. Thf:~ x ' _ Val ' + jill " _ (20)(5) + \(-4)(5/" b, Or, from ~',",~~ +:z.u :

(0 ,

Solvin, 11m qWKiralic,

0'"' (20)' + 2( - 4}.-'

IS - lOr + I( -4)(' or 2t'-lOr+l~ - O

/ '"' 20 ± >/(20)' - "(2)(IS) ... ! (20 ± Iii 1) , , Thu.t, - 0.8:2 0, t, - 9.17., ..n.", t, is lhe rime from the ori(in 10 J - IS m and /, is the lime 10 go hom 0 QuI ""~onci J " tS m Ind relum to thaI poin. _ Al J '" IS m,

v, - 20 - "(O.82) .. +11i 7 mi. v, _ 20 - "(9_H) ... - 16.7 mi.

Ob«' ..... e lhat lhe ~ .re equal. ftl AI J - 2S m," - (20)' + 2(-4)(2!i). or ~ _ *14.1 ml,; aod au _ -lS m, v' ... 20' .. 2(-4)(-lS), or v'" - 24.5 mI}. (Wby h ... the roo! ~ .. +24.S mi. been dis.:ankd?L

Mouminllh.l J '"' SSm, v' - 20' .. 2(-4)(55), from wh ich v _ ±vC4{i. T'I>I' imaginary valli( ofv indicale> Ihal J neve. ruohe. S5 m .... elpeO'ed hom the re5ult ofpan (II ).

, - ' o P.,.id,

II • _ 4 rn/I'

• '- '

~IO ==::..:..::' ===-1 ,----------",1 FIt. lot

x

3.19 A body falb fluly from rest. Find (_) its ~ltratioo. (!o, Li.e distance it falb in 30, (d its speed after fallin8 1(1 m, (d) the lime requi.ed to teach a .peed of 2S mil, Ie) the rime laken 10 fall 300 m.

, 1_) Choose y dowlI ward n pooilive. TIlen _ ... 8 .. 9.8 mN. (b ) For I ... 3.0 •. JI ... Vol + !1Il' .. 0+ \(9.8 m/s')(3.0 . )' .. -HJn. (e) Lenina y .. 1(1 m, "'e have

~: .. v';' 20y .. 0 + 2(9.8 m/,')(7(I m) - un m'I" or

(oil Leninl v/ ".,.. cqul lS mI., ..... hive

2!im!I _ O+19.8m/.')t or

(t) Now .... Ie'}, _ 300m and we have

}' .. Val + \ .... ' 300 m" 0+ 1(9.8 m!.'jot' " J.lII A ball dro;>pcd from . bridge .,rik .. ,he wa1 •• in S •• Calculate ._) tM opee<l wilh whi"" it stri kes and

(') tho: ho:i&I" "f.ho: hritlV.

, Ch<>o:a$e y downward n posi!i~ . lben Q '" II '" 9.8 m/s'. We an: given u._ (), and I '" 5 I to strike lhe w.le •. Let u - vt' (_) v - v . .. III - 0 + (9_8 mtr)(~ I) - -!2..mb (b ) }' _ Vol + I«' ... 0 + \(9.8 m/l"j(S .j' ... lZlm

3,21 A ball illhrown venially doOl'flward from the edge of I hi'" dill with I n initial •• Ioci!~ of 25 ft! • . (,,) H".. fa.( is;t m<Win& aft.r 1.5 .? Ib, How far has;t mo\'Cd after LS.?

, (_) V - ",, +111 - 25 + 32(1.5) or v '" Zl1llI (wher. G '" 1I .. ]2 h I.' ) Ib, For Wistant aa:clcration v-. '" l(~ + va) ... \(73 + lS) .. ~9 ft / •

• ... v...; - 49( 1.5) - 1l.lfl. Or • _ Vol + !",' - 25(1.5) + 1(32)( l.S)' ... 37.5 + 3fi ... ru1I.

.1

40 a CHAPTER 3

l .ll A . 10"" is Ihrown downward wilh indial s.p«4 8 ml> from I hei&l!l of 2S m. Find (. ) !he Ii .... il cokes to reaclllhe """nd and (II) the . poe<! with ... hidl il mi kes.

I Cboose dow'''''ard posi\i .. c. d _ , _ 'l8m!.'. We a.e given II. _8m/ I. (.) One might di . OCIly wi ... J - 1101 + 1.01' . o. 2S m " (8 m/I)t + !(9.8 mls')t' for I: bUI il i> caoie. fim to lind Ih. final velocily; (II) II: _ II; + 2<1, .. (8 m/s)' + 2(9 .8 m/l )( 25 m) .. 554 m' /I'. o. IIf .. 23.5 m/!. Returning to (.), IIf _ II. + a' )ields 23.S mi. " 8 ml! + (9.8 mll')l. Of'_ J.,.lli.

l .ll A baU Ihrown ..... i<:ally upward ",Iu.n! to iii surting poinl in 40 . Find il! initial~.

, Le, u. lah "p a. positi .. e. FOf the trip from bettinnin , to end. J _ O. g .. - 9.8 mi.', , - 4 • . N"'e Iha' lhe start and II>< endpoinl for lhe trip a.e lhe same, '" the di$placemcnl is ..,ro. Usc y _ 11.1 + !<v' 10 ~ "" 0 .. 11.(4 . ) + !i - 9.8 mls')(4 .)' . from wh>ch II. - !'l. 6 mis.

l.l4 An a"ti ai.craf! WI.II is filed .. en itally upwa.d with In initial velocity of SOIl mil . NcglOClinS friction , compule (g) the maximum beigh, il nn .each. (II) 1M ti me taken 10 ",.eh lhal hei&l!l . and (d the in>lantaneous ~locily al lbe end of 60 •. (Ii) When will il! hei&l!t be 10 km?

I Toke ~p u po$ili~ . Al lbe hipSl point . Ihe ,..,Iocily of lhe Wlell will he ..,m. (. ) v; .. ,,! + 2<ly Of 0 - (SOIl mi.)' + 2( - 9 .8 mls'ly (IT y - u.um (II) v/ _ " ,, + /II O. 0 _ 500 mi. + (- 9.8 m!.'jI o" - lli (e) II, '" ". + III Of IIf .. SOIl mi. + (- 9.8)(60 i) .. - 88 mi l Be<'a"'" II, i. ""gal;"". and bec.o.use ... .., a.e laki,,!! up .i posib~ . • M ... Iocity i. directed downwa.d. The WI.U is 00 il$ .... ay do\I.·n 01 t _ {!(h.

(Ii) Y .. II. ' + ""lor 10000 m .. (500 mi.). + I{ - 9. g m/.')t' Of ~ .9f' - m + 10000 .. 0 Th~ q..ad.3Iic f" .mula.

-b±~ ,. ~

gi" Cl I _ UJ and lli. At t _ 27 •• II>< . hell is al 10 km and ascendin&; al I _ 75 s. i. is al the !ame bei&l!l bul delCending.

3.l!I A NII ... 1 bas i. d,opped from I balloon thaI;' 300m 01>0, ,, Ih" ground and risin. al !l mi •. Fo. 1M bal . ~nd (. ) The mnimum llrigltT reilChe<!. (b) ill p<»ilion ~nd "" IOOTy ~. dlc. ~inl Ideascd . and (e) lhe lime bcfon: il hits the ,round

, Th< ini lial velocil~ ol lhe bag .. , hen .elcased is th .... me as IMt ol the ballooo . 13 mil upwa.d. Let ul choose up a. I""itiv. and lake y .. 0 at lhe: poinl of ,eleow: . (_) At the hi@.hcst poinl. P, _ O. From uf .. II~ .. 2.oy. 0 .. (1 3 ml_)' .. Z( -98 mls'}y . Of y - tim. The muimum heigh' i, JOO + 8 .6,"~. (II) Take lhe endpoint 10 be il< position al , _ 5 I. Then , from , .. " ,J + !Qt '. y - (0 m! '.HSsH \(- 9.8 m/_' )(h )' .. - SS m. So ill height is JOO - S8 - ~. AI$O. from ", _ II. + /II. " / _ !J mI • .. ( - 9.8 m! i' .HS 5) _ - J6 rnh . II i! men,n, downward at J6 mi •. Ie) J uSI befo,"" il hil' lhe """nd . III< bag'. di. place menl ;i - JOO m. y .. 1101 + 1m' beoom ... - JOO m .. {13 m/ s)i + !I - 9.8 m/.'p' . Of 4.91' - 131 - 300 _ O. Th< quadralic fo. mula Ii''''' ... t..lJ (ond -6.6 •. which is nOI phy, ic.>Uy meaningful) .

3.26 A Slone il .h.own ,·.nicilly upwa.d ... ilh "elocily 40 mi. at lhe edge ol. dill having. heighl of 110 m. NegiOClin! . i1 .M lan"". compule Ihe lime ,""qui.e<! 10 mike 1M ground al Ihe base ollhe dill. Wi!ll what ... Iocil~ do .. il "rih~

, Cboose upwa.d as po$i.i,·e . u _ - S" - 9.8 mil' .• , ... 4Oml. , J_ .. _, - -110 m. Firsl obtain Ihe final ""Iocity; II' _ v! .. 2<1, .. 40' + 2( - 9.8){ - 110) .. 37S6 m'/s'. whence II .. - 61 J mi. (downward mot ion). Then . " om v _ t'. + /11 ,

-61.}_ 40+ (_9.8)' " , - .l!!..l.i

l.n The hammer of a pile dri~ ..... ;k ... Ihe pile "";, h a Speed of 2S fl / •. From whal heighl . !!<we lhe lop of the pile dkl il fall? Neglec. fOO;"'" fO.cel.

I !)(ownwa,d is po$ili'''' . We .,.,;ume il falls from rest. S<) t'." O. V _ 2S fI!l. ~ .. , _ 32 f. I. '. We ... cd not

KINEMATICS IN ONE OIMENSION J 41

(15 ft ls)' _ 2(32 ftl.')~'

lbcn y -il.fl.

3.1:8 A blosebloll is th ro" 'R StraiJht upward with a speed oIJOm/ •. la) Ho ... long will it rise? (6) 1-10 '" hi gb will it rise? (e) How Lonl after it kaves the hand "'i ll it ~Ium 10 tile: slarting point? (of) When ,",'ill it. >;peed bt ]6m/.?

, We choose upward u positive , Q _ - r _ - 9.8 mI.'. v. _ 3Om/ •.

t.) At the highest point ". 0: $0 for time t" ~ach highesl poim. ,, - ". + at yields 0_ 30 mI. + (-9.8 m/.')r. or 1-.M!lI. (6) Y _ " . 1 + lar' _ (30 m/.)(3,il6.) + \(-9.8 ml. ' j(3 ,06s)' - ~. [or ,,' _ ,,;+ 201)' yiel<h 0 _ (lO mi. )' +2( -9.Rml.'»)". and y _ oI()m l. (c) Withon' cakulation : Sincc lho timt up equal. the time down . ~ doublc lilt time up to ~I 6.12. fOl tlte round trip. With calculation: For the ~nal di'pl"""ment. ,",'e ha'"e), _0. SO

yiclds 0_ (30 ml.)1 + !( - 9.8 m/ ")r'

Thi. i. a quadratic eqUlltion but easy to 001..., . Onc soIution;s 1 - 0. but this rorrespon<h to y _0 ol ;t lea"es the band . l1>c: <>Ike, $Olution i. not O. $O "'"e can dividr out by /, Ita';nJ!

O"lOml,-(4,9ml.'jt 01 ' - !Uli

(of) Recall ing that tlte 'pecd is tbe: ma",itu"" 01 the ,·clocily . .. ·c mu" con.otkr the p"'.iblc "clocilies: " - ;to 16 mi •. II - U. + '" ykl<h ;to 16 mIl - 30 mi. + (-9,8 mi.); aDd ooIving for " ' . - 1.,.!U. ,_ - lli·

1.29 lbc acceicrati<>rr d"" to gravity at tM .urf."" of M.rs is rouply 4 mI.'. II an amonaut {1ft Mars '"" crc to ' ''''' a wn:nch upward with a speW of 10 mi •. ftDd (. ) bow Ion! it ... ""Id risc, (h) bow high it ... ""Id go; (e) its opc<:d at /_ 3 0; or>d Id} it> di'Pi"""me:nt 01 I _ J s.

I Let I" chooot "I' .. "",iti"e, l1>c:n " equals _ 4 mi.'. and p. equals + 10 m! •. (.) By equation of motion Q - (11- ~.)/I . " 'C ha~e

" '''''-;:''em

" ,", - 4m s --'-

(6) By equation of moIion ~' _ ~~+ z.... ~ k","" ... 0' .. (10)' + 2( - 4)<_,

(t) By C<juation of moti<>rr Q - (11 - 11. )/1, " 'e ha'"e

_ 4 _",- 10 ... II, - =l.!!!l.! J

(of) s .. "01 + lru' ar>d thUi

. ," (1 0 x 3) + !(-4)(9) " .30 - 18 - ll!!!

3.JO lbc: aooole,ation d .... to v a';!)' on the moon is 1.61 mls'. [f" person can Ihm .... aSton( 12.0 m limight upward on the con k. how hiJ!h iboukllhe pe""" be able '0 lhrow " . t{lftC on the moon? A_me: that the th rowing speeds ar" the .. me in the two case.<.

, On can~ .. e can wri •• ~i. _ 2&.(12) w~ilc On ,Ioc moon v~ .. 2& ... "' ... l1>< ,11'0" ;"1 vet.,.i' ;e> .t~ U,,, ... me. to the Kwnd e ... r....;on a.n be divi<k<l by tho firs! to give h .. _ 12lg.lg .. l- 12(9, !ll/ 1.67) - 7l!.!!J.

3.31 A proton in • uniform ele<fTic f\cld I1It)Vcs.w..! • lI,aight tin e .. 'ith roIISt_nt fl<nlcrati{lft . St.ning from res! it .ltaim a velocity of 1(0) km/s in a dillal'Kle of 1 em. (.) What is its am:lcration? (h) Whit lime b r(ljuil~ to ~1OCh the &i . en .eloci!)'?

, (_) ". " O. II '"' 10" mI. in .. _ 10-' m of di'Pi"""me:nt . lbc: n II ' - v~ + 140: yicl<h ( 10" m/$)' -0+ 201(10-' mI . or Q _ 5.0 )( 10" mil'. (.) II .. II. + QI yields ]0' mi. " 0 + (S.O X 10" m/. ' ),. 01

, .. 20x lO-',.

3.32 A bottle dropped lrom a blollootl leacbcs the s rOllr>d in 20 •. DetermillC the keiaht of thc balloon il I.) il ... ·a. at re:<t in the ai, and Ih) it wa •• ..,.nding with a spoed 01 SO mI. when thc bottle: w"" dropped .

42 0 CHAPTER 3

, ~ upwar<l a. posilive for bolh (. ) and (6)) . .. _ -, _ - 9,Sm/.' , (.) u. _ O. To find Ihe height. leI y M the di>plattment at time t (rememMr the y _ 0 poinl is at the Inlloonl and we .re Jiven . .. 20 ., Then y .. ~01 + lar' - 0 + 1(-9.8 ml.,)(20 il' - -196(1 m. The hei"'t ;. 111 - l2!i!tm. (6) Here the boetk initially 113.$ the velocity of the balloon SO u.- SOmis. Now. y - vol " !<tt' .. (SOml .)(20i)" \ (-9,8 m{,')(20.)' .. - %Om. Apin the height - Iy l - 2!iQm,

3.33 A sandbag 1n1l.'1 ;. dropped f.om a InUoon Ihat il a""ol><!ing with a volocity of 4(lftl • . [f lhe sandbag ,cae"". the &round in lOs. how high was the balloon when the Inog w., dropped? Negkct air resislaJ>CC .

, Tl ke down" 'ard a. positiV1'. Then Q" , _ 32 It/.' and Vo - - .10 It/I,

, .. V .. + 'ar' _ - 01(1(20) .. \(32)(20)' _ - 800 .. 6400 _ 5600 ft

TIle balloon was ~ above Ihe SfOUl><! .

3.34 A "0"" i. oboI 'Iraight upward with a speed of SO Itl. from a lower 224 It high. fil><! the >peed wilh which il mikes the "ool><!.

, Choose upward al positive ... .. g _ -32 It/.' . v. _ 80 It! •. The presumption i, that the .tone just mi .... the ed", of lhe I,,",,'er on the w'y down al><! <trikes lhe ,round 224 ft Mlow. We <an lvoid III . eferen ... 10 lhe time in the problem by selling 1 " -224 It in the equation

u' _ v~ .. 2111 _ (80 fl /.)' + 2( - 32 It /.')( - 224 ftl - 20 736 ft '/.'

AI><! u _ ±144 It / •. WhCTC the minus ~'" gives the ph)"-icll oo/ulion, The speed is lul -~.

3.'u A nut come. loose from a bolt on tM bottom of an elevator ., the elevator i. mo>inS up the shaft It l.Om/ f. TIle nut strikes Ihe bottom of th. $/Ialt in 2 f. (.) How fir f.om the bottom of the shaft was lhe ele .. tor when tM nul fell off? (6) 110111' fa. above lhe boelom wallhe nut 0.25. after it feU off?

, Here lhe nul initially hal the V1'locity of lhe elevalo,. SO choosinl upward .s positive. v._ 3.0 mi., Also Q " - '1 _ -9,8 mI.'. (_) The time to hit bottom is 20 5. so Y .. Vol .. !QI' '" (3.0 m/ f)(2.0 'J .. \( - 9.! m/.')(2.0 s)' _ - 13.6 m. The bolt"'" of the shaft ;' [3.6m below ",I""e the el .. 'atm "'U when tM nul fell 011", (6 ) The ""'" displacement 1 i. for' - 0, 25 s. 10 Y .. Vol + !~" - 0·0 ml.)(O.2.5'J + II -'I.~ m/s')(O.;!:'i . )' _0.44 m. n us the nul is ._. it. <tan in! positiOn. (This ma~es sense If We rememt>e. Ihe init;al ~elo<ity.l The heigh! abov( Ihe bonom ii Ihus OA .. 13.6" HJlm,

3.2 GRAPHICAL AND OTHER PROBLEMS

J.J6 The ".ph of lUI object', motu,n (alonl . Ii",,) is ohown in Fil. 3-2. F"md the ini",ntar.eou • •• Iocity of lhe object at poinu A al><! 8 . What i. the ob;e.crs aV1'ra~ ~docil y'1 I" a.a:eltration?

, 8cc.iusc the ...,Iocity is giV1'n by the lIope. t.x I dr. of the (angenl line. we lah a tangent to the cu,,'e It point A . The tan",nt line is the curve iuell in lhi' casco Fe. the triMSIt .t'o .... n . , A. we ha.e

... m ----O.!iOm/t dr h

This is also the velo<ity.t point 8 and at every mhe. point on the .traigh1·~ne vaph. It follO\ll'l thaI ~ and 6, - v. - OWroli.

' .. "

KINEMATICS IN ONE DIMENSION D 43

J.l7 Rd., 10 Fi., ],,3. Find lhe inllantane ..... velocity at point F '0' the ""}oct .... bose mO/ion the ",rve ,eprescnts.

I The 11l1~nt at F" the d.ulled liDe GH. Takin, triangle GH1, "'e """"

oU _ O_lS " ·I!m

Hence slope at F i,

' .. , " ,

" , ' D

• , , ,

• • , • • " ' .. ",.""

J.J8' Re'er bKk to Fi,. 3.3 for the moIion of an obiKt llonJ Ih ... a,i •. Wbal is lhe inslantaneous velocity of tM ob)ect (I) II poIn. D? (6) al poinl C1 (e) al poinl e~

I (I) Poinl II i •• muimum of lhe .. _.,._/ CUn'e. The.efo,. v _ d> /dI_O. (61 W;tMul lhe exact .quation fOf .... fullClion of r 0fIC cannOl ~I • p,eciw anS".'. The best we. ca" do is to draw lhe langent line 01 point e and Jet lhe slope in tbe .. me way a. in P,ob. ),37. This yield. Ihe ',.. .... e'

vc " 4<1 - Um1> . , (e) W. proc«d as in part (hI, but here the tan~nl I;ne h .. I nepti"" ,I""" and the an,we.r Ihoold bt

u~ _ rb l ,, _o. !3m!J d, •

J.Jt A prl ".,k •• ....., I n ea.I · ", .. 1 "rect. and II,aph of he. diiplacc_nl from hoone" shown in Fi,. ]..4. Find he, .""' .... el<>city for the ",1101. ti_ ;nle,...0.1 ~n IS we.1I IS her instanlltleOUl velocity at point> .... . B.

""c. I The average vel<>cily is .. ro, sil>l;C th. displllC':'rnenl \"«Ior " zeU), The instamaneous ""lnrit;'. arc juot the 1I0]:fl of the CIl"'. It ea.c:h poInl. AI A IIIe v'lOdt) is 40/1> " 0, 1 m/ mln cUI. AI B it is ze.n. AI C il is -65/S .. -Il ml min east. 0, +13 m/ min .... est.

Di""""" '01' ( m )

~

>0

, - >0

,

, "', 1m",)

od malarial

44 a CHAPTER 3

3." Rd~rrin, W Fi,. 3-4. find (_) ",'erase velocily for 1M Ii"", ;nle",a1/_ 7 min 10 /_ 14 min; 1M inslanlaMoos lIClocily al (10) / - IlS min and (t) I - IS min .

, (_) Ii _ ( - ll - 40)1(1. - 7) _ - 9.3 ml min 0&01 ; (b ) 1M ... "'" a, al poinl C. - i3 ml min .... 1; k) 1M oJope is +ll/(19 - ].) .. !j .g ml min .ut. NOlo Ihal ""plive velocily elt$l ""'" ns motion is ",,51

.~ ,

, "

3 .• 1 "The vaph of. panicle', motion aIonK 1M ~ uis is KillC n in Fi, . 3-~_ Estim.ale lhe: (_) Iverage . elocity fOl'IM interval hom A to C ; inslanlaneou. ,-e]oaly . ' (b) D and I t (e) A..

, (.) I) .. (U - 0)/ (8.0 -0) - 0 6Orroil. From the: slope al u rn poinl (b) 11 -~ and (e) ] .) emi •.

JAl Fill''' J-6 shows lhe: .doaty of a panicle It$ il moves alon,'M I uis, Find il • ..,..,k. l tion I I (. ) A and I I

(b ) C.

• (,n/.)

'11 ,

It.) ",.,.

, "The a=le.alion .1 any t;me is the: .Jope of II>< ~.ys , ,, curve, (.) AI A 'he slope i" from ",-M,. Ihe Ii ngenilinc: IhTOUJh A cuts 1M coordinl" . ... " ~ _ - 7.0/0.73 - - 'I.6m/ .'. (b) "The.1ope and IM.e'=. is .ero.

J"u For 1M motion dum~d by fig. 3~. find lhe: loccle rllion I I (.) B . nd .1 (b) D,

I TIKing . Iopes al Band D we find Ihlltl>< =ek,alion" (. ) - 2.4 and (b) il.!!:!l£.

3.... A ball is tII ..... n lICnicaJly UPWlrd with I velocily of 2Om/ . from lhe lop of. I""", . hl";n • • heip, of SOm, r!i_ 3-1. On ill relurn it miue:< II>< lO"r and finally stri kes IIIe .round . (_) WIIal li me I, ellpsa from lhe inslan! the: ball was Ih ..... n unril il passes Ihe e<lse of 1M lo...-e. ? Whl' velocily It , does;\ have al Ihis time" (b) Wl!allOl.ll time I, is .equired for 1M ball 10 r.adI the vound? With .. hal velocity II, does il Slrike Ih. lfOund?

KINEMATICS IN ONE DIMENSION a 45

'L .. ~~. x r -.. ,

e,

",

.m

, (II". For Ihe coordinar. ,yscem ,1>0<0-.., in Fi~ . l-7. Y" Vol t !Qt'. BUI at 1M .dJ. of the roof" .. O. and rhu, 0 .. 201. t Ie -9 8)1;, from which I. - o. indicaling Ihe in,lanl al .. ·hioh Ih. ball is released . and a lso I . - !..Il!!J. whi<h is rhe rime to go up and retum 10 lhe .dge. "Then. from V " V. + Qt. v, .. 20 + ( -11.8)(4.011) _ - 20 mls. whi<h is lho ""gal;"" of Ihe initial ,·.Iocity.

(II) - SO-2Ol, +I( - 9.8)fi os t, -lli v,_20 +( _11.8)(5.8) ·_37m/s

).45 Refer 10 Prot>. 3.44 aOO Fig.l-7. I .... Whal is 1M maximum heighl ab(r,-. ground reached by rhe ball' \I~) Poin" P, and P, are I~ and 30 m, "'p"Clively , holow rhe lOp of 1M \(WIer. What lime inlerval is •• qui • ..:! for ,ho b;oUro I.ovol f. om P, 10 P,? (~) It is deI.i .. d lilal after passing rile edge. ,ho b;oll will reach .he I,ound in 1 5. Wilh what v.locity mu.' it be thrown upward f.om Ihe roof!

, I.) Maximum heiJlII ab(r,-. ground : h. ,_" t SO. From v~ + 2<ly_ " . O. - (20)' , .-- . ~.-- , - 2(9.8) ~

Thu,.h.Xl.4m. (6) If I, and I, are Ihe tim ... 10 ..,ach P, and P" . rspectively. - 15.201, - 4.9.; and -)(I .. 2Ot, - 4.91;. Solving, I, _ 4.723 S. I, _ 5.248s, and the lime f.om P, 10 P, i. 1,- I, _~. Ic) If"" is lhe de$i .. d inilial vdocily. Ihen -Vo is the V¢loxily upcln pauing Ihe edg •. Then. awlying , _ V. I + lal' to , ho trip down rhe 10 ... .., •. "' .• find - SO. ( - v. )(l) - 4.'1(1)' . or v._~.

1..... A m ... ,un. a, a speed of 4.0m10 10 overtake a sta!>ding 1>10 •. Wh<n he is 6.0m bc:hil>d the dooJ Cat t _ 0), the hUJ moves forward and rontinutS u;lh a tOIIJlanl ~Ierlliion of 1.2 m'~'. ( .. ) How long doI" ;1 lah for lhi' man to gain the door? (6) If in , hc beginning he is 10.Om from the doo . . ... ill he (running at the !oil"", speed) ever ClII<h up?

I At I _ 0 leI II>< man·s po<ilion I>< the Qrigin . .r ... _ O. "The bus OOQ. is lhen at .r .. _ 1>.0 m. Thc cqua.;o.,. of morioo for tl>< man and rlt.e bu. D..,

Now

Th ..

.r~ • .r ... t V...,I t !Q~I' .r • • .r .. + v..,t + IQ.I' v ... . 4.0m/. v ... . 0 a. " 1.2 mIs'

x~ _ 4.1.11' .r • • 6. 0 +0.t\t'

When .he mon ca.oh", thc bu •. x~ _ x. , or 4.0. _ 6.0 + 0.6t'. This .... n he Ittxp.eO$d '" 3l' - 201 + 30 .. O. Solving by Ihe quadratic formula .

2O ±V400- 360 10±VlO ~ ,< I· 6 • 3 .~,4.4s

Nore lha! there ... Iwo po6iliV¢ l ime oolurions. Th;, can be understood IS follows. The fim lime. I, - 2.3 s. correspond!; to his first reaching the door. This is rhe ..,.1 ......... 10 II>< problem. I/""",ve •. rM "<Iualion . .... have ooIvcd ··don't kno ..... he willllop running and board the bus: lhe eqll&liom h."" him conlinuinl runnin, al con" ... r speed. He Ih ... II"'" on past rhe mil: but since the bus;' IoCaIe,atinl . it e_cntually buikh up a Ia.ge' velocity tban the man and will catch up with him . I , _ 4.4 s.

46 J CHAPTER 3

(bl If {he ini{i.1 p""i{ion of {he bus i. 10.0 m. {I><n k ~ _ k. yield, 3/' - 201 ... SU _ 0, ,,"iclllla. onl~ romp ... ">(I' S. Thus 00 ,cal {;"'" ex;'" at ""lIiell II>< man ,.Irn.S np.

3,47 A ball;'; III. own ,irH1Sh{ upwa,d ",';Ib a s~ed u from a height h _b<ln {he !".>unod. SIlow Ihal 1M I;"", laken for {he N il 10 >Ink. Ihe g .... unod ;s

"( ,pii"") - 1+ 1 ... - . Ii I"

, Assum;ng II>< pasili.'. di •• ction 10 Dr up"'OId, Ih. unifu,m """"Ie .. , ion C<lWllion i< - h .. ", - Ii" 12: Ihi. c.n I>t ",,",,.inen Ln Ih. form,' - 2" '1/ - 2J. ' 1/ _ U. Solving fo. , u.ing {he quadrali, fo,mula gh'e. Ihe desi •• d on.wer . Since ,",' ... ek pasili'·. ,·alu •• for lime, (he ... roo! mu.1 be rno .. n

3.411 A N il i. dropped from Ihe .op 'If . building. The ball lak", 0,5. I'l fall plIl' !be 3-m \e nglh 01 . window"""" dist.n"" from lhe lop Q/ . he building. I.) !low fa~{ "'''' !he hltll going as it J>ilSSCd {he lop Q/ lhe window' (bi How fa. i, lite IO!> "r Ih. wind/:tw lrom lhe point at ""hieh the 1».11 .... " ! d.opped?

, The ""Iocili., ",.1 Ih. "'I' .nd ". at Ih. bC)!1'lm <If the "'indow . , • .-.1.I<d by II>< foll"",ing equalion.: 6 _ (I', + 11. )/2 _ 3/0. 5 .. b, '" '" ... p •• 12, and v • .. " . + 8(0.5) .... ". - '" .. 4.9. Eliminalinl ". Drt,",'""n 'h<>e 1,",'0 up .. "ion, )'eld!. 1' ... 3.55 mi •. The di>l .n"" ne.<ltd I'l r •• eII Ihi' <p«d i.

3.49 A truck i. movin~ f<><word at a CUnllanl speed of 2 I mi •. Tnc dn,·c. SnS a staliona.~ ca. di.ectly ah.ad a{ a distance 01 110 m, Aft ....... acl;.,., lim.' Q/ <l.1, Ite awlie, Ihe brake •• "'hiell give. (he truck an acttw"ation ill - 3 mi.' . ,.) Wbat i, II>< maximum all'l.,.ble <l.r {Q amid a rolli";on. and ",·h.l "'"",,.,. ",ill lhe lruck hav~ """,.d befme Ih. bra ke. (.k< hold? Ib ) A"um;ng a reac1;on lime of 1.4., !>ow fa. b<bi<>d th. 'a, -..ill.he lnl<k SI'lP. aOO in how many s«ood!. from (he lime the dri ... fil'\{ saw II>< c~r?

, Tnc lout di~pla""""'"1 " I Ihe lruck r""" Ihe inSI""1 lhe driver sees lhe car is X. o. <l.1 + x~. ,,'he •• XA is the di>placemenl from lhe p<>im 'lf ~re.alio" 10 Ihe poinl 01 ",SI, We a .. 8iven II, . 21 m/ •• OO II>< cIt""lcralion i. II" -3ml, ' , x~ Can be obnined f,om Ih. equal;"" ,, ; .. II'; + 24.rA . ... ilh "J _ 0. ThU! ,,~ .. - ,,~ z... "" ~ .. - 121 ml.)'I( -/oml.' ) _ 13 .S m I.) T'l find the rr.a~imum <l.1. "'e nme Ihal x _ • 110 m .nd x_ • • (21 mi .) .l.1 _ , ... x., "" 110 m .. (21 m/_) ~I_"'" ?3.S m. anod <l.1_. " Lfu. Tnc di".!ICC """,.d Drlo •• b .. kinll r.I>rle-d i> Qf """~ JUSt ". ~I.~. fbI If <l.'. IA., then X" f21 m/. )(1.4,) + 73.5 m. 102,9 m. The di • .,ncc I'l II>< atr " jn!{ 110 m - 102.9 m _ ll.m. To fiOO lh. lime."" need 10 know {he lime / during "'hich (he Iruck _",<re" les. W. have '" _ " . ... Ql.

";Ih again "I" 0 and,, · -3 m/s'. Then 0 _ 21 mls - (3 m/5' \lo"d / - h . The 1,,1101 ,ime ;, / ... At _ 7 + 1.4 .. ~.

l.51 JuSI 3,. car .. arn. I'l """,,",.ale f.om .e.1 wilh ."",,",.alion IA mi' " a bu. moving ,",;{h COIISlan{ .~ed ill 12 mi l plIsses il in • pa.allcl lanc. ta) II" ... k>ng helo.e Ihe ca. 'l,·cn.k .. lhe bu,? Ibl 1\"", fa" will the ca. lhen be going? fl) 1·low fa. "ill {nc ca. lhen bav. gone?

, ",., ca, \l3nl wilh inilial "elocily ze.o.nd acalc •• tion a," 1.4 mi.' . "'hil. lhe bus h., ron>l>"1 HklcilY II ... 12 mls. (b) Bo{h tr. .... , lhe .ame diSl.""" -" in lime I . $0 se{ a,I ' /2 .. "./ I'l give / - lli. (bi n.: ~nal "I""i' y ,,' lit.< c., "" _ a,I"~' Ie) Tnc ."er""", ""'<>cily 'lf Ihe: Car (0. the bus'llixed vek><ity) lim .. 171 yields J( - i!l!.m.

3.51 A mookey in I p:r<h 20m hi", in a t«c: drop< a coconul di «<tly 300.-C you. head .. you run wilh .peed \.s mi . Drnnlh Ihe l",e, (.) How f.r behind you dot.lhe COIlOOul hit lhe ground? t~i If {he mon key had .cally wanl.d {Q nil ~OUf IoU. !>ow much ... Ii •• o!H:Juld Ih. COOOftul h .. ,. be.n drow<d?

, Th. lim. fm Ihe coconul 10 fall 20 m i. given by 20 . 81' /2, <II 2.02 •. Di!lan"",, " (l.S mfs)(2.OZ s) -1J!J m. Si"", ~OIJ are m....-ing", a hed speed the monk .~ <hOIlld h .. ·" drQpp:d rhc OOCOnul 2 O:! i carlie •.

3.52 Two ball! afC drowed 10!1>< ground from dilkrent heigh". On. bltll i, d.oppcd 2 s ~ftc. lhe <>Ihe. bul lh.~ both strike the ground al th" SlIme time, 5 saIl .. Ihe fir>{ i. dfO,,~d. (.) Wilat i. II>< d;fI'"",ncc in {h. I><ighll 01 which Ih.y ,",'ere dropped? (b) From "'bat htighl wa. lhe firsl drowed?

KNEMATICS IN O NE DIMENSION a 47

, n.. I;"", 10 fI ll from llIe veate. height ~ , i. r _ 5.; Ihe lime from lhe lesse. heighl II, ill - 2 _ 31 , Using , - ",' /2 , "'t find (.) the hoi&hl difference h , - II, - , (5' /2 - 3'12) _ 78 m . nd (II) II , - 9.8(25)/2 -11l.m.

3.!J T..., boys "an runnin, OIraighlloward eact. other from tWO p:.;nlslbal are 100m . pan. Or>e rUM with a speed of 5 mi •• while lhe other IIIO\Ia at 7 mIl , Ho ... clooc Ir. they to the1low<:r Ode', stan ing po;nt when they reach eKh other?

, n.e twO bo,.. "",. 1 at lhe same place and time , n.. Ii"", fal lhe II""",. one 10 lra.'.1 a distance ~ is ~ /5, while lhe other bo)' h'" I _ (100 - ~)/7. Equaling lhe lime, yioelds ~ _41.7 m.

.I 1 •

J (., ", ... ,.

3.54 Bod)' 1. fig. 3-8, is reltas.td f.om.m althe top of Ilmooth tncli ... d pI ...... nd II II>t same ill$ll11l body 2 it projMt<I u~.rd from the foot of ttte plane ,,;th such .'.I<>city th.lthey "",e, h.ltway up Ihe plane. [)et<.mi ... ,.) lhe veloci,y of pmjMioa and ,bl the velocity of e..,h body when the) IM<'t.

, ( .. ) In 1M common lime I . body 1 tr .. 'els tM distar.c:e

I I . i-(Ojl+i U .. n9)r'

!. - ~.,I +! ( - , ~in 9)1' , , Addinlth ... twe> «[uation, 1i"V /_ ~.,I or I_I/ ~",. Substitulinlthil "olue of I in tM Ii"" equalinn an<! IIlIving feu v",. we nblain " ." _ ,I sin fI_ Vih , ,II) V:_0' + 2Usinfl)~ eu v, .. v'risinfl - Viii.

I uj _ ui., + 2(- , .in fI)i-",in fI - 81 sin fI_ 0 ••

3.55 Two lrains arc headed 10 .... rd each OIM. on the same Irack "';th equal speeds of 20 mI •. When Ihey. re 2 kin "pan. Ihey!ICC each olhe. and begin 10 decelt.ale. (.) If IlIcit d«o:ltrations ar. unilorm and equal, how I .... mull llIe loa:leral;,;m lie if tM tuin~ arc to ba.ely avoid «>Ili$.'O<I' (II) II only 0 ... In.in slow> with Ihis """" Ie. ation. how fa. will it go before collision ocru .. ?

, (.) EaclI train Stops in ](0) m. on using ... ' - ... ~ • 2Iu yields. 101' .... _ 20 mI • . an acccleratiOl'! 01 - 0,2 mi. '. (II) The dectleral;ng .uin tr.ve". di5lance ~ .. 20t - 0. 11', wkile 1M <>Ihe •• rain tra~el. (201)) - x) _ 201. Beau .... he~ ooIlidc. limes are the sa"",. Eliminating I and .avinl for x by .he quadratic lonnllia yields "",.

3.S. A ball afle. lIavinll f.lIen f.om ,est undcr the infl ... ,...., of gravity fOl' 6. craohes through I horizOI'".1 ~Il$$ pille . the.eby l<>sinl two-thirds 01 ill .el<>cily. If i. then reacbu the pound in 21, lind lhe Mi&ht 01 lhe plate Ib<we tM ground,

, from U" u. + QI . • M . el<>city juS! before st riking the Ila .. i, u, _0 - 9.8(6) .. - S8.8 m/s. 100 ",.he velocity af.e, passing .lImuSh glass i, ( ]/ 3) ... , .. - 19,6 mI • . Thu. - II .. (- 19,6)(2) - 4.9(2)'. 0 ' II ... 5!!,1! m.

.1

48 D CHAPTER 3

3.!7 An illClin~d plaM. f ig. 3-9. malH an Ingle (J wi th the horizontal. A ''''''''~ OA cut in Ih~ plarIC makeo an anglt II' with OX. A ~hon WIOOth ~'~nd., is lru to .Iid. down the I"""'~ under the ;nfl""noe 01 va~;ty. >lan ing from Tn t at tile point (~ •• ,.). find (. 1 it. downward acc<'le. " tion _\ott! tile "00>'<:. Ib ) tbe limt 10 ""."h 0, .n<! I ~1 lIS •• Iocity at O. ~t IJ. 30" .... . 3 m, ,,._ 4 m.

,

! ,. ~,8 noll'

o':~::----Jc----------.c------------",~~------------------cx • .... ,..

, ,., 'The downwant oompoocn' <If I 1" .. 1101'0 OY;' 8 .in IJ; ""!>N. tbe oo..lIwa,d oompolKnl 110111 tl>c "00>'<: is"., sin fI sin .... Si!>N

'00 <in fI _ 0.5 ~ _ (9.11)(0.5)(0, 11) _ ).92 mis'

,b, I • Uol + 1«' . wllc,~ I • (~~ T y~)'" • 5 m an<! u • • 0, Thu ••• - 1(3.92)i' o. I • L.m1. Ie) U.O T (3.92)(1.597). 6.2600/ •.

3.511 A bead. fil. 3-10. is f.ee to slide down a .mooth wi", tiptly ... ~tched between poinu P, I n<! P, on I ~.nical circle of .. <liu~ R, If tbe bead "am from ""t .1 P" Ih" hiv.e-I poinl on the ci,de . find t_, ils "clocily v on 0"';"';0! 4' P, a nd ,bl 'hc ,ime to 4,m" ot P, ~nd , h"'" , hot ' hio ,jme io ,he ... "", f.". ony "ho,d d .. wn fn)m

',.

I

I I '., I ' I

~, I

.... J. l.

KINEMATICS IN ONE DIMENSION a 49

I I_I The alUkration of the bud do;>wn the wire i. g ens 6 and the length of lhe "ire is 2R cos 6 He~ . .. ' • 0' + 2(g cos &)(2R cos B). OT " • 2(v'iR) cos 6.

'" , . !: . 2v'iRcos 6 _

2 @.

" gcos6 Vi ,,'hidl i~ tl\o !.ame r.gardle>o of .. 1\0", P, i. Iocaled on Ihe cin:le,

)..59' An obj.ct i< forttd to rn"". along ,hi:: X ui$ in such a ... ay Ihal its di<placemenl i, gi~.n by • • 'AI + lOt - lv' "'he ... i, ;n m .,..j, i. in . (_ ) Find ~.~iOlK for the ""Iocity -< .nd ..",.,I • • at""'.f. I. the acn leration alRSlant~ (bl What ~'e tbe initial po>ition and tbe inilial velocity of the obj.ct ? (e) At ,,'hat lime aOO di"ance from lbe origin i.lhe ,'.Iocil)' zero? (d) AI ,,'hat time and Ioxation i< tbe velocilY - 50m/s?

, Remembering Ihal aU units are SI and thai

(_) ,, - i _ (20 - JOr) m/,; " - j - - )(I mi.'. A=leratioo i. OOfI".nl . (b) AI' _0. x _ 30m_x • . j _ 2Oml. _ .... (e) From lbe velocily C<ju.li<m. for .. " 0, 0 _ 20 - 301 and, - j i , Substiluting r _ i . into the di'Placemenl equal ion. x • 30 + :!O(i)- 1 ~(j)' - J!l.Lm. (4) Set1ing ,, - -50 mil. - 50 .. 20 - IN. and 1 _ 2\. , Then. x • 30 + 20(211 - IS(2l)' .. .::.tim, (N .... ,ha, • """'parioocr <>f Ibe di.pI~mcn' C<juation wilh x - x. + ~ .. + lar ' ''''''1d han made tbe u~ of cal""lu, unne«: ..... ry.)

3."" A partkl. moving along Ihe x axis h.,. • ,'elOOIy si" en by " _ 41 - 2 SOl' mi l. for, in secoOO, . Fir>d its acttl<ralion 01 (_ ) r _ 0.50. ar>d (6) ,_ 3.0.,

I ,, _ .wl dr - 4 - '. ur. gt~'ng (a , " .. pO cml.' and (b) - H .lI em/i'.

3.61' A ball is ..,I •• ltd from ",,, al It>< .~ <If a <l«p nvi~e. A\.$ume lha' air I'Mlance gi'·e. il an acttleralion of - 6; . ... here y is measur«l posili,·. d"..'nwa,d. (Th i. negative acnle"'ion is proportional 10 it> ~d. ;; Ihe po>i!i ... con"anl 6 <an he found by "perimen!,) The ball "as I 100ai accele,ation of - 6; + g and so

'" is Ih. differenlial equation of moIion, (<I) Soo..· b~ differentia tion and ,ubililuliort Iha,

"a sol ~tion 01 (1) for an a,bi llary ,'alii. of Ibe conl11lnl A ar>d Iha! (2) g:i~c. r - 0 for / - O. (~ I Since .1,. O. ; _ O. prove .hal A _ 81b' and lhow lbat as / --+ "". j --+ gi l>, Ihat is. lbe ,'el<>city ..,ache. • lim:Iinll ~al,", sueh that the negati"e ",'<elc. a,ion due to 'ir "~llan«: .. ""Iy off>etl the po>itive a<ccleralion of ,cavi ly 000 Ihu. j _0. (e, Assuming tbal b .. 0, I ._J. find Ibe di".nc. fallen and tbe .p«d reached after 10 •. (d) Sho .... Ihal after I min Ih. ball ",'ill h.". e"""ntially . eached in term;nal velocilY 0198 mi •.

I (,,) Oiff •• en. i"ing (2) once. )' - -~k' - " + gI l>. Oiffe'entia'ing on«: mor •• f" b'A~ -". Multiplying OUT ul"'''';'''' for j hy - I>.nd adding g ,;' Id< h'k, -". Thu< (1) i< ""i<fie<!. Sun.'i""ine' _ 0 i"'~ (2 ) .... , ",calling Iha. ,' . l. ... e g.I)' . u. (b) Since Ihe ba ll is ",lea>ed from rest. j _ 0." _ O. U.ing our CXI" ........ for j from (") . ..... h",', o _ -11k + &I b. which )'iel<h k • g 11> '. A. 1_ 'Z> Ihc fi", term in j be""", .. infinil<simal ar>d j _ g l b. Thus j must appro,eIl .ero. al can he seen directly from tbe up .... """ for j . (e) [f b _ 0, I I. aOO uling g _ 9.8 mI.'. "'" ha,'e k _ glb' _ 980m. Then •• , _ 10,. r " (980mj(, - ' - I) + (98 mfs j([Os) · ~, ,, _ j .. (-98 m/.)t-' + (9.8 m/.')1 (0.1if) -~ (d) At , .. 60 •. j _ (-98 m/s)t -· +98m/. . , -0 .. 0.0025. so; - \18 mi •.

3.61' A ball i< Ihr_" verti<:all)' "I""'a,d frOll! Ihe "';gi" of ax", ( Y ICgarded + u"",ard). " ilh inil;al velocity Jt,. A,",uming as in Prob. J.6! an alUle'ali,," - b}' duo 10 ail ,,,,ilt.nee. "'" write

j . - b;-8 (I )

[N"'e thaI wt.u j chang .. §ign . IQ does -bj. llen« (I ) is 'alid for Ih •• rip down as ",'ell as 6. lrip up). (,,) Sbow . hal y • Ai' -" - I) - II Ib)r i, , solulion of (I ) for any value of k, Ib) P,-""e that

A- -Hfo+f)

C ;rpvnghted matertal

50 a CHAPTER 3

(~) "ssumins that h _ O. 1,- > and ... _ SO mf. , find the height and speed It ! _ ) • . (d) How lon, does il tal:~ 10. lhe boll 10 aUlin il' muimum height Ind .. ha' is the height?

, (,,) j _ - H~-" - gf l!, j .. h'k~ - ". Multiplying j by - b and subttactinu y;ekli j. which PfO"e5 y is a 5Olution of (1). (II) AI , .. 0, y .. j o _ - bk - , l b. Sol"in, 10. ~. k .. - r:;. + llb)! b. (e) Again b - 0.11, ~nd..., ar~ tQId Yo- SOmIs. Then k· - (SO+ 9S)fO. I " - 14l!O m. AII _ h , we have )' _ - l.sao(~ -" - 1) - (98)(3) _~. Y _ +14&-0' - 98 _ 1I.6mh . (d) for mu;mum h~ight , y _ 0, which y;elds 0 .. 14&-" ~ - 98, O. ,O b ... LSI. Th ... , ,. i..l1.I, Sub$tilulin. 1_4. 12, inlo the), equation yields)' - - 148O(~ -· ·" - I) - (98)(4.12) -~.

3""" A rna .. . 1 lhe end 01 a SJlring ri>.lleJ up and down &<:<:<>I1ting to the equolion y _ llsin l.St em. """'~ t i. the t;m~ in ~ and tI>< OOIllpl<1< •• s""",n' (_nskl 0( ,I>c oir>< '""",ion, 1.51, ;, in ,ad;An>. (a) \\/ha. i,.he: v<:Joc:ity of lhe mass It I ,. O. 75 11 (II) AI , .. ).0 11 (e) What is the maximu ... velocity of .be mass? (Hin,,' To elj>' ess the angle in "'S,oes, m.ltip!y it b~ 111l/ .... )

, The velocity ~ _ ily ! ill .. (U)(8) "" I.SI an/I ~ remc:mhe. that I.St is exprelM'l in radi.", . (I) ~ _ 12"" 1.13 _ U9!!U. (II) ~ ,. lZ""" .5,. - 2.5 cml}. (el Maximum OCCUR whe.e the cosine ;$ ± 1, 50 ~ .. i ll Oan/s.

opvr hIed IT ria

D CHAPTER 4

Newton's Laws of Motion

4 .1 FORCE. MASS, AND ACCELERA llON

• . 1 A for« acts 00 a 2·kS moss aDd ai>'";! an accel~fa!ion of 3 ml" . Wllat """,Ie.ation is produced by the ... me force "'hen acting on • mass of (.) I kg~ (bl4 kg" It) How la.1I<' is 1M fQ''''?

, We first find 1M force F Il~ng F _ .,u, (one: dimcII$ionJ. F _ (2 kg)! 3 mis' , _ ~ T'hon noting ~ - Fim . .... hl'-' for the dilre rcnl ma>Se> m. I ka . d. (61'11)/ ( 1 kg).~; m. 4 kg. d • .L1..!nL£. n.e 0 .. " " .'" .,. thus (/I) ~ (6) Li.rn.lI: It' (U!.

4 .2 (Fin in the blank.) The rnW" of a 300g object is __ (,,) . Its _ighl on carl h is __ (6) An <!bjen that "" igll. 201'11 on unh h'" I m"" on 1M moon e'lual to __ Ie). n.e rna" of an object that weighs S It. on earlh is __ (.tl.

I (. ) )001 (!o ) .. _ ()OOg)(9!!O cm/ , ' , _ 2.'14 . 10' din -~, «I m - ... /g - (20N)/(~,8 mfs') ­Z,{loHg; mm is the Wlm. any"he •• , (liJ m _ ",I, _ (5 Ib)/(32.HI /s' ) • (I. ISS >l ug.

4.3 A , .. ullan! • ., .,,,,,1 force of 7.0 Ib iOC'IS on "" oo)t:cl IIlaI ",. igh • .\I) Ib on carl h, ""'hat is the objeec's """"Ie. at""" ( .. ) on e.nh? (It) on the mOOd ?

, (. ) We u<;c F .. """, whe re ,,~ recall (hat F . tand. lor (he ,nultan( of.1I forces aCli ng on the mass m. To get m " 'e ....... e thlt m ..... /g .. (.j() Ib)I(32.2 It/ . ' ) .. 1.24 !Jug. Tbe n Q .. Flm .. P .O Ib)/(I.24 IIUI) .. 5,6:\ ft ls' , u.) T1le ao:«: le. ation on the moon i. the .. me sin"" the .... uUant force is ~till 7 .0Ib and the mass is tM!08-m.e an) ..... h. re .

4.4 A hoNenta! cable )l\IUS a 200-k, til" Iiong i horizontal tTlll:k. The ternion in Ihe ... ble i. ~ N. Slaf1inJ hum .«(, (.) h.". long " ill it (ake the can (0 reach a s.pce<1 of 8 m/ . ? (bJ How la , ... ill il ha,~ gone ?

, Ass"mi", no friction, tbe (ension in the cabl. is the on ly hmizontal force . T1len from F," m4 . ..... get g, " (SOO N)I(200 kg) .. 2.S(I mi.'. W. no ... "'" kinematics to solv. the problem, C., Y . .. II .. + ".1; .nd"':_ .he tart ... rt. f ....... ' <St . II .. " 0, Then , .. !I. la • .. (~m/ s)/(2,'w mi. ' ) - l.ZJ,. (" Letting (he Slan inJ point be the oripn ...., hne -< .. " .. ' + I"" .. 0 + h 2,'w m/. ' )(:U . )' - 12,8 m

• • 5 A 9O).kg ca, is goinJ 2iJ mi •• Ionl a 1t".,1 road . How large a ron.tant retarding force is "'quil'<'<l 1<) Slop it in a di<l._ 01 30m?

, fler . ... sUrI wilh the b nem.tio<:al e<!u"li<)n Iha t allows u. to find the acule,.tion D.: u; .. H,'. + z...-<, ...... r. H. _0 when % .. 30m and u .. " 2Om/s. Solving ... nbhin .. . .. - 6.61 mI.'. Finally we ""I, .. lor I"" retarding 1o"", using F. .. ma, .. (900 k,)( - 6,67 mi.' ) .. - 6000 N .

• • 6 How m""h force does it ta t " to pvc. 2(Joo()'kg locomotive an aca:leration 01 1.5 mIs' On a le,·. 1 trad ",th a OIXfOOenl of rolling frict ion of 0.03?

, Refe. to Fig. 4-]. F - " . N " ""' , N .. "'g . F .. ". mg + ..... .. (t.()J(20 ((lO)('.8) + 20000( l.S) .. 5880 + lO{(lO _ 35,880 kN

,

51

52 D CHAPTER 4

4.7 A 12.11-& bullel is oculeraled frum resl 10 a speed of 7OOm/ ••• il tra~b 20cm in a !'In ban.l. Assuming tbe acccle. ation to be ron~anl. how large was 1M ocule"ling force?

, From kinematics W(: ba~e v= _ vi. + 20 ..... FOf "". Cllse v .. _ 0 , and v, _ 700 mi. when -" - O.20m. Sol\'ing W(: ,et a. - 1.23 . 10" mi.'. We 1M" obtain F. by noting m - 0.012 ~g and ~Mng F. .. ma, - .I..!...8..kri.

4.8 A 2O--kg ctate hanp ~I the end of a long rope. Find illoculenuion ",h.n Ih. t.nsion in tM rope is (a) 2SO N, ( /to) l SON. (~) .. ro. (a') 196N.

, "The e. ale is &ttw on by two vetllcal fo.ces-the ,ension in the rope, T. up ..... rd and 'M weisht of lbe crau. w" mg. downwa.d . Noting thai .. " (20 kg)(9.8 mi.') - 196N and IIMng T - ... .. ma" ..... gel: (a ) ~, .. 2.1..rn.lI:; (/to) Q, - - 2.3 mI.'; (~) D, " -'.8 mi l'; (a') ~, '"' O. Nole that neptive """,leration is downwa.d for OUr <ase.

4.' A ~:Hg t"mk 'Iiding &<'0<$ a floor 'lows down f.om S.O to 2.0m/ . in 6.0 •. Assuming thai the force actin, on Ihe trunk n constant. find its maptitude and its direction relo,ive to 1M veloci'y vector of lhe ''''nk.

l U ning tM -" oxis M olona 1M direction 01 """ion . .... have lOO-lhe magnitude 01 lhe resultant 10= F. .. ",a •. T ... find a . ..... use th. kinemalinl rtlotionship tI . .. " .. + a.f. with v .. " ~.OmJ., ". - 2.0 mI •. I .. 6.0 s. SoI\'ing .... get D • .. -O.S!) mi.' . TI>en F. .. (oCO k,)( - 0.50 mi.') .. =1!l.I!! . NOlinllhal Ihe • ..,ult.anl flJllX in the , direction i. zero since Q., .. O. we have OUt .... we •• I' i. 20 N in lhe di.ection """"",Ie to tbe Velocity .

4 .10 A r<sultanl 10"'" of 201" Ii'-e •• body of ma.. m . n .tukration of 8.0ml " •• nd a body of m .... ",' an ",,"I.ration of 24 mh'. Whot occelerali"" .... itlthis lorce cause th< I...., ~ 10 kC(jui~ if lut~ned togethert

, From 1" _ m", ",ilh 1" - :!ON ond" - 8.0 mi' " \II~ St' '" - 2.50 ~,. From 1"" ",',, ' and d' _ 24.0 m/s' ..... get m ' .. 0.10 kg. Combini"3 lite tlllO maMe$ yields M _ m + ", ' _ l.ll kl and 1" _ MA y;'kb A, _ 6.0m/ .' .

4,11 An 11000~g ~ar " • ..-els on a s"aighl highway "';Ih a "",ed 01 30 mI. 'The drive, sees ~ red light ahead and "1'1'1;.., he. bfilkcs. whiclt u.n a (:QlI$lanl braking loott 01 4 kN (.) What is the decele.ation of the art (II) In how m.ny seconds "itlthe ¢II' lIop?

•• U

, (a) Use Ne"'ton ', serot>d I . ... of """ion. Take a ret.a.din) fOTee to be Mlltin .

_4 _ 10' _ 1100<1 .. - - 3.6)6m/,'

Thu •. lhe Meele •• ,i"" is 3.636ml .'. v - v.

( II ) Q - -- ",hue v <'<juol. ,ero aft~r ' 1S«Orld<. lind " . <'<jualf. 3Om1l. , O- JO -3.636 _ _ _

, /- 3013.636 - llli

"The ca, ,,-ill come 10 a . lOp in~.

A 10rDO <li1O N ",'CO an object of unknown m ... an aca:le.alion <li 20 fl l.' . Whal is lhe object'S ma .. ?

, F .. "'" F 70 N 7O kg ·ml.' ; .. ", m- (20fl l. ' )(O.30S mlfl )- 610mb' _II.Skg

4.13 A boy havinl a ma .. 01 75 k,lIoI<I< in his hands a ha. 01 flour .... ighing 4Q N (Fil . 4-2). With what fo.ee does Ihe fIoo. push up on hi, f.el?

, For the boy 10 be in <'<juilib.ium. the fIoo. mu" push up on Ihe boy', fe<:1 "';th I foott 1" «lUll and opposite to the combined "'eight of the fIou. and the boy. Let m equallhe ma .. of lhe boy and ... the W(:ishl <li tile IIour'

1" _ mg + ..... 15(9.8) + 4Q _ 7JS + 40_~

4.14 Apply !'lewton', third liW in the followln, situation: Two drivel"$, one owninll large (!KIiIlK IJld the olher ownina a $11\.111 Volx."'alCn. m.ke a bel . The VW o";ne. bea thal lti, ca. can pull as hard as.he Cadillac. The y chain lhe 1_ reaf humpcn loget"'" in a I"", .mply p;i.l<ing lot. Eacb drive. ,ets into his ca. and app!i<s futl powe •. TI>e Cadillac pull> the VW backward all "". r the lot. Th. driv •• of the VW 101C' claims

NEWTON'S LAWS OF MOTION D 53

m _75 kg

j w_ 4Q N

F

fI&,4-1

that hb car WIIS pullin& on the chlin ai hard II lhe Cadillac all the lime_ What dots Nell1Qf1'i third "'w WlY in this~? Asoume Ih.tlhe clIain us r>tJliJible mau .

I Newton', Ihird la ... SA)"$ Ihal lhe VW owner i, right . EadI car must pull on lhe chain wilh lhe ... me nL3",iludc of force, 0fIt being lhe a.<1iQd and 1M OIM. 1M ~a.<1ion . The mOl;"n of the VW is a ron"'quence of tJlI1he fo-<= acrins on iI, 001 just lhe fo"", of the chain_ Thue OIM' forces indulk , in .... rticular, Ihe frictional force Mlween lires and road , "' hicb is quile dilfe. ent for lhe VW and lhe C.odilla<:.

4.15 A 2·lJug mass pulls hOrilOll'aliy Qn a :kIllg mO$' by mea", of a lightly 1!relche<l ",ri", (Fig. 4-3). If at one inol. nl the :Hlug mW hO$ an .=Ie"non tow.rd 1M 2-illlg ma.u of 1.8fl/ . ' . find the nel force on lhe 2...tug maSS and its acttleraliOfl allhal instant .

I "The >pring pulls on uch mOS$ Mlh 1M WI"'" fortt F bUI in oppmile direttions. U. ing Newton's oerond I .... ,

F _ m,~ , " m,~, F _ m,"," 3.1"!!" 1.8 It/.' .. 5.4 Mug ' ft /s'

Uoing Newton's second 1a ... on the ma<s m"

F _ m, .. ,_ S.4lb

m,

:z.. , _ 5.~ " , - ll1!JJ.: m,

4.1' A 96-lb boy i. i1anding in an ele~.to •. Find the force on the boy'. feet ,.-hen Ihe elevalor (.) stands still (.!o) rnoveo down .... rd .1 a """,Iant . eloo:ity of 1 ft /. (c) a<Uleratu down"-ard wilh .n a=I • • alion of ~.O ft / .', and (d) ..::cckrales upward with an a=leration of 4.0ft/ . '.

I (OI) Find !he: mus of thr boy in (he proper unil, thc slug:

m_ .. eifhl(w) .. \l6 tb _ 3Ib . , 'lft _ 3.lug =10 .. 1100 of gravity (g) 32ft/s'

When an .te.-ator sundo .till, lhe for"" on the boy', f«I, """"rdinl lO Ncll1""" firsl la .. , is equal and """"'"te to his weishl. "The .nswe. is IMrefOfe i!!Ul:! upward. (. ) A comI.IInt velocity melns zero ..,.,.k •• lion . !'or ... TO ..,.,.ltra~ion lhe force of the ~Ie ••• o. ""lance. the unh', downward ,,1¥itllion.1 force, th.1 is, llIe .. eiShl of . M boy_ As.in by Newton', ~t .. II"' , .he ........ r is~ upw_rd. tc) When lhe ..,.,.1o .. lion of 1M ete.alor;' downward, Ille boy'l .... iShl exceeds 1M upward (om: £ of lhe elenlOf, yieLdin, I net downward fo",," F. From FiS. 4-4

F _ "'" _ 3 $lug " 4.0 ft /" _ 12 lb . 'Ift " ft/. ' F _ 121b

By Newlon'~ second lIw, Ille 800r Of' M elevltor pushes upward on 1M boy', f« 1 M\h I fOfce of 84 lb.

54 a CHAPTER 4

I m-3 slug

-1 ( lie }". 4-4

(Ii) \\'hen lh~ ~1"valQo' accelerate. aI 4.0h/.' up""a,d. lh" foru E ocring on lh. boy is l!1"a,e, than hi . ... eighl .. by an amount f .. 12th:

4.17 An .kv~lor $tarl, born T." .. ilh a connanl upward .ettle •• rion . It """"'S 2.0 m in Ih. fi~t O.60s. A p,assenge r in the e~valor is holding ~ 3-kl park,Jlg. by. '-ertiCliI Siring. What is 1M 'enu in Ih. sllin. during lhe ~Ieral;n, proteM?

, To obtain [he 'ensioo T in Ih" Stri ng ... ·c apply ' .... i;<cond law 10 {lie "' .. 3.0 kg pack.,. : T - ..... ",a, . with'" .. mg ,,~, The acceleration ~ of Ih. pa<:bge is Ihe ~"'" OS ,ha, of th< d • ..ator; it is obtai ... d lrom II>< dilP1a.ccmenl formul. y" U .. I -+ !Q,l' . .... h. '. t'<\< .. (I and y" 2.0 m. when / .. 0. 60 •. $.Qlving ~ gel ..,. _11.1 mb'. Subs';Ming inlQ our equation for T. Yo" It"l T -~ .

• • 18 JUSl as her par.>ehu" open<. a lSO-lb pararnuli$l i. fall ing at I .peed of 160 f, / •. AIle, (1.80< h., p....w. [he chute i. fully <>PC" and be • ..,et<l II ... d,oppe<Iu, 35 1' /0. Find .hoe a,·craic re. arding force cxc rted upon .he ChUIU.1 during th is time.

, Wc choos.c OO"n ... ·ard .. Ille posih,·c,. di,cc\iOll. "The a'"Cragc a~lerotion of the <hutiSI 0V1" ,"" O.l!O-. intcrval io

(35 - U;(l)ft/o .~ , , ' 1 _ _ _ 1 .... 1 . , 0.800

Nat we usc ; 1: F, _ mii, . or ... - t _ .... ,. with", _ ISO lb . 1ft .. ",/g _".tII .I ug. and t _ thc a"c,aie ret.ordinl force due 10 thc chute . Solving we gel t .. 881 N.

4.1' A boy who normally "'eigru. 300 N on a ba'hroom scale cr<>U<bes 00 tbe ~Ie and ....Jdoenly jumps upwa.d His companion ltOlicc> Ihat Ihc """Ic Icading momentarily jumps up 10 400 N ~sthe boy ~ngs upward Estimate the boy •• maximum acceleration in this proce ...

, The maximum upward loroc exened by the .calc on the boy is 400 N, The net lorce on ,he boy is (400 - 3(0) N Ind Ihis «Iuak mQ. Using 1ft .. JOO/9. H yields Q" Um.I.i.

4.:tO Shortly after leaping from In airplane a 91_S- kg mon hal In upwa.d f(ll"« of 215 N eUrled on him by ,he li. _ Find the rCiultant loroc on 'he man_

, 1'h< .e. ullan, 10"'" 00 'M man i. tl>< "«10' 111m of ,...., for<:tl-ll>t ~ight '" .. mg .. (91.8 k,)(9.8 m/.') " 900N dO'/inwaro. and the WoN fOltt upward. Then the resultant fOfi:e i~ 900 -125 _ 67' N IIolllnWird_

4.11 To me ...... e the m ... of . box . ~ pu!oh;, 1100, . s.mo<Kh surface. exening a nel hori:rontal foro: of ll(llb. The accclenuion i. """"",,d 10 be 3_0 m/, ' , Wha. is thoe rna .. of ,he bo.?

, Usin! f; _ ma, for tl>< hori:ro.nal direc,;"n wo go. ISO Ib x 4_45 Nltb .. ",(3,Om/s,) and 1ft .. 22l kl.

4..21 A book si .. on I boI'izonlal,op of. oar a. II>< car accele,atCi horizontally lrom , ... 1, If the Slatic coooflicitnl of lriction betwecn oa, top and book is 0.45. ,,'hat is the maximum acceleration lbe oa, oan ha"" il the book is not .o .Iip~

, When the book of rna .. '" is ab<:>UI to slidoe. t l>< friction f. wng, Friction is th~ only boriwntal f<Xcc acting, Ihus f _ ma, l~niRg II .. OAS yitlds a _ 1'1 _ 4.41 mlf' ,

NEWTON'S LAWS OF MOTION a 55

4,13 f'nwe II>< following for a ca. moving on a horizontal road: The magnitude of the oar's """"k,..,tioo. cannot cxcccd /It, wl><r. " i. tl>< o(.<"Ili""'nt of friction bel"'""n tires and r~. Whal i5 II>< simi lar . 'presoion for I"" """"kral;on of a cor going up an inchne who>e an!k i. II~

, Friction between tire. aOld road ... pplios II>< f(>ltt mQvinl\ the car. !oO f - ...... 0.. a horizonlal road F" _ nq. lO f_. - pF¥ .. ,.,.,g. tM.oIo •• ,,_ "' ~, 0... 1M inchM. equalion> of motion p.o ralkl and pe'pendkular t" II>< ourface are I'F" - nq sin IJ _ ""'_ and FI' - "'8 CO> 9_ 0, Solve for s : ,, _ . _ (p CO> IJ - sin 1Jl3'.

4.14 A S-kg m .... hang> a1lh. end of a cord. Find Ih. tension in lito cord if lito """"Ieralion of lhe m&l$ i5 Is) 1.5 mI.' up , (II) I.S ml" do"<>'n. and (.) 9,8 m/o' down,

, Choosing upward as posilive we have T - ... _ mil, • .. Itore T i. II>< I.n""'; in lito cord , '" _ S k, •• 1Id '" _nq _ 49 N i. 11>< weii/ll of Ih. mns. Is) T -~ (II) T - !LU::!. (e) T - 9.

4.25 A 7O£I..N man SlaJ>ds on I !ell. on l ito Moo, of an el"'alo" lbe ",ak records lhe force il •• • IIS on whatever is on il. Whal i, lhe s<>ale , eading if , I>< devalor ha. on "","Io,al;oo of (s ) 1.8 mI.' up? (lI) 1.8 mI. ' do ..... ? (~) 9,8 m /.' down?

, Apin choooing up" .. rd .... pmilive arwl le tt in, N •• p.e""nllh. fOl"" of tlto """Ie on II>< man. we have N - ... _ ""', . Not:ing IMI '" '" 700 N, and thaI m _ wll _ 7], 4 kg, .. c wive for N using the valu.s of " , given: (s) N - m.t::!.. (lI ) N - .llll!.. (t ) N - \! .

• • 1.6 Usinl , he scale describe<! in Prob . 4 ,25 . a 65.k, OS""",,u' ,.·.igh. him""U on lito m<:>Of\. wl><re & _ 1.60 m/s'. Wh . t 00... the ",ale ",ad?

, Since: 1_ is tl>< """"I • • alion of fre. fallon II>< moon. and "'_ ~ the for~ of gravity on II>< moon's lOrlace. we have "'_ - m&_ - (65 k&)( 1.60 m/.' ) - ill!.l::!, By N . .. tOlt·, 1i .. 1 law "",Ie reads~ .

• • 27 A rough . uk of thumb ".t ... IMIII>< frictional fOr«: bet"'",,n dry con<lete and a $kidding oa .·, tire, i, aboul equlto nine-Ienlh. of the ra, ', w.igh!. II t l>< ' ki d ma.ks left by a ca . in comin& to re" are 20m Ion,. aboul Itow fast wa, the ca. going iu" before II>< b. ak .. "'.r. awlied?

, Si""" Ifl - " IFNI, and since FN - W on level g.ound . .... have for I' _ 0.9. f _ - 0.9W FInd II>< ca .'. """,Ie.alion f.om f - ""' . which i. f .. (W I&)a with f .. - 0,9W. Thon " .. - 0.%, Si""" .. is 20 m, U"" .. ' - ,,~ _ lou with .. _ 0, to give " . -1l8 '~ _ I ~ . ~ mI •.

4.1II If II>< coclf\ci.nt of friction bel .. ..,.n I ca,', ,,'h.,., b and a roadway is 0.10. what i5 the least d;".nce: in "'hich II>< oar ran aculera" from , .. t to a I;lC:M of 15 ml.~

, U,ing" - I'K ("'" Probo. 4.Z3 and 4.Z7) in the ki nemalical fo.mula .. ' _ ,,~ ... lou .

.. ' - II~ 15' - 0' . --- -- "' 16 4m 2.. 2(0.70)(9,8) _ . -

. ,29 A cortitanl fora: ..,.,.I • • al .. an .lect.on (m - 9.1 • 10"" kg) from , ... 10 a 'P""d of 5 . 10' mi. in I d;>tance of O.lIf}cm, Det.rmi .... this fora:, How many Ii ...... la'8e. than nq i, it ?

, First find tl>< constant accel.ralion from v' - " ~ .. 2ax; tl>< n " .. 25 x 10"/ 1.6 • 10" ' _ j 56 x 10" ml '" to

F .. "'" '" (9.1 x 10-") (1 .56 >< 10" ) - t o . 10 " N. Thtn Fl mg _ a lg - 1.56" 10" /9.8 '" \,6>< 19' •.

4.31 Tht 4,Q.kS I><od of a sledge ham ..... is moving at 6,0 ml> .... 1><" it . "ike'" .pik •• driving it inlO a 101; the duration ol ,he impact (or tl>< lime for the sledge hammer 10 st"" after contact);s 0.0020 •. Find ~ s) tbe lime average of the impact fora:. (lI) Ih. dislance Ih. spike penelralr< tl>< log.

, (s) Tho a"erage f'lKe is a conSla nl fooce that wouW effr<1 the jlme . ... ul1 as II>< actualtim.· varyin, fora: ""er lbe I;me interval in.'olved, Fo. a constanl fortt W(: ran uor v, _ v", + a,11O find t l>< corr •• ponding con"an, acceleralion Q, . Here v", _ 6.0 mls. " . _ 0, and I _ 0,0020 •. Solving ... .., I\"t " , _ - .lOOO mI.'. Tbc:n F, .. mil, y;.1<k 1'; .. -IZ kN . H.re F, "'pr....,n" II>< for~ of tl>< >pi ke on Ih. hamme. · I><ld. Tbc: r.action foroe on lhe !.pike is lbe impact fora:, "hich is 12 kN, (6) TIt< <pike mO . .. Ihe .. me dj".""" as Ih. hammerhead in the ,ime in1erval in q""' lion . For the hamme""'ad w. ha~ . .. '" <'",I'" 14," " (6,0 ml.)(O,OO2(I s) ... !( - 3000 m/.,)(0.0020.)' .. 0.006 m or 1< _

0Jlmm.

atanal

56 D CHAPTER 4

4.3 ' " A body of ma~ m mini .. along y.w that at time I it. position is y(t) .. QI'" - bl + c. where Q. b. care coRStantS. (Q) Calculat. 1M ICUI."lion of lhe: body. 'b) What is the: force actin, \HI it?

, ,,,) ", _ 4'y ldl' , We fim obtain ~, .. d, l dl - 1"'" - band Ihen 4',ldl' - l .. r ''', To obtain Ihe force we ha~e (b, F - ma, " 1""'1 - '"

4.3r Meosu",ment> "" a IDg object movin, alonl 1M ~ U;' ohow i .. poWtion (in centimeten) to be Ii"en by x _0.» - 5.0.' + 1.51' . ,,'here I is the: lime in KOOnd$, Fioo the: net f .... o:e that acted on lhe: object during the time fOt" "him this upr.ssion appli~.

, Bc:cau>'e F .. , _ ...... "'. mUSI find Q;" _ 4u ld, and u _ b I4,. Doing lite difre'enlialioons. u - 0,20-10,0. + 22.51' . a _ - 10.0 + 45' . .., F ... " 300(- 10.0 + ~5I) dyn " 000300(-10.0+ 45,) N

4.ll' A !i(~g maS$ vib'lling up and do"'n at the end of I .pring has its posit ion siven by y _ O. ISO";n 31 m fOl r in .. coMls. Find lhe net foo-t. thaI acts on lhe rna .. 10 give il th;s mOlion.

, The a=lention is d' , I d,' - - 1.35 . in), ml, ' ; w F. , - 0,0S0(-t.35 sin 31) - -0.0615 sin 31 N. The minus ,ign ir<!icales. ",.toring foroe.

4.34' A body of mas.& '" mO"e, along X . uch Ihal .t lime I in position ;. x(l ) - .... - fk' +)1. ""h.", "'. fJ. yare OOMtan". (.) Cakulate lbe Icnkration of tbe body. (b) Whal i, the force acting on it?

, (.) i_ 4<U' - 3fJr' + y and f - 12,,1' - 6(lt

(OJ F. " m.I' - 12m",' - 6m(Jt

4.2 FRICTION; INCLINED PLANES; VECTOR NOTATION

4.35 The b,caking "rcngth of a steel cable i, 20kN. If one pulls horizo.Hally wilh Ihis cable. """al;' lhe: muimum horizontal a=leral;\HI which <'an be siven to an g..um (metric) body . ",I;n! on a rough oorironla l surface if the: coefficient of ki nelic friction is 0.15?

, Let T be till' ubi. force. Then L F '" mu ~(!$ T - ,..mg _ ma, For lIIuimuIII ar.uleration T .. 20" HI' N, SO lhal 2.0" 10' N - 0.15(8000kg)(9.8 mN) - (8tXlO kg)oo . Sotving ~ ,et Q _ 1.03 mIl ' .

4 • .16 A lO- kg ""'~n ;" putled atong lhe: level , rour<! by I rope inclined at XI' abo>-. lhc: horizontal . A fridion fora: of lON OppOSe.lhe molion, Howla.&e il the pulling force if lhe wagon i. moving ... ith (a) con.cant >peed. and (h) an a cnlerolion of O.4Ilm l. '?

, leI 1M pulling fora: of Ihe rope be T. Ur.inS 1: F. - ""'. we have for OU r cue T COS Yr - 30 N _ ""',. where '" _ 20 kg. I.) For «, _0. T .. HA1::!. (b) FOf Q. " 0.40 mI. ' . T "~

4.37 S"PI""" ... >hown in Fig. 4-5. thO! I 7O-kg box is puliN by. 4OIJ..N force alln angle of)lf 10 lhe horizonlal The coefficient of s.liding friction ;. O.SO. Fir<! the "",,ler. lion 011"" box,

, Because lhc: box doe, nOlI mo.e venially. 1: F, " ""', - O. From Fl., 4-5. ~ see Illat Illis equation is Y + 200N - mg .. 0, BUI "'8 _ (70 kt)(9.8 m/$') - 686 N. II follOW$lhal Y - 486 N.

We <Oan find Ihe friction foroe Kling \HI I"" box by ""riling! - "Y- (0.50)(486 N) .. 24) N. Now let u, ... rite 1: F, _ ma. for lhe: bo • . It i. (346 - 243) N .. (70 kg)o-. . from whidl «. _ ! .41 m/J' .

,

4.38 A. 'hown in Fig. 4-6 .• force of 400 N pu1hc:s on l 2S-q box, Staning from ",.c, tbe box achie~es a velocity of 2.0ml. in a lime 01 4 o, Fir<! 1"" coefficienl of oIiding friction betwe<:n box and 1\oor.

NEWTON'S LAWS OF MOTION 0 51

, (llj(1J.1, N

, W~ mu.t findfby usc of F .. ma. But fi ... t we must find a from a moI:ion probkm. We kOO\\-' Ihal ~.- O, v, _ 2 mI. , / - 4 •. Using V," u. + s/ gi'-"

~I- V. 2m/. 0" I' . _ _ ___ .-", m ~ , , , Now we can ",ril~ L F. _ '""., where <r, - " .. 0.50 m/s' . From Fig. 4·6. this ~q nali'," i. 151 N - f -(2~ kg)(O.SO m/ s' ) , or f - 2~S N. We now wi>h 10 use I' _ f l Y. To find y, ",e " 'rile L F,. .. mQ. _ 0, since no "",nical motion 0ttU .... F.om Fig. 4-6. Y - 306 N - (ZS )(9.8) N _0, or y .. 55 1 N.

"". I ~, I' -- ---~

Y 551

4.39 A 12-kg box i, r.leased f.om the lOp of an incline Ihat i. 5.0m I""S and makes an anJle of 40"\0 the borizonla l. A 6Q.N f. iction f<>ftt impede. tt.< motioft of tl>o bo~, (_) What -..ill M 11>0 "<celeratioft of tile box and (II) ho", Ioni .. ill il lake 10 reaell lbe bottom of IIIe ind i",,?

, In Fig, 4-7 ",ell", ... the Ihree forces actin, on 11>0 block: tn.. f. ictional force f - 60 N: lbe normal fOf'tt N, ",h;"h is perpendicular 10 In.. incline: and lhe "'.iJln of In.. block, ..... mg .. (12 kg)(9,g m/s') " 118 N. We cboose Ih. x oi. along 11>0 incline with downward as posilive . Usin8 L F. .. '""" we ha,'~ ... 5in -10" - I .. '""" or (118 N)(0.642) - (60 N) .. (12 kg)/l •. Sohin! we ha" . ~ ... 1.31 ml" . To ~nd 11>0 I;me 10 reach IIIe botlom of Ihe indi"", starti ng from rest, we U,", ~ _ U",I + la.I' , with u .. - 0 and ~ _ 5.0 m. SoI"';ng "i. ger , .. (7.1>30') '" .. .l.1!U.

4.40 For 11>0 situalion Outli""d in Prot>. 4.39, whal is 11>0 coeffici.nt 0( {richon Mtwnn box and inclin.?

, Alain ref.mnS 10 Fig. 4·7 w~ ha~. from 1: F, .. 0, N - wws 4()" - 0 , Or N .. 90 N llIen, r«alling Iha' ,I\(

"""fficic:nl of ki""lic friction i. gi"",n b~ 1'. _ l IN, "'e hO\'e 1'. "0.67,

4.4. An indiDed plan<' makes.n angle of 3()' with the horiwruaL Find the COrmanl force, appli«l parallel 10 Ih. pia"", .. qui.«Ilo ca""" • lS- kg bo~ 10 slide (.) up the pia"" wil h lHX<:lellllion 1.2 m/s'- and (b ) down Ih. illelin. wilh aca:leralKm 1.2 m/,' , NeJlect friction forces .

, Here ..... ...sum. thaI the .. axis i. aloni lhe indi"" and """Ii'. upward. If P i. tl>o conslanl force .. fe" «I to, lI>on from L F. _ m<I . ..... ha¥. p - ... <in.\(l' '' ""'., .... ith m _ 15 kg and ... _ ,." _ 147 N ,

(. ) For Q, - !.2 mIs' , P - 't.Ul:!. (II) a, .. - 1,2 mi.'. p. &W

4Al A 41».1 block otiginall~ m()\ling at l20 om/ . wall. 100m along a lablelop Mfo .. comins 10 rest, Whal is Ih. "",,/!ki.nl of friction bel .. 'ee . blod and tablet

58 D CHAPTER 4

, For the block 1: F, - 0 yi~lds F" - W .. mg lJ>d 1: F .. _ ma yields -pI'" _ ma; Ilclla. I' _ - lI l g. The ~niform aa:xlcnllioo . from ~' - tr~ - 24>:. is II - -1.2'12(0.7) - - 1.03 mi.'; then p .. 1.0319.6 ,,~,

4.43 110 ... la'1" a force parlIUd \0 a 30" indiM i. needed to give a ~.o.k& box an toteJ~ ration of 0.20 mI.' up the i""hne ( II) if frielion is noeiligible? (6) If the coel!icicnt of fric!ion is 0.301

, (II) The compo""nt of the weight down the i""li"" .. "" sin)O'_ 5(9.8)(0.S) .. 24 .~ N .... hilc lhe e~lemal force "f' the pia"" is F. So F ... - rna ~ F - US - ~(0. 20) hom whicll F _~ I.) A friction fOfU _ pFM mll$t be added to 24.S N do""" the indinoe, F.~ _ mg cos :lO" _ 42 .4 N and pI'" _ 12.1 N • .0 F is largeT by Ihis amount ; th ... I' -la.l..t!,

4..... An 8.o.kS box is ..,lca~ on a:lO" indine and aa:xlcrales down the incline 01 0,)0 mI.'. Find lhe frictional fOftt impeding il. motion, How large is lhe coefl'icitnl of friction in this siluation?

, The componenl of lhe ... ·eight down lhe iocline .. 8(9.8)(0.5) - 39.2 N. NO'A' 1''' _leads 10 39.2 - / _ 8(0.3) • .0 lhal / _ 36.8 N. The oonnal fora' f<jW Ihe oo.nponenl of W perpendicular 'o the incline. 8(9.8)(0.867) .. 67.9 N. Th...,for~ p _ 36.S/67.9 _~.

4."5 A borirontal fortt P i. n.rted On. 2O-kS boo in order to slide it up' :lO" ;m:tiJle . The friction force retardinl tile """ion is 80 N. How large must P be if lhe a=leration of the mavins boo is to be (a) uro.nd (It) 0, 7S m/,'?

, Her~ .. .., cI>oooe the " nis along the indi"" with poiiti~~ upward. All tM forces on tbe block are.oown in Fil' 4-8, From 1: F, __ • we hay. Pros 3O"- ... sin)O' - f - _ .. where m " 2Oq .... _ "'8 _ I%N, and f .. SON. ("I For u. - 0. P - iIQU:!, (61 For a, - 0.7S mi. '. P - nll!.

..... 4.46 A horizontal loroe of 200 N is r~qpir~d to .;au .. a IS-kg block to , tide: up. ZOO incline with an acoclcration of

2S emls' . Find (,,) the friction fOfC~ on Ih. block and (b I 1M cocflicienl of friction.

, The sit ... lion is lS!hown in Fig. 4·9, We asain choose: Ihe % ui, along th~ incline. and tM, uis perpenttiruJar 10 the incline . 11Ien from L F, _ ma. w~ g<:1 PCOi 20'- f - '" sin 2fr _ (I~ kS)(O,2~ m/s') and from .quiHbrium in Ihc, di.e<1ion .. .., 8'" N - P sin ZOO - ... 001 20" _ O. NOlin, luI P _ 200 Nand .. - "'8 .. 147 N. "'~ sol,·. the equations. yidding f -l.J::U!. and N - M./:!. respecn""ly, finally 1'. -fIN" ~ ;. .h~ cocflki~n! of friction . Nor. how P COntribules to the normal fol'tt .

4.47 Whal is the smal1 .. ~ 10."" paratlel .o . 37" incline needed 10 keep a 1000N weitht f.om widin, down tbe incline if Ihc cocfftcic:nts of lIalie and kinetic friction are both O.JO'?

, A. ~,u&l , ~ Ihe .. and, ues I"'rall .. 1 and perpendicula. to ,be incline . let F be the unknown force. f 1M friction fo=. and N , I\( ""rmal 10«<, If Ihe weighl is aboul .0 slide down IIIe incline . then tbe frictional force il maJljmum and in the up"'.fd direction . 11Ien

f_. -p,N , .. N - ... COl 3'7"_0.

NOling lhal ... .. lOON and p, _0,3 . .. '. I.' N _ 80N andf_ " 24 N; and finally F - l61:! is the minimum force needed along Ih. inc~"".

4.48 Rtfuringto Problem 4,41. indicat~ ... hat parallel fo,,", is ff<jU;.ed to k«p lhe weighl moving Up'M indiM at COl\$tanl $pet<!.

NEWTON'S LAWS OF MOTION a 59

, H~Te _ haw: ki ... ti~ fric:t;.", down th~ i""lin~. ~nd f • ",N. N is Slill SO N (why1) and sin<:< 1', _ 0.)(1. , ... ~ hav~ f - 24 N. Th~n !: I'. - 0 yields f - ... yn 37' -f . 0: and lIohinl. ",,'e grt F - l!ir!I.

4.49 For the .ame conditiOft!l a. in PlOt>, 4.48 3$$Umc: (hat (he fo, ..., F. up lhe incline . is 94 N, What is the ac:ttkration of the oi>je<1' 1f the object 5tan5 nom ~t, how far wi ll ;t move in ]O.?

, A. be'o",. f. " ,N _ 24 N and i. down the incline opposing the mOfion, "Then Uynl !: F, . ma .. _ ha,.., F - .. y n 37" -f. map NOlin, m " "'t - 10,2 kg "'. soh". gelling a. " 0.98 mi s' up the i""line, Nex, _ .... the kin.m.ti~ fonn ula % .. 1t",1 + !<I,I' . with ""," O. Seningl . lOs and using ou, Q., .... grl % .. b.

4.51) A 5-kg block . .. ts on • .lO" incline. The coefficient O)f SI.,i. fri<"lion betw«n the block and i""line ;. 0 .2l). How large a horizontal fOil:'" must pU;h on the block if II>< block ;. '0 be on the ' '''' ge of .Iidil\& up lhe i""line?

, LeI P be the horizon,.1 fo • ..., . We choose.r: and)' axes a and .\. 10 ,n. ind ine. Since tn. block i. Of! tn. ""'II< of mowing. it i. in equilibrium: 1: F • • O and l: F; - 0, Her. _ haw: maximum static frictional fo= down th~ incline . PCOI.IO" - ... sin.lO" - f_. _ 0; N - ... cos 30' - P sin.lO" - 0. wn.", Nand f .re In. no.mal nd lriction foo: .... respe<tiw:ly . ... - "'K - 49 N. Substituting f_. _ " ,N in tn. first equation .• nd pulling in all known quantitin. we get

O.866P - O.20N .. 24.5 1'1 N - 0,5OP _ 42.4 N

'J htsC two equatIOns ,n lWO un knowns ~an be ooI"ed ,n a v.TtCty ot """)'$. W~ multiply the first equation by 5 and add to the second equalion to eliminate N. Th;' yields 3.83P _ 165 N. Or P _~.

4.51 R~work Prob. 4 ,50 lor incipient motion down In. inclin.,

, The only diffe",nce 110m Prob. 4.50 is thatth~ ma.imum frictional 10= i, up the incline . We tMTelo", chang< the .i", in lronl of f_ in the l: f. - 0 eGuation . We then get for {)tI' two equations:

O.866P + 0.2ON - 24.5 N N - 0.5OP _ 42. 4 N

We again multiply tile first equation by 5 . but n .... sublract the >i«Of\d equatioo to .liminate N. This yields ~ .IllP _&I. t N. or P - l.2&.!!!.. Note that in Probs 4.50 and 4.5! tn. numerical vah'" 01 N a. e different , reflecting thei, dependence on P.

4.52 In Fig 4-10. the 8- kg ohject i. subject to th. lo. ce. F, _)(IN and f; _ 40 N, Find lhe acceleration 01 the ob;ect,

, ,

, Ne"'lOn', second law in component lorm is Fo. + F" " ""'. and F" + F" _ ""', . or )(ICOI4O" + .t(ICOS 10'_ 80>, and )0 .in .I(l" - 40 .in 7lJ' _ Sa,. 5.oIvinl. a _ 4.01 - 2.3J mi.'.

4.5.3 A 7-k, obj«, i. subjected 10 ' ''0 fo~s. Y,. 201 t .!Oj Nand F, _ 81 - SOJ N. Find the a=l~ration of rh. object. , F _ F, + F,- 281 - 2OjN; • .1. t· _ 41 - (20/1)j mI. '.

m

4.54 The forces F, and F, ,ho" 'n in Ri. 4·10 give the 8'IS object the ..,.,.\e"'tion . .. HII mr.'. Find F, and F"

•

60 D CHAPTER 4

, OUI mo!i(lrl equations ta ke the fOlm F .. + F,.. 8(3). F" + F,.. 0 From the laner. f; >in 4()" ­F, sin 70'; from the fonner. f; = 40" - 2~ - F, COS 70', Dividinl these relation. we have !an 4()" •

F,>in 1Q"/(2~ - F,cos 70') , Uling the~. F, -~ arid F, - (lin W / oin W)F, - ~.

4.55 Find the force nu<kd 10 Jive a prOlon (m. 1.67 x 10-" k!) an .<ukralion Z x 10'1 - 3 x IO"J mI. '.

, F .m8 - 334x 10- "1- 5.(11 x IO-"J N

4.56 A 200-& objea is sub;'ned .o a fo'~ 0.3OI - 0.4Oj N. If the objea "arto f,om r ... , . wflat "'ill be the ""kl<ily ... no,o( tbe object aftcr h 1

, , , • • •• +1 • •• + - )· - ..,...10.:lOI - 0.4Oj) - 9I - I2,jm/o

m O . .u.oJ

4.57 If tbe object of Prob. ~.S6 ,tart<'<l al the ori,in, ,"'hat wao ito location al lhe end of the 6-1 ~riod?

" " , .- .. + r •• + !I'. - 1+' + 2m )' - 2(0.200) (O.:lOI- 0.4Oj) _ 271 - ~ m

Thus lbe ob;ea ... as found atlhe point (27 m. - :16 ml ,

4,3 TWO-OBJECT AND OTHER PROBLEMS

4.58 In Fig. 4-11. find lhe .<ukration of the cart thai is ..,quired 10 p..,vent block B from falling. The """fficient of statK: fric tion he.,.."" •• he: blo<k and the cart is 1'. ,

, If the block is "'" 10 fall. , '''' friction force. / . mu", balance lbe block', ""eight: /. mg. Butlbe hori.",nal mo'iOrI of . h. block i. gi""n by N _ ItI<[, Therefore .

l. _E-N •

Since tbe m .. imum val"" 01 fiN is 1' .. we must ha,.., g "'- g il'. if the block is II<)I to fall .

• / .

r-':

( ) () t . N

FiI·4-1I

4.59 A paS$Cnge. on • la'ge.!Up ... iling in a quiet se. hangs I ball f, om the ""iling 01 her cabin by mr:am of a long thread . Wheneve. tbe ship accelerates. she ""'es thaI tile ~ndu lum ball la", behind lhe point of ~nsioo and SOl the pendulum II<) longer hangs ,·en ically. How large is the sIIip·. acceleration "'hen the pendulum r.lando at an ani!.le of 5' to tM , ertinl?

, S« Fig. 01.-12 . The ba ll i. acceleraled by the fo"", Tsin 5'. Therefo.., T sin S' - ""'. Veftically 1.: F. O. so r =5' - mg, Solving for ~ _ g tan S' gives ~ - 0.08158 _ ll!!6m/ li' , ,

"'.' 4." A IlCClangula, block of ma!$ m sits On top of all<)ther similar block. wlUch in lurn roilS on a !\at IIble, The

maximum poo.sible fnctional force of one block"" the other i. VJm N. Wbat is the la'geS! poo.sible accele""ion which <.an bo given . he ",,",'e, block withoulthe u[>pC' block .Iiding otr? Wbat i. the rocfflritnt of friction bc:1"'e<:n II>< ''''''' blocks?

, F_, - 2,Om - rna .... . SOl u_. - ~. Aoo 1' . / Img· 2.IJm /9.8m .\!l!l,

,.,

• • l .. ,,' -

NEWTON'S LAWS OF MOTION D 6t

4.6. A bIock ';rs on In i""line I. shown in Fi,_ ~13(.). (0) Whal mull be lhe frictionll f""'" bc,lween block aJ>d i"dirK' i( the block ;s ll(Il 10 slick llong Ihe i",,~ne when Ihe indi". .. """'Clerating 10 lhe right al 3 mi.'? (b) Wh.al is lhe 1eM! value 1', can ha'~ fOl!hi. 10 ~n?

, Re>oIve llIe~' ~I><lthe 3-m! . ' aa:elo:ralion inlo """'pont"" fN'rpeDdioruLar aDd parallel!o Ihe plan" [Fig_ 4-13(6)1. Wrile F · "", for each direction, O.6m& - f · 3(O.8)m and F,. - o.8m$" 3(o.6)m. whidl yield (0) f. (3.48m ) N. F" _ ('I.M ... ) N

f )A8m (6) 11.,"-;- _ __ .0 . .16 f " ~.M ... --

4.62 In lhe lboe""" of friction . would llIe 1>10<:\ of Prot>, 4./i, a<:<:elerate up Or down lhe ir.cline ?

, Do", .. ral 1.48 m/. ' n'lati~e 10 indirK'. since 10lal a.cee1e"'linn down indine is then 0 .6, by Prob . 4.()I(o)j .

•. ~ n.. indirK'd plane UIown in Fi,. 4-14 has an aeoeleral;"" a lo lhe righl . SI>ow ,hal the bio<k will!Jide on lhe pl .... if 0 >, tan (fl - ..-), where '" _ 10" 8 is ,he """lIlcient of ".Ii< fridion for lhe """'llKIin, ... rfacn .

• -• X f'I&. ~14

, If the t>1<xk is not 10 $lido . i, mu" ha •• the ... me a=leralion Of lhe pla .. e . Hence

From Ih .....

.. , f"""IJr - N oin lJr ...... f sin a -+ N em a - mg . 0

f - ... {a cos a -+ g sin 1Jr) N .... (gcosa - .sin lJr)

L. a COS "' -+Roi .. a • • +g Ian Q

N gcosQ - a sinQ , -"lana

Now lhe ntUimum val ... 01 f /N in Ihe aboence of ilipping is II," tan fl. Thus Ih. ;>culeralion. mu,' ... tisly

,,+gtanll ::s; tanfl , - ollnll

If a ;' , Ian (8 - 1Jr), the block "'-;U .Iido.

"' tan 8- tan 4

' " _,13n(8_ 4) ] +lan81.n~

4.64 Obje<fS A and B. nch of ma" m. are conneeted by I lighl i!lexlen.ible cord. They are constrained 10 m~ on I lrictioniesl. ring il> ' vert;""l plane, I, .hown in Fij . .t-IS. The object. aft rdt ... d from ",.1 . ' ,1>0 pOISition. shown. Find.he tension in the oord jlUt alttr ,tle_.

I At the m""",ot of . elease. II i. constraint<! to move horilOlltally and B ~e""'ally. SO lhal lhe lwo inilial

62 a CHAPTER 4

acttlefation$ Ife tangential as .ho1IIn . funhermore. the two acttkrations ha~. the same maJllil...x. Q. sin« "Iherwi~ the m ,d would h ... to ,"e'ch. Thu •. the horiwntl l force ~""tioo fOf A mel the vertical fOf"CC ~uaho .. for B. 0' the indic.tt<! plHil;om. I ••

Eliminating u. T _~_ mg loin 4~' V1.

4.65 If the , ystem in Fig. 4· 16(a) is .i,·en an acnlenlion . 6nd the for"". on the sphere . ........ m;". no frioct;"'n .

, From Fig. 4·16(b). 1: F ~ .. II , (()5 ](I' - ... - ""' .. ," 0 I nd 1: fj", .. If, - R, sin .vr - ...... Th\l$. ,i><' acting fortooe$ I fe

• -

• R, _ __ _ 1.15 ... ~.,.

'0 )

• • ( 0) R, _ R , <in.vr + - II _ (t . 15 ... )(0.5) +- Q - ... 0.58 +-, , ,

o -

'" -

R,

4." In F". 4-1 7. mau A j, 1Sk, and mas, 8 ;, II kg. If they a'e gi"..n ao upward w;;.I • • ation of J m/,' by pullinl up on A. 6nd Ii><' teruions T, and T, .

, Fim "1'I'1y Ne,noo·' ..,rood la .. to the . ystem I I a ... ·hoIe to lind the fotoe F, accelefating both mU$M

up""Jld.

F," (",~ + "'. )oJ .. (IS + 11)3

Since F, b the fr$ultant fOl"«' , F, " T, - "'~, - "'.r, Ind the 'e",ion T, is the own of the ,...;gn .. of A .nd 8 pi," 1\ .

T," mAt t m.g t F," 15(9.8) t 11(9.8) t 71-\"1 t W7.8 + 78

j matmal

Similarly f(lf mOSS B only.

f; .. moll .. 11(3) .. 33 N

To check, for block ~ only.

NEWTON'S LAWS OF MOTION a 63

t· Fla. 4-17

T, .. m", + f; .. 11(9.8) + 33 .. un.s + 3l - l£..U!

T, _ m~t + m~a + T, - 147 + 45 + 140.8 " lll,..lt:!

4.67 Referrin,lO Fi,. 4-18. fir>d Ihe aa:clcralion of lhe block> ar>d Ihe lension in the C011nectin~ mint if lhe ,ppt;CiI force i, F ar>d the frictional f= on Ihe block. art n.g1igibk .

, Apply F .. "'" lO tach bIod in lum to obtain F - T - m," and T - m,a Solve lor T and .. to oblain ... F lim, + m,) and T .. m, F I(m , + m,J.

4.61 In Fig. 4-18, il F .. 20 N, m, _ m, .. J x._ and the aa:cl ..... tion i, O.5Q m/.' _ what ,"'iII be the ten""n in the COnnett;ns cord if the: frictional form 011 tile two bloc~1 are .qual? How largee is the frictional fortt 011 citbtr block?

, ~.", . 4-18

, Write F .. "'" for.;oct, block using! a$ the friccien for~ on each block. ~ ... obtain F - ! - T .. m,.. and T - , - m,'" Use the pvcn value> and .oh·e 1o fir>d T .. JO N . ad , - 8.5 N.

t." The device di"llramed in RI. 4-19 ;. called an ArlOli>Od', "",rlIin~. In lenns 01 "" ar>d "" wilh m, > no ,. ,.) how far .. ill m, fall in time' after the $)'$tem is .elea~d? (b) What is the le",ion in tM ligllt cord Ihal COnnK11 tM lWO mas.se.~ A .. umc lhe pulley tn be f. i<lionl . .. on<! maulc$S.

, ,_) boIale 1M forces on each ""' .. and ... rite N ..... ton·. second la ... . choo!ing up as ~ive ; T - m,t - m,d ar>d T - m,J - - m,d. Eliminating T givC$ " - (m ,-"' ,n/(m, + m,). Now UIC y -Ill'n 1o find tM di>lano. faUen in lime t. ,b, From 1M above equatiofD, T .. 2m,,,,:.!'I(,,,, + m,).

64 a CHAPTER 4

•• '71 A <:<I. d paWn, OVer a frictionleM. ~te.s pulley (At....,oo's mamine) has a 4-kg block tied to on. end and a ll-kg block tied to the OIher. Compute the IoCCCleration and the t./Won in the cord.

I Uoina the formulas o:Ieri~ed in Prob. 4.69.

12 - 4 , •• --(9.8) .. 4.9 mIl

12+4 T _ 2(4)(12) (9.11) _ SS.II N

~ + 12

•. n for an Atwood·, machine (Prob. 4.69) wilh massn 10 and 12 kg. lind ta) the velocities.t the.1I<I of 3 s and (Ill the distancn moved in 3 s . te) If II the end of J 5 the SIring is CUI,lind the di.tances moved by the IT\8.S$C$

in the nut 6 •.

g .. ( "" - m,. _ 12 -1O(9.1I). 0!19 m il' m,+m, 12+tO

Si""" the aoxeleration i. DCNIMant. the common speed Bt lhe.nd of:3 . i. U" v.+IIl_ 0+ (0.89)(3) _ 2 67 mi •. Mass 2 move. down and mass 1 mov<S up. (II) "The dillancc: moved by.ach m.a>I in :3, i.

, .. Vol + 1«' - (0)(3) + \(0.89)(3)' _ 4 m.

(e) If the .. rinS is cut. the massn fan b«ly with initial velocities v",_ -2.67 mi. and v .... +2.67 mI •. witb up taken as pmiti"". for maSS 2. lhe displacelllCnl in 6 I is then

y, .. v,,; - Igr' .. (-2.67)(6) - 1(9.11)(6)' _ - 192.4 m

i .• . . a downward dill.".,. of l2L!m. Mass 1 Irav.h upward I di ... """

d' " v!" _ (2.67)' _ 0. 4 m 2f 2(9.8)

hefore cominl to a stop and then falli.., downwud. "The tUnc of trav.l upward before comin, to a .. op for mass 1 i.

v .. 2.67 , ----- 0.271 .. S 9.8

It then travels do"WJ\ .... rd S.73 , for a dillance

.r - il( -s)t'l .. 1(9.8)(S.73), .. HoO_ 9 m.

"The lotal di.lancc: tra~eled by rna.. I is then d .. d' +.r _ 0.4+ 160.9 -Iillm .

•. n In f il. 4-20, the ....,igh .. of th. obj«ts ue 200 and JOO N. "The pulleys arc n5<!ntially frittionless and massless_ Pulley P, has a $lationary axle but pulley P, is free to move up and do"WJ\. Find the tensions T, and T,. and the IoCCCleralion of e,..,h body .

C righted IT £na

NEWTON'S LAWS OF MOTION D 65

I MnJ B will ri", and m .... A ";11 faU. You can ..,.. Ihis b~ DOling thaI I~ forc:cs actina on pul~~ P, Ilte 2T, up arn:I7; down. Therefore T, - 2T, (the i""rtialeso. object transmil. tlte ten' ion). T";~ as lar,e a force ;s pulUIIJ upward on B as On A . Ltl ~ .. downwlrd acceleration 0( A. Tlxn la .. UpI'Iard aca:lcration of 8 . [AI the ami bct .... et'n P, and A ~nK1heM by 1 unit . 1M ",,,,,,,nl$ on eilhe' side of P, eath sbortfft by ! unit. Henr;:e , ! .. s.I,_ .. (ja.t')/(!a A , ' ) - ~.laA '[ Write 1: F, " ""', for each ma .. in tum. taking the direction of motion as positi"" in eadl c&>c . We have

T,- 300N_",.(\s ) '"' But'" .. ", I, and $0 III. - (2OOf',l.8) kg and "" .. (30019.8) k8· FurtM., T, .. 27;. Substitution of thne '1011.1« in the two e<[uatiom aUow> uS 10 compute T, and then To and d, The results are

T, - llil::! T,_~

.,7.} An inclined plane makin, an ontle of 25" ";tll the horilOll,al has a pulley at its top. A )o'k, bl« k on the pi .... is oono>ect.d to a ftuly hangins 2O-kS bl«k by mean. of a cord passing oyer the pulley . Compute the di51 an~ lhe 2O-kg block "i tt fatt in 2 S 51amng from rest, Ne"ecI friclioor .

" , The silualioo is as sbown in Fig . "'21. We IIpply Ne'"'ton'. second I ..... to each blrXk ~panrrttly. For bl<Xk B we choose downward as positive. while for block A we choose OUr" uis along the incli"" ";th tbe positive "' .... UPWlrd, Thi. choi« allows \IS to use Ihe sa"", symbol , a. 10.- the a<:<:ele.ation of each block . Then fo.­block B, .... - T .. "'. d. whe re "' . .. 20 kg and .... _ 1% N and T is the tension in the cord. Since tIM: pulley is frictionless, the sa"", ten5ion Twill .xi§! on both $ide> of Ih. pull.y. Then fOf block A, T - .... r.i n 2S' .. "' .... ... h.re .... .. 30 kg ar>d .... _ 294N. We can eliminate lhe lension T by adding the two .quation$ . ... hid! y;'kh .... - .... >in 25· _ ("'. +m. )d. Su bstituting in lhe k""",'Jl valu~ ~ sol"" . gettin8 ~ _ 1.44 mi.' , TI>e e<[uation lor fall from ... t is y 0& ~ .. I + \",.'. witb v., _ 0, Substi1uting in _, .. 1,44 mI.' Ind I _ 2 s. ,"," gel Y - Wm.

4.74 Repeal Prob . 4.73 il the coefficient of friction hel~~n Mod ao>d plane is 0.20.

I The ~Ultion for block A is no ... T - w, >in 15' - f _ ", .... TI>e bl«k B equarion is the sa"", .. helore : .... - T _ "' .... Adding the: 1"'0 e<[u.tions .... g. t this lime w. - "'. >in 25" - / .. ( .... + ",. )d. As 000" as we obtain / we can wiv. for •. To obtain / we lIOIe lhat / .. !"N. whe.e !'. _0.20 i. the coefficienl of kinelic friclion ao>d N is Ihe no<mal fOKe .. erted on the block by the incUne . Noting 1: F, _ 0 fOf the dir«tion perpendkular to the incline. we have .... CO!I2S' - N _ 0. or N _ 2(i6 N, TI>en/ _ H N and $01>1nl for lhe aa:cleration a " o.38ml. ' , Apin u$ing y _ v.,, + '.,.t' ";Ih IJ .. _ 0. a, - 0.38 m/s' . and, .. 2 s. we get y - \!.H.m,

4.7S In Fig. 4-22, Ihe two too..: .. ha"" identical m ..... , 40 kg. Both ,xperience a .tiding lriction forc:c w;,h 1' - 0.15. Find the a<:<:eleration of the boxe5 and tIM: teMion in tbe tie cord .

I. -,

66 a CHAPTER-4

I U';nll - "Y. ",IIt.o Y _ "".mal 1o"", . .... '. fiDd Iha' Ih. lrit,;on I",,,,,. 00 ,lit ''"''1 boxn • ..,

I. - (0. 1 ~ )o(O,87 "'& )

Bu. m - .l()kS and ~ IA - 59 N a"" I. - 51 N, Lei II> apply!: F. - ma. 'O .""" block in 'urn. 'aking l/I.e direction of moIion '" """I;"e,

T -59N_(40 kl)or ... 0.5"", - T - ~1 N - (4(! kg)oo

Solving 'h .... h'ffl.quaiioM fOl" ~ Ind T gives g _ l.08m/.' aDd T _ .lQil:!,

4." Two bod;"" 01 m . ..... m, and m" aro rtka""d from ,I>< """'ion u.own in Fig. 4.23("). [f ,I>< rna .. of ,/I.e >nIOOIh·\oppc<l,abk i. m,. find 'he ..,,,,,ion 01 'he floor on ,I>< 'able ",hile 'h.,....o bodies • •• in mOl:ion, A...,mc '"al,he ,.ble does nOI mOVC,

I F.om FIll. 4.2J( ~ ) , ,he I",,,,, equa'ions for ,I>< bodi •• a •• Body l : !:Fw, -"', - T - m,. Body 2: 1: fi.,. - T _ m,. Table: !:F ... _"' _ T_...,_",, _O. 1:1\..-T-I-O

, --",rio<.

- ~ T

- QI. ,

,. ) .... "'2)

" " . .., ~ I ,»

,.here N and 1 arc ,he v.nial ."" horizontal (frictional) component< of lbe: fnrce exen~ by lhe Ik:>or On ,I>< 'able:. [We aWJMC in FiS. ~ .2J(b) ,M,lolt."" ril"'lep u.a •• load equally. This don not aff«l OUr iUlalysis.[

FlO'" 'ho fin' , ... " C<jU'O';01 ....

aDd . finally .

"', m,g .--------m, +m, m,+ m,

"' ,m,g I - T - m,_- --­m, +m,

. ( "."') f.; _ T +m,g+m.,g _ ---+m, +m,8 "' ,+ m,

4.17 Th •• e identical t>k><:b. each of ma.. 0.6 kg •••• conn«lC<! by light strinp .. u.","'n in FiS. 4-24 ........ mc ,bal Il><y lie on a """"'-h . horizontal surface and a •• oWf"ftI '0 bay. an """"I.ra'ion 01 ~.O mI.' undoo. 'M ""ion of. fom: F. Caleul ... F and ,he two 'ensions.

, Let II>< 'eftSio:Jrn in ,he eo«! be T ... "" T ... • eopect;vely .• "" leI u. wrile N ...... on·.!oeCO"" law'", each block ... ""ralely. cllOO$ing """,i"" '0 'he righl 100eloCh, Since Ih. 00f<h a.e incx,.nsible. we know 11I.,IMy

NEWTON'S LAWS OF MOTION a 67

have t~"me aeceiclation .... 'him .... dellO'lC b~ Q. Then tOl' block,)..t , 8 . C. ICql«1ivdy (ltuing ... _ m._ m. _ m, ),

F-T .. _ma T .. - To. - "'" To. _ ",Q

(Not. how the lension, appear ... ith opposite signs in adjacenl .. qual ions.) T Q wl. 'e I I><,.. .qu.ti<)ns IOf ~ ",,'. add the equal;ons. ond t l>< lensiool~ c:aroo:el in pain.. lea,inS F - 3mo .. 3(0.6 kg)(4.0 mi.') .. ll.!!!. ~ t.nsions can "" ... be oblained by sub>lituling back inlo Ihe individual .qualio,,>:

T .. _ 7,lN - (0.6 kgX4 .(lml~') - !..H.b: T ... _ {0.6 kg)(4.0ml. ' ) _Ul!

4.73 Three bloxb wilh ma>>es 6 kg. 9 kg. and 10kg a •• comlCcle<l 0$ mo...'n in Fig. 4-25. 1"hc cod!lcient of friction be'''"en the table and Ihe It).kg block is (1,2, Find , a ) the oo:«:lera\ion of ' he sYSI.m and (hi Ihe teMions in th. cord on Ih. left arid in Ih. co.d on the righl

,

" • 0,1

, (al Leltl>< I.n""" in th. COld 00 li>c I. fl be T, and 00 lhe righl be T, . ~ pull.Y' a ..... umed 10 be lrictionlc... W. "PI'ly N .... too· .... cooo law t<> .""h 01 the , hr •• block •. (['oosing the po!.itive 5CMC of th .. ui, for net. block con.ist.ntly. Thy, .... dloo.c do"", wa.d a. posih"" fOI bloxk C. 10 the right pooitive tOl' block B, and up ... ard .. posiliv. Ia< block A. Th. lrielional force <>n block B;' 10 II>< l.tI and can be obtai""d from f - /J,N. wl><re /J. - (I, ZO and IIIe oonnal loltt N .quab ,I>< w.ight .... .. 98 N lrom .... nical C<!yilibrium. Th .. f - 19.6 N. FOf our Ih ... C<!ualiom; W( h."e

... ,- T, _ m," wl><re " i, th. """"Ie.alion. m, _ 9 kg. and "', _ 88, 2 N,

T,- T,-f - m.Q T,- w. - "'.Q

",'h.re m. - 6kg.nd ... - 53,8N. AI "';Ih earl~. proI>ltms involving coni. """""cting block •. tile tensions in adj"""nt C<juarion. appear with "I'P"'it. ';g"'. Addin! the Ih, .. C<!uations .Iim;nates t~ I.nsions complet.ly; "', - f - "' • • ( .... + "'. + ",. )d. Note that thi' is .quival~"t 10 a oroe-<limcnsionaJ problem inV(lhinl a oinale block 01 ...... "'. + m. + .... """'ed on by a foru w, to II>< ri,ht and fon:eo f and w. 10 tbe Irh. Sub.tituting ,1M: k_n ma$$CS ...... igh ... o.nd f gives Q - o.J9m/s' . (hi Su~titule Q hack inlO lhe equation. of motion 10, each bloek . ""'" conv.niently t~ fi .. , and third. 10 obtain To'" tll.t!, T, . 81b:, n.e remaining.quation ean be used to chcd tile re$lll,.,

• • 79 [n Fig. 4-26, th. roeffid"nt of ,I;ding frict ion bet .. , •• n block A and the ,abl.;" O.ZO. Alw. m~ - 25 kg. m . ... 15 t g. Ho* fac ... ill bl""k B drop ;n 11M: fi~ J. ahe, II>< SY'l<m i, '.I ..... d?

, , I -

r-' ' , • " .. "" f'II ... 16

68 a CHAPTER ..

, Since, for bIod A . ther. is no motion \'.rUr.ally. the nonnal fOT« is Y .. mAl " (25 kg)(9.8 m/s') _ 24S N. Thon! '"'" y _ (O.20)(24S N) _ 49 N .

w. must fint find 1M aocekratlon 0( tilt s)'Stem and then .... can desc"he ils motion. U:I us apply F .. "'" to eadl block in 'urn . Tlkin, Ihe motion dir«tion as positive . """ ha~e

T - 49 N .. (25 kS)or .. , " - T. ( 1~)(9.8) N .. (L~ kg)a

We tan eliminale T by addin8 the 1WO equalions. Thon, solvin, for Q , we find ~ .. 2. 4~ mI. '. Now we can work . moIi"" problem "";Ih ~ _ 2.45 mJs', v. - 0. 1- J •.

,-Vo/ . ! .. I'

as Ihe dist"""" 8 fans in 11>1' firsl 31.

J - 0 • 1(2.45 mll')(),)' - J.LQ.m

4" How la,!" I horizonlal fOftt in addition 10 T mUll 1"'11 on block A in Fi,. 01-2(; 10 pve illn otCCle,ation of o_n mI. ro"""d llIt~ ? As~m •. I. in Prob. 4_79, .hal ,, - 0.20, "'. " 2S k,. and m. _ 15 kl.

, If .... e were \0 remw fig. 4·26 in .his c.asc, ""e s.hoold show . fom: P pulling toward the kft on A. In addition , the ret",din, friction fora:! should be tevened in difCClion in 11>1' filllte. AI in Prob. 4,79./" 49 N.

W. w"l. F '"' mQ fOf each block in tum. taking tl>l' mocion direction 10 be posilive. W. h.ve

P - T - 49 N _ (25 kJ)(0.75 mi.') ... T - ( 1 ~){9_8) N _ ( 15 kg)(O_ ;~ mIl' )

Solv. 1M lall cquat;on fo. T.nd substitule in the previous equalion . W. can Ih.n solv. 10. tl>l' sinsle unknown. p • • nd find it 10 be ll6l!-

4.81 Th. 1WO blocks shown in Ft,. 01-27 have equalllllWCS. Tho rodficienlS of Slatic aoo d)'1l""ic friction .. e equal. 0_.10 fOf both bloth. If the .yslem is pven an initial .~d (I( O.90ml. 10 Ihe k ft , how far will it move before romin, to ... 1 if lhe inclines are quite long?

, Apin" lI$<ume that tl>l' pulley i, frictionless and the ,ension in the ror<! is 11>1' same everywhere . We apply NCW1Ofl', seeond I .... 10 each block to get rile otCClc,.tion of IIIe bIocu. W~ c/Ioose our ~ axis for .aclt blod II to the inclines and ChOO5C 11>1' posilive ",noc to lhe I.ft. Thon we hlv ... sin ~3" - I, - T .. "'" . nd T - .. sin Y!' - I. " ""'. whe,. '" is lhe common mISS and 101 '"' "'g. lhe rommon ... w." of blocu A and B. Tho frictional foroe. on 1M lwo blocu.", dctermi .... d from tl>l' equilibrium condirion. perpendicular \0 th< ;1I<'Ii"" •. TIl ... for block A, /. .. p,N .. with normal fOT« N, "'" 00$53"; and fOf block B. I. _".N" .... ilh N, _ .. _Y!'. W. eliminal. 11>1' I.n ...... by addill8 our two equ. liano 10 yield 101 . in 5)' - I. -!, - .. sin 30' -2m .. : or ~b!;rjrulinl for 101, /., and/. , "'I sinS]' - /'."" 00$5]' - "."'1 _)(l' - mg sin)O'_:zm... Dividinl our by m and soh'inl . ..... get ~ .. g(sin 53'- P ..... 5.)' - P. COl 30' - sin 'JI1')I2 _ - 0 69i 1111' . W. now apply lhe ki .... malicll equation V;" vi. + z. .... 10 .ilher block wilh 11 .. _0, 90 mil . or, " - O. 69( m/o'. and 11, _0 10 S'" It -I!...a:l.m .

•. 82 [f th< blocu in Fi,. 4-27 ar. momentarily .t rest. whal illhe smalle>1 coefficient 0( friction for wltich rhe blocks will temain ar rCSl1

, Tho lendency 10 motion will be 10 th< left sinee Ihal.1ope is lleeper. For minimum ooefficienl of friction the frictional fOtttS .... ill be rlleir maximum value (11)1' ~rgt' of 'lipping). Thu •. I, _ P,"'I COl ~3' and ! . " p,,,,, C05 30', boIh ading 10 the righl. Thon the .qualions (I( equilibrium.", Block A: "" sin 5]' - I',mg- 00$ 53' - T .. O. Block B: T - "" si n 30' - 1', 1118' 00$ 30' .. 0 Adding the !Wi> equat;on. yield5 ",,(.in 5)" - 1'. _53' - sin Y!' - 1'. C05 JO') _ 0. Dividi ng our mg and ,.arranging lerms. we gt'l /I,(COI ~J' +001 'JI1') - sin 5.)' - sin 30'. and jl, -2JW.

NEWTON'S LAWS OF MOTION a 69

• .83 A blimp is descending wi lh an ao:celeralioon ". How mudt h.allasl must be ;'uisioned for lhe blimp 10 rise wilh lhe s,ame ao:ce\er.'ioa d? There is. buoyant forCle acting upward on lhe blim p ... hich is equal 10 lhe Wfcighl of Ihe air displa~ by lhe l>Iimp; USume lhallhe buoyanl fon:e is the same in bolh casa.

I From Rg. 4-28, the equalions of motion lire ~nding: "',g - F, " "',". Ascending: F. - "'>6 " m,d Addinlli"es (m, - m"lg .. (m, + m,)oo. B UI m, - m, ""'. Ihe ma .. of lbe diKII,ded h.allasl. Thcrefore.

mg .. 1m, + (m, - m)I<> m "~~d)m ,

r

I. ( .. ) Dcoo<!nding (h) AO«IIdini

• .114" Show Ih.al ,he acceleralion of lhe Clenler of rna .. in ¥rob. 4.83 docs nol chanae ... ben ,he h.allllst is ej<:<:'ed. U .. litis fact 10 confirm the value of ", found in !'rob. 4.83.

, CbooM: "P as """live. Since r , on the: l)"Item of blimp and hallasl is the same before and after lhe h.al1asl was thrown OUI. r , .. "",,_. we mlm have ,,_ .. -". before and afi." . Now. mea ... red from IIOO1e re(ercoce level. T_" [m>y_ + (m, - m,lT_IIm,. Thcn .1'_ .. [m,.I'_ + (m , - rn, )y-..]fm ,. 8u\ y __ " . .1'_ .. - g. lind y_ .. ,,_ .. -Q : so - Q .. I"'," - (m , - m:n II"" 0' (m, + m,)oo .. (m, - m,la' a. before . yic:lding lhe same value fOf m " (m, - m,) .

• .15 Three blocks, of tna!Se!I 2 .0 .•. 0. and 6.0 kg, aNan~d in Ihe or<.kr 10 ... ". middle. and ul'J'C'. ,npecti'·ely. arc conn«lw by flrinp on /I frictionless i""lined plane of fIl'. A force of 120 N is applied upwa,d alon& lhe i""hne 10 Ihe uppermost block. ,," ... in, an upward """",men! of the: bIocu. The connttlinl cords are lighl. What is 1M ICCtleration of thc: bIocu?

I Thc ';Iuarion is depKted in R,. 4·29 wilh F .. 12ON.

m ," 2.0kg m , " 4.0 kg ,.,., Applying NeWlon'S wocond law 10 each block , we hay.,

F - T, - m",.';n fIl' .. m>" 1; - m,z";n fI1' .. "' ,Q

70 a CHAPTER 4

Addin& lbe .. cqua,ion<. F - 1m , + m, + m,J& >i n f:tO' .. 1m, + "" + "',)o:r: 1:!O N - (12.0 k&)(9.& m/.-)(O.866) ­(l2.0kJ)o; Q - LSI mls'.

4.86 R. f.r '0 ~ob. 4.~ . Wb~' lrc Ih. ,.n,ions bel",...,n Ih. upper and middle block •• and Ihe tow.r """ middle bIo<b?

, Conlinu;nK from Prob . 4.85. ",bol;IUI. Ih. val"" of Q inlo lhe indi.idual block cqualio .... and .01 •• for T, and T,.

Fur block I : ~ - (2.0 k,)(9,8 m/s' )(O.866) + (2.0 k&)( \.51 m/.') .. 2:O..!!.l:!! For blod 1; T, " I:!O N - (6.0kIK9.8 m/.'KO.8(6) - (6.0 kIK1.51 m/.') _~

Th i. can be che<:kcd by .ubslilUlinl imo 1M equalion for block 2.

4.81 A skit' V'" do .. '" a hillside . .. 'hich .... t •• a" angle 9 wilh rcspc<110 1M borironlaL. If 1'. is lbe coc:l!icicnl of sliding lric1ion bel .. '...,n skis and !lope. show Ihal Ihe !>C'ttic'81ion 01 1M skier is" - g(sin 9 - 1'. 005 9).

I R.ve"" ,be di,enion "r molK,n in P,ob . 4.23. and apply 10 ki ... tic ,.,Mr lhan ""tic friction.

4.88 Ref., 10 Fi/!.. 4-.lO. Find T, .nd T, if tM block. arc 10 acccleratt (If) upward al 6.0 m/.' and (6) downward ., 0.60 mis'

.... I T,ea~ng lite '''''0 ma ..... and Ih. ron"",,'ing ro,d .. an i",1".d objen. T, - 1 ,Os _ 1 ,Oa: then ,,,,I"'ina Ih. !(lG.g m .... obi.in T, - 0.8& _ 0.80 : " i, lite .. me in bOIh expres,ion •. '_I Q _ 6.0 m/,'. '" T, _ 6.0 + 9.8 .. J.lJ!.l!l: aod T, _~. , •• Q _ - O.lIOm/ . '. '" T, _ 9.8 - 0.6 - tll::!.nd 7; _ 7 4 . A. a ch.ck on lbe ans"'c". i",lale lh. 1OO-g rna .. and obo.e,,'e lhal T, - 0.28 - 7; - 0.20>.

4.89 "The ro,"" hnldinl! 'he ''''0 ma_ •• hov;n in Fil! . .t.-.lO " 'ill break il lite lension .xceeds 15,0 N. Wbat ;. 1M maximum up"'a,d accele. alion one can gi"e lbe m~. w;IbouI Ih. cord brelkins? Repel! il lbe ",englh is only7.0~,

I NOl •. T, > T, from P,ob. 4.88. Con,ide, the Iree body made up nl bolh m~.nd lbe m ... leso cord be,,,un ,It.em: T, - 1.0 ~ .. LOa: fn, T, " 15.0 N. Q -11.mLi: In. T, _ 7.0 N. a _ -2,8 mli'. (l1>< ,~'cm mUSI be actCicra!ing dCI,.'n"'ard. Ii""" T, could 1'001 • .,ppon ,he 9.S-N weigh!.)

4.'" A 6.(J..kl! bIoc:k res" on •• mooth frieli""I ... lable . A "ring .,:ached '" Ih. block p._. ove, I lriction~ !",Il_y, ... ! • ,n.kZ m",. h.nt' from , ... "rint •• d", .... i . Fi!. ~·JL (dl Whol i. , ... """""' ... ,inn _ , Cb. Wbal is 1"( I~n""" Tin Ihc Olring1

, Cd ) This l~'pe of p,oblem . • , ... n in Prob . ~ . 78 . can be "en.d" if il ..... 'e in one dimenoion . Thus, Ne"'lon'lotrond la'" takes lbe lo,m

F - ma m,g _ (m, + ... ,)0:> l(9.8) .. (6 + 3)0:> 29,4 .. 9<1 If .. 3.27 mIs'

Cb) Applyinl N,,"tnn'$ se<:ond law I" malS m, .10 ....

T .. m ,a" 6(3.27)-i2&.M

NEWTON'S LAWS OF MOTION D 71

'kg

Fic·4-J1

To check Ihii. apply N.wlon ·, scrond I . ... to m, aJonr .

3(9.8) - T _ .1(3.27) 29. <f - T - 9.8

4.'1 A 6.0.1:, block ...,Sll on a bntil""t.1 surface . Its ~ffk>rnl of killCtk friction is 0.22. The block ii ronnCClril by a ~'in, passing 0 ... ,. pulley to a l .o.kg m . .... as in Fi,. 4-32. f.) What ii I .... accrlcratioo a? (b) What ii the lC ... ;O' Tin lhe ~ring?

, , ~

m, T

j.I,N

, Wc IrQI I .... prublcm as if il .. ·c •• in 01lC dimension. (.) For the _plom a. a .. ·bok.

--' j'

m,

m,g

IICt F _"", m_m , + /n, _ 6.0+ 3 0 _ 9.0kg n"tF _ m,g _ I', N _ nur N_m ,g "'" _ m,g - 1',m,8 9.0. _ 3 0(9.8) - 0.22(6.0)(9.8) - 1.68(9.8) Q _ 1.113 mi l'

(b) ll>c fo~ Ih. string • •• m on m, i,

T - I' ,N + m ,d _ 0.22(6.0)(9.8) + 6.0(1.83) _ 12.~ + ]/198

4.t1 SUP!""'" that blocks A and B h.,·. rna","", of 2 and 6 kg. rcspccth'ely, and a..., in C"Ofltact on a '""'lOI.h ""rirontal surf."". If • horizontal lOT"" 01 Ii N pushts t .... m. calculat. (.) the aca:\c,ation of the systcm and fb) I .... fo~ that ,"" 2·kg block ex",,, on , .... 0 ' .... ' block.

, (.) Se. Fig. 4·33(.). Considcrins the blocks 10 roo.-. as a unit . .w _ m. + "'. _ 8 kg. F _ Md _ 6N • _ O. 75 rnfJ'.

, .. , cd , ~I , ~ - " ,., ,M nc·4.33

72 a CHAPTER 4

(b) 1I .... e now ronsider block 8 to be our syste m, the only lora: actin, on it is the fora: d ... 10 blo<:k A, F.,.. Then sina: tIM: acceleralion is lhe same as in ~rI (_I , ~ h .... F .. - M.~ - til!:. However. we muSI consider also the case: in which "'e ",ve..., blo<:ks A and 8 u in Fi,. 4-3J(b). A, belore. COfIsiderin, the blo<:ks .. I unit"'" have d _ (/.75 mi.'. No ..... bo .. ·.v.'. if "'" conside, block 8 a, ou, s)'Stem we have lwo fora:1 actin" the force 1'10 Ih. righllnd lbe fora: F .. to the 1"'1. Then F - F.,. _ M.d and 50Iving ....... 1 F .. - LiI::i in magnilude and point. 10 lhe I.ft,

W. could verify theSt ..,sults by consid.,ing block A to be: the .)'Stem for the two C3sn .

• . 93 In FiB. 4-34. the pulley is JlsulU(d mu<I.", and frictionlns. find the acceleration of the ...... m in term. of F if there 1$ no friction betv.'een the surfa« and m. Repeat if Ihe triClional force on m is f.

i or!--

, Note T _ 1'12. Newton', law fo, the block gi"". T _ """ , hen.ce Q _ F( 2m. WIlen tric!ion is in "olved "'e have F/2 - f .. """, 50 Q _ (Fi lm) - U 1m),

4.94 In H,. 4-35 ..... ume that th.,e 1$ roegligible friClion between tbe blocb and tabl< . Compute lhe t.nsion in the cord Ind the aculerllion 01 m, if ",,_300,. m, - 2008, and F _ (/AON.

",.4-15

, Writ. F .. "'" for ead! block . let Q be: the aca:1< "'tion of m,. The ,,""le ration of m, i l then GI2 (compare Prob. 4.72). Then T _ m,G and F - 2T _ m,(G I2). 0... finds" - Q,U mi.' and T - !lJ&.H.

4,95 How large mU$' F be in FiS _ 4-36 to aj ... til< 700-8 block I n ,..,..,lero' ion of )(Iem/. '? Th. coc:lIiclc:n, of friClion between the two blocks and abo bet"' •• n block and labl. is 0. ]50.

"', 4-36

, Coov.rtin8 to S] "nit •. iooIate each mlM and note tbe '<>re<', that .ct on each . Vertically only the weight. and normal forces are invol .. ed. FN _ 0.28 on tM "I'I"'r block and O.~ on the low., 0""', Friction fa,ceo arc : ,,(0.28) bet .... een blocH and ,,(0.9g) ot the table . F - "'" lor the blo<:h: T - ,,(0.21) _ o. z" I nd F _ T ­,,(0.28) - ,,(0. 9g ) - 0.. 7G : altcr eliminatinl '~II$ion T bet_en the .. , F _ 0.9.1 + ,,( 1.3,) - 0.9(0 . .10) + 0.150(1.3)(9.8) - UU!.

4,96 Assume in Fig. 4-36 that the coefficient of lriction G the lame at the top and bottom of Ihe 700-8 block. [f G - 10 an/ s' .... hen F _ 1..10 N, bow larce is the: coefficient 01 friction?

, Followina Prob , 4.9S. F .. 0.9.1 + ,,(1.3,) ; tlw:n we ~ .. F _ 1..10 N and G _ 0.700 mis' '" find" _~ .

• ,,", [n fi8. 4-37. "'hen m is 3.0 k, . the """,[nation 01 tile blo<:k m is 0,6 mI.' . while d _ 1.6 mIs' if m _ 4.0 q, find the frictional forre on block M as well .. ito mau. Neglect the mass and triClian of the pulle~

, Apply NCWlon'. second law", m for each CaSt: 3(9,8) - 2T, _ 3(0.6). ~ T, - 13.8N; and 4(9.8) - 27; -4(1.6). 50 T, - 16.4 N. Applying Newton'S second low to M for each case . 13.8 - f _ M(U) and 16.4 - f ­M(3.2). Sol"" 10 find M - !2!I""" f"~,

NEWTON'S LAWS OF MOTION a 73

•

• •• In fI,. 4-38(a). block 1 ;. one'(OIInh tht leBa!h of block 2 and \IIeip.s "",,·(oonh IS much. Mume lila! !he", is no friction ~I~en block 2 and lhe surface 011 which il moves and thai the ooe~nt of sliding friction between blocks 1 and 1 is " , • 0.2. After \be: .)'Stem is "'Ie....,d . find lbe dislanoe block 2 hI> mO>'ed when OIIly oDC·founh of block I is Mill on block 2. Block I and block 3 have lbe .. me m .....

I From Fig. 4.38(b) . \be: equations nf mntion a,e

L Ii · T-", .. ,,, ,,,,,, Solve Ute first and third equation!. pmultan..,...ly to get a, " (&12)(1 - ".j: from the 5econd equation . d," (&14)", . 'O>eB lbe <fupI_ments of block, I and 2 arc gi""n by x .. ' ''''. i .• .•

At the insunlthat one-founh of block I ",mains 011 block 2. %, + t - %, + (//16). wh.", I is Ihe length of block 2. The,efor • •

,

•

., f'~~ ~ ... T

, I

(a ) Confi,u r.,ion al • _ 0

(0)

I .,'

.. , ., -

R,. S""

~ "

1 • 'I

,, -..... , , c!.JL.,

•. " A dinnoe, plale , ests 011 • "bleclolh. "';11\ il5 « me, 0.3 m from \be: edge of tbe "ble. TIle labLccJo\b i. suddenly yanked horizonlally "';th • constant acceleration of 9.2 mls' IFI,. 4-39(Q )1. The ooellicienl of .Udin, friction ~\ ... e.n \be: tablecloth and the plate is " . _ 0.75 . Find (_) Ibe """"lerltion. (6) lbe velocity . and (~) \be: di.unce of Ihe p1Qre from tbe edge of tbe table . when lbe edge of !be tablecJo\b passes undc, tbe emte. of lbe plate . Assume that lbe tablecloth jw;tlits th. tabletop.

74 a CHAPTER 4

... ~ .. , PI ... C====;N~C=:===~~~

I ~ .. . N ~ ..... ,

(.) (.)

, ' _ I fl'Om fiS· 4-)9'(b) , the fum: equation for the pll te is P"'I - ma" or a, - 1'1 - (O.7S)(9.1I) _ 7. 3S mf.'. ~ pll te .ti",. since a, is less Ihan 9.2 m/1'. (' ) AI lhe .ime lhe «IS" of lhe tabledo!h is II lhe cenl~. of I"" plale, II>< <iOlh and lhe plate are alII>< Arne disl"fIOe from II>< edge of Ihe lable :

x, _ .t, 0.3 + 1(7.4)1" _0 + 1(9.2)1'

Solving. r - O.SS o and u, _ 0 + (7.35 )(0.58) - ilim/.!.

(r) ~, .. 0.3 +0(0.58) + 1(7.)5)(0.58)'. Ui.m

4. 110 In lhe I"'lIey 'y.tem ohown in Fig. 4-40 , Ihe nwvablt: pulleys A, B, C are of mau I kg each. D aJ>(! £ I , e fuced pullcY" . The " rinp are .ertical l nd inalen';ble. Find II>< Ie"""" in the SIring and the accelc'"lions of tht- 'mionlcS!> pulleys.

, Write , ,., Y •• Ye for Ihe positions ,,' .he cente n <of lhe I"'llcys A . B . C II lime t: a •• a • • a(" are lhe aca:leflllions at time t.

~'" " ~ """" i ... i ~ ~;o"

-.~ y y !. , , l' A "

, , -, , ~ , • ,

""",. ;c", "' ' i ..... ,

, f}!, , , C

-, .... ....

NEWTON'S LAWS OF MOTION a 7S

Following the I1rinl from the end.t lhe ~nte. of A 10 the end It lhe «ntu of 8,"e gc'1

<1. - y~).;. Y.';' 2y.';' h';' lYe - Y.) - ro.mant

Take the ....:ond time,derivaliw of this equation to get a • .;."_.;. U c _0, Thete i$ iu.1 one SIring . nd, thu., one len.;oo T. The fo.~ equations are

T.;."" - 2T - ""'. T.;. m, -2T _ ""'. "'1 - 2T - ""'~ Substituting'" - I k,.nd $OIvin. the four equations fot lhe fout unknowlH d •. a •• dr., T. _ obtain

UQl' A body of m&$$ 400 kl ;1 suspoen<:kd .t the lower end of a lilht vt'n ical chain and ;1 being pulled up venially (ou Fil' 441), Inilially lhe body is 81 re" and tbe pull on the dlain i. 6000g N. The pullge" smal~. unifonnly .tlhe ralc of 360&' N ~. each mete. throop whidlthe body i • • aiKd . What i$ the velocity of lhe body .. hen it has been .aiKd 10 m1

, AII;me t. lei Y be the heilhl (in meteB) of lhe body above its initi. 1 position . "The pull in 1M chain i$ then T _ (6OC.Xl - l6Oylg and Newton'. occond I.", liv","

Thi$ equatiOtl may be chan,ed into one for j _ v (Ibe ,..I<>city of 1M body) by...., of the identity

.Iv tludy .Iv 2; .. 2- _ 2- - . 2v - _ d(v' )/dy tis d.Y d, tty

2OOd(v') _ (~ _ l6Oy. 01 d (u ') _,(2II _ i.ly)dy

" Let V he the vt'lociIY ' 1 heiJht 10 m, Then, on integraM,

v' .. , ll8y - 0 9y'~' _ 6128(10) - 0.9(100)]- l!i08

V - .;. v'19Oi" " +43.2 mls. The choiot of the .;. si", to< V (upward mOlion) ."""ld be checked , For O s ~ s 10, the .... , foroe ,

(S600 - 360yIl, is pusitiw:, . nd 10 1M aocele.ation i. l'O"itive, Then, since the body " . ned from I'CSI. V mull be positiw:.

CHAPTER 5

III Motion in a Plane

5.1 PROJECTILE MOTION

S.1 A m.rbl~ " 'jlh ~d 2Oan/ . fQlb Qft' the...t", of a lablt I!Orm high. flow kmg does it take to drop 10 the 1\oot1 How fa. , hQrimntally. from lhe IOble «18<' does "'" marble OInk. <be I\oor'!

, Cl!<.>oM oo..n,,-ard as posili~. with ori&in It edge of table lop.

u",-" o-20cm/. ",, _0 G, -+S'-+9IIOcmJ.' ",_0 To find t ime of fal l. , .. ~ .. I + I,. ' . or 80 em .. Q + (49(lcm/.'),' ; ! .. 0. 49 .. Tho: horium •• ' di~l.ntt i~ IIO'I.n from .r .. t'", r .. (20cm/.)(O.4().j -Itl!£m.

5.2 How fasl mu~. Nil be rolkd along a 7(I.an-high table 10 Ihal when il rolk off the edge it will strike the: Il00. at this ... me diS1a3Ce (rocm ) from the point diJtttly below lhe table edge?

, In the 1>0';1.0"'31 problem, X" U,I gi".' U, " 0. mil. In the 'enicai problem. <hoosin~ do.'/I as pooilive. It . .. O. y " 0.7 m, and Q .. 9.8 m/. ' , U.., th,"", ,·. lu~ in}' _V. I + ar'/2 10 give r .. 0.3730. Then u, " l.85ml •.

5.3 A ma.ble (r.""ling at 100 emls Tolls oil II>< tdge of • le,d table . If it hit< II>< ftOQ' 30 em away fl'Qlll the spot directly bel"", Ih~ «Igc of Ih~ tabk. h()1ol hi,h is lbe t.tle~

, Th;s is .l"oject;l~ problem " i,h .... - lOO"m/ •. Fo-r ,be />orin)nt.l mohon:

f . _ .... ' , - 1),305

For lbe v~nical moI;on :

" .. ~ .. ' + \~I' - I) + \ (980)(1),30)' - iLl..m!

!." In an of(linary t~le,ision ~I. Ihe elttlron !>tam ""Mi", of eleclroll'l 'hot h""ironlally l ithe lelevioion screen "";Ih • 5p«d of ab(>u, S" 1(/' mi •. How f •• do<-$. Iypical eleCfroo faU a. il """"'. 1M .""ro~imllely o\IOcm I,,,,,, tI." el<-ct,OfI 8~n to the 'o<"I"e.n1 For corn!"', i..,.. . """" ,., ,...,,,W. "roplet of ... ,"' ,h<>t ",,",on •• lly.t 2 m/! I.om a hose dfOp a. il mo:w<:. a horiwmaJ di>lanc= of 4Ocm?

, FOI 1M .len,oo, tM horizo"'al problem );'1<1> ,ho rime 10 hit tM SC'.«n :u . .. z lv, - (I.40/(S" 10') .. S)( 10 's, "Then in th~ ~eniCliI pmblem.), - "<> ' + Itt'n, SO Y - 0 + 4.9(64)( 10- ") _ 3 I" lO-" m, For . dropict ,'" 0.40/2" 0.20., and SO y .. ~.

50! A booy p'o;«-ted up...-a. d from 'M Ind "",,00 at all angle of 50" with tM horizootall>as In initial spe«l of 40 mil, How long will it b<: befOf~ il hils .he g'oond~

, .

.1&.5-1

, ChOO6C ~p_rd., pu>itiv •. and pl= 'he origi n at Ihe launch point (Fi,. 5-1) .

" .. .. ". cos 50" _ (40 mf>l(0642) .. 15.7 mi. II .. .. v. sin 50" .. (40 m/ s)(O. 7(6) _ 30.6 mi . u. __ g _ _ 9,Sm/s' a. - I)

To find the Ii ..... in ai •. """ ha"e )' .. "",1 -ls" and ~~)' _ U at th~ ~nd of nigili. 0 .. (30.6m/ .jl ­(4.9 m/. ' )t ' . or ".IIl' - .10.01. "The lim ",Iulion t .. 0 CtlfTupon<l> to tbe Itartio, point. )' .. 0, 1M oerond MlIulion i, I'lOl uro and;" obtained by div~ing OUI by I, 4.9 •• 30.6 and ... il:!..!.

~.6 In PfOb . S.5. bow fa. f.um tbe .. aning pOinl .. ill ibe body hiltbc gfOUnd. and a, ",'ho, ~ngl" wilh ,be horizon;.I?

76

MOTION IN A PtANE I a 77

I TIle horiwmal disl.n<.: lra,..,ltd , Or range R, is obui ... d from R _ v .. ' _ (25. 7 on/,)(6.24 . ) _ !!Qm. By ' ymmeny il OIri k ... al 5Il" (wilh ... gali,..,.r axis). Thi, e~" al"" be _ n by nolin! that v, _ - v.,.. II. - v .. , SO tan 1/ _ - Ian O. aOO lherof()fe 180" _ 0 _ 0,.

5.1 A. body is proj«t«i downward al an ansle of 3(1' ",,'ilh the horizontal from Ihe lOp of • building 17(1 m high . Its in;tial spe<:d .. 40 mi •. How long will il lake before slrikina the ground?

" . I

, , , , • ,

I ~ dO"'n"""d as poiiliv. and omgin allhe 101' .<lge <If building (Mg. 5·2).

II .. _ II. (:0$ 3(1' - (40 on/ 'WO.AA6) • 346 mi. "., • v. sin 3(1' • (40 m/.)(O,500) _ 20,0 mi.

We can ~ve IOf rhe time in dilJerent way •. Melt.od I:

Y-I' .. ,+ !8'" or J70 m - (W.Oml.)r+(4.9 m/,'J"

We tan sol,.., lbe quMiwit to yitld

- W±(JOO-+J312)" ,' . I · _~ , .. (Wt k«p ooly the posihe solulion; lhe negative ti me rorre'pond. 10 a lim. before' _ 0 " 'hen il ,,'o,dd ,,", .• been al grouOO level if il "'er. a projectile launched so ." 10 reach Ihe staning position and "docily al , _ 0.) MClhnd 2: We avoid the quadratic, First filKl II, jU$t before im pact:

II; _ 1I.l,. -+ 4)' or v: _ (W.O mi.)' + 2(9.8 ml.')( I ro m) ", • ±61 on/s

For oor ca~ v. _ 61 m/,. Next ·,.-t find.:

",.,· .. +gl 61 mI. - 2O.0m/s + (9.Sm/. ')r

5.8 In Prob. S.7. fiOO out how far from Ihe fOOl of th. building the body will Slrike and al whal angle ,,'ith lh. ""rirontat

I .r . v .. ' • (34 .6 ml.)(4.2 .j • H1.m. We n ... <1 Ihe angle ,h.l lhe "«lor "elocily make . ... i lh Ihe .r a, is jU.1 before hining the grooOO. We ""oid having 10 direClly d«i= Ih~ qwtdranlthis an31e is in by solvi n~ for lhe acute anJI. 9 madt " 'ilh Ih~.r nil (posili,'e or n~gal;"~: abm'~ or bel",,') .

tan /:I -I ~J" 1.76 '"' 0_60,4'

Sin<.: II, is n~pl i"~ ond II, i, posiriw. ' hi!;' clearly th~ an31~ belo .... 'he posi,ive -' axi~ ( .... Mg. 5·2) and eq ual. tbe .ns 1e we a re looting 101'.

5 .9 A. body it projeCled f.om ,1>0 ~und al an angle o f 30" ... ',h ,iIe horiZOfllal ., an in;t;al .peed of 128 fl / •. Ignoring air frielion, dclc rmi ... (., in I>0'OI many seron<k il ,,'ill st rike Ihe ,"ouOO. (b) how high i, .... ill ~. aOO (r) ,,'hal ils ronge will be.

I (.) Find "~ and Ihen d" .. mi ... Ihe time for a I. eely falling bod)' p. ojee,cd up"'ard ., 'hi, "clorily 10 return to Ih. ground : Ihal is .• , "0, From Fig . Sol.

v'" _ v sin fJ . 128sin 3(1' _ 6oIft l . S, _ 1',, ' + 1st' fbI Si~ lime of a~~1 i~ ~q ual [0 [imr of d~\Ctm . [he prtljeaiic [ClIChe!. mUi mum llcighl H at /-Z s, Thus,

H .. " 0, r -+ !gt' - 64(2) + 1(-32)(2)' - !i::!l!

... JPYflghted IT ria

78 J CHAPTER 5

,

fla,5-3

It) The projeClile ua,-e!s with a COnstant v.locity It, _ It ros 9 in 'he z dir«t;on. II rumts 'he ground in 4 •. The range is

S, .. v,/ .. (It ros 9)/ _ 128(ros Ja')4 _ 512(0.866) _ 443 fI

S.IO A hooe lying Qol , .... ground >1I00I. a stream 01 .... Ier upw.,d a' an angle of 40" '0 , .... horizonlal. The >pe<:d 01 the w.ter i, 2Om/. Il$ it leave. , .... hooe. How high up will it mike a ... all .... hieh i. 8 m away?

W.ll

....... , Selling coordina\e~ as shown in Fig. 5·4. with v. " 2Om/. and 9. _ 40', we ge'

v .. _ " . ros 9._ (20 mMros 40' '' IS3 mi. It .. _ It. ,in 9. - (20 m/,) sin 40' - 12.8 mI.

~ .. It,.', aDd "'''ing .r .. S m we find the time to hit , .... wall : 8 m .. (lS.3 m/s)l, , .. !ding I .. 0.52 •. To find ,he .... ighl at ...-hid! i1 hits t .... wall, W(: use 1 - It .. ' - !gl' . wilh I" o.S2 •. This yield. 1 _ (12. 8 m/s)(0.S2 $) _ (4.9 m!.')(O.Sh)' - Ul.!!:!.

5.11 A NseNIl Nue, hits a home: run Nil wilh a velocily 01 132 fl /. a1 an angle of U' above 'he boriwrnaL A fielde, woo has a reaoh 01 711 above I .... ground ;. backed up against ,he bic:acl>c:r woll. whieh is 386 ft lrom home pla,e . The NII .... M 311 above Ihe ! .ound "'hen hit. Ho lO' high above Ihe/\elder', ,I""e does 1he ball pa~?

, ,

'. ,,~- - -- - - - - - - - - - --$

- J II ' __________ --;"'::;:; - (,roo"" fIs. 5-5

, The silualion is depjcled in Fig. S.S, wilh Ihe origin aDd z axi> 3 ft above tl>c: ground. W. must find I .... value 01 r on Ihe N U's trajectory corresporwling 10 .r - 386 It. Then W(: can •• blrKI Ihe 1>c:w.1 of lhe /\elder'. glove above tl>c: z axis , i. • .• 7 fI - 3 11 _ 4 ft. To find r we not. thai

v .. .. V. CO\ 0. .. (132 f1/il COS 26' '" 1\11 ft /. v.,. => v. lin 9 . .. (\32 Itlli) sin ')1,' .. 57.9 Itls

"The time 10 . eaoh .r - 386 ft i$ given by z .. ~",I, or J86 ft .. (1 19 ft{1)r, and I _ 3.24 s.

MOnON IN A PLAN E I a 79

Then y .. V.,I - igl' .. (57.9 ft /I j().24 I) - \(32 ft /,'j().24 .)' .. 19.6 ft . Height above ~ .. 19.6 ft - 4 ft " ll.&1I. [Note: The trajectory equation y" (tan e.).o - rx '/(2v~ cos' e. ) <;:an al50 be ~ 10 ~nd , .1

S.U A ball is Ibrown upward al an angle of)(l' 10 Ihe I>orizontal and laDds on 1M lop edge of a building Ihat is 20 m . .... y. The lop .dg. is 5 III above tbe Ihrowing point . How fa>! WlIi tbe ball thrown'

'.

~201m_ ",. ,.

, The situation il depicted in FiS. 5-0 "';Ih e, _ 30". We can u$C tM Irajectory equa!;"" y _ Ian 90% - 1%'1 (20icos' e.), $Ctting % - 20m and, _ 5 m. Then S m" (O.58)(20m) - (9.8 m/.')(20 m)'/(20: xO. n). Or v. - 2.l!..mb. (I"ote: If,..,.. didn"1 'emember 1M trajectory equat;"". )'0\1 rould mil 501v<: the problem by llsinglhe x· .... .., and y ..... .., equatioM.)

S.13 A projectilc: is fired with inilio1""locity v. " 9!i mI. al an anglc: fI - Y:I. After S. il stri kes the!op of a hill. Wbat ill lhe .Ievalion of lbe htl abo>'e lhe poinl of firing? AI wh al horizontal di M •• ", from lhe gun does lhe projMile land?

1M

FiC. $..1

, ... 11. snuallon 1$ as sh""' .... ,n 1'". '·7. V. " ~~ mi. ; 1I. "::1J". Al any lime I, Y. V.,I - Igt' . ,,·he •• v., • v. sin fl. " n.s m/s. For'" 5 .. .... g.1 , " (n .s m/I)(5 .0 s) - !(9.8 m/I')(5.0 I)' "lli.m. The horizonlal dj,tance. %. is giv.n by %" v .. ,. v.cos floI " (61.1 m/s)r. AI 1.5 s ...... ha •• %" X!.1.m.

5. ' . A ball is Ihr1)Wn with •• peed.f llIm/. al an angle of 37' above the horizonlal It lando on lhe roof of a building at. point dilplaccd 24 m I>oriwntally from lhe Ihr\W'ing point. How hlsh abo,·. th. Ihrowing poinl ill Ih. roof!

, The velocily rom...,nen!s art v .. - 16 mls and v., - 12 m/s. In th. bonzonl£1 problem. I " xlv, - 24/ 16 -1.5 s. Then in Ih. 'enita! problem "'~ ]>a"" y " u.,.' +~,'t2. iO, -12(1.5) - 4.9(2.25). 7.0 m.

5.1.5 A hil b,...,balilcaves Ih. bal ,.jlb a ,·.Iocity 0( I IOft/s al 45' above 1M horizonlal. "The ball hiu lhe lop of . screen al lhe 320 .. 11 mark and bounce. in!o !he crowd for a boome run. How hi"", abo.~ the " 0.,,><1 i$!he lop of the SCT«n? (Neglect air resinance .)

' J, "' V,I u,"1I0ros4S'~ lconuanl 320_IHI(0.707)1 ,.·tl lI

J , .. v.,.t+ !<II' _ (1l0 sin 45")(4.11) .. !( - 32)(4. II )' _ ( 110)(0.707)(4.11) + I( - 32)( 4.1l)'

_ 319.6- 270.3 J, . 49ft (heish! of SCT« n)

5 •• ' A ball is th rown upwa.d from. point on "'" sido of a hill which . Iopes upward uni formly " an angle of 28' Initial veloci,y of ball: 0. _ 33 mh. al I n angle 9. _ (6' ("';th respe<110 Ihe bonzonal). AI what dislance up 1M slope does the ball slrike aDd in whaltime?

, Th. siluation is dopi<ted in Fig. 5-3. V. " 33 m/ • • nd 9 ... 65'. The lfajc:ctO<y equal ion of lhe ball is

Y . .. (Ian 9.).0 - rx'12v~ cos' 9. ) .. 2. 14.< - 0.0l5.'

80 a CHAPTER 5

.... w The equation for the in<line i> y. _ (tan 28"»' _ 0.5.lr, At the ,'aluc of x for which the bal1lt.ilS the incline y. _ y" o r O,~.lr _2.1 ..... - O.02s..- ' . and O.Ols..- ' .. 1.6u. wh;m yield!. i - 64.4 m. ~ di<la""", alort,lhe incline. S . • )'$ x " .~ COl 28" or S - 1L!.m, 'J"he lime 10 reacb any x value;' givcn by % _ " .. 1 _ (!!" COS 9. ), .. 13.91. So roo- % - 64.' m. , -~.

5.17 A projectile io to be.1Io1 al »m!1 <m:r le.el ground in >u<h a ""'y that i. will land 200 m from'he ",,,,,tin, point. AI ... hal "nsle Shoukllhe proio«.;ie bc $l\Qf'

I In the horirontal problem. X" ~ .. I givcs 200 _ (»=9)1, wl>cre 6 io lhe anile wc ....,k, In the .enicol problem. y _ ' .... 1 + "" 12 gi'"" 0 '" »Ioin 8 - 4.91. But 1 _ 2001(50 COil 8) , and 10 50 sin 8 _ 4.9(4/cos II). -n.;s Ioimplifi<s to 2 sin 8c .... 11 _ 0.7,,",. and 10 .in 28 _ 0, 784, from which 8 -ll.J:.

S. f8 A. ","",,'n in Fig. ~.\I .• b.all is Ihf1)'a-'n from thc top of onc buildin, tOW<l,d 11111 bttiklin, 50 It ''''''y. lbc initial "ek-.;ity of the ball is 2Oh!. at 40' aoo.e the ItoorirontaJ, lIow f., abo',e or ""low it. oriJinalltnl ";lI lhe ball <lrike ' '''' ol'l'<"$itc ",.II?

• " .. _ (20 fl /$) COil 40' _ 15,) fl i . OJ .... (20 h (1) loin -'10' - 12.9 his

In the horilOftlal nl(l(ion. " .. _ ",. '" Ii, _ IS, J hi •. lbcn x _ fI,l Ii"cs 50 ft .. (1 5.3 h!.jI. or , - 3 n $. In tlte ,,,nical moIion , t.kinB do .. ." as positi"e.

y " "".' + jQ," - ( - 1 1'~h/l)(3,171) + 1(32.2 h /s' )() .n.)' .. 130h belo ...

5.19 '" ) Find lite range.t of a gun -..'hich ~ru a shell ";lh muzzle "ck>cil)' U at an angle 01 elev.tion II. (It) rind lhe angle of ele,'atioo 801 a gun ... 'hich firfi. wll .. ith • muzzle veloci'y of 1.2 km!1 at • "'Iel on ' he ,..me lenl bu, IS km dis'an'. Sec Fig. >10,

'.

~,~ .• - ! ric. 5-10

, I_I Le, I be lhe time it takc. the >hell to hit the targel. lbcn, % _ OJ .. ' or , _ % / u"" Coru.idcr thc vcrticll """,,ioo alone. and take up ., po;i'i"e, When tlte shell s'ri h $ the la'lI<'l. ~cni<al <Ii$j>lKtmcnl _ 0 _ " .. ' + I( - & )r' Solving ,hi. equa'ion gi.e. , _ 2",.lr, Bu, I '" % It,,,,, 10

X 2v ... - 0 -'. ,

"

MOTION IN A PLANE I a 81

(6') From lhe range equalion found in (a).

sin 28 .. g~ _ (9_8 ~ 10- ' km/ s' )(l5 kml_ 0.102 ~~ (l.2km/.)'

whence 29 - H" and 9 - ~,

S.zt A rifle buHellw I mu.zzle velocily of 680ft/s, (a ) AI whal angle (ignorinS ai r res.iolana:) ohould the riOe N poinled to sive the muimum range:? (II, Evaluale the maximum rangc .

, (.) By Prob. S, 19(a ). the nnS" is • maximum when sin 211 _ l. Or II _ 45· .

~~ (68Oft/ s)' (6 ) ~-' -8 - 32ft/I' - 14 4SOft_

5.21 A COIf b&U In"es the golf club at an anglt o( fH above Ihe IIorizonlal ... ith a "clocily of JO mil , (a) HolY hilh does it 101 (6) A...,ming Ile",,1 lairway, detcrmine how far aWly il hitllhe """n<I .

, (a) Wilh lip poiIilivc . v~ - )I) sin (;1.1' '"' 150 mi. In<I . 11 maximum bcight, v, equal! uro, Thu~,

0_6H _ I9.tih '" .. --~ 19_6

(II, By Prob. 5,I9(a).

5.22 Pro,.., that "Iun -...ill sbooI thr~ ti ...... as high when i!5 angk of elevalion i, fH as when it il JO". but will <any !lit same 1Io01Om.1 dista~e_

, We assume Ihat". is IIIe same . t bo1h an&les. Maximum height is given by the rondition " . _0, 'Then. ,,; -,,~ - 2gy y;elds y_ .. ,,~/(:z,), Noting tbat " 0,- _ .... ":n 9 •. "'e bave ~nally y_. - (v: ,i.' 8)/21_ 'Then )_(6O')ly_(:IO') - ":n 60'1":0' JO" _ 3, To show that the iKH'iZ01ltal range: iI the .a.me. " 'e noIe Ihat in lhe range: formul. IProb. 5_19(a Ji COS 60' lin fH _ COS JO" sin JO' since 30' and 60' are complementary an&les.

5.13 A. shown in Fil. S- II. a projectile is fired "'itk a horizontal ,..,Iocily of 330 m/s from the lOp of I dill' 80 m hip. (. ) How 1on8 will it take 101" the projcaile to llrike Ihe Itvd smund at the ba>c of the cHfr! (b) H"",' fa, from the fOOl. of the dill: ,,~II it 'lri k~' (e) Wilh what ~elocily will il llri ~e?

rF'·:;:::::· i'~ • •• L~ __ "","-

I • .,.. !-ll

, C.) TIle horizontal.n<I vertical motiono arc independent of cach otltcr . Consider ~"t llle ~eni<:al motion . Takina do ..... III positive "'e ""ve y - " .,.r + 1_,/'. or 80 m .. 0 + 1(9.8 m/ l')I'. from ... hid! I '"' H~ s. Note that !he initial ""loci!), bad zero ~enical a::lI>Ij)Onent . and so "0_ 0 in the ""niralll'lOlion. (b) Now <><>rWde r !he IIorironUI motion. For it , a _ 0 an<l SO 11, _ .... _ ~" .. 3JO mI •. TIle", ~";ng the val.,. of I loon<l in 1_). :t .. fJ,1 - (3)0 01/.)(4.(1"') -1ll9.m. Ie) The final velocily has I borizontal component of 330m! •. Bul its ""nical velocily al t _ 4.{14 . is siven by t·", " v.,. + ~, r ... v'" _ 0+ (9.8 m/.')(4_04.) .. -10m/I. TIle I"<:SIIll.IInt of Ihe", two components i. labekd It in Fig. 5-11; w~ have

u - V(4JJ m/Il' + (3)0 mI.)' -lli..m.l1

Angle 9 mown is si~en by tan 11 - 4JJ13JO to be ~

alenal

82 a CHAPTER 5

S.U A .Iunl Rio,;. movin8 al 15 mi. parallel 1o the Ral pound 100m t..1ow How la'g<: m .... tbe horizonlal dista""" ~ f,om plane to 1If~1 be if I _ k of fIou , .~I~a>ed from llIe plane is 10 .Irike llIe ta'g<:I?

, Following the r.a .... procedu~ al in Prob , S.23 . ...., Y _ ,,"" , + !<r,I' to gel 100m _ 0+ 1(9.8 m/s')t'. 01 , .. ~, h. N"", U~ X _ ",I _ (1 5 ml. )(4.5 .) "~. oinoe the laCk'. initial wlocily is that of the pllne,

S.U A ba..,ball is Ihro"," wilh an inilial ,"<'I<>city of loom! s al I n onJlt of 30" abow 1M horiwntal. U"'" far from 1M Ih.o .... ing point will 1M baseball anain ill original level'

, By 'M rang<: formula .

S.U; A cart;' moving I>oril000tally along a Ilraigll1 Ii"" wilh constanl . peed lOmb. A projoeclile i. to t.. firN from Ihe moving cart in $UdI a way Ihal it wi!! ,etum 1o 11K can aft •• 11K can has mO>'ed 110 m. AI whal speed (~Iatiw 10 llIe can ) an<.! al .... hal Ingle (IQ 1M hori',,"tal) mllSl the projoecli!o t.. fi.ed?

, To mowe horizomally .. i lh Ihe cart. Ih. I'.o;'<lil. must t.. 6re<1 venically wilh a 8i",1 Ii"", - -" ' ''. ­(110 m)l()() mI.) .. 2.67 I. lbt initial "el<>city t'. must .. I;'fy ,. _ t'" .. ... ' 12 . ... ilh Y .. 0 and t _ 2.67 I; Ih ... 4.91 _ ~ •. and <0 to. - L3 1 mil at (I - !!!t.

S.21 In FIj. 5·12. ~ panicles from I bit of . adioacti"" materia l enle, through slil S into the ~ be',"'""n I-..Q la" c pa.allel """.1 pial"". A and B. connected to alOUfU of IIOIIIIC, As a ,,,,,ul! of tbe uniform eJectric IIcId t..tw""" lbe pllte •. each panicle hIS I romtant aca-Icralion ~ _ 4 )( lO" m/ .' normal 1o and I_aro B. If ". _ 6" Hl'm/ . and (I _ 45'. delerminc h and R.

, U ••• 1M e~ric force takes lbe plaoe 01 Jravi' y. 001 otbe.....u.: the anaJ)'$;' ;. 11K ......... Cho<>$ing "I'~'ard IS posil;" •. i.e .. the di .. <lion from B to A • .... e """e • reAu!ar lrajoeclory problem wilh ~. _ 6" 10' mIl; 6. _ 45"; SI, _ - 4 x U)" mi.' ; d, _ O. Thtn x _ ~ .. t _ ~._ 6 .. _ (4,24 " ul' ml')i; and similarly

Y. "..-' + .)<",' _ (4.24" HI'm! .)t _ (2.0 x IO" m/ . ' )I'

' '> - v" + a,I .. (4.24 " 10' ml,) - (4.0)( 10" mI. ' )!

AI the /tigbrot poinl . y - A. ~, - O. and I - 1,(6 )( U}- ' •. Th ,.., h _ (4.24 " 10')(1.06)( U}-) - (2.0x 10" ) " (1.06 " 10- ' )' .. o,ns m. l1>o:: hori'Qrnal ranle R cQrr .. pon<b to x at 1M t; ..... the " particle rel~"", to plate B. By .ymmet/y. t~";" 21 _ 2. 12 x 10 ' • . lbtn R .. ( 4.~" 10' ml.)(!.12 " 10- ' I) _~,

I ~

S.2I A ball is Ih,."" .. n up"'ard from the t"l' 01 • J5·m IQ..-el . Fig . S· ll. with initial '~I<>city "0 " BO mil at an an"" 9 _ 25' . (. ) Find the Ii ..... 10 rcach the VO~nd and 1t.. dislance R from P 10 the point of impa<!. (lI) Find lhe magnitude and dire<lion of the ,'.I<>city al the moment 01 impa<t.

, (., At 11K point of impa<!. y " - 35 m and .. _ R. From y - -35 - (8hin 25")1 - 1(9.8)1' . , .. LIlli. l1>o::n .. . R _ (110 coo 25")(7.814) _ 566." m. (lI) At impact. "~ _ 110 sin 25' - (9,8)(7.81 4) _ - 42, n mi. and " . _ ~ .. - 110.". 25" _ 72,5 mil. Thll$ ~ _ (42. n ' + 72.5')" _ 114.18 mI. and tan II _ - 42. n l72. 5, or II _ - 30.$4' .

MOTION IN A PLANE I J 83

,

". I •

5.29 n.., ~fT.n~ClIlent in Fig, 5·14 is the sa""'.> that in Fig. 5·12 , .. rept that" plInicln enter 'lit S from t1O'(> sour=, A, l nd A,. at angle. B, aoo B" ~Cli'·dy. u. and ~ .Ie th, ... me fo< bo1h gIO"I">. Gi" en that u._ 6 x Itfmfs, Q _ 4 " IO" mf>', B, .. 4S' + I' . B, _ 45' - I', !.Itow that all panicln ore "focu",d" .t • ,;ngl, point P aoo find the v.lue of R.

, Here .... can u'" the horizontal range formula R .. 2u~ cos B. sin B.I", Noting that for the fiTSI " panide B • • 9, _ -16' ar>d for 1M >eror>d a panicl, 9. _ 9, .. 44' , w, .. e Ihallhe lVon.", rom~ment.ry angle~. Thon COS 9 , sin 9 , " ,;n 9, = II, Bnd the: rong .. a", I"" same. Indet<! R _ 2(6" IO"m/s)' (0.719)o(O.69S)/ (4 " 10" mi. ' ) " !l1!l..m.

5.31 Apin referring 10 Fig. 5·1 4, find It , - It ,.

, Fo. the ,.nical heighll " ', hl"e v: _ vi + 2of,~ wilh v, .. 0; Q, _ - IQI - ~)( 10" mis' , Thcn

h, - it, " 0,016 m. 16 mm

1 • • . ,

=

, ,

llJ· 5-14

5 •. \1 A ball i. Ihrow" upward wilh initia l ""Iocily v. _ 15.0 m/ • • 1 . n angle of 30" wit h the horizonlal. The Ih.ower stands 1\(a, Ih. top of a long hit! ... hid< .... pe< """"' ... a,d ot an ongl. of N. When dof> the hallll,ik. the

"","'

, w. <boo<e the launoh poinl at Ih. oriJin (S<'<' Fig. 5.15). The equations of mOtion " f the b.11 arc

, _ ". ';n 30" - kr' .. (7.5 m/.)l - (4.9 mls')l'

~ cquatio. of the straigJIt line ind ine: i" .. - x tan 211' '' - O.36ob-. W ....... n' ,ho limc at ... hictl the (a . ,) v.l .... for the ball satisfy this equation, We thu> WbotitulC the lime upr~ for ~ aDd x,

.1

84 a CHAPTER 5

,

t'lz.5-15

(7.5 m{.)t - (4.9 m/.')t' - -0. J6.t{( 13.0 m/.)t]. 0. 12.21 _ 4.91' . The ooiution. a ••• - 0 (co"."""",in,lo x _)' _0) and 1-1&.1.

5.32 Referring to Prob. 3.3] dtte,mi ... 00... fa. down tl>o sla,x th. ball <lrik".

, Referring to Fig. 3·15 ",e n""d tl>o , and)' components of I~ 1"p\~menl to posilion A: ~ _ (1).0 m{.)(2.49'J - 32.4 m; )' - - 0.364.t - - 11 .8 m. lb(n L .. I +)' .ol:iJ..m (or directly f , om ~. L_ x/ros2O"_~.

s.n R.ferring to Prob. ~.31 indicate "';th .. hot velocity 'M ball hit, .

, fOT Ihe velocity of lhe ball juS! ~f"re impan . ~~;

,,~, .. It. sill JO' - " _ 7.~ mi. - (9.8 ml. ' 1(2.49 . ) • - ]6.9 m/o

Th ..

~IO\II positive I lUis

S..l4 A bom~r . FIg. 5·16. is flyi"3 level at , ."..cd It, . 72 mI. (abool 161 mi/h). II an el."alion of h '" 103 m. When directly over the origin oomb B '" released and strikes tile truck T. "'!rich ;' moving . Ion& a level road (the X &Xi.) ",ith con<tlnl speed . Allhe inil&nl the bomb i, "Ie,oed lhe truck is at a disWlCe .ro- 125 m from O. Find lhe ",luc of It, and lhe time of flight of B. (A .... me Ihot the lruck is 3 m high.)

, The equation. for I and)' motion of the oomb • ••

x _ v ... , . v, r" (72 m/* )' - Yo + It", ' - Is,' - h - 1&". 103 m - (4.9 mi. ' ), '

The lime for hiuing II>e truok wrres.porKH 10)," 3 m. on

) m" ]03m - (4.9 ml.')t' .00 , --, "

, - i..ll.J

Then I .. (72 m/ .)(4.52 I) - 352 m.

I - X. 325m - ]25m •• ~_J_ v,·-_ .. ~ ( 4.S2 5

o t±::::..:::=

MOTION IN A PlANE I a 85

S.15 A proj<"<1iie. M" S-Il. is fired upward .. ';th velocity II. It an ande 9. (.1 At ..... It point Plx. ~ ) does it strike 'he roof of the buildins ..... d in ,.-hat ti ..... ? (6 ) Find.he magnitude Ind direct;"" of. I t P. LeI 9 .. 35" . ... .. 4Om{s, .. -)If, and h .. IS m,

, Fint nolC that (usin, notalion i .. II .. Y" II,)

i . " t'. cos 35" .. 32.7661 mI. >< _ t'o sin 35" _ 22,'U3 m/ s

and, from the equation of tl>< roof,

f" h - Jr tan .. - IS - (O.5773S~

(. ) Eliminating 1 f. om f " yo! - 4.91' by Jr - io!, "'. hl'"t'

( ' )

(f'.) ' .9, ' y- 7' x--:,­. '. (I)

for II>< path of lhe projectile . Equaling f in (1) 10 J' in (2) and imcnin! numeric.1 vaJu.". 0 ,004564%'­l.mS58x -+- IS _0, from "'hich x _ UAro. 11>cn y _ h - (12.28) \an" "~, The Ii .... to ~trike is &i>'en by Ill!! " 32.7661/, or /-2,11U.

(6) At p, i ·:to·32.766ml. Y _ Yo - 9.81 .. 22.943 - (9.8)(0.315) .. 19.U8 mi .

Thu~ 11 _ (i ' -+- y')'~ " ru.mu and tan tJ .. iii -0.S88 , 0. tJ .. ~. where " is lhe angle that . mak." wilh X at P.

~ . h · ".nD

,

'-----'---- '-------'-- --""---.x nc.5-11

5.36" In Prob. S.35 . angk 9 can ~ adju .. e.l . Find the valu<: of 9 fur ",him ,t.., projc(1ile "likes , I>< roof in I minimum time.

, Again y .. j.,t - !S,' " (1I. ~in 9}l - \gl' '"' Equaling Ihese lwo upre-lOion> for J' and eliminaling x by x _ i o! . (lIo COS 9)1, we obnin Ihe followinl eq,,,,"ion for lite Ii ..... of .. likinl: !gI' - lIoIeos 9 tan a -+- oin 9}l -+- II _0. Or. ""nllhe addition formula sin (9 + a ) .. sin /I 00II or +CQS /I oin a.

\g,' - [":0 .. sin (8 + a)} + h _ 0

For a minimum I. we mu-ot have dlld/l _ II. Dilf •• entiahnl (I) ",ilh reopoc' to 11 and "'(lin8 d' / d/J _ 0, we obtain

-L':"" a>!O (/J -+- or)} _ _ O

which implies thai (since ,- .. 0) a)o$ (" + ,,) _ 0, o. /J - l!!l.=..s.

'"

This u'$u!t mea ... Ihlt \be: prot-lOtil. should bf aimc<l in lilt- dilec1ion of mitimum disllllt«, jUM as lhough llIe ..ettlelltion of pavilY did nol ex;'\. Howeve., , . a';1)' unnotlte iJllO'cd i. this problem, If w. seck to dcICrminc Ihe val ... of I ... by SU~ilul;nl /J -+-" .. 9(f' inlO ( I ) and IOlvinl, "'e obtain

I .... • ~. - "vi - ?&~ <:0&' a

.~.

86 a CHAPTER 5

wtlidl i. comp!cx if v. < Viiii COl G, In other wo. do, if ~o< Viiii COl G. the projotctik n~v~r .uch<'$ th~ T<><>f. wh .. ev~. the valu~ of 8, and the co"""pt of a minimum time become. meaningk&$.

S.:n With rel~'~nc. to FiX. So lS. the pro;ecrile is ~,ed with an initial v~locity ~o .. )5 mI. at an angle 8 .. 2)'. The truck is moving d ong X with a con .. ant speed 01 IS .. ./$. At .he inltant lhe pro;ecrilc is fi . ed . the back of III< truck i< at ~ .. 45 m, Find Ihe time for lhe pro;ecrile In strike the b.>ck of lhe truck . if Ih. lruck i. v~J}' ta1\,

I In this case. the projotct ile hits tl>e bac'k of the trudt at the moment of ov~rtakinl it . wh;"h i< the moment at wtli<l! ,he dis"'''''" of lhe back of I"" lrod . X, " 4S + 15" equal' the horilootat distance of Ihe projltdile. X _ ( ~o COS 9)1 _ 12.221,

" 1"32.2Z-15"~

r ,

..... 5>18

S.3S What wi ll happen if lhe truck of Proh. 5,)7 i. ooly 2,i)m tall?

" "",

,

, At t .. 2.1>14 S. when Ihe projedi1e overtakes the hack of the truck . its height is. noling ~. sin 8 ..

,

1l.67 m/ o: y " (13.67)(2.61 4) - \(9.8)(2. 6 14)'" 2.2S m, i.e .. 2S em abcro'e lhe lOp of tbe trlld, Since tbe projedile tra,..,11 f"'te. horizontally than does tM trud , it is dear tllat lhereafte, the pro;ecri le ,emaiM ahead of the back of Ihe trl>Ck, and s.o .... v • • hih Ihe b.>ck.

The pr01«tile win Teach (for the second time) a h~iJhl of 2 m in & tot.lt ime I, gi.'en by 2 " (13. 67j.r, - \(9.8)1; . Of " _ 2.635 •. that is, 2.(ill - 2.61 . .. 0.021 • after ,"'ertakin. th~ ba.:k of ,he truck. Thus the projectile hi .. the top of Ih~ truck of. di"ance of (32.22 - 15)(0.021) .. O.36m .. ~ in fronlo( the rea. edle.

~.JI Rderring to Prob , 5.37. fin<! a value of Ito , all o'h~r conditioM remaining the .. me. for which lhe pro;ecrile hi .. the truck II y " 3 m,

I The lime taken 10 o\'e .. ake the t>ock 04" II>e truck i! given by

In""ning II>e nUrMrical val ... ~ of lin 9 an<! ros8, ...., rotain the follo"';ng quad.atic ~uatioo for ~.' u~4.5S) - v,.(60.3) - l512 .. 0, Solving, ~o - JS,l mls.

, . .,. The motion of a panicle ;n thoe XY pl .... i, live n by x _ 2S + 61'; Y _ - 50 - 201 + 81' , Find the foll"",n, initial '"1Ilues: Xo. y •• i • . y •• " 0'

I Atl_O,X_ xo _2Sm;~ _ .'I , _ _ SOm,

'"' .... _t,, _Om/. u, _y _ _ 20+1br

'"' ~ .. -Yo--20ml. u. _ (u;' + .:")'~ _~

MOTION IN A PLANE I a 87

S."- find macnilude and direction of •• the ilCttlcration of the particle in Prob, ~.40.

, ", . ii, _ 12 m(,';", _ ~, _ 16 mls'.

a. 16 4 tan8 a"" __ _ _ ' d.123

implies that 8, _ ~ lbove the pooitiv • .r axis. d - (12' + 16')' ~ -l!l..ml£.

S.41 Writ. an equation for the particle·. path (fioo y a. a function nf .r) in Prob. ~.40.

, We ehmin". ti"'" bet .... en the tWO equation. as folio .... : ,, - ( .. - 2~)/6. Olio. If .. - 25)f6)n Then substituting into the y «tWltion, y - -50 - 2QI(.r - 25)f6J'~ + 8( .. - 25)/6, Of 6y _ -SIlO + S. - 120( .. -25)/6J'~.

5.43' A particle moving in the XY pia"" has X and Y romponl'm. of velocity given by

i _b, +c ,!

... he",. and yare "",,..ured in meten and! in """,nm. Id) Wh.at are the units and di ..... TI$ioTl$ of the «Instants h, and b,? of c, and c,? (b) [ntevate the above relations to obI:oin -" and y as funcl;OIIS of time .

P)

(t) Dcn<Xing total aca:lcration as. and total v.locity I. ' . find expressions for the magnitude and direction of a and of • . (4) Writ •• in tu"", of ,be unit ve<:tors.

I la) Inq>edion of (1 ) silo .... that b, and b, mu.t represcnt velocili .. in ""'len p<! • ...::ond : unit· dimensionally c, and c, must be Im/s'l, thus arooJcration •.

,b) .-Xo+b,I+!c,I' Y · Y. + b,l +jr,l'

... he", -"0.:11> are ,he .... 1,.., of .. and y at ! _ o. (t) Differentiating (I) with respect to !, Ii _ c,. y .. c,. Th ..

• - (i ' + j')'~ .. (e; + el)" .i' c, tan<> _ _ _ _

Ii c,

.... be'" <> is thoe anJle a make .... ith X. NQle that a is ronStllm in magnitude and direction . For the velocity,

v _ (i ' + i ' )'· .. Ifb, + C, I)' + (b, +C,' )'J'~ • b,+C,1 tan ___ _ b, + e,1

.... h.re fJ is tbe angle . makes with X. (4). _ (b , + c,l )I + (b , + c,l)j .

s...... A panicle mov ... in the XY ptane along thoe path giY~n by, . 10 +.h + 5.1', The X romponent of yelocity. i _ 4 mil, i. conslll/H. lI>d at rOo 0 , .. .. ... - 6m. ( II') Wri te y and z as function. of I. (b) Find y and i, the romponenlO of aooolcrition of the particle.

' C.I ; - li+Hlu: i_4m/.at alitimesand"0 _ 6 m. Then

-" - '" +i l - 6 m + (4 m/s}l,

y" 10+ (18 + 121 ) + 5(6 + (1)' - 208 III + (2S2 mfl}l + (80m/" )I'

'b) , - 0 f _ 160mb'

S.4S A ball. B" is fired upward from tbe origin of X. Y with ini~.1 velocity v, _ l00m/. at an angle 8, _ 4JI1'. AfttT I, • 10 I, .. can eollily be !.hown, tho ball ;. at point /,(z,. ~, ) . """ue z, • 766.0 m. y, .. 152.8 m. Some: time later, an<Xher ball, 8" i. fired upward. also from tile origin. with velocity v, a' angle 8, _ 35' . (.) Find a val ... of v , such that B, ... ill pass throu! h the point PIX,. y, ). (b) Find the ti ........ lHm B, mltSt be fired in ordet that tbe two balls will collide at P( .. ,. , ,),

I (. 1 let (-"" y" I,) refer to tbe coordinates and time of 8, and (x" y" IJ to t"'* of 8,. Si""" B, i. to pass through P( .. "y,).

X, " (If,ros.JS)f, _766.0 y, _ (v,';n 35' )/, - 4.91; - 152.8

Elimina tin& I"

from which V," 1!l5 69 mi •.

88 D C HAPTE R S

(6) In$<!ning ,I>< "aloc of~ , in .. , _ (~ , «>S 3S')t, _ 7/i6, Om. yo'e find I, _ 8.848 •. He""< , ""i,h ~, _ 1(\S,M mI. and fI,. ]5' , 8, paw:s Ihwugh f'(~ " y,) &848 •• fler il i. ~,W . Bu, 8 , arri~<. a' thn point 10 . after $I.Irting ' ·Ience, if the I .... a,e to collide. the firing of 8 , rnu>! be dela~w 10 - 8,848 - Lllli.

S.M>' The mOIi"n of a panicle in the XY pl .... ;, given by

-< - 10+121 - 20<' ~_ 2S + 1 5,+3O<'

Find valoc~ of ... . i ,,; y", y. and 1M magnitude ~nd diItt!i<:>n of ' 0,

, At 1_ U, -< - -<. - 10, Y - Y. - 25. Differentiating "". get i _ 12 - 401 , y - IS +601 ; i . _ 12; Yo _ IS,

" " - (i~ + YfJ"'- 19.2 O._ '.n '~ - 5]']· '"

ahove P'J"ilive -" uis

S."", Referring 10 Prob. 5.016. findi,!, and a . ,I>< .«10' accele'alion .

, We ditfelenli.te the expression) for i alld j 10 get i _ - .j(), Y" 60. Acceleralion is thu$ amstanl and a _ v'0iiJi + iiii' - 71. The direCI;OO of a ;' gi.'en by

/I - Ian ' I : I-~

5,48 Referring 10 Prot.;, 5.016 and 5.~7. ".Ie ..-hNl>cr or 001 1M motio<! i. aloo& a maight line,

, No: Since the lio>tS <>f Yo and a "" ""I roincide Ihe palh must be paraholic. as in pro;'<lile moIion Sec Prob.557.

S.W Can ti>< motion of a ""nide Ix gi"en by

.• _ 5+10I+17r'+ 4,' Y - 8 + 9r + zo.' - 61'

if .h. ""nicle;' .CIW On by a <QfISlan' f""",~

, i _ 10 + .~I + 12r':)' - 9+ ..or - ISs' : i - J4 + 241; y _ 40 - .l6r. BUI 1M acceleralion i. 1101 ronSlan •. as il muM be for a ron.lanl 10""' . II."..,. ,I>< mOIion is imp<>Uiblc.

S.2 RELATIVE MOTION

s.so An el..-ator is m"vin. up...-.rd at • ron.laBt . peed of ~ mI •. A ligltl bulb falls oUI of. _kel in 1M a.ilinl of 'ho cle~al"'. A man i ~ , he: bu ilding .... alching the: cage sees Ihe bulb n.. for (4/9.8) $ a!>d then fall for (~19. 8b: al ,. ( ~N,g). Ille bulb appe",. 10 be al re>! 10 lhe maB, Compule '~ vt"locily of ,he bulb at 1 _ (4f9.8) S from t he , -j . .... of ~n "b!.<"·er in thoe elc~al(>[".

, Taking up as posilive 1'_, .... u _ _ - ", ... ,_ .. 0 - ~ _ - 4 mls. Allcrn3li" ely. in lhoe i ... niaJ f.ame of 1M ele.'atol.

u - u. + as - 0 + ( -9. 8)(~) - -=.i.!!!JJ .S S.S I A sltip i$ "a~eling due east . t IO km lh , Wh.t must hc: Ihe speed'" a s«ond sltip heading)(l" ca .. of DOI'!h if

it ;s aj,,·oY" doc north from lhe fi "t .hip~

"

,,---- ..

nghted mater~1

MOTION IN A PLANE I a 89

, Y, _ velocity (If finot ohip relatiV<' to the eanh; v, _ V<'locity of S«OfId >/lip ,elati"e to the eanh. Let v" _ ... I •• ive velocily 01 oc<:<>D<! .hip 'Q fi" •• hip, 1'1><" v, _ V" + v,. (Fig. ),19). " 'I><,e v" ;, due Jl()rth. ThU$ v,oinW - v, - 10kmJh. and v, - mm1l!.

5.52 Durin, a ra;n>1orm. raindrops .... OMc .... ed 10 be slriking I"" grou nd at an angle of 35' ,.,ith Ih. '''rtical, l1!e wiD<! .peed i1 4,5 mft , Assuming lkall"" I>oriWfllal "elocily romponenl of 1M raindrop;;' the sa"", a\ 1M $pC'ed of tbe ai, . whal is II>< ""Mieal ,·doci.y rompoMm 01 11>< ,aind,,,,,,,~ What i, .h"i, sp:ed?

, Le. V - ""Iocity of , aind.or> ... Ialive to earth. v. ,. wirn! veloci,~ , ". ,. 4.5 mI., " • • ,,~ . F",m Fi! . 5·20,

"- .. ,-,-v' ·"m3'·-~

,, _ "v; + v; _ 1,M ml1

"

. , • • , FiII·)'20

5..53 A "".,boal;' pointing pcrpcndirula, '" Ihe ""nk <>f a <i,·c, . The """Cr can pmpclthe boat .. ';tll a speed of 3.0 m/' " 'irJr _p«' '" .~ ",tIl". 1'1>< river h s a current of •. 0 mI •. I_I Conmuct a diagram in which ,he lwo velocities I •• • "pr-o~nted .. v«lo,,", (b) Find I"" "eclor wltich .ep.e~nts the boa1"s V<'kxity .. ilh respeel to Ihe ..oore, fe) At what Ingk: i. 1hi' veclor indi .... d 10 lhe d;reaion in " 'h ieh Ihe bQat is po;nl ing~ What is Ih. boal'S speed wilh rc<pc<1 10 Ihe launch poinl ? (d) [f the riv.r is 100m ,,·id<:. <leterm;n. oow far do .. n",e.m of Ih" launch point the rowboal is ... h. n it re...-hes the opposile ban k.

- - 1 --~ .. - -

,., '" J.". ),21

, (_) Sec R,. 5.21(d). (b, c) The v~'1tr ~ilh'~ to th" sho,. is gi"en by v., _ v. + Y • . Sin.ce v. 100 Y~ are ptfJl"OOirul •• , .. 'C ha'·. ,,_ - u. + u. _ "3' + 4' "ll!!b. "The angle 'I' "'o"'n in Fig. 5·2] (b) is del~lmined by t~n 'P .. u. / " •. FOI tM 'pte'" liven. "'e IiOO 'I' - u,r. The bo.at moves ~Iong a line dirr<:!«i H.I' do...-ns .... am from "Mraight """"":. (d) L. lling D .. d;SI.= do...-nSlream .... " h",'e D/l OO m" to./". _ 4/3. w . hat D .. Ulm.

5.S. A swim", .. can .wim a. a spted of O.70ml' wj.h re,!","' '0''''' " " 'cr. She .. an" '0 e .. ,.. • ri ... r which i, SOm wide and has a CUrrent o( O.SO ml' , (_) If W w;,hes 10 laOO on .he Otbe. bank al • point di.ecIly acroq .1>< river fr"'" her ,,"ning point. in ... hat direaion mu" shc .wim? How lapidl)' will >he i""",.", Iter d;S.an<c from the near bank? Iiow long ... ill;t 11k< her 10 emu" fbi [f. "'. in'tead decide> 10 ~ in the , hott",! poosible ti"",. ;n whal direction mUI' ,h. lwim ? Ii,,",,' rapidly ... ill she in""",,, "". d;" . n<c from Ih. ""ar ban k? How long ,,·ill illake her to cross? 110 .... fa. "",",'nSlream will me be wbr:n . hc Ian"'?

, (_I Let " be the "el<>cily of Ihe current. Vr he ,h. Hloci'Y <>f 'he swimmer " 'i th r~'ptct 10 Ihe wal0r. and " be tile "elocily oI.M ,wimmer wilh resp:e. to tile ,hoJc, Then v, _ 0. + Y,. For a direa crossing . Y, mu, l be ptrptndicula. 10 v •. "Thcrefore sin {J _v,/v •. ~ 1110"'" in Fig. S.22. W. are gi~e n the ,·. lu ... ", _ o. SOn,{ •. u. -0. 7Omls. aOO d _ SO m; " 'e find {J .. 45.6' up>lf.am f. "m !he direClion " "'sighl ""'0»." The , wim"",. will iDere,"", hr. dill.nce from th. ncar shore 111M . ate v, .. u. cos {J .. 0.49 mf •. She " i ll crOSS the ri".r in a 'i"",. _ dfv, " 102,;

90 D CHAPTER 5

\ j

• ',] .. ~.,

(b) T o muimiu the romp:",eot of her vdocity JIC''l'''Ddicular to> the river l>aok. the ,wimmer ,hoold head s!rai,ht IoCI'<IOS the SUeam. She will moss in .. time' - d/~w - 7\.4., and 1M will land I di.tantt ~,' - (~</ ~.)d" 35.7 m dovo'mtnam from he. stanios point .

5.55 An armore.:l car 2 m long and 3 m "ide is mo>.ina I t 13 mls ,,'hen a bullel hils il in a di rection making an angle arelan (3/4) wilh the ca. a, sun f,(Im the ,tntt (Fig. 5-23<1). n.. bulll't enlers.,.... ed~ <>f 1M oar at tile come. and pas.ses out at the diagonally oppo$ite rome •. Negl«ting any interaction het .. een bunet and cal, tiDd the time for 1M bullet 10 CroM lhe aOT.

, Call tlte opeed of the bullet V. Beca .... <>f lhe rTIOCion <>f the ca., the ",,\ocity <>f the bullet .elative ' 0> lhe ca r in the direCl;o" <>f ,he !cnSlh of Ihe ca. is V 00$ 9 - ]3 , aDd the veloo!)' in the di.ection 0>( Ihe width of lite eo. i. V >in 9 (Fi, . 5-1Jb). n..n. from, .I~.

Eliminale V to> find 2_(VOO$9 _ lJ), 3 - (V >in 9),

' - ~(-' -'l-"'-!ill> 13 tan 9 13

, (. )

, ,

( v ... , I

(»

( (

V_i _ l l

5.56 Rain . pouring oown at an angle ..... ith the venical , h",. cumtant opeed <>f IOm / •. A woman .um .,..imt the rain with. opeed o>f 8 mIl Ind ~ Ihal 1M ' "in makes ao angle fl wilh Ihe ""nic.al. Find 1M ,.Ial;o" Mt ... een .. and fl.

, From the ve<1or diagr.m . Fig. 5·24.

hied IT ria

MOTION IN A PLANE I a 91

5.$7 V~rify that the trajectory of Prob, S.46 is a p.lrabola by cboosing roordinalC a.CS parallel and I"'rpendicular to the ""Mtant """"I~rahon ,·«I0r.

'-'--'-",'-----,

I Figure S·25 lob ...... lhe new coordinale .ysl~m ; from analytio JeO""'lry"~ have the relarion.

• .• , 2 , y · - .. .. n +yroo __ ~ .. + ~y Vl3 vB

HelICe Ihe eqUlltions of motion in tile prime<! coordinates arc

VTI .. · - J( IO + 121- lOt') + 2(25 + lSI + JOt')"80 +66t

Vi1 y' - -2(10 + 121- lOt') + 3(25 + lSI + 3Of' ). SS + 21/ + 1.lO1'

We ",,0 now solve (I) fOT I in I.,.,... 01 ..... <1<1 .ubotilute the inuit ;n (2). obtain;n, an equalion

y· _u" +b .. '+€

fOI' dtfi"ile ""Mlants Q. b.~, Finally. """"pleting tile squar. "" the right 01 (3). we tra""lolm it to

Ol (2)

y '-fJ -K(x'- ... )' (4)

... hieh is reoognized as the equation of . parabola. with "ertex It x ' .. ... , y ' _ fJ and "";th axil p.lullel to ,hot y' axi •.

5.58 Qbst ..... e' 0 drops a 1I0ne from the ,hirtieth Hoor of a sky..:,aper. Ob... ..... er 0' , dt"",ndinll in an elevato' at ""nSlam Ip(W V - S_O m/s. passtS the thirtieth IIoor just as the SlOne is released. Al the time I - 3.1h after Ihe >lone ;, dr""""d. find the pooition. the "elocily. and Ihe a=leration of the stone relative t" 0_ 1Mn find the pmilion. lhe .'elocity. and the acccleration of th~ ,lone relalive '" 0 '.

I For 0 , the pmition of th~ >I""" il gr.'en by

.,' x - x. +vol+T

,,-here x .. 0 al the Ihirtielh 1\00, " i th the oo· .. 'n ..... rd directi"n as the. """iti"'" direction, TIl ... , at 1.3.0.,

... 0+ 0+ 9.8 mi.' 2" (3_0.)' .. + 014 m

Abo , ~. v.+ a/ gives ~ '- 0 + 9_8 m/l ' )( 3,01_ +29 m/ s. 1M 1"""leration of. truly fallins body. a, ....,o by the obH"'.r 0 who is OIaliunary with respecl to the

earth. is known to be the COnstant ,ravitational acceleration. (Indeed. tN' underlies the validily of lhot two ~alculalio", immedialely aoo.'e .) y"" tb ... Ilave a - +, _ +9,8m/'"

0' me ..... re. polilion z ', reliled 10 " via .. ' _ ... - VI_ Hence, after 3_01. z' _ 44 m - S.Om/ .)( 3.0 • • +29 m. ThaI is. the >l0!\C is Incated 29m hel .... obser>'CT O' " tbe end of 3.01, 1M ,tone'. v.locity relativ. 100' is v' '"'v- V; hence , I t 1 .3.0 •.

v' - 29m/.-S_Om /. _ +24m/.

Since V;. ""'mant • • ' .. . . and~' .. +8 - +9.8 mi.'. ~ ..... er O· ....,. 'he 5I(",e to have the.arne do"lI ward aca:leration "' thl!....,n by 0, ( In gc!\CTal. kttlerations are the same in all inertial frames .)

92 0 CHAPTER 5

S.9 A lrud: .. IIlIvelin, due non h and desa:nding I 10 petunl &rade (angle of 00pe " lan - ' 0.10- 5.7") at I OOMlanl speed of 9Otm/h . Allbe: b.a .. 0I1be: hilllbe:re i5 a genlle curve. and be:yond 111&1 Ihe road is le",,1 Ind be:ad>.lO' cast of !be: nonh . A lOuthboond police tal" with I radar W!UI is trlvoling II 8Okm/h oloa.,lbe: \evel road at the, b .... of Ihe lUll. app<oachinl tbe: lruct.. What is the, .. locity ,.ectaI" of lhe, lruck wilh relf>CCl to tbe: pol~ rar?

, We u" coordinate "" .. with i eISl.' nonh. and I '·e"icaUy upward. We let ' f he, tbe: velocilY of the, lruck with rcs.pec1 to Ihe around and~,. he, lhe, velocity of lbe: police car wilh r •• peeI 10 lhe, vound . A<X>:lfding to the infOl"ml tion gi'·.,n . ... ., have

' r - Of: + (90 km/ h)(cos S. 7")j - (90 km/ h)(sin s. 7")t and

' " - ( - 9:1 km /h)(lin .lO'}I - (!(I km/ h)(cos JO'It + 01

Reducing lbe: ... we find ' f _ (lJ9.6j - 8 .941) km/ h and ' " _ (- .so.Of: - 69.3j) kmlh . The: velocity .r of the truck wilh rclf>CCl to the poti« car is given by

I f - ' r - ." - (010.01 + 1S8.1ij - 8.91) km lh

SoW A bi rd. in level nighl al rollstanl a=lcrlllion" relative 10 tbe: Vooond frame X . Y (Fig. 5-26). Ie" fan a worm from its be:ak . Whal i5 lbe: path of tbe: worm. IS .. en by tbe: bird?

, In tile bir<l·, nonine"i.) roordinate .)'Stem X ·. yo (f Ig. 5-Z6). the "'IUlliO<1 0( lI"oCKion oIlhe worm is

,.' - - & - .. - oonslanl •

Thus tbe: 8Oteleration of tbe: worm i. tollitant. and in path is a Slnight line (>4JPposin8 Ihat it was droppe<l from mt ). The: ~ of the line with respect to lbe: horizonlal is 1l1li (J '"' , I ll •.

, " ;

71 •

- -, ; a x '

0 , F1c.5-16

5.61' Refer to Prob. HrO and fig. S-26. (I) O¢termi"" tbe: palh 01 ,lie wonn as !Ccn from IIIe ground. (b ) Verit)" tblt tbe: lwo descriptions of tbe: palh I re "'IUivalenl.

, (ll ) In tbe ground frame X. Y. lbe: worm ha. ron>tanl aca:Jeration j _ - 8 and an initi.1 velocily .i._ v •• where V. ;" Ihe ",,"cd of lbe: bird al tbe: ;n>lanl tbe worm i. released (C&11 Ihn t ime 1 _ 0 ). Hence

.r -.r. +v.,r Y-JI> - Ig,' ( I) Ind tbe: palh is. paraboll . {b ) Lec us >4JPPOSC Ihlt at /. 0 the twO coordinate fnlmcs coincide . Ac time I. O ' will have Idvar.rcd I distance Vol + ioo.r' alon, lhe X uis. 10 thaI tbe coordinates (.r. y) I nd (.r'. , ') of the ... mm in tbe: tWO Sl"'tnlll are felaled by

~ _.r ' + (vol + ioo.,r')

1bc path in the X· . Y' sl"'em is obtained by subslilutin,lhe expreWons (2) into ( I ) :

.r ·+....,.+IQ.,r' .~ . .. ...".. ,._,. - Igr'

which is I waight Ii"" 01 slope , I"". as found in PTob . 5.60.

MOTION IN A PLANE I D 93

z

• y ,

$.62 A helicopler;' trying to Ian<! on a oubmarine deck ...-hic:h i$ n>O¥ing south at 17 m/o. A 11 /llla .ond ;' b\,ow;ng into the we3ol. If to tbe .ubmarine = '" tbe- helkople. is d<$«nding ~rtiu.lly at S mis, ""'at;' il> SJI«<I (.) relatiyc to the _ ter an<! (6 ) relative to th. air. Se.: f ig. S-27 .

y ..... _ .. . _ +w_ . 17j + {-5jk.07j- S. ) m/s y _ _ y __ + , ___ , _ - y __ _ ( 17j _ 5. ) - 121_ (_ 121 + 17J - SII) mi .

atanal

CHAPTER 6 lID Motion in a Plane

6.1 CIRCULAR MOTION; CENTRIPETAL FORCE

6.1 A O.J.kg rna .. anacbed to a 1.5 m-Ionl string is wh irled around in a horizontal cirdc: II a speed of I> mJ •. (_, What is lite centripetal a_lelation of the mas.s? (b) Whlll;S the tension in lhe string? (Neal«t gravity.)

~.~ .. (6 m/ft • 0l!.m.IJ: R 1,5m

(bl The temion in 1M ming cum the centripetal fon:e required \<> ke-ep lbe mOSS in circular motion. This fOT«;, r - m~ - (0.3 ka)(~ mi.') -Ul:!..

6.2 A .... all ball i> fastened to I 'Irin~ 24 em Jong and ~lIded from • 6. «1 point P to IlUOU I eoninol pcooulum .• s shm"" in Fl,. 6-1. lbc ball describes I 00,; .... 1111 ci.de about. «nler ' emeally uDder point P. and lhe SI. ing make> an In,le of IS' with tile venical. Fir.d the ~d of the ball.

, T 00115' .. mg ""' TsinlS' -­, Si""" . .. 24 sin 15' .. 2.\(0.259) .. 6.22 em,

"' Ian w_ 6.22(980)

and hef>CC

v .. 4Q 4 cm/ t

"' tan 15' .. -•

603 In the Bobr model of the hydrogen a'orn an declrflfl is p;ctured rotllin, in a circle (with . radius of 0.5 X LO-oo m) about the positi"" nutlc..s of the atom. n.. centripetal fo.", i$ furnt.bed by tM ~lectn. attraction of tM positi"" nllclc", fo.- the "",.ti"" electJOll. Howlarp u. this f= if tM electron is movin, "";th • '!'«'I of 2.3 x lit mJs' (lbo mll$S of an ~Iectron i. 9 x 10-" kg.)

, FOfU " (9 x 10-" .g)(2.3 x IIr m/l)'I{~ .O X to-" m) .. 9.S x 10- ' N

,... Find the maximum ~d with whKh an automobile can round a curv<: of IJ)-m radi ... without slippin, il tlte "",d is unbonked and tbe coe~ent of friction between the "",d and the tires is 0.81 .

..

, Fi .. t draw a diavam showin, the fOlCC"l (Fi,. 6-2). If "'8 i. the """igilt of the automo:;,b;lt , then the normal fOfU is N .. mg. lbo frictional force ",pp/iel the ~nuipt(al f= F,.

1'.- _,.,N .. 0.81 mg

A"" . ""' 0.81 mg '" SO u' .. 0.81 x 80 x 9.8 - 25.2 roO

MOTION IN A PLANE II 0 95

N

mg

'05 Who, is 1M m.IOI imum ~~IO(ily. in miin ~r hoour . for an aulomobil" rounding a I~~'l <\Irw of :!O:).lI radi"" if II, MI ... ~n tim and ,o..dMd i. 1.0.

m. ' , II,IIIA- ­ ,

. ,- JOmil h , ~ - VII",' - VI.O x 122 II I.' x 200 II - 80 fl i ... 80 III. x 4:ifii$" 54.S mlih

'.6 A car is lfov,ling lS mi. (S6mi lh) .. ound a ~,·.I <\I"" of radi .... 120 m. Whal is Ihe mini",u", value of 1M ~f!icitnl of Slali, friction IItt ... ""n lhe ti'" and lhot road .. qui~ 10 pre"en. Ihe car from .kidding?

, ,, ' (lS 00/.)' II, l!: S; - (9,8 m/.,)(I20 m) - 2JJ

6.1 Traflk is oxp«t.d 10 ~ around I all"" of ,o.di ... 200 00 at 90 ~m l h, W1111 sIIoukl be: lhe "alue of lhot Nnkin8lngl~ if roo d<~nd<~ i> to lit placed 0" friction?

FiI;.6-3

, s.~ Fig. 6-3, '"' m.' , _ _ _ N";nlJ ,

Di'idin! ,lit >C'CQnd by Ih" tirs. "q"'''ion .•• n IJ .. ,,' fT/!' Subs!;,u\< lhe data. dWlging kilo .... ' .ll p<r hour into m~I""~' se<Xlf><l:

(90 km/ h " 10000ml km)'

.l60hfh ,anlJ_ 2OOm><9.8 m/.' _0.319 9 - 17 T

,.. A. indi~aled in Fig. 6-1. a pia .... ft)ing al con".nl Il""'d i. banked al onglt IJ in order 10 8y in a hori>onlal cird~ 01 nodius , . n.. a.rodynamic lift fortt acts 1C'nt"r.llly up"'aTd al ril!J>. I ngl" 10 t he pia",,', wing> and fu .. la, •. This lill for~ COfffSponds to ' ho I.,,!.ion pr()Vid<d by .h, sIring in . conical p<nc!ulum. or tilt roor",.II""", of a ban. e<! road . (<II Obl.in the "qualion foo- ,he "'quired banking angle IJ in l.nTlS of v, " I nd 8, (b) W11a1 i. lhe requi re<! angJo for " _ 60 .,/ . (216 .",/ h) ..... , _ 1.0 km?

T ",."

, (<I ) A, in Prob!.. 6.2 and 6. i . • an IJ .. It'I'6'

"" lJ _ t.n- ' - l:l.Lt (1.0 " 10')(9.8)

,,, C JPYnghted matanal

96 D CHAPTER 6

6.' A a. ,~. around a CIIrv. of .adius 48 m. If I~ road;s hank~d ., an angle <>f IS" ";IIl Ihe l>ori>Qrniil .• , .. lUll ma.,mum ~d in kilomcl.n per bour may ,lie ~ar Ira",,1 if tiler. is 10 "" no lenderlq' 10 ,kid e~cn on .. cry ,Iiwery p;I"emcnl?

, U ... tile njualion 10. banking of lIip.",ay>=

" lanfJ __

• " lanlS" __ 0/8(9.8)

v _ 0.268(48)(9.8) .. 11.2 mI • .. (1 1,2 mll)(OJ))1 km /m)(3tOh/ h) _ <\!.l-l kmll!

6.1' A «nain ca. of mM, '" Il .. a maximum frictiQnal furt:<' of 0,7 "W """ ... ~n it and p;I ... m<nla. il "",r.d$" curve Qn • !\al road (I' .. O. 7). Uow 1.>1 can II>< a. to. moving if il is 10 wcceufully nc,olial. a ctlr.c of lS-m radios?

, The "",nlripelal force ("'~ 'I') mUSI bt $OlppliNI by IIIe lri<lio",,1 I~. In IIIe limiting Ca .... ww'/, - f ";,h f _ 0,1 mg. Thu •• tI ' _ 0, 7'8 and II _ 10mls,

' .11 A clate sito Qn 'M !looT of a bmca •. The roeft'kien, 01 friction bt ..... en 'M en,. and 'M floor i. 0.6. Wha. is .M ..... imum r.peC<!lha"h. bola, a n go a.""nd a curv<: of radius 200 .. ...;.IIoUI au.in, .h. cra.e .0 slide?

, Al in other " unb.anked<ll,,'e·· problem. (e .8 ,. Pro!>. 6,5) .

v:". _ ,."', _ (0.6)(9,8 m/ . ' )(200 ml " 1176 .. '/.'

'.ll A I»y on a bicy<:le pedal. around. rim. of 22-m radius a. a ~d of 10m/ •. TlIe comhin.:d mb> of ,b<: boy and ,I>< bkytk is i(I kg. ("I What ia ,he "",,,.rip<Lal ["""" exened by II>< pa~emcnt on lhe biqc"? (bl Wha, is Ihe 'JI"""'O f<lT« • •• tlc<I by the p;lvemcnt Qrt t~. bicycle? s... Fla . ..,., .

" ... ' 80(10)' , '.1 F; -- ----,..N " n Ibl N - "'I - 8Oj9.8) - llit!

F, m - 80kg ,_22 m

'.13 R.I.. \0 Prob. 6.12. """'a, i. the ~n~lelha'lhe bkytlt m~k .. "i,h 'he 'c"IeaI?

, For II>< t,;cyclt n<>l to fall. lhe ,o.que aboulll>< ~nlcr of gravil~ mu.1 to. lC:fQ--Sooc a.'p>. 9 ~nd Io.-..·hi<h mean> Iha, Ihe ""clo. lore .... ned by Ihe ground muM ha"e a lin<: <)1 a<:lion ~"8 Ih.ou'" IIIe ~n'''. 01 ""vi')' n",.

,. '" '''n f} .. ~ . .. 7S4 - 0,46-13

6. 1~ A t ~ ul m .... 0.28 oJ" 12 em from.he ~~t .. 01 a p!>on"l!.aph .crood '''''''vi"8 .1 33; fj>m. (,,) What i\ the

MOTION IN A PlANE II a 97

magnitude of t"" centripetal forte on 1M fty? (6 ) What is Ihe minimum value of tM ~fficient 01 stalic friclioo between Ih~ fly and Ihe r«Ofd r.,,"i~ tQ pr.~enl th~ fly I",m uidin, oI'I'!

"" (33'33 min-') ~ ·T ,",bf' - z,.. 60s/ min (12)< IO- ' m) - O. 41 9m/,

u' (0.2>< 10- ' kg)(0.419 m/t)' 0-' F _ ,.",_m__ _2.'l2>< I N r 0.12m

(OJ 2.'l2>< IO-'N

F, - 2.92 >< 10-' N S I',m: 1'. ;,. (0.2 >< 10-' k&)(9.8 m/$' ) -!.lH2

'.15 Find (.) Ihe !pe«l and (6) Ihe period Q/ a spaceship <»biting a.wl>d lbe moQrl . The moon', ,adi .. i, 1.74 >< Hl'm. and Ihe actel~ralion d ... 10 gra";ly on tM moon is 1.63 ml'" (A .... me Ihal Ihe spaceship is Q. biling ju", abo,·~ tM moQrl', . urface.)

, (G) ~ • V~_R __ v'(I.63 m/ .')0(1 .74 >< tit ml • 1.66 )( 10' m/s _ 1.68 kwH

(6) The orwmf ••• """ Q/IM ()fbil is

d. 2.~R _ _ (6.28)(1.74 )( Htm) _ 1.09 x 10' km

SOl 1M pt.iod is

d U/9)( IOOkm . , --- - 6.5><IO' , - .I.llIl..rnin

II 1.68 km/.

. ... AI !be e"ualor. llIe d'ec!ive val ... '" Ii is ""aUe. Ihan allhe pole>. 0... reason fm Ihis is lhe ce"lriptlal a<ttieralion due 10 Ih~ ~anh·. rotalion. The magnitl>de of lhoe ~nlriptlal ."""leralion mil>! be suOlracl~d from ,h. magnilud~ Q/ the ."",,'~.alion d ... purely 10 g. avily in o.der 10 obtain Ihe drective val ... Qf , . (G) Calcula,e lhe fraction .1 diminUlion '" Ii at the ."uato. as a resull of Ihe ea"h', ' ota,ion . Expr ... you. result '" a pe""nrag •. (6\ H_ $hQrt _Id Ihe earth', ptriod of rolalion h.~e '0 be in order for obje<U al Ihe equalO' '0 bt .. w.iJhII ..... (Ihal is. in o.der for lhe ell'ecti"" ,·.1 ... '" Ii 10 be ..,,,,)7 (~) How _1<1 the period found in part (6 ) <Qmpa .. wilh thaI of a $.lreUile slimming Ihe surface of an ai., .. earth?

, (0 ) Usin, R • .. 6.37 " 10' m and T _ 24 h _ 86400 s . .... find a _ 11 '/ R. _ 4:r'R.lT' • 3.31)( 10" mis'

TIIe .. fo •• Q/g - 3.44 >< 10· ' . Since Ii .. - , - Q. Ihe fractioo.1 dimu";li"" is (g - ,..)1, - at, - 0.344 If, eent. (6) In ",de. Ihal g •• _ O . .... ne«l Q _ Ii _ 4,,' R,/ n Solving f", 7; . ..... find T, - l,~v'R./1i _ S.(6)( 1 J -

84.4 min. (e) $i""" an o.biling Soatel1ile hI> "Ill _ rn, and , •• _ 0. ill period equal. 7; .

'.11 A particle i. IQ slide .'0Il! Ihe horizonlal offilla. palh Qn Ih. in~ '" Ihe funn.1 "'"",n in Fig. 6-<>. The '"'f''''' of Ih. funnel i> fnctionl~ .. . How fUI m""llhe particle be moving (in t~rtn' of, oad /I) if it i> IQ • •• cule Ihis molion?

, , , , .......

opynghtoo IT £na

98 a CHAPTER 6

, The fun"l'~ is C9,U;" l le"11O 1 road N nked al an ~ 'J(f' - 8. Hence 11'/" _ Ian (90'- 8) _ 00I8.OfU_ 'lI'00I8.

'-18 An automobile mov .. ar<)Und a """'. of ,a.nus 300 m aT a amslant opccd of (IJ mi . IFi,. 6-7(")1. (.) Calculate the /e. u)tanl ~hln~ in velocity (magni!OOe and direction) when the car ~ aroo.md the a~ of fH. (b) Compare lhe magnilude of lhe iMlanlaneous """",Ie.ano.. oIlhe ca. 10 the magnitude 01 the a"",age aa:e leration over the fH a~ .

, (.) Frum Fl,. f>.7(b) • .<\11 .. !i!lJnI and .<\. mat .. a 120" an&le wilh YA. (6) The in.tanlaneous """",leralion hu magnitude

.' "' ... - .. - .. ll.inl.£ , ,., The lime average """",).c.alion is 1 - .<\./ .<\,. Since

we ltave

' •• ,.,.,-.1.

ds T &8 300(,, /3) 5,. &,------ __ S 11 11 60 3 .. .,

a .-,, ---U 'mis' ... , S" fl

' . · 110 .. ,.

, . J(JO ..

t. ,

,.

" .. (>,

'-1' Wb ile driving around a ~u .... e of 200 m ,ad;us. an engi"l'e/ 001 .. that a pendulum in lhe car hanp 01 ... angle of IS' to the ""nica]. Whatlhould the speedomeler read (in kilometer1 per hour)?

, ~

, T"n 8 _ ~; T """ 8 _ mg whe •• T _ 1."';on. Thus Ian 8 __ O. " _ V,,lan8 _ 23 m/. _ 8iUkmlll , . '.lI The 00& ohown in FiB. 0.8(,,) has ju>1 1001 it> lOOting Mar the t"l' of the slati""ary bowling ball . It Uide.

down lhe ball without .~.bI< I"""ion. S/>o:rN that it "';11 I ... vc the ., .. 1""" oItM ball al the anp. 8 .. am:os 1- 48'·

",

•• ,..---~,; ~h

-, ", ". ...

ted malarial

MOTION IN A PlANE II a 99

, n.. centripetal 10=: is pven by

mv'l, .. "'8 COlI 11- F" , 0, At angle 9. the dec . ...... in pot.ntial ene.g)' . "'811 _ rngr( 1 - cos 9). mu", .'1",,1 the ;l\Cfe_ in ki .... t;., ..... ,g)'. m,,'12; hence.

3mB em 9 - F" - 2mg

At the iMtant the bug Iose$ contact wi th tile ball . f ;" " U and 0) yield. em II .. i-

'.11 A 1110·111 pilot i$ u.~tinl a ""nical loop of ...mIlS 2(0) II at 350 mi lb, With ,,"'hat f~ do<!. ,1>0 !Ca, pTC$.$ u!""ard ag.inst him .t ,he bot,,,,,, of ,1>0 l""p1 , ".' 1" - "'8- ­ , ".' ,.- ... , Fir>! we cllanae milO1 pet l\Qur to f •• 1 per 5«QIld :

5ubstitut. valu .. ,

3S4J mi/ h .. S 13 ft l~

f ' .. l~ Ib )( (51J Itlo)' .. 11iI'1I11 " iLtl!l 32.2 It'" " 2000 It

, .2:1 How many g', mu,t the pilo» 01 II>< preceding I"'oolem with"and at the bottom of tl>o loop?

, From tllc centripetal lcul.,ation lonnula ~, _ ,,' I , . Substituling .

" nl.'

Dividing th;. , .. ult by If (32.2 ft l.') . ... obt.;n

l32ft/1' ~c .. 32.2 ft/.' .. !:.!J:

0,

1"

6.13 The dtsig .... , of a roller coo.t ..... isMs the rider> to nperience · .... · .. ightleMM" ..... they I'OOnd tl>o lop of one hill. Ho .... fall m01.' tl>o car be going if the radiu, of curvatu re . t the hilltop is 20m?

, To "perie""" .... ightlessn .... the gravitational fOfCC "'8 muSt e"""tly e'lual .be "'qui",d cen.ripe.al for« mv' I,. Equating the IW<I and ... Iving for " gi'''' l!mil.

6.14 A huae pendulum am5i", of a 2IXHg ball at the end of . cabl .. 15 m long . 1f the pendulum is dr. "" n back to an lJIgIe of 37' and '.Ie .... d. what maxim um lorce mUM.1>o cable wilh'oUnd •• 11>0 pendulum owin~ back and funh?

, n.. maltimum ten5ioo will occur" the bottom ""hen Ihe cable m\t$t furnish a foeu mg .. mv'/,_ To rueh the bon"",. the m",~ lall, a diU""", II .. (IS - 15 COS 37'1 .. 3.0 m. Its 5Il<'«f th~,~ ",i!l be v .. (21h I'" .. (lot I''', n..rdor~ the t.m;on ... iII be T .. 2((lg + 200((,g-)/ 15 .. :ru&.I::!-

6.2 LAW OF UNIVERSAL GRAVITATION; SATELLITE MOTION

6.15 Two 16-lb ""ot J.PIIc"'l (as uoed in track mrell) ore held 2 It ap;ofl , Wkol is the 10''''' <Jl annc:tioo bet-..'een them1

, In American enJineerin, unito. 16 I b~ 0.~97 slUI and Ne ... ,Ofl·, law of gravitation ha. 1he form f .. G[("' ,"'"I/0I ' 1. with (; .. 3.44" 10 ' Ib, fl '/slug' _ Thus.

f .. (M4 )( 10 . 1~~::'t·~97 'IU(~ ~~.491 oIu~) .. . _12 )( W ' lb

righted makrral

100 D CHAPTER S

6.16 CI;a,late Ih. fom: of anrloClion belween two <JiO..k! .phe ... of metal spa<:<!d $0 !lUll IM;r ""'nl.~ Ire 40 em a~n.

, In 51 ~n;!5 the g.a~;talional ",",,-ant h .. Ih .. valu .. G. 6.67)( 10-" N· rn'/kg'.

F • G m,m, • (6.67)( 10- " ) 90(90) _ 3 38 )( 10-' N " (0.40)' .

6.2'7 Comput. the maU of 11>0 eanh, as.surn;ng;1 to} be I , ph .. ", of radius b:l70km.

, u,t M be the mass of 1M canh. Ind m tM ml$$ of a ",,"ain obj«t 0f1 Ih .. eanh', surface. Th. weigltt of the object ;, "qual 10 mg. IT;' 11$0 e<jual 10 tl>o gra"';ta1iooll 10= G(Mm)I". wbere,;s tbe earth'. radius. Hence. mg. G! Mm /"l . from which

M .~ .. (9.8m/>')(6.37)( Hfm)' - 6.0)( 10>" k G 6.67)( 10- " N· m'lk,' g

6.28 'The .""r~ <.W:mily of solids "",ar the sulface of the earth is p. 4)( ]0' 'q/m'. On Ihe (c"""" .... umption of • sph .. rical pla".1 of unifClfm dc:noity P. calculat. the gravilalional C<)nSUml G.

, 'The mass of 11>0 , ph .. riell e.nh;' linn by m. _ pV - 1"pR~. lnsert thi' valu .. into g. Gm, /R; and soI~. for G, oblainin! :

G • ..2!.... 3 ><9.8ml.' - 9)(10- " N'm'/k,'. 4,.pR. 4,.)( 4 )( 10' kJ/m' )( 6.4)( 10" m

lli, cakulat;"" most ""nainly ' ...... " .. I;m.1 ... G .• i"", P (and 1><"", m.) .t. untkr .. limatcs.

6.~ A ma", m, _ I kg....,igM on<~ix'h .. much on Ih. surface of the moon .. on the earth. Cakullle tht ma .. m, of the moon . Th. radiu. of the mOOn i, 1.738 x 10" m.

, On the moon. m , w.ighs 1(9.8 N).

Gm,m,

W,'" ""T I 670- n Ixm , 6(9,8).6. xl (1.738)(10")'

m • (9.8)(\.738 x Hf)' • 7.4 )( 10'" k , (6)(6.67 x 10- " ) ,

.... 'The earth', radiu, i. about 6370 km. An objcCl that h .. I mus of 20 k, i. tak.n to a hdgltt of 16(1 km .boYe Ih .. canh', ,urface . (.1 Whal is the objc.l'. mus.t this hei&!lt? (b) How much do« tl>o objoect .... eiJh (i .•. , how large a grav;talional 10= doeo it experi."",) at Ibis htiJhI?

, (.1 The maSS is!he $1m. as thaI on the canh', surf"",. (bl A.long ....... are OUI~ the .uth' . .... rfacc. the ....,igltl ( f""", of gra"';!y) vari .. inyerscly .. the "Inare of the disla"", from tbe center of tl>o eanh. loxleed .... GmMI" . ... 1>0,. m. M are the m ..... of objoect and eanb. re~i""ly, ond ,i.lhe di~ to!he center of 'he earth. Th .. ",J .." - , V, i ... """ G, "', M are co ...... n' in 'his prnblem. For "'" <*00..., .. '''' - 6l?O kin and,, " MJO km and ... ,. (20 kg)(9.8m/.') _ 196 N. This Ii""''''' _ J..86..lli. Note Ihal .... e oould use lhe floCl Ihat .., .. mg and tbat ... ;. con"anl to ~Dd lhe a.a:eleralion of 1l1 • ...;W.t the two h"igh!5. llIal ii . " /r," rl/rl.

'oJ! 'The radi ... of the .. anh i> about 6370km. whil .. lhal of Man;, about 3440km. If an objoect weighs 200 N 01\ unh. ",hat would ;t w.igh, and what WI,luld hc Ih .. a.a:eleration due 10 sra>'ity OJ! Mars? Mars h. a mU$ 0.11 th .. of eanh.

, N""'ton'. I . ... of lVavitatioon ... - Gm M/" . gi"'" ,""" w," (M,IM,j(, ;/, j). Leuin! ! rd .. , to . arth and 2 .... f .. to Man,..., hay • ...,. O. II(6370/ 3440),\200N) - lU:!.. 'The ooo.l"' llion is JOlt"" from "", ,,,, . g,I". o. g,. (75/200)(9.8 N) - .l..!.m.l£

02 Th moon omit. the canh in an appro~imatcly circular path of lIdi ... 3.6 x 10' m. It tl k", aboot 27 days 10 complete on. orbit . What is lhe rna" of the ... nh a. obtained from these data?

, 'The lU.vitational aura.<lion be''''«TIthe eanh aDd mOOn pr<>Vides the centripetal force: the,efore. m>l/ . .. GMml , ' . .... h.r. 101 is till: earth's mass. Tbcn M. u', IG. w', 'IG. Now w. 1 rev/ V days _ 2.7 x 10-· rad/ •. , • 3.8 x 10' m, and G - 6. 7 x 10"" in 51. Solving 10. M, ;\ ;. 6.0 x 10>" k,. (Compare Pro!:>. 6.21. )

MOTION IN A PLANE II a 101

6.3l '"'" wn' . m .... is aboul 3,2 )( 10' TimeJ lbe earTh', m ..... Tbe wn is .boUI 400 I;me< _ f •• f.om lbe earTh .. tbe moon is. Whal is lbe .alio of lbe ma",illOd< 01 lbe 1"'11 oIlbe wn on lbe moon to 111.01 of lbe 1"'11 01 the eanh on lbe moon~ IT ma~ be alo'umt<ilhal Ihe sun_moon distance i, eorulOnl al>d «Iua! 10 lhe wA- earth disc"""" .)

I leI m denote lbe moon ', m .... M, lbe . un ·, rna .... M, lbe earTh ', rna ... ,_ lhe cenleH<><>enler di".""" from lhe sun 10 lhe moon . II>d ,_ the cente.-to-«nler dktanct lrom the earTh to lbe moon. We let F ... denote tbe ma",itude of the gravitational fOTtt uerted 00 tile moon by tbe . un. and F_ deno1e the ma",itude 01 the vavitatioo.t I"""" e. me<! on the moon by tbe earth . Tben F_ '" OM,ml "_ and t'.- _ GM,mlr:-, "" that

6.34 E.cim.te lbe size of •• o<ky ophe.e .,.;.h • density of ,Hl glc:m' fmm tbe surface 01 whicfl you could jllSl boo.ely Ihrow a ..... y a 1011 ban al>d have il neve' ~lUm. (A$Sumc: your \)(,1 thruw is 4O m/ •. )

, Tbe csca~ spce<! u. from I ophe .. of radius Rand m .... M;, liven by the energy..",...."'.tioo cqllalioo

I , GmM -muo--2 ,

Substitution 01 M _ p lllR' and solution foo- R gives

R. ".~ )G •• p

If P '" 3 )( ](I' kglm' , then R _ o. n )( 10' t'. , "'Mre R i. in m."e" al>d " . il in mele" ~, s.t<:Ond . Eslimatin, the hi&he>! 'p«d a ... -hio:h • human can throw . golf ball .. aboul 4Om/ s. we find R " 3)( 10' m - mm,.

6.l5 Ne'A'on , wilhou. knowledSe 01 Ihe numerical val ... of tbe gravitational COnllOnt O. w~ neW'nhelesl able to ealculale tbe ralio of lhe malS of lhe SIIn 10 the m ... of any plane •. prov;de<Ilhe planet has a moon. i_I Show that f<>r a atalla. Ofbiu

wlw<,. M, ;I the ma$S of.he . un. M, tll.o ma$S oI .he plan .. t, R, tll.o dista""" of tll.o plane. from the j,IIn, R* the disurIC<' of Ihe moon from tll.o planet. T_ 1M period oIth .. moon around the planet , and T, IIw< period of lhe plane t .'OIInd the , un . ib ~ Jf tbe plan .. t is thoe eart~ , R, " LSO)( Uf km . R_ ", ),SS)( 10' km, T~ - 17) days .• Dd r. -~ 2d. rs· Calcula!e M,IM,

, ( .. I Applyin. N ..... 1Dn·' 5eC01>d 1.0 ... al>d the law of In,-;Iatioo to each omil , ... find (up."';ns cenlripetal fQfCt in tt.m. of period , T. usin, v '" 2_~RI T).

4,,·' M, R, .. OM,AI, T; R; ...

,.-Iw< ", m i. tile m",~ of t lw< .. ttllite, Sobi n, lhe abo<o'e «I~llioM for AI,I M" ...., obtain

10$ .s.-sire<! . (b~ lnsen ins t~t &iotn numerical \'.1,..., "'t fil>d

•. _ ( I SO" 10'\'( 32;3)' _pO" 10' .1,1, 3.!jj " 10'1 \J63.l

6..M (.1 Find the <)fbil.1 ""riod of a sald;I" in • eiKula. o.bit of ",diu" aboul • spheril:al pI . ... t of maso M, (b~ Fm a low-altitude orbit (, .. " ~ ' show that fQ' I lio.n .ve",~ pla""tary defl$ily (p ) the Ofbil.~ period is indepen.s.-nt of th .. size of I"" pla""t ,

236 a CHAPTER 12

--or , , ..

f'Ic. ll-li

Pd -lAlli, where IA - InrI'}IJ il lbe moment of inc Mil of the bl! 1000ltbe pivot . SillOC Ihe WIIUKC from A to C is//2 , _ al~ hne v. _ ( ..... )12. By combinin,tM eq .... tion •. _ filKl m[( ..... )f2j.d .. [(ml')/3]"" whlcb implks thlt d - 1J11. [NOIe tbe recipnxil)' "';Ih Prob. 12.36: • blow to tlw: elKl (~.p.) <;ausn rotation about 1M ~.p, (end).]

UA Refer 10 Prob. 12.39, I.) Whoal is II>< period of mcination of tl>< rod wben il ill .... pende<! from A 1 (b) Whal is tM len!th of II>< simple pendulum hning 1\\(0 $lme period? The length)'Q'l obtain \\(ore s.hoold be tbe samt ... 11>1' dillan", you obtaine<! in Prot>. 12.l9 . The ""nlcr of pel"Cll..ton rclal;". to A is.1Io ca lled 11>1' u nltr 01 0JriJ/mi"" ,elati"" 10 A.

, 1.1 The mcillalioo frfijllt"ncy is !i>"en by (I« Prot>. 12.38) ,= v~ - 2" Vat:

T~" 2.1r f!!. Vii l!o) The period 01 a simple pendulum oIl.nl lh L is Jivcn by 2-~ViJi. "Therefore Ih. length of. >lmple pendulum ,"'ith 1\\(0 .,..,ination period T~ is L .. ll/1, ,",'hieh ill eq ..... 10 11>1' dioIl nce found in Prob, 12,39.

U .• ' " rin, 01 m_ M and ... diu, R i . hun, from I ~nifc cd", . to that the ring tan . wing in ill OW" plane as I phy>i<'.l pendulum. find Ih. puiOO T, of unall mcillitions.

, The rin, is~"II in Fi,. 12· 12, with t1>l' kn ife e<!,. It poinl A . W. mw;t find lhe period T, 01 omaU oocillation. in ,he p1ar.c: 01 the pape •. Talr.in,lhe ofiJin 01. coordinate l)'Stem II 0 , the equi~brium posilion of ,1>1' ri",', ",n,,, • . "';th the po<ISili"" 1 His ...... ,.,.ng ,ow.r<lthe vie".r, the """""nt 01 ioe";' I"," MR'. By <1M: parallel.axis theotem. the moment of ioe"i. 11>001 the knife ed,e is Jiven by I,~ _1", + MR' _ lMR'; hence. G;'_2R' .

-'" L' ,'. . , ,

Subslitulin! this Ind D _ R in Eq . (I) of Prob. 12,38, "'r find .. , I i2R T, - - - 2-~V-: " ,

UAl Refer 10 Prob. 12 .• \. 1_' S~WQK 1hal an mnl~ rinl is pi"Dled from In axis PP' lying in the ring plane

ROTATIONAL MOTION II : a 237

-- L' ,--- , ;

L, Fl&. ll-IJ

and lan,enl 10 II>e ci"",mfe"'ntt (Fi,_ 12·13), This ring cln .. «ute ~U .. ,ions in lnd OUI of Ihe plan<', fi J>d Lb. period T, of those Wl:llI oscillations, (b) Which oscillation ha. 'he longer period? H.,.,.. much I"",e' ?

I (., Now ,be rin& is pivoled aboul Ihe ui, PP' .. ,oo-.·n at ri&lt'" By lbe parallel.ui. Ihw",m, I" , .. 1, ,, + MR'. As 5hoOl" in Prob . 11.41 . I _ (MR')12, so I, .. " (lMR' )/2 With G~ .. 3R'12 and D _ R, (I ) of Prob. 12.38 give> T, " l lv, .. 2.~V3R 12;. (6) 7;/T,. VI _ US47. The period for ""';lIatio ... in Ihe plane: is 15 .5 pe.cem Ion,e' , han the period for oscillation. abou, Ihe ui. PP'.

UAl A uniform ri&ltt.an&le iron i. hung ov" a Ihin nail so lhar lh. i.Of! pivot. f •• dy al ,he btnd (fig. 12· 1~)_ Each arm of lbe iron h!l5 mass m and leng'lh I. find the period T of -""all oscilla,ions (in ' he plan<' of , t.:: iron).

I The cent •• of m .. !oS 01 Ihe system is located on Ike an&le biK<:!o,. at a dillance D .. (IV:2!4) from the piVOt A . The moment 01 ineni. / aboul an uis Ihrou&lt A and perpendleula. CO lhe xy plane i. gi"en by (mf)!3 + (m!')!J _ (2mI:)/3. The"'fo", the square of the gy.ation radius i. given by 0' _ JIM .. [(2ml' )/Jt(2m) - /' /3. We can n.,.,.. apply (I) of Prob. 12.38:

, .m.' ~ /'/3 f2J2i T _- .. 2/, -_ 2ft - 2,n/ ~--v Dg !(M)/41g Jg

U.44 figure 12·15 ",p~nlS ' Ihree-dimensional obj.cCl (Il0l ... ~rily or un iform (\cn,ity) " 'hose cen~r or lO_ is at point C. The axis ZCZ' passing through point C h .. been ok.,.... al random orienlltion. The body had IYflItion radius Oc(ZZ') aboul the uis ZCZ ': Lhe 1I011Iiott Gd ZZ' ) mo k .. uplicil the fKl lhat 11M: g)'raliott radius for an axi, Ih rough C depends on the panicula. cl>oicc (ZZ') of u i •.

Suppooc Ih.t • physical peDdulum i, constructed by pivoling 1M body aboul Iht axis PP' , whir~ is parallel to ZCZ' It • dillOn« D. Pr""e 11'''1 Ihe fr."uerocy v or small oscillations about t<juililmum is sivC'n by

I J Og v- 2,T G~ZZ')+ f) '

---"1'

! J.." 11-15

I 1lIc desi red . e.ull follows II onoc from (I) of Prob, 12.38. II i. secn Ihal Ihe f .. "ueney of pendulum <>$dil ation, i, the >ame 1o. any dIoice of ui. PP' on I <ylindc. of . adius D ecnte"":! on ZZ'.

U.~ Figu •• 12· t6 depicts 11M: K,,~. fN"d~/~m , U$ed CO ""' .. u ••• 1M: aruluation or ",,";Iy wilh hi&lt "'-""ta<)'. It OOIISist. of I ri&i<I rod on whirh . bob is """'nted. "The mass of lhe bob is ",t!icienlly 1"80 so lhal lhe ecnte. or mass of the peDdulum i. fl i.ly III I.om lhe middle of the rod. The bob is """,nled on. o.iidc; ~y mo";ng the bob and I""n dampi.! il in pooition. I"" Iocaliott of I"" ecnter of mass of the pendulum can be adj u.led. The.e I .. I .... very p.-cci5<'ly honed knil • ..,dgcs """,nled on lhe . od. TIle peooulum can be .wung from

238 D CHAPTER 12

"

++-_"1--- c"" .. , ....... K .. r ....... l

••• ':"'Lhd==: (:f !X':~.' 10, - ­I ; 1 , , .1 ,

[If _ Rod

f1&. 12-1'

knitHdgt' 1. and then rc,'eroW and IWlln, from knile-.edse 2. If ~ is lhe C.llion radi ... of the pendulum aboul an 1ll.,lhrough its <:enler of ..w.o. whal are 1M: up"e~ for tho: p<riods of wWl OIIcilLation. about knifc-edge 1 and knife-edge 2?

, From (1) of Frob. 12..\/!,

T,. 1. .. 2-~ ~G! + V: ~, Vt ,

UM Ref~r 10 F.ob. 12.4S; can T, " T, for D," D,?

, Yes. Con,jdcr T for an arbitrary V and fiud~. T .. 2.~ v'(G! + V ' )/( V8). If Wi: pkol T again'l V ror D between 0 &nd.." Wi: note Ibal T _.., .. V_O &nd T _ 0> a. V_"'. Thu. T has a minimum v.I"". T _. for some V .. V". Th ... for any T ;> T _ . tho:re a.c IW<I V ~alUCI . ODe "caler and 0"" less IILI" Dot.

U.3 ANGULAR MOMENTUM

u.n Find lhe roIalional e"",gy and an",lar morncnlum duo: 10 lhe daily roIalion of the Ut1h aboul il$ Illi •. DOl. : M." 6 x 10>' k,. R." 6.4 x 10' m. OJ .. I /M400 rcv/~. Auume ,he eanh 10 be. uniform 'p/w:re.

I K." /.,'/2 .. [(2Mr)ISl[(2Jr 1M 4OO)'lf2 .. 2.6 x 10'" J. An,ula. momc:nlum .. /., .. [(2M,' )/ S](2Jr/M 01(0) .. 7.1 x 10" k&. m'/s. (Th' equivatent unit J .• is abo IUCd for on,ul u momentum. especially in the I tomic domain.)

UA Each of lhe whcc:ls 0<1 • <:enlin f"".·wheel ""hide hal a ma .. of lO k, and a .adi ... 01 c.alion of lOem. When the ca. ill goi", fCH'Ward and lhe wheels arc luminl at 5.0 ,.v/ •• what ill tho: roI&tional ki"",ic e",,'gy !.Iored in Ihe four wbccb? Whal is tho: ansWat """""ntum of the ""hide about an u i. pontllel to lhe wheel Ill;' and !hroulh !be ce-n!u of mass? Is the angular momentum v«cor direatd toward the driv .. ·• ri&ht or

""' , K." 4(/ ... '/2) - 2(30)(0. 3O)' (10Jr)' - ll!!!U. The angulor morncnlum ~ee!or too- each -..-heel ;. along Ihe Ii"" dc>cribed, 50 101. 1 L .. .f{IOJ) _ 4(lO)(O.09)( 10,.) - )4(l tg . m'(~ . Note 'ILl' '" mUS' be in •• dian. pc. SC<;Ond. 1lIc loul L will be in lhe ume di'«Iion as the angula, .-clonty. Ilia! is. to lhe dri~e.'.lch.

UM In • r:ontlnOll pIt)'SiclJ«1ure dcmonwatioDS •• 1~lu",r ~I$ on a SlooI Ibl can rolale frccly .bout a vcnical uis on Iow·friction bearings. 1lIc l«IUtCI holds with utendW arml tWO dumbbell •. each of mas$ .... and kickllhc floor s.o as 10 .. h;'ve an inilial atlJular 5pted OJ,. 1lIc lecturer lhen pulls in the dumbbell!., 501h .. 1 tho:ir di"al'lct'S from the rola!ion uis decrease from !be iniu.1 ~ah .. R, !o lhe final ,'.1"" R,. Dctcrmi"" ,he

244 0 CHAPTER 12

11.67 A g!>dent volunteer i5 sining .talionary on I pi.no $1001 willi her feet otr tM ftoor. 'Thc: >lool can tum freely on its ule.

(a) 'Thc: voIuntee, i, hande<l I """rotatin, bM:ycle ... heel which has handles on tM uJc. 1I0klinltM ule vertically with one hand. ohe ,rasps tM rim of the wheel ";!h the other and spill> the wheel c!od; ..... (u ",en from lbove). Wh.lt happell>to the volunteer .. ohe does !hi.7 (6) She llO\II' graspl the ends of tM vertical axle and turns lhe -..heel ""til the ule if, horizontal. What happens? (~) Ne.n ohe giva the routina .. heel \(ltM inslrvctor."ho lums tM allc until il if, verrical with tM wbeel rotatinl dod"; .. , as oecn from above, 'Thc: ill>ttuCtor ""'" h.ando tbe whed baek \(l the V<IIunteer. What happoll>1 (d) 'Thc: vo~unteer 1JaspI the emb of the ule and turn. th. uJc until it if, horizontal. What happ"DS """,1 (t) She continues lumins the ule until it i, vertical hut with tho .... heel rotating oounterdo<:kwise as viewed from abov •. What is 1M .. suh?

, S;""" the ule of the piano .tool i. frictionle ... tMre are no vemeal \("ql>eS exerted on tbe >lool·voIunteer .)'Stem. 10 the vertical component of angular momentum i. oonserved, (1"hore ar. horizontal mrqun; the", result from fotte" that tkl- floor •• erlS on the ~ of tbe stooL )

(<lI'Thc: iniliallngular momentum i, uro. 10 the final .... ular momentum mUSI also be uro. "Therefore, lhe YOluntee.!-pins count.mod ...... (6) 'Thc: alll"tar momenlum of lhe .. heel i. now horizontal. Tbc voIunleer', vertical component of angu~, momentum ma,t now be zero. >0 >he Stops >pinning. (e) When lhe ,.beel is handed hack to the volunteer. lhe l)'St.m of ,.heel and volunteer has a downward vertical anrulu momentum. an contributed by the wMel. 11M: volunteer remains .tationary , (d) Si""" the vertical componenl of the 10111 a"",lar momentum must not chaoge, the voIuntee. mu.t route dockwise. (t) The ... heel ·, angular momentum i. now upward. The volunteer must therdOle have I <lc>.....ward vertical angular n:omenl\lnt 10 kUP III. tOUl angul •• momentur.l I'OIntJn. <!Own. SM mull IM .. ton IpIn clOckWISe. (Her.1"<' rot. IS !WI""

os fosl os in p&rt (d). )

11.61 A IOJ> consi.u of a unifO<m disk of maSS m. and .adius,. ri&idly anad>ed to an Dial rod of ne&!igible ma ... 11M: top is placed on a .lOOOlh table and ",I spinning aboul iu ui. of Iymmetry ";Ih 'nau~, speed ... , How mlldl W'Otk must be done in ",lIinl the lOp ,pinning? Evaluate )'Ou. result for m. _ 0.050 k, . '. _ 2.0 em. and w. _ 200Jr .ad/ s (OI 6tXXl rotations per minute).

, ~ m<>mcnr of i~rtia 4 of ltoe ' Op is 8inn by 4 _ !m.,;' The """"k requir.d ' 0 "". the lOp spinnin8 wilb angular >p«d .... .. eq ..... 1 to ,he 'pin kinetic energy !""':. f or lbe gi~.n nume,wal val_. we find 4 - (o.SO)(0,0SI)(2.0" 10- ' )' . 10 - ' k& . m' , The 10Qfk r.quired is (U,5)(lU- ')(2OOJr)' - l.211.

n." Rek. '0 pfQb. 12.6tl. The tenter of ,be di,k is. <Ii ... """ d from ,be 'op', poinl of """"oct wi,h.be table. Tbc lop ;, observed 10 pre"" .. It •• di!)' aboul lhe vertK:aJ a . .. ";th Ingular op«d "' .. AMaming thOl .. , ~~ .....

wrile .... in tennS of ' •• d. "' .. and g. Evaluat .... for d - 3.0em and g _ 9.80m/s'. with the other quantities os given in Prob. IHi!! . b your mull COnoi5lCnl with 1I>e assumpeion .... « ",.1

, For ",,« ... tl>e Inlula. ,peed of steady p<ffC"';on" appmximated loy

",_td,!. • G~",.

where o! _ I"J .... _ \,;, With tl>e numerical val_ given , .... find

(9.80)(3.00" 11)- ') 1 "' . .. (0,5)(2.00)( lO- '}' 200.v .. V ol rad ii

The ralio of Ihis angular .peed 10 ... i. ("', I",.) .. 3.72 " 10- '. "The assumption Ihat ("'./"") « 1 ;, fulfilkd.

11:70 As ",own in Fig. \2·24 .• .otid o;)fIical tOJ' of mass M, hcilhl h. and radius R ;' spinnin, aboit itllymmetry axi, 00' ...;th spin angular ~ed w,. Tbc axil 0 0 ' maka an an&le " ";Ih lbe vert;""l,

We nolc that for .""h a system tbe tenter of m"", of the tOJ''' Iocal.d along 00' .t a dil1!ntt 3JI /4 from the verte. 0, and the momenl of inertia I about Ihe . . .. 00' .. given by / - d,MR' .

I.) Find Ihe a"."tar op«d "', II which .1>o'0J> plcoesses about the ven;""l. (6) Consider a lop lot- ...-hir;h A _ 10.Oem and R _ 3.0em. The lOp is .pinning at 5800 rotalions per minute . V.ing g - 9.80",,"'. eval"ale

Ihe p<ffC"ion angular s.pud w,. , (.1 Under the ",""mpeion thu "'.« ........ e un .... ume thaI

MATTER IN BULK a 251

I For a force F and displ~men! J,

It _ f _ "'& _ ( 1,8 k!)(9 ,8 mi.')

J J 0,02

F _/u . (882 N/m)(O,1)S m) f.4-U N m_4,SO x,

13.14 A Il).k& m .... is iUpponed by I sprin& ,.,hIKe con~lanl is 12 Nlcm, CO<np.ue IIIe donplion of llIe spring . , •• f.'!!l. • • p . 8,17 em

Note Ihll Ille mi. ed units for A D«'d not be con"cned Nnce "'& in newlOn, .... nce" llIe """" ..... in It,

13.25 A \oo.d of 100 Ib is 1P!'lied to tile Io"-cr cnd of a ' teel rod J ft kIntIlnd 0,20 in in dia~ler. How m""h will lbe rod "retch? y .. ),J >< 10' lbl in' flU >!eel.

, 'IA Y -

IlLfL " IlL _ LF .. (3fl)(IOOlb) .. Z,9><W'ft

A Y :reO, I in,)(),3 I< to' lb/in') f>.L • 3,5 >< to-' in

Note !hat!he lIni .. of L and A need not agr« as 100, as tile units of A """ Y 'V"", sinee (di"~') cancel!. belW<:cn Ihc>e tMl last quantilie<,

13.26 A wire w""'" ~"'" 5eCtion is 4 mm' i, '!rclched by 0, I mm by I ""nlin ""Cighl, How far will i wire 01 the same rnalCrial and IenJ1h stretch if its crosi-s«1ional a~a is 8 mrn' and llIe .. ~ ~i&hl is attached'

, If the male rial and IenJ1h are fixed as .. ell as the Mrclching wei&hl , Ihen F, I, Yare fixed in tile ",lllion f>.1 - (F1)1(iI Y), Then ,

",. ­A

A t.J. COnS! " iI , ,,"I,_il , t.J, ii , 4 mm'

t.J, . - t.J'.-, , O, l mm A, ~

t.J,.O,{I5mm

13.17 A steel wire is 4 ,Om IonJ Ind 2 nun in diameter, How much is il clon&a1ed by a iUSf>Cnded body of m ... 2() k.? Youna'. modulu. for steel is 19(iO)O MPa ,

, UI f"L lie ,he donplion , Then, by HoMe',II"',

... IIe", Y i. Young's modul .... , The clongation;'

f. y,,"L , L

,,"L " .!.!:L " ",&L _ (20)(9.8)(4,0) _ \.213 ><: W , m _ I m mm Y iI Yil ( 19(i><: 10'):r(0.001)'

lJ.2I A c:tlppe! wire 2,Om IonJ and 2 mm in diamc!C! is Inc!ciIed 1 mm. Whi! 1CTWOn n 1lCC1kd? YounS'i moduhls for cop~r is 117 600 MPa. , , 6L

- ·Y­, L f"L O,all

F - Y T A· (117,(0 ><: 10')-,- :r(0.001)' -l.8:!..1l!

0 .29 A wirc;' stretched I mm by a for« of 1 tN. (a) How flf would II wire of tile WIle material and lenllh but of foor times th.t dillrnc:leT lie ""etched? (.) H"", mucb work i. cIofo.c in metchin. cadi wi",?

, ' oJ 'The .Iongat;o" is invt'ne\y proponional to 111e ~I lfca, Ind so f" L .. (1)(1)' " G mm. ,.) 'The work done in ot'.t~hing ,he wire in the I..., case. is W _ f x , Since the force . anes linearly ... ith dinlnee ,

10)0 + 0 w, - --, - (o'OOI)-W ,

W, _ - W, .!!.9lLl1

" O.JO Gi""n. 2,O-m length of ot«l wi", with I .O-mm diameter, about how mudl willtbe wi", ",etch ut\der I

S,D-kg load? Y .19!iO)O MPI.

, Since temile modul .... is !IIresi/mllin, .... ha •• f"L / L _ ( f I A)/Y _ 1S(9I1lJl l ,,-(S ><: to-' )'( 19!i >< 10")1 -),2><: to- ' Ind ,," L .6,4 ><: 10-'m,

252 a CHAPTER 13

13031 AWroUmately how large a foroo i!. requirood tn ~t .... teh • 2.Ikm-di ....... ter ~tul rod by 0.01 p"ttent? Y- I'1S000 MPa.

, Solving Hooke '~ la ... fm F. it is F _ A Y(.1L I L ) _ .~(o. O I )'( 1 9S x lo")(lO~') _6100 N.

13.32 A platf""" i$ suspended by f()\lr wires at it. COrne~. n.. wires are l m k>ng aoo ha~e a d"iamete. of 1.0 mm. Y""ng·' modulus fm the material of tbe: wi ..... is 180OXlMP •. lIow la, " ill the platform drop (due to elongation nf the .. i ..... ) if. SO-k, Ioa<J is plattd at the center of the platform?

, .1L _ (LF) / (A y), ... ht,e L _ 3 m. A _ ,.(1.0 x 10-' m)' .. 3 . 1~ x 10- ' m'. and since eadl wi,e supporu one~uane. 01 tbe 1000d, F - f(SO kg)(9.8 m/")1/4 - 123 N.

(3 m)( l23 Nl _ • .1L - 0.14 x 1O-' m' )( 1.8 x 10" N/ m') - 65 x 10 m 0.65mm

U.33 What is the minimum diamet., of. bfaa rod if it is to >Upporl .400-N 1000d witl!wt nceeding the elMtio limit? A .. u ..... ,hal ,be "n'$$ {or tile .10$1i<: ~mit is IN Mh.

, To find tile minimum (Ii ...... t .. , a nd he"", minimum e. OOO-$<ctional arn, "'e ..sume that the for"" F _JOQ N brings US to the el .... io li mit. Then from the meu. FIA _ J79 x 10" Pa. we get A .. (400 1")/(379 x UJ' Pal _ 1.0554 x 10-· m' . """,n

,., . ~D ' ,.-­•

D' _ 411 _ 4{ 1.05S4 x IO -'m' ) _ 1.344 x 10 -• ,. .~

D _ ';1.344 x Ur· m' _ 1.16 x 10- ' m - I..l.tmm

U.Jot A No. 18 ""PP'" wire hal. dil mrier <>f 0.04 in and i$ origiuUy 10fi long. (.) What io the greate .. load that can be support.d by this wi", "';tOOut e:lcuding it. elailio limit? (h) Compute the chanae in length of the wi.e under tlU. load. (e, WhI t is tltot mnimum load that el" be ,uppon.d "';tOOut ""'aking the .,.;,.1 (4) Wh at is the mnimum elongation' (As.sume that the ellstie limit stress i$ 23 (OJ Ib/ in' and thallhe u!timate .. ",ngth ...... is 49 tXXIlbl in' .)

, (.) F IA _ 23 (OJ Ib/in' . so F _ , • (23 000 lb/;n' )A .

,.D' ,.(O.(I4;n" , _ • . , , . -- _ _ 4,.x! tn , . (b) Ai _ (F IA )(11 Y) _ (23 (OJ lb/ in )10 fl Il7 x 10' lblin - 0 om fl It) F' I A. _ (49tXX1 lbl in'l; F' .. (49 (OJ Ib/ in' )(4,. x 10- ' in') • tL1.!!! (oil AI' - (49 (OJ Ib/in')(iO fl l 17 x 10' lh/ in') _ 0.028B fI - iU:!fun

U.15 A st. el pia"" wire has an ultimate sr.ell!!th <>f about J~OOOlb/in'. How I.rae a load can a O.5-in-<liameter steel wire hold before breaking?

, F - (35000 lb/ in')A. A _ ,.(o.~ in') _ 1.964 x 10-' in' F _ (3~(OJlb/;n')(I.%4 x 10-' in')_~

U,36' A wir. of oriJinal'."lith L and ~ionaJ a •• a A. i. ",etehood. within the .Iailie limit, by . 'Ire" t. Sho'" that the density of "ored elastic . ""rllY in the "r.tched wire is i'n y.

, Let the tOOl\ delormation be flL. so that r _ Y(&L/ L). by Hooke', I ..... Apin by Hool.' , I. ", . tit.< stretchinS foroe at deformation -" is gi"en by F(x ) - I (A Y)I LJx. Hence the ""red tlM!i. cnc:'n is

L&L Ayr " ·' r' W _ F(a) ,u - - .r,u _ _ AL , lY

But. ncJ.lectill3 a tiny cha"s •. the con .. ant vol ume of ,lie wire i. V _ AL; benet:

W e V· rY

N<>te thai the .... ' n dem.ity i. independent <>f the wite dimemions.

U.J7 IXtermine the fractional change in volume 1$ ,It.< pr<$Su", <>f the atm"'phere (0.\ MPa) around a metal block i$ ",dUttd to um by placing the block in VaCU um 1be bulk modulu. for 'he metal is 125000 MPa.

MATTER IN BUlK a 253

, B _ - top Or AVIV

AV flp -(-0,1) Q~,n- ' ~---- '"'~ V B mem

13.J11 Com!",le lbe .0lufM chan,e of a solid copper cube.,j(I mm on each edJe. whem subj«t~d 10. p' ..... '" of 2O MP •. The bulk modulus for <>:lPP<'r is 125emMPa. , - VAp -,,----B

NOle Ihallhe ""ito of V and flp need not avee bta .... flp l B;' djfMMionieoo,

1l.J9 "The p<ffiure in an nplooion ch.mboer is 34~ MPa. Wltal """,Id be Ihe pc","",1 mange-;n ~olume of. pin:e 01 copper sub;e.:leil 10 lbi! pmsure? The bulk moduh.s fOf roppcr is I:J.B GPa (_I:J.B X 10' Pal.

I The bulk modulu. is defined •• 8 .. - flp l (flV IV ). whe re the minus sign il i"..ned becau .. AV;' .... ,..Iiv. when flp is pooil;VC.

1" 1 flp J.4S" 10' ,., - - 100- - 100 -~ V B U8" 10'

llAO "The rornPfl'$$ibilil~ 01 waler i. S )( 10· '· m'IN. Find lbe dea._ ;n volu fM of 100 mL of Wllef ... ben subjected 10 a pre .. ure of I~ MP •.

, We nolo thlllh. rompreSlibily. t. is ~mply Ih. rttip"""al of lbe bul k modulus . Then flV" -VI< flp .. -(100 mLI(5)( lO- lO m'IN)(IS)( W· N/ m,) = -0.75 mt.. The dcaea.. i. Ihlll~.

13.41 How larS" • p<eSSUr. must boe applied 10 ... 1 •• if ;1 is In boe oompr ...... d by 0.1 pe.cenl? Whal is lhe ratio 01 Ihis p<c .. u.e 10 almoopberil: pre"" ••. 101 kp.? "The bulk modulu. of .. ale.;' 2100 MPa.

I The volufM OIra;n fl V / V. - 10)( 10-' ; .. I B _ p l (A V IV.) 10 find p '"' (2, I " 10')(1.0)( 10- ,) _ 2100 kP •. DividinK by lhe al""""""ric pre."lfe. Ihe ralio is li.

llAl By .. bat fractKln ",illlhe voIum. of a .... 1 bar increue'$ Ihe air is •• acualtd from. chamboer in Wl\idI it r .. u? Slandard at""""""riI: pr ... ufC - 0.101 MPa and B fOf it •• 1 is 160emMP • .

I fl VI Vo _ - flP / B _ O. 101/ 160 em .. 6.)( 10- '

13.43 WIlli increa ... in pr ..... '" is "'qui,ed to deaea .. lhe ",Iume of 200 L 01 " a", by 0.00. pcrcenl? Find flV. (B_ 2100MPa)

, flV - 0.00004(200 L) - 0,008 L flp - B( - flJ) - (2100 MPa)(n:LL) - 0.(18.1 Mr. - 8:!../i.fI

13.41 Compule Ih. oompr .... bilily of &l ycerin if a p"' .... ,. of 29(1lb/ in' ca ...... a volume of 64 in' 10 d«ru$C by l)( 10-'in' ,

I "The: rornpreSlibilily ;' lhe reciprocal of th. bulk modul .... . Thu,

"

V ;.,,- ',.,' t ___ -"- _ " 162)(10-" " Ib t.p v. (29(1Ib/in' )(64 in' ) _'_ ._ In

1l.4S Two pa.al.,1 _,.., oppmile forces .• ach 4000 N. are applieillan,.ntially 10 lhe uppe' and Iow<:. {....." of a nobical me"l block 2S em on • side . Find IIIe ontH: of ..... ar and the diopi"""fMnl oIlhe uppe' surfatt ,.Ial;ve 10 lhe I"",., .... face . Th. IM a, modulus {Of 1M melal ;, IlOG Pa,

, We u>c lhe al'P'OW;male form S .. Ff(A.) (Prob 13.20). ,.;th S - 11)( 10'" Nl m'. F - 4000N, and A _ (O.H m)' _ 6.25 ~ 10- ' m' , So)";n! f ...... W. ,.1

_ (400) N) _ )( _, ., (62S)( 10- ' mil(S" lO,oN/ m' ) 8,0 10 .ad

"The dis.placcfMnl of lhe uPJICr surface j > p".n by d _ L.,. whe,. L ;. an .d,c 01 the nobe ; d '"' (8.0)( 10-')(25 em) _ 2 0 X 10-' em,

254 a CHAPTER 13

13M "The sMar modu!u. fOf • mcu.l it. 50 OCIO Mh. SuI'J"l'$e tMt • Mea' force of 200 N is applied to the upper wrfaa: of a cube of thi. mclal lhat" 3.0an on _ II edse. How la. wi11the to!' surfaa: be d;.p!aced?

I Shearing Slrain _ ALIL _ (F /A liS _ 2OO!lL '(5 )( ]0'")]- (~ )( IO-"jll. '; IOlve lor AL with L _ 0. 030 m. I1L - ].3:) X 10-' m - 0.133 I'm .

13.47 A bl<x;k of ,elatin i. 60 mm by 60mm by 20 mm when unmesocd. A lorce 01 0.2.(5 N i. applied tangentially 10 the uppe. IoUrfaa:. caU$i"l. S-mm ofuplaa:rnent .elative to the lower surfaa: (Fi •. 13-2). Find (_, the sM.rini ww. (b) lhe sMaring SlTain. l!Id (t) the shear modulus.

(0)

,.)

r----' . toO """ -----<JI-r-r J • ~ "..., .-_-CCcrrc _________ ~DTciD

. ," ,,, .. n - tan - ;; - 20 - !!..U

F/Ji b8. ~ ,hear modllluo - S _ -- _-_ miNim' t." /I O.~

13.48 Two .hccu of aluminum On an aircraft wing arc to be held together by alwnin\ll!l riveu of ~ional arca O.~ in'. The shearing $I .... on eadl rivet muSl not c_cad one-tenth of th. clastic limit for ahuninum. How many riveu are needed il cacl! rivet .upporu the ... me fTaction of I toul sbca"", force of 2S OCIO lb? AHurne thOl the .lutie ~mit s"m is 19 OCIO Ib/ in' .

, F /A _ ~(19 OCIO !b/in') _ 1900 Ib/ in' muimum Sl.e .. allowed for cad! rivet . This mealll • Iohcarins lorce of F _ (1900 lb/ i:n')(O.2S in') - 415 lb/rivet. Number of riveu - ~ 000 Ib/(4n l1>frivct) _ 52.7_ or S3 riynJ.

13M A 6O-q motor >i1O on lour cylindrkal ",bbc. bIocko. Each cy~nde. h"" • height of 3 em and • trooHectional a.ea of IS em' . "The oltcar modulus fOf thi. ",bbc. is 2 Mr •. (.) If a sideways force 01 JOO N is applied to the """or. how flf will i\ move side ... ys1 (bl With what fr"'l~r.cy will the moIor vibrate back and f<)flh sideways if diSlurbcd?

, (_) We ...... me tMt the she.r force it. distributed evenly among the lour cylinder>. Then fm a gi~en cylinder F - 7S N.

F 7SN ~. AS - (IS X 10-' m')(2 X IO' N/m')· 2.S >< JO- ·.ad

The displaocmcnl it. then J _ L • • (3.0 em)(2.S >< ]0-' rad) _OmS ern. (.) Siooe the shea. lorce on cath cylinder;. pr"l"'n ionallo ~. it i. IIWl proportional to the borizontal displaocmenl d. F _ AS~ - [(AS)I L)J. Sintt there are lour cylindeR. lbe toul nlCmal horizomal force. FT. it. Ii""" by F,- - I(!AS)I LJd. This (or« jU$l b.ol~nccs the elwie restorinl lorce. or effective '.opri", .. forcc. of the l)'Stem F, - -I(.AS)I LIJ· If lbe Iohcar force is .. """",d . tbe >)'Stem oscillates wilh an cf(ectiYe force oonstant k _ (4AS)l L _ i .O X 10' N/m. AHWning that the ma!iOCl of the cylindeR arc negligible, we ILave

f - ..!. I! _..!. /4.0 >< IO'Nfm _ l..l.1Il;. 211 V;;; 2Jr" !ill kg

13.50 "The twioting of a cylindrical shaft (Fig. 13-3) through an angle /I is an uampk of a $heari", strain. An

anal yoil. of!be si!UI!ioo """""' !IIac Che In&le of!...rn (in radi3ns) i. v'cn by

'" 8- ,.SR'

where , _ appliHi !Qrq~ I. ienglh of C)'li""'"

R a ,..tillS of C)'li""'" S. shea, modu lus fOf ma!erial

MATTeR IN BULK a 255

[f I !orq ... of 100 Ib . IT i$ 'Wl;"d !o the t IKI of a C)'lindri","l Sl •• 1 shaft 10 fllong and 2 in in dla"",!e, . !h.oo,h b.ow many <kgt«I .. "iIl!be shall !...-iII? A$$Ume ,hal S _ 12 >I l(flb/in'.

I We I •• Ii'cn S a 12 >I 1000 Ibl in'. R .. 1 in. , _ lOO[b · fl _ 12001b· in. 1_ W IT _ 120in. Ulin.!be liv.n formula .... 6nd !ha!

2r1 2(l200lb· in)([2(1 in) _. ,. <R' · (. '",.f)( . ,,_7.64 >1W , ad .. o.~n·

" .. " I . >I I" ,n lIn

IJ.51 An en';"" deli, ... 1-*0 bp a! 10) rpm 10 an f!..fl solid· iron dri'·e shaft 2 in in diame!e •. Find the up. of !...-iII in!he dri'·e wft. A$$Ume !ht S _ IO>l 100Ibl in'.

, First caLcullt.!he co.que . lhe n apply Pro!>. 1l.5O .

.J ft· IbiS) P • 1-*0 h,,\ 55O""""iiP • 7.7 >< 10' fl . Ib/.; '" • 2Jif • 2..~(801"h;od'.)(] min/lilh ) a 83.78 .ad' s.

p _ Tw. P 7.7>< ]O' fl· lb/ . . r___ _ 919Ib · f1 _ 1100l 1!>-,n. w 83.78 .. dls

Finally

CHAPTER 14 D Simple Harmonic Motion

1'-1 OSCILLATIONS OF A MASS ON A SPRING

11.1 A spring make. 12 vibrations in 40 •. Find the period.nd frequency of the vibration.

, T • elapoed time ~ 4O'_l..l..i Vlbrations made 12

, _ vibraTion, made _ ~-2..;.l!!.l\l;

cl ...... d time 40,

U.l A SO-8 maloS lIanp al Ill< end of • Hooke"" sprin,. When 20. more .~ ..:\dod 10 I"" end of the sprin8> it ",elch .. 7,Ocm mO« . 1_) Find lhe spring oonstanl. (II) If the 20, at. now removed , "MI will he the period of 1he motion?

I (_) Since lhe spring is linear,

• l!.F (0.020 kg)(9.8 m/,,) , I

- - - .. . 8Nm t... O.07m ---.. , r. 2,.. ~ .. 2,,~o.riSiHI . 0.84 $ Vi UN/m --

1'.3 A spring if, ""'tched 4 em when a mOM of 50 . is huna on il. [f I 10111 of 150 I ill hung on the srrin& and the m .... is otartw in a .... ,.,;""1 oociU.,>on, wlto' will tho period of tIM ~Ultion be?

, Fim find Ihe spriJl$ CQ",,-anl k;

To find the period uS<'

k .. (SO &)(1180 an!.') .. 12 2SO dyn/cm ,=

T .. 2. .. /i a 2. .. )1;:0 -llI(O.II01) - 2&!ll

14.4 A body of weight 27 N hiUlp on I long spring of .""h .tilfncss thaI In U tll fora: of 9 N ,tretches 1he ",ring O.O:S m. If the body is pullN downwud and re\eased. wlLat is iu perind?

, The SPrilll <OlI.$tant is"" 910,OS .. 180 I'l m. and $0

T - 2:. @_211 ~l7/9,8 _ fJ:dl Vi 180

1405 A 3-tb weiillt lLa"lll at tbe end of a sprin, whioh h'" A _ 2S Iblft . If tbe ,"'eiill' is displa«d sli&l1t1y and released. with what frequency will it ,ibn.le~

, Usinl'" - 3Ibf (32 .2 ftf.' ) _ 0.093 slug V<H

f _ J... I! _1.. I 2S 161ft -lil..I:i! 2..~ 'I;;; l1< VO.093 slul '

W.' A """" m suspended from a _pring of constant Ie hIS. period T. If. m ... M is adde<I. tile period becomH JT. Find M in terms of m.

, The period .aries •• tbe square root of lP1e rna ... Th'" lbe m ... m"" i""",ase ninefold. matin, M _II!!!.

W.' A OS·kl body perform. simple harmonio motion with I ffl'qlltt>q' of '2 Hz and an amplitude of 8 mm. Find IIIe maximum "eIOOlyof tbe body. ill mui:num acceknuion. and ,lie ma.imum rHtorin, fora 10 which the body is subte<ted.

, W. are gi""" freq""lIoC)' v and amplitude R: v _ 2 Hz: '" _ l1<v _ ~11 radi o: R _ 0.008 m. Then " ... 1\;1,. ,,.

HYDROSTATICS D 273

15.1.5 find the ralio of a $)'5101j(: blood prtl4\lre of ]~ (in mmHsl 10 .. IIr\CIOJlheric: preI$UIe . Standard .. lmospMric l'fC'Sure is 1.Il1 x ]0' Pa .

, Pr .... " '" d"" 10 0.120 m 0( m.mory _ Prh - (13 600)(9.8)(0.120) _ 0. ]6" 10' Pa. R.tio .. O. 16/ 1.0\ .. ill.

15.1' If the blood ve!KII in I buman hein& acted .. simple: pipes ("'hich they do noe). whal would he the dilfercnoe in blood pr"'u", hel",een 1M blood in a l.80-m.taIl mao', fed I nd in h.is head ... hen be is stlndiO,? A .... "'" !he $pCcltic .,..vily of blood \0 he 1.06. ,

1lI .17 Whal ;' the: pres.'lure d"" \0 lhe ",.Ie, 1m; henealb the ocean', surface if we awunc the mc:an demiry of se . .... ter 10 he 1025 kgl m'7 II its compre14ibilily is lhe ........ as thll of pure "," Ier (bulk modulu._ 2IXlO Mr.), by wh.1 pcrotnl bas the densily cbanled;n goinl from lhe "" rface to tbis dcprh (1 mi - 1609 m)?

, top _ Prh .. l0l5(9.8)(]609) .. ]6.2 MPI . By de~nition of lhe. bulk modulus. B .. -tlp/(llVIV ). But . for a filed mass ,",

"fIIetdor<: B _ 6p /(<lop / p), OJ

<loV _ _ <lop

V ,

15.11 A cytindrical tank J It;n di.meler Ind 4ft hiy. ;. filled ";Ih wate, " \Iose """Y.I dcnsiry is 62.S lb/ f1 '. find lhe puge prcsIure It the bottom 01 the tank .

, U.., lhe equation for pressure II a dcpl h 1 in I liquid. Isnore limospheric: pre .... r • . P~" d.l" 62.S lb/ft' " 4 fi _ 2SO Iblft'.

l5.lJ A lank contains. pool of mertllt)' O.30 m deep, coyered wilh I llytr of ..... ttt lhal is 1.2 m dup. The deMi ly of watu is 1.0 x 10' kglm'.nd that of mercury i, 13.6" 10' ks/ m'. Find lhe p, .... u'" en rtcd by I"" doubk larer of \iquKIs l llhe botlom of lhe tank . I""". lhe pteHute of lhe almosphe,e.

, fim nlld the pressure al the lOp of lhe memory pool. FOf I poinl below tbt- surf_ of lhe mercury 1m. mly he regarded .. IlOUrot 01 exlernal prC$$ur. P' M' Thus

P ... " P. _Ih __ .. (1 .0 " 10' ksJ m'}{9.8 mls' )(1.2 m)" 12 kP.

"file pteMUJ"C P .. exerted by 1M memory column itsell i$ found in. he ... me m""""" p_ " p_,h _ _ (1).6 " 10' k" m,)(9.8 m(,')(O.30m) - 40kPI

"file 10111 preowre II lbe bottom ;. thus 8..!f!.

1lI .lO How hi,h does. mercury blll"Ofl><'c, stand on. da~ .. bt-n a'mospheric: prC$$Ut. is 98.6 kPa?

, • P_ 98.6" 10' Nlm' ." 0--- - 7.....,mm P"",, (13.6" 10' kglm' )(9.8 mi.')

C!I.lI Two liquilh wh;ch do noe relOC1 obemiQlly I re placed in a hent tube . I. >lKn>n;n Fil. 15-3. Show Ihal lhe heiy.ts o f Ihe liquids abow tbeir surface of xparl lion ar. in'· .... ly p. oportiorullio lhe i. densilie • .

"

FIe. 15-J

, "file pressure II the inlerlace musl be 1M ... me . o.kulal~d .-;. eilher TUbt-. Sin«: \10th lubes arc opcn 10 the alma.pbcre . we must ha.e p" It , .. p,gh , . or 1t ,l h, . pJ p,.

274 D CHAPTER 15

15.11 Assume .ha •• hI: .wo liquid$ in .hI: U·.tI.pe<!.ube of Fig. 15-3 .'" ...... er ar>d oiL Compu.e .he densi.y of.he oil if.he "" ..... ands 19em above.he inlena«' . nd 1M oil "aods 24 em above .he inIOna«'.

I We .""I~ the .e ... lt of Prob . 13.21 :

('0) ,,~ P .. - - P'--2~-(llXXIkglm ') - 792 k &lm' II.. 4 em

15.23 A uniform glall lube i~ bent into I U.m.pe ~""h '" that silo".,. in Fig. 15-4. Wa.e.;. poured into .he tube until i. 51aods 10 em hi", in """b .ube. Ben •• ,.,. (.p gr _ 0.879) is Ihen added "0 ... 1) 10 lhe ,ube on 'he lefl unlil.he ..... e. ri~ 4cm higher (In lbe righl. Whal!<ngth illhe col umn of benzene ,,'hen .he ~it".,ion;1 ",,,,,I><d' (Wate. ar>d benze"" do not miJt .)

I An 8-rnt column of ,,'ate. Nlance •• he benze"" column ; ... by P.ob. 13.21 .

h. _~_ 8cm _ t.l.ml P, /P. 0.879

( . .. ~----

Iko,." ... " . to< •

'0 ___ \ f" •. 15-4

1!!.l4 The ma''''''''''e • • hown in fig. 1S-5 "IC' mer<:Ury IS its fluid. If a.mospheric p .... ure" lOO k!' ..... hat" tbe prt'SOIl'" of lhe gas in Ihe conlai"". ,hown on the left?

, AI poin" A and 8 p ..... u ••• mUll be equal ,i""" the fluid i. no. moving, Therefore P - p. + PI"-100 kPa + ((13 .6){9.8)(0.12)t kPa _~.

15.15 A mercury barome'er ".ndo a. 762 mm. A S" bttbblc. " ' ...... vol""," is 33 em' " 'hen i. is a, .h. bo" ODl of. lake 45 .7 m de.p. me. to the luffacc. What is its volume al .he OlIff""" of.he I,k.?

I In term. of lhe weight density. PI. 0 ' ..... 10 • •

P ... _ - AA!I + P_ - pgy + p(t; ),,,, .. - pgj45.7 + (\J.6)(o. 762)1 - 45.7pg + IO.4pg - 56.IPS

For .he bubble. Boylr ·, I"", .. a, .. that pV _ ron,tanl. alluming thaI ,he ,empc.atu.e .tays fixed. Tbcn.

Y _ .p-y _ _ 56.IN" 33 - 1lt£!:!i p_ JO,4N

lS .UI A small unilorm ,Ube i< ben, ,nlO a cirde of ,adius,"hose pia,.,. i. venical. E<jual volun>« of lwo ftuids .... hose densities arc p and 0 (p :> o)~!l half the circle tIC. Fig. 15-6). Find lhe angle .hal 1M rod;u. paWn, Ihrough the ;nle,facc mak .. "";lh the v~nkal.

HYDROSTATICS a 27S

, Of !he ulemal for.,.. actin, O'n tbe IWU fluid oegnenl. ""Iy lilt two Vo1:ight •. NV and 03'V. have tO'rques about tM "",nte. 0 ; the IO'IUS exened by the rontaine •• re pu.ely radial. Thu •. for equilibrium.

0 .. pgV, .in (.45" - /I) - "8V. !in (45" -+- /I)

o .. p(sin 45" COS /I - COS 45' .in /I) - o(.in 45' CO!I /I -+- COS 45' !in /I)

0_ p(1 - Ian /I ) - 00 -+- tan 6) ,-. tan6 _ _ _ , ..

15.17 Rework P.OO. 15.26 by . equirill,lt that the ~\lid p.ess •• es be equal O! tM inte.I..,."

, 1be presoure at the interf."", muS! be the same for tbe heavy and li",1 liquid. 10 ho'", equilibrium. Rele rnn! to' fi~_ 1}-6, ". <;:akula'. the pre .... r •• , ,1Ie in!erf&« due IO'!he lIe.vy liquid and.." it equal 10 lhe prcswre due 10 Ihe li"'lliquid_ (n.. added OO<ItribuliO'n O'f llIe gas in ,he reSt of lhe tube ;' the _ klr both liqu,ds and can be i&nored.) 1ben pgh . _ ogh , . .. here It . and It, are the heish!> of the re'!'«'liye liquid> above ttoe inte.fac.: . We h."" , from fig. 15-6.

It , _ , "'" 6 - , "'" (90' - 6) _ «cos 6 - si. /I) It ,_ , coo 6 -+- , !in /I_ ,(coo /I -+- sin'l

Substituting inlO the p .... ur. equation and canceling g and , fO' r both sides. we ha.e p(roo' - !i. /1) _ O(COO /I .. sin 9). '" di'iding by roo /I. p( I - tan /I) _ 0-(1 .. tan 9). Soh;ng fO'r Ian /I. we KCI Ian /I w>

(I' - 0)1(P -+- 0).

15,28' Water stands II a doprh It behind tnc, , 'en;...! fllO< of . dam. Fi,_ 15-7(<1), It ex.n, a .... ullanl horizonl. 1 fO'ftt O'n Ihe dam. tendinl 10' !.lido it lIon8 il$ foundation, , nd . lorque . «nding 1o ""enum Inc, dam aboul!M point O . FIT1<l ,<Ii "'" I'oorl"",al IO'ftt . (1)) ,he .O'rque . boUI O . aT1<l Ie ) ,he helgl<' . ' w!l;<h '~e .... ultan. 10'"," would h.,. to' acl to prod"", , he $lOme to'rqll<'

r ,.J ,,'

, (<I) F~ur. 15-7(1)),, a view O'f Ihe face O'f the dan from ull!uream. The preHure .1 depth y ;'P" pgy. We mal' negl,e\ , he atmospheric preHUU. !inee i! ICI. on ,he ",he, l ide of the dam allo. [The conuruction '.II<"' .. n in Fi! . ] ~. 7(c ) may be u..,d '0 jU>1ify the ""gl«t of " """",heli< """",,r.-l 1be fO'rce api n>! the lhadc:d strip i. dF _ P dA .. pgy/ dy. The 10101 force i.

(b) Th. IOrq"" of Ihe forc. df ' Oboul an a, i. IhrO'ugIo 0 i •• in magnitude. d , . (h - )')iiF - pglyjlt - y)iiy, 1be totallOfque abou, 0 i.

[ pgllt' r_pgl y(lr - y)dy _ _ , _

(e) 1111 is Inc, Might abo\.., 0 at \\'hirh the lotal lorce F \\'O\Ild ha"e 10' oct to p<oo"", Ihi' !"r\jut.

HF_ T f pgllt' 16 h

H -'F - pglh'/2 - 3

15,lt' A conical cup.' _ (b - z ) tan n . • em op<'n-<:nd·oo" .. 00 a omoolh lIal ... rf""", . "' 1Ito\\." in fig. ]5-8. The: cuI' n ,0 be ftile-d to. heigh, h \\i,h liquid of deMi,y p . 'lihal will be the lifting fo= On ,he ('Up~

276 a CHAPTER 15

• f'II. 15-3

, 1m"""" tha, the iMide owface of ,he gjp ronsisu of an infini,e numbe. of infini,esimal rin,-shape<! SI.po (Fi •. 15-8). The I"ns>I'. PI:) actin, on ,be veni ... 1 face of a $1rp does 1\1)( co.uribu,. '0 the ~lting force, since;\ acts MrUonU':Il~ . Th .... lhe infinilesimalliltin. for«' if. only ,he p<nsuTe fO<cc on , he horizontal face of 'he SI.p:

dF, .. p(:) dJ\ - Pf(1I - : )(2, .. d,) • Pf( ~ - z )l2J«b' - :) '.n "'I( -d. ,an ... )

In,egTaring ,0 obI.in ,he ,otalliltinl foroe . ~ ~,

F, - -2"1'8" tan' <> f (II - :)(b - :)d. - lI",( 1Ht' -~) ,an' ...

1.5.lO' Redo !'rob'. 15.29 ... ing 'he .. tigh' of ,he "uid and the force i, eun. "" thr na, $urface.

, 'The '01al I~m: elened by ,he pressure of. Sla,ic ftuid on its ooma;"". i> equal '0 the ftuid', ... igh, . Hence. for this problem. F. - F, ...... ... he •• F. if. ,be downward foltt on lhe pia". surface; F, if. ,he ~lting 10m: On ,he ClIp (by .ymmeny. ,he horizontal p<e>$IIre foren on the ""P cancel ,; and .. if. ttoe ~igh, of ,he liquid. Now.

,., .. ",V .. '" f If" dz - 1'8"J< ,an' ... [(b - z)' dz .. (prJ< ,an' <>,(b'lI - bil' +~) CorIHq".ntly.

as befort".

1.5,31 A. ohown in .'K. 13·9. a .. eishled piSlon holds oompr....,d P' in • tank. The piSlon and i •• weights have. rna .. of 20 kg, "The Cf'OSHCctional area 0( ,he piston if. 8 em , Who, is 'he ,,,,.1 prnsure of the , .. in the IInk1 Wbal ,"",uid an ordinary P'e$$u •• gauge on ,he tank read?

n, . ....

, The IOId ~re ..... re in ttoe ,ank will be the pre ...... of 'he a,moophere (abou, 1,0 x 10' Pa) pi,,. ,he p'O$$ur. due to the pislWl and weigh ....

p _ I.~ x 10' Nl m' + ~~~::~l:' " I,~ x 10' Nl m' + 2.4~ x 10' Nl m' .. 3.4' x 10' !I1l m'-&ill

A preuure PUle on 'he toni: wouid read 'he dilferencc bel"ccn the prC$$UIe inside and """,ide lhe tank: gauge r.ading _ 2.45" 10' N/m' _~. I, ••• do ,he pruou.e due to ,he pis,on and weighu,

15.ll Rd~r '0 !'rob. 15.31; ... ume 'Mllhe ~.n 0( the ,ank bonom i. 2Oem', (.) Find ,he ,oW for«' on lhe bonom of ,he tank and rompare i, wi,h ,he ...,ight 0( ,he piston plus ,he fo"," 01 annos"he .. on the piSlon. (The ... iSh' ~ lhe oompressed ai. ;" ... glisibl •. ) (b') How do )'00 nplain lhe ••• ul1?

HYOAOST A ncs a 279

1.5.2 PASCAL'S AND ARCHIMEDES' PRINCIPLES; SURFACE TENSION

15~ A bydraulic lift in a RMce SUotioll has a lar.e pkton 30 em in diamete r aDd a ....,11 pi:l.ton 2 em in diameter . (. 1 What 10"", i. required 011 the: =all pmon 10 lift a IQad of 1.500 k,? (.) WhIt is Ih. ~ .... re iloCfeue due to the force in lhe: confined liquid1

, (_) pasar. principle: says that the: p=r. chanse i. ~nifomlly ""n...,ined throup.oullhe: oil. so Ap _ FdA, _ F,/A,. whe:re F, and F, Ife the forocs otI the: "",aU aDd OIl the: lar,. pi"OrIS. ra.pecti...,ly. and A, aDd A , ". lhe: K1pKtivc areU. Thl,l$.

F, 1.500 x 9.8

>«2' / 4) - .II{lO'/ 4)

MuUlpl)'ln. boIlt side. by .11 /~ and ""lYing lOT F" ,... obtain

(0,

F, " ( ISOO)(9.8)(l') _ ~

'" 1; M , top - - .. -- - lIN/cm _ 2lOkPa

A , .~(2'/4)

15.«1 A hyd",ulic tift is 10 be \I5ed to lift a tnxk washing ~ lb. If lbe diameler 01 tbe lar", pi"011 of tbe lift is lit . ... hal gauge prt"Mur. in lb/ in' must be . pplied 10 the: oil?

, The "'''8<' prnoure in lhe: oil am. by pascor. principle. 00 .he: bon"," of tbe larcc piston 10 prodlOCC tbe fora: tltat lilts lhe: load .

F ~~ 6370. top _ - _ -, • - - _ 6370 Ib/It'_ - _ 4-4 .2Ib/ m' A .II. n(i}. 5,) 14-4

CU I In a hydraulic press the Jarec p"on !lao trOi5HeCIional area A, -lOOem' and the s.maU ",stOll has CfO!.S-SCdionl1 . re. A, _ 5 em'. If I force of 2SO N is applied to lhe: $III'U piston. what is lhe: fora: F, 00 tbe large pi"on?

, By Pas<.:<Il'~ principk,

prnoure under lar,. pi .. "" .. pre,",,,,, under smoll pilton

A, "" F, - - fi .. - (250 N) - J..Q.ll! A, 5

15.42 for tbe Iystem $bown in Fi • . 15-14, lhe: cylinder on the: lelt,.t L. !lao a mass of tiOO kg and. aoSI·SCdional Ifel of !OJ em' . The piston on lbe right. II S. bas croos-w:ctionol un 25 em' aDd ncJliai~ ""';&11" If the: apparalus i. ftlled with oil (p _0.78"cm'). MIa' is thc: f"fCC F required to hold the .ystem in eq>;ilibrium a • ..... , ,

,

flI. 15-14

, The pre .... r .. It ""inti H, Ind H, are equ.l lincc they Ife" the same level in a .. nJle connected ftu;d . Therefor. ,

prC$lur( I I H, " preuure at H,

(prnou re due to) ( p,noure due to F) ( due Ii f '1)

left pioton - aid ";ghl;Utoo ... pr ... u.. to m 0 '"

(tiOO)(9.8) N.. F ... (8 m)(780 k.1m' )(9.8 mi.') 0.08 m' ll " 10" m' ...

SoI~;ng for F givn it 10 be l1.l!.

r hIed IT ria

280 D CHAPTER 15

15.4l A block of wood wei&hin8 71.2 N .00 of spmJl< 8ra~ity 0.15 i. tied by. win, to the boItom 01. tank 01 lW.t~. in ord.r to ha"" the bloc~ totally immelVd. Vihat i. tM tension in tM string?

, TIM: block is in equilibrium under the Ktioo of three IOf~M weiJht "' .. 7l.2N, the te"""" T. and tbe buoyant force B. with B .. '" + T. We can determi"" B since B .. P,AV., where P. is the dcn$ity 01 water and V. is the voh""e of 1M t",oIly immer>ed blod.. '" .. /1.& V •. where /I. is the <!wllity of the block. TlM:n .. I B .. p.l pL " (IPP) . .. 0.75, "" B .. ... / IL 7~ .. ~.9N. Finally, from Our equilibrium equation. 94.9 N ... 7UN + T. 01" T - Z,1..11:! .

15.... A mcuol NlI ~;ghs 0.096 N. When sU$p<'ndc<t in wain il has In apparent ~iJht of 0.071 N. Fioo lhe okn$ity of the metal.

, TIM: desired d.n";ty is Ki~~n by p -mlV. But since the volume V of the Nil is II ... the volume 01 di~d wlte •• the buoyalll fo,oe is Ii""n by B "/I~_gV. ThIlO .

.. (""" )p _ .. (0.096 N)( 1 )( 10' kg/m' ) .. 384(J k. 1 , P B (0.096-0.07I)N ",m

u,~ A block of maleriollw a dcnoily p, and fIoals lhr«-fO\1rths sub ..... ' •• d in I liquid of unknown dcn$ily. S"",", tbat the okns.ity p, of the unk""",n liquid is given by p, _jp, .

, By "rd!i ..... oo· principle. /I , V" .. p,(JV/4'jf . .. itich g:i\~1 /I, -Ip,.

l!I,46 ",. d.m.ity (Of ice i. 917 kJ/m' . • nd lhe Mppmximal. okn<ity of lhe scawale, in wbic:h an icebe', floats is IIl25 kclm'. Vihal fTactioo of tM iceberg is be"..lh lhe ..... Ie' SIlrla<o:?

, By Prob. 1S.4S. Irrr.ction ... b ..... 'ACd .. p,/p, .. ~ "!U2.

l!IA7 A bIod:; of wood N . I ma .. of 25,. \\1M" I $-, ..... tol piece .. ith • volume of 2 .... ' is auache<!lo tM bollom of lhe block, the wood bartly fIoal$ in wate, . "'''hal is 1M volume V of th. wood?

, TIM: dI"«tive dcm.ityof the ')'Il.m i, JO/(V + 2)g/cm '; I\(nce. by Probs . 15.45 and 15,46.

. JOI(Y +2) I .. fr.cllon submcr,cd - 1

SaMng. V _~.

l!IA " ~ of wood weiabs in ...... units 10.0, in air, \\1M" a he.~y piece of melal is ... spcnded below it , the ..... uo1 being wbme'l¢(! in walt • . 1M · ..... ci&h' .. of wood in li r plus metal in walcr is 1".001. The '"v;e;gh' " ... htn both wood and ..... 'MI art ... bmcrge<l in ..... ,t.;, 2.00g. Find lhe vom ..... and ,he den$ity of ,be wood .

, Since in both ClI>Clll>e "'.1-01;' submerge<! . tl>e diff.r.""" in _;Jht is only tl>e buoya'" fonot on II\( wood, 01" (12.00" 10- ' )(9 .8) N. Th.refore . lhe volume of \1\( wood can be found by eqUltinSl1\( weighl of di>placed waler. 1(((l(9,8)V. to this. Then V ... 12" 1Q-0 m' . It. rna.. is 10" IO- ' kg and ... il> dcn>ity is (1 0 )( l o- 'll( 12 " 10- 0

) .. 830 h Im' .

15.... What is the minimum volume of. block of wood (d.n$ity _ 850 k&Jm' ) if it is 10 hold. 5O--kg woman .ntirely above the ..... t.r when sI!e ".nIb on it?

, TIM: wornan', weiJht ph,. the blod ', we;,ht mllOt be equolto 1M buoyant fo"", on lhe iUSIlwely ... bmcrged bloe~. SQg + 850Vg" 1000V" which leads to V _Q1l.l!£.

l!I,SO A man who$c weigh, i!. 6f>7 N and ~ <knsil~ i. 980 kg/m' <:1In illSt fIoal in wat. r wilh his head.~ lhe $urlac:c with ,I\( help 01. lif. i..,ket which is "holly ;"' ..... ..-1. A_nUnS ,hal ,he voIu ..... of his head is n of his wholl< volume and Ihal ,he sp«:iJl< pavily of II\( ~fe ;'cht is 0.25. find lhe volume of the life jac~ct.

, Tht man', volume is

. "" V. pg .. ('180)(9.8)" 0. 07 m'

&j ... linS ,1>0 buoyant 101« to the w.ight of the man pillS ItIt ~;,ht of the life iKket.

p._ ,IIW + V"l - ff>? + (O.25p_18V,1

ngntea IT

HYDROSTATICS a 281

SoNinl .

(667{p _ ) - (141 IS)gV 1.1667 - (1 4/tS)(9.8)(o.07) , V._ 1.1.753' - (c.I.7S)(9.S) _ QOOoIm

15.51 An irregular piece of Itl(tal " weigh$" 10_001 in air and 8.001 when wbm< rled in ... ter . (., Find the voillltl( of the metal and its deMity. lbo' If the .. me p;e~ "f metal "'ei~ K_SOg """"n im"",rsed in a partialliu oil , whit is the density of the oil?

, B"",yan1':';-~'~-i(1';'~"i-:'~)g N • ..,jlh 8 _ 9.8 m/l'. This eqllal, the _ ight of tIM: d i$plACCd wale. , P. Vg. p_ - l (O) kglm' .ci~'" V - 2 >< 10-om' . The deMity - maso/V - (10 >< 10-' )1(2 >< 10· ' ) -in oil;' (1 .30 >< IO-')g N. The _i"'t of II>c displaced oil;' p. (2 >< 10-*), .

15..52 A heaker panly fiUed ",ith _ter b.as a total ma .. of 20.00 g. If a pica: of ",ood havins a density of O-IOJ"em' and volume 2.0"", ' i, fIoated on the "~'ater in the beaker. h ..... mud! will the beaker " weigh" (in II1amo)?

I The sale mllot lUppofi both the original 20 & and the (0.80 gI"",' )(2 em' ) .. 1.60-8 bllxk. 50 the •· .. eighl .. - 2Ui&_

15..53 A be.ker when ~rtly filled ..,jlh water has lOIal mus 20.00 , . If a piece of metal ..,jlh deMity 3.00 "em' and volume I _OOcm' is suspended by a Ihin Wing to lhat it i!. submerged in the ..... Ie. bYt does not rest I)CI the bottom. of ,he beake., how m..m docs , I>c heake. then .ppear t" weiSh if it is resting on a said

, The len';on in the Ihu ad i!. e<j""llo the ,.-eight of the metal ~s the buo}"nl ror(:e_ The buoyan, fon:e..,jll he (I x I c.I- ·m')(I(O) kglm' )8''' 10- ' , N • ...mcK, .. 9_8 ml,' _ The wei..,1 of the melal i!. 3 )( 10" , N_ ThcrcfOfe . the 1h.ud exert. an IIJIward force of 2 x 10- ', N_ 1I.r>ee the scale supports the 101<11 weight I .... the I<nsion in lhe thread . Ther.fore . the appaKnt ... eight .ead by a saIc will be (23 - 2) , . or~.

Eq";""lontly we ... " VI ,to. ' e<ul, by "<>ting ,h .. if ,to. "'~'.' u ..... an ~pw~td buoYSd' romo of 10· ' , .. 9.11" 10· ' N 00 ,he metal , by Newton', third law ,1M: metal .. crt. a like for"" downward on lhe wale • . Th ... ,he $CI.le hal""""s tl>c weight , 0.196 N, plll' the downward fOfU of the me1.11. 0.0098 N. for • lOUlof~_

15.54 A ...tid cuhe of material;' 0.15 em on eadt edge. It floats in oil of deMity !OH" m' ..,jth one' lhird of the bIod: O<It of the oil. (a I What is 'he buoyant fum: on the ""be? (h) Wb at i!. ,he den';ty of the maurial of the cube?

, (a) The blod i!. in equilibrillm so P. V. g _ BF .. pJ2V.13)g. Sin"" V. _ 4.22 x 10· ' m' and Po" !OJ k"m'. Bf _ 2_2 1 x 10- ' N_ lbo) Sir>ee BF .. p. I', g . .... find P." 2pJ3 _ 533 h i m' .

l5.5S A o:ubical (OWI" block is I.SOan on each edge. (a) What is lhe bYoyanl fora: On it when it is Sllbmer..,d in oil for whid! p .. 820 k&lm"? (h) What i!. 1M tension in the "'i"llhlt is sUPl"'rting the block " 'ben lUbmelJed? Pc." 8920 k&lm' .

, la) The buoyaM fora: eq .... l. the "'ei,hl of the displaced liqu;d . OF - P .. Vc.8 • ..,jIb the volume V .. 3.311 x 10· ' m' and p _ 820 k" m' _ Thu. SF -1!.!!Z!1::!.. (h) The forces act,nl on the block aK the te"""" T up, BF up. and the block ...... ight Pc. Vc.S -8920;3.311 x 10. °)(9.8).0.195 N "",-i"1 oo...1Iwa,d. Then T _ 0_295 - 0.027 - lUM..I:l!.

l!I..56 A balloon having a maSS of SOl) kg ",maiM SUIopCndN motion I .... in the air. If the aiT density i!. 1.29 kglm' , wtta, is ,he volume 0( the balloon ,n cubic melers?

, By e<juilibrillm .nd Archimedes' principle. ~ _ t. 29sV, or V _~_

15.57 A cylindoo\ wooden buoy . of height 3 m and ma .. 80 kg, f\o;tu ".rt;"ally in water. If its spcal\c gr"'ity i!. 0.80. how mIlCh wiU it he deprc$$Cd ... hen . body of rna» 10 k! is placed on it> IIpper ,"rface?

I By Arcl!imcde$' principle, thc .uh=.~ed hei! ht, 10 , of the IIn~ bYoy is Ii".n by

10 _ p- 3 - (0_80)(3) _ HOm P._,

282 0 CHAPTER ' 5

Unck. loading, 'h~ wb .... rg.d lM:igh. is di.eclly prop<>nional '0 'he tOlal w~ighl 0. mau .

.;+0 .... 1!O+1O 10 to - .- -~ 0. M .. iiO h ... iiO 2. 40 .. \lJ!l.m

t5~ A luk conlainl w.le. O!1tOP of m~mlJ}' . A cube of iron, IiOmm along ~ach ~d,., is lining upri&hl in equilibrium in lhe liGuids. Find how much of i. i. in each liquid. 1lte dcMilies of iron and .... mll}' are 7.7)( ](1' kg/m' and 13.6)( 1(1' qlm'. resp«l;vely.

".

M ",""ry nco ISo IS

, Lei '" .qualthe diil.""" ... b ..... "'d in "",.euf)' . Then (0.06 - x) equal' lho di>!.OrI(e protruding inlO lhe .. .,.te •. The net 'enicll force due to lbe liquids is prA. - p ,A. . wll..-re p,. p, a •• lhe preuures al tIM: lowe. I nd upper r,,,,,, of lho block (sc~ Ma. IS-IS) Ind Ai> lhe foa area of the block. Fo. equilibrium we hne III..­wei&hl otlhe iron "',. p,A - p,A. (p, - p,jA. The pTeSill.e di llerence is j ... t p, - p, . p..g(O.06 - %) + P,008'''. Then (p , - p ,)A '"' p.g(O.06 - .. ).-I'" p..,pA. !NGle ,hatlhe 1 .. "0 t."", on lhe n&ht ~p~scnlll>c weighl of di.placed .... 11.' a nd mucul)'. 'esper:ti'-ely; .1Id lhe exp.mion Od lho ldl is lho buoyanl fo.«: . Thus A.ohimedcl· prinripk holds .\"¢n """'n lWO (0. mo •• ) liGuids Ire displ_d. TIlil ;" Iru~ for Iny shap( object, 1 W. arn now !GIve fOf l from 11M: equilibrium equalion. notin, Ihat "',_ p,gV, _ p.&(o.06 - "')A + p,,.g .. A; or canceling lhe g. we hl\'e

17.7 " 1(1' k&! m' )IO,06m)' _ (1,0" 1(1' k&lm' )(O.06m - .. )(o.06 m)' ... (13,6 )( 1(1' k,/ m' )x(O.06 ml '

and 7,7(0,06) _ (0.06 -.<) + J),6>- 0.40 - 12.6:r

% _nOll m -n.mm - deJMh $II~ rged in mtrcul)'. lnd~ prol.\I<k, inlo "",Ier.

IS.s9 A slender boornot:entOus rod of len,tn U fIoal. panly imm ..... d in "'at~r. being >upponed by a itrin8 fOSlel>Cd 10 one of in ends. IS pictured in Fi,. 1S-!6. If the specifif; JI".vity of tIM: rod is 0.75. whal is lhe frar:-rion of the "n,th of the .od tha. exr.nds oul of lhe wate.~

,

r". ISo I.

, Since the buoyant fom: &CI5 Ih.ough 1M cem~. of "avity of tke di>placed "'at ••. lhe rondition fo, ro!llionll equilibrium is. taking nKOm~nt$ aboul. point 0 II"", the line of ar:-rion of T.

0_ L T,,- "'/ ('0$ 6 - F.(U - j) COS 6 _ P .... gA.(U)(f <Os 8) - " __ ,,A..« 2/- j) <Os e

- (_,I "_,,A 00$ 6)(,,: _ 4/" .\- 4 P ... I') P--,

"'here A _ "'oss $«Iion a! ar ..

ed mak:f~1

HYDROSTATICS a 283

f rom this . x, - 4U + 1.001: _ O. ')r x _I. J/ Di0C8.ding.he oonphysical root . "'t .,,,: .hat or>t·half.he rod u •• n<h out <>1 'he wa.e.

SlriC1ly s-p .. king. lhe above WIUlion is ooly app",,,imalt .i<I<C Ihe "Ole •• urface does 001 all . he rod pc:rpel\llicula.ly . H""",,"OT. rhe e .. '" will be r>tghgible if A is ' ..... n.

IS.6CI The w.,ight of a balloon and lhe gao il conrai " . is 11.12 IN. If ,he balloon dis-pill«S 1132 m' of air and .he ... ·.ighl of I m' of air i. 12.3 N . ... hat is Ihe a=!e •• tion "";th .. -him II>< balloon boa'n. to ri .. ~

, The .quation of motion of tl>< balloon ;,.

F. - ... ( 1I12}(12.3) - ( 1. 1I2)(10') , ,0--_ . O'2_ ~"lmh· m (1.112)( 10')19.8

15.61 A .mall block 01 ,,~, of den,ity 0,4 x 10' kg/m'. i. s-ubmtrged in ,,'at., al 0 drplh of 2.9 m. FiOO (.) Ih. oocdeU.lion of II>< block lo,,·a.d ,I>< .ur/au ,,·hen ,he block i • •• !ea .. d .• 1Id (~) ,he time for Ih. block. to ,ueh Ihe ",rfa", . [100" 'iKmi'Y.

, I.) Sy Ar<:himede>' princip[., ,he "". upwa,d 10"'" on IIIe block is r - P. _&V - p_,V, ,..1Ie ... V is Ihe ""Iume of Ihe block. Then

QO'!O' P . ... ,gV - p_ SV "' (P--_ I ) ' 0 (...!... _ 1)(9_8) _ 14 1 mi" m p_ V p _ 0.4

'" ,j¥ = , ___ \I~ __ o"" • 14_1

15.61 A body of den";ly p ' is- d.""""d f.om .c" al. hci&ht It inlo. l.ke of dcns.ity p, whe.e p > p '. Neglec' aU di .... pative efleer. and ca[cu late (.) Ihe 'P«d of lhe body jWII before enrering 'he lake, (b) tbe _Iera,ion of 'he body " 'hi [e il i. in Ihe lake, 000 (e) .he maximum <kp!h 10 ,,·hioc:h.1Ie body ";nk> belo ...... ,uming to float on Ih. ,urf .... .

, ,.) The .pew jUiI befor. en.e. ins tan be determined f.om rome .... '"'lion of me<hani<al ene,gy in IT.., fall. 01" di.ect[y lrom the kinemalioc:equation.off .... fall.yie[din!vO.Viilo. (b) The bouyan. force of lhe lak •• B, is grule, Ihan the _i&ht <>I ' he body ..... since p > p' _ Choo!.ing upward '" !""iti"e, we MI'~

,·.I><re B _pgV In" p 'V '"' v _ .olum< 01 bod)I

Then. con«l;na V on hoth >ide>,

(p - p·)g .. p·a up., .... d

,<) To find ,he ma.imum deplh "'. h ... v' . v: .. 2IIy, wi,h v. ,he ''Cklcity al y .0 (from pan d). u .. 0 is ,he .. I"""y.' mllJlimum <kplh, and y is .1Ie ""gat; •• disploament!TOm tl>< ... riace '0 the maximum drplh, Then

'"' IS.63 A soap bubbl. h'" I .adiu. of Scm. [f the soap ",[ution ha ••• uriace ,elision TO' 30)( to-I N/ m . ... ~t i. the

gaug. PIC»U'. wilhin 1he bu_~

, Coruid< •• I><m;.pl>< ... of II>< bubble (Fi,. lS-i7), The .k....-n~ord force 01 "'ri~ ,~~ on e,..,i., or t l>< tWO bubble , urface •. inside and ""Bidr. i, 2.", T. For both ",riaces lhe 10101 force F is F _ 2(2." ,T)O' dpA, .... he ...... i'lhe •• e. of lhe fta, ci..:u la. 10'" oIlhe hem iophcre. ( Sec, e,g. , Prob. 15.36.) Since A "' ,,,r.

4T 4T . (30)(10" ) brT _ 4p(,.-r) - _4p 4p ___ _ .t.!.fj

, , 0.05

15." Find on e.pr ........ fOf lhe heighllt Ihal I liquid of drn";'y p"";l1 ri .. in I capillary lube of , adi .. r if 1he ,uriacc ,ens;"n of ,he liquid is r and the meni ...... m.kn an angle 6 "'ilh lhe lube, as >ltown in fig_ 1S-18_

, r i~. f",« pe' uni, lenglh aOO poin" in difle,enl direC1ion •• "",IId . he tube, as shown, AI ~q"i!ibrium .

HYDRODYNAMICS a 291

" A",.a ' I' -t _ "., L" , ', r: -:- J." - -,1 I 1

1 ~ ~'~"'. " !Xl \

Y· " ~ ,.].,,/:" "

16.n 1"1>< ~ninl ""a, the bonom of the '"<'loSel in Fig . 1!>-9 has lin "Tea Q . A d;"k is I><ld opinst 'he """n inM 10 k .... p tile liq~id, of deni'ty p. from runnin8 0~1,

\., With ,.·hat net force <10<. t l>< li<j~id P"'$S on the disk? (6, The di,~ i. m,,,,,,d aw.y from II>< openin, a slLOM distance. The liquid !.quirtS OUI. ,uik ing the disk if>(:lasti<;ILIly. AfIOT Shiking 11>< disk. tl>< ,..arer drops vertically downward. Show tbo, ' '''' r~ . , efled by t"" ,...Ier en t"" d;sk i. twitt tl>< force in pari • .

--, ,

.1&. 16-9

, (.) The hydroslJllic p.e»u •• on Ille i"';d< $Uno"" of l ILt dil t is gi"en by p. _ P" .. + pgh. 1"1>< air pre\SIL.e of Ihe outsid< of the disk ;s P. - p_, Since tl>< disk has .... Q. Ihe net out·.,a.d rOT",,;S (P. -P.)a- l!S!!2. (6 , Once II>< disk i. , em(Wed, II>< nuid quickly al1ai ... a (rel,'i'e ly) "cady no ... . Torricelli'. lheo, em implies that the nil opced i. u " V2if,. Since friOlion i. being ;~,ed. thi>;, the exil .p«d """'"" the entiu opening. TIIC,dou th. rnaM ftw: '" .. pow .. ",,"I/iiir and the flu. of rigllLlO"a.d mom<nlUm i, (poOt lOt - 2pgh ... If tbe wale. Iosn it. entiu right"'ard fnOmen' um ItS it Slrik" ''''' disk, ,he disk must """,.h rnomc:n'um "I I"" •• 1< 21'1"". Th., i< . it ,.,ill upe""""'" • righ,wa.d fNce Jpsh ... .... hi<h;s "",ice 'he hyd."" .,", fo. ce f<>und in p'" • . NOI. that the force d"" to the .tm<)$p/leu c. "",,1s on Ihe left and right of the disk.

16.1II A Hat piate m(W .. no.mally to. 'ard a diw:harging jet of water .. the rate of 3 mi •. The ;'t di!<:ha' g<"> ... ·ale. 01 the •• 'e oro, I m 'l~ and., ~ 'J><e<l of 18 mil, (., Find ,lie: ro«;e on ,he pia'e due .o the je, 000 (6) compare i, with ,hat if the plate were ".tiQn •• y.

, We do pan I> first, With no otbe, information " 'e ~ume 11>0 piate ""'" the forward motion, bu, there ;. no botI....., bark . i. . .. the walrr ~pla,,," along 'he pi'''. at rianl In31 ... ,0 , he Qrig:inal motion, Then the force r>OTmal lo tbe plate equals the time rale of cIIo"ge of momentum along,he di.<cti<ln of II>< ,."" e, ;Ct . or F _ (poro)v . where the term in pa.enthe~l is t l>< mOlSSllime hitting the pi"'e and g i$ 11>< ~~ional area of the jet. We are given ~ .. 18 ml~ and aOt .. 0 I m'I •. jJ _ 1/XlIl kMlm' f - (1000 kg/ m' )(O, I m'I')(l8 mls) -""'-",

In paM .. tM piate i. mO\ling t"",'aN the $Iuam a, 3 mls . Two thinp ~ff«l a dlange in the momentum Change/ time. F"rnt if the liquid Igain splasb<. ot righl an! I,,. to the plate it has pi<kOld up a "cloc;ty of 3 m/s opfJ<I$it.,o the j.f. di.ection . 1"1>< tot~1 dlang. in f ....... ud vdocily i. therefo •• n(Jt u _ L8 m/~ but _ 18'" 3 _ 21 mI •. Scoond. tbe ma .. of ,...te, hitting ,he piate per ICCOnd i",,03"" from au'O a(u + 3), Noting

292 a CHAPTER 16

that a" (0. 1 m'/$jl ~ """ Ilan for tM fora:

F ("), 'j' 'j (lOOOkg/m,)(O.! m'ls)(2! m/$j' "C('~1

= p - ""'~ 1t"'~ = .,~ u 18 mi.

16,19 A pump draw> .... 1.' from •• eoer;oi, and send . illhrough I IIorilomall>ose. Since Ih ..... te. "an,.t I"esI

and i~ set into motion b~ the pump. tlte pump mu.t deh • • power P to the ... ole, when the IIow rat. is ¢r • • ~.n if lIuid friction is ... gligible. A ... '" pum p is \0 be orde.ed "'hid wiU pump .... t., through the sam • • ystem It a rat. 41 ' .. 241. What mu ... bt 1M pow", P' of 1M new pump? A .... me that friction is ,riU ... Jligib!e .

I Sy 1M II<Ork ....... 1Y theorem.

KE imparted to 1M water KE vol. ... a,er , , PK . .. X ,,~x~ .. u

hme \'(II. wat.. tim.

The m"~ lIux 41 i, prnpon ional \0 ~ so thai P ~ ¢r' . Thu~ if 41' .. 241, tM ... w powe, P"" SP.

n.lO The U tub<: 0' Fig . 16-10 ront';", a !en,th L of • ..,ro-';srosily fluid Show lhat the hid column osciUate. lit. • r.i mp!e peoKlu!um of !ength L/2.

Fi(. 16-10

I Let p denote Ihe densily of the fluid , M 1M ml .. 01 the fluid .• oKI A tM CToss·sectiooal a ... 0' the lube . !f one end 01 the nuid "'llumn i, depr.,sed. dir.canc. ~ btlow Ihe equ ilibrium le~.1. 'he Olhe, end mu .. (",uminK incompr .. , ibilily) ri ... on equal di>!"nce~. The weight d ... to the height difference , h, is a . .. toring force. f _ -",A(h) - - h . Thul """ hl.e SHM. of f.eq ... ncy

,if ' ~2"'A' ""', " 1 - 211 V"M - 2tr ""M - 21! Vp;4L - 2.~ "Vtn 16.2 VISCOSITY, STOKES' LAW, POISEUILLE'S LAW, TURBULENCE, REYNOLDS

NUMBER

)6.3) A numbt. of liny sphe. es made of " .. I with denSoily p .. and hl';ng.arious radii, ... ,e released from . .. , ju.t uoKle, the .... rface of. tank of wate •. ",hos-e density i. p.

(al Show thai the " ""1 .. a';tational force" actina on I sphere (the combi""d effect of ""';lIIt and buoyallC)') I1as magnilude (41113). :{P, - p)g. (/01 A$>uminK Ih.al Ihe fluid /low around each descendins .... he'" is lami ... , . fit><! Ihe lerminal speed u or I spI>e.e in te"", of, .. p" p, and lhe ~iKO>ily ~ of lhe Wlte •.

(weighl ) - (buoyant force) _ p.g (~ I!'~) -1'8"(j I!r.) _ 4; ' !(p, - p)g.

(/oj When the ",her. is deoccndi1l3 0' terminal .peed, the ""t fo.", must vanish, SO we can <'<IUlIle the downward, "nel ,.a';totional fo...,'" to the upwlrd ';KO\IS draK. whitlt by SIMes' law .. M-~,,1t. Thus

SoI';n! for lhe terminal ~. we obtain

1',32 Deorribt an ide.liud rxperiment fur defining the """fficirnt ol";O()I)$ily, " .

" J

I Fi,u,e 16-1! ..t.ow. two very lar,. pa.allel pl.les A and 8. ~paUled by. dist.na: d. The space btlwecn them i. filled ... ilh fluid . A OOMtant fMec f mu .. bt applied to plate 8 10 keep it mo.i1l3 II • ronstlnt speed v. with 'especl to pla,e A . If I thin lamina of fluid II. unifo.m dist. na: y from plat. A moves ",ith I sp«<I

HYDRODYNAMICS a 293

,.. .. '< H • ...!.... f-------'~~-------=:':':-· 1 '=1:13:1:: - ",

, -, j"= - ",

j j ,-• • ' .. 11

v(y) • v.,yld, II>< flow is lamina, _ n..11 it is touOO that " .. the shea. mess on thoe liquid . is proponional to void aOO tl>< proponionalil)' conStant is tl>< viKOsity. '1_ Th ...

0," '1vJ d ( ., From (1) it is !Cen that the 51 unit of viscosity is the pajclll· Jrt:Qltd or P"iJ""iII~' 1 PI - 1 Pa" _ 1 N · I/ m'_

16.J3 A rolalin,-<:ylindtr ~Ier is emplo)'«l to measure tile ~flititnt of vi>cosity of alStor oil at a temperalu ,e of 2O 'C. T11c r&diu. of the inner <ylin<lr. is r, " 4,OOem. aOO tl>< .. dius of II>< Oule. q~ndtr is r, .. 4.28 em. n.. inner cyhn<lr. i •• ulmKrge<! in lhoe oil to • depth 10 .. 10,2 an, When lhoe ouler C)1indtr i, roulin, at 2O.0tev/ min. tile lonion bal • ..,. relds a lorque T .. )'24 x 10-' N · m. FiOO Ihe v;"""ity of lhe <aSlor oil.

, n.. qlindrical .pace belw""n lhe oule, and inne, qlindtn ollhc viscometer i, r>OI a bad _wro. imation ' 0 tl>< idtal ftal.plal .. $)'$, .. m of Fig_ I().I I_ H .. ..,.. by (I ) ol Prob, 16,32. ~ - o.(dlu.) ,

n.. .. "ned .r-ta of Ito. inner <ylin<lrr i. Q _ 2",.,10 , Thus )'Uu """,,e o. _ F (2.TT,Io. wllete f i. tM dra! toroe applied to the inne. qlir>der by tile nuid, ..-him is dri"'l1 by Ille outer cylindtr, TIIi. force . ... u]" in a torqlloC T • r ,F. which can be rud on the Kale ol the lo .. ion Dala..,.. In te"", olthis torque . the shear SlreM is 0 , _ TI(2:r.jA),

Si..,. ,be inn." c:yli nder;$ at r.,t. the rela,;ve .peed u. of ,to. t .. -o qlin<lr .. is 1Ii,-en by L'._ "",. where ... is the an~ular speed oltlle O<Ile. C)'linder , ,>.1Id tile diSlance between th .. cylindt .. i. d .. r, - r, .

U,i", tile .bo1.'e v.l ..... ol 0 .. v •• and d. )"OU obtai n

1nscning tile numeric~1 valu", gi>fl

T(., - ") ,. , 21< ..... ,.,10

(3. 24 " to -> N . m)(O.Z8 " too' m) " "" DO ~ .. - 'LZLJ.J

2. .. ( 20.0 revl~:,::" ' ad1tl'V)(4.<IO" 10-' ml'(4,2!j " too, m)( 10,2" 10-' m)

16..M /10'" la.1 " ill an ~Iuminnm $ph<tl' of radius I mm falllhrouglt ,,-ate, ~t W'C once ito t~rmi ,,"1 Sf>Ce<! has bun ruched? Auume lamina, now, I>!' y (AI ) .. 2,7; ~~ .. 8)( 10- ' Pl .1

, Substituling the d.11 in ( I) ol Prob, 16.31. we lind V" 4.6m/ •. (AC1ually . the auumplion ollaminu ft"'" fo. this proolem is not r-talil ' ie; s(t Prob, J6,51.)

16.35 A 'ypi<:al ri~erbor"" Iii, panide hal a r.diu, of 20 11m and I deMity ol Z )( 10' kg/m'. The viscosity of ..-.te, is 'PPfO. im.tod by 1.0mPl . Find thoe t~rminal spud .. ith " 'hi<:h...rn a l"'niele "';1I ... nle 10 the bottom of. mution~ ,"O!ume of .... Ie •. (Unl ... the .peed ol internal nuid moIio", is sm.Uc, Ihan Ihis senlinl spee<!. lhe oi lt p.n",l", .. ;11001 ... ,,1e to troc t>oItom.)

, U,;ng (I) of Prob_ 16_31.

", • Z(2ti " IO-"'~~~~O "-I~~~" lO' K'I.80) .. 8.1 )( 10"' mI •• 0,87 mmls

16..36 Refer to 1',00. 16.35_ Snpp' .... tna!)'Un 6lkd. I·L oonle of $(km' c."" S«\ion with ,,-ate. from a mwJd)' ,i"" .• uch a. the 10>0-0' Mis,;uip!'i _ Af'er all int~rnal motion> of the water it ... lf had itoppc<l .• bout how lang would it t.ke for all the ,;It ", seUIe 10 tile bottom?

, Since tt.. heip.1 of tt.. bottle "ill be aboul 200 mm. and iince all lilt panicl", ( in particul.r. tlK>\e a' tM

294 D CHAPTER 16

lop of Ihe bonle, which go· .. ern Ihe ""erall ""nling lime) take Mgligiblc time to arnie'"e the terminal speed,

. 200mm . ""e .. l1""nling llme - o_ ,-""

.~' mm s

" .37 A tiny glass sphere (densily 2tiOOkllm' ) is I •• falllh'ough a val of oil (p - 950 kg/m', ~ _ 1).21 PI). In 100. i. ;. observed 10 drop 4Jcm. Ho," la,1I" is .he sphe",?

, Subs.;.u ••• he da •• ar>d " _ 4.J x 10-' mi. in ( I) of Prob. 16,31 10 ~nd " m 0.52 mm.

16.311 In a ceruin cenlrifu", .he liquid ;. rouled 01 2O . evl. al • • adillS of 10"", f. om .he • • i. of . otalion_ Tiny spherical particles of .. dius band dell$ily HI!O kg/m' in» dilule .... Ie. lOIulilm (densily lO00k&!m') '.e placed in lhe cenlri fu", . Find lhe .ermiMI speed wilh ",h;"'h lhey ",,1I1e oul of I"" w/ulion, Ignore 11>< .ffttt of gravity. ~ . _ _ 8.0 x 10-' PI.

I The: paTricIe Meds. ""n.ripelal fortt m",' , _ 1(4Jrb ' p)!JI(20 x 2Jr)'(0.1O) _ 6. 7S x 100b' N 10 ka:p il mo:wing in a circl._ A liquid (wal<r-.rompmcd) p;lrticle aloo ""eds a cenlripelal force 10 keep il moving in a circle. Since ooly lbe surrounding liquid an s.upply Ihe ""ntripe •• 1 force. partiel'" more den"" Ih.n "'aler ",ill p;I" Ihrough 10 the O\Iler N!< of llIe cenlitug •. This io Il>en ""'1' s.imilar 10 partidC$ falling Ihrough .... aler in • gravi ••• ional ~.Id . "The aClual Sla.ic force supplied by lhe " ,,'.r in Ih. ""nlrifuge is just I"" d«li". '" buoyanl Iorce" ar>d must equal II>e centripet.l force on an equi"alenl volume of wale. (using I"" same , • ..oning U Ihol for Ardlime<les' principk in a gravitational ~eld), Hen"" , SF _ (100011020)6.75 x 10'1>' _ 6,61 x 100b '. The ,emainin! (OK<;S ... pplied by II>< vi""", friclion force (6. .. ~Iw), Hence. 6. .. ~b" - 0.14 x 10'1>' . wilh ~ _ 8.0 x 10-' PI. yic:ld! v _ 9.3 x 1O"b' mls.

16..39 The viscous force on I liquid p;I$$ing .hrough a length l. of pipe i~ laminar I\ow is given by 1';, " 4l1~LH~ . wl><re ~ i •• h. liquid vi!OO5ily and v~ i>.1Ie ma.imum velocily of .he liquid ( i.e .. along lhe """ .. ,,I axis of lbe pipe). Find an • • pre .. ion for v~ in a horizon.al ""gmen' of pipe in lerms of p, and p, . • 1>< pr ... ufe a •• he bac~ and fO""ard end! of lbe P;pe. ond in trrrru; of ~. L. and r • • he radiu, of Ihe pipe .

I In Ileady 8"", .h. vi"",u, force is balar>eed by lhe fOTte doe lolh. pr""'ore differ.nce al.be back and fronl end$. Thus p ,lI" - p"n ' _ ~Jr~l.,,~ . Solving for ,,~we h."" v_ "' I( p , - p,)"II(~ ~l.).

16M Wh~1 i. lhe 1"""""'''' d,,,,, (in mm Hg) in Ihe blood O! il p......-. Ihrough • apilla.y I mm long ar>d 2 I'm in radiu~ if lhe speed of lbe blood through lhe center of .he capillary;~ 0.66 mml.' (The vi ...... 'y of whol. blood is 4 x 10- ' PI).

I By lhe "'SIll! of Prob. 16.]9.

_ • _ (4}(4 x 10- ' Pa · .)(W- ' m)(6.6 x 10-' ml» _ 2600 Pa _ (2600 Pa)(1 mm llB) _19 S H p, P. (2 x IO . m)' 133 P • . mm8

16.41 U,ing Ih. ", ... It of I'l'Ob, 16,39. fir>d an expression fa< .he volume now "lIe of fluid Ihrough a ""s men' of pipe. In laminar now lhe .,.."age IIow veloci.y "'." a cros/i oect;on ;s v_12.

I "The volume flow ... e. H. is lbe prodllCl of.be CTOS&-~iooal arn of lbe pipe ar>d II>< ."",aSe 110", ",,\ocily, Th , .. 1/ .. (Jrr'v~)12. U.i"lllhc .'pre:ssioo for ,,_ from Prob. 16,)9. _ ho"" H_ 1lI"'(p, - p,))I(811L). This;s known O! PoU~~iII~'$ Ill"".

16.42 How much pow.r u deti"" ,.,.. .t lbe hack end of lhe capillary of I'rob. 16.40 in pushing Ihe blood lhrough? Allume p, - 10.0 kPa

, P _ power .. for.., x ,'elocity _ (p, ... , ' )( ,,~ {2) _ p, H. by Prob. 16.41. Thus

(1,0 x to' Nlm' )(3.14)(2 x 10-· m)'(O,26 x 10' Nlm') _" P - 8(4 x lO-' PI)( IO ' m) _4.1x1Q W

l ' AJ AIIUmin! all el"" remains.1Ie .. me. bo ... .....,uldlh. now rale. H. change if lhe radi ... of. pipe j . doubled? How ... ould .1>< "" ... ~ • .....,$5A1)' 10 pu>h Ih. nuid Ihrough mange? How .....,uld lbe fluid velocity change?

I Referring 10 Poiseuille's 1.", (Prob. 16.41), and ... uming thaI p,. p ,. ~. and L f.m.in COMlant . .... IC. Ih.1 doubling' .... i ll i .... e ... H si" •• n·fold . FOf Ihe po"-er P ..... h1'-e from Prob. 16.42 P _ p,JI and lhe po_r increases s.il l"en·fold as welL lbe ,-elocily "an.. a.,' (""e Prob, 16,)9).nd therefore only quodrupk-o.

GAS LAWS AND KINETIC THEORY a 327

I (,,)

(II) llIc alomic ma .. of hydrozen is 1,008. 50 Ihal 1 mol of hydrogen (HJ coolaiM 2.016 g. Or 2,016 )( 10- ' kg. llIc de.,.ily of Ihe hyd'''len ;" lhen

nM (6,13)(2. 016~1O-·) : " P"'y· 10-' oo l,24 k!!,m

(e) The atomio ma .. of oxy,en;" 16. 50 Ihal I mol of 0, COIlI&i", 32,. or 32" 10-' kg. llIc MnsilY of Ihe oxnen i. Ihen

_ ~M _ (6, 13)(32 X 10-' ) _ 9.6 kg{ • P v 10- ' 1 m

28.11 A oomprCWJT pumf'S 7(1 L of ai, into. 6-L lank with t .... lemp<'ratun: rt"maining unohanged If .UIM ai. is originally at 1 aim ..... hal iI the final ab$oJuto pre .. un: of Ihe air in Ihe IIok?

I Usc Boyle', I . .... 5i~ tile temperllUR i, con$lal\t.

p, Vo- pv p. _ 131m V .. 6 L

1(76) - p(6)

:to.' A O.025-m' lank COIllains 0.084 kg of nilrogen gao (N,) al I pUle pressure of J.1711m. Find tile lemp<'. llun: of the '''"' in degrees Cel';w; , (p_ "\,013)( 10' Pa).

I We note lhal the maM of 1 kmol of N, i, 28 kg. From 1M idell gas II .... pV . IIRT, wilh " ... 0.084/ 28 kmol . P ... ), 17 + 1 _ 4.17 aIm .. 4.17 )( 1.013 )( 10' N/ m' . R .. 8314 J l kmol . K. SubSlilulin,.

4.17(1.00 )( 10')(0.025) - (0.084/ 28)(831 4)T and

.. 28(4. i1)(1.0i3~ 10')(0,025) " ,"' .... , .. _ 4""K .............. 0.064(831.1)

28.1' A panially inHlled balloon conlain, 5(X) m' o f Ilelium .t Z7"C and I-31m P'C$$u,c. Whal;" the voIumc of the IIelium II lJI altil ...... of 18000 fl, wi>ffl, Ihe pressuR ;" O.5atm and Ihe lemp<''''lure is - J ~

I f o< a confiDed gas.

p,V,.p,V, T, T,

v ... SOO(Z70) • ....,_, , 300(O,S) =-=.

1O.1l An ai, bubble reLea .. d al the bonom of a pond expando '" four lim .. its original volu ..... by In.. lime il reaohcs Ihe surface . [f atmospheric pressu re i. 100 kP .... h3t;" In.. absolule p . ...... re allhe bollom of the pond? A'I'ume COfI"anl T. , ,

Po· P V. • (100 kPa)(4)_~

• 28. U A pre!oSIIre gause indiell1e& tile diffeRn0e5 between atJl"OO$pheric p<C$$llrt" and pr=u,e iR5ide In.. lank. The

pull" on a I.OO-m' o~y,en lank read< 30 atm. Aft~r .ome uSC of th< OXYlen. the lJIu,e rea'" 2'i atm. How many ""b~ me,"", of oxy,en 01 !III.mal almospheric pressure were ~d? Th~re is nO temperaturc ohange during tM time of con.umption.

I Si""" 1M t~mper.tu'~ i. fixe<I ,.e ~ Boyle', la,. to >(INC In.. problem. llIc lotal p'C$$ure in Ihe tank hao been rcduoed from lilo 26alm. AI the I,tler pressure . Ille '"' orilinally in Ille tank would occupy l! m'. Si""" 1.00 m' .emains in In.. lank, tn.. amOUnl of po u...:! ,.,.. " m' al 26 aim. At the same lempe"'lu,e and at atmospheric prC$S<l.e, lItis....,.,1d occupy a volume: 26 tim." a. lar,e' l.OO..t:J(.

2O.U An air bubble of volume V, i. rele.oed by. fish 01 a deplh A in a lake. The bubble rises 10 In.. surface, Auume conllanllemperalure and standard a lm05pheric prC$S<lre abo. e lhe lIke; wttal i, ,n.. volume of lhe bubble iUS! befo. e louching In.. surface? llIc don';ly of th< _Ie. ;' p,

I EqualingPo V. at deplh A '" pV alille .urface leads 10 Ihe equaliotr (p + pgA)Vo. pV. f,om ,.hid! V - (l+pgA /p)V. ,

328 [} CHAPTER 20

20.14 If 2_1212 g of a mon.tomic gas Ott\Ipies 1.491.. "'hen the temper-Iolure is O'(:.rod the preSoSUre i. 810.6 kPa . ... hat is the 13$?

, pV", (m IM)RT. "''''',e m _ rna .. of P' and M _ "",I«IIlar .. ..,ight .

0, 9 0-' , (2 .1212)( 1O-' k8)(8.314J l moI · K)(27J .15 K)

(810.6)(1 Pa)( I_4 ~ I mj - M

M _ 3_ 'N" 10- ' kg/mol _ )'1'1 s/mol.:;> gas i, ""Ii"m

lO.L!i Use the gtMul gas La .... to compute Ih. densily of melha .... CH •. '1lO'C and 5-alm pressu. e. A kilomole 0(

metha ... is 16.0 kS_

I pV _ nRT. with T .. 20 + 273 - 293 K. R _ 8314 J l kmol· K. " _ 1 kmol, "rod p - 5 'Um _ 5(1.013)( 10') N/ m' Tbrn

5( 1.0(3)( IO')V - 8314(29)

1O.16 Soh'. Prob. 20.15 without finding tile volume

I U .. pV_(m fM)RT.ndp .. m IVtugiv •

,,'" V _ 4.8m' (" ) ".0 d _ - - -- 3_33 kglm' V 4.8

.. pM .. (S )( 1.0L3 )( IO'Nfm')(16kslbtJOl ) _ 333 k I ' P RT (831 4J/ kmol. K)(293 K) . Il m

2O.n A lOO-ft' volume of nitrogen at 21 '(: and ISlblin' is compr._d to fill a tank thot i1 initially empty and ha. a \'Olume of 5 fl' . If the finalt.mpe.otu,. of t l>< nitrogen is 11"(;, what is the abs<llutc pr<"SSU •• in II>< tan. '

, Use the K.hin ""ale and the 8 .... rlI l gas !low in th. form

15( IOOj p>(5 )

ili+T7 - m + 17

21.18 In the pr.parlltion 01. ~akd-(lff :!I).mL ,ube a, I"", ltmpcralure •. 0 .... drop (50mg) o. liquid nitrogen i$

accidentally ~aled ull' in th. tube. Find t l>< nitrogen (N,) pl ..... re .. ilhin II>< tube .. -I><n II>< tube W',II\$ to 27'C_ A.,u"", ideality. Ex","' you. a ... "",. in . 'm",phe.e> (I . Im _ 101.3 kP.).

I U .. pV", (mRTI/ M. M is 28.OJ N,. U .. T _ :lOO K. "'"' 5)( 10"' kg. V", 2)( 10- ' m' . and R _ 8.3!4lrJfkmol · K. Ji~;nl p _223 kPa'" 2.20alm.

21.19 A car Ii •• i. filled 10 a gaug. pr ... "re of U p'i (lb!in' ) .. -I><n t l>< t~mperalu" is 20 0e. AIt.r ttoe car hu be<:n running al hilh speed. II>< I;r. I.mperature noes 10 6O'C_ Find th .... w gauge pr ..... ,. wilhin II>< I;r. if II>< lire's voIum~ doc$ not chonge . , T, . 313 K

p, "'p"r." [(14,7 + 24) ~1293 K '" 44 _0ps; " 29_3",i gauge pr ... ur.

2:0.10 In. di.~1 engine. llIe C)-lindt. compf"CMC' oi. from approximalely "andard p.e .. ure and lemperalu,. 10 about ""e·,ixteentb II>< origin.l volume and a pre""re of aboul SO .Im_ Whal is II>< tempe.atur. uf Ih. compre$S<d ai.'

I U .. (P, V,)/T, - (P. v,, )f To 10 find To - (P,f Po)(v'{ v,,)T. - (50)( 4 )(273) - ill..K-

10.21 On. wly 10 rool • gas is 10 leI it upand_ Typ;cally. a gas at 27 'C and a pr ... "re of ~'Im milhl be npandcd 10 atmoo.ph.ric pre .. ure and a \'01 "",,, 13 lim~ Lar, •• , Find II>< new lempe •• , .... of lhe p._

I As in Prob. 20.2(1 . T, - IP,I Po)(V,IV.lT. - (-4)( 13)(:lOO) "' 97.5 K ( - 176 'C).

zt.12 A .. rtical tighl cylinder of I><ighl h .. 31)'00"", and ba'lt .... A _ 12.0cm' i, $inin, open under ,1.Ma.d I.mpe.olu,e and p.C"SStlr • . A H)' kg pillon Ihal filS liynly inlO II>< cylinder is _ placed into 1toe cylinder and alLo....,d 10 f. lllo an «]uilib. ium I><ighl wilhin i1. Whallhe n is ltoe heighl of lhe lIi"on and whal is th. prffiu.e within lhe cylindc.~ A,;sum" the fin.uICmperalUre10 be Q'C_

, P ..... "re in II>< cylind.r is inc.eaoes by o.p "' (mg) / A "' (S_O k8)(9_8 m/r)I(l2,O)( 10- ' m' ) "' 41 kPa. gi'in~

GAS LAWS AN O KINET IC THEORY D 34 1

20.1. A VQlnme of )Om' of ai, " 'lh SS% humidilY al 2O'C is passed Ihrough dJyiR~ ~quipm<n! Ihal ...,"""'~s all tho moistur~ . How many ki!Qgl1lm~ of wal~' a", ",,,,,,,,,d?

, Usc lite ideal gas law and T.ble 20-1 to calculate the densilY of lhe ..... Ie. "'po. al 2O'C:

Mp (l8 kglkmol)(OSS"UlkN/ m') 00 , _, , p.-. - 14 k ... m RT (8.~14 kN· m/ kmol K )(293 K ) .

Hence . (30 m')(0,0145 kglm ' ) - 0435 kg .... ale • • e~d,

2:0.15 If the ai, in a . OOm ha, I dew poinl of II "C. whal is ill ",Iali~c humidity al 21"C!

, By the a.gumen!> of Probs . 20,71 aoo 20.73.

"'P al 1I 'C U4 R.II. " .. -.~ •

• vp 112] 'C 2.50

lQ.76 11 I w.mple of air al 68 ' f and ,,% .clali~c humidity is o.Iowly rooled. oonde ..... tion ",,11 OCCUr ot "",,"I lempe.aIU,.?

, We again",", Ihe ...,o",nin! of P.ob.lO.73. from Tabl. 2G-1. , he ",u •• ,ed vapor pressure 1168" f ,, :WOC it 1.31 kPI. (0.SS~2.3 1 )" 1.27 kPa. which is ,h • .,.,u ... aled ".po. p' ... mc., (appo~. ) I O"C "~.

28.77 The •• Iative humidity of •• oom is 75% al 21 'C. If Ih. tempe. alure falb 10 19'C. whal will 1ho •• Iali"e humidily be?

, We I,ain u,", lhe reason in, of Prob. 2O.B . From Table 2G-llhe .. turated VlPO' pfUlure of WI ... ,'opo. II H'C is 2.U kP ... !ivin, an .. <1 .. 11 press .. ", of (0.75)(2.88) " 1.1 6 kPI. F.om Ihe lable. lho Sllurated va por pre .. u", at I9'C iI 2.20 kPI, Th. ne .... ,·al ... i, lhe",r",. ",llIi, .. humidilY '" 2.16/ 2.20 " 98%-

10.711 Whot i, lhe de .... poinl if Ihe air has I .. Iative humidily of 60% at lS"C?

, f",m Table 21).1 , h. , alu.lled "apoo- p ....... '" of ""0'" v.po. al 2S· C i. 3,26 k PI: he .. "" P '" (0.60)0.26) : 1.96 kPa, The de .. point i. Ihe «mpc,.' ure al which Ihe co ..... pondi"l! ,'apar p ....... '" iI the $l.lu.aIO(l "'po' pr<$$"~' fro," lhe table. then. de .... point - 11'C.

» .79 If the normal lapse rate (dec",_ of I~mperalur<: wilh allitu<lt) ptevails .• nd II>c "mpcrature at !round Ic>-tl "21"C. whal " illlhe air tempe.alu,. be It heighl 0( I 100m?

, The norma l decrea5e of lemperatUre wilh altilude i. about I 'C per 110 m. (lIOO/ I IO)(1 'C) _ 10"C decrease. 2] 'C - lO"C -ll.:!;; . (Note that 294 K-21)4 K ;;, a """II ~,unt~ de=1$C on the Kel~in ..,ale,)

lOJIO , . ) P;1ou of light pl.""s mU$! be o;3.",ful to nko.al.'~ Ihe Io:>a<k on wann da)"$. Wby mUSI pilou k:. ""g fr<>m 0.- landin! at high elevalion (fur eumple. [)en, ..... CoIo",do. 0. Muico City) be particularly ca"fuI? ,I>, Compare 'he density of lI.,e . i •• , O"C 10 Ihe de~ty al )(I "c. A$!Ume itkn'ia.1 pr.sou,n

, ,.) In relalively .pa.rw Ii • . Ihere will be Ie" lifl on tM winp 0( 1M .irpla ... . Othe. fac:lol"$ being equal. loke..,ff .UItS will be longe •• d imbin, . ales will be ...... 11e •• 000 Ihe <It""'''t durin, landing "pp'oo.<'I>n"';1I be m",e rapid in the Ie .. den,", ai, .t high elevalioru. aoolo.- on warm da)"$. ,I» AI ronOlanl pres ... "e. den,;ly and temperature are inverwly proponional p,/p , . T,/T,. With T, _ 273.15 K (t, - O"C) and T, • 303.15 K (t, _ 30 "C). "' ... find pJ p , • 303. 15/273, 15 • 1.11. That iI. al 0 'C tbe <ltnoity is II percenl hi,her Ih.n il is al 30 "C.

11.11 Compare lhe de~ly o f ,he .i •• t Logan Airport in Boslon ("Ie ..... hon Om) I I O"C 10 ,ho <ltnsily of air al Stapleton FJeld in DenV1'r (elev.ti"" 1600 m) al 3O"C. AI rons,.nl lemperalure. lbe almoopheri< prmuIC p obe)"$ appro~ima"ly lbe equation p . P .. -·~'" if 1he .k:valion l "expressed in metcrs. 100 P. iI lite almosphcri<: pi"''''' al 0 m.

, We begin by delermining Ihe v",", .. re at Stapk:IOfl FICId with the help of the hydrosUliI: i>othermal prolile, Using Ihe $ubscript> I and. to "f" 10 Logan 000 Stapleton . .... ha~e

opynghtoo IT £rla

342 D CHAPTER 20

W. no ... <.\<t<rmi"" the <'\<"';')' ... hieh """ n p<md$'o p, Ind the gi~cn 'eml"'ratu, e T, [f. "" 'eml"', atmn "-.r. C<jua[ (7; - 1;1. tit<: dc"';'y ratio P. II', W()u ld "qual the pressure , .tio p. l pi (by Boyle', la .. ). Si""" T, .,. r. . "'" mu~t UK the m" re general ..,Iationslt ip (p ip TI - ron, tan, (si....., the go. com!""ition is .ssumed to be ,lie same a, ,lie tWO Iocatioosl. ThaI is.

e: • .:!;~ .. .:!;t " ~ .. " ", p, T,p, T,

.... ,th r. .. O'C .. 273. 1~ K. T. .. JO'C .. .JI)J, IS K. and l. ~:, " 1600 m. " .. " oblain

e: .. {~. 1 5)t " -""" .. (0.9010. · -· ' .... _0 ,74 p, \303 15 ,. -

Uodcr Ih. liven ""ndilion! and .slump'i" .... lhe <it",il)' is 26 1"'",.nl lo" ·.r al Slap!tton Field lban at Losan Airport, Equi"a lenlly. Ih. densilv al Loiln Airport i, 35 pcr",nt high • • Ih an '''''t ~I Siapleton Field .

20.82 Ikrive lhe la ... of almosph • ..,. from 1M Ma. ,,'e ll- Bolllmann >lali" ieallaw " 'hi<h staIn l ~a' ,h. numbe r of porl;';I •• ( in an eq ui librium <"""mble) wi.h eM'!), E is prop<:trt;o.,al '0' '''- Assume ""nStant tempelatUle "".r 'M height> beinS e(msid<~d. (S<-e Ploh . 20.79·1

I Sir>ee Ihe onl~ t.rm in the enerlO' of air molecul •• that depend> on lhe ,·"n ieal M"hl.:, i. the " .. itational poten, ial """'8Y. mg: (m.asur.d from grouoo zero). "'Mre m is lite ... ·"rag. ma .. of a mol«U H: 01 .i •. rhe partic!< densily n. (:) obeYI (Max,,'.II-Bolllmann): n.(: ) '" t -t ~" '''1 .. , -"', and 1n.(: lJI1".(0 11- ~ '-, " r I I Si""" 'M <lensity of air . p. i, pr<1pOrt ion.1 ti' It •• " . ha, . Plz)! PIO) - t .. ··' ..

Finally ... corntanl temperalu, e. PI:)" p(z ): "" p( z) .. p(o "' - - ' " . ... hien is lhe la ... of alrn"'Ph.r.,.

10.13 find the uniform temperature al ,,'hleh the . alio of the densili •• of mercury "apor at t h~ KJP and bmwrn of 0 2b-m·high tonk " 'ould be l I t . (A"urn. an ideal I'" ..... Id be obtain.d .) (Molecu lar ... igl.t of me ,cury - 201.)

I From 'he la'" of 'Imo~pbe,". pd p, _ n p 1{ ~mgh )I(k Tll 11.""" 1f2 718 - e' l' (( - ... (9.8)(2)11(* T)} wher. m .. 2011 N, .. ] . 34 ~ 10' " kg. r nvert and ,ake na,ufOl logarithms "f each side to /ind I .. 19,6mlk T, SoIvinl for T yields T _0.47 K ,

11M A , .. of dust pan;';l .. fill> • 2.(1·m-high lank. AI equilibtium (27 'C). the densily ,,[ parti"" at the '<)1' of the t.nk is l i t the density al tho OOllom. Find 1M mass of 0 Iypical partic\c ~nd find bo ... mo.)' time. mo •• ma";ve it i. than . ni trngen molecule .

I Proceeding a. in 1'",,/:1 , 20,83. (mg/r )/(H) .. 1 This givn m .. (kT )I(gh ) .. 2. I )( ](I !.I <g, Th. m~s.s of • n:trogc n molecule is 281N~ _ 4,65)( ](I' " kg and ", , he mas. of the du~ particle i . ~ lime, !reato r.

I Figur. 21).7 lhows a Ihin ,lab o. air at altitude t. F,,,- cqllilibriurn of Ihe .Iab.

1p(:) ~ pel + /lo: )IA· pgA /lo t ", ~- - pg

" Fmm ,he ideal ,as "'W.

GAS LAWS AND KINETIC THEORY a 343

",'htrt m i. the avera,," moitcujar mass for ail . lllus

[",oring the variation of T and , "';th altitude.

[ "--""{'" .. p kT

211.116 U,,;n& 'he Dulonr-""' i' ta ... . < .. i"'4'e ,1M; high-<cmpcr.,",e 'f'<"i!i< he . , c,of>K;'y (in Jf~" K) for ~ra"iu,,' metal (M . nil.)

I From tk Dulong- P. tit law the molar heat capacily at COfI'lla", "olum. C~ .. 3R. Thus. €v " Cv IM .. (J(8J j 4))1238 .. lOS Jl k, ' K.

10.87 1M Ou~-P.tit I . ... was u!o«l 'arly in this century to dctermi M the molecular _ ights 01 orys,alli"" <OIi<ls A cenain pure metal has a sp«ific h.al of nol/ kg ' K at high t.mperatures. Whal is the moIc:cular w<ight of the metal?

I Again use Cv " 3R and M .. Cv/cv " 3(1tl14 J/kmol' K)/ I230 J/ kg ' K) .. 108 ki/kmol.

20.811' Show how th. equipaniti"" theorem IP.ob. 20.3!1) lead!. to the law of OulonK aDd Petit.

I At high temperatures. it may he ~ppo<e:d .ha, .. ~n"'lIy all the int.mal e""rgy of a metal is due to th. "bration. of the atOrll$ about their equilibrium I"";t""" in the ,rystalli"" lania: . If ..... pKlu.e eoeh 010m a. oort""O'ted to ito "o;ghoo" by <prings. ,hon.he . tom will h,,,,, kiM';" and poton,i.1 ono~o. 0lon8 ,h,.., mutua lly pc~Dd""'la. di.ection..........;. modes in all, By II>e equipanilion theo. em. ill lOIal energy .... ill be 6(lkT) " llT. Jiving a mola. energy of E .. N~(JkT) .. 3RT aDd I molar hea' capacity 01 C., .. dEldT .. JR.

:10.89" lbo Dulo"ll_ Pelit law holds true 'urprisingly _II for so~d. if the tempc..,tu •• ;. high e ....... p. To p<.dict the behavior 01 the beat capa.city al Iow<:r temreratu,es,. q lantum·mech~nical modol mw;t I>e u!o«l . One wdr mod.1. ori!inally used by Einstein. a...,m .. that all ato,," .. in a solid .ibuto at the ........ frequcncy ~, The tOlOl energy of . solid of N aloms is then the sa ....... the energy of IN one-dimensional ...allatoD. n. COIl'C<"t GUUltum-mcct.anical CKp<cssion fot the , ... rag. ene .gy of Ih;' collection of ...allat ..... ;. ( E ) .. 3Nhv[i 1- 1/( ..... • - I)]. wh • • e fJ .. I/H, ond It • 6.63 )( 10-'" J . s. Show 'hat this modol gives the mol" heat capa.cilyas

(a ' r""

e .. 3R r) ( .... - 1)' ,,'he,. a " hvl k.

, W.it; ng A. for AvolI"dro·. numN ' . ... .., hovo. fo. one mol< 01 moteri"l.

( E ) .. JAhv[~+ .'. " ~ _ 1 J (O J

e d(E ) JAil I - I) (-h~ ... ,., ('.)' ... '" - ""dT - " (~'. ''' - l)' H i ! _3(A k ) AT ("""" - 1)'

Sincc Ak . R. the gas """,tan,. Eq , (2 ) become.

e .. JR (~)' (.::: I )' (3)

whe •• " .. ba.'. put a _ /tv lk.

20.90 R.r... ' 0 Prot-. Xl &<I . o., '~rm'n~ ,he heh.v ..... of e in 'he t, m,' T» R . • 00 in ' he I,m" T« R .

I Fo. T» a, "" ha\'e a l T« I. so E<! . (3) yields

C " 3R(91T)' [I+( 9 / T)+ " ') . _JR (9 1T)'!I+ .9 / T] __ lB: ([I+(9 IT) + -" ] - I)" (€lI T)"

which is tho Dulong_Petit .... ult .

Of) rghtoo matmal

350 a CHAPTER 21

AJ;'N' OC

' ... ~ p. ------ I,,"h<,m

: A<Ii.~., ,,,

11.26' [n a p_V dia,ram (Fi,. 21 ·2) an .elial>atic and an iJoIherma[ "",...~ for an it:kal Bas intelWCt. snow that the absolute "alue of the sJope of the adial>atic is r times that of the ilQlherm. Hence the adiabatic "",,"c is ",cepcr beCituse the speciflt; I>cat ratio r;s veater than I.

, Denote tl>c intenectioo point by (P. , V. ). Then the ;liOI.hermat cum: is given by pV _ Po V • • 01

P .. -P. VoV · '

",hile the adiabatic cu,,"e is described by pY ' _ P. V:. or

P .. - PoV. V · '

( 0)

(2)

The slope of the isot hermal <11 ...... at (P., Y.) i. found by difl"e",ntiatins Eq . (I) and ..... laol;n8 the deri .... ti~ .1 V _ V.

(") _ - Pov. v -· I - -Po dY .. .,. V.

Us.ing Eq. (2). the slope of the .eliabalie cu,,"e at (P •• Vo) is

Equation> (3) Ind (4) show Ihat

21.1 THE FIRST LAW m' THERMODYNAMICS, INTERNAL ENERGY, p_V DIAGRAMS, CYCLICAL SYSTEMS

lLl7 What is the change in intema[ ener8Y 01 0.100 mol 01 nitrOlion , .... it;, loUIN Ir"," JO to 3O'C at 1_' co",tant volume and (1llconsllnt pr~ur.?

(3)

(.)

, Inte rnal ene'l)' for an idea[ las illinearly related to the temperature of the "" wilh 6U _ ~C. 6T", the .... _. to par" (_, . nd Ib) il ' l>c sa~; 6U .. (0. IOOmol)(4.96ca[/ moI· K)(20 K) _ 9.92 •• 1 - :!.l.i1,

21.28 When SOL 01 ai, at STP i. isothermally romprcued '0 10 L. how much hell mU$t flow from , he gas? (p_ - IOOk P •. )

, Since the proceu is isothermal. and "" .. ,ume an ideal, ... 6U - 0. Then 6Q .. 6W. But 6W" ~RT!n (VJ V,) " P, Y, In (VJ V,J. The wbt.titu,ion mabs use QI tho ideal gas law. P, V, " ~R T,. ar><! recosnius that T i, conltant. In 'his ..... 6Q .. (I )( IO')(S )( 10-') In LO/SO - -!1050 J ; i.e .. t1!UJ. !lows 0111.

21.19 An ideal .... in a cylinder is compn'SSCd adiaNlicalLy '0 one·third il5 ori,inal V<IIume. During lhe prot ..... 4S J of work is done QrI ,he sas by the comp",";nS ase n'. (. ) By how much did.he internal ene", 0( ,he,... chnse in the P'ooc,,~ (b) How much heat nowed inlO the gas~

, In lhis ca ... 6Q - O. 10 6 U .. - ~ W _ -( -.S J) .. ill; (II) the heat now in the adi.batic 1""""" ;,. u.o.

11.30 In elch 01 ,he following ,itultiom. ~nd the rnange in internal cne,1Y QI ,he .)'Stem. I .. ) A Sl'ltem absorbs SOOcal of heat and .. the sam. ,i"", doe$ 4OOJ .,f WOTk. (6) A ,)"Stem al><olbt. JOO~al and at the same time 420J of work is <kine QrI it . (e) Twel~e hur><!,ed calories is" ,emoved from a ps held at OOll5l" nt ,"O/ume.

, I .. )

(b ,

'"

THE FIRST LAW OF THERMODYNAMICS ] 351

flU .. flO - fl W .. (500 cal)(~. J84 J /eal) - 0100 J .. 1700 I

o U - flQ - fl W .. (300 eal)(4. J!I4 JicaJ) - ( - 420 l)-lli!!ll

flU " flO - oW .. f -12OOcal)(4.1 84 Jfcal) - 0 __ ~

NOle Ihal tJ. Q is pooili'·e ,,·hen he.1 i. added 10 Ihe sy<lem and fl W is I">"il i,·" ,,·I><n Ihe ')"Slem does work In llIe r.'·e~ US<"$. flQ and a W must be lahn neg3live.

21.31 Re.t..,ive tl>< ,e.ul! of P,ob. 21.25 ... illk Ih. (jr;t law of thelmod)' n. m;C; .nd lhe- fae" Ihat C. _ C" _ Rand C,IC .. " y for an ideal gas. ,,'he,. C de note. mol"1 heat a p.<ily .

, UI lhe hUI I\ow be denoled by aJl , and 'he inlernal energy thange by fl£ Since !he ",octU is adi''''''i •. flH _ I) and Ihe rdore AW _ -AE. Du! AE .. "Cv(T. - T~) . ... d fu"bemo.e R " C. - C .. .. (y - J)C •. He...., " 'e ran up"" the . .... fIY .hange a • . ,

"---(T.-T~) (r - I)

BUI the ideal gao I . ... )'i.ld, nRT _ p V. so we oblain

AW _ _ tlE_ -p,V.+ p~ V~ .. P~VA-P.V. r- I r-l

;n 'ir~me.1 "';111 the resuit of Prob_ 21.25.

11.32 FinJ tl W a!>d fl U fm • Ikm ..... be of i. on a. it is heated from 20 to JOO ·C. For iron. c .. 0.11 cal /g . "C and the volume coefficient of tbermal expansion i. tJ .. 3.6 ~ 10 '''C-'. The mass of Ihe cuhe i, 1100 g , The volume of the ..... he is V .. (6 em)' .. 216cm' Usi ng IA V )IV .. tJ fl T. "" ha>e

fl V .. VtJ AT .. (216 ~ 10- ' m' )(J,6 ~ 10- ' 'C' )(2IlO ' c) .. 2.18 ~ 10 • m'

Then. assuming atn>Osflheric ",e .. ure 10 be I ~ 10' h.

AW - P flV " (I ~ 10' Nfm' )(218 ~ 10 Om' ) - 022 J

~ut tM fi .... t Law t. lls us Ih.t

flU .. tlO - t:. W .. ( S2 000 cal)(4 .184 J/ eal) - 0. 22 J .. 218000 J - 1),22 J .. 218 000 I

No~ ho ... "e,y wnai l lhe work of upansion .goi""l the alm"'!'h.re i, in comparison to tlU and tlQ. Often A w a n be ""g\c<led ",I><II duling .... ith liquids and $01id>_

21.33 A Qlbic meIer of helium orilin.lly at O"C. rId J·a tm ",,,",ufe n cooled ot con"a.t r . ... ure untilll>e volume ;,; O.7S m'. How much he.1 w" remO\'ed?

, Usc lhe so""r.1 gas I . .... 10 find t l>e fin.1 tempe •• ture . Then"," the fi "t law of I""""od)'n.mico,

p, V, p,V,

Th"

-----T, T,

I 0.1S m aT,

flQ _ flU + flW ; and for an ideal .... tl U _ m<,. fl T. ThklS. tlQ _ m<"" AT + P flV " ~c. tlT + P tlV. Notinl that al STl' . I kmol OCCUpiC1 22A m'.

I (1.0]}~10')(O, 7S- 1) tlO-

22A( ])(205-27)) + JISI --~. II -6.05--!j.2',pl

11..M Th. volume of I kg of .... ... at 1(l()"C ;" a_, 1 ~ 10 ' m'. The volume uf l toe ,-aI"" formed ""toe n il boil. a, llli't<mpc. a,ur<: and., , tanda,d aunosp/>eri( pr .... u'" i, 1.671 m'. (_) How moch "",,"k is don" in pu<hinJ bad; lhe a!mosphe,"~ (b) How much is tM in<.<asc in ,I>c inc.,na[ .nefgy ,,'hen Ihe liquid chang ... to vapo(?

352 a CHAPTER 21

, (.) n.e ,,'O.k done by the ... ate. is gi~en by

l!. W _ P_ l!. V _ ( I.on x 11)' N/ m'II{I_611 - 0-(01) m'l - 1f!2..k./ .

(6) ACWfdini to Tabl~ 17·2. t .... latut heat of vaporization 01 wate, is L _ 5-10 kcal{k,. Thi. is t ........ at per ~ilog.am tlult must be added to vaporize 100--c ,."ne. at I coMtant l"""W.e of 1 atm. 11>( mange in t .... internal en. ,IY of 1 kg 01 ... ater ....... n it is boiled at I aim i. ' .... ,do •• giW" by

l!.E .. l!.H - f>. W - Lm - f>. W - (5-10 kcol{kg)( 1_00 kg)(4 . UI4 U /kcal) - (169 tJ ) - ;:Q2!ll.l

2tJ.S A tank contain. a fluid that is $tirred by a paddle wheel. lbc: p""'cr input .0 ..... padd"= "" ..... 1 ;. 2.2~ kW. H~at ~ tnmftrr~d from the tint at t .... nte of 0.586 tW. Conlidering the tint and the fluid il the i)1item. dete,",iM , .... mange in t .... internal energy of t .... iy&tem per hour . , f>.E - Q - W - - 0.586 - (-224) - L6S4 kW - l_6S4 U /. - S954 Uih

or about tMllb..

n.l6 A spring having a spring cvnstant S N/m i, compr~ 0.04 m. clamped in thi, oonfiguratKm, .nd dropped in,o I rontaine, of acid in wh i<:h t .... opring diooolves. How much potential e".,lY;' stmod in In. sprin!. aOO ... ltat happen, 10 it wh. n the .pring dissolve.?

, The 'Ilring in being compru.~d acquired elastic potential en.rgy of amoun! U._ - lh ' - \(S )(O.o'1''' \!J!!:!l. When tbe .pring is dissolved. thi. o.de",d potential ene.gy is convened into dilO.de",d potential and kinetic en. rgy 01 tile ')"iIem. o.·erall . • ""11 i. oon~~.

n.37 0". pouOO <X f""l, having a h.at of oombil<licJn <X 10000 Btu/lb, ...... bumrd in an "ngiM that Tai~d 6OOO1b of ..... t •• IIG fl. What pe",. nllge of the heat .... u transformed inlo u~ful work?

, .. ...,. ...ork done by .n&iM em"",""Y _ WOTk C<juivalcnt of but I1.JWlied -,",(""~O'"'"'c"<'"",)".~ (l0000Btu)(778 fl. Ib/ Btu) - O_oo _t.tl!

1I..J8 A sample containing Loo ~mol of tbe ".arly ideal gas belium;' put through lhe: cy.1e <X operations 5110" '" in Fli. 21·3. BC is an i50thennal. and p __ 1.00 atm. V~ _ 224 m'. P. _ 2 00 Itm. What a", T • . T •. and Ve?

•

" fit. 11·)

, Applying ,he: perf«1 l as la ........ /Lave

T P. v. (1 .013 x ]I)' Pa)(n.4 m' ) .- " R ·(l .ookmol)(8. 31 ~ x l O' Jfkmol' K ) - 1Zl.K

Becau~ ,ho: procns AB ~own in Fig. 21-3 is isomelriC. T. _ (p./p.)T • . With P. " 2.oo.tm _ 2p. , .... find Iha, T. _~. "The process CA ;. i!Obaric. 10 Charles' law applies. Thu. VelTe _ V. f T. , 10 ,hat Ve " (Ta T.W •. But BC is an ilOlhtrmal pnxcss, 10 l/Lat T<"" T. _ 2T._ lbc:,cfo.e Vi' _ 2V~ _~.

11,39' Rde. 10 Prob. 21. 38 and calculate tho: """I; OUlput duri ng the cycle

, Th. """I: done by I .... p. along AB is uro. >in« dV .. 0_ lbc: ",'O.k done on I .... p . along the ;..,the:"" BC;'

[ OC j"" (" ) f>.W .... - pdV_ ~dV_RT.ln...E - RT. ln 2 - (S.314 U / kmol- K)(S46 K)(O.6931)_3t50kJ . ~ ~

THE SECOND LAW OF THERMODYNAMICS a 361

, ... ..., Q. - Q, T. - I; ~u .. ,~"<}' .. --- . --

Q. T. 500 - Q, 590 - )90

• "" "., 5OO-Q. "l69kcal .. , Q, " 331 kcal delivered 10 lhe ';nk

W _ Q. - Q. _ 169 kcol _ 169(4.1114) - Zl.!I.M-

22.14 A Slum engine operating belw-<en a boiler t~mperature of 220'(: alld a ronden",r lemperalure of lS"C del;'~rs 8 hp_ If iu dli~~ is 30 pettent of Ih" for a elmo! engine operati", between 1he1c tempe.ature Urn;tr. how many alone. arc aboorbcd cadi oecond by lhe boile.? How many aloric$ arc cxllau~tcd to the ronde....,r each >CCO<ld? , octual efl'\cicnq .. (O.lO)(CarlKlf efficiency) .. (0 ,30)(1 - :) .. 0, 113

BUI from the rel.alion

.. ...,_ OUlpul work em .. ",ncy ..

inpul beat

( '""' ) . oul I wo.k/. (8 hp)(746 W/ hp) 4.1 114 W 'npul helt!. _ ~ .. .. 12 7 kg!!ll

e ency 0 . 113

To find the e~gy r.}eclcd 1<;1 the roJ\dcn"'J, we U$C the law of """""vation of energy:

input enugy _ OUlpul ...,rk + .-.joCled ~nergy Tho,

.-.jcaed energy/ . .. (input energyl.) - (outpul work/ s) .. (inpul energy/. j{ I - (efficiencyJt

.. (12-7 kcallo)(1 - O_IlJ) .. I Lll;GI/i

22.15 How many kiJosrlUll$ of .. at~J ,It O"C """ a frecKr wilh • roeff\cienl of periormance S make inlo iee cubes II 0'(: wilh I.....,.k input of 3.6 MJ (one kilowan·hour)? Usc Table 17·1 lor dala.

I By 1'Tob_ 22.3. the coeff\cient of performa"", i, Q.lW,

S.. mL .. m(80 kcal/ kg )(41S4 1/kcal)

HXlifJ 3_6_10'J

Solvina. m " 54kg.

22.U A refrill'rator remo"", helt from • fre.rin, chlmber .1 -S 'F and di!dlargcs il .1 95 'F. What;' il' maximum coellicit-nr of perionnl1>te1

, For I ComO! refrigerator. Q,/W .. V(T. - T. ). U", aboolule temper.turel (Rlnkine). n.en

Q. .. -s +460 455 W (9S + 4<>0) _ (-S + 4<>0) - iOO .. 4.55 coefficient of performance

22.17 A freeKr has a coelftcient of poerfurmaroce of 5. If Inc temperature inside Ihe frecKr ;" - 20 '(:, .. hal;" ,he lemperalure al which it , ejo<u hut1 As.&umc an ideal _)'Stem, , Q. I; - .--

W T. - T.

12.1 ENTROPY

22.lr Give a mathemarical definition of emropy atld diSCIM i1> relation to Inc """,nd I.", of thermodynamic>.

I Any thermodynam;", ')'Slem has a Slate function S, ""lied the ~ntropy, By Ihis is "",anllilal S-l ike p. v . and U-is oJ"'Y' lhe lime fOf the 'Y'tem when il;' in. liven equilibrium .Iate . Entropy may be defined .. foll".,.,_ Let a')'!Iem I I at>oolu te lemper.,ure T under", an infinileiimal ~,.ibk: proem in .. hich il l biorbs heat QQ. Then the chan .. in entropy of the ')':Ilem n Ii' .. n by

AQ .,.­, for infinite';mal>

C JPYnghted matanal

362 J CHAPTER 22

Not •• ka. dQ is !lO' .he dilfe,emial of a In", fu nct;oo. En.ropy .. ; 11 have Ih. uni" 11K. The CluusiUf rqlUllio~. dS _ JO l T. I>oIds .mly for ""e~bk p~. H" . .. cyc •.• ino. S;, ,, ~t.t.

function. the c",T("'~ oh.ng" acwmpanying on i" .. ·.r<.ibl. pf"1)<"tSS ran be calculated by imegrating J Of T &lon3.he pa.h of an .,bin 'l}· ,.."",..ibl. p<t><ess ronnu'ing tbe initial and ~nal SUleS.

"The import. nce of Ih •• nl"'P)' f .. n<bOll ;s uhibil.d in the following {Ofm of .he J«Ond /a", of Ih"modynumic~: In "ny PI'O«.JS. 1M I<>/a/ mlropy of 1M .y"rm "nd lIS .~ ''''~nJjllg. ;'''''''''lt~ or (ill " rrl~TJ;blr pM""") dorJ nOl rl!",,~. Th. ~nd law appli~ 10 Ih •• ys,.m alon. if IIIe s)"!lem is i<olat.d; Ihal ;,. if il in no "'a~ in .. ",<., with ih ",,,ounding>.

ll. 19 De fin •• nllopy in '., mS of o, (\e, /di<orde •. and di~" b-rieft~ .

, TIle """,nd law of lbe,mod)"namics indira, .. Iha, enlropy i •• mealu,e of i" .. 'en.ibilily. In""ersibilil~ is aSwtialcd.OII Ibt molt.:ular level. " ,Ih lbe ir.c", ... of disorder. MoIt<ul., ')"!lem. lend . as lime passe •. 10 become eIIaolic . and it i. uI",mely Ilnlikdy Ihal a more ""aniud Sial<, once left. " 'ilI .~ .. be regained. Ano<btr . fully equivalenl. definilion of enlropy can be gi...," from a (\elailed moIc<"\Ila, analysis 0' Ihe ')"!lom. I' a !yllcm ClIn ""bie..., Ihe same "ate ( I.e .• lbe same .al .... of p. V. T. and U) in Q dilf ... nl ,.ays (dilf.renl amnSCmenlS of Ih. molecule,. fOl example). Ihen Ibt enlropy of lhe: "ale is S _ k In C. whe •• In is the logarilhm 10 ba .. t and k i, Boltzmann's CQnSlanl, 1.38 )( 10-'-' 11K.

A ~'I. Ihal can OCC\Ir in only one way (one . rrangemenl of its molC<:\lI~. for e.ample);' . Stat. of hi&fl Older. BUI a SUle th.I can occur in many ways;s . mon: di<o.(\e •• d stal •. To as<Ociate a "umber wilh di..,. (\e •. Ih. d;<OIde. of .... te ;, laken p«>ponionallo Q . Ih. number of ways Ih. ~t.l. can I)00,I1. S.causc S. k In C, Ibt .n,ropy is a m ....... of diso.de •.

Spontaneous processes in ~)')I.JlIS Ih.t comain many molecul", .I",ays OCC\I. in • di~ion from

(stll.lhOl can Ui!.l)_ ( su.'e Ihal can nisI) ,n only a few .... ays In many .... )~

Hence ~YS'cms ,.btn left 10 Ih.m .. I,·e. retain Iheir oripnal "ate of orde. or el .. ir>eru .. lhe:i. di..,.de •.

ll.lO 1'0" he:al enpnc. 0"". 0,,", .,..-k. JoS '" JoE '" 0,';""" lbeengine mum. I" it. oripnal IIale. Th. r .... law of IMrmoo)"namia lhen gi~ .. for tbe "·,,.k do,,", by Ih •• ngin< pc • .,..-1.

W _ Q ... ( 1- ~:)- T_ AS .... - w' - T_AS .. ~ '0 wtlc:T<: W ' i • • h. work ,hal """"Id be dol>< by. Comol engine opelahng bet"'","n II>< same t .. o lemperatu, ... and oS ..... is ,be en'ropy chang<' of Ibt .. n;,'."" (in Ih;, ca ... lbe .nl"'P)' rhang<' of lhe hoi and mid ",,,,,,,01 .. ) during one '1'<'1 •. Dcri". (I) and ,bow iii sipt;~ca""", for lhe: second law.

, Fo, any engil>< ope .... ,ing bel",,,,"n Ih. l.mp" .. I .. '" re .. rvoi .. , ,,'.!>ave W _ Q_ - 0 ..... $in« AU - 0 ""or a cycle . (II.", Q_ is dc~ned .. positi". "'lien h .. t leav~ Ih. engine. II< is usual in .ngine .nd .efri, • ."OI" problem •. ) In addilion lhe: chan!", in entropy of Ihe b<Jt and roId .... rvoi .. a .. Ttla.ed I" Ihe he:al tra""e ... 1 constanl lempe.,Iu", by Q .... - T ... oS ... : Q .... _ T _oS ..... Then. noling ':"S .. ~. AS ... + l>.S ..... we. ha.e T _ l>.S .... _ 0 .... - (T ....Q_)I T .... Th ... W . 0 ... - Q"", - Q_[1 - (T...JT _ )) -T .... oS""". SiOtt Ih. efficiency of a elmo! cycle ;s W' IQ ... _ I - (T ....I T ... ), ..... ha,'. the: .. mainde, of 0 ... r .... It' W _ W' - T _t.S ..... Welhu! ... . hal fOl. gr.·.n Q .... w s W' t::>6S Z. 0. 0 .. in 10"'" of ef!icicr.cy . ~ s r,' t::> 6.~ z. O.

n .ll When 100 mins aTt tossed. ,Ilc,e is one .... y Ihal all can rome: up head>. Tho .. arc 100 "")'1lhal ORly one lail ;. "p. The.e a •• abool 1 )( 10'" ways l!>al SO he:ad> can come up. One hund •• d coins are pI .... d in I box wilb OIIly 01SC he:ad up. n..y arc wken and lhe: n Ibt", arc SO he:ads up. Whal ...... lhe: chan,. in enlropy of the: wins ClOuscd by lhe: shakinl?

, From P'fob . 22,19.

AS • • (In C , - In C,) _ (1.38 x 10-" 1IK)!ln (I x 10"') - In ]00)

- ( 1.38 )( 10-" I I K)(17 In 10) .. 8.6)( W-" 11K ... iog In 10 _ 2,J01.

n .22' n.. num,",. C of $1010$ accessible 10 N alo,"s of. monatom" ideal sa. wilh a vol .. me V. "'h~n Ihe .ner,y of ,"" gas ;s Ix,;,.""n C and C + dE, can be r.hmo.'n 10 be C _ A(N)VN C""·~. whe .. I"" fado, A(N) dependl only on N (.) find I"" r.!fopy S as • f""",ion of V and C. (iI) U~ng Ihi. en!f"PY function and Ih. de~nilion of lhe: Kelvin I.m.,. •• t",o. liT - (~SIS£)v . • h"", Ihal £ - l NtT.

372 a CHAPTER 23

while Ihe ;nslanla ..... ou! "aDlV~f$t: v~locity is *. I:~A sin 11:(% - VI) + J1

Comparing Eq •. (2) and (3) . _ .. ~ that

" '" - .---3% va.

a, dcsi~.

2331' Give the ,~ ..... ral mathematical 10m! of I wa" e "ave ling al C01\5Ianl s.pe<:d. withoul di"';pal;on . along the x uis, and ~w thaI ;t sati<liel the standard .... ve equation.

C»

(.)

, A wave y(%.I) Ira,"ding along X wilh an unchanging form and l1"'ed " is given by y - /(% ± t"), .. he", the m;nu, and plus signs rder to wave propagation in the positive and ""gative X dire",ion$. 'o<peelivdy.

"These ful>Clionl sati.fy lhe one·dimensional .. '4 ..... <q""riDn

if, I if, <1% ' - ;;; iJt'

as can immediately be sun by Kiting w '" % ± til and using

lOT eithe' choi"" <>f w

D .l!! Sketch lhe profi le of ,he wa'"': 1(% • • ) .. A • . . ... K~ at I • 0 s and p . 1 s. using A .. 1.0 m. B - 1.0 m- ' . and ~.+2.0m l •.

, Sec R,. 2l-S. No,e ,hal 1"'ak·\<'>-1"'ak di" __ divi<k<! by lirne in'e",ol e<jua" v.

,.. m ,

flJ. :tM

23.36' Verify by partial diff.,enti.,ion Iha' l l>e wa"e funcrion 01 Prot> . 2),)5 sat;,fic$ 1M one.dimen';";'nal wave e<juation.

, We begin ",ilh the proposed wlution

Dif'lerenriatinl with respeel 10 %, "'e find that

~ . - 2118( .. _ 1It"' - ·'· - H~ ;.

Dif'lcrenliating with ""peeI to I, " '. find tha,

~ .. 2118,,( .. - wlt - "·- H"

"

" )

(3)

")

WAVE MOTION j 375

support: lllhe other cnd, a I(mion of 100 N i, aWliC<!. If I tran"'e",, blow i. ,Iru<ok al onc cnd 01 lbe lube. how Jong doe, it lake to reach the OIhe:r coo?

, u _ J!joo ~ Ioo _ IS.3 m" ,.. 0.30

s _ ~, S.O _ 18.31 '"' 13." !tefn 10 Prob. 23.43. Whal frcq ... 1'oC)' 01 vibralinn muS! be applied 10 lbe lube 10 produ<,<, a standing ,,'a"e

with lour stemenl' in lhe: lenSlh of lhe: lube? (Thi, f.equency is called lbe I",,,,/r h",m(HJi( Irtq~~cy. )

, We h.~. 4{.l.n ) _ 5.0 m. or .l. - U m. ll>en V" u/l, - 18.3 /2.S oo1J1:Y.

13.45 Slanding waves .re produ<:ed in I rubber tube 12 m long. II lbe lube vibrales in ft"" ""g""'"1> and lhe: ""I<>cily of 1M Wi ve io 20 m/I. whal is (.) th . .... av.length of the ..... ves. (b ) 1M freq ... 1'oC)' of Ihe " .. "e,~

5(i) - 12m .l. - til!!

'" 20m! •

• - -- - !.Jllli 4.8m

U M A . tring ha, a length of 0.4 m .nd a mm of 0.16" If Ihe ten.ion in Ih" string is 7O N. " '\\lll a •• lbe Ih,« Iowal freq ... ncieo il produca when plucked?

,

v .. 4 18mll .. m.u. , 0.8m

lbc 5eamd and Ihird hlrmonic frcquc:ncies a •• IMn 2v , - .l!.!!!i.lil; and lv, -lW.l:ll,

13.4'7 "The third overtone prodroc:ed by . vib.ating >lrinl 2 m long i. 1200 Hr.. Whal are Ihe Ireq""""ie, of Ihe lower ""enones and of the fuOO.menla!1 Whit is lbe v"locity of pro-pagation1

, The third oven o"" is lhe: founh harmonic v,. ", _ 1200 Hz implies that v, _ 1200/4 .. J!!!lJ:!;. Then v, • lv, .. !l!:!li\i;. and v. _ 3v, .. mli1. II - .t,v ... (I m)(I200 Hz) _ 120) ml' (0. II _ .I., v, .. (4 m)(300 Hr.) .. 1200 mIl).

U .-48 A 16O-cm-1on1 SIring hu two adjacent reson~ al frequ.ncies of 85 and 102 Hz. (_ ) Wh al i. lhe: fundamenlal freq ... ncy of Ihe llrina? (6) What i. the !cn!th of a segment at the: ss. H. reson."",? (e) What io the: speed of the WIVe! 0<1 1M llrin,?

, (.) ll>e I~OOamenlal i!./, _ u/(2L), tM /lrh harmonit freque"'Y is "t, - 85 Hz while Ihe ,,"xl ".nnoni<: frequ.ncy io ( /I + 1)/, .102 Hz. From these...., obrain I, - l11:!! and /I" S, (b) For /I _ S, IM,e are five segmen .. , rhe length of each i!. 160IS - Hm!. (e) From 1, . 17 - 11112(1.6)]. ~ _ 54.4 m/J,

13.49 A venieally ~nded 2OO-cm length (II string is given. tensioo equal to 1M weigll! of an SOO-g mass. The "rin, i, four><! to resonale in thr« '",men" to I h e<juen<y of 4SO Hz. What i. the m ... p'" unit 1e,,! lh of Ihe Slrinl?

, In lhe third harmonic. (3.1.)12 .. 2.00 m; so lhat ~ _ fJ. _ 480(1,33) _ 640 mi •. W. IMn use the .. Iatioo II' .. T i ll 10 lind I' .. ("'1)/11' - 10.10)(9.80)]1640' _ 1.91 )( 10- ' klllm

13.50 "The equation for I panicutll Slandin. wa .. on a "ring i. , _ O. I' (>in 1< COS JOOr) m. Fi OO Ihe (d amplitude of vibration at Ihe antinode, (b ) disla_ hel_en 1IOdn, (e) wavelength. (d) frequel'oC)' It) speC<! of Ih" w",'e,

, (.) By CQltlpari<on .,.;th Yo _ .-4 $O n [(2Jfz)ll ]cos (brfl)"'" have .-4 - 2.ll.m. (b) When the: ""umenl of lhe: .,:"" _ 0 , II , 211, .... _ have nodes. Si"", Z .. 0 is a node ..., ha"" for lhe ",,1I node 1< .. II and -" - il lS -~. (e) "The wave",nllh io ""ice thio, ... 1 .. (2,. )15 - LZll!I. (d) Sina: 2,,, _ 300. I _ ISO/" _ i1.lJ:Il; 000 Ie) ,, - ),f -J(2,.)/S](I SOI"j -!lQ.mIj.

J7ti a CHAPTER 23

U .51 An <)flan pipt 1 II long" Opt ll .1 bolh ends II Ih~ "elocil) of ' 0'1[101 io 1100 It / ., ,,'II .. are Ihe frequtooes of , toe f,, !>damen,al and of ,"" fi lS' h.() "vert"D<~?

, A, (:1111 be II«n ff()m fig_ 2HI. ~(A/2) " L h<>lds f<)f an open orgiln pjpe as " 'ell a. a , uing. n. .. ,. Ihe fund.men'al frequency " V," u12L .. (1100 f, /»/2 fl " S~ tlz. l1>e firn <)\'en one i. l!>e 0«000 h .. monic. aOO V," h ," 2(SSO) .. J..I..OO..LH Si milarly v , _ 3v, _ 3(5SO) .. J.@l!l;(second<werlone),

I . .. tI,

>< --fuM."",. 'al

xx-1',,,, ,,.·w .... '

., - I .

U.52 A d"""d org~n pjpe 2.S II long i. _nd<d, If Ihc ,·"'oci,)· (,f Ille $<lund i~ l l00ll/s. "'hal are lhe "'OOamenl.) freq ... ""y and II><; firl' 1"0 o",r",r>eS'

, For a "ckr$W" pjl'< (t ha, is , ol'<n a, one coo) (:!n - 1)(1. /4) .. I" " , can be .. en f,om Fig, 23·9. Hence v,._, .. fUr - J Ju/ 4L .. fUr - J) v" >0 <>Illy ttoe 00d harmonic> appear. For Our c.",. A, .. 4L .. 4(2.5) .. It) h , " _ v,A, or 1100 _ v,( 10) and V, _ l!!!lil:, The fim overto.e i. lbe Ihird ha,monk. $<l A, .. 4L/J and v, .. ,, / )., " 311 /4L _ 3v," 3(IlOj " mlll, Simil.rI~. v, _ lv , _~,

c~ ,-

I ,-

il :.5(,

1 FiI·23-9 FiI· 23- IO

1J.5J A sounding luning l<)f k .. -no.. f,eq""!K1' i. 25t> HI i~ Mid ~,'e, .n emply "",. ,uring c),hllde •. S« Fig. :!J.IO. The !<lund is fain. , bu, if jill ' 'he righ' am""nl of " 'al" is poured inlo 1M C)'linder, il b«<>m« loud. [f .he oplimal amoonl of ,,-ale, produces an air colum ll of Icngt~ 0.31 m, whal ;. ,b~ speed of _lid in air 1<> a fi,,1 apll'o, ;ma'ion?

, Th~ loudeslsound will M llea,d al re>OIIane<:, ",lIen.1Ie frequency <:>f "ibralion of 1M air colu mn in 1M cylinder i. Ihe SlIme as Ihal of 1M lun,,,, fork Si""" 'he ail col~mn is open alone eOO and clooed al 1M other. "'e COItoClude Ihal Ihe " ',,'.I. nglh of Ihe vibra,ion is lou, times .he lenglh of Ihe column: ;. " ~L _ (4)(O.3l m) _ 1.2~ m, l·I ... .... ha,-.: >S$umrd ,hal 'he "W"'c;l r~>onanl <»<ill.,i.,.. of lite air column io ;IS fu"d"",~,,'~1 UK,mal;on, Since Ihe frequency V " 256 Hl. Ihe ,;<lund Ip«d,, " vA .. .!l1.!!!1i, Thi. i. an undt"sli""'le .ince Ih. displacement anlir>od. lor Ihe pre" u' e ""de) ,.-hk h i,localed a d,scance of

378 ] CHAPTER 23

23.W A unifQrnI 'lrina: (1e",lh L. !i"" •• den';ly 1'. and lension fJ is vibfaling wilh amplilude A. in il' nIh mode . Show Ihal its lOla! ~""'8Y <>I ...aUal;';'" i. liv~n by £ _ ",',,~!!,L.

, The displ"""""nl'" Ihe "ring is livu by

,(x./l - A. sin "'; COS (2"'~.' + 0)

'" Ihe lra",'·e, ... '",!ocily is

~, ""'x ~ - -h".A. SillTsin {2:<v.r+ OJ

Th( .Irin,', 1<1131 ene.1Y of ...an.lion is C<jllai lo lhe muimum kineTil: ene.gy (NOlo Ihal QI/ poinT> on Ihe. >tTin, achieve lhei. m .. imum kineli< e"".sy althe ...... ti ..... when, - 0 fOt aU -<_ J SiJl<e am - I' rU. we have

The ma. imum v.iue occurs when sin' (2.~v.' + OJ - I. "' ..... find th. t

, '{" ,,"''' E - i (L~v.A.)" sin'T<U

n.. av"a,. value of sin' j(n,u)/ LI OV(Cf any nurober'" hal f "Y"!., is Ii"cn by \. '" lhe illlegral has lhe val ... L /2. Therefore E _ 2", '",,: .,A;(L/2) _ ,,' v'.,A ;/lL • .. desired. [C>mpaIC wilh lhe r'5'l11 lor . tra".ling "'a'·e. h ob. 2J,Z9.)

23.60 A laUl squa.., ""rob,."" &Scro 0)11 a ,ide is I.".""d 10 a rilid fra"" on ill edges. When lapped lighlly II il' cent". il si"'" <>f( • 10<><: <>{ aboul 200 Hz. A ..... millS ;1 10 be reS<>lIalinl in the mode shown in FiS. 23·12, " 'hal i5 the ..... "" . p«d in , he ""mbr"""'~

, The ,.""","1 r .. qucnc:i.-s of. reclan,"la, drumhead al. ,iven by the f<>rnluia

, '(m)' (")' f_- - V - + -2 L. L,

.. ,here m .rod n are lhe respediv. numl>crs of half' ''''avclcng,hs lui fil inlO ,h.e dimensi<>ns L. and 4. N'"'' 23-12 shows lhe m<>dc m _ " _ 1; hel>CC.

givins p - 2+0 m/I.

'II alenal

24.1 SOUND VELOCITY; REA TS: DOPPLER SHIFT

DCHAPTER 24

Sound

l4.1 Hetium it a moooalomic gao thai ha. I den";ty of 0.119 kglm' at a praou." of 76cm of meorcury and a tempoJalur. 01 O'C Find Ihe sp<~ of mmpre$$ional wa,'a (WU M) in helium . 1 1m. tempcratuu and pru, uu

, u. VB/p, .... he •• 8;' I"" adiaNlic bulk modulu •. FOI' an ideal ga •• 8 .. rP (..,e ('lOb 21 , 2~ ), and y" 1.67 (Of • monatomic 8as. ~n

~1_67(L013 x 10') N/ m' """ _ I. V · _~

O. I19kg/m'

1.4.2 Usinl''''' fact that hyd.ose n g3~ wn,iu, of diatomic molecules "";th M .. 2 kalkmol. lir>d (he sp<~ ul wulld in hydrogen al 27 "C

, For ideal I''''' ' 1/ .. Un> )/1' I'" .. [{rRT)1 Ml" . SO

" .. [(1 ,40)(8] 14 J/ kmol - K)(JOO Kl/12 k"kmol)] '~ .. UZI mil

1.4 .3 From the fact ,11<11 the molecular "'ci&l11 of ""y,en molu"lc, illl kg/ kmol . find .he 'P"ed of _nd in ox)' • • n '10 "C.

, As in Prob. 24 .2. II"' [O.40)(8J14}(271 )13W ' - lJ..lJ.!llj

U.4 The ,..,k>ci'y <>f ""und in a ronla;"". 01 hy.!, ... " . al -13 'C i, app'", imat.ly .wooft/~_ Whal woold lhe \-doo ly be (in fl/s) if Ih~ lemp.t. alUre of Ihe hy(hogen ,,·e.t . aisW 10 121"<: wilhoUI • cha",~ in yolum.?

I Sin~ to - l(yRTl/ M) '" wc hayc. "s->uming COnSla"1 y. v - ,.,,\'T/T.,- (4000 ft / s)V(.wo K)I(200 K) _ S6§7 ft f1.

Z4.S v,'hat is Ih ..... e<I of sound in oi. ",hen (h~ lempe..:"u •• is H"C? "The speed of _00 in oi. al O"C i$ 331 m/,.

I ,,% Vt - '1273 + I; W if "" - .peed .. I _O"C. to _ v.,( 1 + 11273)'~ _ 331(1 + 3SlZn)" _ J~16m/$.

Z4.6 By !>ow mum must the lempe,alu~ of ai. nca. 0"<: he changed 10 <aus.. lhe speed of souoo in il 10 change by I pe.cenl? , (27} + I) " - (27.\" ' )

273" "0,01

N01iRg 11 + ./27J I'~" I + ./!Hi>. we gel (. /S46)(I00) .. 1.0. which I.ads to I _ ~,

Z4.7 A «nain gas mixtu.e is compo.scd of two diatomic gaon (molecular weight> M, and M,). The . atio of Ihe maue. of the 1" 0 80s.., in a gi"en volume i. m,lm, " r. Show lbat Ihe . peed of _00 in the gas mi .. ure is., foU""" if ,he p...,. ar. ideal :

~1.40RT M, + ,M, ". M,M, 1+,

I to _ h yp 111')"'. b"l I' .. ",IV .. (m , + ... ,)lV. and (Oallo,,., law) p _ p , + p, _ (" , + " ,)((R T)/V )I. ",ith " , _ m ,IM, 000 n , _ m,IM, . SO that pI p - 1(m ,IM , + m j M,)RTi I(m , + m,) _ [(11M, + . / M, )RTI/( I + .). with, - m j m, . Subslilute pi p and y _ 1.-liJ into the up,ession fOf to 10 ~nd Ihe lequiled I.soll

Z4.B SUP!""" (,",'ilh Ne,,-ron) Ihat Ih" romp .... ional \'ibtation in a 8"' " 'ele i>01 h.,m.1 in cNlractel .. th., tban adiabatk. Find Ih. e.ple"";on equh·.lcnl to (}"p I p)' ; for Ihe speed of >ound in Ihat •• s..,

, Fm pV a eonsl.nt. "'lpV) - p ",V -+ V "'p " O. 0 1 ",V l llp _ - VIp. BUI flV l flp" - V I B f.om the de~n;lioo of B. Th~Ief()~. B - P and" .. (B I 1')' ; .. (pI p) ' '. Thus " .. /to ,~ .. y"

'"

380 a CHAPTER 24

24.' A looo.Hl wund wave in air s,ri k .. ,he surlac<: of a lake and pene'ra'" ;n'o Ihe water. Whal a,.., 'he fre<jucocy ~nd wa~eleng'h of ,he wave in waler? Assume Iha'ihe speed of ""und in warer is lSOO mls.

I n.e number of c<.>mplete wa'·eS pasoiRg .n~ poinl in air aoo in " ale. in unil li"", i. ,he same. 50<' ,_ 1000 Hl fOf both "",dia. n.e,..,fore.

1 _ ~._IS<tl m l' _, <_ w , 1000,' .t.c<..!!!

14.11 An "n<lt""a'.r ""nar w "ree OJIC'ra'in8 a' • frequency Qf WkH, di,..,cl> it. beam lOWa,d lhe .urfooo. Wha'is lhoe .... '·elen.' h of Ihe beam in lbe air ab<m: ? Whal frtque""y "",nod d". to ,he "",ar _= doe. a bi. d ftying .t.we ,he .. ·ate. he • • ? Assume ,hal,,;n a;, " 33Oml •.

, A, indicaled in Proh . 24 .9. "emaiM constan,- in Ihi. e .... a, 60 kHz. The wawtenglh - "If -1301(6.0)( 10') -li.mm.

24.11 o.fine p;t<:h. ioudnr .... q""Uty, d .. d~l. "''''rlH-''''''''' n"1tIl". imerjr'rll«. ""au. Dop"u' ejf«:,. "'ptrSOIIi(: wlOOty , .hod waw, and MIlCh ~umlH'.

I Pil€h i. 0 ..... nd charaCleri"i<; Ihat <ltpen<H on Ihe f.eq",,""Y of Ihe fundamenlal. Higher pilch "",ans higher bequency.

/..J>udM53 .den 10 Ihe , trenglh of lhe audi,QI)' Stns>lion produced by 0 ..... nod . It depends on lhe inteMi'y anod frequency of 'he «mnd_

n.e qlUlU", (or limb .. ) of a wund <ltpends on lhoe number and ;n .. ",i.ies of the ","e"one'_ D«iM (dB) is Ihe unil of inle",ity-I .. el. fr, of "",nd. One dB i. len limet lhe log of an in'ensily ratio of

LUt: L T1>c: equali<m for fr i. fr - 10 tog (1110). wllc:Te t. _ 10-" W / m' . The ,...,..me""iotI lim. of a room i.< , he Ii"", required for lhe ""und level.o fall 60dB after I~ ""und

"""rce: i. shUI ofi"; Ihot is. for lhe intens.ily '0 dtt .. aSt by a foClo. of a million, Inler{rrtllC' is Ihe wperposition of wund ...... "<:5 10 produce: cither constructivc 01" dt'$lructive addilion. &tU$ are HUCluatiool in """nd in,en";ty ,hal oa:ur when IMre is interference be",'ceo ".'-0 sounod waves of

"'Iwo! intensily &00 slightly differen, f. e'!ueocy. 11><: Dopp/u rjf«:1 il 'he app,aren. change in fr~,! ... ncy as lhe 00II= of a """nod and an OMe"'" of 'he

sound m","e ,clati"" 10 elch OIhe •. A Ju~'S<Htk II<'loc/l)";' nne Ihal i~ grta'"r than ,he local velocity of wund, A shock "'all<' is lhe wave ....... iOR acrompanying obj«1> 1,,,,,,ling a, ,upeooni<: $pUds. MIlCh nllMlHr i5 ,he ratio of the ve locity of an object. or of. shock front. 10 'he local velocil~ of ..... n<! .

14.11 Two dosed or,an pipe. """nded .im uhollCou>l~ gi"" fi"e beals per serond be,,., .. n Ihe fonda""",al<. If lhe ,horter pipe is 1.1 m Ion,. find Ihe length L of 'he Klnger pipe . ASMrme Ihal u in air _ 340 mi.

, Each pipe is oa:upied 1»' a "uart~r.wave (ref., ' 0 Prob, 23.52) ; he"ee ,

34(1 mi. 340ml. 5 H,_ ~ , - ~ . _ ______ _ • ~Am 4L

Solving. L - l.J.JI.m,

24.13 Two """n organ pipes. one 2,5 II and one 2.4ft in kng,h. are "",nded ,imul"'ncously, Ho ... many beats per Soeoo!Id will be prudPCCd be",Un Ihe fundamen,al,ones if the ve\o(;ty of lhe wunod is 1100 It l.?

I E!'Ch pipe i, omJpitd by. hllf·w.'~ (~ Pmb_ D:ll). S()

1100 fl /s llOO fi l s V, - v,_ 4,8ft - 5.0h -~

:1-'.14 FouT beats per second are hea,d when Iwo luning foru are ..... nded .imul'aneou,ly. After aUIOChing a ...... U piea: of ,ape to ooe P""" of lbe Stoond tuning fml, , he 1"'0 lunin, forks a,.., IIOtInded "pin "nd 1-...-0 bea .. per second are heard, If Ihe fi~, fork ha.. fre<juency of 1801b. whal muil lhoe original f'equency of lhe second fork ~ave been?

I The f,..,que""Y of 1M oerond fork m .. ' be higher 'han Ih., 01 'he fif$! lork Or adding lhe lape would ha"" jtt<:TN5~ !he number"" bea" . TherdUK. ~T - 180 _ 4 <>T v, -~.

:1-'.15' Some of Ihe low ke ys of 'he piano have 'WO IlrinSS_ On • partieula, key one of lhe llrin&> is luned ror1'C<:lly

ELECTRIC POTENTIAL AND CAPACITANCE a 431

V- q/e si"'"

q,; - q; 3>< 10· · F 4>< 10 ' F

Sub'i.i.u.ion of thb in tho p<e\i OVi Njuatioo givos

0.75q; + q; _ 61'C

Thon q; _ 0.7Sq; -2.S7I'C.

:16.107 T.., capacilOlS. e , _ 31'F ~nd C, - 6I'F. all: ""nn..,t.d in ~~ and d"" •• d by ""nn«1ing a banery of \/Ohage V _lOY in series wj,b 'I><m. Tho~ ar<: ,I><n di!«>ft""",ed from II>< banery. and .be K>osc wir •• are connMe<! togotber, WIla. i, .be final cflaJge on eacb?

, The capacitor> afe cbarged in se~ 5<> .hey originally have equal CMfges. Whon ,ho l~ " ·i.,,. are reconnecle<!. ,hoy neuITalize eath otho. givin8 lero final charge,

c,

C, @J C, C, C, <, C,

c, C1, 'Ij W C1, .11 " , .,~ Ah« llJ. Ui-3.\ .,~ Mt., F1t. Ui-J<I

:16.1011 Repea.!'rob. 26.107 if after being di"""nn..,te<! from the banery. the capacito ... aro diilOOfl ".,<led from o..,h o.M,. Thoy are <>OW recon""".e<! n .hown in Fig. 26-3J. WIl.t is t .... final charge on ach?

, Tho original charge on each is Q _ C .. V _ 20!,C. After bein! oonnected as olio" .". Q, + Q, _ 2Q _ 401'C. AI50 V, - V, and 50 Qde, - Q,/e" Solving'~ tWO eq uations simuhan""",:IIy gi>'fl for tho • ..., charges 26,7 "C <In tbe 6Ji F and 133 ~C On the 3 1'F.

:16.1" If , ..., capacito ... C, _ 4 " F .nd C: _ 6 Ji F Ire ori~inally COnne<le<!1O • bauery V _ 12 V .... silo" ." in Fig. 26-34. and 'Mn di"""'ntCled .nd """",,,.,<led as siI()\\'n . ... ·h.t;. tho final charge on nch capocitor'

, Originally, Q, - 48JiC. Q, - 72I'C; henc<, Q j + Q; - 72 - >IS - 24 "C. AI",. Vj - Vi 8i>'fl Q j/C, ­Qi/C,. Solving .imuh.neou>ly 8i ... 9.6 and 14,4 I'C,

C JPYnghted matanal

CHAPTER 27 D Simple Electric Circuits

27.1 OHM'S LAW. CURH I::N'f. IlESISTANCE

TA BLE n ·, It .. ioti>;,ia 41'1 lit ze 'e OIMI T~",,,,,,,I.'" C....eidno •• (CI •.

_1~riaI p."'- . Sil"~r 1.6)( 10

C"we' 1.7)( 10 A luminum 2.8 x 10 Tungst~n 5.6 x 10

Om" lOx !(I •

G,aphite (ca. l>o<o) 35(0)( 10 •

" ,"t: •

3_~x I(f '

3.'1 " 10 ' 3.9)(]0 ' 4.5)( 10 ' 5.0 x 10

,

- 0,5 " 10 '

n .1 What is the ,elat;on ~t"· •• n , • .mane. and resiMj'-ily?

, Tbt r'~I.r>« R of . " 'i,e of Iongth L and Cf(>!6-"'Clio",,1 ",.a II i.

,,'he •• /J is" ron>lant ",,1I.,.j ,be ,~si.ti"ity and j • • chara"'eri\.lic of rho ma'erial from which 1M wi", is made. For L in m. II in on' . and R in U. the unil, of" arc 0, m

n .1 How does th • ...,si~.n"" of " rondooor ,-ary with ,em1"'.atur.?

, II . " i •• "", a ,e,.;,I",,", R. at • '<m1""'lu" To. l!>tn it. ,"';\I .. nce R al • lC1!lpt""ur< T is R. R. + ~R.,(T - 7;.) . .. ,here " is the ,..mIN"''''''' ClNfJi~i~~1 of rt'sUUJ"U of t he ma le rial <Jf 11M: ,"'ire . (Xually " ,'ori .. .. ith temper atu« and .... linear ,cla(;oo i, applicabl< only ""cr a small (cmp«aI",c rang<c. n.. unilS of " ar. K- ' (>r 'C- '

A simil. r •• lali<.>fI appli~ (> tl!<: va.i •• ;"" of rcsi>(i' -ily .. ith tempe'~lu'e_ If (>. ud (> .. e Ihe .-m(i,;I;" a(

T. and T respectively. Il!<: n p .. p., .. "p.,( T - T. ). Tabl. 27·1 li,1S (hc ..... Ii';ti.' of • number of cond....-cOfS for T • • 20 'C. as ",.11 as lempe •• lure coel!icienl$

of . csi,lancc .

11..J JIow arc ""m,ol and .... rrcot density .da(cd~

432

I Th •• at. of now (>f .leclriC cha.g. across. gi'·.n ••• a (wilhin a cond uctor) i. defined as the rlrctr;r CUr"'''' f th. ough Ihal .rea. TItul.

1 _ <!i "

OM

Th. '/Ulrje (u''''~1 density J ., a ""'nl (wilhin a condUCIO. ) i,. ~Of .. 1>o!;e di,CCli<.>fI is (he direction of now of <ho. g< allhat "",nl and .'how map'i.ud.;' lhe """.nl th""'gh • uni, gffll pcrpmdkula, 10 ,It. flow diTrC!i"" al thaI poin!. Thus . lhe <u"cnllh,ough an .lemenl of .... dS, a.bit.arily oriented wilh '"1'«( (0 lhe now diJ«lion. i. given by ("'. Fig. 2H) dJ _ J . dS .. J dA. ,,·be.e dA _ dS COS II i, (he " ro;.cl;on of tIS P"'l"'ndirula, to (n.. !low. di.CClion . Th. 100ai ""TTCol ,h.wg!t a ",rface S (. _g_ .• eM» >«lion of Ih. conducto. ) i, ,n.. n

C righted matmal

SIMPLE ELECTRIC CIRCUITS a 433

-L '" /'IS ~ . ,j.5

\

~ F~",,' , h"",

-'.. , / '- • - Pl&, l7-1

27.4 Slarting from Ihe .undard form of Ohm', I . ... . V _ IR, find tho , elal;on belween J . Ihe current densily. and E , the clcctric /\old in a current-oorrying conductor.

, We consider. rondoctor of unilQml .~io",,1 ar •• A .nd lenglh L. ,.". ..... i$tana: R _ p(L!A), where p i, Ihe ,esisli~ily. The current <an be exl" •• ...! IS I_/A , and lhe polenl;.1 drOl' across Ihe resislor;' ,d. t<"<lto lhe .ver". electric field by V _ £ L , Then V .. IR b<com .. £ L -/Aip(LIA)], Or £ _ pi. Often one SPNi~ the coodUCIivily. (I , iMleod of the resi>li';ly. ,,'he ,e (I . l ip . Then / .. oE. Tn;. resull <a n be generalized to an arbitr ary condoctor in the .ector form, J _ (IE. which holds II each point in lhe conductor.

lUi li o-w many electro", pcr:scrond 1"'$$ Ih_gII • lCCIion of "ire "'trying. CUTlcnl of 0,7 A?

, / _ O. 7 A mean. 0.7 Ct • . Dividing by . _ 1.6 >< IO-" C. Ihe rlagniludc of cha'ge on a s.in~ electron, we rei number of elect""" pc. ,"corid _ 0.71(1 ,6 >< 10-" ) _ 4.4 >< 10" .

27.' A currenl of 7. ~ A is mai nlained in • "'ire for 4S •. In Ihi$ time '_I how much charge and (/I ) h"", many electrons now Ihrouf,h lhe wi .. ?

, (<I I q _I, _ (1.5 A)(4S .) ", ~ (6) n.. num"", of electron< "" i$ gi....,n by

,..._ '1 . 337.5C _2.1><10" • 1.6xlO-" C -- -

where < _ 1.6 " IO- " C is the cha' !" of an electron.

27.7 II 0,6 mol of elc<:lron. How th rough. wire in 4S min. ,.,h.t a .. (_I the lOIal charge Ihat paucs through the wire . and (It) the magnitude of the cufTent?

, (.) "The number N of electrons in 0.6 mol i.

N _ (0.6 moI)(6.02 )( 10'-' electrons/ mol) .. 1.6)( 10" elcctrotl$

q .. ..... .. (1,6)( 11)" )( 1.6" 10- ,. C) .. j 711 " 10' C

(It ) I " (4~ min)(6/) ~Imin) .. 2.7 )( 10'.

27.. An elettron gun in a TV '"I !ohC)Ol. out .. beam of electrons. T\c beam current is 10 "A. 11_ many electrons strike lhe TV ",",u each ",cond? II"", muct> <h~rge >lrik .. 1he "'recn in _ minute?

" .. .. ,."." M electIOns peT !.Ccond. n. _ 1/ . _ (\.0 ~ 10- ' CI_)/( 1.6)( 10-" C) _ , ,.". charge Q mikinl the ""ee, obey> 10) _ IT _ (10 "C/.)(6/).) _ 600 "C . . 1"" lICI ual ""0'1< i$ 0 .. - 600 ~C .

27.' In the Boh, modol. tl>o electron of. hydrog.n _tom m"" .. in. cifcular orbit of ",,:'ius 5.3" 10- " m ,,·ith . speed of 2,2 >< 1000m{. , Del<rmine iIJ freq ... toCJI f and ,he CUrt ... t / in the orbit .

466 [] CHAPTER 27

,

L'---l $,,'it.h i, cl"",d the c~""tit'" ... ·ill ,lil(hargc 'hrouglo ",~isl''''X H. fioo ~n upo-es.s>oo fo, qUI. tM charge on 'he <"p"cito, at .O ~ ti"", I.

I Wit h Ihe switch d"",d ... ·c mltSt h,,·c for the I""" e'lu.tion q /C ... III: _ U. Si"'X 1_ rate of dj",harge. I .. dq/ dr. 11I<n. !>Uh$titu,ing in , he INJp c'Iu>lion . ... ·r ge' dq /d, _ - ( I /HCIq , Or dq /q .. -( I / II:C I dl. Integ .. ting,

, ' In - __ _

Q II:C

11I<n q _ Q.- "t«'. l1!e quantit)· He • • i. ,,>lIed .he ,i me «.m<lant "f the ti",uit .• he 10'1"' '. the Ionget the di..,h.,ge time .

n .14Z In ' hc labo, atory, a ,tu""nt charges .1-I' F capacitor by placing it act.,., a UN ba ttCf)'. v,'hi!c dioronn«ting il. ' he Sluden' OOIdS il> two I""d ",·ire. in t"·,, hand.. A .... minl . hat the , ,,;"a,,,,,, ,,( the body bet",'«n lhe hands i. 60 kU. "'bat is the tim< ronstant of Ihe suit> ti. cuit rompostd 01 tbcc capari." . and the uudent'$ bod~? I I" .... Ion, doe!. it take for lhe charg" 00 lhe ,-"parilOf 10 dtO!> to I /~ 01 it> o. giinal ,al ... ? To I / IOO'! (Hi"" In 100 " 2.3U log 100.)

I . .. HC .. (6)( 10')(2)( 10 ") .. 0. I! •. N"... 0" 0 . .. 1' (-,I RC) " O. exp( -I{O. 12). If cAp( - , /O. ll ) .. e~ p (- t ) . then' _ 1). 12 •. If exp ( - ./G. 12) _ 0-01 . thrn HI' (,10- 12) _ 100. "" .10. 12 _ In 100 .. 1-:10 ~ 100 .. H,(Lso ' '' ~,

Z7. I<13 In a """ain dec" on;" <levitt, • 1t).I'F capadtor is . .... rgcd to lOOOV. Whtn thc de,itt il $hut off. the '-"pacitOl i>: d(",harged for !o:lfety IC""'~ by a W-C1l\l~ hl<:<:dcr ~,'(/f of I MQ placW :>eros. its tcnninab. How long doe!. it take for the m a.ge on rhe c;op<>cirOf to d«tc,"" 10 fUll of its original va luc?

I We . N.11 that In 11 .. 2.:10 '''I! '< ; . .. RC _ 10 I We "'ish that Q IQ. _ up (-' IT) .. 0.0 1. From Ihis. '/10 - LIO loS 100. "hich giv<:s • - ,!M.

27.I-M A -IO).I'F capaciH";' ronn«l~ Ihrwlh a resi .. or to a Ntte.y. Fin<! (_I tM "':sOSl a",,,, H and Ib, the . ml 01 thc b. nery Z if the time ron" am of the ci"",il is 0.5. and the maximum rna.ge .... tM capatilOf is 0.024 C.

, 27.145 A 5().I'F capacitOf initially unrn.rged i> connected through. JOO.O rc:sO.rOf 10 a 12· V Nttcry. I_I Whal is the

"",,,,itude of lbe final charge q. on . i>e capacitor ? {b) I low IonJ aft • • the capacitor is ronnc.:l«! In the ba ttCf)' will il be cha'Jcd to iq.1 (e) 110'" 100& ... iII it take for the capacito. 10 be ch • • gcd 10 0.9Oq. ?

I In d .a.ging we h.,·c 'I - q..(1 - ~" '.'). with 'I. _ CV. (d) q. - cy ... (SO)( 10 • F)(12 V) - ~"c. (b) T _ RC ... (300 OJlSO )( 10 • F) .. 15 )( IO"! .. 11n!l. From tM IOfmula the charge rexh"'. • in the lime I \u<h Iha, ~ -~"" .. !. o. ,{ (RC) .. II . - In 2 .. 0.7. Thu>: ,~ , .. O. 7t - 1.!1..t!ru. (d Similarty. , -.' .. O. I aoo ,I . - In 10 " 2. 3. Thus ' .... 2.3. -~.

n . '46 A 1S(i.I'F . ap<><i'or i< waneoed ,brough a soo. O .e:sOuor 10. 4(l.V ballery. (.) What i. the final marxe q. on • capacilor plale' fb) What nIh • • i.nc <:o.mant of the ti, cuir7 (d H<l'" long doeo i. take the charge on" capacitor plale to .e;u;fl O·Sqo?

, 1_) q.-VC - (.j() VII I50)( lU • F) -~. Ib) . .. II:C " (5000)( 150)( to • F) "'ll.mI. (d FoIIO"';ng It.c pr<>«dure in Pmb. 27 . 1 ~5: -II • .. In 0.2 or l/r .. In 5 .. 1.6. Then, " {I .6)(15 msl - U!l!!!i.

THE MAGNETIC FIELD D 469

'" ~ , (3x HI-' m/.)' ,

, . - .. . !'I _ mm .. 1.81 x 10" ml"

:IS.U If il i> m umed that . circular ""lh around 1~ (al1 h I,..,.ius - MOO km) can lI<' foor>d upon " 'hich 11 .. eal1h', field i> horizonlal and conmn! at O.:!OG. how fa>! mu>!. Pfown J,., mot in order to circle 11 .. e.rlb? In ... hat di reaion?

, The ""nlri!",'al (""'" (mo ' )/. ;. furni .... d by It.. for« ",B". Then 0 _ (~B".)/m _ I{ 1.6)( 10 ' ") x IS )( 10- ')(6, 4 x 10")11(1,67 x 10- ,,) _ 3,1 x .10" mi •. It rna$! be 01>01 "'''lword. since It.. field poinu ""flh ,

:tII.t4 Why is Ihe solution to Prob. 28.iJ faull y~

, 'The ~ of the prol0n cannot u=d I~ sp«d of lilhl . Evidendy. It.. tel.limli. «)lLilion of motion 5houW ""~e been employed.

:tII.15 Alpha ""l1id .... 1m .. 6.611 )( 10 " k,. q. 1-2r I .• ""elerated Ihrough a polenlial difference V 10 2·hV. enler a ma,,,,,, i" field 8 _ O.2 T pcrpcnd;wla. 10 It..ir di.eaioo of rnOOon, Calcula'e It.. • ..dillS of 'heir pl-Ill.

, n.. KE of. particle is ronsc .... ed in the m"l"<'ti< ficld :

They 101""" a cirntlar path in ... ·hich

, 'i ", v l .Vq

,.~.~ ~.! l vm __ ' _ l(I(1K)V)(6.6II X 10- " 9 }.32mm q8 qB V-;;- B q O.2T 3,2)( IO-" e

:IS.16 Afte. being ~Ic.aled through a potenlial diflerc:n<:. of 5OOOV, a ~ngly chatgi'd carbon ion """' .. in a circle of radilK 21 em in 1M mag""'ic field of. ma" op«lromctcr . What is Ihe m..."ilude of 1~ ficld?

, Follow;nglhe approa<:h of Prob. 28.IS.

B' . 2V", .. _ 2(500) V)tl'.')( 10-1" kg) .'q (0.21 ml

'(1 .6)( IO- '"q

' .. 1.112 x m-' T' 'S-21m

:til." A panicle wilh chargi' q and rna" ", i$ oI>ot wilh ki""li. energy K into 1M r.,ion bc",'un Iwo plates a, shown in Fi, . 28-2 . If the m...,..li< ficld be''''un lhe pial" i$ Band a • .00..11, how largi' must B be if the panicle i. to miss ooltision .. ilh Ihe opposile plate?

" " " " " " .-l " " " " " ,

FIc·Z&-l

, To just miss tilt oppooile plate, Ihe p.anic~ mllSl mo"e in • circular palh w;lh radiul d SO frum Bqd _ mo; and usinlL K .. ("'0' )12, we h .. e B _ (2m K)"'I(qd),

:IS.I' A co,hodt-ray beam (an ~Iectr"" Mam; m • 9.1 x 10-" k!. q • -~) i> benl in a <in:k of radius 1 em by • uniform field ,,'ilh 8 _ 4.S mT. WIIa';' The spud of the ckctrooo,?

, To describe a cirrl~ li ke t~i,. the palticle must be """,ing perpendicular '0 II , Then

" _ ,,,8 _ (0,02 ml( 1.6 x 10- '"_ ~)t". 5)( 10 'n _ La " 10' mIl m 9.1)(10 kg

:tII.I' In Fig 28-3( .. ) .• prown (q _ +<, m _ 1.67 )( 10- '" kg);s ,1>01 ";Ih ~d 8" Hf m l~ al an a"sIc 01 3ir 10 an z4reaed field B . 0. 15 T. Delerill<' I"e path followed b)' the proton

THE MAGNETIC FIELD a 473

, H~,e .

and Ihe \..orenl' equalion &i= f _ _ (3)( 10-')1(5)( 1(0)k + (4" U.l')(1).766j +1).MJk»)( (Uk)] - (3)( 11)- ,)[(5" HI'jl +(4" 10')(1).766)(1.3)1]

_ 1),lml +O.mN

2:11.:111 The yol ... of el .... can be oblai ... d by us.ins. sp«ially designed vacuum lube iIlUr.I,aled in FIg. 28--7. [I COR"uns a .... ,ed 61amen, F .nd .It anode A -..hich;s maimoir>e<i 01. pooiliye pocenlial re~li'e 10 I'" filamenl by. bauel)' of known WlII~ V, EleruOM enpora~ f,om tIM: I>(aled ft~mefit and are oa;c-I<raled 10 II>( anode. which h ... small hole in Ihe .. n'" for ,he dec,",o,," .o PI'>' ,brough into a region of OOMtant magnel;" field B . ... 1>id! poin .. inlo 1M papt •. The denrun$ IMn mo". in. semici,el. of diamttn d. hillin, 1M dele<lor .. "",own, Ploye Ihal elm. a 81' / (Bd), .

• , Refer to '",. Z8-7, Ne .. 'ton"..."..,.a law implieo that ( .... u')/, - .-vB. ,.here, a Id is 1M . adiu.o( Ihe electron', orbil. Soh'ing fOf 1M op«d, we find I,,"t

eB, eB "0- 0- ' m. 2m.

Sine<: lhe electrons h."" been """,,!<raled t<sentially from '"' Ih.ou,h a potential differeroa: V, Cltd! hu kinel;" ene. gy Im.u' a eV, "" thaI

Combining E",. (I ) and (2), we obtain

" uV "" 0-m.

0)

(2)

2:11.19 If in. "' ... !op«Irugnph "",bon;oM """,e in. circle 01 .alli"" o _ \l,Orm and O»:YFn ions moye in a circle of radius '0 _ 11),4 = . .. 'ha, is 'he m"" of . n oXYlen ion?

, We ..... me Ihat V and B are tbe 101m< fOf both ion •• SO thaI from Eq. (l) of Plob, ZIU8. m a (B' .'e) /(2V) and

m~ .{. (11).4= )' - a - a al33 "'< ' ;' ('Hlcm), .

TIlaI is. the mau of an ox)'gcn iOd is 1,33 tim .. lbe m"" of • ""rbon ion. Si""" ,be ma .. of carbon is de~""d to be ua<11y 12 ~, 1M mau of oX)'Icn is "'0 - L33m < - (1.)3)(12 u) - l!uI,

MAGNETIC PROPERTIES OF MAnER a 519

O'-er the volume of lhe wire. 'The ""'gnelie ~M'IY dU~ oo.uained in a cylindrinl shell of knglh I, in"", radius " .0<1 oule. ,adi ... , + dr;1

Il ' I') ,.';'1 dU~ .. r~ dV " ~ 2;r,ld, " --. r' d,

2/,. ~"" Henre the magnelic enNIY p<' un;1 Itnglh contained in lhe ";.c is

- -- dU .. -- r'd, _ __ - . -U. 'f ~" [ "" ("I) ~" I I ~ ~,~Q' ~;r~' 4 1M

29.ll 'The e"e.nal mognelic field of a spherical objecl of radius R, sU.ro\Indn! by varuum and e'''yin. an (idealized) renlral poinl maen.l;'; dipote, conlains frI ....... ti. e",,'lY U - lot' /2)IiB;R')/(2I'.lI· The "uanlily B,. il Ihe muimum ""' ...... IK: field strength al the >uri_ of the object-that i., the ~al"" of B at lhe object ', ma"",tic poles.

(a) The u .th '. nt.mal magnelic field i, ~ Ih.t of a pure «,me.ed point dipole, 00. i. tile ea.th surrounded by. p<riecl vacuum. Ne'-enhele ............. bIe e.lim.le of Ihe e",,'IY conlenl U in lhe e.lernal fitld is u .. (B:,R ' )1(21'. ). Evaluale Ihit. " sinl B,. _ 6.0)( 10- ' T and R .. 6.4 )( 10' m.

Ib) Com"".e the e"".1Y U 10 lhe lou l ann .... ' u ... ge of electric eM.1Y in the United SIal.,. I ...... 1.7)( lO" kW· h in 1972.

, (a) With B, _ 6.0 )( 10-' T, R _ 6.4)( HI' m, and!'o _4;r )( 10- ' N/ A' , .... I\.iove

B' R' (6.0 )( 10 ' )' (6.4)( 10')' U .. ::.t:.:.. _ _ 38)(LO" 1 2,.., 2(4.~)( 10- ' )

(6) The deetric e"'''lY " ... g.;n llIe Unit., Stale. fo< 1972 ... as

E _ 1.7 )( 10" kW . h _ (1.7)( 10" )(10' 1/.)(.l6OO.J .. 6,1 )( 10" 1

29.2 MAGNETS; POLE STRENGTH

29..M o...c.;be 1>0 ... 1M 10tq"" 00 a bar maglltt pLa.ced i. a magnclic field can be upr ...... d in lerms of poLe strenglh.

, When a ba. magnet i. pt.eed ... ilh its axis P<'1"'ndicuLar 10 a unif""" mag""lie fieLd of inl~nsily B. it e'p<rienca a 10.""" , _ Bm _ Bpi • ... her. m is tht magnel;'; moment, f io tile lenglh of the m."",I. and p ;' the poLe su ength of Ihe ma ...... l. FundamenlalLy. Ihio torque ames from the inleraction bel ... en 11M: rnag""lic induction and Ihe infinite.imaL currentloop$ "<soci.ted wilh lhe o. bils and spin of cenain of lhe magnefs elect.on •. An aLternativ. approach, f.e"I>Cntly ronvenient, is to endow the magnet ... ilh a pair of e"u.L and OJ'I'OSileLy signed /KJ1c al il$ I...., end< . and to anriblHe the lorq ... 10 lhe e"ual and "I"P"'ile 10',," e.ened by Ihe licld B on Ihese poLes. When a pole of ,trength p;, pL~d in a magnetic field of inlensity B, tbe pole '''p<ri.cnces a lorce F _ Bp. In !e""nl. , .. m8 sin 9. ,,·her. 8 _ .ngle bel,.,un .. and B.

29.)5 An iron bar ""'8Mt of Lenglh lOcm and e<aU lo<OIi(>n L.O""" hal a m"-8-""tizahon I)f 10' AIm. CaIcuL •• e ''''' ma",e", m.!""I;'; poLe st .. ngth.

, ",. magnelic moment of. bar ma,""t of length 2J bas m",nilude m _ 21P1 d., ,.,he .. Ipl io Ihe poLe .I,~ .. g.h . In terms of Ihe magnetizalion M, llIe ma"",'ic rnomtnl of the bar magnet is .. .. M(2ad). ,,'hef~" is the o<aU·sectional a,.a . Hence Ipl .. mi ld .. dM. Wilh ,, - 1.0 em' _ 1.(0)( 10 • m' and Itt .. L.oo" 10' Mm, we oblain jp l - (1.00)( 10- ')( 1.(0 )( 10') - 1.00)( IQ- ' A 'm.

29.36 The field of . bar magnel can be con<i<k.~d 10 be <.used by a suriaoe O\I.",nll\ow;nl "" the .... riace of Iht mag""l. If . ba. ma",el is 10 ael like a Sl)knoid ... t>oo.e intetio.- /\cld is 0.3 T. ""'"' Ia,se' .urf""" ourrent must 1\0,., 00 each centimeler Len,lh of Ihe bar"

, In Fi,. 2'#.7 take lo<gmcnl Length / .. 1.0em. Sine.; B" •• ial and - 0 outside , he """,noid, lhe c;rou;I.L la,., y;.Ld< 81 equaL 10 ".,1_ . so 1_ - 0.3( 10 ' )1(4;r)( 10-' ) _ 2390 Aim _ 23.9 Mem.

29.l7 "''hat is Iht .eLalion,""ip between Ihe magneTizalioo (1)1" Ampc .. a" ) SIlrf~ """'n' 1_ in a Ion8 bar mag""t and Ihe pole men! lh?

, From P,ob. 29.]$. p " aM. Applying the a.cuilaL II'" for B 10 tht magnel (Fil. 2\1.7), aO<l OOI inlithol B _ " ,,(H + lit) and that H ca nnot <X><tTribute';nee the ..... no 11"1" cunenh . .... g<t MI _ I_: Then p_ "/~/I.

680 a CHAPTER 36

36.46' Reconsider Prob. 36 . .w " 'ben lhoe IIOuroe ~mi!> a O;OOlinu"". m;'lu'e of wa"~leng1h •. The rtWh>i", po_T. al ..... vtltngth A • ." II\( graling is defined as R • A16, wile", 6 is Ihe small ... t ..... vclength dill'~r.""" fOI" wllkh lhoe $ptCl.ral lines l -\'" and A + !6 are rellOlyoble. find R~. d •• resolYing pow1:. in lIIe mlh orde • .

, ACOOfdin,1O R")'leitlo ', rriltriott. two peak>.re iu>l resolyable .. hoen thoei, angular 5<"paration is ball lhoe anguLa, widlh." oillle . peak. Thi. minimal .. parat;"" is iii"'" by (1) 01 Prob. 34.40 as ,

(69)_ - Nd = 9

Dill'e,.nti. lioo of t ile "atina e1luation. d sin 9 _ mA, gi_

, d=9(69).. _ _

N

ComJl"rin, the two up ........... 10. d COl 9(69).. gi_ R* _ mN. Note that thot resolying """ ... is lhoe .. me for all wav.ltngths.

36.47 A tunsmisslon "aling is uoed with light illcio:knt normal to ito pl."". "The: widlh 01 each slit is o",,-third thot .pacing belWffn slit • . By comiderin, ';ngle-slit dilfnction .• how Ihallhot third-orde, (j _ 3) mwti$lil dill'.action malima a •• miss.ing from lhoe dill'Tactioo pattcrn of lhoe ! T&ling.

, Ai diKUliCd in Prob. 3~US. lhoe intensity prof;]. in 1hoe """.all dill"artion Jl"U.m d..., to N sli .. of ~nite width is thoe product of thoe '"ideal"' N·slit in .. n';ty profile and tile single..sJit pro~le. Acco.din, 10 Eq . (I) of Prob. 36.35. tile j th ",,,,,imum of the ideal N ... lit profile OCCUT1 at an angle 8, luch thaI .,

lin 9,_ ;- (I )

",he .e d ;1 lhoe slit spacin •. Howevt. , if cl><il 51it has width ... _ d f3. lhoen fig. J6..14b [0. (I) of Prob. 36.33) implies that tbe lingle slit profile hu utO< at angles givto by

sin 9 _ mA _ lmA (2 ) . . , 10. m. O. ±I. ±2. ±3. Equation. (1 ) and (2) silo .... that the o""nUl profile mu>I haye. uro at "vtry third maximum of lhe ideal N-sli1 panem. In particula • . lhoe third ... rde. maxim. (j _ ±l) a.e ""TI""",d by lhoe finol minima (m _ ±1) of tke ';n,I ••• lit Jl"nem.

)Ii.. A beam of light of "'·. ·.ele08th .. f.11s on a diff,a<lion g, ati", of line "",ring D at "n!1e of inrilkn<e 4t "",",urtd from thoe roormal to the plane of lhot '"tin,. Sho", Ihal the maxima in \hot diff. action pan.m oc<u. at In,," 9 "'hich arc detcrmintd by lhe e1lultion jA ... D (lin 9 - li n .) . .... he •• j. O. ± 1. ±2. ±l. . ..

, "The: litultioo n indicat.d in fi,. 36-17. The: initial beam direction i, ... umed to be perpendicula. 10 lhoe

htoo IT £na

686 D CHAPTER 36

now poIariled at 3/)" 10 tile oxis of tile analyze" so tile inlen)ily of IIle lighl pa«ing Ihrough Ihe analyu. is 1'. f cos' 9 .. 0_134co.']If - 0_5634_

36.7e Polarized lil!,hl of inl~n';ly I.. i, incident on • pair of Polaroid w eI,. LeI 6, and 6, be Ihe angles bel ... ·...,n lhe incidenl .mpliludt and Ihe ""e. of Ihe finl and second ""...,to, .e'!'Ccti""ly. Show Ihal the inlensily of the transminedlight i. f - 4=' 9, <06' (9 , - e,l.

, The ;nte",ily of the liglu ofle. pa«inlthrouih IIle ~r:;t polarile. io I - I"cos' e,. This light is pola. ize<! in tbe: di rmion of the ui, of the ~r:;1 """"I, ond so ito ""is mlk.-s Ihe angle 9, - 9, ",ith the "";.. of the seoond weI (..,e Fig. 36-23). CooseGuen'ly. the in,ensi'Y of lbe hl!,h. aftu pauing Ihroulh lhe st<X>nd polarize. io

f' - I co.' (8, - 9,) .. locos' 8, co.' (9, - 9,)

36,7\ l'wo Polaroids are aligned wilh Ihei. a. es of transmi"ion making an angle of 45' . They are foUo,,·ed b~ • thi,d Polaroid wbo5e a,is of ".n""i";on ma kes an aJlgIe of 90" w;lh llIe fir:;1 PoIa.oid. If .1IIh •• e are iokal. what f.a<lion of the m ... imum possible I;ght (if .1I PDlaroi<h were al Ihe .. m. angl.) plS$t1through all ' hrec?

, The Tedudion faam for either of the la.ll"", Polaroids is = ' 45' - l IZ; SO lhe ov<:.-aU 1""l<n is (j)' .. 1-

36.72 The a""" of" polarize. and an analyu,.re oriented al nah' angle. to each o.Ile •. A thi.d J'Q!aruid ""«I ;S placed bel~n Ihem with its a. is at 45' 10 IIle ax .. of Ihe polin .." an<! analY'. ' _ (a) If unpollorizedlighT of inten.ity I" is incidenl on Ihi' ')'STem. what io lhe inlen';ly of lhe Ira1t$miltedlighl? (6 ) Wbat is the ;nlen'; ly of the ""lI'lmined lighl when Ihe middle Polaroid !heel i • • emo,....:!?

, (a) lbe lighllhal pitSS(. through Ihe polarize. has an intensity of !4 and i, poIariud al 45' 10 lhe middk shee'_ Thus , Ile lighllMt pas.sn through lite middle oi>eel has lhe inlensity I _ 110005'45'- 0_2SIo an<! io polarized 4S' to the: u io of the analyze •. Th ... lite inlensily of the lil!,ht pas<ing through Ihe analyu, i. f'. I COl' 45' . !lJ1U,. (6) If the: middle PoIHOid sMet is .<mO\"ed, ... e h."" a cros..ed poIariu.·analyu. an<! no lighl l;C'u Ihrough . (Compare Prob. 36.68.)

36.73 FOur p"'rfe<t polaril;ng plaIts are 'I""ked SO IMlllIe axis of ~""h i, lu..,.,d JJ1' clockwise "iTh ''''pee! 10 lhe p.ect<lin, plale ; Ihe lasl pla!e i, aO$SCd "';th Ihe fir:;l_ How much of Ihe inle~';!y of an incidenl unpolaril~d beam of lighl i, lrammined by the: lUCk?

, The fir:;l plal~ t,a!lMllh, 01>( half of tile incident intensi.y. E""h succeedin, plale mak ... a "<"'0' resolution al up. ]If, Iran,mininl Ille f • ...,liOIt cos 30' of the: amplilude. Of 005' 30" _ ~ of Ihe in"'lI'lil~. leaving the preceding plal • . The fraction ofllle initial inlen';ty Iran, min.d by Ih •• t~k i, Ihen Ui)'.!!.,lll.

36.74 A '1uarte.·wave plale io made from I male, ial whOS( indi= of .ef.-action for light of f.ee·.pace wa .. cl.nglh 4 - S89 nm are II , · 1.7J2 and~ • • 1.456. What io the minimum ne<:eSsaty thickne .. of Ihe plate 10' ,hi, w.""lenglh?

, 'The optical palh knglh (P.ob. ~.S3) of the ordi .... ry wa"" in. plale of Thirkn." I is ~ , /an<! the opTical path kngth of Ihe exl.ao'di .... ry ,ay i. ,,&1. Since the lwo rays must eme .... from tile plate "';th a 90' phase dilference, the optical p<ltru. mu.1 diffc. by (t + !).l", k .0, ], 2, .... T'bt minimum .hidnes! thus "li,tiM

"'om 4( I. 732 _ 1.456) - 534 nm

nlea IT na

SPECIAL RElATIVITY a 689

Eq ... lions (11) and (12) ."". wilb Eq<. (S) and (8) <If Prob. )7.1; It.. inwrsioo of lhe y ' .nd :' equalions .. In_ial.

37.1 For lhe silualion of Prob, 37.1. r.uPJ>OSC Ibal al lb. inslanl oris in 0' roincides wilb 0 (al I • I' .0) .• fta~bulb is exploded allbi$ rommon origin . Acco,ding 10 0""' ... ..," in !t, a sphuk.l ...... ·dronl upan<h olnwa,d from 0 al speed •. Show lhal ...... n lhough!£' .. tnOI'inl ,elali.e 1O!t wilh •• Iocily~, oI:...-rv01' in !£' not. an • • actly similar w.""fronl upanding OUI .... rd from 0 ' ,

, The: equa'ioo <If Ihe ... a.efronl in !t i$

Thc: loRntt u.""fo,malion of ( I) i. (Eq •. i~) 10 (8) . Prob. n . l )

[ %' +"1' ]'. [1'+ (u<'!c')] ' VI - (u 'IA + (y')- + (z')' - ( ' VI - ("'Il'~

(~ . )' + (y 'J' + (: ' )' _ c,(, ·),

EquIlion (2) ""prescnl. a ",he"",,1 " ·."efronl upanding OIII ..... rd from 0 ' 01 . peed c.

,,,

" )

ActuUy. lhe ........ ing goes Ibe olhe . .. 'aj' round. Thc: !t and!t· ....... ".., .. mu" _ e'ltClly Ihe ... me kind of wave. by lhe poslulales of specia l .. IIlilily, This (almost) force_ lhe ,e lation. belween 1M "",rdin.les in lhe 1...-0 I",,,," 10 h."" a partiN]" I~ha, 01 Ihc Lo .. nl, lranllomt.lions. See Prob, 37.S.

:r7.4 When inerti.1 reference fnmes !J' and :r roincide. let • n.", 01 ligh' be ptoo"""d 01 Ihe common origin Eath o""'rve, ;" j ... tified in (:OnSiderinS rnm .. 1f 01 Ihe cem •• " I In . xpandin! sphe.e lIf ligh'. Experimenl has ,..,,,,,.Ied Iha' .ltCh oblains Ihe same .. Iue c for lhe ~ of lighl . Thc: plil."" Iran.fonnlhon . • · _ .r - l ",

does not Ii"" Ih ...... ull, Thc:r.fore Iry. modificalion. x' '"' r(x - VI). wlH:,e y is 10 be dele,mi ... d . Thc: principle of equivalence requi res Ibal Ihis equalion hold lor III<- in.'e .... I'ansl""",ation . .r - Yi~ ' - ~ ' I') ... 11~ ' + wl In Ihis equation ..... use tile .~rtion Ibat v' - -v. BUI 101 ~neI"1l1i\y. Ihe possibilil)' has I>oen allowed lhal ,. may b¢ dil'f~,cn' ff"'" I.

If ~ and~' a,c Ih. ;nICn«1ions of III<- opIICf< ,"'ilh Ihc ax;s al Ii"," I and I ', respe<1; ,cly; (. ~ To whal ;. z · /I" .qual~ Ib) To what .. ,, /, equal? (~) U .. Ih. , .. ults of partl ( .. ) and (b ) 10 elimi",,'e ~ and z · in 11 .. Tr ... formation C<)ualions and In ... 10 del.nn;ne r.

, W. nOI. lhal Ih. problem Slalement includes the lacit ISlumption Iha, y' - r ... ·hoo to' _ -u. ,_) At lime ,., ItI" Ii"" sphere o"",,.,..,d by !I:' ;n''' . ..... 1 !hr pooili •• ~ . nis al • """din.!e z' ,. C' ·. Th .... fore .r '/1' _ c. (h) AI lime " Ihe lighl sph ... o""'"",d by !L inle.-..cts 1M positive .. ax;' 01 • coordinale z _ (I. Thc:relmc~. It) Since 'hc equation, oblained in parts (_) and (Ao ) rde, 10 the $31IIe poinl cticnl . ...., DSC ..- ' - n' and z - c, in "'" 'quolions In 8el

,."

Mu lliplyinll Eqs. (I) and (2) ,"'C ha ••

Sollin. Ihis 10' y .... , find

d '" y(n - ul)

n _ rid + u,')

c'n ' _ y(c' - ,, ')11 '

37.5 COI!linuing with 1M ' .... 11501 Prob, 37.4. u .. In. ,..,lalOon for y ... p,w" in lenni of. and z.

, When Eq, (3) i. u,..d in 'hr Tramformation .qualions (PTob . 37 .• ) ..... abt';n

, ' ", I (z - IJ{) VI - (.,!,":)

'"' ) ,. .)

" " ) ",)x + u, , - (II' c-

These equalions ,ela'e arbitrary " .·cn,-ooordin.t. O""'l"IIalion< of """,,,'e,, 0 and 0 '. Uun8 Eq. (4) 10

(0)

'"

,»

") ")

690 a CHAPTER 37

eliminale I' from Eq, (5), ... " I\Dd lhal

(I-on ) v,' X _ + I - (It'le') VI - (It'le' )

S<!hing lhi!. fOl,', "'e fiDd (aher some algebra) lhal

r' . I (, _ VI) VI - (It'le') C'

which i. Ite staDdard '-'""nll (, a,,<form., ion njualioo 'or ,',

37.. A. meaJure<l by Z a ft.<I1bulb goes of( al .. _ lOO km, y _ 10 km, I _ I km '" , .. 0.5 mi. Wh al are the roo,di nal .. L', y ' , z', and " of Ihis ,,"enl as delermined by . second ObscfVC., !£' , mo';ng rel., i"" 10 Z "' - 0.& .k><S the C<)mmon .n' axil?

, From tn. Loo-cnlz lTllnsfOfmaliortl,

I - VI lOOkm -(-0.8x3xI0' kml.)(5x)(l',) .. ' - - -367km

VI - (lI'k') VI - 0.8'

, ' - ("k'~ 5" 10" $ - [(-0.8)(100 km)]1(3 " 10' km!.) ,," __ , - _. - .I...Oll.IJ.ll

VI-(II'/c') V)-(0.8)' . y'_y _ IO km

37,7 AI Ii",.. , ' .. O. ~ ml , a, mtasur.d in Z', a pa>1 ide is a, 1M f"'int . ' .. 10 m. y' .. ~ m. :' _ 6 m (NOIc that this ron>,i1u1cS an ",·"nt,) Comput. ' h" C<>nes.poDding ' ·ol"e. of >, y, I. I. as "", • ..,rcd in Z, for I. ) v - +!OO mi., (6) 11 - - 500ml • . and (e) It .. 2 x 10" mi • .

, F.om lhoe in¥". sc L'''enu "ansfonnat''''' , !'Tub. 37 , I.

Co) X" 10+ (500)(4 "- 10- ') " )(1 + (500)(4 )( )(1- ') .. ~ y _ ill! I - full "1/1 - (SIJO'ki )

C

CbI

C,)

37 .8 AI . .. 1 ml ( in :t). an explo!.ion QCrulS '" -< .. S km. Whal i. lhe time of Ihe ",'cnt fo. ,he 2' o~rver. if for him it <l«1I1'S at .. ' .. 35 . 1S4 km?

, F,om Ih.l<!r<nt~ transformation for x',

35.3$4 " 10' .. (S )( 10') - Itl to·,) v'l_ (u,/ci)

Tlten. fro11. th" LOTcntz tran.formation for",

u _ _ 3xIO' m!.

, 10"' + [(3 ~ IO')(~ x 10'))k' 00!j7 ,_ _ I !!!!. VI - (1I'le')

37.9 A spate.h" of (, est) len31b 100 m tok« 4 /i' '" palS an """''''er "" canh. W,,", i~ il' ,·d"";,y . ola.;,·. ,,, the unh?

, Tlte ob>er>'c, mea,u,e. Ihe length of Ihe .... < .. hip 10"" 1 _ umh - (u'lc' )(m), "'he, " u i, . he velo<ity "r

SPECIAL RELATIVITY a 703

I n.e rMt mll$S loot in the fonnat;"" of I deutron is

(m, .. m.l - m. _ (1.67261 + \.614!12 - 3.:w3S7l x lO"n _ 3.96 x 10"" t, whictl is equi~aIcnt to a r .. t-cnerzy loss

/lEo. 3.96 x 10""c' "3.56 x 10"" J. 2.22 MeV

Hence. by eneTJY _"'.tion. the . ut1Ollndinp mlUl llain nOlClJy 2.22 MeV of enefiY. Tbif, $I.JIW a""",n, of energy . tM hindu., t'l(", 01 thl' deuteron. """,lei ""_e to ~ wpplied to thl' deuleron 10 IUr il apart into an infiniTely separated proton and neuTron .

noSl o.t ... ni ... The "'1l$S ond $peN olin el<'<'lron hI';n! 'inetic energy of lOOkcV 0 .6 I( 11)"" J).

I n.e kinetic e ... Tn is 1( .. me' - ", ... ' . Or 1.6 I( 10' " J ... (m - m.)() I( 10' m/I)' . and m - m. '" 1.78 )( 10'" 1<&. In Prob. 17.49 ..., are Ki~.n :;r" 9.11 x 10' " t !; 10 m - (9.11 .. 1.78) x 10-" t! _ 1.089 I( 10' " kg. To ,et "elocily we ""te IhaT m " mol 1 - (11'1("'). or 1 - (1I'lc"). (mJ m)' _ 0.700. n..n 11/ .. _0.548. and 11 - 1.641( 100mll.

n.5l Find the ${ION thai. prot"" muSi be Ji~.n if iu ",a .. is to be twice iii .... rna .. of 1.67 x 10- n t!. Wl'I.t energy mu" be Jive" the pfOlOfI to adrie.·. Ibis opu<!?

I Use the m.us-increa<e formula .

•••• /~. _,,: ,.,.~/~. _,,: .(._,c)." c' c' c' c' 4

" , AW _ (m - mo)<' _ (Zmo - m, )<' _ (1.67 I( 10·n kll() x 100mll)' ... \.50 x 10- " J -m MeV

n.53 How mlll."fl . ... rgy mllit be gi~en to an eleClrOfl 10 acttkral. il 10 0.9Sc?

I U.., the maso·inctea<e formula.

m_ m. . 9. IX 1O- " k'_29 Il(iO. n k VI_(II ' /c1) \ii-oW' !

A W ... (m - .... )c' • [(29.1 - 9. I )(10' '')](9 I( 10'') ... UI I( 10' " J _ I, 125 Mey

n.54 A 2·k3 object is lifted from the "00. to a \.Ibktop )() om .00-.. tile fIoo •. By how mudl did the ma .. of the object incre_ beau ... of iii increased PE?

I W. use AE _ (Am)l"'. "';th AE - m,lt. "Therdo«

6m. ht: ... ~ .. (2 kS)(9.8 ",10,)(0.3 m) -"'C""'C-C",,,. "7 c' () xIO'm/.)' . -

n.5!l The .... mau of. protOfl it .... . \.67261~ x 10' ''' ka. hllal!oratory spud after having faUm Ihl"Ollp a high difl"er.n« in potenTiaL, 6 V. i. u ... 2)( 10' mil. ( .. ) EV1.hr.lte 6 V. (.) F"tnd IIIe lotal eMil)' of ,he prown.

I e .. , n.. ki MI;' energy of The proton ;'

1(_ (m _ .... )1" '. [ I - I Jm"" - ~hV V I -(~ '/c')

from "'hictl

(II, Th. reS! .... Il)' of llIe proton i. m.,c' - 938.3 MeV. arid so IIIe total ene.1)' is

£ - 938.3 + 321 _ m9.3 M,V

37 . .56 SIlo ... tbol KE .. (m - m. )I"' r. duce. to KE . Imoll' .. hen ~ it "ery much ...wier thin c. , KE ... (m - m. )I"'" ( ItOo _ mo\~' _ m ... ~( 1 -II'/C' )" "' - I] "I - (u /c)' r

LeI It ... - u'le' and expand (I + It) · '" by the bioomioJ theorem. n..n

I '" ),,' '_ 1+ __ + __ + ·· 2 .. ' 8 .. '

,.

SPECIAl RELATIVITY 0 105

Thus thc daired relation is

u, BcCI....e of the form of (I) momenta I re often ~~ed in MeV/~, on the atomic level . where energies are

given in eV '" McV. (Sec Prot>. 31.63. ) (611bc total energy of the el«lron is ilS rei enell)' . O.Sll MeV, plus its kinetic enefl)', I MeV. 1bcn ( I) gives 1. ~ ll ' O' O. ~ll ' + (p<)'. from .. hich pc - U2 MeV. 0< pO' 1.4,2 Mey /{.

n.6) Compuu Ibc tofIvcf1ion facio< between the .... diury "n;1S of momcntum and McV Ie.

I 1 McV Ie • (1.6 )( 10-" I)1(l )( 10' m/'l - S.H)( 10-11 h . mi •.

31.64 Show thll .-w mass is no! ronR ..... ed in a symmetrio.al . porfe<:tly inela$tic coIli<ion .

I Lei tWO identital bodies. cadi with 1\'$t mISS "'0 . • Wroach e~ other at equal ~ II, coIli<II>, and <lid tOSCtber. By morncnlllm _ ..... atioa, \be congJorncrau body mUll be al , .. t . with energy M.,c'. "r'M initill cne,Sf of the 'J"Icm .. as l/.....,'NI - (,,'k')l- He~.

"",' VI _ ("' Ie)" Moe' 11 is seen tbatlhe 6n&l rcst m ... . Mo, acceds the initial feSt mass. z.....

37-'5' In newton ian one<:hanir::I\be relation dt:/dl O' F · v is vllid. where t: is lbe tOla! cnc'lf of I panicle thaI is moving ""'III veloeity . and is lCIed On by I nel f .... cc F . Show thatlhis relltion is &I", .. lid in relativistic medlania. . (Assume lila, Newton', oeoond law i. valid IInde. special reLalivity. )

I A. shu"", in Prot>. 37.62 . tbe panicle' . in>ant.ancous momentum P(f) and illitlntaneoos ene'BY t:(r) .,e reLated by

E' .. c., , , -;. (",,,,,')' Differentia ling Eq. (1) ...-ilh rcspcct 10 Ii .... , we flnd 'hat

d£ dD 2£-_lc', · ;:.t:;

" " dE .. ~.~ dr £ dr

BUI, - "'. - l""o" and £ - me' - \"'I""'. 100 (c.,)/£ " " U!oi"l! t his in Eq. (2) . .. c obI,in

!oinoe F _ d,ldl.

37.'" Sbow tbalthe components of lhe velocily 01 . patlidt olener&Y E and momen,um, I re given by

" " " 0 -• iJp. " 0-, ... 1bcse re)a\ion$ apply in both lhe relativistic and newtonia n domain • .

I 1bc ene'D is &i_en in terms 0I1lu: momcn'~m by

t: - vi'-p' -;. (""",'i' .. y,'?",',','.C,':' .;';,,!),.:o,'.',""I' Dillertn\ialing Wilh rtspeclto P •• ~ ftnd tila!

aE 1 2c'p. ~ r'Y"'.~. iJp, -2Vr'(p!+ p; +p!) -;, (",.,t')S- £ - rm.c' - ~,

Tbc 1"00& for lhe! and : oomponenll follow \be wac proudur ... .

,u

Sinoc III< ""W\Of\;an , eswI is just \be low·ydocity limit of lhe relali,i "iI; re-sult . the above rel atiOll$ bold in lne _""ian \Ioonain. 1bc~ m.oy al", be easily derived from E - p' / (2I'Ir. ) - p;/(lm,,) + p!/ (2m,,) -;. p;/( z",. j. 1bcn aE/ap. " p.! ..... "' ~ .. ele .

SPECIAL RELATIVITY 0 707

$0 lilt- mal~lir field magnilLl<k ., d~",," ,. is

" _ /loi" _ I'.r _ (2)( 10-)(1.06)( 10- ') - 2- 20 nT 8 2. .. " 2:1<. 1.00)(10-' <_-"_"-'_"

F,rully, f~ - ro;8'- (l .6 )( 10' '')(5.2941( 10,)(2.12 I( 10-°) - L. 7'91( 10' '''' N low;lfd 1M beam

J7.70 A la.ge <Yi'IO.fOIl ;s &s;g:ntd 10 a.a;I)krale dtut<:rons 10 450 MeV of kint:tit eM'g)". Thil mean. that lMiI s-pud will become a ... bslantia! f' action of c. H.""" their mass will become ... botantilUy larl'" IMn IIIe .eot rn .... [f lilt mal""l;" fitld ;, o'..,rywlltre of tilt ... m. wah ••. this ... q~i"'lhat lilt frC<jucncy of the osciUatinK potenti.l di/l"e ... """ applied bet~n tile Off. be <I«n~ durinl tM """,Ie.ation of I group of ckuto""". What i, the ratio of tM IInal to the inilial f ... quency~ ll>e .... 1 m~ of a ckuteron .. 3.1" 10' " kl.

, For a particl. duling 01 ... lati ";'lic ","cd in • mlsnetic field. lhe , elatiYiotic fOfU . , _ d" ldt. is .. iU O)ITttliyal,·on by lhe Lo .. nll formula:

" r ___ qy x B (I)

" 1M momentum is given by , - J"'I.-. wbc,e r " II - (It'Ic')r '''· II the ",," id. is cin:1in! ";lh .ngular v.locity " , . .... ha'· •. • j""" the maRJIi.OOe of " i. constant.

Equat;oos ( I ) and (2) imply tha •. in ma,..i. l>de.

", he ... '" .. }'r>I. ;' the ,elativi .. k m . ..... Sin~ the «ute,OM ar. initiaUy non,.Jati.iotic . Eq . (3) im p/ieo 'hat

~_ .l2 • .!. "'~ r, r,

Itt 'emll of the final ki"". i. one'g)" " ,. "'·0 ~'·o

1M ,eM ''''''8Y of the « u.e. on is

'" + m..c ' " . , ., • (33x lO· " kS)(3.00X 10'm/.)' J 86 O' M V

m.r_ 1.60xIO' '' J/I, leV - . XI •

Then .... ith 1(, - 450 MeV. Eq •. (4) a!>d (5) yield

~_.!._ 1860 loW "'~ r, 450 + 1860

"I

" I

PARTICLES OF LIGHT AND WAilES OF MAnER a 709

38.9 A desk lamp illuminales' dnk lop witb violel lighl of wavelength 41 2nm. TlK amplil"do of Ilti. elearoon.gnetic wne;' bJ .2 v Im, Find lilt! number N of """,on •• lrikinllM <k>k per.eoond pet un;1 'rn. ,,",uming tlta\ 1M illumin.tion is oormal.

I From '1_ ("") / .1 .nd 1M energy donsily formula, (E. EiI!2, for a "'ave with ei«tm Mklamplilu<k E. (Prob. JJ. 64)

N .. .l~.E; .. (412"10 ' ")(8.PJ"10' '')(bJ.2')_ •.• " .".pbo (.' lit 2(6,6) x 1(1'><) Ions I m

".1' A ""RIOr i, erpooed for 0,\1 to a ZO()..W lamp 10 m away. n.c ......... hal an opeoing IILaI illOmm in diameler. How mony pboIOIl5 enter lhe .. "SOl" if Ihe wavelength of lhe lighl is 6OOnm? A ..... me lhal ,~J lhe enel'lY of lhe lamp is given 011 .. light.

, lbe ene,1Y of • photon of Ihe li&hl i.

E .. ~ .. (6.6.1 )( 1(1' ''')(3)( \(1") _ 3.3)( \0' " J 1 600 )( 1(1"

n.. lamp """, 200W of ~r. lbe nwn"", of photons emilled pel §CCCI>d is therefore

200 " .. 3.3)( 10· .... 6.] )( 10"' photon.l.

Since lhe • .,jiltion is spherically . ymmctrical. the numbe, of photo ... enierinlille ""nsor per $CWIld is n multiplied by IIIe ,atio of IIIe opetture arel 10 Ihe . rca of • spheu of ..diu. 10 m:

(6.1 )( 10"') :~(~::;' .. 1-53)( ]0" phol"""'.

and lhe number of photon. thai enler the .. mar in 0.1. is (0.1)( L-S3)( ]0") .. 1.S)( 11)" .

•. ll Whal is the IIloOIJIC"lum-enerD relation for pOOIOftS?

I From f'rob. 37.62, .. _ (pc)' + (m ... ')'. BUI • pboton hal zero ... t moos. so I' - pc.

lII.U Wha' is ,he momenlum of a single photon of ,cd light (v _ 4OO,{< 10" Hz) moving through IT« spa.ee?

, n.. momenlum is given by p " M.l .. (h)/e. Hence

(6.6)( 10' '')(400)( 10") p- 3)(1(1" .. 8.8><1O' '" kg·mls.

38.1J Wha' wayelen&!h mu", ei«t.omagnelic radiation haye if . pholOn in Ihe beam;' 10 haye 1M SlInK """,,~nlum IllS an ele<:!ron moving wilh . $p«d 2)( 10' mil?

, n.c requi.emenl i, thll (m,,>,- " (~/A,-. From lhi.,

A ~ 6.&l)( 10' " J. , .. ;;;. (9.1 )( 10' " kg)(2" ut m/s) • :M!!.nm

This w • ...,length is in the x·,ay region,

lII. 1. ... .(...,am of p/>nlom impingins norm.lI~ on • complelely _bomb;.., screen in ~l<Uum .ntt •• p...,,,,ur. f1. Show th.1 ~ .. l ie. WM"" I i$ lhe irrodi.""".

, n.c time .ale of chan,e of momenlum e"uals the fortt , lhal is, Ap/f,.r _ F. n.. fOf(:( per unil ar.a , FlA, if IIIe prc$1u.e, and SO

"'" ~ ---Ad<

BUI W. cp fa. pbotons. ,",'Itich mean. lhal J1 ~ _ C i!J.p . Dd

~ - ...!...~ Ac i!J.r

Si""" imtdiance by dofinition is ene.gy pcr unit .... per un;1 time, we obtain~.

C ;.pvrlghted material

710 a CHAPTER 38

Jl,15 A coI.~ml!ed beam of li&h! of nUll delliily .10 kW/ m' ;. irw;idenl normally on I lOO-mm' oompklcly absorbin~ screen . Usinl lhe resull! of Prob . 38.H , detcrmi...: both the p<emI.e acned on and the momentum lransicrm:ilo the screen durinll liD) .• interval.

, lbt p< ... ur. ;. si mply

~ .. ! .. 3 )( HI' -.lJ!.:.f:J. _ ~ _ ('?fIb') c ])(10' A A

,.., (One could .ctually build an inl<'rpla"",ary sailboat ... ing ..,Ia. pressur •. )

Jl,16 Whit i5 the Ia.SCSi momenlum we an expea fo< a microwlve photon!

, MicrO'Qve fr.quencies JO up 10])( )I)" H z. lbtrelor •. since p" h/A .. (b)/c.

(6. 6 x 10-")(3 x iOn) p - ]xllf _6, 6 x I0- " kg_m_ . - '

311. 17 How m.n~ red pIIocom (A" 663 nm) mllSl stri ke I totilly "fI~ ocr •• n pe. """"nd. It normal incidence . il ,he exerted fo~ i5 '" he 0.22$ Ib?

, We know 111.111 F .. ,¥/bl Ind, in th" r~. /J.p i. n,'il'<" Ihe il\Cidtn, mnme01um, Thus, i f N is lhe numher of incoming pboton. pe' second. f .. Nl(2Ir)/ AI. Since 0.225 Ib .. IN. we have F _ I , and

~ 663 x 10· ' N - 2Ir .. 2(6.tu x 10. ") " 3 x 10'" pboto"./.

311.1' What poIenli.1 dilfc, cn,," must be applied 10 Slop lhe I""col pbotoclcc" ...... . mined by I nickel .urface under lhe action 01 uhravK>lcl lighl of wl vd.n!'lh 2(00),.? The work funnion of nickel;' S.OleV. , N 1240 eV ·nm

cne,IY of pholon - A '" 200 nm _6.2OeV

lbtn. from lhe pbolodectric eq .... tion. lhe . ""'IY of the fUI. 't emilled electron is 6,20 eV - 5,01) eV _ I ,WeV. H.nce I retarding potentill of ~ is requirt-d_

.11.1' The WOI"k fullClion oI5Odium melal is 23eV. What is the longest·wavelcn!'lh ~sht lhal (:;In elUS<' pb<xoclectron emission from sodium?

, A, Ih,.<-hokl , lhe photoo .""'IY jldl equal. lhe ' ''''TJY required 10 lea, IIIe .Itttron 100$<" from tile mctal. namely, 1M work function W_.

"-. -­- , , 1240,V'nm .. ,, __ A ," .. ~

2.leV

.11.21 Wlra! is the work lunct;o" of sodium metal il tile photoelectric threshold ~velenllh is 680 nm?

I

311.21 Will pbotOClcct roM he emined by a copper wrfacc, of work function ~ ,4 e V, when illuminated by ~isible light?

, AS in Prob_ 18_ 19.

Ihreshold J. ... N ... 12.oeV, om ... 282 11m

W_ .f.4 cV

Il(roct: visible: tighl (4()) 10 700 nm) (:;Innot (jec! pboIodectrons trOll! wrpc:r .

311.22 14ht of wavelenglh t-.:)) nm falb 011 a metal havillS phocoeleclric work function 2,V, Find '.1 the enefJY 01 a plio, .... . (bl 1M kir>etie .... IIY of tile m<:>51 energetic phoIoc\ectTOn, and (cllhe "upping potnlial.

t: _ I..: ... lZ40c V· nm .. 2.07 eV J. 600 nm

ngntea IT na

(0)

,,'

PARTICLES OF LIGHT AND WAVES OF MATTEA a 711

,,_ . - £; - W _ _ 2.07 - 2 - 0.1)7 eV tV. _ ,, _ _ 0 .0'1 eV

" V, .. O.07V

J&.2J It takes ' .2eV 10 remove one of the Least tilhtly bound electrons from a metal ..,rfau. WIlen IllIJa\iokt photon. of a linpe fuquency llri ke I "",tal . eleccrons with kinetic eneflies from ze .... to 2.6 eV Ire ejected. What i. the ene'l)' of tbe incident pboton.? , WFa4.2 eV KE..,..2 ,6eV ,,'"

)IoU A phoIon of enerlY 4.0 tV imparts III iU tnergy 10 on tLcctron that leaves a ""'tal ~rlac:c with 1.1 eV of kinetic enerlY. Whit is tbe work function W_ of tbe "",Ial?

, W_'"' hv - ,,_a4.0 - l.I.L2.£Y

)1.15 Delermine tbe maxim um ICE of phoIoelectfOl\$ ejected from I pot"";um wrface by ultraoiolct lisht of wlvtLenJlh 20XI A. What retardin. potenti.l difference is required 10 lIop tbe emission of electrons? 'The photoelectric thtelohold waveLcnJlb for potassium is 4400 A. , . IK 124(JJ(V· A

WOIkfunctJon-W_-A-" 44(0); _2,UeV

IK Itc 12400 "1- ...... KE.. • • 2000 -2.82.6.20-2.82_UlU:

'The retardin! potential mull be just I.rl" enough 10 >Iop electrons ";th k E.... 50 V -l..liY .

.. .14 A SIIrf..., ha. light of wlveltnl1h A, a SSOnm incident on ii, ausi", tbe ejection of pholoelectrons fOf which tbe 1I0ppini potential is v" " 0.]9 V. Sul'J"l'" Ihat radiation of waveLen,th l,a ]90nm were incident on tbe SIIrface. C.lculate (I) tilt <toppin. poten.ial V;" t~) lbe work hu>ction of the w rfa.ce , an<I tf) tbe Ih,eshoId frequency for the ... rfKl'.

, tl) Since t V· K£;_ •• hv - W_. ";th W_ I coMllnt of tbe >urface, ... blve t(V; , - V; ,) .. hey, - v,) Of

• '"(' ') 2<0( ' ') v,, - v, ,+- (v,-v, )- v,, +- --- _0. \9+1 ---"!.ill< • e l , A, 190 SSO

'" IK 12-10

W_ .. r,-- .V" as;) - 0, 19 .. tJ.!Z.£Y

,,' hv_aW_ or V. W_.(107)(t.602 XIO- '"). 4911THz • < h 6,6)( \0- " - -

)1.27 In , .... pIw>Ioionintion of a.om;" hydrQifn. what";!l be lbe m""imum kinetil: ene,1D' of tbe ejected electron when a 6O-nm phOlon is absorbed by the • • om? The ionization ene,1Y of H it 1l.6 cV.

, 'The ionilltion ene'lY is minimum (nel'~ to .. move lbe elmron from the .1OIn. 'Then

Iu: 1240eV · nm K_'",-13.6 eV .. - B.6eV - !J...tY

... 60nm

38.2 COMPTON SCATfERING; X-RAYS; PAIR P RODl1Cl10N AND ANNIHILATION

)I.ZII Suppose lha,. 3.6-1 nm phoIon JOin, in .be + ... dire<1ion collld« bead...,n .,.jlh I 2" 10' mi. cle""," mo>1"1 in ,be - ~ di",ction. If lbe collision is perfectly elastic . .... Ita, au lbe coodiliom afln roIlision .

, f rom the la ... of co .... rqlion of motncnlum.

momentUm befo,e. momen'um afle, • • - - mv ... - -mv 4 A

Bu, . from Prob. 311.13. hO .. .. mv, in this ealoe . HeDtt. MA • mv. Also, fOf • perfealy elastic roIli .......

Itcl hel - + - mv~ . - + - mv' ... 2 A 2

KE before. KE aflc,

712 a CHAPTER 38

Using the fa.;!$ th~, III .., am,,_ and hI~ _ mlt , *t ~nd Iha'

uJ' .. Ivo) - v(.- .. I") The. efOTe " . " . and the elearon Teboonds ... ilh its on,inal spew. Bttau~ hf" _ mlt _ "'''., II>< pholon abo rebounds and with il. original ... a,,,length.

lII..D The CompWtt _qUJUion can be: wrin.n as),' - A .. [lt f(moe)I(1 - (:06 4». The faCIO. hI(moe ), .;:aIled lhe CompWtt "",wk""h, i'lhe .... .. d.ngm ohift that 0CCIlf"$ ... I><n 4> a9O". Eval"ate il lo. Selltcrina from (.) elearon< and (10) Pfotons. [n II>c case of e[.,.lrons, ",ut percent..,.: dlange i. this il the incident .adialion U, I wa.elength 1.-) ~nm (vislble tight) aM 1<0 O.OSO nm (o.s.). H ays)?

I . _nin~ the: .est masses of the: electron (~ . • " 10->0 k&) and proton ( •. 61" 10-" tS) into 10 1m<: yield< Compton ..-.""length. 0Il1!.!ll.A and 1.32 " 10-" mo reopec1ively. For the .Iectron (O.024/~)( 1(0) -4.8 " 10- ' perunl, ..-bile lot II>< o.s.A ~-'.y, (0.ll24/0.SOO)( ]00) _ 4.8 pe,cent.

JII,.lO A photon (A _ 0.400 nm) >trikes In clearoo at rest and •• boonds at an an!1e of 130' 10 it. origilWll dirmion. Find II>< speed and w,,·.length 01 the photon aft.r tl>< collision .

, The!opeCd of . photDII is .I .... yo the.peed of li,ln in '-..cuurn. c. To obtain thoc .... ""I.ngth after collision. we u~ the e<luation lor the Compton effect :

l ' _ l + ! (I - =.) - 4" W " m + (9.1 ,,6;~.7 :;~: ~ 'I~ms) (1 - coo ]30')

- 4 )( 10- '" m + (2.43 " 10- " m)( I + 0.866) _ O . .tG4S nm

Jl.31 SuPJlOSC that. beam 01 0.2-MeV photon>;' SellI.r.d by th •• 1«1rO<\> in. carbon '"'St't. (. ) What is the wa,,,.,ngth a..ociated ... ilh thc>c pohotons? (10) What is the ... ...,length 01 those photon, scaner.d through an Inlle of YO"'! It) What is I'" cnc.l)' 01 tIM: ocamred photo ... lut emer,. 11 an angle of 60' .. Ia~;~e to the incident direction1

l .. M _ 1240 M~V . fm -liZOOlm. £ 0.2 MeV

(b, A' _ ~ + ..!!.... (I - coo tI) _ 6200 + (2430 )( I - 0) .. !K!JQ1m m,<

«, /0( /0( 1240 1' '' ). + [Io/lm.( JI(l - ros tI) 6200 + (24:lO)(l _ ' ) .. Q. II!!! Mev

.I8.Jl Verify Compton', "'Iuation (Prob. 38.29) "'hen IIIe phOlon is hack-scane.ed ( . .. II!(J').

, A .. umins th • • Iectron to be at "'" initially .... e ha.e lrom co .... "·ation of m"",cn1um

• • -- p- -.. ;:

and from OOII .. ",at;o" of •• Iativistic ."'''gy

" " - H .-- +£ , "

"

Substituting thc>c up •• u ions fOl tIM: electron's final momentum aMI final ene.l)' into the momentum-cncra relation fOl ,be elcctroo, .... obtain

£ .1 ), '- ), ) _2Jtc

,,·IM:IKe).' - l_ (2ilc)!E . .. (2il )!(m.c), which iI Compton', equation with COS ¢ _ - L-

38.33 A pboton Mrik .... fro" electron at , ... t and iI SGln.red straight Mckwar<l. If the electron', speed oft.r coli;';"" is M . ",he:", ,, « I , sbow , bot the: .Iro,oo·, kinetic "",:IllY". IraClion ... of the phofOn', initial

e'''''Jr. , Since ,, " <K and " « l. II>< electron i. nonr:lativistic. C<m.crvation 01 momentum yid'" /01 ~ + hi).' _ mu; multiplying by c . (he )/l + (M )/l ' _ mil(" .. (m u')!". The .ncrl)' equation i, (/0( )1 .. - (M )I ~ ' _ (m,,"'1L

PARTICLES OF LIGHT AND WAVES OF MATTER a 713

Adding we ~nd that (1oc )IA a E, '"' (mll')14 -+- (",11' )/(2,,). Since .. « L ("'~')/4 i. netl«t«l aoo ("'11')12" ":E,,

38.34 A pboton "';th A'"' 0.5 nm strikes . fr ••• Ie(tron !>tad...,n and if, ~n.red WligJII back ..... rd . If the .lecllon is initiaUy I t rest. ", hal is;1S .pe<:d after th. ooIlisioo'

, M""",nlum oor""""31ion gives ItO _ "',II. - It IA' , assuming Iha, ,!>t snl1ertJ e!tcttoo is nor.r.I&,i";'lic . A is ,,~n as 0.5 nm .00 A' lor 1!lO' "",".ri"3 is 0.5 nm -+- (V. )1( ... ",) '"' o. 5048 nm. $0 II, - (h l"'t)I(l1 A-+-I/ A 'J • [(6.6.3 X 10-")1(9.1 )( 10-")](3.98)( loj - 2.9 )( JO" mk.

38.l5 A pllo(on "';th A .. 0.) TIm is mO'o';ng along the x allis "hen it Slrites a free d«lrun (initially at reM) and i. "",n.,«1 so as to mo'"<' along the y axis. What are tlte ~ and y component. of the elMron'! ~Ioxity after ooilisioo?

, In the x directi"". oonservotior. of m""",ntum yields It l A _ .... " . aoo in y direetion hi A' -+- "'oil, _ O. From ,he Compton ..... n.ri'" relation fo.r f> _ 9('1', A' a A .. 0.024 A _ 5.0 -+- 0.024 _ 5.024 A. SoI.in, for II. '00 11, from above , ",ith m. tt.. electron rest ma». give. v. _ + 1.46 )( 100m/. aoolJ, _ -\ oIl> . 100ml', the el~ron is non.e(a'ivi.rie .

.l8.)1i Oeri~ the Compton "'luabon, .. ~ng ,!>t rd.tiv;'t;';' npre!>$i<>os lor , I>< <"".V and mOmen'um u l the el~ron .

.f"\J""'- -- __ _______ ... , ,

I For notation, see fis..:J.S.L

oonservotion of z momentum:

ronscrvation of y mom.ntum:

h, hI" -- - «IS f>" P «IS {J , ,

h! -+- mot' - hI"" (P'e' -+- "'I.r') '~

(/) Solv. (1) for P <nil 8 and (2) for P .in 8. Square and add, Jr', ' - Vr. 'J/" _. + Jr'!,' _ c'P'. (ii) Sq .... ", (3) Co obtai!l; h',' -+- It '!" - v. 'J/" + Vr.m",'(f - f") _ c'P'. (/ii) Subt.-.ct : (f - n il!' - lhI(", ... ')!! 1 - _ to). (iv ) But' - rIA aoo.., (f - f')1(Jf') - (A' - J.) /c. ( v) The,.,,,,,,. l ' - 1 _ Ih/(moC )]( I - 00> .)

(I)

1')

JII.J7 [n whal amounts to the in ....... 0( the photoel~ric effect. X-fay phot"'" aro prod"""d ... hen a tunpt.n tl .. ot is bombard«! by accele..ated .lectrons. [f an ""'Y m.<h;n. hu In o=I • • I,ing pot.ntial of 60 kY ..... ha' is ,he sl><>rtn' "'"""!tngch pr=nt in its radialion'

, Since tl>< wotk fuf>Ctton 0( tun"ten is vcry m""h .... all.r than ,ho ."""I.ra'ing ~.nti.l. ...... na)' .uf'l'OSC tIIlt the ontir. kinet;,;, e...,rgy of In elect.(Nt is Io$C t" <n.,. a ~ngl. pftoc"n ,,( ma.imum ene,gy:

A __ Ite _ 12-10 keV , pm " 20.7 r V WkeV ~

JII • .l8 An electron is !hot down an . ... y lube. [u en. rgy just bel"'. ;1 .trikes th. ' .rll"t of the tuhe ;,; .lOkeV. If it klses ollthi . .... rgy in I ';nR-!t colli""" "';th I very ma .. i,.., atom. ,"'1"11 is ,he ....... lenS'h of t l>< ~nR-!t .-.ay photon that is emitt«l!

, E Ioc 1.24 keV · nm -T- ~ '" .. -)0.0.0413 nm - !!.!l.!.A

ghted material

PARTICLES OF LIGHT AND WAVES OF MATTER a 715

I For Bragg ,~nffiion II>< "" ot.cring an,le;. si.~n by sin 8 .. (JlA)/(ld)o. ((2)(1.2))/ • .•. and fJ .. .u:.

38." The anenu •• ion <o<llid .. n. 01 .Iumin~m I", soft X .• aY" ;. L 7) /r;m. Com",,'c the ftacliotl of .1Ine X •• "Y" tnnsmincd by an .I~minum sbee\ US6cm thick.

I The exponentialla ... 01 anenuation is

1. . ~ . ' .'~' ,,., .. ~., _1. .. _ , __ o.m 10 ~' 2.718'

13.S% trwmjllfd

38.. One-half the intensity 01 a oomoaeDWY. x-ray beam ;. . ,moved by an alumin~m 61t ... 5 mm thid. What i. Ih. fraCl;on of Ihis l>ornot:,,,,,ou, beam that would be removed by 15 mm oI.luminum?

.. i (paction remaining) 1 -! - ~ (fraClioo ~m""cd) , , 38.49 SI'Iow that ",h.n I po6itron and an elec,ron (both ~_ntially II 'C$t) annihila", crelting twO pboIonl, eim",

phoI:on ha.lbe Compton wa •• ,.",.h (Prob . 38.29).

I Total ,r,,,,IY befo., annibila.ioft is 2m.c'; aDd all ••• 2({Itc)/A) (mom.ntum come"",'ion rcquilCl that the phO(oo energies be C<juol) . Then. by ..,....,,,,.\ion of """'IY.

,- , ," =,( - T " • h-- 14.1Ofm •• <

JS..5t ~ow Ihat the Ih.eohold .... ""length for the product;on 01 a ~tron~lffiron pai. is hatf II>< Complon wa""tenlth.

I The incido:nt photon mwot ha.e., leut enough ~ne rgy 10 create I pair i1avinl zero kinelic: enerlD' (_ ;Inore Iny kinelic .... rgy 01 the recoil n".deus). Thus.

" • ;s-- Ulllm lm.<

38.51 Aftel pair annihilation. two I-MeV photons move oil in oppxile Iii.CClions. Find the ~;Mtic coerlD' of UlC: elect.on and ~tron.

I The IWO pboIon. ha.e the...",e .nerl)'. and bence the .. me mom.ntum. ",hich _ are told •• e in opp;II$it. direction •. Thus by momen.um "",*rvation II>< electron and pooitron mllS( ha"" had equal and opposit. momentum. H • ...., tllei. ki""tic en • • gies a.( equal. By enerlD' o::c>f\Urvation: 2 MeV _ Um..,') + 2 KE _ 2{0.51 ) M.V + 2 KE and KE of tach _O, 49 McY.

38.3 DE BROGLIE WAVES AND THE VNCERTAlfInY PRINCIPLE

38.5Z A proton ;, O(l(>:I.r.,ed through. potuli.! difrt~nct 011000 V. What is its de Broglie wa.-eleogth?

I We !La"" 1.1I/("'~)_ But m _ ,..., (oon~l.ti";:'t;'), $0 ~ may be found from (m~')n .. llKXlt. SubstiTUtion gives 1.9 1" IO- " m.

38.53 Find the do: B"'Ilie ... ""'enSth of a thermal neutron of ma .. 1.67 x 10' " klt .. ""li", al. specd of 2200mj,_

I Use the de Broglie wave equation.

If 6_63" 10->< 1_;;;;; - \.61 x 10' ''(220:)) _0.18 om

38.S' What is the do: Broglie ",avtl.n,th fo. a particle mavin, with speed 2" 111m!1 il tbe particle;' (.) an electron . (bj • pl'oton. and (t) a 0.2-kg ball?

, We make W5C of the do:fiuition 0 1 tbe do: B"'I~e wavelc",'h:

It 6.63 x 10- " J.. ""'c'c'e':--c'mc-c'" .--- - -mu ",(2 x 10' m!.) m

C ;.pvrlghted malarial

PARTICLES OF LIGHT ANO WAVES OF MAnER 0 111

SUbstitution. T .. • . 25 )( 10- " ' ",. U,,;ng 1M election ma .. and Ihe helium rna .. (4 )( 1.61 )( 10-'" tV . .... e ha..., r.-46600K lnd 1i .. "'§..l.K.

311.6(1 Col\$ider a ~am of clemons si>ollow. rd I cryst.l. as ,00.-" in Fill. 3/H. n.. Cf}ltallpacin." b. '" indie" IM. FtlT .. 'hal de Bro81~ wavelengThs win lile eltctron beam be llrongly fcReett<:! ~ltliglll bad upon ilsclf? For whal el«lIon lin.lie el\trlics? II i. foond by npcri .... nlthat electro," ha'in~ t~ cl\trJic$ arc unable 10 move througll wch a CT)">lal in the dir«lioo shown. Ey.luate tile energies in eV for /) .. 2 )( IO-" m,

, Stroo! . eftection will ottur if 2b - d. because Ihe ~am' . ellettcd f. om ,u.cceW"e pia"", wtl1then m nf""", . n.e' et"'e. 1 _ (ll> )/'" For ""nrelali"istic .1«1 ....... E. _ ( p;)fOm) _ 1t'I(lrn1:) _ (,,' II ' )I(Sb'm ), ,..hich Ii"" E. - 9,,,,, ' ,V.

311.'1 n.. nuclei of atom, hl "e radii of Of(\c. 10-" m. Con,,;der I hypothetica l ,,;tuotioo of . prolon confined 10 a narrow lube with len,th 2)( 10- " m, Wh. t ",,'ill be the de B""I;' w. "olen&tll$ ,,"'hich ... i ll . esonat. in the lube~ T a ... ·hal mo .... nlum docs lhe lon,csI wa.'olength ~pond? If ",Iati" i, tic oll:«Is are O$5 ...... d Mpi&ible. to what eM rg)' (in .V) ~ ,hI. (:Or~~

, This i. I ";mple >landing-wave p<nblem. Fo' ')1n rnetri<: boundary oondilioon •. the allav.-.d l's at< J.,."' (2L)I~" [(4 )( 10-" )1111 m; since 1. '"' Mp •. p. _ ( .. h)12L" .. (1.66)( to " ), Non .. lativistic kiMtic ene,l)'_ E. '"' p';l (lm) _ ",(&2)( 10-" J) 0. ,, '(5 1 Mev). ,, - 1 <Qf'''ponds to IonJ~" ""vdtnJth.

311.61 At .... hat .""'lIY "';11 the nOnre lati";>1;< calC:Ul.tion <>f the <Ie B. O!:lie ,,' ... elength 01" an .Ie<:troo be in trro, by 5 pcrttnl?

, Fo. the nonrelal;vislic cas.e. Ihe de Broglie wavelength i,

A _ ~ _ lie

N ~ \12m",,' K

( K + ",,,,,' )' _ (~)' + (",,,,,,), • and the de Broglie wavelength i,

" " 1. - p; - (2mo<:'K!1 + KI(2m.,c')I) "'

F", our cas<, A_ - I.. .. O.OSl, , A.I) •• .. \.05.

1.05 .. ~I + 2(0.St ~ MeV)

Sol'ing . K _ 0 1M Mey.

311.63 Show thot tho de Broglie ..... elenglh 01 1 particle i, Ipproximately the same a. Ihal of a pOOl"" ... ilk the same ene . gy .... hen the eMra of 1M f'lnicle is rnudlg",ale. I""n in , .. t eMrg)'. , £ ' .. p'C>+ £~ • E~ (')' E P -c 1- ; -c FOI a photon. £ .. h .. (IIe)I.l,.. SO A, _ ("")1 £ _ A.

£ »E •. . " 1 _ _ _ _

, E

yr atenal

718 a CHAPTER 38

\ ' • 0 \'. I<I.V

38.~ An clea.on i. confi""d 10 a I~be of Iongrh L 1M .learon'. pol.nrial .ne.1)' in OIIC half of 1M IUM ir. .ero . ..... hil. 1M polenlial .""'1)' in 1M oth., half n 10 eV_ If 1M clr<:lron ba$ a lotal "".:rar £ .. 15 eY . ... hal will be the . atio of the de BfOIIic ",a •• lenKlh of lbe .lectron in 1M llkV .egion of lile lube 10 thal;n tbe other half of tbe ,ube? Skeldi the wa.'e tunaion "fot lbe .lectron Ilonlthe I~be .

, 1M .... a ... function i$ ,kclohtd in Fil. J8.5, ~ .. hlp .. 1t/(2mK)' ~; with K the ki""lic .""'BY of the eleCTron, 1,11,,, (X,/X , )' ~'" (1$15)'~ .. J..D:,

38.65 A panicle of map m;' confinc<lto I ot>e-dimeMionalli"" of length L Frum I rg1l"",nlO Iw.<d "" thc ......... inlcrp,etation of mille,. $how rh.t rbc ene.1)' oilM palticle an hne only discrele .... 1".$ and delenn;". thcK .alue._

, If lhe pamde ;. confined to a line ""menl. oay frum % _ 0 to % _ L. Ih. probabilily of findinllhe panicle oulside Ihn .egion m",1 be 1.0'0. The.ef"re . 1M w .. e fur.c1ion III mu" be urn fO'f ~ '" 0 o. ~ a L. <ince lhe ""' .... of III gi.e. the probabilily for fiooing thc panicl. al I unain Iont ion . lnoide 1M limilc<l .egion 1M "",".elenKlh of " mull be such Ihat \I' .... nMes at Ibe boundaries ~ ... 0 and ~ .. L. SO thll il can vary co.uinuouoly 10 the ou!!.ide ,elion. Hcn<O/ only I~ wa ... ltn,IM will be po!Sibie fOT whim an inl., •• , numMr of bolf W"vt\enltl .. fil bcT ...... n ~ .. 0 Ind ~ .. L. thai is, L " (nA){2 • ... he •• n is "" in'egc', called lhe qlWllW>l " ..... /It •. wilh ' alues " _ I . 2. 3. __ . , from tbe de Brogli. ,clalion,hip A - hlp "'c then find that I,," panicle', momenlum can ha ... only discrete val..., givtn by , .,

P-j'"21

Since the panidc ;. not lC'Ied upon by Iny fofttl imide 1M region . its polen'ial c"c'l1 will be • """""nt which w. lOCI e<jualto 1.0.0. 1M •• fore thc ene.1)' oil,," body n entirely kineTic and will ha"e tM di",r.te •• 'ues obtai"ed from

£ .. X .. !m~' " ft.. .. [(~h )I(2L)1' 2 2m 2m

IbT is, " , -.'-­• 8m L' ~"'1.2.3,. ..

Thi' .ery <implc p!'oblem ilhm.ales 0"" oill>c basic: fUlu ... of 1M probabiliT~ inlerprel.tion of mailer; n."""y. lhal The eM.1Y of a bound Iy>t.m can lake On only di ..... I. values . wilb uro energy nol OOnl l po!Sible value_

38,66 ....... me IMI the uncenainly in th. pooilion of. panicle is e<juallo iIO de Brotlic wa.elen,lb. SIIow tba, 1/1( uncenainty in ill velocity is equal 10 Or greal.r rh.., 1/(4,. ) limn its .eIOOTy.

, Use the Heisenberg uncenainty principle wilb ,u .. ~ " " I(m~. ).

Since m n romlant, , , -(mtov,) 01!:­mv, 4,.

, , - t'.(mv,) "'­mv. 4, ..

, , - tov OI!:­~, • 4,.

, t'.11,"- Il. ••

.11." If the uooenainty in the Ii .... durin, ...-him an electron remai", in In excited SUle is 10"' •. whal is the Ie ... uncenainly (in J) in lhe .""'1)' of the .. cit.d Ollie?

, Let W be lhe .... 'lIy of the ucited lI.ate .

II 663)( 10' " t'.WO, ,,,- (toW)(lO-)"'· .

4/1 ,. oW .. 6.63 )( 10- '" .. 0.528" 10- '" J ••

Các trang còn thiếu:

102-235, 239-243, 245-250, 257-272, 277-278, 284-290, 295-326,

329-340, 344-349, 353-360, 363-371, 374, 377, 381-430, 434-465,

467-468, 470-472, 474-518, 520-679, 681-685, 687-688, 691-702,

704, 706, 708, 714, 716, 719-737

51",*"" _ sa.....n'. Solved P_ GuoLII$ b«auI&!hey jlfOduc<I resuIIs. Each ."..... ~ 01 ot..-u improve ~ test """"'" and finto' gra<Ies with _ ... ....... .-Gel ~ «l\1li on)'OUt _ ... U .. $ellaum'.1

/I vou don·t". .... 1ot oj 111M but • ."t to •• cl'lln cbo, •. uN this boGIe to; • 8",." up bef_ W,~ • Study quickly and ....... _ Ively • lHm the but II~ !of IOIv1ng tou!Ih proI>IefM In 1Iep-by-.18p elMotl

RevIew _I you .... Iearn«j .. claM by $OMng ~ 01 .... IeG proI:>­... thet test your ..... Compaliblot with....,. _oem Ifl\, Sc:N!.m', Solved Problem G~ lei ~ ptK­lice .t your own _ .,.., ~ ~ 01 .. ~ itnpoI\8nt ~ I8CIIniques you ~ to , .. , ... ,obeo· Inti And 5cI\fIum·. are .. ~~, tIwy' .. P8ll11Ct lot Pf-""lllot gram.. at. Of 1"'0118 ,.",.. ,xatnI,

H you went lOp gradeI and IIIorQugII ~ 011 physIco;. tIM ~ 0Il.I:t\I \001 '- the beBllVtOt )IOU cen .....

C/IaptM:~: . ~tQllntroducllon ' Eqo.oIIiIriJm 011 Concurrent Forces ' Klnemrica in One Oinw.sion ' New\oo·,La ... 01 Mallon ' Motion in a Plane I ' Motion in • Plane II • _ am en.tvY • _ and SImple _ • Impuisa am t.Ionwttum • S\aIics 01 Rigid Bodin • Ro!atlONl Mo\Jon I; ~ am ~ • Ro!allonal Uo!ion II; Kinrio: Eot<v1.1v1gujar 1mpuIM, ~r Momeotum • u.ner in SuI< • Simple HarmonIc Motion • Hydrostatics · HydrodyNrnics • Temperal"" am The<mal ExparoWn • tiNt and Calorimetry . HeIII T_ ' GBILaW!l_ K.-...tic n.-y . n.. Forst Law 01 ~. n.. SaconcILa.oI ~. Wave Mohon · Sound ' CcUomb. Law am EMcVic:..- . EIteVIc _ am Capacit&nce . s;",p. E.a.crnc Cin:ui!s ' The MagleIic Field ' Magnetic p,O!l8'_ 011 MaI\e<S · Ino:Iu<:I(j EMF: Gananoticrw and MoIorI • IncIuc1anaI • E~ W_ • UghIand Optical PMo ........ . Morrors, L--. am ()pbcaI1nWumtn1l ' 1rMrNo .. oca. DiIIracllon. and PoIanralion • Spacial RelatlYlly • Pa_ 01 UgIIt _ W_ 01 Maner • Modem PIIytics; Atoms. NucleI. ~SUo\e Eiecboo b

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