Learning Guide in Engineering Mechanics

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/' Publis'Md .t Distribut.tJ by:

~O~-t~~ . . 8~ ~k:anol' Reye si:si. '

Tel. Nos. 741-4916 74;_.49:20>1. 1977 C.M. R.cto Awnue '

Tel. Not. 741-4956 741-4967 M1nifa. Pt.irlppillff "r ",:

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Joj;" BilJOna [oriu~daip_

' : .. RBSra ~ ~ Asi!oOa~s~: Ar'ociaf:iqaot P,lll1iJIPin.e. ~~ A's lnt.ernational Book'.-Assepations 11.embership: Asian . Pao&c.Pubmd-s ,AsSiatim tAPPA>;.Associatioa aL South

' EMt~Asian Publisher. CASEAP>; ~tiooal Publishers " "'! . ~at.ion

P~ted by i AEX ~OJ11MY.ic. 84-8fi p_ Florentino St_ Sta. ~C!S11 He.ights. Que.ma.City. Tels.:nUl--08, 712-U-01; Fu no_ TII-54-lZ .

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Pre.face "

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. 'The book: entltled Learning , Gulde In Engineering Jfe chanlcs was written as text/ reviewer for. fhe purpose o .f .-present'Jn(. fhe prin~iplEia and concepts of 'E'ngine"ring Mechanics iii a ver1. basic and .$YSte.aatic !lPProach to help the . atud~nta. underata!id and lear .n the ' subJect matters that wJ,11- develop .the 'Jr .. 'orderly ' ~ pfoc~'ssti's ::ot th'tnk,t'n{ . .. :

.. . . 1, . . lU th .. the . pr ese~ted t~pJ cs. 1 , ~ . he u,s.~rs .will 1''a"e ... a_ ..

very comprehen~ive, a :t~df" Of the Auhdamental pr:tnC'iplt['s of Engineering- 11ec11anica wblch are' appl I cable.: ~o wide varJety of prac~fcal . s~ tuatii>ns norally encoun~ered by

" them iin \ heir' day to :.day acttvJtle,s ~ . .. . " . '. .. I . . ' I Extra effort was exerted so that everyt~lng will

b'e presented 1n .!l 'v.ery ,p~dectly~ unde.rslood, .but ,in . a clear and concts*' language. This l'itll eilminate' or reduce ~o a liainimua the !'tudent's attM~de o.r " b~~it of inemp~i.z~ng tbe con~ept with o~t u.n~e~:standJ~g. , '..:

_ The authors wish to acknowledge ~h~J r inde b'fedriess to M,ichael Siongco, .Jef'frey " Bo,rlongan,. , Jl.onaldo , C~tin~lg, .. and Rqm~o Adrlano fo'r tlletr valuable

contributions ln preparing and .;rJti~g tbe manuscript.

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N . c-: De l.a~.ama A.G. Mendoza

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ACKNOWLEDGEMENT '. I ..

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The a'uthors arc parti~ularly indeb~.in the p~aration ofthjs book to . Mr: RomcoAdriano, Mr. Ronaldo Catindig. ~. GeiardosiimSon, and Mr .. Jcfilcy BOrlongan.

, They further wish fo speeia.lly ri;eogniT.c the exerted effort of Mr. AmoldQuetiU! in wtitilig the mS.nuscript

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S~ial mention fo those who briitg tb,e greatcstjoy into Nick's life, r,,ii~, Nikki.Md' NikkJl:, .

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CONTENTS

Chapter 2: R.aoultonfc; cf force~ t - H Choptor 3 : fquil1bnum ci" forco ~ 2S - sa Choptor +: hno~ of Strud~ .s.-86 OqJtcr 5 : frlction fJl -m Chapter 6 : Fbroo Sydcnw. in Spcx:o 110-BO Chopicr 7 : Controidco "-Ce-tlC'Jl"'C of

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ft>f.) Determine lhe. X 'Y conwponGOf6 of ooch of the for-ces ahown in Figure P-2o+. consider ftle 3QOJb lood: f -300 ( J,~ &* )

-t.5012! ( _b ) F.!:I 3GO Iii' t~

.960 ""'. I'.~ 300b. eonaM:.ler fhe. ..fD lb lood: r--tOOl!L r,. --too eoe -t0 eons:Jer the R> 1ood :

- aoq :fil lbs. T!4 - .f

c.cnsider the 6

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T v'rh t llrs ../(-GO)~ ~ (-:aoo)c .!1&!1.1 lb . -e- ton"' m/.,oo - so.gg.

z.ao) Tn Fig. P - !l.10, tfJe x oomponent of the force P iw 1-t0 lb. to tile Jen . Determine P tf} jf6 Y componeni . .

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p. 8.06(1+o)h - 161. 25 /b. Ton-&- P1/p,c Py -+h (1+o) = so lb 111~ The body on the so 1'noline in Fig. P-.2.11 i~ oded upan by o" forcp P inclined ot f20" ~/ ihe hOriwntal .If, P i~ re-solv~d 'into componerrfe par

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Y ." ~Fy~ woosiriatit~ -~lliina0-1oob -s~'- -J..\ ~ .,. ' ' ', I t ~'k/ . _;, . ~fy .; ; :..f~~Jb'. . ~ I ' '." :. ' -;.=---'~"'1r...i~-:+ R J~F,.~ t ~Ff'" "'_J(_-+a66.o~y~ t (-1~00)2

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. . :R'." -"09j.9.7 lb d~'.,.;hdo tl1P 'lef'f ' ' ; - ',._'. ' ~, . , , :: Q '! ' ':6 . '

. , . ,~'d-Qw"-' h t=Jg-. 'P--2.16 . Oetei; .-, ,,.:, l oe"s ~fx>w'"1 ;n fig. 217. ~r~ ' =iEF,. 100~) H.00COGo

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. ' .. ~. \' '- 1 '11 I 'I ~~~~.~ ~(+l'.s) .:t. ;9~(~,~~~) "';2t (14i) , ... , '::rf.,;. 2~,'8~ ~b. . . ' . ' .. . . . t--:--Ml'~r-::..,,,. x_ _, .. :E'l=y_~;;3QO ':" , , , B .... ~7e , 1::i d,owrf tp t~ -Nghf

',:, r_',.' 'I .' . , 'w ~ .' _ ,I ,, 'rt '{{'-.:': ':- ... ........... r:e-.lf. .~'. lan -i '~~O.j!,S'':z89,5.,2 .. .!ll2.p~ 1 .. ' I ... ~ . I ' .! ~ ' ' ; ' ~ ,... ~i.0.2 Deter-mine Jhe re.sultant of fhe"-foul' ,foicee ootrng ~ the.

~~j .. ~, 1," ~ ' ~....... lc, bOdy"

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00000 a (-'9f.t$)2 taf4>- - ~(!Zs2.1.s)fane- t 1....+00 8535'-G to""-& -70116-9t tone- -766()0 "'o ~~io Formc1la :

ton-e,, 1.+t j -f'"t "' 5~!>.18 fOl\-&i -0.~16 . ; ~. - -a1.63

c1- -e; ~ .-c.1er frorn1

I'.' .. 2g~.15 . ' - 511, 61 lb 56-~e

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~) ~~t ~- ~1ll -.r ffe J'C$Jlt01'f iio "fOo lb o~n fo ft-e r~ht of 60 W,ff\ jhe.' X O)ltG. '

ft)I ~f)c . . y p

MJ()lb ,,

to. obf~I"- 0-~()I'\~ paratleJ to fhe i"dir-ed plone. . e.,,~ ~o . .t::f=y' 0 PG06tfi -~ sinao - a.co 00680" . -o ~ P- +ee,0 it>. 6 7.' 600 CJ:/6'3(J - P 51r11r/ - 200 6~ao - ~F)'.' 2 02. 62 lb.

R J~F,.. ,_ f' ~F7" - /(w~.aiy -to - ~e' are- 200 lb~ !Z-40lb wtile -Me- ongle he;tween fhem .~ (I:) rird the. mwltont pull on tro_ bor9~ ~-the

, - I . ~~ betwee:,i Jd\ cf 1he ~ ~ lhe 6ideii Of ihe CXll'\o l . \ . , aoo In order to have o . resultani para.lie? to

)~ bon'k& ~FyO ~Fy 200 sin e- - 240 sin"'<

O 200.si~-&- - 2'+0 si~(60--&} bv), s11'1 (60-&) "' sin 60 cose - sin-c.os6o

O 200 6~ -24-0 ainoo COS9-+ ~40.9~COS60 zoosin~ - ~07 . 8S COS& - 120 &ine-

.3:t0 s inr:r ~T-85 oose-5m C'"/cos~ 201.9~i10

Ton

; ' .. '[ (} 0-+---''--~)(- ~Mo - ~F. L.!i = J361 (~,61}(iy) .. "fOO

L!f' 1. 33 n: Clbove o

.jf~;:;.,~-;;~:~~~~r ,.I~; ... J . -~ ~ ,iff1"t.,f,~"!'.=-i1 .h}rr.,r:~-~1o:, t ~ \ .. .,. .f...t-,il t.-1 )

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. 233 ,)' In fig. P-2~, a force. P inlerseds 1he IC .o-x1s q) +ft. to ltle righl' of o. If it6' incmenl obout /\ ie 110 fi-lb counlercl09k.y..I _.., ils ~.menl ~l J3-~. 4 a _ lr1ongulor lood vory;ng from

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p.-1~(!SO) t Y2(12)(1i:>C>) B 1soo lb (downward)

Me - goo(+) - 600(6) " - 7200 fl -lb. d 1200,1(900 ... +.e Fl Jefl of 8 .

ia~.) The 16 fl ._ wi~ of on oirplone. ;c;r ~ubjeded too 1irt w/c . vorieso from zero of tho t ip to 360 lb per ft of the f\Jse-w;~.

1 ' J lone occor0ina to w 90 )C t'.z lb 11er ft where x is;- r-neosured fl"""' t O:::J ~ r~ . ~: . ."rrorri the tip . .Cornpvl,e fhe ~ultorf 11ij its l6salion from the :.~' . \f 1 'f' ', f

'i(Y'.. . wing tip ; 16 tt:~~( .. Rfwdx' :; "''9ox'1t1dx go,,.df]:. t/'.li I I ;1 Jo -- ' ' " r' ' . 3/2 0

. 'fb(M)a/I - go ae-10 lb . .iv, L

Mtip J J wd-; 16 aoxa.t'g d~ ~ go x.!V.q]" ( !:>/'2 l>

: 90(16)a.tt _ go(~ 3686+ n - tb . 'h ~

d .. 3686~ - o.tSo n . 3&-t

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m'777.7:'7)'7777:>'7''77nj i- thfcKness' '*' dstii;/t.., arle c.on'~tonJ .' ~ , J _l .A.Pr 10(1+). _,; 140 in~ ' ' T .AH - .ic::!12 -, 1~ .. 01 in~

; 16 ' I J ",. .... ' : ' " ,.'.' ' + ! ~/; r;..L.(

~~E:~~: ;::,:):. >.-> '.... .. . ~ ,. ~ .. ..... ~~. .. =.; .. - .~ .. .. .. rt.+' '.r':2~S'.) ; .T,n~ ,_.,~ultont of three poro,1,l~'l . lo'Oqs (one load it.~P,issinq ih Flg.P: z-+sY is .ao lb 9cti'19 t!p .ot 10 ft .' to 'ihe[-1ghf,0CA. Com-

pute the rnognifude ~ po'iit:on of t'he JT1~nci: I~. ' WIO ; 6011:> ,

. t-2'-+-f-z11.'==f ~ --

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R- +o , 60 t F -f - -+-O t 60-SQ - '70lb. F ,. -70 ft> ( do~n'rY'~~)

.tMA .. f\ (10) 1 . ~-70(x)t 41(2) t 60 (13)-;. aoo

:x=Brl from ~.he right-or/\ b '" ... '. ~s.) R;..rer to toe ngure . /\ COl,lple consists or iwb '/ertico,l forCes ;: i:.: . or 60 lb e,aoh .. One ror-oo oct4 up through 1/\ ~- .the oher ocfs

.. ( r .. r ~. thrt?u.g.h D . T r;o.nsform the eovp'7 info:c;i.r'l eq;volent couple no-te:,: .. vin~ (loriwn1ol f~s ooflr;'1g. th'roug~ t 'ls..,f: "' . ~. ' CA.b- 60 .. 2 -120,in - lb ::Cer . " 1 /~;::'.,, _' ~ ,,:>,l 3 ! ' Ce~. r(s) .G:129 'in:db . 'i

. ~ ,.., ' . . ' .... . ~ , f -; -+O lb act1119 os showil

c. --- . . ' h1 , .

IJ . 2'f-6.) Deformirf, lhe resul!ont. mo~nt obout po~nt ~ of the

sy~iem or force1; s hown an Fig. P- 246 . Eooh.squore IS 1 n. on 0 side . ao1b

by resolvin9 lhe forces inlo ils .x ">Y cornponen ) s : MA wo(s4;)(3) + ooo(ll/-lfa)(~) tao(a)

-(eo)(1) 'IOO(if,[4)(3) -100(1/5){-t) t 1oo(a) (1/.f:i' ) - 1oo~k:f)(-+)-~oo(~)(~)-1oo(~X1)

' 4!M"' ~ s6C.8 f1 -lb cw

Z_.7.) The three-step pulley shown in Fig. P-2-+7 is subj~cted to the given couples. Compute l~e vo)ue or )he resultonl cou p )e . /\)so de t errnine lhe forces ocli'ng ot }he rim of 7he mid-

die puHey lhol ore required }o bolonce. ~he given sysem. so(n: ~c ao(12)-

C c 30R (-too -100) ft-lb - (30/,Q) R

!3 '" 120lb d1"reoled verlicolly up ot /\ ~ down ot 5.

2.so.) The con! i lever truss shown in Fig. P- 2!50 corrieG o vert1col load of Q400 lb . The truss ,s supported by bea-r ings o) /\ 9.., f3 which. exerl }he forces Av., /\h, I/; Bh. The four forces shown oonl1h.ie two couples which must hove oppos'1 le mornen~ eff ecls lo prevent rnove-rnent or lhe trus-s. Determine rhe ma9n1~ude of lhe -supporting forces.

6 . Sh

~400lb A.v.

in order to constiiute. o c.ouple Av "' 2-'tOO lb (upword)

2"'(6) - Bh (4) Bh "'Ah "' 3600 lb

2-51.) I\ verf 1co) force Pol II ~ ono}her verti~I F ot /3 ir"I fig . P-2s1 p~uoe o resulfont ol 100 lb down ot D ~a ce1:1F1iePcloc~ise eouple c or 2o6 lb-(1 . Find }he mognifude

~ direction of foroes P ~ F . /".C,.!ZOOlb-fl ~~ =~B :1~lb ,._ ;i'- r

(2001b-f1 lo 4'

R1oolb

~MB" 100(1) =-200 .t P(3)

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p = 300 lb. (downward)

.fMA -200 '"'100 ( +) ~J (a) F ~ 200 lb (upward)

l::f-): ~ 2.t't)_,p[ ~ ) p.,._ "3~

2U.) A Wc.c ~ern oonsisls of o cloc:f\wiae couple of 1-901b41 plus o 210 J>- foroe dirccled up to lhe riqh\ lhrouqh }he "9in of 'X ~ Y oJr.eS al -&ii: ~ 30-. ~ lhe given syslem by on equovolen1 ~le fOrce Ii; cvmpute . lhe inlercepls d' ils rinc of ocl ion wilh ,lhe 'X ~ y oxes.

R,. = F,. ~ 2-t0 cos ao = 2a7. 8.S lb. (to lhe righl)

Ry= JY = 2-fO .sin sc) .._ -120(~)

M C ~ F)I i:,-iy = ~ - 2.31 fl . ob

rned }ho) lhe inlernol slreSes ore detd)rmined from the e

211s) Replace }he sy"lem of' forces shown in fi'g P-~a by on oquivalenl force through 0 % o couple oding Jhrouqh A ~ 8. Solve if thB forces of Hie couple are (a) horizonlol ond (tj.) verlicol.

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361 fb 0

p p

0

F,. = 141.+(V.n) t 22+(.JS) -:s6t (o/Ji3) ~F.,. a 100 . ~tbOo }he righl) .tfy = 1f1.+(Wz:)-!l2.+(1/,rs) -.361 (o/.J6)

~fx~ -300.S6{means Gl"'rw,.ord) R ,. /(100.01)2 t (-3oo.'!J6)-i

316 .79 lb (do.vn lo the right) n+tpn~h = Fy/.:i:F~

f

e/C tori' 30t'.),s6jioo.og e-~ = 71.se

o .) C =~Mo 3F "'361(o/.JT9)(1)t361(.a.413)(1) t 141-41

(1/..iz)(2)-141.+(1/..rz)(s)'- 224(2/-19)(1) -2~+(V..rs)(1)

, !3F c 100.1 . F =33.37 lb b~ 4P "' 1001.1

e= 2.s.03 1b .

260-) The effecl of o cerloin non-eoncvrren 1 forlem i} is fovnd lhot ~x = -60 lb, ~Y- +1601b, ~~Mo., +ao lb rt In o counlerclock-w 1r;e sense . Oererm'1ne lhe paint' o) which lhe re6ultanl in-len;ec)s lhe x o'llis.

~ :ZMo ~ .fy t;i: Lx == +ao/f6o ~ righl of Orig.in

26U Delermine complele)y lhe resullon) of }he forces oc-ling on lhe s)~p pune~ 6hown in fig. P-'262 .

t 1SO ~r,. 750 COS.Bo .. '250 . 89Q.s21b(lo lhe righ) ) .fy - 7SO s 1nao - 1250 )

- - 875 lb ( - means downV'IO R- ~F,, t ~Fy

250 (899.S2) a ( -87s)a R 125+. sg lb down lo r ighl

-&,. }on"' 87S/B99.S2 "' ++.~1 Rd ~Mo 750 (u.s) - 1250(0.s)-280(1.2") Rd .. o " d ojsoR po~es }hro.Jgh lhc

oxle 215&.) De}ermine lhe resullonl of lhe force ~~}etn shown in fig . P - 26S ~ i}~ 'lf" y inlercepl~

It~ [8GV .sinso. aet (V.Jj3)-224(o/.fi) \

1+9.W(to lhe..riglll) Tt"-~-t--t--'"""N:: . -"- ~fy 300 cooao + 22'9(1/..f&) - 361 (/.Jii)

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.a61 lb .224 lb :: s g .61 (upward)

R c ./~F~ 0 t ~Fy .. /(1'tQ.g)2 t (.S9 .61) .. 161 .318Jb ( up to lhe ri9h'l)

ton&,. .EPy/~1">< ... .59 .6-y1~.9 -&" fon-1 .&Q.61/1-.9.9

-0-x 21, 69 ~Mo - 300 .sinao(2) - 22~(1/~)(2)-~1(o/.fill)(1)

-100.6 0-lb (- means Counler CW) lx = 1006/ .s0.61

c 1.67 ft r19lil of O i.y ~ 100.6/1+9.9

- 0.61 n below' o

26+.) Completely d ::>lermine lhe reGullonl w ilh respect ro pl. 0 of lhe forc.e syslem shown in Fig. P- 26+.

IY I :if,." 141.4(Y,J2) t 300 sin6Q i 260~%i69)

- 240sin30 1:79.Z.Q lb .. Fy.,. 1.+1.4fi/.J2) t260(.ru'3) '1"2"40 co~ao

-soocos6o .. 257. 99 lb, R - ./:.F,.1 + .:EFy

26t = 10.000 - 6000 cos 30. = 4903. 95 lb(lo lhe

right.) , -ify d -24000 - 6000 sin so

; -2?000(-meons dONn-. word)

(l

-9->C ton_, fy/1'Sc - ion-1 10007.09/-t00a:s2 -e-? .. fJB.~ ..

LMA .. +f-80 (1/~)(s) t +'-90 (1"6)(10) t 2ooo (s) t 30006) , 'looO (20) t iooo (ao) .. 160081. 12 Q.-lb.

}( .. 16ooe1 . 1~Aooo1. 03 16.0 n. r-iQht or"

268.) The resultonl of four force.G', of which :three ore sihown. in fig . P- 268. iC '" 'loo lb to the left

C '"Mo 1Wlb

+SO -Mp t 110(+)+120(~) !Ar ' ~oo Q-lb (ccw)

MF Fd d 200/200 1 n. obove 0

269.) Repeol Prob. 269 1r lhe reoultont ic 390 lb dirooiod doNn to the rig hi of o slope or .s to 1Q possinq through poin~ t\ . Al $0 determino the x ~ y inleroop .. ~ of the. rnSs-inq force F

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flO I i50 ,' R.,, ~F,.

r.---1--- ,+-"--'' 390(1'/13) - N t uo 1so(3/s) /~ ~fx 169lb ( to !he ri9h~)

II ~ Ry ~fy / .,. "'-..390 - 390(

8...t) .. Fy -120 +160(+/s)

120 --- ~ - 1so (downwor-d )

RF F~ .:r)I ~r7 2 - (160) "" + (-}So)~

P !219, 32 lb (do.Nn to right ,)

-6! = -S15-MR ~h4

390{o/ia){3) + 3'N>(1~)(1) "'; 110(+) t 1!0(2) t Mf Mr ..itJO n -lb cw

b "" -t - -:a.067 fl - obO"B 0

210-) The th-cc fOrcc.s '6hown i1' Fig. P-Jt70 or& required to (t.ll.IGe a ho"i2()0tol reGlJltont octinq fhrou. doteNnine fhe ~ues of P,, T_ Hrilf : Apply MR-~ io ~ier~ R .. then MR- 8l'IC to ftiid P. ~ f.nolly either trtR.-.=ato or Ry -:EY to c;ornp.ofe f .

#F E

; 1---4-1--...._ ........

! 8 I /' ~ - ,, \

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'T ~p

. ,.,_ !l10 ~ :rn

M9 -~ -316(1/.Ji0)(t)t'316(3/-.1ii>)(2) r R(t)

R.,. -t9'J. 6+ lb.(io f he ric:#'t .) "ll "' ~Mc

...,.,.ff(.3) ~ 316(1.Mo)(1 ) - 316(o/.Ji0)(1} -t P(~)(-t)

p- 1'7'4-82 lb t.tq ~Mo ff9.6+(a) c -1(-Ae)( ... )f 316(~)(:t)

.. 316 (1/-11o)( 1) T- - 225-18 lb

Dt} The .lhree. force6 io Fig P-270 creof& o vertiool reculionfoc-ting fllra.gh point A _ If Tie l, conp>fe -the va-IU06 of F .., P . .::E:hte - M"'-

-361(~){_c)t361(24;a)(_-t) -R(e) R .., :t00--19 lb ( do.,.,.wvord ) ~=~M& -100-49 ~) = F(3/.r.a)()- r(~X1J

r- ~sa.n lb ~T =MR p(245)(1-) .. (ua.29){-W.-0)6)- 2S3 -~(~)(1). -f-00.+9 (+)

P - 1cH . ..st lb -

Chopier 3

fquilibrilPl of forco Sy~fems

30fl.) Tbe cyhrder C i"' fita p-ao2 weighs 1000 lb. Oreiw o FBD of cyhroer C 9'j of rod /\B.

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GY i: + wc:;~1o001b 3' Ah -+--

Av

303'.'j 1he u~forrn rod ~I'\. f;9 . P-aoa we~l-w 4QO lb % hOG its Gef'lie.r of _grovity o l 6 Drow a FBO o( '\he rod . tle.91ec~ the. ih1cl

D -i- '._ D

101 @ ~~ ~ -+ tAv "' I e;oc)lbt 6' Ch c: ... G

I! Ev

ao6.) Drow o FBD af pulley'1 E ~ D ~ of ~he bor /\ D .shown ~ Fi P- :a06 . /\uume o II h(n,geg to be smoofh ~ ne;gle.ct t~ w01ght of'

~h bol"'.

309~ The coble~ boom shown in f1'g . P-aoa support o load of 600(b. Oeterm1ne the lenstle. force T In ihe coble ~ iho compressive {Orce C in the boom .

T

t MethOO I (uJ:;hg hor-izontol "'-. verticol Axes) ~Fh-o

Tc:osao C cos +s .Z:f.,, -0

T.sn30 t Csm+s - 600 subst. eq_ 1 io2 (eq_uoto I in terms or T) T .i;in ao + (icosao/co.11,...,,)(.sin4s') = 600

I =- 4.39. ~.3 lb. C "1'39,23 (cos.ao")/cos-+S

C"' 5a7 .24{4 LtL.. ~ Meihod !(using rololiorl o~es)

T l _s C L fv O : Csin7~ 600 .a;,..,60 - ...: &.oJ7. 9'):.? lb

"Q.or,"t

,Lftico : T=60000S6t;)tCcoc;7S' 6CO t;Ct; 60' t .sa7-94.s; (cos 7.& )

T - 439. 2.s lb. # Mel hod ll ( ui.ing force i l"iongle)

.909,) J.. cylinder. woighing -.oo lb- i~ held 09o'1nG+ a .smooth in-oline by rneons of the woightlesr> rod AB1nfi'g.P309.Det. P"He.t-erfe.d on the.eyltd~ 8

-

.

p .

b Fv 0: t1Qn6Q ...00 &1~

lr1e.thod l (1.1Vir1CJ t\"-. '< o~) FBO ol Cylinder e

.~ ~fh c O : Pc.osa. "J1co& sio

P = Ncos~~CX>G,'s Psin'1s -t ti .sin a~ - ..00 @

.cubct- 1 +o 2 [Nco.r.s (~1n12s'.j/eo&t;. t HGin as

- "tCO -.- N - -HS.60S lb p ~ 11 coi;as/cas2e

= -tte.60.S(cosasoYco&~' p - .378.36 lb.

ti~ . -+19 .605 lb ~A..-o: P-te>.:;os;6~+n~,o p "'400~661' "\18.60.S c;o46()' p : .378.!!15 lb-

Melhod J: ( using Foroo T r.:.ongle)

p = ~79 . -25 lb .. H = +is. 605 tb -

a10) f. !10!' lb bo,c is held at rect on a smooth plone by a force P 1ncllned ol on angle f} w/ the plane as .shown ih fi9 . P - 310. If e-" -ts det - the volue of P ~ the normol pros;svre H eicl"'ted by

!hop/a~ ~

FBO of the bloc.K p

Mettoc::I I ( Us;;ng ~ Y axec:) .f.,0

300 H sin 60' - P s ;n.+s' @ Fi.o

PW' 10' H c.oc; 60. P H co~60/cos t1j @ ~tx;t 2 .n 1

Me thod [ (U6ing Rololed "1.ttr.) 300 tfamcso' - tioos60/co~

p

aoolb

fv"'0 ri6in4S' 900a1n75 H 409,907 lb

1'.Fh 0 0 : P s rlcoce' -300c:O!: 1s P "I09,eo7(eoc

lllli.). Dolermi ne lhe mognituclos or p ~ f nece~ry to keep the concurronl force '"r"em , shoV >

f aooCQ1,6a t Pcos .,o- 2oococo1os f aooco56o" - 1 33, H.S (cosao~- 200 co&105

F 86.s7 lb .

Ja'la.) fig . a1a repr-e4;en\s ttc c.oncul'rent force ~e.m ooHng ot o j~nl of o bridge t rvss . De lenr>;ne the voluOb ot P ~ E to mo"1n-tain C71!"-1.aoo~is"t Fcos~ p - 16'5.4tt lb.

314-.) 1he r ;ve IOl"ce& s hown ii\ Fig . P-31-1 ore In equilibrium .

aia.) The :aoo lb force ~ the ~lb forc;c s~ in fig . P-3\5 ore to I: h~ld in ~ui libriurn by o third force Footing ot on unkno-wn orete~ w1lh the ~zonlol . Oet~ne the volues off~&.

-t001b

.. ,.. ('-~ ~-o

300"

/ :a17.) The ~y.sfom of !W*fed ~ dwwn in f'.'-4 P-317 support the 1ndicoted wolghts. C()mpute ~ ienGile fon:e io eodl"'fC A C0661> c b 22307 lb.

1 p

using Force 1 nongle

~/ 200 / \

by aone low

'" -$.- -~ $1060 Siil .?A0 ain~ c - zaa.01 /b

c

v by sine law

p -'--. - _Q_ ~ Pl~ &ltl4!0' p ~23.Q7{s 1 r179')

S tn-4$"

p . 3)4.719 lb

Fv O

CGtn7s" P~+5 p 2~a.01 n 1r.

ax; +s

p " ao-t . 7 1CJ 10

""' .,, Lt cv; f n

Pf"tlblem 3 19 .GOlu f;on . ~Fh O

+oO ~& 200 ~ ~ 6o ~Fv =o H 1 't .sin e- = soo

N BOC - 400s tn60 H . 453. 6 ~ 4-S

aita:) The Fink truss shown in fig. P-322 iG ,supported by a rol-ler a l II ~ a hinge at 8. The given loads ore n0rmol to tne inclined mer;nber>. /Jek"mine fhe re.oofronc, a+- t4. ") B. l/1hf : F

323.) lru f ruGIO shown in fig. P-323 i~ .Guppodecl by o hinged of /\ ~ a roller ol /3. }. lood of ~OQ'.J lb ii:: oppl18d. ot c. Dot. the reootionG at A" a.

c

'"""'" 2000 lb

~M/\ 0 30 Re. fZOOO (c.os:30)(1s) l 2ooo(s1n3C"XR--.v " 1199.38 lb R,..,4 .o RAh1 t IV..ve. RA I (113~. os)4 t (1199,35)e.

R ..... " .2106 lb

Tone- = R11.v 1199-.3 5 RAh 1731Z .os

'-& a ton_, 1 1~. 3 5 1 732. OS

-e- = 3 -t. 70 ~

R.11. " 2100 lb down fo the left a i -e- = 34-.7 .

.32+.) /\wheel of 10in rodius corrie~ o load of 1000 lb, os s~ in Fig.P;32.+ .. (d) determine the horizontal force P applied ot tho cente,r which 15 necessory fo sforl the wheel over ihe .!I-in . blocK. /\Isa find the reootion ot the bloc!\ . (l:l) ,r the force P' rnoy be inclined at ony ong\e w1th the hof"irontol, determinS: ihe rri1ni.rnvrn voluc of P to start the ""heel over trc block.; the

. ongl~ thot P m9kes w1lh the horii.ontol; 'It.., '\he reoohon ot the btocK. .

~ It)

sine- ~ a/n - SAo ,. o.s -e-- :so

o) pz i:la

zfv1a .... 0 sP" 1000(1090s:90)

P-" 1732:051 lb :::EF>< Ro

pc Ro cos30 : 1132.o&r Ro .. 2.0oo lb. .

b.) P iG minimum i f i~ wilt be .1 to ~o

hence, -e- .. ~o

.Ma~o

10 Pmin = 1000(10 CO$ 30) Prnin " B~ lb .

..:tFi< =O Ro cos so = Prr)in COSfio

Ro e[Pmin(a:>s60lJ/cos30 = [8610(COS(;O")]/cos .90

Ro = .soo lb .

~.!!l.r Determin e the omounl ~ dir ecli.on or fhe .smollest force p required to tori !he wheel 1n f 19. P-312s over block . Who f i6 the reodion ot lhr: block ?

P. c;oa~P Py s;i11"tP

C-OS~ = 1.74 ~1rfo=O ,B=+1-+1

.tOOO(o)-P~(b)-fY(o)-O .sin & .sin 71. -.1 : ~ MOO (t.89)- Pcoso

326) The cylinders in f19. P-326 hove the indicoted weights ~ dimenG"ions. /\s'1ul'T'ling smooth contocf .surfaces ~deter-1Ti1ne lhe reoc tionG ot A , e, c, ~ 0 on the cylinders.

ff>O of small cylinder, l10(:>1b

~o

:i.fv O 200 - P.c .sin so" .

. Re .;

329.) Two weightles'> bors pinned toge-Jher oi> c:hown in fi9 P-:a!20 s>upport o load of 350 lb . Determ1ne the force P 'llo, F octinq respectively o long bo~ AB ~ AC tho~ mointoins eqvilibr;um

of pin '}--o' ~;.jpcoi;~~,, ~~ ton o< ..!L ; ex .ae .66

10

l's.n"<

8, (+)fcos;O

h2.) Oet.ermine the reochonG for the beom shown in Fig. P-332. 30011>

"!!i" 100 lb rt . -.s ..t'.MRr OP

~(4)-100(1+)(9) tR1 (10)-300(16) e d R, = 1.Sso lbs. ~fyrO

. R11Rj = .aoo+ 10~.._) -t ...oo Riz - .s2q lbs.

3:Jaj Determine fhe reochons R, ~ Rl? of the beom Jn Fig. p-333 looded w1th a concentn;iled- lood of 1600 lb ~ ci load varying from z.ero to on intensity of 400 lb ped1.

f600ib -iOOtYn

I

O::MFc o (t! 12F, - ""W0(12) () co

2

F1 - eoo lbi:;. ~r., - o

f1 t fa e i4co{12 )]j:z Ft {f 'f

33S.) :rhe roof trocos in Fig. P-~ is supported by o roller ot A ~ o hinge al B . find the values of' Hie r-eacfions .

eoolb

:tMaco so R..... = .S00(1o) t 600(10) +goo( is)

RA "' 966 .67 lbs. ~MJ., cO

:soRev'" eco(15) t 600(10) t .soo (Ila) Rev" 9::E.33 lbs.

-2'.fh =O l

Con6idering the loWer beom.

I ~

~MR1 0 10~ - 1600(H-) - -teoo(4'-) 0

R11 - 38-40 lb. ZMRit O fOR1 t 1600(+) - '4000(6) o

R1 1160 lb. 338~ The two f~ n. beams shown ~ fiQ . .a-160 on the page 69 are to be rnoved 00--i zontolly wrth f"eQpeot to eoch other 7' lood P shifted to o new position on CD so that Gil three reodions are ~ual . How for oporl will R'1. ~ R:!!!l then be? How for w-.11 P be from Df

:::iR 960 .. R ::1120 .,, R1 Ra =-Ra

~eririg ihe lower beom,

y

f,,O Re =R1iRe

320 + S!lO

P,.c - 6-4 lb

~Rot o Re('() ,.. R2(12)

y .. 320(12.) 6-fO

Y "' 6f! .

considering fhe uppe.I"'. beam , .

!9601b 11->< - x f Ac ~MRs 0 0

P(x') ~ Re; (12) P(x') 6-90(1~')

960 x = 6'f0(1!Z) 'l' = 6+o(1a)

960

}(. ""en. . Ra \._Ra '1s 60 oparf .

S,.... olao P '1s en. rrom D.

3.99~ The differeniiol choin hoist .shown in Fig. P - 339 oonsietlO" of two ooncerdric pulleys riqidly f'oste.ned toge.1hel"'. The pulley from form _two sp~t~ for on end le"' choin looped over them in two loope. In one loop is mounted a MOYQble. pulley Guppor-ti~ o lood w . liE'-91eofin.g friction~ friotion doter-mine ihe mo-x1rnum load w that can just be ro'1seol by o pu'll PopPliOd as .stic>wn. >

~Mo-0 wfa (DI~) Wfe(d/2} t P(O/~) .

w0,4 - w~ = ?o/2 w/+ (o-d) "'PD/~

W-4PO !i(O-d)

w- !2PO (D-d)

tH) for the 1em of' pulleye 6hown in t=19.P-3...o? d o1er-m"1ne the rotio of w to P to mointoiri e

342:) The wheel foods on o jeep oro_givcn iri f1~. P-a42. Deter-mine the distance ')( so thot the reaction or the beam at " 01& twice oi; 9reaf os lhe reoohon ol 8.

R.1..-~Re (Gpcoo fied oonc:li4o"or"l) .:201YV.-O IBRe ~ 600(x) t !ZOO(x t"t) 1sRe soox + eoo - ~Fv-0 R..., -t Re =- 600 t '.lOO 3Re aoo . . Re= 800~ -- -

.avbG-titufe '1. in 1, IS (904/3) BOO(>

346) A boom /\8 is suppo,..fed i"n a tior"1zontol posHion b-f a hinge /\ 'I..., a roble whiol) runs from C over o smol\ pulley al D as sho"'{n In f1~- P -346 - Compute the. tenG1on T in the coble k the hor'1zontal . '/it...verh'Col components or Hie re.aclion ol /\ ... Neg-lect the .size of the pulley ol D.

T ton-& - 0,4 G 2 -e- a 6~-"t-3 ~tvl" O flOO(fi.) I 1oo_(cs) - T (.sin63.+:i) (4')

T" 279.[?JR. lbs . .=;;fh-0

Rh "'Tcos6a.4:1 a ( ~79. EJ~) (COS 6'3.4-!3)

Rh 1!ZS.OQ lbs-. ~FvO

Rv i Tein6a.+a "' ~00 t100 J

94g,) 1he f~rte .sho-'I in f ig. P-3-te 'n; c;upported .n p!l(()fli ol A"-. B . Eooh rn~ber weighs so lb per r. Compu~e the hor'1zontol re--oction oi " ' ol the hori"Zonlol ..... vedicol Cot"4)0Mnl~ or lhe. t'!!OC..tion oi B.

Longlh of ro .[8f't&" Fi5 - 10('1.

:.:Ma o .....,(112) = 500('4) f (,f)O(,r;). 2000(11.)

M 12-466.61 tm ~l'h O

Ah 6h "2-t6S.61 lbs. ::t.fv O a., 600 + .soo t 600 t 1.000 BY " 3700 lbs.

9-+9.) 1he truss shown in f19 . P-:a"'t9 iEJ supported on rolleNO al I\ \...., o hinge oi B. Solve for the compononte or ihe reooions.

6004b ~fh"O

Reh HOlbs . ~IAe 0

2+~ I 24-0(16) c60'.l(lt) t +o()(36) R" .. 7-f lbs. ~MAO !I+ Rev 6lX>(11.) u +o(10) - -400(11)

.R.,, 200 k .

a5o) Compuie the tolol l"CO

.tFn"'O Reh" 300 tbs.

MeO .a6 RA. 600(.so) + 1a.oo(!20) - soo(!:Zo) - soo(20)

ft.- :: 100.s .71 lbs. ~fv O Re" t R ... - 500 t 1!ZOO + .SOQ

Rev ~ 121-+. 20 lbs. Re /(soo)" t (1e1't.28)" 11.1so. 79 lbs.

ton-e- .. n1-1.aa/!!oo . . tt " 76.12 :. Re :c12.00.79 lbs up to the Jeri ot 11S.12

31>1.) The beam shown in . figure P- 861 is supported by o hi~ ot /\ ~a roller on a 1 lo2 Glq:>e of /3. Deterrri1ne the resultant re0chonG ot A\., 8.

1~'

::EM>. O 16 (~av) "f(1 ~)

Rav ~ 300 lbG. ~fh .. O R.-.h - Rel\ - mo lbs.

:EMeO J6(R,..v) = 400( '!-)

~v"' 100 1bs.

By Rot10 \s.., Proportion~ Reh 1 Reh - 300 (1/e) l

F~ of ths pulley,

a!Meo r (a) . aoo(a.)

T: !IOOlb&.

Lfv O Re 200- rm:io'

a !100 - 200(11inao') Re 100 lb&'.

f6D of tne beom,

~-o Rc(9) " 100( ..+)

Re"' so lb6. .ER,, O

R,.. ..... t Pc 100 loo. R..o.v - 100 -.sc> "' ao lb&. ~ f'h o : R.--h - 1 c:oe zoo

RM ~00 GO"I!:()' 173 .20 lbs .

Ro. (SO)

Ri =Ria (1-6e/e) -~[1 _ 6Co.e6)l 1no lbs. 18 18 "1 P~ Rh (1t 6e/a) ... ~1~[ 1 ~ 6~~56)] - 1790lbs.

35-t--) Comp..ile the total reoclions of ,.-.. i..._ 8 on the fn.;w Ghovvn in fig. P- as-+.

'klOO/b

ion~ 1%0 . .,, 11' j -e- = !l6.S&0 ~,.... .. o

eoRe.v t s up to f he ri9h~ of ~ " 48. 36

3ss.) Oeiermine the l'eaoi10ns al /\ ~ 8 Or'\ the fink. fruss- .sho.v in fig . p-3ss . Members CO ~ f6 are respedively perpend:cu -!or to AE "BE o1 their- midpoints .

lo------60'-------~ ton -e- t%o ; ~. 26.!>6 ' ,.4.0 , BF :~ 16.7ycas 26.lifi /\C a6 - 1~/cos26.s6" AO BF f9,76 ft

l\C~ e G 16.77 fl. ~M&"O R,..c.os:a0(6o) a 1aoo(....,)i ao>(+1.2S)t "f-000(16"17)

R ..... .S361 .~7 "ti .5360 lb&'. ~ ... -o

60 Rev : 1!ZOC(1.s) t ~000(1e.7s) -.000(+1.:w) t '!000 COG'16.s&'{t1flll) Rev 61?11-.67 "Ii 6130 lbs . ~Fh O

Reh "t1n26.G6 - R.-. e1nso ..0004'1n2~.s6' - .sa61.27.sinao

Rah a92 .096 ioo

~6.) The conti lever I truss ~1-iown in Fig. p-350 is &up1X>ried by 0 hin-ged ot /\ "'i 0 strut ec. Determine the reoction ot "t .. ""e.

1oooll:>

Re .. .34-6+.1 lbs .

By Rei;olving t~ forces lo ita eqJ1vo -lent for-ce triongle,

- I

By .s11ne low,

~-~ 101nao0 s 1n9o'

357.) The uniform rod in f19 . P-357 we1.9hs +20 lb"*-, has ,fs ~rit of grov"1ty al {). Oefermine. the teMsion in the cable ~ the reoc-iionra oi the Gmooth surfoce6 af A~B.

ti~,,..,

e'

~Fh=O : T "n cos-ts :. T .. 2 S4.S6 (C.05-46')

T -= 1BO lb

~lllo=-0 ,

MA .. O 2T t

zf'hO P t T s;an-e- ., ~ p = -f00 - (1!Z96.7'1)Sin36.B7 P 371.< .. tD0(9) ~ R-. C()f;~(t2) 1oox - 219.Bt(o::J630-X1{l) - 1eoo

x - '"'t.82(1.

By Sine. Low, Ra. Re - 300 ~ 5'nao- .&otl106 .

i16o) Rererring 1o Prob 389~ who-I volue of T ocfi~ o+ x s f\ . from 13 w i ll lteep the bor tl0rlzof"11ol ?

.z:Jr1c - O

AE 0 s.7s0 . Co& 60 G ~ ,AC - + ,.99 '

g ;76 I

00 = "t.3':/ - 3 .. 1."9f4 . ui: - 6 - 1.39 = 4-61 n

T (_.. 61) = 200(1.39 ) T = 60 .3 lt>S -

361.) RefelY'l-ng to Prd>. ~.,if T 300lb '6i.. " 3fl. d~fenn.oe H-.e angle -G- o t whioh the bar - ,u be 1nolrned io the hon~onfol when tS in Q .poGt;on of' equil;br~um .

0 (n.(eo&&-tsin9-) , HfooGO{ts.o .. )) / 1 t ton60 1 i 10'?&>

._y=y llsnlj Aooly1ic 6t>omeiry .

..!J:!J1 - m( lf.-7'-1) for RA. 11 fan60 12 (CDG9 s10Q.) - x

~ -0 = fon(6($)(x-o) ~ c Jtfan6()_. eqRA-

fo(- Re. 1 ~-1!ZSfl& 7 -~{'x-12a:1S&)

_y = 12(cos-e-+sin&)-~ - eq,Re. .Mo =:o 0

"ll = 12(cos&-~s1n"6-) 1 t fonQ>

-' ; 1!Z fQl'l 60 ( COSfH Sin&) ft f on60

~11ga"66-Gn&) -.3~1 ~ ~r9(X)6&- - tt(CO.Sei&,no-)J L- 1 +-fan60 J 1 t ton60 J ffi0066-a..&- - 600SD- - noos&- - =->(COG&Glll6)

1 t fon60

33COS~ - 60(~6in&) 1+fon60

1 +on60

33aJ6&6ttan60) " 60COS& t oos~ ~33(1-tfon60) -60] &9,.s1n~

~{1 ttai60) - 6Q] = tao&(60) f~ "' ().5026279

-e- = "16. 6Q6391

C hopter 4

AnolyGtG or Structures

40f.) Joint 8 of /ha iruss shown in fig . P- -t-02 ,s euQjeded to the fore~ eJterted by the hree melTlberG AB. BC, 'ii., BO. Mem f,B'l+c,80 ()t'Q. in the .s;omc;i crlroight line.buf 8C iG inclined of anon-gle of-8 dEigrece wifh this s!roigh~ line . how ihat the force in DC must be xcsu . 6enerolo~e lhic 1'6sul "'i then ehow thol the for: ces in ~ember CO, OE, EP, Fr ,til,HK, "'-,J I p ..03~ Deiennorie fhe forces in eoch bcir or ihe truss shown in fig P-403 . Hlnf: fi'r'Gl deiermine wh'1ch bctrs ccirry no loacl us"1n9 pr1n oi pie de"sbpeol if'\ Prob . .77Pcoe6'0 f3f 0577P(O.'!>) BF ,. Q.289P - T

40+) Ostermine #le forces in the rnembers or the NX>t t ruSG shown

o.nsider-;ng the tJJhof& flg,,..re, ~Fv. :o : RAv + Rov = -too+ 1oosin30

RAv t Rov 1SO- -0 "rt:.:._::=----....-+~c-~-:_:..'""'~llC"-;-R __ oh__ ~Fh O: Roll ~ 1o0oc::os ao

@pL 8.

1oot> RoH - 6 6.6 lb. @pt.A.

100lb

~Fv "'O RAv ~ Aesin 30 RAv o.sAe-@ ~fh-0

1001b AC " A8 cosao 1fv O : BDsln~ + Af'>Stn0 100 +1oolOlnao0 f:-C = o.866 AB-@

o.s BO t MA0 " 100 tao if Ae:: 100 lb BD-+Ae=100(2) .. ~oo- RAv"o.sAe

fhO: BD.e) ~ . .sin-+ a.B6b(100) ~ $

@pt. a

Ae 10001b

ir eo :zsoo1b In 1 ae -2000-0.e(~~oo)

0.90) = ~17.!i lb - . - C

407.) In the coni!lever- truss shown in Fig. P-407 ~ cornpute + force in rnemben; .Ae, ae, ~ OE.

10'Xllb

A

@pi. 0

@ pt . A -'"'"c:.c---=-""" ~"'0

ABGtnaO c1ooos1n60" t\B == 11se lb - .. - T

@ pf . B.

Us1~ R.ototion of' A,os, Lf'v-o

se

~ +- .fv O OE Gin 60" 100() PE 11s+lb - - C 1000lb

-409,) Cot'l"lpute the force in eoch member of the Worren tru~ shown in Fig. P--100.

~lb. 30001b. conoiderin9 the whole' ngure :

--E~aO

@pt. /\

-~ ~-o RAV = A6s1n60 ' '

.-

/\Be .+~.50 o:ii66

R .... v(20) WC0{19) I "')

RAV c ~.50 lb.

(@ pt. 8 2000\b

~ rv "0 : l\6sin60' .. 2000 ~ ecsincso' l3C "

@ pl 0. .l(F\, O De&tnBO' .. cosina>' t .aocxJ

PE !lO!l0(0.966) t 3llOO ~~ ~

CD/ I \oe OE! - S49+. 2 ~ ~90 lb c

411.) Determine fhd force in memberG /\8, N:,., ~o.co. '-, ce of the rontilever t~s ,sho-Hn '11'\. r.a . P- +11 If the lc:xds lf/Sr-6 opphl9CI of C \._E 1nleod or ol l3 \_O, specify whioh member(; would hove their internol force ohonged.

100 ton&c ~ 30

-fT- - 33.d.>.

@ pt . /\, r:::Ae

~ AC !

@ pl e .i:'.Fv o . ' aoo ~-o ABalflaa-6'" 100

ec

@ptC, l"lco ~,,.,,.,

..... c ' Ci!

,. ec ~oo lb "'e 1eo.2e 1b ~FhO ~OJSaa ...,. aocon.~

/\13 - eo 180.2& lb - .. T I' tono .icua

10 -& ea.12

oe . ~ i::etS.!lf' w JO' ~Fv-0 CO.olil~ ,~ - BC

co - ~olb--T I ~Fh O

C~ COGOG6&.12 tl'C "" ~so coss~.12 t 1so

er: "' aoo lb-.. C ./

~FhO /\C ..... ~ 006 .... ,.q

AC- 1~0 lb

-tos.) Determine the force in each bor of ihe kuss shown in Fio i:>-1-0.s cau1>ed by ~ining the 11.o- lb toad oi oonstont velocity of 8 ft per SOC Wha~ change iA iheoo f o_rces, "if' any results from placing the roller GUppor a D t.., the hinge support at A?

B

~Fv=o Av+ Ov "110+ 3/5 ( 120)

bvt Av =ov .. 20v "'12.0t 3/s(1w)

2 Ov e 192 Dv " 96 lb. '" Av

ot A: ~~-Ge /'.c

~- l\C A6 6 11 iO

AC 12e 10.(1ent:1dl'\) AB " 160 lb .. (comproi:sion) ot o:

BO .. 160 lb (com~io"') co s eo(&Ao) - Oh bv~ . Oh - 961b

; CO 32 lb ( iCl"lSIOC'I) at C: ..-eFyaO

/'w +!).., ~ BC BC 192. lbGension)

l11tercl-onqinq hinqe ~roller . .su.pport will rot change the forces in each oor e"G"J>t f oc- AC '*. CO

AC = '9'; t 9,/1o(Ae) = 224 lb (fen;sGW"I)

CO " BO(Q/-IO) " 126 lb (tens.on)

4C9.) Determine the force in rnG.nbers AB,00, BE,~ OE or ioo Howe rcx>f irus& shown W'\ fig_ P- 400.

p .iMH,.0

,a..., ( 40)- 60d.il0) - 1ooo(20)- '4-()0(io) = o /\v "' 10.so rb. ~..1!!::-..._-=~~..;:;;...-=-11--=-~" ~ :t:M" o

H.,(o40)--400(30)- 1000(20)-600(10) =o Hv = 9.50 lb.

oi B, fB

~~ ~fyO, ~ BE s 1n6 - 600(S1t'l 60') O

:. AB "' 2100 lb (~ion) BE IOOO lb(a:mprc~ion') K. 18ZIO it, ( tom;ion) _D,. ~ 0 ,

BO 9E.~c;o GCO(cos GO') 2.100 SD 150 lb (eom~ion)

,..10-) Determine ihe force in eoch member of the Prutt roof frvss- 8~

'1CMI\ O, "800(M)1800(1'1) ~ 180o(e) = Av (3~) A .. .. 'l7co lb .

~A. L r=--"c ~"' AB -~

3 s ...

/\.B +SOO lb ---C = FH AC= 6 H 3'6001b--T

ot 8,~fP Uii119 Hew AxM, ~fy o, BC 18001bC "' FG

/'I! e

41~.) Compute fh8 force in eoch rnsmber of the Irv~ ~ in fig . p.,. ,.,a. . If the loads oi B '!..., D ore

1 sh,r~ ecl 'iooring the whole frgur-e . ~Fv o

!=V.v t R'"" ~ t 200 t 200+ aoo ~AV f R,.,, ~ 1CSOO lb -- ,sin63,.o;s

M - .s95. 9 t "" 985 lb-- -6 :E.Fh o

A C c 1\0 C0668.4 S .

.. 9 9 15 COG l>a.+s IC~ 44

@ pl E~ [ , ~fv -O oe ""' lb - .. -T

.+'3.) Determine the Ghown P- -+13.

@ pf. /\,

@J pf. r I

~I ~l'V -0

Rrv- >ii Df' / or .re(1~0) OF -1609.97 :::: 1610lb- -C

force in each member of the

coi;;1ne LOW :

o2 = b + c - abcCosA 6 1 ,. b 11 9 11 -2b(9)~A- be a OG +Ct. - aoccoee be e 0 + 9- .11(~)(9)CC$1l10 b - 1:a .oa '

n l , 6" (1.J)St + 9 - a(13.o&'K.9)cos/\ I\ !l3.3P

c:ont.

M.) Oe~ermine the f0t"C8 in member AB, 00, g._ CD of the truGS Ghown '1n f i9. P-""14.

0

slooe ot Ae ~ 1 slOP8 ot BO ~ 1/3 s;;lo)'O ot CO c 4 / 3

~Mti = 0 . Av(36) c 3oof...21) t 300(16) t QOO( O) Av 600 lb

ot A, t

AC ~FyO .

1/.(1 1\S = 600 A0 048.-'.3 lb - .. re

..tFi< o AC c 1/ ..(1 AB

/\C: 000 lb-.!... T .( ot e,

~F,. ~o

3/,f1o BO 1~ AS BO 632.~ lb - - C J

..:f?Fy O

~ 1'16 ~ V..t1o BO t BC BC 400 lb - - T )

o t C,

~FyO 300 -1 ">/ s co .. ec

C D e 125 lb - -C _)

+ts.) Solve for the force in members FH; DF. ~ DG of' the trvss shown in fig . P- 415 .

D

aoolb 900fb @

;iJ v ~f,,.~ fwv = fH G1n.+s

fH c 900/s1n4e FH" 1a7i:?.79"" 1~7olb - .. - c

@ P.f f,

~F~o ~+'

17:) U1>in9 the method of 'eGlions, de.termina the force in rnem-t>erG BO, CO, ~.Ce of fhe roof lrvss showf'\ in F19 . P- m .

it-.... ~1o1c o

BO

JV.v (1~) " 0 0(9) so .. 11l0(11l)

9

eo - 160 11:1 - - C

~F=v o RAV " 3-1; CD

GO" .13/'9 (1'10) co .. '100 lb-- c

,.

"MoO l=V.v (24) ': ce (.g)

CE e Hl0(!24) 9

CE c :3!lOlb- -T

418.) The Worren truss looded 01> ~ in f9 P- -t1a is SuflPO"-led by a roller oi C ~ a t\1nge ai 6 . 0y ihe m&thod of Geolions,com-pvie H-e force in lhe members BCr DF, "1_s CF .

considering th8 -...hole fif)ure, .:ffh:O : !:{GH 6()() lb

..:fMc"'O 6()(J ('10) t 000(1 o) 11ooo(ao) 400(~~ RGv(..o)

100Clb RGv 900 lb .

6C - ++7.:21 "'4+9 lb - -C

.100?lb

~Mi:O OF(ao) - R6(flo)

or c eoo1b --C ~Moo

CE(:!O) t 1000(10)-t RGti(!ZO). Rc;v(ao) a_s eoo('ll0) - !000(39) - 600(20)

QO Cl! 1001b - -T

) lJGe fhe method of &9CfionG to determine /he rorces in memba BO,CO, "', CE of lhe Worren TruQi; .

toCOlb aooolb - ----1.P

\ ' \.,~

Bo(g.&6) -

4f0.) Oet~ fhs f ON:B in the members Of, 00. lio, J:6 of thc!s Kowe in>" Q\own in fig . P-~ .

.!IE~ O ! Of(9) t100(1R) Of c~lb - -C

,Jtf..,-0: ,~ t ~ 06 flOO 06-16001t>- - -- c

'"'o: DP + -4fs oo e6 moot ta(~) - e6

f:6 4001b -- T

4lf.4,) f(lf"' the tNGS '"'1oMl in rig. P --tK, determine the f()r'C.O j) Br by ihe rnettiod of ~ts.., then check ttiic: ~ult us1ng meihod of GeOfions. Hinl : .To opply ihe. roelhod of secfionc:, {ird obtain ihe ...olue of BE by inc;pection.

consldel-tng ttle ~ figure., al!f" O

Rott1~1b ~F.., 0

RA.v I Rov ~112'00 RAv I Rov 3c50C - G) ~oO

R.-.v (19) ~(9) + 1200 (12) rv._v 10011>

Mot hod of Join ts

@ pf. "'+ NII

in 1, Rov .. 3600 - 2000 s 1600 lb .

I\ N:;.

!El'v' O ~v '4/11 "'8 e RAv

AB-{s(~+ "'~- 2600 lb

BE1~1b. BC = !24

@pt 8,

~Fito ~Bf'+ ifft eo 1200 t jy-sJIB Sf s %[1'J.OO ~ ~(~)-.!1,5(~]

f3F = 2500 lb - -.:..C ~Ftt"'v

k\6tlod of Seolioo ,

:~1 .. ~1~ ~-o

~ft, 8F t 1C00 EF SF ...--t-(aeoo- 1.jl

Fiio CE t 3/s oec o/.Ja-OF

cc. = o/..ss(s.02)- 3/5(2) CE -1- Klp1; - . - T

-4!16.) Show !hot ihe method or joinls connot determine the fOr--ceia il"I oll bors or lne Fon fink truss ~ Fig . P-~6- Then use i he rnelhod of sections to compute the forc.e in bors FH, Gli,\.., EK.

~lb

cos30' " 3o/c 0. c e 3 -4,&f- I

f0t1 ao o/;;o o ~ 11.a2 '

iti 1.s' i X2 ~1s'; x'9-::r:z,s '

ccnsidering the wt-de rlgure, a'.iO f t! ,.

P>.v Rov 2RAv = 1600

i

..tM11 -u RAv(-ao) .. eK(11.a~) t 1oo(ao) t iJOO(atJ-1,.S) t .200(ac:>-1.!S) t 200(ac-22.e)

El< 690.94 ~ 699 lb - -T 43t.) Dofermine fho fbr-oe in the memberco Or, 06. ~ EG for ff-ie Pori

~M,. o R.,e(t6) ~ .89()9. 31 C.!591()11:>- -C ~Me O ,

R.-.v( e) t AJ AD(Q) ~ " 0(19) /\0 3000 It:> - - T

fe ~r ..... -o

~t %BP~AB p0ca-%[%;?~ -~J

ao 5001.+ s;:. 50001b--T

.FttO ec - fr7 1\1!> 41~ eo sc-~(s91o)t +,4 (!000)

BC-6~lb--C

-fU.) Computes the for-ces: in bo~ AB, -4,C , OF, llr.,.OE of the scissors' irv" how"' in Fig. P-.. a .

~1 -0 RAv (60) 1:1(S>)1n(-.o) t 1~ (30}t 1!Z(2.0)

t ~(10)

~J. " ..

, - - - -~>e.i ' , '

01'

.1?Mco P,..v(!.>O) ~ Aa(16) ., o/..fa9 A0(:20)

t'a - [ so(:20)] (s(GO)-e(.14)]

AB - 70.75 "" 70. 9. i

+21.) Use the method of sections to CO('l)pvte the force in ihe members Of. EF. ~ E

lb .M.-.~O f,.. (so) -400(10) t 2COI..~ t 000(30)

200(;30) Fv = 7~() lb. 1an~; 2o/.,o o.&

-tr ... 26.57.

tonu.s1 = o/21> 0" 10 ,

~MoO c e (10),. 1W(1C)

Fv 7SO_, CE "- 1~ lb - --T .!!Mc. o (~..tog BO ol f )

?20(....a) ::i.oo(w) + 000(20) y.{D eo (~) BOc 491.93 lb - - C

MF' ~o (~solV.nq CD oi C) wo(w) t 800(20) - 1/-ie co(..o)

co .. 1118.03 lb .,.. .. -c 423.) Use the rnethod of sections to cleief'rf\1ne the force octinq in members OF, EF, i.... E6 of ~he ~o-we \rt1~s deSGribed in Fig . P-409 on poqe 91 .

0

,. OP I'

~ H E- 4ilO

.1Mp o fl.,. (10) eG (10 \on ao)

E6 c 164-S.4'5 lt>- -T

~M.-.o H .. (-40)~ 400(1!0)t 1000(:.io)t ~('10)

1-1., = 900 lb

Me."0 (Resolvinq Ofol H) 1t .. (20) "t00(10) t Of(611'l.30.)(w)

OF c 1.000 lb _ .. - C i.Mtt O (ReGOlving j:f' o I E )

400( 10) EF 6n30 ( 20) l:'F ,. +00 \b - .. - C

4 2-f-.) Foritie l rvss shown ,n F'ig . P - .+24, do\erm'ine thefoN::e 1f'I Bf by the method of joiritso '*, then check th'1; rc:;:&uli using t he rrethod of .seciions . Hini : 1o oppfy the rneihod of eections

first obtain ihe volue of 0E by ins~tion.

9'

/\to Q -....nale, ~Mo"-0 ,A. ... (18) 1~(12) t 2400(9)

A.,, .. 2000 lb --ot A,

~o Av c4/!J AB

AB= 2~ 10 ---C 8=-,. ~o

AC 3hA5 1~ lb--T

t2001t:> otC, +-N; CF ~F,. ~

lb ,...., 81'

at F,.~l!f CF

. . 1to0 ~F11 -o, CF - '3/15 BF . I

I ' " ! 1.!500 (5/9) ~F .

BF .. !2.!SOOlb--C By Section,

.MAv 2-400(0) t 1200(1t) "' +/~ BF(18)

BF "' 2.!500 lb - -- C

427.) Oe~ermine lhe fC>C"CC in barG BO,CO, "-Of or Hle nacelle trv6G i;;hown in F io. P- -.t1. 'b ua>lb

' , ,. -t- 1t'

e' _L

B . ~M,,-:O, c;.,(i&) t uoo(r.) " ~ 24) t 1200( 18)

Cv " i' tb .

~fy O, 1/,w, o~ t 1600 s 12llc:n fJ(}()

OE "'632.""6 lb - -T ot Jdtnt 0,

. M& o (Res;olvin9 coot C) 1/,m, DE (18) ~ a,'1;(12)Cco)

1), o, CO" 5

4~.) Fo,.... the contilever truss .shown in ft"g . P-429, deter-mine9 the forceG in member' OF. FH. Fl, GI.~ FG.

@. H

:ZOOlb ~lb

Lef\ of A-A zMG=O. s/-129 Of(24) =200(oo)t 200(~) 1

100(w) Of ~ 1256.s+ lb - - T

Lefl of e-a Mt o,

MF o ( ot A-A) .s/.1+1 fH(M>):-400(~o)t ~40) t200(60) ~ MGI:. 400(20)-t ioo(-4) t2oo(6o) 2G0(80)

61=1166.67 lb--C Fi-1=166+. 81 \b--r ~F11 0 : GI ~~FI c'*F'H

Fl 206. 2.7 lb - .. -c L.cn of' c-c

Z:M...,O: 60 FG = -1-00(60) t-400(41-0) t 2.00(20) f6 733.33 lb --T

+30.) The lood.s on the Parker truss shown in fig. P - 't.30 ore in tc.ips. One "'-iP equol.s 1000 lb . Oeterm.11'\e ihe forces in members eo, ee. CE. 'iii.. OE

H

lw ... LMs"'O (Left ola-o)

a.1 CE = 109(29) Ce = 97. 22 Kip'-.. - j ~"'E. " 0

Fv =O Av tJv 30(7) Biii Av Jv

. Av ,. 105 l'.i~

JE.IY1 ol !he: intcrscx;i ion of 60 8l, CE o 27/.f13st 131:(160) t 30(135) 8 110(10~)

BE~ 63. 88 ~ps - -T Lof'i or b-b ~ ot \he 1nief'Ceciior'1 or BO~CEO

160(30) t160 OE t 30(135)=1o!S(110) OE "' 16.07.S K1PG - .. -c ~os (oo) = ~o(2!J) .. s/~ B0(3z)

0D " 1+:9.41 K.ops - -C

440.) For tho fromo !oodod c.s shown in Fig. P-440,,de>tr -mire H1C horizontol ~ -JOrliool cornpononts of tfio p'1n pi-os-

. suro ot B. Spoofy d irnotions (up or down; left or r"ighf} of tho foreo OG '1 t od.s; upon m~mber' CD . .

c 3001b 2

8 BH

Gv +'

D Doi " ~ .55('.)lb

c 3001b

:!Mo "0 B11 (4) = 'E09(6) .M

BH"' -t5()1b ~Fy"O sso -Bv =o

0v "'.S.SOlb (down}

.,M,.. =O Ov(-4-) - wo(a,) + -aoc(G)

Ov " ssolb LMo " 0 A" ( 4-) " :300(6) - 2.00(:Z.).

Av= .:isolb

'l:Me"'O At-t(4)" 3So(4)

t- WO(~) AH" 4!l0lb ~FlC"O

(ten) ... AH" BH .. 4-eo lb Fy =O e'(-~-.3so Ao

Bv ssolb

+'l'I ) The structure shown in Fi9 . P-441 is hinged ot A ~c. find tho horiz.oritol ~ vorticol ccmpbncnis ol tho hingo force ot B, C,!,... A.

s' 1001b

B11 : 17s lb .;EFx O; BH AH - 17~ lb ~F'y -o ; l'w - 100-100 00 0

A" "' :zoo lb CH=- BH "17.Slb

443.) The frdme shOym in fig. P-+,4-.3. iio hinged to. r19id s:up-. JX'rls ot /\ ~ E . Find the a:impoi{onto; of tho hingos forcos ot /\ ""-E ~ tho fon:;ot; in mom~s. BC ~BO.

1tolb

4' 4' BC .

::l'..FH O: BD(.3/50) - Ali =0 eo "'(5h)A~

. 80 - 200 lb - .. .;... c .

.M.--=0: 1W(4) - EH (4) =q E:H ~. ;Wlb

:Z::f11"'0, : E11>-A1-1"0 E~ c AH 8 '\ZOlb

ot isolating bar Ae .::pv'\a"'O: Av(6) -120(4:1, "0

. Av= 6o lb. ~ng 1ho wholo f'romo:

.2'.:Fv =o: Av+ Ev - 120 "'0 i::v = 120-co

Ev.,, 60 lb.

XMA O; 1W(4h OC(B) - BD(4/.:s)(a) ~o J3C"' 100\b --T

. : i 44+) fhc frornc -?~wn in fig. P- tt4' is svpportcd by o hin-god ot /\ &, o rofler ot E . Compute tha,, horizontal ~ vorti-

. ~I . co!T)ponerits o.f1 the. hingo forcoG ot B ~ C o~ 'they ad upon member ft,,~ ' . .

.(' ' .::t:MA"'O: 1'20(11) - Ev(1o)=O

lrolating bar BD Bil B 6 1 l "3' /240lb --tji;,, 0 Otl ~~o"'O: 240(3).- Bv(6) ""0

E'.v" 264:1b ~Fv ~o : Ev -Av - 240 =6

Av "' - 24-0 -t. i=v. Av ' .24-lb.

o~ BO %Me-0

240(9) - 0..-(6)-0 " = 3001b.

Elv = 1QO lb { l)pvvord w I ~peel to bar /'C,}

lsololing bor CE

Isolating bor /\C

t Cv II " .,,

::i!Fv ,.0 Cv t Ev - 0,.. =' 0

Cv ":360 - '2.64 Cv : 96 lb { dOWr'l'NOrO w oih roped to

AC} 1v1oO: C.i(6) t C,..(.3) -E.,(t) eo

C11 = 264(2)- 96(3) G

[ io tne 1en l Ci - -1-0 lb w i th t"C~pect to AC J .;;!M,..O: &i("-) 1 Cv(s) - C..(10)-Sv(~)o

s .. = "t0(10)+120('.!) - (96)(5) ....

BH * 40 lb to tho r ight

445.) Tho frame shown 1n F19 . P - "4-45 iso .supportod by o hin -ged ot E ~ o rollor o~ 0 . Compvto the hori20n~ol ""' vcrti -col components of the hinge for-co cit C os H ocis ,upon eo .

Ov

lliOlotlng bor Ae 2,-0

,..... " ...,, ! 2 ' ~ 4Av rev

2'.M.-. O Bv(6) c: 240(4)

Bv '"160\b ~MeO

(G)Av 240(2.) A.v= eo lb

.Z.M E =O Dv(0) "' 240( 3 )

Dv"' 9olb ~Mo .. O

E ... Ce) = z -40(s) Ev c 1so1b lc;olo~ing bar OB

0

~Fy"O Cv = 1ro-go c ... = 101b

ZMeO c .. (3) " 7o(3) t 90(7)

CH ?.001b.

~.) A three; - hinged or-ch iG composed o( +wo fr.usscs hifl

.2'.Me 0 Av(2.0)-A.1(20)-vt-0(10) .. o

Av - AH a 120 - @ eq. &... @?.,

Av (+)- AH"' 14'+0 - (Av - Al

449.) The bridge shown in Fig. P-449 mns'1sts of tY110 end s=ctions, each weighing 200 tons with ~ntcr of grovity ot 6 , hingod to o uniform center. span weighing 120 tons. COm putc ~he rcoctioni; ot A, B, E, .&.... F .

@membor CO -40' tJtiT , 0 C~ C,_-=--"--"o;.,

Cv 1~T

Ott p

~Mv 50 Dv(60) = 120(30) t 60(40)

Dv : 100 tons :i:Mo ::O

Cv(6o) = 120(30) + 60('20) Cv" 0 0 tonG

.>!MF "0 E,,(so) m !200(.30) t 100(70)

Ev "' 260 toni;

fv "' 4 ton6 -'!Memo A" (so)= 20o(w) - eo(w)

Av= 48 tons .::EMA =O Bv(so) - 200(:30) t 00(10)

Bv = 232 fons 4.SO.) A bi I I boo.rel BC w eighing 1000 lb iG subjected io o w'1nd prcGGure ot 300 lb per n CIS shown '1n Fi9. P- 450 . Nog-

l~ting tho wci1g'hts or tho Guppoding mcmbc~, dctormino the c.ornpononts of the hinge fori:;cs ot /\ ~ F.

Wind pressu--e == 300 ~t>/fl. ~ .300 lb/fi x 10 fl

- 3000\b.

@member ce. c..

G C..

c

~ +

+'

+'

~MeO C11(10) .. aooo(s)

CH 1eoot>

.2:MJ\ .. O 3000(9)+1000(~) t FH(

451.) The frome Ghown in Fig . P- -t.51 u; hinged o\ E ti.... rol lor supportoci o~ A . Dcto~V'lc hor;zon\ o l ~vertical componel"lt of tho hingo fws o t B, C, !..,, o. l'legk~c\ the weigh~& of the rnernbon; .

2.'

It '

~M/\-O Ev (1Z.) "'300(16) - 2"'!-0(10)

f:" ... 2.SO lb ::i:Me c:o ':

Av(1Z) 240(10) - 300(6) Av f so lb

3001b

:E.F,. o Ett - 24 -o

E H - 2."'\0 lb .

@ momber CE Cv ' ~~=o

c

0"

,/" DH (4) c ~"!0(0) 0

" +'

e

OH = ...aa lb %Mo =O

Ct (..-) = 240(4') CH .. 2;...0lb

~t.'leO . IA-(6)=so(6)+aoo(~t)t~1t)

T. Dv .. 810 1.0.

~ Ol OM " 8ti 400 lb ~' ~Fy cO

8v .. .so+ 010 - 30() 0v ,. SliC) lb .

Choptor S

Fr i ct ion

-'15.) A block we.1'ghfn9 W lb is placed Ufx:Jn o plane in-clt'ned at on ongle tr wi'th the horizontal. Discuss who t )'Viii hoppen 1r th6 onglo of friction Is (o) greoter than e-, (b) equal to -e-, (c) less than ir . (o) If )J is greater than fr the. plock will not .slide down instcod it will reforn '11.6 poGif!'Ot:'\ becouse rhe rrict1onol force iG so ,much tho! it will hold the blocK.

, (IJ) If is cquol to J- the bloo/

..507.) The ~oo-tb block Ghown In figure is in contoci wifh o 45 incline,. The coefrrcien~ or fr1"ct.1on i5 0.20 . COrnpute the value of the horiwnlol R>rce. P nccc50g,) The blocks shown in fig. P-,!;Q{l ore connected bH f'lex ible inc"ll f e-n.s.1b/e eords passin9 ovor (r1c+1onlcss pul/;.1:;1s . A. t A !he c;oeffloiont or (rict ion ore ~ o . ..,o Si.,, f1

flo1'ent of' friction i~ 0.10, ~ Jhe pulleys ore.. oGSumed to I . be fr1otronlesG' .

sol. .!JO(i fBD Of 81ook /\ :

~ . .!510.

reo of Blool

bli.) find the leos f voluc or p ~qu;red to couse the sys.-fem Of blocks 6how In r i9. P-s11 to hovo ;rn~r:d;ng m:>f 1~11. to the. left . The coemc1enl of fr1'ciion. '1 ~ 0.1 under oooh blook .

~~ -:... _: rt~P of ~loc,k /\ ~ ~ : 100 1 ' 11 I

_.

-~ ~ F T aOOlb

' ~f~-o : N .1100- p.s1no< T 100 ~11 -o: T tf-Pcos"'\ -o sin .. 1:!11" .sin 79,69' 67.32 t(o.t.)(soo-Psino( COS(

0( -11.31

p " 124 . 8 lb .

;1t.) ,A, ho~neous ploc;.k of we4ghf W rest upon the ln-ollne shown in Fi3. P -512 Ir iho. .. eoe fficicnf of rr:ction i(; o.30, determine the grootest hoi,ght h_ al which a fOrcx:.. P paro llol to tho- incline ~be opPlied so thol the block. will.s-l!de vp the. incline w/oot tppl~ over.

; >-

.s1~) _In f.'g. P-~12 , ihe homo,ge.noou~ l:>loc~ wel9h6' 0oolb ~ ih coe-rfi.oie.nt of p--.,

~f'l! O: Ws1n& R1Ew11+0t- -1 Ri s1n1+.o+ ~005rn&- (R1 t R~~ Gin 1+.01- . . (R1 t R~) " B!l-1.-f am~ (j) .

~ ~~ -o : R.z C

reoo~a: 'l'4 J.V02~a.. o-1-..ot c 270 -;;- off.llf ~ .son-4$0 96 &n 1+.ot- C II OO ,.,,; C - z01,9 lb.

'- iJ _c

~

f ii:t WI\ ~ 598.3 lb

I ~: ' .,,. c

szo) Re fc,,.,.-;rtg to fi'0 . P-~ro blook A we:ighs +oo lb 'B -aoo lb . 'If f - 0.20, under 8 , eqmpute the minimum coefric.ient of frl'ot ~or"\ vnde-r A to prover>~ rnol'on.. .

W~llOO J'A 0.Z .. -(}- 11,3

~ ~pr ;;: ,,,.:::: f OB of fl iF,9 ~o: ti .. +oO t csinao

s:u.) Aopeo~ illut;'. Prob .s17, oGSum1ng thof tt-e .sil"Ui i~ a uniforrn rod wei9h;n9 300 lb Hint : f/rvd 1so/ole fho 6"!ruf 0$ o Freeho

i.r.!JO; Wf1tfa f)l ;.Oi N1 c N2 -

F; "Jl'l'f1 o.2H1 = o.2N,, F, Fa.

W F1 t ff. e o.eN1 ( t) W 0.4H1

.M" 0 W(L-1) t F2 (f..) - Hi(:t.) ~ o

s26 :) ti ladder 2on long we:gh6 -fO lb~ .rs cC>'tcr ol 9rov:fy ,s 8fl' frcrn the bottom. The, !odder 's ploced ogo;ns~ o ver+i c;t;1I wall ro ih::lf H mo~ an orglc:. o f 6J0 vv/ the- 9rotAnd. HcM-far- vp fhe. ladder con Cl 1b0 lb man olirnb befOre tt-e ladc:ler

iG' on ihe "~of sl1pp1ng . The or-ele of (riof ion oi oil con-toct surfo ce- r!O 15

r---~(St'3) t A ,;o\b- IS H I ~b R~

;. I (10, 17.3) -.( t VJ'

@!1/1'11\1b - ~ 1Q'17SX ~~-~.) -m(ll-71.)

{.!l-11,3~) Ton1s (x-10) G>J~. 11\e. vol...e of .!i:

-,,..,-=~'4.!~~ .1..---__J To,.. 110' ')( -na2 -Ton 1SJ'. t 10Ton15' o.o, ~ 11 (rorn{; .. To,, 1s ) 'lr - 20 @~M~ "O w(&-8C.OG6 cylinder .3 fi. in d iometc.r ~ wei9h1n3 aoolb i~ resting on two inolined p lone os Ghow n in. Fi'C:3 .Rsf7 ff the- orgle. of friction '" 19 for- oll contoc\ burfoces com puto the. r1109nitvd e. or the c:oUple required to storl lhe cyl1 d er rotating covnierolool

~"1,sx -Ton-+O'J< t10Tan-te'

x(Ton6s tfon-te} 10 lon

1he rnoxlmum value of the ongle 0( at whioh .sliPPing impends. \ e ~c,i)

.. ---._- --w\ . ::.-- ~ Ton~ s ~"' o.7i!J 2. ,!1~m~tb (.H-~) -m(x --;i,) :I o(o..:J6.2 Y f on ts"x y t 21..Gll'l"'') " 2LrecTon60-ll!i..sina

.

~-o

F+- 11 p Av fCO ~fy O: tf 200 t Av

fie 701.6 lb ~/). O : P AH t- f U

~ +.Rt. :E.Fi.o : R1cos+.s" = R2cos7.s ,, . . r, 7 frn*)c~# ~ Re - g86.4 lb

n' co.s1.s

- .- r We R. ~Fu ~o : We R2 sin'ts-- R1 son4S0 I& ;;J

N, P..t Ws (aa6.4J&-1n7,s'- (141.-H.)sin4>0 We 2.73. Q lb

537.) In Figure ,de/ermine the volu~ of P. just S-Af'f;cient lo (;;forl lhe 10 we.d9e under- the +oo-lb J)lock. The o'"Ble. of' fr:ction is w for oil oonto.ct .surfooeG.

R1_~ ~1_9.-z lb _ p M 411:1.'l &1n&)0 ' 6in70

pk a41.z 1b 536) In Fi9 , _s:J7, defcrm;flr"BD of/\ r----:---,--z'. W'fOO

,.,, .

11.i f.

R& Ai 70 &0"

p _P_ - 381.1 sinao J;;on 10

R% 301.7 lb ...... p ~ 203,f lb .

.sag.) If the wooge deGcrib

feoofA ~t W Fe .~ .. 105

' C'.J*""'~ .,, . . R1 ~oo

. ~~-

I .sin 10&' s1n10'

Rt 601.tlb.

p. "r76 7 lb ='r

.

'

, ""'

" ~ 't ,

- "''-'. Q"t4'C. ~-- ;Ra .&!...--~ 4 ... _,., Ra 1og.01 lb ~F~-o= p,..:Pt-Pi P (1og,&1)&1nao .. {73.a) r;.nwl

P&:; 91,slf> p,.. c 18.!l lb

.st1) Determine fhc fcrc;e P required to start thl. wedge dlown in figvre P &tt . The ongle or friction for off surfaces io COlloot is 15~

rwbf /\

reo of 8 tOOO

'6'11. "0: Rtcos111 -Rtsin11t . .. l!C:l0 R1 Rt COS1~. - 500

llf

!IOOlb ...

: .SIO 1!!" --.+--- eoo l :FycG>: R, cas1!':-R. G1ntll' :20

p

.5+3.) To odjui;t the vertiCol position of o pos1t1on of a rolumn Gvpport ing a 2000 lb lood two~ wodges ore usod os shown in f igure P-s-+a . Detormine the fbrc.o P necos.sory fo \Sfort the wodgos if the angle:,

(.stn70CO$~ -eo&70S1f'\O

W ~lbFlO.o+3tl573~

-s-t1.) /\s .shown ;n fig P -~....., , o Gqvorc threodi=.d 8crew ~s U&e.d in a vie;., to eiier-1 a pres&ure of' !l tons. Jf the- sO

l ..isolb .o'j. ;.,,c:>nloo\ \

.:M11 = 0

P / 16"'- ... .1fl.) ~ 11 (.2 iA. x .1C:L) I.. 12A . laj'P,, p( 1 . .333 f'l ) +9a.a3 { o .1667)

p ~ 61 7 lb . .s.s+) /n fig P -SS+, the. c.O.erf:-101ef\t off riotof'. is 0-20 bol-woe.r- -!he rof 'ifii the, fhoo drum ~ be-tween. oll surfOCC!>' ir.. oonfocl . ~el" mlnl!I the, f'(lin/mum we9hl w to pre.vent downplaf'\e motiol"'I. or the

1~lb bod!:1

ry O ri1 W COG 36&7 ~F ... o : T~ -F1 Weina6.87 T~ -(o.2)(wwr;36,87) Ws1na6.e~"

l, Tz = 0.76 W

.o!!:Qj - o .: ti'L 1000 coca6e1 + t11 tl2 ..... eoo -t n,

..2'.r)!. -o: T, t f11 .f .. 1000 &in a6: e1 . 11 t f'-111 +;~11" - 600

b.Jt T1 1 .sn~ Ji1 -wd,,J1,.ti1; 1.s1(0.16w) t (o.i)wc.os

i/Ylp . 0 (..!1)T9 t Teh)-(2tf~ - (_3 )T1 O

.s rt. - 2T ;5 - 311 0 ..s ('N/:1.) - 2 (o.e2-+~)w -'3 (!Z.ot>) .. c

o 8(;1 t"W 600 ) W I 7o+ 7 lb

C hopter 6

Force S~tc:;m~ in Spoce

f;OU Dciermino tho mogniiude of the resultant. ii' pdint;ng ~ ifi; direction c.osines for the following system of non -coplonor . C()nCum::n+ forc.oS . .300 lb ( 3. --t . 6) ~ -40? lb (-2. +, -5); uio lb(-... .!'J -3).

f()RC E Q>MPonlilfTC> OF " JC y Jt A"aot:>.lb. ii --4- 6 e. '4001b. -Jl. + - !I r. 1/X) ....

-+ !J -3 lnl'AL(~)

ioget OiGtol'lGC(O) use oall2 +y-t:t ~ .. ~.A~ -6....

)( y I(, 0

pu;T, (0)

.,.e1 6 71 7.07

.300/7,91 A.,_/a .. I\,. " 11s.2+ lb. 3()Q/M~1 l'w/-4 . . ,Ay c -153.65 lb. 3

~" a 37.1-+ - 127 298 i 0 - -00.158 lb. :%.y -11.+.0!l lb. . .::Ell - 7+.:t.0 t 2.12.16 .. 20) .. 406-4-4 lb . Rt "' (~>s &z. c +79.9/ ..s20 . .s3 o. gos

ZCOMP -e-+.01 -228.ll.' -166-98 479,g

:. po;nti119 bockwords, \ upwords 1o ~he right

605) Three concurr-ent forces P, Q., to..., F hove o resultont of' ..s lb. d;:. ~ii "16.2.3 ; ~ 6.+g ; ~z 9.74 cos_, 22.46/33. 68 z.~ = Rcos&i. c s ( casGO) "' 2.s

-e-,.. ... s.2 ~Y ~Reas&,. .. s(cosGo'),. 2.s -et1 cos- e .%,/3.3.68 .~ = Rc;o&-e-;r. = ..5 (cos4s;) - a . .54 -Goy 4 7S.6 fy = 2.s - +.37 - 6 .49

-G-2 COS-1 2.3.~7/33.60 fy .. - 8 .36 lb. -8-J. -.~.3.

Fx 2.s - -1 6.2.3 -s.13 . pon I '19 bockwa.-.::1.s, "i dcwn......o.-.::1 . . F,. - 22 .... 6 Jb. to the lefi.

Fz - :a.s+ - 9,74-17. 4-7 Fz - 2:9 .~7 lb.

607.) t\ force of 100 lb is d lrecied fr-om I\ toward 6 in the cube shCM'n in fiq . P - 607. De-term ire the moment of the force obout each ol the coordno~e oxes .

J," -

// / / F/ / /,; ii ' ,; ,

:// / ~ ~ 4 / . A

"

B < ,,..,.

,

' ,,

- ~ ..

fv / v ... I/ 1,

' ,/?'.) ~ da:-tt3:zt-+~ d ~ I0~.3 -R

f,. ~ fL.,. 100 lb -:;- a -4 c; .

45t0~ /\ furce cl' '400 lb ir; iirecied from C toward E in the cube ~ in fig . P- '10 Oeferrnine ihe. momenf of the force obovt eoch of lhe ooon::lino!e oxes .

d 1 2 1 +"42 t:Za d ... g n.

111 .. Fy(+) - fz( ... ) .,, .a26.S(+) -163.'il() ~M11 6!j:z.& lb-0.

My ~ fx(.,..) t fr (c) 163.3(-4) t 10:!t.9(2) ~My .. 079 .8 lb-0 .

~l. ~ -f,.(4) - Fy(2) :. F. 163.3 lb. ; Fy 326.S lb ; Fa 163.3 lb . c -163.3(4t)-.326.5 (2)

Ml! .. - 1306.2 rb- 0 . 611.) /\ fOt'Ce P. d irecied frotTl f toward 8 in the cube s'"'""'1 1n Fig.P-601 . cour;~ o momet'lt My 1600 lb-0 Oelerrri1ne P 'ill, als moment obout the x '- z ollCS .

d' -.'+s .~c d,5.38:> n.

-*'"'1t Pa ( z) 1600 Px (2)

p,. eoolb . 2L L :. P1077 lb.

4 lJ.385

: . Py = 600 lb. ; P.e "'tO lb. 1-MJ -Pr.("4) t Py(~)

: - iOO(+) t (>00(.:Z) M, " - -tOO lb -fl .

%"11! -P. (+) . c - 900 (+)

M2 - .3:ZOO lb-fl . 612.) A force P i' directed from o point A ( .... 1;4) to.o./Ord o poin'f B (-3 .+.- 1). If ii COIJlOCS o momen~ Mir 1000 /b-tf, Oelerm.1ne ihc! "" men~ of P oboul lne X ~ Y oxer; .

~- ___ sP!. ____ 71 d' 1t~'+3 /' : I ,,' ' , t I I n / ' . ,,,, d. Q.11 :-!-:.-:~ ...f_ .&.. &.. - 13-

: G.11 7 3 9

: . P, nyg,11 ; Py & .311'9.11 ; P.1: .!1~.11 Mar .. p,.(1)t Py(-+)

1900 7/g,1, (P) t .s(+)%.11 p. ~11 lb.

p,. 700 lb-; Pr; 3001b; Pz - s00lb ~M~"' -Pr(1) - Pt(,.)

~ --!?00(1) - 700(+) M11 " -1700 1t>-f1 .

%.My Pz (-t) - Px(} .!500~ .... )- 700(+)

:,t.'\y - - 80 lb-fl . 61+.) The .sheor-le9 derrici-. .shown in Fig. P- 61+ suppod" o ....ed ical lood of 2000 lti . oppliec;I ol /\. Points B.,C. "'Dare in the. .some hOri-zontal plone ~ /\.O.~ D o~ in lhe XY pl9ne. Determine the force in eoch member- of \he derrick. .

i.---20' iY 10'--J,..(1o,1s.o) d,...1 (11>-0)2+ (1s-0)2~(0-(-s))c

Isolating fro~t View,

" 0

Mq.c,& 0 : UX>C(10)- Oy (c.0)~9 Oye1000 lb.

LJ._. ~-~ : J2.I... 33.&+i' 1s ao o

'. l\D - 2:236.1 lb.; Dx 2ooo fb ; Oi:. O

r O,..e 1S.708 .jl . d~2 =(10-0)2 t(1s-o)* t (10-0)2 d .... c w.616 fl. 1 \Ill 2 d...., ~_[10-(-wu t(1s-o)

fsoloting Top View,

10' c,. c

.iMe"'O c~ zM.cO

- 0"(10) 1 Bit (1!5) Go Bit [2000(1o)J/1s B>< 1333. 3 lb .

.L&._ c C ~..Qi_ . Ci! 20. &16 10 1 !! 10

Ox (s) -c~ (ls) .o Cit =[11000C1J/1i; c 1(,66.7 lb.

l'\C : 13741-. !) lb .

AB" 2-494 . .:a lb .

b'l!) .)' The fromoVW"Ork shown in fi9 . P-615 con.sists of throe me ber.s /\!::>. /\C,"" /\0 whose lower ends .ore in ~he .some hori ~onto! plone . i\ hori:r.oritol f()rCe of 1000 lb octing poroll~I to the 'X axi& iG applied o1 /\. Determine the force in each member.

lsolote front Vio.w,

*Mo,.o,c V O ! Dy(6)-1000(6)0 ay-10001b.

.oil.~ e,. 6.'106. 6 3

"6 "1110 It>. ; Bll .500. lb.

~ (6-a)' (6-o)' (o-or d..e 6.7o& ~(a-o)~ +(6~)s t(o-(-3))c

d.t.c 7.3!1, d,;.(3-o'f .. (6-of +6:;-0)%

d .-.o B.367'

l.atote R igh~ Side View,

..i.MoO -tOOO(B) t Cy(9),.0

Cya62S lb /\C Jd_ 7,u ' U '

/\C a 765.'- lb

zlAc O 1000(~)-Dy(&) o

0 1 37Slb .

~,+ AO .s~ .Q.+ lb.

616.) Refe rring to Fig.P-615. replace tne 100010 force by a verli-col downward lood of Q()(){) lb. Determine the force in eoch mernber under this revised looding.

ly A(.a.6..0) ! I .

e (6.0.0)

d...s 6.706 fl . d ... c -A~-o)~ t (&-o)'' t (o-(-3))1

d...o -vf.a-)1 _. (fro)1 1 (- o)1 . d>.0 S.367{\.

l&olole fronl View,

.Mo.o.c. 0 Sy(6) - 2000{3) wo

By 1ooo tb. ~ -~ 6 .708' 6'

lsolote Jl.ighl Side View, y ...

.tMoO ( '2()00 1000 )(5) -c,. (6) 0 :.Mc. - o

Oy(e)-(200:>10C.\'.l)(a) o Cy =621' ~-~

7$& '

AC 76J5.6 lb.

(Jy 37S lb . _t,,Q_ ... ~ a.367 '

AO .. s22.9 lb . 011.) The pointi; 0, C, S., D of the c;on I i\ever fromework iohown in fig . P' 6 17 oro olloched too verlicol wall . The 0-lb 1ood i~ poro-llel to the z ox iii ' "" the 1200 lb lood is vedico1'. Compute::. ~he f'on::e in eoch member.

!"' I D 0,6,0)

.1

d-'fl ..k~o-o)"' t(o-o)~ -t (o ..! (-+))d .:. 10.n' d,v:, .. /(10-0)" (o-o)"+(+-o)" 10.71'

'

d.-.o /(10-0)2 I (6-0)'t + (0-0) 2 11.(,6 1

' ~ r .. , J ,,. o.ci...,.,.,} . o;,? _..q _____ _

c .,,..'!,,,"'' I ,,/'' co.o.~') ..

1l!001b

lc;olote. Ff'Of"lt View, 1.solote Top 'View, ~lt1c. - 0

e,. 11'00 lb .

.. ,

Mc.o,e - 0 MeO Ae - 161S.5 lb-- c i2o0(10) - o. (&) o ROOO(+) - ..otoo

61&.) The unsymmetricol cont"tlevel" f rort1ewof'k .shown in Fig. P-61!, .suppo"t' o vel"ticol lood of' 17>1b of A, Pciont.s C tt... 0 ore in the (;()rne vedicol plone whil" 0 iG .af+. in fl"Ot'I\ of ' -fhi.s. plone. CJmpuJe the forc;e. in eovh rn:>mher.

~

.MG,o~ 1100(e)-8" (10)- By(~) o e,(10) t Sr,(a) ;. 1"6a::> - .. -ff)

.....@__. -.fu!_e~ 6~ e.77.9 e 6 -+

:. Br 6/15 e,. - 5Ub6. lo 1 , ~(10) 6/., e,. (3) 1::1~

a1 a 1000 lb. :. B;r" 1200 lb. ,'. /\e " 17SS lb - -c '-.

d...e..{S-3)1 t(o-(-,))1 t(+-0)" 8 .1't.9' d ... c /l.a--0)* .. ( .. -a)' t(s-o)* 1a' d.\o /(e-)2 + (.+-o)2 t(o-(-~))' 9.1~'

IGClote 1op Vievv,

.1:Mo :0 c,.(10) -1d00(~)-aoo(3) ~o

C1 = 8"1Clb . -1lQ.... B~

12 ---g-

f.C 1%0 lb-- T ~c tJ

1000(+) - 86a(9)- o,. (10) '0 0.. 160 lb

/\0 160 9 1"8 -a-

Bi .: 000 lb . AO= 1133.3 lb- -T 61g,) Solve Prci:> 618 1 t the 11c:x> lb lood in Fi'g. 610 acts horiron-killy Ou~wor-d from/\ 1'n the direc.hon rr-0m E toward/\ .

from Prot>. 'fl.. 610 dAe = S.775 n., d,oc, " 12 ft.

d"'o = 9.168 O.

lsolole front Viow, Cy

c .. " "

~lr'\c:,o 0 0"(10). By(3) - 1100(-4) O

B" (10) +By(a) .. 6000 - - .M- _fu_ .. ~ Jk_ 8-77!1 !I '" ..

:. By 6/!1 B,. - sub8. to 10Bit t fD/s e,. (a) 6000

13,, g,. 6800 a,.-.sootb. :. ey~lb;

r AB 877.~ lb - - T \ ea-4lb.

lsolote 1op View ,

.a!t.10 O - C..(..o) t S,.(l.) +0E(3) - 1700(_1) c O

c,..(10) ~) t 1!(3) - 3"400 c,. - eotb.

IC 00 ~ -0-

' /\.C 12olb ---C zlt'lc;O

- o.(~)-S.() B;i(.9)+ r]Q'.)(6) "'o D.(~o) .. - .!oo(+)+

lliOlate Front View, _..,

1l " .

IW

lsolale RiQhl ~ide View, ll(;W

I\

%McO, -0,,(&.ti6) tBy(.+) .. o By= 6.66/+ Oy --

zMs "'0, ~

lsolola f ront View,

:. By. 199.Q lb . ..!:__ ~ -~

1~.tu: 10 H .i'H :. Cy 200.1 lb ~Fy O 0y 19g.9 200.1-eaoo

Dy -4()() lb .Ma O Oy (xo)-0y(6) O

1Co [zoo.1(elJ/'400 'l.o . +'

_.t_-~-~ ,Y.._ 1-t.ii-2 6 14.1-.2 6

c- 160.1 lb. . , ~ -120.1 lb.

lsolcne Righi Side Yie,,..,,

800

G,r

~Mc OJ) -Dy(, 20) I 800(') - By(t"t) "0

- ...oo(, o) t .+eoo- 2799.6 O - 2-400 -4()() ilo - :ZOOt . 4

~o 0.991 1 ~ .JL_ 10 o.rn1

Oz ['4CO(o.9g7}]/10 Oz 39.6& l'o

/\0 i..t60)1 I "IOO' t(39.09),_ /\0. +32.7 lb--c

ou.) In fig P - 6U, 1f P 1~lb ~ lhe COON:f,noles ~Doro Y..,,S ' ~Lo 2 i1, compvlo \he forc.e in eoch leg of t~ tripod .

p d..a ./(0-0) 1 (too) t (o - l-91)1

cl.-.e .. 12. 006 ft dl\G ... /(6-0)T. .. (10-0)~ + (v -o)1

d...c : 1+.1'1-2 f1 . o...o ~ /(o -(-.!))1 t(10-0)1 (~-o)&

dAO = 11.356 f1 .

l~lol& fl"()n t View, ~1:1

4!'.Mo O lJ Cr(,.) t 1200(2)- ey(1o)o o .=;==---c:..L...::~ -- By(10)-Cy(-4-) .. .2-+00 --

Z:fyaO,

GUbs. (i} fo@. 400Q;Cy(13)(1o)-Cy() 2~

1~000 - Cy(~~) -Cy(+) = 24()0 -Cy (ao)-- - 9600

Ct 3zolb. :. Sy.a ~ooo -(320)(13) ~ 36e lb .

.5

c,. t()yt Sy-1~00-0 32Ct Dy +368 - 12000 - Dy~ -512. lb.

/\B = .366 AC 320 12.SOCO ~ 1++-H ~

/\0 471 .3 1b C /\C ~ +.s2 . .s lb-C ----6.Q_ .S12 11.358 10

1\0 - .s01 . .s lb -c 6%3.) Oeterrn1ne the mox1mum safe vertical lood W thot con be supporlocl by ihe tripod srown in fig . P-623 without ex~ -ing o compressive lead of 2~lb i'l any member.

w

o(o.,,o) dM ./(o-ft)y+(o-o)"t( -o)i

1 do>. 7.~3'

- , B(-a,c,-.s) dos ~o-fz)}2 +(G-0)2-. (p- (-e))1 -;r' doa = S.06::t '

"-I ,...__,...,,......_...__,-,-__ --~--1--1.. doc -/(-0)"1(

.,. if. DA., 2400 lb.i\ 24()() ....6.a_ ~ _&_ -~

7 .... 3 2 ' ....

:. ~ 6-+IA-S lb; Ay '!Q:M-36 lb ;Ait 1262.91 lb. lsolole Righi !;1N, lsolote Fron\ Vie#,

Mc. .. 0

w 0

A"'(

Isolate Tep View.

.,..' 1d1.8

c

z"1t. O 1331.:s (.5) - Air (o) o

,... .. _,. 739-61 lb.

checl

toke f,...,,...+ View, i: - c F ~ 1 , , I

to/ 1 f1c M,,,8 O 120l(10/1) 1 1000(1-) - c6o) ~o C 0001~-r

ZfyO.

fol

Top View, EAe-o SJ Oi (a)-0.( )-C rololl!l

leolob Fron!

~l.1~o

dee ~1' t (~o-e) i (10-0) .. . ~us deo 2 /(0- (-10))2 t (20-0)1. (o-(-i6))1 .[6oo lsolote Fron I View I lsolol6 R.gh side. View'

, iO

..&.:.. "

" .c.Mo 0 ; 1000(110) - Ay60)-E,.(s)o /\y('J.) i E,. .. 000 - -

.&M& O; (b(&) t1()()()(ao)- Ay (10)-A. (s) o ;..1(1o)t/\ .. ($) - o,cs) 20000 1\,1 (-) t A.,. Ox -+t/ao D1/10 0,. Dt/2 - &0b.s.iD@ 8~d~o 8E ,.S1.5.4lb . !.y(4) tA.Ca) - Oy eocx>lb~ usrng lfy 1.s~ c -t. ~(u:>) /\y (z) -1 A - Oy/2 : 4ax:> lb Ey == 37.5 lb !lr(2.) ' (~) 'IOOO Using 0,, -40/.ioEy 'IO/so(a7s) A~(+) I 1()0() - Oy 8000 Dy "soo lb.

(4;\y - Oy 1000 --@ 80/JiO ~/20 : . BO 61.V! lb.

.... ,._,_ -5

~A(t) -~ - 0000 -~ =:Gr- " 7000}

U~rig@ Ay(,..) - Dy 7000 f..y (+) 7000 .soo

Ar ~ 7.500/+ .-. Ay 1875 lb. 0-ing Ez 10/10Ey =#15(375) " 2SOlb .

/\ (2) 1000 A ~ !ii()() lb.

02. [00o

~32.) ~ boom of n_, di$rric"' shown 1n F"ig P-63a. 01G rotated back 30 from the Y:Y plane. Dele.rrri1ne tht- forees in BC \...BO~ ~he compone,ntG of ~h~ beoring reoction en the .i;oci -1T.3Z/6A :. OA . 10 n. 0 _ - ----- (w,o.o)

. (O:o,u.~) ax.~ .. 10/c;&. . .CA ,. 20 n. de: a./Jo-w)tt (~-a)"t lo-a)' " wfl. e~ OA t x doo I (to-o)' + (2()-0)it(17.320Y dl&.2.flfl. l:: .. -10 g,(i, .. 18'"" n. dee /6t>-o"f+7i,)1t(o-~t7.!>t)) 1 2.6 .~6 fl . 'OC' ~10too60 . 11.~2 n. lsoiot6 F~\ 'l/iew, BC ~ 577.316 :. ec " 816.3 lb.

t&.'20 2.0 a

...., I

:

1nEq. Cy t Oy 1732. 0., 1732.- .S77.316 - 115+-'&+ lb

00 1154 .60+ eo ~ 16.3u lb . ,oP . i .'l

___ .. '!~- ~ .. -- . H.~e 2t>

Cy Ay ~ "'

Ref ern'lg to Froot View: ~fit ~ 0 A. - c. -o~ ... o c,. ..511.3K; (10/~) ::E:Mc.o.o O.

-A, (io)2coo(16.6'}f. o A" 3732. lb

.tt-1.-.o G (Cy 0y) 10- 2000(9.i;t;) O

Cy ' Oy 1132 - -~)' lsolae P.igtr~ $Ide View ,

.~-:

s l> ~ ---- _ ,... ~--------~ (;

0t u.32, "Y 1'T.;'lt i '1

~ 2.0'6 658 t-577.~ A,. 9661b

: 268,6!!9 lb . 0, 11s+.f>&+(10)

20 Ci. .S77.31G (i7.:?i.2)

1.0 " ~ 577. 31il lb. Oz 11.s,.:6s+(11. ::1~)

2.0

Oz = .999 .

Choptcr 7

CentroidG &, Centors or Grovity

10s.) Determine t he cenir-Old

101.) Determine the centroid or !he quodront cf the ellipse !Ohown in Fig. p-101. 1he equo lion of !he ell ipse ,s if. + 1. c 1.

~ o" . ii'

b .Ci ij

9~ c 1 - >.h)

c .1L. [b """"'''] Ot1

!J IO." !.( = t,j/"'" = h/b" ~~hx%"

]: = b (nt1) bvl ihis ~ is nt:z

reffered to(o,o)

b - b(nt1) - b(nte) - b(m1) --nt'2 nt2

x =

709.) Determine the y coordinate ot the s pondrel described i0 Problem 709

J\y =fob Y.e y( ydx) = 2 r: y11 dx b H fl ll [. 2nt1 l b

a Y2 Jo h ~n dl< = h~b2n ~nt1 -Jo h~11n [ b21"1t1 - oJ hil. (b'SQt1-tA . 2b 2nt1 2 ( !Zri t 'I

_ ~ c h(n+1) 4 0 t 2.

710) /.acote the cen~roid of the oreo bo1.>nded b~ the X ax i~ ~ the s;ne curve y c or;;;1n 1f'}(_ trom x " 0 to x = L .

/\ "l yd,. I' o sn~x 011 -= o7f (- cocoT~),~

-., -ov.-[ cos9f,,. ~ c.oo] -.9..h..(-1-1)

l

/\y f>12 f1 ~dx ~ 11 J: ~11 chi ~ ~a C .s1n11 TYi. d)( - }4 Cl~ 'ff r'J)I -Sin 21/\lL

QL + l..

1 ~=~ -~ ... lo T 4 B f

711-.) The dimension~ of ttie T-section of o ~r;;;l - iro~ beom are Ghown in fig. P-71+. How for is the oontroid of the oreo o~ the boes.

~ 11 1-- /\y = ~Adx T [1(9)t1 (G~ g .. [ 1(B)(4t1)] + [ 1(,)(0.!!)J e 1+ :i .. 43

~ .. ::i .o7 in

715.) Determine the coordinates or the centroid of the ar-eo shown in Fig. P-ns w'1ih respect to the given o~es .

Aij : ~Adll flH6)(9) ~T(a)i] _ij - [~(,){9)(4159)] f-[2T32l(~ t 9]

4U4Y - 307.23

9' 7 .+ 7 in . 01 "l1.1''t '1 - [V2 (4i)(9)(~.r;)] t [lleT(,) 2 (a5]

.+1.1"1 ~ , o bent wire shown in Fig. P-111. The wire '19 homoqeneovs "-> of uni form cro~ ~lion .

~ rs""' _ a an :so 2.e6S in . ---;c- 4 01' .. '*

.s1n:ao . ~96Sin ..:. :g 1.+a~in. r 2(2113,.. c~,. ~&?)) t .t1 ~ f2(oal

no~ Deter mined the centroid of the lines -trot form the boundo -ry or the ~hoded orea in fig. p - 719.

[ 1~ t- f180 t 6 t ,Jff + ffe] X 1:2(6) + Ji90 ( ~h)( o/.[i) + f72(-f7%)(V-{1) .sni(.~)('1.n-) t6)1

y: ... s.~6 in . . 49.39' g .. te(,)t6la> .ft90[(~){Y"'5} 6] ,Jn(~)(Y{f) + ffe("%)( Y./1)

+Q.397 ;r ~61.659 g" ..s.+1 in .

7ro.) The cen1roid of the shadod ar-oo in Fig-P- no "11;1 required to lie ()11 the '( O/llli . Oe.lermine the difonce b lhol will rullfi ll th"u requirem~nt .

[ ~(-.)(e)J[ne] [Y1lb)(6~[ o/sbl es. "333 in 2 in b 1

b' ~ +2.6665 m2 b ~ 6 .$3 i11 .

n1.) Refer to the T aection ol Prob . 71+ shown in Fig. P- 71"1- . To wrot value shou Id %e 6 in . width o f the nonge be chonged so thot the centroid vf the oreo Is ~.sin . obove the boe ?

(1(8 ) t b(i').'i>.9) 1(g)(H1) t b (1)(0.!!1) !lO 9,. 5 b s "IO t Oll b I

~b ~o b - 10 in.

I 722.) L.ocote the cenh-oid o( the shoded oroo in F19- P-71111 cn:x:il..d by cuitlng () semicicle of d lorneter- r from 0 quo .... ~er c;rcle or

r~c -"';"1 ~ ('"~")(~~ -(~" )(%) Jf- ~. rh - r;{2 Jl!!.r: . sri -1' -~ - o.637 r

9 ~2 12T 1"2 q ~'AX~.,')-(Tr%)(r1i)

Tr/a~ r~ - 11""'% - 1G"r 3 - $Tr3 ~i "' 6 .S7Srll

't$.6 ] ~ l).7~C w O.JJ"\-9r

n -

..-a

~a.) Locote the cenfroid or the shaded area in F1q . P-72.s. y

,

a

. 5

[ Y2(1-1;)(c;) ~ +.s ()- !('}1x .. 1.e(1.s)(c)(V.6.s)t+.!i) ~ (.._s){')(~.as) -: lT~ .. )(#)

j" 24.-43142. :i

7'.) w ;lhout osi'"9 int&grohon. compute i for the on90 bovnded by tho l< Ollio, ihe. line. x 12, tt.,. the CUl"VS H .+X t %- x.%e .

l'I ~';ldx Ax = !LAd-. A ~,. { ..,.,. +l)- bhl~e bh(~)-lbh \(b~\ nt1"1 tnt-t.,"/ (i1tif) bhn ie - b'h _ b1 h n11 2 ntF

~ b~li(ntt)- tb .. h f(nt&) ~ ;e " b~6v-si

nH (nt&) ~,, b(t1s 2 l'ltt ~~ bh (l}4)-(~[~

PJL. -~ .. bhttin t1) - b n"' ~ ~(

' _l_ I . [ 1 (' ) tU( t )1 ~1~)(1)] ~ t (' )Q3.s) t 12(1)(7) ti2 (1)(0.ll)

3()~ - 171 ~ -.s .7 in. ~M,.;, 1 (') (7,a) t ( 1) (7.!1 )(-3,GS)

~~M 7.a.+ '1n~

n1.) Two 10 in,-1.S.3 _ib chonn&I~ ere welded to.9El-1her oi; :shO"-Nn 1n fig . P-731 . Find the moment ol O"GO or the upper ohorine-1 obout the hof'izontol oeo~rO;dol ox\Q l(o of the. enf1rc, Geofon .

oreo of 10i" - i.s.:9 It> +.47 in~ 2 (-4~7) !i +.+1(~ t 10) I +.'" (-')

9 .9t g 67 . ..sSc;"I' .,g - 7,&6 in

7a11.) A 0 rd in 0 bridge, fruGS i oom~ed or the elemenl Ghowti . in Fi'e . P - 1a2. Ref o- to Tobie. v 11 2 !Or- +he prope..-fies of the onglss ~ loco re \h~ ceotruid of the bl.ii It up cec-t ion.

1ax1

[ 9(1)1 "(l'..)1a.1s t4'.1SJ ~ ,,~)~r,to.s) 1 11-( 'fr.Y._o) + us(14-o .,.,.) + .s .... s (1.ce)

!K;, s; 5 - 2_79-347.s !j 1

7,,.-.) /... ro9h~ frion9le or 1:"1cJei;i b~h '1c; rofotecl obou Or'\ o"lt i9 c:oinoid1"ng wifh 4id& h lo gsnero~e o ri9ht ciroulo,.. cone .

-. .. _ .o.

7.57.) ~ivt. V +f~ Tr

A 11. ;: . le.n9-lh of ovrve -1f . 2~~.Tr

I\ - +Tr 1

voff""'Obol ... d air 'il A 'tlT . .'l.J4. 1.bh

v '5. v ~ b'Th

1.s.o.) Determined Hle vo\um$ of' the ell ipsoid of revolution ~fed by rotolinq ori oll ipr;e obovf o ) "tis no.Ja- oxi~ ( Pf"Olofe. ollipf;()id) ~ 'I>) 1'5 minor- axis (oblofe ellipsoid 1ol\e the lorqu- sem~o,.ii; oG o

~ the. smoll&-- .ge.mi oi

b)

Vp~V.lE 2T. M . " ~-+b . Tab 5iR" ,_-

y =+rab'"

.tioat ~fb' 1 ~fo .. 1- ~% ~ y.f ~ a}'bA (b' - ~~)

A.1. (b 11tdy Yoe>-Arc -~l .19_. if ob 4 ft t ~f.:o~(~"-'j~Jy Y -fTo11 b - Cl~il r~l, -,.rs ti> o}b'fb1-W$ +b- bfa]

A; 4 o/I [ tb%] T~h-~ = 4 alli

T j"~

.9T l"IC.) /\ circle of rodius r lie1> in the 'XY plone wi\h ,+~ cet1s wi II .9enerote o do1.J9hnut - shope ring co lled o torus, prov.1d&d a i- groofe.- ttion r. Gompule the .surfoc& area ~ volume.- or + tor'IJ.S .

V '= !Zf.ll .lr~ V21f'r"o A Jo2ilr A "flfcra

7.+1.) /i. 6'0 pipe ell;>ow hoG an i11~erno I dlo meter of +in . fhe rod . of curvoture of lhe pipe '~ oent&"i1.Y1e IG 6in . find ~he interriol vol. of' too e lbo.v

V Arco x centroid ~d1sfonoe. teve rc:ed - i()~ x c; x (oo' x l/180') V = 7g . 96 in:'

7~.) find fhe volumo of the .sphef"icol we09e rormed by rofolir"19 lhrough on on9le or +.!I' 0 .semicircle of rod1u.s r oboul 'd~ bo&e diornetet'

I\ l~"l- 11.7' ~ . (irr/ /\ =Tift Toto/ &iurfoce Arco + n r "'

A.,.,..- ~ 0 Tr 2

7~.) Compvfe the .svrfacc oroo "} '

7 .. s.) ll.e oreo conioined between two concentric -5erri1c;reles of radii 1.s i~ . ~ 3 in . iG f'ototed obout on o)C ;s 4 in. owoy ~ porollel- to !he boce diometerc of the oomicm?les. O:imput1f."f~1] '\Q.(;0:3 g - 16. 75

~ "' 1.4-85' in. v. 2T. (1..+et>t-t) . 10.603 " _aGs . .+ in~

~

7+6.) Ooterrnine ihe surroce oreo ~volume generoled by o complele re'a i~ or the Ghc:lcled orea of Prob . 72~ .

J .s.oa ,n . v .. n . 0.09 . [ +.s (G) t '/2(G){u,) - T()%] v G 779.92 in:11

A ~ !Zr. 9- le~lh of c"1rc-unioibin9.., Areo /\ 211' . .:i.9'-[ '1(11}{ t 6 t s t J,+ 1 .~J

A ~ +92.S7 '1n.z .] for let19th [3." t :r~ f~-. .. ~ B .. a 6.s) ...+.s(o)~ ~(,) . ~ .s'l-c~(-J,,,:" )(Y..rn-)

. . + T (~ ((; -~) 1'}.9 97 ~ - 70 .20

~ ,, 3.~; in.

1s1.) Deter.mine the ceritroid of o hemispher'C! of' radius r. ta-King the axis of' .s;ym:mefry as the z oxi~ .

z

754:) A. un; form wire is beni info the .sho~ shown 1n F19. P - 752 . The .stro igh~ segment~ lie In h$ i.-z plons, ~ the 9 in . length makes on angles of 30' wilh the X axis . The -sem;c1~ulor ~og-

. rnents is m ihe 'lC - Y plane. Locofe the.. oen}er ol g~+y of Jhe w;rl!

(H11(.+)1g)x ~ -.(,.)(-+)., e(e++~:ao~ :i'.s")l 14-1.91'1

i w.s;,:a+ in . /~: .. .,,g -: T('+)(~r1)

'16.,s,, g .. ~fl . I :g 1. 204 in .

'26-.s&' z '(a) t e ('!-.sin ao) :;lG.&6' z ~ .9'4'

.Z 1.28 in , 7~.) Locofe the center of groviiy or 0 s teel r ivet hoving 0 C!,j-lii'idricol body 1 in . in diomefer ~ 2 1'n . long with a hemispher10ol heod of 1 in . radius . Use ihe resvlt of Prob 7"11 .

I! ~ i 2" !, j. a () ' [ % T (1 }~ t T (o. 9)~(t}] ~ - [(%)T(1)1][u % ~1)] I T(o.!J"f(4t)(1) :3. GG.S ~ "' G . s4-S f - . . . ~ 1 .7&; on. o- "' )

-t .so ( 16 - 11 ;: ..s in from base ;from ~- 75S l" 1-'5.) A body consists of o r 1qht cir-cular cone whose ba~e is 12 in .

~ WhoGe olfi i ude ji; 16 in . A hole e in. in d 1"ome.ter ~ + in. deep been orilled from the bo&e . The Ol\iS of /he hole ccimc ide.s w/ ~he oxic: of lhe. cone. . locote the centroid of' the ne-t volume . Use fhe re.sulj of Prob . 7~ .

756} Dcfermine the height h or lhe cylinder movnled on the heft>;Gphe,..;cal bos:e sho""1'I in Fig. P-7-56 so tho} the com~ite body will be in s foble equ-1libr;um on ils bose . Hinl : /\.s lon9 os the cenler of 9"0"iiy does r>ol lie ob

Choptcr a

Momenta of lncr"lio

904.) Oe\ermi"6 ti'~ momen~ of ine,...\ia of o triangle. of boee b ~ oltdvde h wi\h re$ped \oon o'llico through ~re opeic porollel ~o the bos9 UQ:. tho tf'oni;fcr f onnvlo ~ ih& n9Gvlk of illu$ . Prob &OZ.

'IJ":. /ah

b

r. " bhY-'45 I .. f" t l\d" t:>h/~ bh/2 (2/3h)2

tih~!O t 2bh/g I .. bh8/

SO!J) De.lerd1he the fl10t11enl of inedio of ihe guorler circle ~ 1n F(Q . PSOS with rei.pecl io lhs _g,..,.," oxes .

r r~ for &ttl'~cl9'

J f p 'd/\ J 1: p (Tf'#d/)

(ff,.dP .J e -ff]~ -tr-+

r4/.+ J: Vr( - ~.te-)d.,. - r~ [+ - .su11fd8df 1'" f0T prJp""~fr !,'-,~ (cos~de-)]; ~r r~ (CC5~~d9-) r -+/+ S,: ( 11 ioMe-2 d

eo1.) Show the moment of inertia or o sernicirt;le of radius r 1s 0.11 r w ith respeci to o centroidal 01

a10.) Deiermine the rnomsnl of inertia "., the rodius of grrolioo,w/ respeol to lhe Y Ol

J-1,. +ly ly & "1)1 .. S"4 in~

L1h .s+ 270 in~ "',. _lJ/A "~~1%l) 3,97.31 in.

817.) Deiermine. the rTIOl'T\cJA:.;;...df' :r Jpd/\ -~Ill ~'1'3of .... .,,.

A 1r/+(do2-dt~) lr/.ot(~- ~) ,

J : tJf' tr'] r-o T f\.

.A1e.111n :J Th(~-~) 102.11n. ' K. "'JJ/A ,. ~1021/~71 = 2,55 in .

910.) II hollow square croG$ section CO'lsists &' on 8 in square. frooi which is subtrQcted o conoenlricolly ploced ~quore 4in by.,. in. fir'\d ihe polar (TIO'Tle~ or ine:dia "" ihe p:h-- radius of gyrot.'on with ~pecl to o Z. oi1i: po~ing fhrough on& of fh~ ovl.s;iae. ~

l. I, iAdi 1., l 7 +Ad

I., ~s-. 6(e)(._,)' 1;u;s.33 ;.,;4-

1 . Iy, ~~e)3-. e(e)( ... -{ c 1.:36.S.~ in.+ /\ e-... t - ...e in"l

J, r . t lyj ... 134;5.33 ~13"5.3a ., ~730.6'0 in! I,.t." ~4)6 + +(-1-/(2.UF 'l7'f.'33 in"t "-. ~ Ir ~~4) + +(""'~t2)2 = 277.33 in~ = -.J ~1741,46

I\ ~ 6 ,73 in. J1 '" lllC t]~ = 277."!3 u11 . .33 ".s.5't. ";,,~

J T J,-J2. .. ~730.~c:;-~."'" ~z11

&1g,) ~let-mine the moment of nerlio of the I - 6ect"i01 shOWT1 in f~. P-82.0 w/ reGpe:;\ to 'it~ cen\r-Qidal o>U .

, .

.n-11 .. 2_S

l\d ~Ay ~ :y e.(e)(e+) .. ~(e)6)

1 = ~~.n 'i,. .!!i 1 2(,e)(t.5'f' t}fl)3 2(e)(t..&)s.

1f. "

l'tt 1.(e)(t) -32,n~ 1 ... 290.r.1 in:4 920) Determine the moment of inertia of the oreo shown 1n f1g .P 02D with respect to '1\s cen+roidol o~.

Ar : 1) H2(1) t 12(1) = 3oO in~

/\tr 9 b'\y 30'9 i2(-.)(o.s) t 12(1)(1) G(1)(13.!1)

y =~.7 in. r,. :"()..(1)3. i2C1)(s.2)~ -i112~ t12(-t)(1.g)"

12 ~ t ~. 6(1)(1.0)1

12

I.,. .. 95.s.;a 'm:" I,. ~ 1 (12) 3 ~ 12(1)3 1(,)a 103 ,n~

12 12 12

021.) f ind the me.men-I ci' ineriia about the indicated J< Ollis for the shaded oroo .sho.....n In F19 . P - 821 .

I,.:'.[,. tM~ r., e~tt + s(1o) (!1)11 ~666.67 in~ -~' iO

-

1~1 = 26".i;;1 - 11,0.~3 = g~.14 in~ 022.) F1'nd ihe cenfroool mornenl" d iner lia of !he ~rope:e.oid ..;;howri 1n Fig P- 02.

Ar 69+,)( ' /2 ) c 72 int j,. -[ ";t)3 + 2 ( '4)(" )(o.s)~ z t '1(6f..

12 t r.. (GXo.s)

I,. = 190 in 4

823.) /\n equi\o\crol ~iong\e has itt; hose b honxontol. ShoN tho\ the c.en\r01dol mornen\ cf' inertia with ~pect h rori-%00~01 ti.. V&f'\1eol o~ ore equol .

"lit

h /b1 - (tY2){ o.~b i, ,. b(o9'6b)3 = 1.90+l'10- &b+

""

b 1 c o.~b(tl/~}3 1 .o.Mb (t!fe)3 = 11.801- xiOt b+ 1il 12 92.4-.) Compute &e rnomenl of 10e-lia with re~t k>an 01t'1c; possing \hraugh ~ oppo&ite ope"8G

fl!A 12.5(6)3 t ~(12 . .!5)(

fn::>m Tobie vtll-.2 Prop. of Shvch,.1rol Ar-olysiis Chonnel Size /\rY!O i,. TY x :;

12'20.7 &.o.., 12e.1 .s.9 .1 o J,. ~ j,. i A~4 1,. (128.1)2 ly"fr~Ada r,, =[.:..91~03(dh o.1tJ ~ r ... iy 12e.1t11.J '1f3.9 t6.os(cl/2 to.1)2]

d = 7.c:;e in

--------1~--------.

__ J __

~ .

---------i----------

929.) Oelerrnine the du>lonce d ol which the two 3 in . bys in. rec-iongles shD-Nn In Fig. P- 829 should be spaced so ihcrl f > J,, .

r ~ I~ f, 1 Ad2 ; [ 6(3)3 + 8('3)(c:l/2 t1s)2] 2 i,. ~-:. i2 Ir ., 1,. :. [3(8)3 t o]'i., ~[e(a)3 + e(aXd/2 t 1.1J)2.P-12 17 120 " 18+2-1-( dh ~1 !l) 2

c;1 .. 1:20 in. 030.) The hor~ le~s or four 6 by +by 1h ifl. ongles ore connected too web plo+e 2.3'h in bv .5/,6.11"1 . to fonn ihe p\ote &i ong\e girder Shawn in Fig. P- 030. Compute i~ volue of' i . .

y s 12 ;n, .i. " 5A6

1,2a.s)3 .. [ c;~~-!5t l.(o.!i)(1t.1s)i] 1es eoch 18 in by 2 ;_. in ru; .i;hown in Fig . P - 031.

j,. ~ 1 [1\12s)' t 1a(u.s)(a.375)t] ~ [~I 8(1~.75)1]+ t[i~i)S I 3(1)("1.'79)9]+

- i ... I, 1.::&+.65 in. ly 1\1)1 + [a.2e(1e)-a}t [1~23 +1(e)(~.s)1] ~ n , I[~ 5(1)(1)t].+ f .1 y .301g.ss 111 .

Usng Tobie vu1 -2 Properf1e~ of slrvcturol Sedions s11+ 111- : Areo 11 io~ ; 1. 1t.6 1n;J1 ,9,6 in ; ; < a.o.s in 1' y , .~ in

1. [1(-..)'.lA1 t [11.6 i11(G.2)] [(1e(t .i.slV\2 -t 1e (Ms)(e.a1.a)~J 2 I,. 1'e1.99 .n.

ly rH.(.)'],/12 I [,9.

+ 3/(a.s)(c;.1~9)1]+ " i.)I .s91.6 it"l .perheso of Structural G:c\. 10 15.!1 lb Area 4 -+7 ot'l" 1. ~o'll'l~ ~ 11 2..ain~ 5i o.64 in 1oi., y o

AT 10+4."47 14A7 in2

A1y :a.Ay Ho,4.7y 10(7)t'4+7(13.6)

y 9.o+in .

/Vea ~ 101n< J11 ;)39.2 tn~ !&,. t 0.1+ on.

i,. .. 2..3 t -t.n(-t.od)2 t :139.2 t 10(2.o+) 2 T11 476.0l in:"

Q3!1.) Two 10in 19.alb chonn

I= MK' .. Md% = Y2Mr1 + M(r)2

I = s,h Mrz

063.) By using the tronsfur fon'l'lula 'bi._ the resoult or Prob. B62 9 defermine the moment of inertia of o homogeneru~ sphe~ of'

mo~ M ~ rod'1uG r w1\h re~pect lo a tangent . I .. Mv.2 I Md2

I ~ 7(.s lv'l r 2 866.) 8y using the tronsfer formulo ~ ~he resvlt of' Prob.663. de-termine the moment of inedio of o rod -..dh re&pect to on axi" through the cenler of grovity perper"diculor to the rod .

p7 I c [ t Md2

YaML2 :: i t M(L/2)2 f = 1ML2

11

06S:) Sy using the troniofur formula%., the result or Prob.961,de-termine the moment of" inertia of ihe rec"tongulor porollelepiped shown ,n Fiq . B-2Q. with ~cpect +o a med&on line otfhe Z face. Take tho median line porollel to ihe )( Ol'

812.) Determine the moment of inedio or o hollow steel cylinder w'1th res~t to its geomefric mciG. The cyhiider- iG 1fl long ~ hoG on ou+side diorneter of 30 ~on inside diameter of 2 fl s+eel weighs 4Qo lb per ft~

. l/ 1f(M-1t)(1) 3~Q27_n-a w .. ..wo 1b/ffa (v) --31~ ~o lb/_w{ (a.927 f({)

w = 1g2..+.23 lb IS\, 1ft. I -Y1 M(R t r ") = "Jt(192.U.~) [(1.!>)2 t(1)'l)

'32-~

; I .. 91.11 ftlb-sec 973,) By using the method discus~d in Prob. 070, detern;l1ne tha rnoment of ir>edio, w1 \h ~Sped to the 9eornefr1c Qxi G. Ol O cylin -der of rodiuio R from which is drilled o concentria hole of ra-diug r . Denote the mass of the resulting hollow cylir'lder by M, "-. the mo~ per unit volume by 'Y.

h 'v't\ m1 for the ctlnder R< V21 rn2 f Or- fue hole

r I ;": 'Al MRt tvt ~ ~v :(': Yt[m1R2 -m2r2]

M.,'f,y v, ~vR'h -rn1 >ilt'i:l..~h 2-lfr'Zh-~ , 'lltrlh

M '6lfh(R2 -1 ~) ;= 1/2 [1SR1h(~)-~n-~h(r~)1 ! = ~ ~JTh [R4 -r]

, =1h glfh(R1 -r2 )(R1 +r3) . ""/t M(R2 +rt)

974.) A slender ro:1 6 n long rotates about on O)(iS perpendicul to it ot o p01nt 2ft . from one.end . The rodweighs 40\b. Carnput~ the rnorn.snt of 'inerlia obout I-he oxi ... or rolo\ion .

tt'..f--:s '-' ... ::::J W'4Clb. I = 1A'-z ML2

~ "42(~i.t)(6') +f+-0/.;2.i)()t ::r ~ +.gj fl -lb-.sec2

87S.) f()(' 0 hollow cost-iron sphere or W in avtside . cliometer ~ 16 in- inside diameter, compute th'': . R=-10 in . 0 "16 r .. ein

w "' 4~0 lb/ft3

\/ 4"% [10-a:_ 9ll] ~ 1.193 fl3 (t1)'

w c 4.SO_ lbj( (t.183 fi') .. .532. 3.5 lb .... ~ ./Vtil

- /C~~.""'t-..(1_++_)"""(..-32.,,...,-=-ct>) .r -'3~-'35

K. ""7.35 in. 876.) l)e~ermine the momen1 of ir'lerfio or the COE>t -iron flywheel Shvvn in F1Q . P-07S with resP

..

Choptc:;r fO

Rcctilineor Tronslotion

100.2.) 0.1 G\ cerloin .s;trdch of trock, !r01n.s run oi 60mph. for bock of o stopped irain .should a warning iorpodo bo plocod fo "ignol ori on coming froin ? /\ssurno lhdt tho brokos oro ap-plied of once ~ rotord ihc. lroin ol ~ho uniform role; of 2fl p:Jr soc,2

Givon: Vo" 60mph O Q -2fl/SOG"

. yr'-vo' 2os -et/ " -2(2)S

Vo = 60 x s2so ,.. esfl/soc :96

to05) A sW1CJ s droppod ~ o wol1 t..., Ssac iofar lho G0"1d ar ~ ~ -.. hcol"d- 1r 1t1G "IOloc.ity or SO.id K: tim n ~ goc I whof ~ tho depth a lho""4?

l,.c J!OmG \l!s ~ tltlOfl/.-=

t. -~ b- lllOslono io6' ClfllOd ~-t, - 1,iilne ''" tlbc;.$OIJll1d iobo t-cordj(iJJ-~ " d ~= 2+

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Learning Guide in Engineering Mechanics