Chap2 Solutions Sp07

8/6/2019 Chap2 Solutions Sp07

1/15

a PROBLEM 2.5b21W)N The 200-N forceis to be resolved into componentsalong lines a-d andbH. (a) Detennine the angle a using trigonometryknowing that the

component along a-d is to be 150 N. (b) What is the correspondingvalueof thecomponentalong bH?

a'

SOLUTION

F= zoo N

(a)Using the triangle rule and the Law of Sines

sinft =sin45 .150N 200 Nsinft =0.53033

ft=32.028a + ft + 45 = 1800

a = 103.0~(b) Using theLaw of Sines

Fbb' = 200Nsina sin45Fbb' =276 N ~

PROPRIETARY MATERIAL Ij) 2007 The McGraw-Hili Companies. Inc. All rights reserved. No part of this Manual may be displayed. reproducedor distributed inany form or by any means , without the prior writ ten permission of the publ isher, 0/ ' used beyond the limited dist ribution to teachers andeducators permitted by McGraw-Hili for their individual course preparation. Ifyou are a student using thi.vManual, you are using it without permission.

7


2/15

PROBLEM 2.16SolveProb.2.2 usingtrigonometry.Problem 2.2: Two forces P and Q are applied as shown at point A of a hooksupport. Knowing that P =451b and Q = 151b, determinegraphicallythemagnitude and direction of their resultant using (a) the parallelogram law,(b) the triangle rule.

SOLUTION

Using the Law of Cosines and the Law of Sines,R2 = (45Ib)2 + (15Ib)2 - 2(45 Ib)(15 Ib)cos135

or R = 56.609Ib56.609Ibsin135

151b-- sineor e =10.7991

R = 56.6Ib Y 85.8 ~

II'

PROPRIETARY MATERIAL ~ 2007 The McGraw-Hili Companies, Inc. All rights reserved. No part of this Manual may be displayed. reproducedor distributed in anyf01w or by any means, without the prior written pelwission of the publisher. or used beyol/d the limited distribution to teachers andeducators permitted by McGraw-Hillfor their individual course preparation. Jfyou are a sllldent using this Manllal, you are ILvil/git without permission.

18


3/15

PROBLEM 2.25

Q

Member BD exerts on member ABC a force P directed along line BD.Knowing that P must have a 960-N vertical component, determine (a) themagnitude of the force P, (b) its horizontal component.

A B c

SOLUTION

(a) p=~sin350%ON-- sin350

or P = 1674N ~(b) PyPx = tan350

960N'=-tan350or Px= 1371N ~

PROPRIETARY MATERIAL.


4/15

I_S~jll.-r +C 6 .SOin. I

l/.'"11 lit.

18-'.'-\L~~_ ~~ . /';l.~ "'.l~ .3

PROBLEM 2.35Knowing that the tension in cable BC is 145 lb, determine the resultant ofthe three forcesexertedatpointB ofbeamAB.

..I.",",

SOLUTIONCable BC Force:

'&= -2/16

84F = - (1451b)- = -1051bx 116Fy = (1451b) 80 = 1001b116

100-lb Force:

Fy = -(100 Ib)~ = -60 Ib4

Fy = -(100 Ib)"S= -801b'V=-40lb 156~lbForce, Ib).!3.= 1441b.!..'Y F, =(156 13R

- - - -) 5 _ -60 IbF = -(1561b 13 -

andRx = 'iJ\ = -21Ib, R = "IF. = -40 Iby yR = )(-21Ibl + (-40 Ib)2 = 45.1771b

Further:40tana =-21

a = tan-I 40 = 623021 .Thus: R = 45.2 Ib 7' 62.30'"

PROPRIETARY MATERIAL. ,e, 2007 The McGraw-Hili Companies, Inc. AIl rights reserved. No part of this Manual may be displayed. reproducedor distributed in anyform or by any means, withoUi the prior writ ten pe/mission of the publisher, or used beyond the limited distribution to teachers andeducators permil/ed by McGraw-Hill.for their individual course preparation. ffyou are a student IIsing this Manllal. you are !/Sing it wi/llOutpermission.

37


5/15

B

/'1//;/

7! ,!OUIb:.,A

SOLUTION

Free-Body Diagram

Law of Sines:

(a)

(b)

PROBLEM 2.43Knowingthat a = 50 and that boom AC exerts on pin C a force directedalong line AC, detennine (a) the magnitude of that force, (b) the tension incableBe.

Force Triangle.FAt:. o

/bO2So

FAC _ TBc 400 lb--=-sin25 sin60 sin95F = 400 lb sin25 = 169.6911bAC sin95 FAC = 169.71b ~

TBC= 400 sin600= 347.731bsin95 TBC= 348 lb ~

PROPRIETARY MATERIAL. '~2007 The McGraw-Hil i Companies. Inc. All r ights reserved. No part qfthis Manual may be displayed. reproducedor distributed in anyform or by any means, without the prior written permission of the publisher. or used beyond the limited distribution to teachers andeducators permitted hy McGraw-Hili for their individual course preparation. Ifyou are a studen/using this Manual, you are ILsingit without permission.

45


6/15

E jH.1m~ PROBLEM 2.53"J I In a circusact, an aerialistperformsa handstandon a wheel while being.\ i~_


7/15

PROBLEM 2.59For the ropes and force of the river of Prob. 2.46, detennine (a) the valueof a for which the tension in rope AB is as small as possible, (b) thecorresponding value of the tension.Problem 2.46: Ropes AB and AC are thrown to a boater whose canoe hadcapsized. Knowing that a = 25 and that the magnitude of the force FRexerted by the river on the boater is 70 lb, determine the tension (a) inropeAB, (b) in ropeAC.

SOLUTION(a)

TO \b

(b) Then T.4B= (70 Ib)sin300

For T.4Bto be a minimum~'B must be perpendicular to TAC:. a + 10= 60

or a = 50.011I

or TAB=35.0 lb


8/15

.

T

SOLUTIONFree-Body Diagram of Pulleyand Crate(b)

(d)

T T T

T TT ,...

PROBLEM 2.69Solve parts b and d ofProb. 2.68 assuming that the free end of the rope isattached to the crate.Problem 2.68: A 280-kg crate is supported by several rope-and-pulleyarrangements as shown. Determine for each arrangement the tension inthe rope. (Hint: The tension in the rope is the saIne on each side of asimple pulley. This can be proved by the methods of Chap. 4.)

+ tLF'v = 0: 3T - (280 kg)(9.81 m/s2) = 0T = 1.(2746.8 N)3

T = 916N ~

+ t~ = 0: 4T - (280kg)(9.81m/s2)=0T = 1.(2746.8 N)4

T =687N ~

PROPRIETARY MATERIAL. 'IJ 2007 The McGraw-Hili Companies, Inc. All rights reserved.Nopart of t it is Manual may be di~played. reproducedor tlistributt 'd in anyjilnn or by any means,without theprior written pel7nission of thepublisher. or usedheyondthe limited distr ibution to teachersandeducators permitted byMcGraw-Hili for their indMdual coursepreparation_ Ifvou are a student using this Manual. you are IL~il1gt without permission.

73

.'r--;' '-1 ,--' ,.--- I-((I') ib) (r) Id} It)


9/15

11 PROBLEM 2.76Determine (a) the x, y, and z componentsof the 1900-Nforce, (b) theangles OX,0Y' and 0=that the forceformswith thecoordinate axes.

.f

PROPRIETARY MATERIAL It>2007 The McGraw-Hili Companies, Inc, All rights reserved. No pa, .t o .f this Manual may be displayed. reproducedor distributed in any form or by an)' means. without the prior written permissioll of the publisher. 0 ,. used btyolld the limited dish' ibutioll to teachers andeducators permitted by McGraw-HillIor their indh'idual course preparation. If you are a studelll using this Mallual. you are using it without permission.

80

SOLUTION

(a) F'x =-(1900 N)sin20sin70= -610.65 N

F.v= (1900 N)cos20F.t = -611 N ...

= 1785.42N

F; = (1900 N)sin20cos70,= 1785 N'"

= 222.26N

(b) cosOx = -610.65 N

F; = 222 N ...

1900N

cosO, = 1785.42 Nor Ox= 108.7 ...

) 1900 N

cosO: = 222.26 N

or Oy= 20.0 ...

1900 N


10/15

PROBLEM 2.83A force acts at the origin of a coordinate system in a direction defmed bythe angles (Jx = 43.2 and (Jz = 83.8. Knowing that the y component ofthe force is -50 lb, determine (a) the angle (JY'(b) the other componentsand the magnitude of the force.

SOLUTION2 ( )2 2(a) We have (cos(Jx) + cos(Jy + (cos(Jz) =1

( )2 2 2cos(Jy = 1- (cos(Jx)- (cos(JJSince F.v< 0 we must have cos(Jy < 0

cos(Jy = -0.67597

(b) Then:

-501bF = -0.67597F =73.968lb

And F'x=Fcos(JxF'x = (73.968Ib)cos43.2

F.y=53.9lb ...

F'z= (73.968Ib)cos83.8F;=7.99lb


11/15

" PROBLEM 2.91Two cablesBG andBH are attachedto the frameACD as shown.Knowing that the tension in cable BG is 450 N, detennine thecomponentsof the forceexertedby cableBG on the frameat B.

SOLUTION

BG = -(1 m)i + (1.85m)j - (0.8m)kBG = )(-1 m)2+ (1.85m)2+ (-0.8 mfBG=2.25m

TBG = 450N [-(1 m)i + (1.85m)j - (0.8m)k]2.25m= -(200 N)i + (370 N)j - (160 N)k

.. (TBGt=-200 N ~

PROPRIETARY MATERIAL. ' :9 2007 The McGraw-Hili Companies, Inc. All rights reserved. No par t a/this Manual may be displayed. reproducedor distributed in any/orm or by allYmeans, without the prior writtenpennission of thepublisher,o/'used beyondthe limiteddistribution to teachers andeducators permifled by McGraw-Hili/or their individual course preparation. lfyou are a student using this Manual. you are IL~ingit without permission.

9S


12/15

-" PROBLEM 2.93

Detennine the magnitude and direction of the resultant of the two forcesshown knowing that P = 4 kips and Q =8 kips.

SOLUTION

p = (4 kips) [ cos 30 sin200i - sin 30 j + cos 30cos 200k]= (1.18479kips)i - (2 kips)j + (3.2552 kips)k

Q = (8kips)[-cos45sin15i + sin45j - cos45cos15k]= -(1.46410 kips)i + (5.6569 kips)j - (5.4641 kips)kR = P + Q = -(0.27931 kip)i + (3.6569 kips)j - (2.2089 kips)k

R = ~/(-0.27931 kip)2 + (3.6569 kips)2 + (-2.2089 kips)2

R =4.2814 kips or R =4.28 kips ~Rx _ -0.27931.kip=-0.065238cos8x = Ii - 4.2814kipsRy _ 3.6569 kips = 0.85414cos8y=Ii - 4.2814kips~ _-2.2089 kips = -0.51593cos8= =Ii - 4.2814 kips

PROPRIETARY MATERlAL ( ) 2007 The McGraw-HilI Companies, Inc. Al l r ights reserved. No part of this Manual may be displayed. reproducedor distr ibuted in anyform or by any means, without the prior wrinen permission of the publ isher, or used beyond the l imited distr ibution to teachers andeducators permil1ed by McGraw-Hili for their individual course preparation. If)'ou are a studellt using this Manual, you are using it without permission.

97


13/15

PROBLEM 2.106r.\ The support assembly shown is bolted in place at B, C, and D andsupportsa downwardforceP atA. Knowingthat the forces in membersAB, AC, and AD are directed along the respective members and thatP = 45 lb, determinethe forcesin themembers.

SOLUTION

See Problem 2.105 for the figure and the analysis leading to the linear algebraic Equations (1), (2), and(3) below.(1)

(2)

-(%)FCA+(t)FDA= 0(14.6)(2)B.4 11 3" FDA

From Equation (3): FCI=(%)FDASubstituting into Equation (2) for FBAand FGf gives:

(3)

From Equation (1):

Since P =45 Ib

L:~)C:i6)(~)FDA+ (~)(%)FDA+ (~)FDA - P =0(838)r - FDA=P495(838)F. =451b495 DA

or FD.4=26.5811band FBA=C:i6)(~}26.581Ib)

or FB.4.= 23.51b


14/15

PROBLEM 2.113A 16-kg triangular plate is supported by three wires as shown. Knowingthat a = 150 nun, detennine the tension in each wire.

SOLUTION

wdDA= J(4OO nun)2 + (-600 nun)2 =721.11 nundDB = J(-200 nun)2 + (-600 nun)2 + (150 nun)2 = 650 nun

I ') 2 2doc = \j(-200 nunt + (-600 nun) + (-150 nun) =650nunTDA = TDAA-DA

= TDA [(400 nun)i - (600nun)j]721.11nun=TDA(0.5547Oi - 0.83205j)

TDB= TDBA-DB= TD!J [-(200 nun)i - (600nun)j + (150nun)k]650nun

7' ( 4. 12. 3k)~DB -131-13)+13Toc = TocA-ocToc= Toc [-(200 nun)i - (600nun)j - (150nun)k]50nun

=Toc(-~i - 12 - ~k)3 13 13W=Wj

TDA+ TDB+ Toc + W = 04 40.55470TDA -TDB- -Toc =013 1312 12-O.83205TDA- -TDB - -Toc + W =013 13'3 3-TDB - -Toc =013 13

(1)AtpointD LF = 0:i component:j component: (2)k component: (3)

continuedPROPRIETARY MATERlAL. ,~ 2007 The McGraw-Hili Companies, Inc. All rights reserved. No part ~t'this Manual may be displayed. reproducedor dist ributed in any fiwm or by any means, without the prior writ ten permission of the publisher, or used heyond the limited dist ribution to teachers tlndeducators permitted by McGraw-Hill.for their indil'idual course preparation. Ifyou are a studem using this Manllal. YOlltire using it without permission.

]23


15/15

PROBLEM 2.113 CONTINUED

Setting W = (16 kg)(9.81 mls2) = 156.96NAnd Solving Equations (1), (2), and (3) simultaneously:

TD.~= 62.9 N ....TDB= 56.7 N ....Ttx = 56.7 N ....

PROPRIETARY MATERIAL r t; 2007 The McGraw-Hili Companies . Inc. All r ights reserved. No part q{this Manual may be displayed. reproducedor distributed in any fOlw or by any means. without the prior wri tten pennission of the publishl!r, or used beyond thl! l imited distributiUllto teuchers andeducators permilled by McGrrIl Hif/for th"ir indMduaJ course preparation. 1/) '011are a student using this Manllal. .\VJI/are using it without permission.

124

Documents

Chap2 Solutions Sp07