.,

·.• ,o '

PubliiMd.A Oi11ribut6d by:

.. .. ~ok•t~~ IH N~~ Reye,, Sr."st. ' ·

. Tet.'~.' 741-49·16 • 741·49·20 1977 C.M. Rec:to Awn~

Tel. Nos. 741-49· 66 • 741 -49-67 Menil., Philippinff- '

• ~ • "-e

) ~oj,'• B~ for"th~~ ·

PEPA's IntematiOaal Book'ASsoriatians ~'bership: Asian , .. l:'aaf;ic.Pu~ ..\sscw=i•tion 'tAPPA);.Association al. South

EaSt Asian Publishers (ASEAP); lotem.ational Publishers ~tioa <IPA> .

P~ted by i REX ~Plf1MY.k 84-Bfi P-. Fiorentino St.. Sta. ~es;. H~ts. Quemn.City. Tels. 71241-08. 71241-01; Fax 00- 711-54-12 .

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, , 'The · : book/ ~fitt tled Le'a .rnJng Guide In . . > •

Jlec~-~flJcs rra,a ~rJtt~n as . text/re~Je.,er . for · EngJneerJng .o:l ",P!esen t.J ng .. , t'iie pr Jnc Jplea· and the purpose EnrJneerJnr ·Meehan.lea '-Jn ,. . conc'epts · ol

.'~pproac1i · to ' help" the· t ' a -. v~ry, bas Jc. an,d ~yate.matJc ·th · · . . . . . a udenta understand d . 1 .

e . s'!bJect . matters that ""JII ' d . "~ an ·. earn P.ro~essu .of " thinking . . e.~.e. lop .theJr . orde-rly .

. . .. . ... . ... . , . · " "

" " f'.'J. t.h .. the . presente d topic th · , ·, . very compr~hen'SJve s 'fuify .• o·l th s'. ... 't '" e 11~·~.ra wJ ~I ~ave a . ol EngJneering lie'chailJcs h e .. U{!~amental P".Jnc·Jplll'.s· var)ety . ot .·pr'actJ:cal sHua~J! ch are· applJ~able,: to . " .fde . the• , f~ .their day . to 'day· actl ~: t ~:=~ally encountered by ·

' . . . . ... '

Extra - eliort " , 1 " • " ' ... b'e pr-ese nted .Jn~ ::: exerted so . that - e~eryth,Jng "ill Clear". and . . i .. .. . y ,p~rlectly . unders l-ood bu' t Jn

cone se· Ian ua · · · ' " . ': a reduce to ~ ~inf t g ge. This ~Jll el.lmJnat~ or memorizing t'h .. . ,mum., h.e .!J.tu~ent 's ·a. tti:~ude .or ,- haliU t

e c onc:ept wJth out u.nderstandihg·. .. 0 ·

- The autho_rs wJsh · to ac~n~ 1 d .. t . " . " .. to JUcba·eI Si - .. " e i .e. hf:dr Jndebtedne s·s C . ongco, J e ttrey Borl . ~tin"lg, ... a nd Romeo Adr·Ja l . . on~an, , ·:~.onaldo

·contrJbutJons Jn p~.eparJng an;o . 1 ~~ - their valuable· wr ng tbe manuscript.

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N · c-: De 1 aRaaa A· G. Mendo:za ·

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ACKNOWLEDGEMENT . I \ .

Jbe authors arc perti~ularly indeb~. in the p~eparation of this book to .. Mr: Romc<>'Adtjano, Mr. 'Ronalda· Caiindig, . Mr°. Getardo Slimson, and . · Mr. Jeffrey BOrlong11r1. '

They further wish·io speejally rceogni1.e the exerted effort of . Mr. Aritold·QUetua in writilig the m8i11.1sc'ript

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spcc'inl mention 10 th<>Sb who bring the grCaicst.joy into Niel' s lU:ti. Nico, Nikki M<i<Nlkka ' . · · · ·

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Equilibrium of foroo ~ /\no~is;: of Strvcturos: Friction

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furn:; SysterJWi! frl SpOoc ' Controidif I\ Cenlc:iO; of '

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Grovity Moments cl" lnortio Reclil~ ·ironsbtm

·Chc)pter 11: Cur-vilioeor. TrooGlotion

Choptcr-12: :· ~ion O;opla "M: w~ '*' Energy ~Or- 15: lrnpU&Ge ~ MOrnontum

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213.) ~h~term1~0 .the' ~i.>l~d~f :'or f0e eohc.u~t lr:i f.19· P- 21a. ly

ioo lb '

.- 0 . . ' 0

;efx ·".° _--'fOO. COS6P• t 3~ C0S4'5 -1<><:> COS ;q ·

~f=>< .· "" -:161 :07 lb '. ' ~Fy a ~ ~1n.~o0 t ~00 .sin+s•:.. lloo sin-:ro ..

.. ~Fy "' +s8. 8.+ lb : A,.; /~F1-'f-.,. ~Fy2 " /i""t--1...,..lS:....1--,.0:-7~)2;;--t-,(-:+-=!l-!-:8'-: .. .s:-:+;-';):22

. R-= f.06, 01 -lb ~P to +·he 1ert.. \ ,:.",::~·;;.!:;:•-·-, I

-B'-x ; ton_, ~a-s-+ , :- .70, 6S • . 161 ~9..7. --

£1+.) Defei-m.1n-6, the res~Jtqnt' of the concurr;ent system of

fprces ~/->own In Ff9 · P·- QH· · ~FY.• -of900C-OG30• -soo6(~) taoqoo0s30•

- - -r(3,66 .03 lb.

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; 11~,b :· : _ .,:!fy ~ '.f'~,6iil~·~ ,~ -~~:sinaa~- 1000. - ·s~i) . -.,·: -~r. .... .,-··,:...1.!5o6'itf .' . j ' . ', ...

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R. ·• . ./~Fx7 -t 2<:F)'~ ·. : JC.,:+066.03)-z. t ( - 1500) 2

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· · " , · R'.'" · -"Q9J .9.7 lb . dow". to· the 'leff- · .

::'. ,· "; :tfx':• .. tO,i; ~1 · 15_?%~;03 ;., 17,~:i:~ · ; .t;s,.), f1iyj, . the·/ esulfor:if: of . the oor.cu~t· farce ~yE!tem ·s/iov:-il) :~ ''/

" ' • A , F~·-· P""~1:S .. · .. ;: .: :· ' •. : . · . ,,·:;-. ".:~'-. -, , . ,· · .. • _ " v · ' :€'~.: ~-~oocooa0• -'aoi::? ~t>O··t 3qp,'coe ,~· . / , . .. ·:

\ ' :. · . . . ~f· _.;;. ~~6; ;5-g lb . ' . ·:· ~lb . ,1. • . '. ')< ·' . .. ..~ . .. . '·o ,· ·• . •

~Fy ~ . ·~~ s10 3o - aco .sin 6:Q . - a!='o .Sin.~. , ~F; ~ :-:-4-1~;62 :•/b. . . . · ~· ·_d . ion~·l ·.1-1.0 .. 62 k '6j., 61 ~ .

2. ~6.;_7g . ' . .

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... · ,R,, .. v'ir,<"'.-t ~r;-·~ ~v(dq6 .. 795 "· '+- (-:.i-1.a.~~) 2 • , · \

·· .. ·:· · . : · · · f<.,,,·,476;ea ·1b'.: .-,do:...:1;1 t~ 11:ie· rr91.1+.: ~-) /\ wncurrenL foi-9=,'~Gtei:n '1s'~n·' ii'\ t=lg '.p_-. ~16 . . l')ete':'~ rnin.~ ... the ·r~~ltoiif. · . · .. ..,. . · . '. ~ . .. · ., . . . y' " ~/~ '- : ~F,_ .;_ 4';Jo qos~· t ~oo(i-).:. 3p,0 cos ~b~:.

" . = .-31:6 .. :+1 lb . . ': ..

-. .. , •·· .-~--~~-~'----'X · 1~Fy" "'KJO~in 30° - llob(~) ;-a.x>~in 60°

·60· , ,:ff.ry · ~ ..:.27e.a·116: · . .

", '390/b R. V~F'i" t ~Fy" _;J(~n6.+i)2· -1 (- ~1~·81)'" R *' $es .. 27 i,; :· d;;~· to_•the r-~h+ ·

· . · ·-e:x· • ton'' · 1l.1~,SJ/a16•+1 · ' · 3-i',-79,•

!Zp) O,mpufe: 'fhe value dflbe ~lta~f ·or ·fhe' c~~rrent j . ' , ' ·, ces . ~f-own 1n f ig . 217 .

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y ' .:iEF~ p 1~ LL_.)· t-i-00 C.OGOO: - acob:fsi;0° -2do~~ \[ICff • : . . " . , . '

' ~FJ< _,;, 1++; 43 lbs. . '

- - --."--,111c---,..-,--,-.,.....- x · ~Fr.1oo{_~) t ~s1n6if:.~61_~;ie;~ -. ~00 .-i .... +5"

~Fy ~ -~·~.9 lb. . ,

R ::{1"'2':fE.'"""F,....,/•...-:.t-:::i:'""Fy.,...,.2,-_ :._/ (1++.-t-?Y.' t- h :a6.89):z. '

. R" 1+~-07 lb down to the r 1,9hl . · -&-ox =' ton-1 .;1~ .,e9/j:tf;13 7 1+.. 3-3° , .

218.) The body shown '" fie. P-~ a1e · iG' o0ted ori ·~ t·h~ fbrce~ . LJ.it.ermine fhe resuUonf.

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Fco& e- t · !Z40 coG ao~ ·;; -?oO· Fco5~ M :202,1.s ,

'd~ .. fy a F~•n;f7 -a4Q ·sin.aei · @

"., ;,;ry = !Z~2. JS tan& - 120

p_ = P + ~Fy a. 300 ~

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~ .. .:,. (:t!~.15.)~'.fOCf.:e. - ~.(~IJ.i. ;:t,s) tanJi. . ·:f. 'f+t<ld{ , '.~3!;3-~ fon~-8- - 70116-9"- ·lon6-..:.·1s6C:k,_:.;: o · · . ' ..

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f'!J~IC F~b~: 'I ...... ; . .. . . • . '

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'fbn-fia ·.i-·.:...·ci)£.16 ' ; · <&.a.°- -ai~·~3 • · · <;·

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t= '!" 2M".t5 . ' sri,61 lb ' . "·

·.· 'itt_) R,~f · ~-L~ ·if. ·fhe rO&Jila,,~ .at 6C( w1fh. .jh&' X o)ric;. . . .

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·_ lo obi°"'- o .. reGultont pe>rallel k the· inclit>~ plo~~ e'y',;0 .

~y'"'O ... ·" . , . ·• ·'

PCOG1ii -.~ s;nao•- 200 ·cc&ao'1 _ ..,_o ·..::+ P "''- -+sg, g 10 '.

~F·./ - . 600 ~a<) - P Slfl 15:0 : 200 ~~~ao • -- ~F ' ·,, .ng' 8 .. . lb

• . , . . .,.. <. 2. 2 . R :J~~,. -z. -f ~Fi..... - /(29~,'Bi_)'L -t 0 ,. -~q.e. 82 lb .

m:)~Tw.o' ~ on q:poo~to · bankG of a conol pull 'a bqrge. , ~inq ~ro~lel ' to fhe ~nks ' by means: of ' two. hqr1zotol.rn~ '" t~ tens10n "': ~. ~ QfY:, !ZOO' lb~·. :Z-te lb while- toe- ongl~ . refween fher;n 1<G 60 . f,nd 1he. rowlfant ;pull on the b£?r0e. ~. -the

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,<h, , ), ' " \ ! " ':.E.f I ;:, -~COS 33• -'240 c;CS 27 .. -381.§8. /b, • :~ • . • .. )C • • • \ • . • •

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~°";· "I" · · : ·R -·~6.* + E Fy~ "' . 381 .. 58 lb. ' ·~j\:~~ ,' "~6,) In fig. p1~·i~6 . ~~~uml.~ b'1od.,wlse ~menfa os pc;;~ifivo. ~;.:::.1.,,' .... Wn;ip~to.,. fh.e ,inom~~l .of. .force f ... +~.lb~: of force P=3611b. ·

.::. · o' ·"""·1 •IVltnfs A,13';.C;~ D. . . ~~\ ~· 1"'1- ~ ''·. • ~-r··' ".·· · · 'A · ,. · · ' :" .'{ · co·nsider forc-e f

\ '{ . . ~ .. . . . ;~. F •:• . M ..... ~-fy.(1)-f><(3) . ·. . · ·

.,i;1.'' 1; · _-;.v · :./: · = :... -t.5o(3/5)(1) .-+so("1'/~)(.3) .'

.I\• . , r· C ·MA= - 13s6, ft-lb .~ , Me resolve al pl.- 1

t': . p\ - i Me -= fx (6) = +so:(+/s')(6) = z1&>fHb

[) .. 1-. e Mc resolve al pl. -~

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Mo reGOlve al pl 2 Mc' . fy(s)"' +-'O ( 3/s)(.!S) ·/Mo.:"-'fy,(1) t f>< (3) Mc .- 1350 ft-lb •

= -+50('4-/.5)(3) - 4,50(.ak)(1) Me· r~sdve a"l' pl.+ I ..

Me" -Py-(1)-= -; 3o0.37 ft - lb. .-:Mr . s1o·ft - lb. eonsiaer. force P M.4- rv'solve of pt. 3

Pl< ;. 361 '(~/ffe) c 200,. 25 I b. Py ~ 361° (:3/ffe) "' 300· ~7 'lb .

t;fA 1' Py (2) .- Px (3) .

= 000.37(2) - 200.2s(3)

M,., .=o

Mc :re.solve ot p1 · .3

Mc·,. - Py(4) = -:-300·37.{4) .

Mc :=-1201."1·8 Q -lb .

Mo .reool~e at pt.+ Mo =PY..(+) = 300.37 (+)

Mo = 1201. 40 ft - lb .

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~; .. · ·~. -.. , . 227:) Tw~ foix:es P· ~ Q · pa~~ thlfough a, :po.int A·w/q;::rs ..i:J:-1 :":' · , ' '\';' ~" . .' · to }he r1g~t. oft ~ 3 fl. p~ve a mornqi l. cen for. O. fbrCe f!:is, · ' .~.

WO lb .. d irecfe<(:J yp>.)6 }h~ r!ghf Ol• .3G0 whthe ho~i~n{al. :'~: ' . '. . _';;,:

force ~ is 1~ l? · dir-epted up to· the len al 6r;lw/ lhe ·horl- : · : .. ~ t~ .. ::· < .. .' .. t.ol'\.loJ: Oeler-mine lMe. moment· of' the' r-~suJ.fo~fof'.fhe~ : : ·" ;·:.

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lwo fo~ w/ ~~f>ecl. · )G> ·O .'. , ,. . .. - ,. . /·I. ,;,, '7 . . ~"100 ' ' . . . M~ ·-P~'(.f)'tP~(!3)-.Cy("t)-rl,K· (3) : I

• ~~ ~ T . · . :- ... , ,. ;: (,700 . .sin ao~)(-:t).• (~~cosao~)(3).-

... ' (too sin 60~)(+)-(100 cCs60~)(3) I ' . •! M: " " ~~7~,'79 a~·1 ti (~cQ~l~ ccw)

1---- -L--

• • + • • • • ' '.,,: :·:

,· 228.) Without ·c.omput;lig · }he mogn'1tude of the resuiton-1, del . . . :-"'here the res.ulfan) of. lhe forces show~ .' in F1g .·P-228/ inlet"-: : ' i.k sects ·me ~Ifs Y 011es. · · . . · · · · · · , . .

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~fV!o = -:-361 ( ~sJ.6,1~(.s}t:~ (.900 .co.s. ·~·)-"soo(.s•n <t\$")(:z) ·:, . • . .co ·1001.23 ;i=..-1b. . . . ' ~f'x ' = ~oO coS"•t-.s• -t- .361(~.61) = 65~.d~ 16: . ,,.

,~F{· =. . -5?0 .sin +9°' - ;36.1 '( o/~,61) = 153.31 ' lb: , ;". 1

ii<;,. £.Mo/~F/ ~ 1001 .. ~/1.53·.31 ,;; 6·~ In. to. 'the )en of'O ·, iy, = ~Mo/.:e.F,. .. 1oo1.2:y6s3,gQ c 1.!;a fn obove O

y . ~lb

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0 . ..._ -'---'--l--'--"'1-- X 5• 361 lb

~9.) In f~· P-~29, Find lhe y oooh:linate ·of pOinl A .so tha'l: lhe 361 lb force wi)1 hove a · clockwise moment of 400 .f}-/p

aboul O · /\ lsd delerrnine· .Jhe x Bf) Y intercepts of the action· line of l~ fqr:'ce. ~Moc a61 .. ("!Ai.61)(~,..) ..,~6t (a/a.:~1)(~) .. ·wofHb

y 4b £Mo c 4-00 fl - lb ;. 300 YA - -f<>O lb • .

~'!:=~!!0'- ~ ...... 2.67ft. " : ~Mo =~F_y i>< ... "fOO .. a~.f (%.61) t,)i<

:Y~. 1

ii<"' 2.oq Jen or o . , .

0 -+----"'- --=>'- ~Mo = ~Fx ~.!f ' = .3~1 ~%,61)(ly~ ... "f-00'

i Lii • 1.33 ff obove 0

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de"' 1.!1.180

reGolv·e oi . 0 :'.- ' .t.MG • f(3/.J1o)(16) t f .(1/@)(12) = Fdg

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•. •• • ·!·. . • . • ·$·"· 1~;~7·· rt : '.· . . . . . ' 2a1:) A force P passing lhrou9h" 'p:bii1h A· ~· B in. .,Ffg. P-~31

~S Q cloc')f..;vi$e. rr'lOl'Tfo~t Of 300 n-(b obout 0 . 9Jmpute the· . r.

. vOlue of P · \:: ··,·; :-._, . - y / j • when re'solved ol "·'

"" Mo= P)l.(3). "'?0<:>.fl-lb , . . px:, ": 300/a =:;1'00 lb. to )he right.'

0 when res0)v.;a dl-· 13 ' • . Mo· py(6) ;. .300 fl -lb

, Py = 300/6 -: so lb. . · P=./P,.2-tPy2 . =/100,•+~o~ = 111.Blb

232.) In -, Frq p- 2~1, . lhe rnomer.11 ·oro ceda'•n force: Fis, 1sof1-lb.n\ocK:.wise.,.' obo.uf 0 ~ gq n - lb coun}erclock.wlse obQiJ.t /3.

"' . Ir 'its mdmen h. about . A , is zero, .delef'tm·lne llie foree ·

refecv,.t~ the f'igure· 231 .

when .we resolved ot /\ Mo == 100 • Fic (a)

Fx =;. 60 )bs.

w hen we resolved of .. 8 . Mo= 90 " - F~ (a)t· f~(6) . fy == ;+s lbs. ----f =/F,_ 2 -+ fy ~ c /60~ t 4-9._

F "' 7S lb.

ton·-eT = 4 o/6o -&· = f-on-1 ..-~/60

-& ~-36,67 "

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. 2:9.)' I~ fig. p-2~.1, a .~rce.: p inlersecls lfy~. X o~1s q) -i' ft, to t.l;ie rig?t ~fa . . If ,1tG .. ~t obout Ai~ .t7b fl,- Jb <X't.mlerGl~k.~f ,: . . ~ ils tno.menl:· abQU( .J3. is 40 ff-lb clockw1s· e · d -i. . • "t'· ... · : · ·· · l .,,_ l. ·· · ' · · · · - ~ · e 1erm1ne 1 6' y , · 1n ervop ·. . ; ' AT .

3' . .when_ ~~olved 0 } · pl. c • .Ma• Py(.2) .... P.y • Me/2

--;:::+-'---:----="''-:-· ,..:..B- . - P,x ... "f<>'fl-1% 8,. ~ 20 lb.

, Mr." Py.(+) t B-.(3) '"170ft-lb " 2olb(4fl) t P1< (3) ;·,,, 170fHb

. p}'. .. 30)b.

soi.y ... -t(w) ~ Lx· 2.61 ft below o

.236.) ;\ parolle~. force sxslem_ o.ds. ~D }he le~er shown in f(g: P-2P6. Delerm1ne, Jhe ma .. , . ...1_11.:. ·1· .. . ·: · ' · · · · , , · _9n11uuo :<) pbs1 11on of l he · rea Ii · l , . •. , . . u ion .

30}b. 60}b.~lb. 40lb. ' '

J;_;_._l ;3' ! 2' t .. ~ ·· · ' . R • -30-60 HO . ..:.W ? -1·10 lbs.

£M" c .30(2) t 60(s)-: 20(1) t (11) +o . :;~ 660 lb- (f. .

MA"'R(dfr.A)

dfr.A ., .6,6p/110 ., 6ft. from A . . \ ' -~

~~7.) Determine }he re~ulton) o( }he four- ~orallel forces be~ ing . on }he rocker orm of fi'g . P - 237 . ·

~· R • ' so +<oho -~o· • cso 10:

1 ~Mo·." - .so(6) t "f.Q(~)- 20(3 ) t 60(a) "" . .:r20.0 fl - I b (cw) .

~Mo= R (d fr. o) c;Jfr.O ... 200~o "' -4-rl . -right of 0 .

' :zaa.~ The beam J\ 13 in fig. P- 238 suppods o I d h. h . varies from on 1nferi.s-ity ~f· .so IQ. per n . ~o 20:~b· w. ~ ·. Calculate the mo ni lude ~ · 1 · · · • P~ 1

• : ' · 11 · t · R 9 , "') posi JIOn of the re6ultonl food. •r · eplo~ lhe given loading by 0 uniform! d . 1 'b ·t·--' load of .solb . n

1 . .. y 1s r 1 u c;Y·

per', . pus o - triongulor load . c:-, vory1n9 rrorn

" .. ·· .~i

'? 'I

'' .11 ' ,,

d:~ 3686+·~ ... . g,60 ·fl . . aa:10. . #

\_.i' .;•· . i" · ... ! ,.2-te-) The shoded oreo in fig . P- 2-.0 represents a ~feel ploto

~: . ~~E!~.!;'~:~~ ~:~:E~~~::~~~!0~~· ~[. or191nol ·plate minus the weight of fhe moterrdl cut a·,,/alj . ~' Rep~'Sent the original weight of ·the plole . by Cl downW'Ord

r,.~· force oc~inq of the center of the 1ox1+ in . r.ep.tongle . P-.epre '' sent thewelght ~f the maleriol cut owoy by on .upward foree

'acting · ot the center or the circle .. l ocate th~ .po.sit1i:?n of. the

· resultant of these two forces with re.gpeot to the left edg9

12 -

•• . r.

~.I. 'l

' ,.

·;;• '1' ; . . ·• ' '

' ·" . '· . - ; .

, 241.) .Locate I~ 'omouril ~- pOGitiori:.·of '~ resuiton'l or.'the · : · loode acting on ,lh'e fink ., .. Ir.us~ -shO.:....n in fi9 . p'- 2-"A · · · '"!OOlb • • • -r.1 • I

R= 2oot·300t_m•ao0t2oot 2000

R ... . a-+00 .lb ( djreqfe.d downwqrd} £Mc= R'(s)

8' ~· ''\ .200t aoo t "'fOO f~ t 2oo

200011:1 R' "'· 1...00 Jb ( d 1reofe? do-Nnword)

~M.c • 1"!00(5) "' 7ood Q- lb . ,

Rdc = :i:Mc - de -= 7000/_,;400 "' 2.06 fl r-ight of . C or

d .. 12.'06 o. right or!,\ . .

24 2.), Find 1he values of P ~ f so lha} Jhe four fOrces· shoWrl 1n f1q. P-2+2 pro~uce on upvvord resultan,l of 300 lb ociing · at + fl. from the left ~nd of the bar.

F ~lb

::_ . ! ~:t ~Mp= -F(~) ~ zoo(s) 1' 100(~) • ·aoo(£)

f ~ 200 lb ( do1Nnvvord) . ~MF"' aoo(1) ~ P(3) - ·1oo(s) - 2oo(e)

P .. -tOO lb (upward)

. . -;~

:•' .~· · . ·:·{~

· '· ·~0 . , :,i{ I ·'II

. l ;·~;1!;~

"· .~; . -

'.·.· ·

" ,·\

....... :

,. .. -.

: ~ I

. '

2~7.) The Jh~ -step-pulley . shown in Fig. P- 2+7 is subj~ctecl . to the given' couples . Compute l~e vo)ue or . )he resu I tonl cou p)e. /\)so deJ~inine lhe forces ocHng ot lhe rim of them~

.l

" " ·': •' ~ .··. ' · ,J

; .· ' '

t-dle · puHey lhQl ' ore req~ire~ fo. Pci10~0e0Jheg,l'ven ·.sys}~:; sol'n': ~c .. ·ao(1z1.-+t9(16).,.6d(e) . ' '

= '~760' itf :ri> (.;.im~ns .eou11fi::t-.cw} , · . . · .:.;,,;.::'--:::::: __ . ·· ·f= :~ 76%~ . = ·. 6a.a lb:(Cdupl~of

0

'thls . m~c.n .. .. ''"' requ1f-ed) ... , .

·:. . \ . ~;

. ' . \ ' ' , ' . " ' 4011:1 :

~'··: ~: '." .2~8.) To dose o ga~e ".'91ve· ·il ·1'.s necessary R:iex~rl tw~ :for...: ~\;r: . : :, .. ~. :of 60 lb df opposi}e ' sicie'G' of o hcin~w~eel 3f1 irf dio\:h~.:: · ;.~ ~> · . fer. Th~ugh '_ on·pccident l.he ·wheel fa pro~ :~ '1he V.ol~e .;· )'_' .~ must · b~ closed . by lhrusti'ng . o bar' }hrough ·a s/o} j~· lhe

•• :1 yqlve . .S}eni ~· exe.:--h"ng 0 (Oree 4 ff. ou} rr-6.~ lhe cenle~:: ; ._Delerrrune ·}he_ for(je · r:-equJ~d ~ cir.ow,, a f~· booj d.r~grom · ·of the· b.or'. · · 1

i.·.

6()1b .

f '.'-tlqnq w·~~I ·a ~t. in di~.melei. ·~ .

. ·t''·'\. I,.">?.:

" ...

. ' .

,4P" 60(a)

P="48 lb: ..

249.) f'.9 · ~-21:9 re.pr.esen ls }he bp vi.ew of d spe.00· · ~u­. Ger v.:h101J I I>; .geared for 0 ,. fo.lJI"' fO One rdduoffQf\ in S~ ,

The ~or:-que mptJ} at the hor1:z.onlol shoH C is · 100 l,b.-N.T forque output of }he /;lorizon}ol shaft O; because of' .the s-·

. peed red~d1:on , is "'!CO lb ~ f't . Corr:ipu te }he lorq_ue .reoct1on · o) r~~ mounting bo)t,s /\.~ B h9ldlncrthe redtJCf:3r ro lhe

floor. ·Hird: The torque reodrQn is caused by the unbci "­lonced . torque~ whioh is o ; coi:;p le. · "

,.; .• .. \ ·' I

15\

c

C"'30R (-100-100) ft-lb -(30/i2)R

B"' 120 lb d fr·eded verJ ica lly

up ot /\ ~ down .o t B.

2so.) The con ti lever truss shown in Fig. P- 250 corries o vertical load of 2400lb . The }rusg '1s (;uppor)ed by' bea­rings at /\ ~ 13 which. exerl the fOrces ;\ v , /\ h. ~ Bh . The four forces shown con1 1tu~e two couples which must hove opp:>s'ite momen~ effects }o prevenf movement or lhe truss .

Determine }he mo9n·1tude or the .supporti ng forces.

S . Bh 1n order to const itute o couple

Av ., 2'foo lb (upword)

2400(6) - Bh (4)

Bh .. Ah "' 3600 lb 24001b Av.

2-51.) /\ vert.1'c6)1

force Pol /\ ~ ono}her vedi~I · F 'at f3 111

Fig. P- 2s1 pr;'Odvce .o resu liont of' 100 lb down at D ") a Geufl•oPolOC,..,;.,ise couple c of 200 lb-0 . Find the magnitude·

~ d;·reo-tion of forces P ~ F .

G~lb·rt. tr "' ~~

4' I 3 ' .. :::r' R-1001b

(2oolb-H

IP +' J R•1oo lb

~Me., 100(1) =-200 .t P(3)

A 3'

p = 300 lb. ( downward)

~MA 3-200 ""'100 ( +) ~_f (a) .f ~ 200 lb (upward)

l=l-): + 2.VO....P(~) p -~ ') 11{)

16

f B f ·

.tM.) A kTc.e ..y.;fein oonsisls of o clockwise couple of .... solb-A plus o 2"fO JJ-foroe dircded up to lhe right }hrough lhe orqn of X ~ Y o-xes al -&,. = 30•_ ~ lhe given eys}eni ·by on equovolenl single f'orce ~ cvmpute . lhe inlercepls a ils line of ocJ ion wilh . lhe x ~ Y oxes .. ·~ ·

R" = F,. ~ 240 cos~·

= 207 .. 85 I> .. (to lh6 righl) Ry= 'Y = 2-ta .sin 30.

- 120 (u~)

M -C"' F.,, i:Y i.y = ~ - 2.31 fl . obcNe 0

2>Z85

ix =~ - .... rl . lell of o 120

2Sa-? In Fiq- P- 253 o G)IGfem of fbrceG roducos loo downward ver li'Co l fOroes or -t«> I> lhrough A plus o oounlercJoc)(wise <XJUple of im IJ-ll. De~ioe }h6 snJe fON;e, Jhol w~ll pro­

~ on equiVolenl effecl _ ·

0

!:----;-- - - -,A 1- II

l

I • 12 I u x

R-F= 100 I> downward

Mo:-MR. .4oo(-4-) - 800 = 100X

x =2n right of o .

254.) Rework Prob- 253 if the .sysfem reduces to 0 left ­

ward honzanlol fo~ of~ lb )~ pain} I\ plus 0

cfodtwiGe couple of 750 Jb - ff ...

R = F = .300 lb lo )he. left MR "' Wlo

300y =.300(2.) - 750

y = - a.s fl ('"'> meonG belovv o)

256 .. ) A short compreGSion member corr-ies on eccefl tric load P = 200 lb 51)uoled 2 in .. from lhe o'll:is of the member:

OG shown in fig_ P-1255 .. In strength of moler1ols a is- ,er:.,

17

rnec:I· lho) lhe inlernal slre~es ore d etermined from the equivo\enl oxiol loo<:J ~ covple inlo which P rnoy be re­solved . Oekrmine )his equivalent oxiol load ~ coupl.e .

:ioo ~· P~wo)b. . w '1lh lhe oddi\ion of o pair a \ opposHe

200 axial )oods eoch aqvo) lo 200 1b ) we ge) f ,. 200 lb ( dOWf\V"Ord) c - ~oo(~) 9 400 lb-in cw

2s6.) /\ ver )icol ·short /\B ii> .s ft long ~ bolted ro o r igid sup­par) o) its lower end /\ . /\l i)s upper end F3 ·,s ~Hoched oho· r iz.ontol bar f3C which is 2rt lonq. A.i !he end or C ·, ~ opp\it:d

0 force P - 1BO lb. Forc;e P is perpePdioulor lo lhe plor'le con­

)ain;n9 poinlS /\' 13, ~ c . Delermine lhe lwisl inq ef'fed or p on )he -G'haft AB ~ lhe bending effe/;l al point !>. .

Tvvisling effect " 100(2) ;: 360 fl.- lb (1

Bend1'ng effect - 1So(s) ==goo fl-lb

2s1.) Repioce rhe system ~r forces ading on ·u-ie frorne in fig . p-257 by o rosullon) R o) /\ ~ o c ouple' od;ng hor•'zontolly

lhrovgh /3 ~ C. ~ 2011> I

f '-, I\

1 - I 3' l

R "' ~F,. ao- 20-60 .R = - .so lb (~) meons downward)

~Ma =C 60(4)-20(1),. f ( 2)

F "' 110 lb thru J3 ~ C os shown

18

f

298~ Replace }he . Gy~lem or forces .shown in Fi'g· P -2ss by an

~qu1volenl force lhrQIJgh 0 \, o covple octinq lhrough I\ ftJ B. Solve if lhe forces of lhe couple are (a) horizontal ond (b.)

verlicol.

f

I

.' I •

:141.4 lb £F,..,. 141.-+(1/a) t 22+(2/.JB)-a61 (o/Ji3) 2!fy. G 100.()fj !bOo lhe righl) .ZFy = 1:+1 .+ (1/a) - 22-+(1/-19) - .361 (o/$6) ~Fy = - 300. 56 (<-> meons downword)

/ R 0 ·/(100.01)2 + (-.aoo.~6).:z r--;-+--4c---1--J 8

/ 1' 1' ........_ ......._ • 316.79 lb (down lo the right)

361 lb o ;n4ton-&,. .. C.Fy /.:tF>< ei<· ton' 300.s6/joo.og

-e-,. = 71. 58. /\

p

0

p

F

o.) C = =EMo 3F = 361(Q/.Jj3)(1) t361(sA13)(1) t141 .4

(1/.a)(2)-141.-f (1/-f2)(3}- 224(2/,ra)(1)

- ~2+(V..rs)(1)

3f c 100.1

R E ""33.37 lb

b) 4P = 1001.1

P"' ·2S.03 lb .

:60·) T~e effect of' o cerloin non·concurren l force coyslem .1.; ~oftned by lhe fol1ow1'ng dolo : :EX - +golb,~Y .. -601b,ond

£Mo ~ .360 lb ft ~vnlercloc'kwise . Determine lhe poin) 0 ) w/c lhe resultonl 1nlerseds the Y O'X is .

0

zMo = zf,. ly . i. y = 3601'90 .. 4 fl below 0

261.) In a cerloin non -concurred force. Gystem j} is found lhot ~x .. - BO lb , ~Y = + 160 lb, ~ 2'.Mo" -+BO lb -rt in o coun lerclock­w1'r;e sense. Oererm.1ne The poinl' o) which lhe resultant in ­lersecls }he X o"Xis.

~ ~Mo ~ zFy Lx

Lx "° +00/160 ~ r ighl or Origin

19

I \

-l

l ~

,i·;l!J ' '

' 1 ·1

.262~ Determine cornplelely lhe resuHon) of lhe forces oc-1ing on lhe sl~p pu11ey .shOwn in fig. ~-.262 .'

1750 ::fr>' · .. 7!50 cosso· i 2so · · • 899.SZ lb(lo )'tie .right)

£.fy G 1so s1n30· -12so , . ..,,_975 lb (- means down'/llO

R .. ~f,.11 t ~Fya.

= (099.s2)a • (-a1s)~ ·

,B = .125+- SQ lb <)OWn Jo righl

-&.,. " lon-1 87.S/899.152 = .t;t.12.1° Rd .. ~Mo• 1so(1.25)-12.so(o.s)- 2so(1.!Zs)

Rd .. o :. d 0 0 j so R po~Ges lh~gh the oxle

.268.) Determine l he resuHonr of lhe force system shown in

f ig. P-~65 ~ '1ls "~ y int ercepts

.~ l:y .£fl< u300.s jnao t361(ll/.J1a)-n+(o/,,f5)

~ 1+g.g g_ (to lhe..right)

....+--l-'',..._,,_-1---''""=- __ _ x_ .tfy " 300 ~ao + ~~(1/./6) - a61 (a/..Jiit) = .sg.61 ( upward)

..A.+----'f-4---1-'---' 224 lb R"' ./~F'/.'2 t~Fy"" 1'

361 lb .. ./(11·9,9)'1 t (.s9.61)"-... 161 . 9161!?... (up to }he r ight)

tone-)< = ~ry/.:EF>< - .s9 .61,/1+9.9

-(7)< ~ fan-I .!?J9 .61f14g,9

-B-x " 21.69•

~Mo "' 300 sin .30 (2) - 2~4 (1!~)(2 )-361(2/-113}(1)

= - 100.6 Q~ lb (-meons Coun~er CW)

!.)( - 100.6/.sg.61

" 1.67 ft. right of O

~.Y .. 100.6/1+9·9

= o.67 fl below o

20

26+.) Compleldy dderrnine )he resuHon) wi)h respecl lo pl . 0 of 11~ force sy'1lem shown in Fig. p- 26+.

1+1:4Jb

,. 0

· .ifi<" 14-1.4(Y.JI) t 300 sin6Q i 260~3"$9) - 240 sin.30 • +79.]9 lb.

£F:1" 1-+1.4 ('/.Jl!) t260(M3) -t240 cosoao· -3ooc.os60° • 2s1. 03 lb.

R " ./:~.F,. 1 t ~Fy¢

·l(+1g.7g)f2 t (257·0:3)'2

B = s.+4.68 lb(up ~o lhe r19hl)

ton ex =· :i.fy /:E Fx

-&)(,., }on-1 2s1.0.3/419.7g

-&x ... 2B.2!5 • .

-LMo"' 1.+v1-(1/,/I')(a)t sooc.os60-C4) t .300 .i;in60(+) t 260 (1%3~ - 260(.!V.,Ji3)(+) = 1779.19 ft - lb cw

~R d"" ~Mo/R· "' 177g.1g • 3 .27 ft.

5+4.68 > 0

265.) Cornpule l~e resuHon) of lhe three forces shown in fig P-

266, Locale ils 1nlerseclion w'1lh lhe X 'rs Y oxe.s . Jy I

,. '

~f7'., 390 (1~3) t 722(o/-1i3) ~ (sinao) - 810.47 lo lhe righl

£.fy"' aoo(-7(a) - 1~(iz/..fi§)~aoo~· t' I .__.._-l._.L.-,~'-1.-.J.- _ lL. • -.s10 . .s \-moons downward).

o 1n lb R .. ./;EF.,,.a + ~Fy¢ tor,&J< ,• .:Fy/~Fx •/(810.1-7)~+ (-s1o.3)2

-e-.,,. = lon-1 510.3/810.1'7 R • 957. Q7 down lo right

-Er)l = 3!2.19°

.£Mo= 390(12Aa)(2)-3go(•Aa)(s)+ 122(2/,ffe)(-+)-300sin 30(a)

~Mo= 1121. 97 0-lb CW . ·

,~ l.,c = 1121.97 .. ~.2 n . righl o) 0 Ly . . 9 10·3 -

-&,, • o tic Ly .:; 1121 .97 -= 1.3 8Q. obove Q R 910.47

21

111;1;· r '~. 1; I 1.. I

rh,i 1''1 ['

1j1:;' I u•

•I' h": ji;,1'.

l~;l I

. •

iiii·· .,, 11>: 1 !!;hi ,i,.I •! ~.

\l~'i I 1111·!:

'1 ~~ · ·' ·1iW1 }

1l1j 1 •••

II' ;:1~ I' ·1111·•' I~ I

;j '

il111~ ~I

l:l·ij

·i1!J 11

1

11 1 I

: !j~ I:~

266) Derer-mine lhe resuHont of }he lhree fOrces ochnq on )he dom shown in fig . P-266, ~ locale "1ls intersedron w"1lh lhe base 1416. for good design , lhis 1nler~ec1ion should occur w/i"n

lhe middle third orlhe base. Does ·,~ ?

--=-==--=--=-- - -- -p ... 10,ooolb.

6 '

l\i----

"$-~ = ton·1 Z.Fy/.%.f-x

.lfi< " 10.000 - 6000 cos 30.

= 4803.9{1 lb(lo lhe right) ,

~fy• -24000 -6000 sin30·

::: -27000(- meons clown­""ord)

a· 11 a R .. ~f,. t~Fy

=/(:+B03.65)a. •(-rnooo)a.

R = 2742+.02 Jb.(down · to the ~~-:=:u---.-..-rz-~

r!ght) ... lpn-1 27000/-400.3.85

-e->C ~ 79.91. i.~ = -24000(11) + 10000(6)- booo(+)

....:: -22~000 n -: lb (- means ccw)

~Ma"' zfy .,c b xb • 22Q000~700o ""' BA·+ ft .( from t.he lefl of f3 ~

berice w 1lhin lhe middle third of lhe bose)

267.) . The Howe roof lruss .shown ii"\ fig . P - 207 carries !he g iven loods . The wind loo.ds ore ~rpendiculor to the in­clined rnernbers . Delerrn"1ne lhe mognilude of the resu l -ton l. its ·incli"nolion w·1lh the hor1.:zontol . ~ wher'e ·,\ inter-

.!?ecls /\B.

3000lb. '1000)b.1000lb

.if,. - 2000 t 4490(1/..s&J - "'t()03 • .S2 lb

~Fy .. -~ ~ 2000 - 10QO - 4490( 2A!f) _:.:,12Q07_.03JQ.(- meons down ­

""'orcl)

Rll • zF-x2. + ~F1<Z.

=./(4003.92)'! .. (-10007 . 03)~ ~D ..J077~_.17--1Q.(clown lo the

right)

22

-(}-JC "' ton·1 Z.fy/zf5c - · ion·1 10001.os/-t00a:e2 -e-,. .. 68.2·

~MA "' ++BO (1/~)(.s) t +4-90 ( o/.f6)(10) t '1000 (s) t 30006~) t '10o0('l0) t 1ooo(ao) .. 160087. 7'1 o.- lb.

x "' 160097· 7'1/10007. 0.3 "' 16:0 n. ~ight of"

268.) The resu I ton l of four forces, of which ihree 'ore Ghowri.

in fig . P- ~68, i6 ot a couple of 4SO lb-ft : clockwJ'ge. In sen.Se.

ff "eooh squore i'" 1 fl. on a sido, determine -the (Ourth force

completely .

!Y 1101b I

/ 0 /

180 b

I I

1~0tb

_L.

~Fx - R>t

F11t110 "'1so(3/s) • o f11 " - '100 lb . (-meonG to the le<l)

£F.,. ·Ry F7 - HW t 1so(-t/s) = o

~ therefore

r . f)( - '100 lb to the left

c~~Mo

+so· -M,. + 110(-i.) + 120(2)

Mr. • 200 0-lb (ccw) MF• Fd

d ~ 200/~oo • 1 ff . obove 0

269.) Repeot Prob. !268 ·,r the resuttont ic 390 lb d ireoied dONn to the right oi o ~lope of' .s to 12 possing through point /\ . Al­.so 'de~e<'mine the x ~ y intercepfo of the. missing fo~ f .

•"/ 110 I 15()

I

I

,' ~h •zfl'

390(1ajia) -·r,. .t 110 t 1so(3&)

1>/~~ .:Efx • 160lb Uo lho ri'gh~.) , IT•""' Ry · zfy

x li)390 -:390('5/ia) • Fy -120 +1so(+ls)

1M - - - ~ f):' • -1so (downword)

R .. F 2. .. F = 2::FJC + ~fy~ - (160)~ + (-150)~

F c 219, 32 lb (dawn ~o l"'ight .)

23

I

-Oj = -S.15-

Mtt · ~~ 390{.Yia){3) + 3'J!0(1~)(2) =110(+) t 120(2) . t MF

Mr~ 1'"10~ ON

b. .,. ....,oJ{eo - 3.~l7 n. f"i9h• or o Ly> "'90/460 - . :a.067 fl . ()bOV'S 0

210.) The t~ forc:;es -shown· in fig. P-no ore- roqoired to c;x:uJGe a ho"'::iioofol ~tont ocfing fhrouqh point A- If F = .

316 lb. doterm'ine the volues of P,, T. Hii1t = /\pply MR""'~ io defer~ R .. then MR - ~IW\c. to find P. ~ f.ilolly e;lhcr Mrt­.:Oto or Ry - ~Y to c.orr-pufe T.

1:IF

! a

' 'n \P

£a.fe -~ -316(1/.t;o)(1)t316(•1/.,it0)(2) s R.(1)

R .. 4t"-6+ lb.(to the right.)

M1t · ~Mc 4f'J").6+(3) c .316f'Ai>)(1 ) - 3"16(o/"1o)(1)

t P(2/"5)(-t) p- .... M.82 lb

t.1~ ·· ~Mo

199.ff(a) r -T (•Ae)( ... )" 316("4io)(i;1) -t 3'16 ("Alo)( 1)

T- - 22s.1e lb

or) The .three force6 in Fig P-n' ere.of& o vcrtiool reca.oliontoc­fing flu-ough pc)Wlt f\ _If T j40 1<nown to be 3611b~ pom~fe the va­

lu86 of r .,,. p .

.:Et.41; ~MR

-36~ ($)(c)t a6l (~X+) = R (~) R.,,. .:'f00.-+9 lb (do.Nrlvvord)

MR. = £til'le

..,00.<f.9 fi) = F (3/9){!z)- r(1410¥1) F· - ~sa. n lb

~l ="1~

P(24s)(-t)• (153.1t9) ("'k.o)6) - 1Z.53 -~(~)(1) ... -f00.+9 (+) p - 419~ -.s+ lb .

Chopter 3

E.quilibrit.nl of force Systems

25

'

r \

I

'

l.'

304:) Th& cylinder C ir" fig . P- .ao2 ""ei9hs 1ooolb. Orow Cl fBO

of cyl1'nder C ' of rod /\B.

Cir ~ + Wc;•1oOO lb

3'

Ah -4-· Av

303) 1he un; forrn rod II"\. .F i9 . P- aoa wei911G 4!20 lb ~ hos Its Gel'ler or _grovity o t 6 . Orow a FBO of '\he rod . lie_glec1 the. ih101<.ne5S of the rod ~ o65ume all COr"toot ,s:urfoces to be,srroo\h.

ao+) The fr'Ome ~hown i" fi9 . P- .304- Is supported '" pivots at

J\~13 .· Eooh member w~1ghs .so .lb per fl Drow .. a ' FBD ofeoch

l'l'&mber ·. /\

8 v

BOS .) ;\ 600 lb lood is suppo""ted by a cable which runs o -ver o pulley~ iG fos\ened to !he bor OE in f ig . P-ao.s. Oro o fBD of bors /IC ~OE ~ of the pul ley . Ass~mer oil hin­

ges to be .smooth ~ ne_gleoi t he ws1c;ilh~ or eool\.

bor.

26

D

Ev

goo.) Drow a FBD of pull--~ "' 11.. D ._. f th b /\ D . -,~ 1.. ""> J o e or shown '" Fj9

P-306. /\ss.urn6 oll hin9es to be smooth ~ ne_gleot the 'NC1ght of ~h bor.

soa) The coble~ boom shown in fig . P-aoe suppor+ 0 food of 600lb - Deterrn1'ne the lens.de rorce T ,·n the coble ~ tho compressive force C in the boom .

t Method I (u~ing horizontol 'vertical Axes)

~Fh •o

Tcosao· • Ccoe+s~ CD

6('.)0lb £f., - 0

Tson'.30· t Cs1M·s· - 600 © svbst. eq 1 io 2 (~uote 1 in ~enns of" T)

T ~i~30• t- (i'cos.ao/c.os4"')(sin4s') = 600

I= 4.39 . Q.3 lb.

C - 439.2.3 (cos3o•Vcos-+S·

cc 5 37 ,.2,!.S JP;.

1< Method I(using rdotion Ol'es)

T c ~fv-O : C.sin7$> • 6~.s;noo· - c - .S37.9't.5 lb

27

J l

l I I

l I

''·

I'. , . . , "

.ifh =O: T"' 600 c:Os 60' t C GOG 75• .. 600 cot> 60 t .S.37· 94.5 (cos 76.)

T "' ~Cbg.3 lb .

•Meihod JL (uSing force triongle)

~.) /\ cylinder. weighing "400 lb. i" held 09a'1nGt a smoot h in­c line by mOQnS of' the wo'19hi less rod A0 Jn fi'g. P·309 · Oet · pt, ti ~-cried on the.cyhdcr: 8~

Method l(v.:;lng ti'" Ol'~) FBO of' Cyhhder B

. 4. zFh"' 0: Pcos·~· c ,Hc,osl91i

p: N co:--ar~~· ©

P.sin2S• ~ Ii sin3,.• • 4'00 @

_,bc;t . , to i

-4001b

~fv •O : · /1~n60· • -tOo & 1""5"

N ~.+1s .60S tb

>lf:f".i'l •Cl : p • -t00 COG6'f.• +rt c.05 60"

p "'~O C.0566° + "'l-18•60& COG60'

p"' 378 .36 lb.

Method ]I[ ( tJ&in<;! fol"Ce T r.ongle)

~~ v [N c;or..3s• (r.1n lls·j/~a,.· i Hr.in as•

p=.378-35 lb ..

.ti = "'1-18. 60.S lb· - ...00 - N -11s.~9s lb

p ·= H co~-as/co.s2&0

= -HB·60S(cosaGY006llt0'

_E - 370.36 lb.,

a10~ /\ :soo lb b°" is helc:I of rest on a smooih plone by a force P anclinod o~ on angle f7 w/ the plone as .shown in f iq. P - 310. If tr"' -ts 0 c:let. the volue of P "'> the normol pres:isu~ H exerted by

1he plane ' "f 28

'I

FBO of the block

p

Method I . (us;ng tt'"-.V oKe~) ~f.,aO

300 • Nsin60• - Psin+s' © zfh·O

Pcos11i - Hco560.

P = H cos60;/cos1s" ® .subst. 2 in 1

Method C (Uoing Roioled A¥."5) 300 - tfsin60· - Hcos~/co~a· rJ N,; ~.907 lb

p

30C?lb

.:Efv = 0 : H &in:+s• • :SOO sin 7s •

H • -to<J, 807 lb

~fh 00 : p: t1cos45· - 300COS7s'

P = 'lo9.eo1(C064!1>·)- 3ooc.oco"7s"

p "' 1212. 1.312 lb .

p .. 212,1.32 lb

Me1°hod I ( using Force Trion9 le)

~ _aoo z _rt . _ P_, !J()D ~ .s1n-os" ,.,nio.i 6IO;;id

P N"' 409.007 lb .

p = 212.132 lb .

311.) If the value or P in Fig . P-a10 i6 180 lb, determine fho an9 1e

-e- ot w/c . ... ~ mu"t . be 1nchned with. the. smooth plone to hold the 300 lb t)())( in e<:juilibriu111 .

Resolving the fon:-os to 'dG equivolenf fOrce t::.,

<><. : 56.+4·

:. -e- " '.3.3.ss ·

29

,I~ • ,l i~tl

lh

aHi.) . Ootermine the magnitudes of P ~ f nece$SOry to keep the concurrent force c;y•lem , shown 1n fag. P - 312 in equ11ibrium .

~f',,. • .O

.aoo,i;1n60· t .Ps1n oo· ~ 200.s1fl1os•

p- -133.24!1 lb

~fhTQ )

f • aoo cos w· t PC£>S ao'- 2oococo1os· ·

f • :aoocos6o' -13a,2"1-.s (cosao~- :zoo cos1os'

F ~ 86 . .s7 lb.

B1a.) · f 1g . .:i1a l"epresents tte concur"rent force sysi em 00Hr"1C3 ot Cl

joint . of a bridge truss , Determine the voluos or P ~ E . to mo·1n­

toin eq1iilibr'1um of the forces.

us;ng Rotated ·A -,.es 'Method

.;:Efv • O

;;ioosin1s• t F s ·1n7s' = -+00 s1n•5° + 200 sin10!l'

· F =··412.435 lb

.:£Fh • O

pt 400 COG 4'!;;0 ~ 200CCG .,~· i 300ccs is' t FCDSJS

p - 165-4++ lb .

314) lhe . f';ve fo~s shown if'\ Fig. P-31"1". ore In equi librium . O:im·

pute the volues of P ?», F ·

p

F

zFv ao

f = - 126. 33 lb

~fh "' O

p t AK>O cos 7s0 t tl<X>COG6<>' - f cos:so' i aoo o:>sao' ·

.s;u b~t;t ... te. the vol'-'e. of' F

p+.,ooco&-;is" t aooo:x.60• •-126. 33 (~ao') + ::100 oosao •

P "'" - .sa.129 IV .

30

:s1&.) The · 3':'° lb f ~r~ ~ the 4()() I b force shown in fi'g. p-315 ore to be held 1n equil1bnurn by a third fo.....,..~ F +· t 1. ·...... oc ang o on un,..no-wn a119le -e- with the horizontal. Determine the values off~&;

• . <fOOlb

·~oolb 3d ·o .. i& • Q.

F

:E.Fv •O

400sin:ao' = fs1ne- (D ~-o

300c<KJO~ao· ~ FcoG~ © .subst . CD to eq. ®

3QO .. 4<X> cos ao' +(.oteo s1nao' ) /cos&-) sine- l .

-e- '" -76.935° ~ 76,935 belew x-axis

F "' 400.sinao" - 400-•n ~· · sine-

F = 20S.31S lb

316.) Determine the values of fhe ongles O\" ~fr 60 that the for-ces shown in F1'g . P - 316 will be in e.quilibrium . ·

..Olb .. ~ Wlb~lb

by cosine law

w« ~ :ao.2 t-fe"- 2 ( ao)(40) C06&­

cos -&- "' c>.a1s -e- e cos-1 o.07s

-(7 "28·96° "' Q9·

•by sin low

20 - .30 ~-~

.s1no<. '"[3o (sin 29°))/ao Sino<. <= 0. 7272.

o( " 46.6.S 0

31

II

I

,/

' • 't1 1

/ a11.) The system of IW*ted c;Ot-ds show" ~ ".'~- P-317 support the indicated 'l"e1ghts. Compute the ~le forc.e in CXJOh cord ·

~fv •O

0G1n-t6·, a oo1'.'C61n60·

e • 3()()t- 100(s1n60•) G•n+s·

a =- 914.162 1b.

£Fti-=0

A " ccos60- • ecoG+S· - JtOO(CJ:JG6o•) t 91+.162(coos'")

I\ k 946 . .f-1 lb-

319.) Three bars , hi119ed at A "°' D '-> pil"W'l6d ot /3 "C os ~ in f ig . p-318, form a fwr-linK. tne(lham&m. Determine the.value

of .p that w i II prevent rnotion .

0

usi~ ~1ot~ 1v<es Method

~-· ~fy~O

/\S'llUJ• = ~ san4S•

/\ • 163. 1.!"9 lb

.£Fh = 0 sut>sJ. A

c •'UXJ ~16- t /\C066(>.

c = ~23-07 lb.

32

J

y using Force. T l"'1on91e

~ .·/ \

by . sin~ law

~- C ,. A srn60' iiio--~· .etn.ote'

c c '2'23-07 /b

c y br s ine law

_P __ " C & _Q__ 61n 7s • . sin,+s• .stn60~

P." !l!l3. 07 ( s 1r175-) s 1n45•

p - 3£>+. 719 lb

£Fv c Q

Csm1s · "Pco6 .,..5 •

.P c (lfl3 ,07. & In 76 ° cos ... s·

p " 30A-. 719 lb

Problem ~19 so•..ition .

~Fh " 0

400 COG (;)- ~ 200

ff "60°

~fv =o H t "I"()() Sth $- = soo

H ~ S00 - 400 &1n60• ·

H = 45.:1.6 ~ 4-s4 /b .

a19.) Corcts are loop--' , ~ around a srnoll ~pace,.. . cylinder& e.aoh · h ' · seporofrng fwo

weig ing 400 lb ~ poss os sL- .. . F ' p-319 , over r. · t ' rXJV<n 1n 1gure

1 rrc 1onleGs pu lley5' to . ht " Determine the o~le f} L . Hl8 . I .wet9 s 01 2oolb ~ .+oolb.

""> porrno P~'Sc.Are N bei th · ders ~ the smooth ,__ . t

1 ,..._ · e cylr'1 -

11onzon o sur roce .

33

- I

.aita:) The Fink' truss shown in f ig . P-32£ jc;; ,Gupported by a rol­ler of /\ ~ a hinge .at 13 : The given loads ore riormol to tne incl 1'ned rner:n ber . Oetcrrrline the reoctlonG of /I~ B. 1(1nf : Re-

pla~ ihe ioodG by thei_r re~ultont. fOOOlb

~Me •O

60RA .. 0ooo(a1n60')(10) RA .. 4iS18 .S lb "'"46£0 1b.

z fv 0 0

Rev .t R>. = BODO sin60•

Rev • BIXl061fl60 - "t618.8

Rev • ~.309 . .+ lb

4fn•O Reh· a00o eo&60'

Reh .. 400o lb .

Re4 " Rev" +Reh" " (~309,+):z t (40oo')2

Re c/( 'l.309.~) e t ( ..coo) a Ra "' 4618 .e Jb ~ 46£0 lb

t an-<r • ~309. 4 4-000

-e- '"' ton-1 Qa09A· 4000

Re " -4620 lbs ot 36° wi th the horizontal

34

32.!l) ~ 1ross; shown in Fig . P-323 iG '°upported by h . ...,,...,.. at /I. ~ Q roller at a. ,A load f ' 0 •• ~ the reootion'° at 0 2000 lb i" applied. at C. Dot.

A~ a. D

• in-.o' 2000 lb

~M" • o 3 0 Re• £000 (cos30' )(1s) t ~000(101n3dX<fO)

Re • 65900. 76 30

Re"' 2199. 36 lb ~ 2200 lb

~fh. 0

R .... h = 2000 eos .ao• f(Ah - 1732.05 lb

~Fv •O

R>.v = Re - 'l.000 su1 ao• RAv = 1199.3e /b

R,._ 4 ·• R.Ah 1 t JV.v Q.

R-- ·./ (1 73£.os)4 t ( 1199, 35)2

R"' = 2106 lb

Taf'l& • R,..,v ~ · 1199.35

R"'h 11a~ '. os ·-e- " fa.,_, 11~. 35

1732. Os

-e- "' 34.70 •

. R ..... c ~106 lb down t o the lef'+ at -& ".3+.7.

·35

I I

I ..

1 •

11 ':

.. ,.

I. ii i 't"

l 1. ,11

'. ,. '· 1"

I

,1;

I,· 11

' 1•

.32.+) A w~I of 10 in rodius corrte~ a lood of 1000 lb, ~s showfl in Fi<;J ,p.:32-4-.. (a.) Determine the hori:z:o.ntal force P applied ot ~he center which 1S necessary fo siort the wheel over the. s-tn.

block.. A lso° find the reoolion ot the block. . (b) ·,r the force P moy be inc\inecl ai oriy ongle .,...;1th \he hor·1zontol, de\ermine the rri1nirnvrn volue of P to start the '<"heel over tre block.; the

. ong\e· thot P makes w1 \h -the horizontal ; ~ the reoctlon ot

the biooK.

s in&-= a/n ." -e-- 30•

o) pz Ra

zMo u.o .sP " 1000(10(;0S30')

p .:= 1732.:651' lb

zf)I no p •Ro cos::10° " 1732.o S1

Ra .. 2000\b.

b.) p i6 m'1n irnurn if i~ w i ll be ..L. to Ro

hence,

-e-" "o' ~Ma ~o

10 Pmin = 1000(10 COs3o')

Prnin " 866 lb.

~Fi< =: O

~a cos 30° = Pmin COS60.

Ro e [Prnin(c.os60·fl/ cos 30'

= [B6G(COS60°)]/cos 30'

Ro= .soo lb ·

36

3~.s.)' Determ1'ne the amount ~ direct1'on of lhe smallest fon::e P required to s tort' the whee l 1n f ig. P - 3as over block . What ;6 the readion ot the block.?

FY p, •COG<><. p

Cos13" 1.;4 Co,;71.+t r % fi " +1.+1 • b. o.6tf1.

£OOO(a)-Px(b)-Py(o)•O .sin~ • s in 71"t1" '; \-&- •tl:'11t "'

2000(1.a9) - Pws<><.(0.6t}-Ps1n<><(1.89)ro 0 • 1.&9 fl. -&~ 71."/1•

~Mo "' O

Pcos<>< (o.6+) ~ PG1no<.(1.s9) 2 .3790

(o.lrt) P(-sin.._) t o.6+ tJf< co~"< ~ 1.2.9 Pcos<>(. i 1.99 .9f. sin"< • o

~(~.G+c.o.s .... t1.99 S•n~J = o.64Ps1n~:1,99Pco9>(

dp ~ (o.6tPs1n<>< - 1.99Poos0( ) _ 0 do<. (0,,6+.COG<><f 1 . S~ S lri&) •

1,99 ¢cos°' "'0.61-y/s1 n<><,.

.s1no< COS"(

= lone<_ ~~ 0 .6t

~ • tan - 1 1.99 0°64-

o( = 11.29 · "'"' 71 . .s ·

2000 p R s1n9o' s-,-n-71-.!2-~-· .sin 19.71°

p"" 1994 lb ot 71,3° with the horizonto l Rsm9D

0

.. 12ooo(sin1s'-n•)

R .. 641.6 lo ~ 64'~ lb

37

I I I I ' , I

~

11,il,

I I

I

I

1 , tt ',

'

. : 'I,

326) The cylinders in fag . P-326 hove the indicoted welghts ~ dirnenGions . .t\sfOuming smooth contact surfaces T oeter­rri1ne the reoctions: ot /\, 8, C, 11., D on the cylinders.

FeD of .small cylinder,

f~lb

Ro

2'.fv • O

200- P.c .s1nao" .:

Pc =4-00lb

~Fh- 0

Ro = Re cor;.30•

Ro • 100cosao· Po· a6+.41 lb.

FBD of tho big cylinder,

~· ~Fv=O

'400 t Ro &111 ao• .. Re

-4<>0t "400 Ginao' • Re Re- 600lb .

~Fh•O

RA :; Re coc;ao• 'Rit. ., 400 COSl50"

Rio. • 346.41 lb.

327.) f~ P ~ F acting olonq the boq; .Shown in Fig . P- 3'27

rna•ntoin equilibr'iurn of pin /\. Determine the volue of P \.if r41;1n ... 1·, , P61t1'-~·

9 ·. 6

c ~···.::f.M(;.•O

1801b lo oheOk :

l&.(f) i;insa.1 - 100(6) t ~(12) F"' 390.1s lb

~Me"'O

1s(P)s1n6a·+ ... 100(9) - .300 (12)

P = - 1-+7.61.I lb (- meon6 compres­

sion)

Fcoso sa.1 - Pco406l5..+ - :aoo aO

S90•15 (cosse.1) - ( - 147.6.r)(c:.os6a . .+) - -aoo ., o

38

329.) Two weightless bors pinned together os s;;hown in Fi9 P-329 support o lood of 3SO lb . Determine the force P ~ F octing

respect ively olong bors AB~ AC tho! mointoins eq1iilibr;um

of p;n }---•'••~~ fon O< •...@... ; o< • 39.66 •

10

~lb

~Me·O

e' (+)fcosO<. t (2)fs1n<>< = .ssO(s) I -+ f~os 39.66)t 2 f(sm.!!&.66).. :5!!10(8)

___L_ f "' M<>.1 lb. - ~Mc-o

2.Ps1n~ • -+Pco&,B • 3&0(10)

2P(e1ns6.a1•).+ ~P(cosS<S.31•) • 3500 ·

p e 901,39 lb (tenG1on)

s119.) TwQ cyl1"nders /\' B, we83hing 1oolb ~ 2oolt> relilpecfively,

~ ~nnected by a rigid rod curv~ parallel fo Hie smooth cy­lindricol· .&urfoce .shown in fig . P-329. Deferrn1"ne the angles 0(

"-> -&- that defined the pos1t10n of equilibr1.um.

39

--&- tO( Ir 9().

e- .. 9()-«.. © ~Mo•O

-tool'cos<>< • :Joorcos&-1oocosot: aoo cos(9o-O()

100CoG"\•2oo(ctJ~·ts1n~s1noc) 100 OOG<>< • :'ZOOSl•~'tO &•nO\

~ - fOt"I"'( • ..:.:1;....00.;;__-'-~0( ~&•n«Jo0

<>< .. 26. 33' .!!+.

-e-- 90-C>(.

- 90 -26•.s3',s..i. '

-6- • 6.:1°U'.!t,e" .

_,.~32.) Oeterrnine the reoctions for the beom .shown in f1~ . P-~2.

..t'.M~, c 0 p . ~(4)-100(1,..)(9) t R1 (10)-300(16) • o

R1 = 1.seo lbs.

~fy:O

. R1 • R: " ~ t 100~-4-) t 400

R2 • .s20 l:Js.

'133·) Determine

looded with a from z:.ero

the reachol"\S R, ""'-. R2 or fhe beom . in f ig. P-333

concentroled· load of 1600 lb~ a lood varying

t '

~MFc•o(t! 12 f1 - ...ao(12) (4) • ·o

2

F1 • eoo Jbs .

~f\, -o . F1 t Fa ~ !4oo(12 )]/2 r, -fr -4<>0(12)=1/2)- soo

Fe ,1600 i>'.

of 400 lb per fl .

11 a' r' 1~' P., Re

~M11-t-O

16Ra " 16 F.2 t +F1 t 1600(a) •

16R2 "'1600(16) t S00(4) t 1600(3)

Ra • 2100 lbs .

~FvcO

R1 t Rii ~ 1600 t goo t 1600

R, = 1900 lbs.

334\.) Determine the reoci ions for !he beam loaded o6 she.vi"\ in

f i9ure P - 33"1'. -£Mitt =0

1sRa •12o(3)t16¢(6)(1a) -t 60(6)(6) --r

Rz - .sss lbs. £F.,.•0

R~ R1 1 r<'a • 120 t 60(6) t 1so(6l !l

R1 - 930-sss

R, - sn lbs .

40

:as? 'The roof fruGS in fig. P-~ is supported by 0 roller ot A~

a hinge al B . find the values of:' the reactions .

~Maco

30 R... = .500(10) t 600(~0) + 900(1s)

RA"' %6.67 lbs.

~MA"0

BoRev .. eoo(15) t 600(10) t .soo (120)

Rs ... -= 933_33 lbs. ~fh~o

Reh · o : . Re e 935.33 lbs

336 ) The conhlever beam sha.vn in· Fig. P - 336 ·,s built into ~wall 2rf . thick. so thot it resf.s qgoinsl- poin!s /\~B. The beam ·113 12'

long'!,... weigh- 1oolbperfl . A. concenfroted. food of !lC\'.JO lb is opplie!ld oi the tree encl - Compufe the reactions ~ A ~ B.

aoooJb .

101

1cclb/rf .._, 6' ...,, 8

~M,... -o 2 Re • !looo(tz) + 100(112)(6)

Re". 1s600 lbs.

I<... = Rs - 12000 - 100(12)

RA ~ 15600 - !2000 - 1200

R,.. e 12400 lbs .

337.) The upper beorn in fig . P-337 is S1Jpported by 0 reoctlo11

at Rs \.., o roller o1 I\ whion separates the upper ' fONer

bt90":15. Dete.:niine _the volueG o( the reactions .

'4<lO()lb

~

10' Ri

600/b

f 10'

+' r~

ro'

190('.)fb l +'

-t'

.:l!MRs'O

101<,.. -6C0(14) - 1900(+) "0

R"' ~ 160? lbs

~Fy=O

600 t 19CO = R,., t Rs

Ra E 9oolbs .

41

considering the lower beam.

...ooolb 4' l

%.Mit1 •O 10P4 - ·1600(1+) --+000(4). 0

R2 ... 3840 lb.

%.MRiz •O . . 10R

1 t 1600(4) - -4000(6). o

P.1. 1760 lb. .

he two n n beams shown ~ fi9. a-160 on the page 69 are :as.) T ed h~i zontolly ..,..·,th reQpeoi to each other.,,. l~od to be mov . . . CD that all three reochons p shifted to o new pos1t1on·1rR ~ ;:' then be? ttow for w·,11 p are equal . How for oporl w• 2 .?I .

be from DP P•960lb

.. 11•-· l lc~====Y===~c~l§.;~,~~lR=2~~====;1~

aR • 960

R • 3!ld-= R, • R:r =Ra

~eriri(j ihe Jower . beam. Re '( t, 12-y

zfv • O Re "'R1tRe

• 320 t320

Re ~ 6'4-0 lb

~M~ ·O

Re (y) =R2(12) y" 320(1;1

6'40

y .. 6ft.

~Mit:s "' o p(i<) .... Re (12)

P(i<) ,. &t0(1£)

960 )( ., 64-0frZ) '>' = 6'f-0(1a)

960

;>( ""en.

. . R2 ~Ra ·,s 5f1 opari · ~ oleo

p ·,s en. f'rom o.

42

1:139~ The differeniiol choin hoist shown in fig. P-339 cor)s·1stCO" of two concentric pulleys 'riqidly fostened together. The pulley

from form .t-M::> sp,-oc.kef1; for on endlesG- chain looped over them in two loope. In one loop is mounted a MO,v'qble pulley suppor­ti'!g a lood w . Ne:gleoting fr10tion, friction d6terrT1ine ihe m0-

x1rnum lood W that c.tln juGt be rc)1~eq by~ pu'll Popplied as shown. >

&Mo-0

wfa(O/~) •Wfa(d#) t P(0/12) ·

w0/4 - w44 = Po/fl w/1' (o-d) .. Po/2

W-4PD !l(D-0)

W • !2PO (o-d)

;HO.) for the Gy6 fem of pulley1;1. ~ho-Nn in rig. p -3"f0. determ'1ne

the rotio of w to p to .mointoil'l ~uilibrium . Me,g\eot O'llle rnc­t1on ~ the weights of the pulleys .

W "'3P t 3P t 3P

w • 9P

Wjp :: 9

&tt) If eooh pulley 6~n in Fi9 . P-340 .....eighs 36 lb ~ w • 12olb,

find P to rnaintoin the equilibrium . 3W1 • 36 t 7!10 3P - a6 t W.,

W1

• 252Jb . · p.,,(~6H252.)/a p. 961b.

43

l I ! 1.

I

I I I I

"

3.,.11) The wheel Joods Ot"\ o jeep ore_given in Fia· P-a .... !2 . Deter­

mine t he distonce 'JI so that the reaction of the beam a1 A ·,s

twice o' .9reot o.s \he reootion ol i3.

RA·2Re (speo1f1ed condition)

£MA •O

ISRe" 60D(x) + 1200(,.t-t)

1s Re • 900,,. + aoo -- ·<D ~r,.cO

RA t Re = 600 t ~oo

3Re c aoo :. Re= 800/.a ---© .eu~tltute '2 in 1, .

1s (e<XV'a) ~ eoo(x) t goo

. )( :: +fl.

s.+3) The weigh+ w of 0 t ro"elii;g or?ne '1c; 20 t ons aof1ng as s~n ii' f ig · P-340 . To pt"event the crane f('()m fippin~ to the r'~ht wh­en . corryi\19 0 lood p or '20 tons' Cl covn·ter w-'1.,ght ~ is used . Det. t he vol~e ~ po~H1on or~ '50 that the crone .....,·,11 remci•'"' ir. eciu"lll- ·

brium bolh when the ma,,.imurn load P ·, .. opplied \i..... when ~he I~ P is removed .

wbsF1 \ule . in !2,

1!20 t .s6? • !2'20

~ = 20 ./ans

From eq. <1),

'Qx = 1!20

Lirni Hng Conef1+ions, when P=O i R2=0

when pc20 i R1 • o .z~1 cQ ( ..,.,he,n P =O

~(><) = '20(6)

-6{1" ~ 120Tons -· - CD ~Mi:t2=0 (whe-nP-20)

~()(t5) = 20(1) t 20(10)

~x t~~') "" !l!ZO Tons--- ©

x = 110/Qo = 6 n.

44

3-16~ A boom 115 is supporfed in a hor"1zon+ol posi tlm by oh A \.... a roble wh - t'i fi 1nge . . ic ronG rom C over o smoll pulley I D . Ghown in F•e· P-346 . r~ t h . a as ho . . \..Urnpu e t e tenG•Or\ T in the coble ~ the

r1zonta l ~ verhco I oornponenls or the t. I I t th

, reoc ion 0 /\ Neg:. eo e size of the pulley ol D. .

T

0'

a'

ton-e- • BA c IZ

e- - 63.+.3"

~M" •O

100/ti

2oo(d.) 1 100.(6) - T (.sm63.+3· ) (+)

T = 279.afl. lbs . ~Fh·O

Rh • T C06'63.43°

• (1219.!!>£)(cos63.4'!3•)

Rh ~ 12e.02 ibs. · ~Fv • O

Rv t Tein 6a.+a· = 1too t 100 Rv " 300 - (a79.01Z)( .sin 6a.4:::i")

Rv -= eJ0.06 lbs. .

a•n.) R<3peot prob. 34'6 ·,r fhe coble pulls the boo An . t . rv..c.i t ' t h. ~ · t · . ,,, u 1n o ci r-- IOI'\ o w ion J IS inc lined ot ::io•· a ... -. +he . . I d

, t . I vvve hor' :zontal . The oo s · remain ver 1co . .

~M,.,, 00

.ofT • 1ZDc{_r1..)c,()l;ao~ i 1oo(6)coc;ao'

T ~ !Z.16 . .so Iba.

~fvso

Rv +TGm~ = ~00+100

Rv • 300- 216. s (sin 6 d)

Rv = 1H!.SO lbs .

s1nao• . )(/+ )( • !l'

~fh"O

ton&' • 6/+eo&.ao•

fr " 60°

Rh • T cos-&

45

Rh • 216.s co.soc•

Rh .. 100 .2s lb .

I ..

"

]I! ,i I

349.) · The frame s~ in fig . P-3"t0 ·,s i;upported in pi>10/G at A 1,.,. a. fooh m~bor,weighs .so lb perf1· c.ompute theho~1zontol ,-e.­

oc.f10!"\ oi A. ~ at the hor'rzonlol ~ verfico l oomponerih• or the

reoction ot B·

h Leoglh o\ Fo e .f8c t 6 ~ . ro: 1on.

~Me, " 0

Ah(12) = soo(_.) t (l:Jd.o) t 2o::>0(12)

Ah .. i.z...a .61 lbs.

£fh - o Ah 0 13h =12-t66.67 lbs.

~fv cO

&, • 600 t .SOO t 600 t a.oOO

B.,i = 3700 lbs.

3-49.) 1he trus s . shown in f ig. P -349 is supporied on rollers o\ /\

~a hinge at 5 .'. ·Solve for the components Ot the recidions.

6001b ~fheO

Reh - 2-t0 lbs .

.:EMe." O 2"'!-f<A I 2-40(16) .,600('1e) t ~(36)

RA .. 7-+0 lbs.

~JAt. •O

2-tRev • 600(1£) t R1-0(16) - -400(12)

Rev" 260 lbt;.

'350) Compi;ie fhe to\ol reactions a l A"-. B for the truSG sho'Nl"I in Figure

p - 3SQ.

46

~Fh"O

Rah= soo lbG.

.£Me•O 36 R-'. ... 600(so) ... 1aco(~o) - soo(120)- soo(l2o)

RA= 1oss. 71 ibs. ~Fv •O

Rev t R.-.. • ooo t 11200 + soQ

Rev "' 1121+.~e lbs.

Re .. /(.soo)11. t (1z1+.12e)" ., 12ao.79 lbs.

tan-e- .; 112M.aejaoo . " . & c 76".12•

,', Re ::112so.79 lbG up to the lef'l ot 715,12·

wn in. igure - ss1 as suppo"ted by a hi~ of /\ 361.) The beam -sho · f ' p . ~ a roller on a 1 to 2 slcpe ot /3. Oeterrn'1ne the re 11· t . at !-. ~ 8 . su on reochon;S

11z'

~M..,•O

16(Rev) • "t00(1z)

Rev:: 300lb6.

~fh .. O

R-.h·R~n •1!5o lbs .

:EMe • O

J6(~v)"' 400(4)

R....v " 100 lbs.

By Ratio ~ Propor-tion ,

Reh • _1_ Reh· 300(1/2)

P.av 2 "' 160 JbG .

Ra -..f1so)::i t (soo)2 .. 335.41-1 1bs.

0 1fJO. 21 lbs.

:asz) !'I II n . d. ' pu ey -tr ~ . 1n 1ameter ~ supporting a load or 200 lb ie mo..in-

ted at Bon a hor12onia l beom ( Fig P-3!52) ·The b . ti . . eom IG supporled by a

·~at "~ rol~9"5 al c . tie.gleoi1n9 the weight or the beom, doter-

rr11ne the reac hons ot A ~ C . ·

47

l I I I .I

I

1· ~\

f,

' Ii'

1:1

F80 of the pulley,-

A!Ma•O

T(a) .• !200(2)

. T = ~(){)Ins. .

~fv•O

R.e • JZoo-Tsin3o· a gO()- QOO( Ginao•)

·Ra· 100 Ibo.

F00 or the beom,

.w. /\ +' lRe·:,~

Pl~v ~M,+.-0

Rc(Q) " 100(+)

Re" go lbs.

~fv"O

!<,Av t Re a 100 lbs.

c

fRc

Ji..v a 100 - SO <> SO lbG .

~fh cO : R.Ah • I cos BO• RM • !200 cos:ea·

• 11:a .20 lbs.

R>. • (so)2 t(17.3.W)~ "' 190.27 lb6.

ton¢ = .50ji7:a.~o "' 16.10 •

RA • 1so. :27 lbs. o~ 16.10° up to the

. right.

:ss!!.) The forces oct"1n9 on o 1-n \enght or o dam ore shown if\ fig . P-353. The upword ground reaction varies unif'ormly from on

inte;1siiy of p1

lb/rt . ot /\ to p2 lb/11 ol 8 . Delerrnin~ P1,.,,p2 <!+.._ ol ­

so the hor1z.onla)' resistance to .slid1.ng.

p.10.000 1b

Resistance ~o s lidir19 ,

%fh • O f.: 10000 - 60CX) cos 30•

f = ...,_go3.9S lbS. ~ 4900 lbs.

~F" , 0

R "' 606o .srn :.o• t ,a"'l{)OO

- 27000 lbl;.

~t.'1a=O

P4 11>/f1 R,., = a-t0:XJ(11) t 6000( + ) - 10000 (6) x .. e.++ n.

e., 9 - 9 .++ = 0,56 n.

48

,.

Pi = R/e (1- 68/e) • ~1000 [1 - c;(o.e6)l "' n2o lbs . 19 18 "j

P~· RJ8(1t6~/5) .. '11000[1 t 6(0 . .SG)l • 1780 lbs. . 1g- 1B ·J

· 3S"f'.) Campl.lte the foto l reactions ot fl i.., 8 on the truss shown

in Fig. P-~.

20 1

10001b !looolb ::!<J()Olb

·ton.a-' 1%0 . " Y1 ) -e- = ~6.56° .::fM""O

BO Rev t ac2+o(s1nu.si>)(10) 1200(10) = ::z::zi<J(COG!26·56)(6l>) 1

:a()()0(6o) • eooa(4<>) i 1000 (~)

P.ev ~ 46:27.49 lbs.

~Fh - O i RAh "' 20CO t ~!240 sin 26. e;cS0

.. 3001 .18 lb&-.

Z.Me-.o BO RA" = Q000(10) t 1000 (6o)t12ooo(~)t aooo(21j) + 2240 (co526,66)(21>)

· t 2::z40(s1n ::Z6 • .s6)(10)

RAV .. 3376.09 lbs .

R,., • ./(?oo1 . 7s)a. t(:a:a76.09)2 .. -t.517.57 tbS.

tan~ " 3376. 09

3001 .75

-t%: tC1n ~i 3376, 09 3001. 75

~ = +e.3s0

• • R.-_ "' -+.517.57 lbs up to the ~i9h~ at -e;, .,,. 4e. 36 •

:ass.~ Determine the reaci1ons at /\ 'B on the fink. +russ .show in F•g . P -35& . Membe.rG ~ 0 ~ f6 are res ped ively perpend icu -

lor to AE ""-..BE oi their rPidpoinls .

49

I I : I l

; I

i.--- - - --60' - - - -----! ;..o· :o- BF .? 16.1vcos 26 • .s6

A D'" BF,. 19176 ft . ton & : 1~0 j {)- • 26!:16•

AC "BG = 1!:;/cos~6. s6· AC c BG : 16.77. fl .

~Ml'>cO R,.. c.os:ao· (60) = 1200(~) t aocx:>(+1.2.S)+ +DOD (16.n)

~MAtiO 60 Rev = 1200(1.s) t aooo(1a:1s) • 4000(.....,.2s)• 4000 COG!Z6·sl(tM16)

Rsv • 6 1.!\<l-.67 =>1 6130 lbs.

~fh·O Rah • -tOOO s1n26.G6° - R" s1nso•

c 4000.s1n26·~6~ - .5361·27 s '1n:ao'

'Rah = 892 .o96. fos .

~6) The canti lever I truss shOV'Jn in fig. p-3.56 i s ~up~ded by 0 hin­ged ot /\ ~ o strut 8.C . Determine the reoction ot 'A ~ B. .

1()()()1b

Re ... .346+.1 lbs.

By Resolving the forces to its equ1vo -

lenl force triongle,

By s ine low,

..&.- ~ ~ ic;1n30' s 1n90'

p_,._ "" 2000 lbs up 'lo ihe r '19 ht ol 6().

50

357.) The uniform rod . F. P

f in 19· ·357 we 'gh

o grov'1ty ot (j. D~ . J • ' s +20 lb~ hos 'ds ceflt · ermine ,he tens10 · th

i1onG ot the ~mooth r n in e coble ~ the reoc-• sur1aces of /\CJ... 8

W" ac:Jlb HQt'>45 "'\ •

/,_B~-~M~CO..!>'

2'

.o!Fh"'O : T ~ H C05-"t5. © :. T • 2s+.s6 (cos-46·)

T = 190 lb

9Nsln4a• t 6tlCOG+e'

but, sin~· =co.& 45• 2T . t 2520 • 16 Ns1n.+&•

T • 8 N cos-+s· -1260

" B ti s1n-+s" - 1260 © subsi . eq :2. in 1,

N cos+s· - 8 Ncos-t-s· -1:260

- 7Nco.s."'te• • -1260

H .. 254.56 lbs.

359.) A bor AE • . ' /' /I IS In eCj ·j'b ' fl u1 ' r 1um uncle fh orces shown in f ig p r e ocf1on or ihe ri:_,e

~' - 359 ' De.term ine P., R ' T.

-400lb

=£MA'"0

R (16) t Tcoe.~(e)-1 T~1n~(s) 'o. 600(4)

t 400(9)

16R t 8Tcose t 6T .sine- v 6000 . © ~Fv'"O

TcoG~ t R = 600

subci SCj. 2 in 1, R ~ soo - Tcose- @

16(600- Tcos&) teTcos&t6T . .

1

/ . e1n& a 6000

6~600- TqOG.36.S •) 7 t 81 COS.36·07" t 6Ts n •

T

I ,36. 87 "' 6000

= 1ffiS .71 lbs . From eq.2 ,

R ~ 600 - (128.s,71) cos .36.97•

R • 4~8.57 lbs . (upward)

51

..£fh-O

p + T GI0-0- "' -iOO p = .-f00- (1~·71)s"'::J6. 07• P "'371.1-£ lbs. (to the lef1)

/3tW.) A 1~'1 . bar of negtigibte weighf n?GlG in o hon:z:onto \ pa­sitian on ttie ~ plane s hown in Fig. P - 359. Compute the cJeonce -x ot w hieh lood T .. 1oolb should be plaoed from pt.

8 to Keef) the ~ h()ri%Ql'liol .

F~p--a59, C p-360 ... 361

~-o

By Sine low,

~= ___&!-- ,, 300 -904:s- .s.nao· .son'I06 •

JV. - ~19.61 ft>G.

Re "'155-!Z<J lbs -100x t-f1J0(9) • R,..ro$:a0•(t2)

100)( == 219..61(~"')(1~)-1800 x .. -+.92f1 -

:a60) Re ferring ;to Pr-ob· 359 , who\ volue . of I octi~ ot J< ~ 3 0.

frorn J3 wi ll ~. the bor nor-tzof"liol ? ~

;:.G 1 2 s.O'IS· .51(l 7 b .

~c - s.7sn . cos «> .. ~ . NJ - + . .39'

e,.7S I

Oo = "t.39 - ~ "' 1 . :39f~ ­m - 6 - ·1. 39 = ... 6 1 fl.

£.tf1c "' O l ( -4.61) : ~00(1.39)

T = 60 . .3 tbS .

52

0 r-rvv. 399 ~ It T ~ 30:1lb 351.) Rer~·ng t ~ . angle -e- at which the bar · 11 be . . ,,.. )( - 3 n • d~tennioe ~l-.e · i · . · w • 1nol1ned -to fL -u is 1n a .po6•t;

00 of .

1.b . • oe> honzon~ol ._ __ equ1 1 . num. wroc-u

O ( 't2(00&&tsone-) t 1. Ion / . / 1 f ion60 ., 60!,_CD60-t s 10&)\

1 t for)6l:> 'J

l151°9 Analytic 6eomei ry.

..9:91 - m(x-x,) foe- RA. ~ -o = ton(6d>(x-o) ~ .. )l fan60- . t'.D RA, .

foi-Re. -~ ...Y - 126.n& = - · ~}. ,, ) · JJ11""'° lx- 1..:GOS&

~ the coordinoles of . .-l "' •. o _y = ~

x fan60 : 12 (rose- isin<>-) - x

)l : 12(cos EHG1o<J.) 1 t ton@

from sq. It'\ ,

..Y = 12(C06& +s in&) -x - ·· eq,~. -' : nfoo w (cos~t s ine-) 1 t fon60

LMo=O·

~ 11~CDS&t&tn&) -.3~1 r 1 +-fan60 J ffiC06&-•~ - 60069-

1 •ton60

~[ <JC06& - n(C06&• s ino-)] 1 -f ton60 ]

~ 0?69- - 36(C-OE>&-t 6lrl6)

1 • ton60

33 COSi} ~ 60(~ 6 intt). 1-t ton60

53

. I

Chapter 4

II' · J ..'Analysis of Structures

11:: ,.

.i

'I· ····· I';::

. ~ 54

-4<:1~.) Joint B of lhe fruss shown 1n fig. P- -+0.1? ·1s euQjected to the fOrces e)lerted by the ~hree memberG AB, BC, ~BO. Membe · · M~ BO ore. in the some siroighf line, but /3C iG inclined of an an­gle of-& dE'.9rees with this stroigh\. line. Show that the force in fjC must be :tero . 6enel"'Olt'ze fhis result ~ then show thol the for. ces in m,smber CD,OE, r::F,Fl,tfl,'HI<., ~.JI<..' ·,s also zero.

Ans. Ai o Joint suQjected to ihe. action of three members but no oiher load,

· r H' two or the members ore coll1neor,

~a 0 11

J . the flXCe in the fh1'rd mernb~r muGi' be

~ _______.: z.er-o. ~C I! 6 I K

.• p

'fQs) Determine fhe forces in ecich bc.ir of the truss shown in Fi9 p:. -403 . Hirif: ffr-st ' determi'ne wh·1ch bqrs c.orry no load us'1n9 .

pr 1noiple devebpecl ·11"\ Prob. "402. P ce -cF-o

+ F

.

£fvc0

P"' co S1n6<:>· t OEs1rnso·

G1hce : CO"' DE·

p c QCO s 1n60°

CO 00 o .e,77p =OE - C

co =ao "'o.!!!J77P

~Fh~O

BF =- BDO?S 60•

J3F ~ 0.577 P cos 60 •

f3f ... 0.577 p (0.5)

BF" O. !l.99P - T

55

I l

·I \ I

·1

40+) Oeterrroine the f()l"C96 1n the rriemberG d(' ths l'OOf trusG

showr> in fig . P-~-o;ni1der'1ng the- u>hole figure, £fv =O : R,..v + Rov = 100+ 1oosinao•

@pt- a.

100lb 100lb

. P.Av t RtN • 1!30-... -6)

~fh··o: Re»1 -1ooca;ao0

@pt./\.

P.oH - S6,6 lb.

~ ... --o RAv " l\BStn ao•

RAv •o.sAe-@ ~fh - 0

AC " AB cosao· f':.C = o.966 A0 - @)

'1f A0 = 1o0 lb o.seo + OJ5M .. _ 100+~

eo+Ae= 1.!JO(~) .. :Mo-© RAv "'0·5AB

. !<Av c a .5('100)= .so lb . "'fh•O: SOcos-eo• :Aeco&ao• t 1oooos30•

eo ~ "e t 190--© ~,AO• o.B66(A0)

.,o.e~,C1o0) "'96.61b-T ~.s in.+

ASt 100tAB"' 300 ,..,e =1001b -c

.. eo - .Ae + 100.= 1200 it> - c @:Cr o:

:li!fh"'O AC ·CD a 96.6 lb· .. -T

~fy 0 0

0C =i00 1b·· ·-T

"406.) The con ti \ever trvss in fig . P- 406 ·1 s hi~ <::it D ""--E. Find

the force in eoch member .

1cxx>lb

@ pt./\

.£P..,•0

ASs1nao" = 100 : Ae-~lb-· · T

lll!rn · O: AC. AP>COS"°0

56

AC~ 200ocOS.90. I /\C = 173Q lb -- .. C ·

@pt. 0

if' BO• ~SOO lb in 1

BC -eooo- o.e(2000) 0°96"6

~C"'866lb---· C /

. @ pi.C CD

I! ...c .

b

~Fv "O: 8Cs1n6d+BDs1n-ao• • 1000+ABS!n30•

o.eBDto.B6'66C = 1000+ 200?(0.~)

o.seD+o.~ec,. Qooo -· · - (!) ~Fh =o : BOcosao· w J:>Coos6"0" t ABoos:ao•

o.e66BD•o."el3C t ~000(0,966') - --·@ 6y Elirnino1ion : 1 ~ r;

+ (o.s.eo t o.aGGec = ~ooo)'o.s ( 0·9~ 00 - o.!!)sc ~ 173Q) o .96'6

o. 2!3 eo + o~ ec - 1000

+ 0.7!!J ao - o.+.s~ ., 1.soo BD • 2800 lb ~- --T

CDs1n6'0°• ec81060• t 1000

co• 966 (o.B66) + 10oo .. Q0.20 lb __ ·- T o.e66

~fh•O

COcos60· t A.C t BC ()Ot;(it). - ci=. CE"" 2020(0.<:>) + 1732t 8$. (o.!J) = 3 17.?l lb -· . - C

407.) In the con1ile'<'er truss shown in .Fig P-...,...,.., t i r. . · ""', conipu e "T

1orce 1n memben:; AB, 1;3e, ~OE. ~ @ pl·" .

~Fv•O

AB1>1nao •1ooos1n60 10001

b

AB " 1 7S,e lb - .. - T

@ pt. s.

U61~ Rototio" of A")(•S, Uv-o

£fv•O

OE. '"'" 6'0° • 100Q

PE• 116+1b -· -· c

57

. SE

"' . ..a.e ao

1000 lb

...ae.1 Col'l:'pule the force ir'I eoch member of the Wo"ren tru(;a

sho'l'ln in rig . p-~e. '3()00 1b. coMidering the whole' f1qu"e:

~Mi!!•O

£fv•O IV.v " ABs1n 6C) ' '

l\6c 4250 o.966

R .... v(ro)., 2000(19) t -1<XX7(10) +

::iOOO(!j)

p_,._v .. 4-~50 lb.

@ pt. e izooolb

z fv "o : Aesinoo· " 12000 ~ ecGines¢' 13C :o4910Stn60° - :2000

s1noo·

/\£> "' 4007.62."" 4910 lb-C / BC "' ~~ lb - .. - 'T

~h .. O : BD = A8 C/JS60° t l?>C ocs6o0

l?>D • 4910(0.5) t :2500 (0.!'>) :tfh .. 0

A0cos60· "' AC

/\C " 4910 (cos6d) ,AC,- :245.S lb - .. -T

SlZ: ~·

/ BO ;.. 375.5 lb - .. - C /

pt. G £fv ==0! eC 6 1r'l60• t C051r'l60° "'"\000

2600(0.066) t co (o.066) = 400o C O"' ~01~ .94- <;:1 ~o201b---T

~fti co:

BC c<f.> oo' t /\C -= CAJ CX:foq:l t CE

2600(0.9) t Q.+55 "' '20~0 (o.~) t CE

Ct= .. 1300 t ~45E> - 1010

Ct= .. !ZH~ lb. - · .- T

58

@ pl. o. £Fi,,•O

OE:61n60· "'CDs1nro· t 4!000

PE " 20QO(O.B66) t 3000 ~~ ~

co/ I \oe oe - .5:+e+.2 "" ..!34eo 1b c

411.) Determine f hti l'. · . . . iorce 1n members /\0, Ac, 130,CD, ~CE of' the conh lever truss .sh<M'n in. FIB. P-+11 . If the lood 1· _ ..:i oi c · . . s we re opp l6U

"".E '~"_tead or o~ 13 \_D, specrfy whioh memberG' would hove their internol force ohcmged.

100lb tone-"' :10 3o

-e-. 33.69.

@ pl . /\ ,

~~e ~

@ pt. 8,

~ · 19·'"' ec • !200 lb /\e"' 1so.2s lb -~

00

~Fv ·o fl ~rh ==o ·

ec Aacosaa.~· "' aOC06118-6'•

,._a - ao - 1eo. 2s lb - .. T (

~= ~ · Oe" f3~g· '.20 -'CJ I •

~i=-..·o

C0 61nsa.1!2· - BC

ionO • fa.93 10

-e- c ,93.112·

co- a.90 lb - .. - T I ::S:fh •O

CJ'O c COCOG.s~.12 t AC

,,. 200 coss a.12 t 1.so

C~ " .300 lb~ .. C /

59

AC-a Aacosaa.89

AC- 180lb

. '

I

~ I

"tos.) Determine the force in each bar ti' the \russ shown in Fio P-"1-0S covood by \if\inq lhe 12.0-\b \ood oi cons\on\ ve\oci\y of a n per sec . Who~ . change in ih66B fo;ross, ·jf any. resulh; fr()IO p\ocing the roller Guppod ot D ~\he hinge support ot A?

8

~Fv=O

Av+ Ov = 12.0t ~/s (120)

but Av = Ov •• 20v = 12.l>t a/s(120)

2 Ov .. 192

Ov • 96 lb. = Av

ot A : ,..f>~ · , ~~-£96

,.,c

~-AC ,. Ae 6 -ll- ""10

AC • 12S lt:> . (tens1~) A0 " 160 lb .. ( CO("lprcss1on)

o\ 0 :

BD • 160 lb (c;ornpres1>1on)

co = eo( e,fo) - Oh

bvt. Oh· 96 lb : • co - 32 lb ( ~c:ns1on)

at c: £Fy=O

/"w +O""' BG

lnterchonqinq hin¥ ~ roller support will rot chonqe \he forces

in each bar ell°'Pt for /\C 1*. CO

AC = ·gc;; t 9/10 (AS) = 22:4 lb (teos1on)

co c so(e/10) · = 126 lb(icnGi<>O)

BC • 192. 1b(tension)

60

:·' i

409.) Determine the fr....ro · L..- • , . "' """.' in rn6'mucrrs AB, BO, BE,~ OE of -the Howe roof irvSG shown " fig. P--409. ·

0 £MHoO

Av("+o)-6Cd..30) - 1ooo(w)--400(1o)= 0

1\-.i :: 1oso rb. 'k'a::=:::-;::1t-~-'?'i~~-d--1~~H .. ~ £MA ;. 0

H..r(..io)--400(30)- 1ooo(:u>)-600(~0) :o

at I\,

:. AB• %100 lb (~ion) /'C • 1820 ib ( tooGion)

Hv = 9.50 lb.

~~. :%fy•O, 1;00

BE s1n60• - ~(sinoo") •O

.BE - ~oo lb(c;vmprcssion)

BO• BEQJG~· t f,()()(et:>s<OO') "' 2.100

-4' ) . BO"' 1500 lb (eomp~iori) 10- Determine 1he forc.e in eoc;h member ci' the PrcrH f I

in fig. p..410

. roo rvss eho-Nn

lb o+ 0,~fl' U6iog Hew Axos .,. Z:Fy • O, 0C• 1s001bC = FG

~ ec; . <L • • in rne Of'1g1nol o><es

%l1/b4•0, 1800(H)t1!IO'l(-t4i)t

1900(6) = Av (32)

Av = 2700 lb .

M A . ~

Av=~~ AC 3 s ---+""

• • /\8 • +soo lb - -·-C =- FH

AC= GH ~ 3600 lb-·-T

AB-BO~ Of' "' +soo lb - ··-C

••• CD 3 2163.'33 tb -··-T ~ O&

~Fi< = O

AC c o/"13 CO t Cl=.

CE. = :a600 - Y~ ( 216::1.33)

00. : t=.6 c 2400 lb - ··- T

DE " O

61

41f.) Compute . the force in eoch member of the \ru(;S shown in f ig. P-:-.+10. . If the loads oi 8 ~ D ore 1 .shined vertically down­word to odd. to the loads ot C ·~ J: 1 will the~ b6 on"/ chonge

in \he reactions? Which membel"s , i,f ony, ""ould undergo o

ohonge in ln\er-nol force? 4001b

(l'.)nsidering the whole figure •

:a;:f ... ·o R>.v t ~FV " ..,.00 t QOO t 2()()t 800

R.-.v t P,pv " 1600 lb ..:-©

SQ)lb

@' pt. e, 'IOOlb

!Ct=h=O

~80 "'~ /\B

~Mr"o ·" / R.-.v (oo) '"100(-40) t . .+C(a:x>) t ~ad) + eoocao) G eeo lb

i;ubst the volue of ~ ... " in 1,. Rrv ~ 1600-BSO .. 720 Ii:>

(

@pl~~··:.:::..

zfv•O R>.v cA8.sin6a.4g·

"6 ~ .,993. 9 1 '<:: 99$ lb- .. -.C ~fh-0

/\C c A0 COS68.4S •

c 996 COGGa.+s· IC .. 440.ee ~ 440lb -· ·- T/

BO " 98% ~ -49~.5 """'1'9~ 1b-.:1C

.....!..-,.. W +~lie = 400 tac 'W '<5

~ (-t9Q) t {s--{9f3s) - 100 t ec

SC - 701 .o+ -=::: 700 lb - . -:;.I @ pt . c,

~fy "'O

J:>C "{e co + czoo 700 = ...L cot czoo

"9 - coe 1118.0?> /

~ 11120 lb-.. - C

C~ - 440+ ~(1100') .. 1141-76 ~ 1440 lb-.. -T

62

@ pt . E,

+ ~r.- -o . DE = !100 Jb _., - T

@ pt. f I

~fv"O

DF I ~

~rv - ~or OF = ..Je(7!2o) / · OF • 1609.97~1610lb-.. -c.

.j.ta.) Determine the r · h iorce in eac member of the crone Ghown P-+13.

@pl. A,

coi;1ne LOW:

o2 = b2 t c2 - .::zbc Cos!\

6 2 = b2 + 9 2 -2b(9)cosA-©

ba."' o-.i +ell.- :zoccosa

be= 60+ 90_ 2(6)(9)cos120'

b ... 1il -08,

In 1 , 6"1•(1a.oa)Lf92 - a(13.oS)(9)cos/\

Ar. ~3 .39··

~b c.ont. ()t'l 8, ~fv aQ •

eoGinac) • ecs1r-i60•

£.fv=o

J..Cs•n23.a9· .. s200

/\C"' 13()99.64 ~ 13100lb -·'-T

~h-9 /'

/\C cos 2a.a9• = A8

AB" 13100o::>s2:a.a9• ,. 120.23.-49 lb

AB"' 12000 lb- .. -C I' @k BO ~Fh•O

/\8 • BOcosao' + BC cos 60 •

1aa:io c o .966 eo t o.~ ec -<D

. o.eeo • 0.966SC -"@ .

EUbsri . 2 1'11 , 12000 "o.s66(o.:!,6 ac) + o.sec

EC= 6000.!16 ~ 6000 lb-.. -yc 1n 2,

BD .. . o.&66(6000) "'10.392 lb

o.s ~ 104ootb-!.'.'.c

63

+i4-.) Determine \he force ~russ $hown ·,n fig . P-""'14.

p

in member /\6, SO, !iic. CD of the

slope ot A'e • 1

slope at BO .. 1/3

slope a t CD " "4./3 at A ,

2Fy•O

,

Y-li A6: GOO .

Ae" 048.~3 lb - ·.-r C

i:.Fx =o

AG

3/..fto BO " 141 AB I BO = 632.40 lb -··-C

.;i; Fy ~o . * /\B ~ 1/"10 BD t BC BC ~ 400 lb - ··-T ·

)

900 lb ~v

Av(3G)"' 300(2.1) t 300(1e") t 900(9)

Av= 600 lb .

ot C,

~fy "' O

300 t 4/!EI co = ec · .J CD= 125 lb -··-C

64

+16.) Solve for the force in mernbers fH, Of".~ DG or the truss shown in fig . P-41S.

0

@)

~ ~fv•: rR:-FHV = FH s1n4s•

fH "' 900/sin-4!1°

FH = 1127l2.79 'I: 1270lb -·· -C

@ pt. i=,

~FH •o .

~ ~

Ya FH =· ::i/,.io DF

OF• -1Wfr210) ('3)..W

OF " 946.'6"" 947 lb-· ·-C

(@ pl. 6 ,

~· ~rv •o

+foDG t F6 .. 900

.. /.e 06 t 6CO ., 900

1::?0 "'::ioo ( s) 't

06 = 37.!' 1b -·· ~r

@ f,

~vco.

conGidering the whole, ~Fv "'o

R.iw t R11v "' 300 t ~ t 900

R>.v + Rtiv .. 1500

£.M,., "'0

R>Jv(:a6) = aoo(9) i :aoo6e) t 900(21

Rttv .. 900 tb.

R.-.v = 1500- 900 .. 000 lb.

Y-ta FH - f6 t Y..rro OF

Y-12(1210) • t=G .t J4w(941) f6r:: .59S . .B6 ~ 600lb

65

411'.'> UGing the mS-thocl of .oections, determine the force in rnem­berG ao ,CO, ~.CE of the roof fru~S shown in f (g . P--t17.

BO /'ll!l,

. i ·.!c

E

,, . ~t.'lc•O

FV,v (12) = BO ( 9)

go "' 120(1& 9

~o .. 160 10 -· .-c

~i:v . 0

p.,...v "' 3/,5 C O

.-

co .. b/?J (HlO) I

CO"' !ZOO lb-··- C

...:;Mo ' O P-Av ( !2.4) :: er; (9)

GE = 11ZO(~~ - 9

1lz' 360lb

CE ~ 31ZOlb - · ·-T

41EJ-) The warren . iruss looded os show" in F•9· P~+1a is suPPof"'-!ed

by a roller ol C ~ a ri1n9e a* 6. Sy the rn&thod of seolions,oom-

pu-\e \he force in \he ~ember-G BCr OF, ~Cl'- · cons·1derin9 the ~ho\& flgure.,

.:tfh:O : ~0t1 - 600 lb .

%Mc=0 t;,00('10) t 900(10) t 100o(~O) .c -406(~+

JtGv (-40)

RG" - 900 lb .

-'fOOlb 100Dlb

66

1f..; oO

o/.Ji BC .. -+00

ac • +47. 21 ""4+S 1b --·-C

.1@.lb

£MJ!·O or(ao) - RGV(2o)

Of" BOolb -· ·-C

LMoao

CE(~o) t 1000(10)-t RGH (!lo}. RGv(3o)

ce~ecd.:ao) - 1000(10)- 600(:20) !20

ce .. 1001b -··-T

me or secfion6' to determine .+1Q.) Use the thocl r- .

ao,co,. ~CE of the Worren T row. GOOOlb

B ~--··· -""B;:;...0_--1

'

C. CE 10'

'f<lOOlb ~Mc·O

Bo(e.66) =-3ooo(s)t •nso(1o)

6P • .32soo/e.65 BO " .37.stZ .9 lb- ·· -C

~Mo=-0

( 9 -66)CI: = +750(5)

Ci: "' 257.SO/ei.66

Cc c 27+:2.!5 lb - ·· -T

com:idering ihe whole f'igur-e , ~M..,•o

Rv•(ao) •'1<J00(1o)t~e)t

!2000(1!1)

J:l.vE w •t-7!50 lb .

sin60•(•) c e e-e.66 n.

~Me.•0

Rvt=(1s)- co(1o)a 3000(10)~"!000(6) co· _a12s,9"'1o

CO • IZ1~S l b ....:. . · - l'

::!!,Mi; aO

-co(e.66) + 31svJ(e-GG} c aooo(s)

co .. 1.S000-(32500. 11+) -~-6'6 .

co " 1Zo20 .~ lb:::: ~o:zo tb -··-T

67

, I

.+20:) l)etennm 1he f OCce. in th& mernberG Of' 06. ,.,..1:6 of U\8

ttowe in>S:~ GhoNO in fig. P-.ov.IO·

!f.l•!ltCJOI>

£MG•O: Of(9) • 11.100<12)

Pf = !l900 \\:> - · · -C

.2:f .... •0 ~ 12£>0 t OIAJ 06 • a100 D6•16001b - · ·-C

~f'h·O ~ Df t -4{900•1:6

:zsOo t -4/9(1.500~ .. ~6 f:6 "' .+oOOlb - --T

<4!M-) for 1he truGS ~ 'in Fig. P -<tff, determine t~ f()f'Ge in Br by the rnettioO o(j~ts ~ then chec\( Hi~ result using me-l~ of section~. Hint: .To' opp\y the rnelhod of secli~, fi~ ob\olO the

'(()loe . of ae by incopection. cons"ic:Jen°ng the whole figure.

..e'11•0

Rott .. 1!/.00 lb

~F,,.=0

RJ.,v t Rav = ~.it<)0112J:X> RAv t Rov "3600 - ·©

~o·O

-Wt>t> ~v (19) = 1,,,.00(9) + 1!100 (12)

Method of Joints .:

@l pt. A~. , /'13

" " N:-

1'f,,~O ~v

-./!J/\S ~ S:V.v

/\B=[s(~+ J\B= !ZSOO lb

~v-2000lb

in 1, Rev c 3600 - 2000 =: 1600 lb .

BE. •1200 lb

- 0C = 12-400 lb .

Z:ft1 -o Rotl ~ a/9 eo

68

ao • s~c1200) ao ... woot> -

-=N-0 -f'/:5 BF t 1~ • l:F

CtfECKEO: using Method of Secfl()r"I • @

::E.Ftt'"C

3/.sao=ae ; BO.· ~a(1roo) BO~ !2000 lb .

~F '" 4fi.J(~) t 2..:.00 t 1200..:. gdoO

EF .. 3!lOOlb

bF ... JV+[ata00-1.~001

BF - al500 lb -· ·-0

-MS-) In the Fink tru" &hown in Fig. P-~, the web members

~ Et= ore pec-pendicola~ 1o the inclined members ot tne4r midpOin~: IJIOe the mehld of' 6e0fi0os. to determine the forc.e in mernberG Of PE~\i._CE . . . '

{tr-

£ .. ~ 10 x

x c ..s.&9'

..liiiO .. y 00~

Y"'1t.S'

£FV•O

itFA.v·-Jf8" -1t2t~Hl+1 - g~

...,. .... - fc;¥ .... tc.

~~ .. o ~-S9)t2(2.s)t 1(12.s) .. ,.6 "(n.10)

OF •.s.9139 :w.s.92 ~pG -··- c ,SVa(): 4A;oe t '1- • weoF 't !2 i 1

OE .. fl tt.ipc; -· . - T

69

~ I I

I I I i

I

, I

I .

. I I

fij.i•O CE t ::l/s PE ... -i/ra· PF

Ct:."' o/{S(s.02)- '3/5(!2)

CE e + Kip~ -· · - ·T +26.) Show !hot the · m~tnod of joinls connot determ"ine the f~ ces in oil bors of the fon fink truss i9 fig. P - -+.126. Then use

the method of 6ections to compute' the forc.e in borS f H,Gti,'11\EK.

?£)01b

cos -ao' = 3o/c c - :3-4.64'/

ton~·· o/ao a"" 11.:a:2'

ton60· c 17.a2 -'-t)

u= 10'

coosidering the wrde r.·gure,

R.-,v • l<ov EK

2 FV.v c 1600

i:t-.v = eoolb "'Rov

In fig :2, 30 • Ae t b ·, b•10'

. AE • 30-10 • 20'

TI "'° " ""& I!

In Flg._-a sanao· =DE

f\ E

oe. = '20(.s1noo·)

R.>.v ~Mi:~o

fV.v('2.o)tW0('.1'.1·S-20) - rt-1(10) t 100(20) t~(w-1.!J) + Wo(12.o - 1s}

fl-\~ eoo(w)+ !loo(!2.s)- 1oo(w) - :zoo(1:.z.s) + 2oo(s) 10

FH = 1100 lb -· ·- c H

zMA=O " ,co , '.1()() (7.!J) t 200( 16) + '.100 (12.;;..s) • GH ( A.i=/)

l"l.fl \ I ( \ ) ~I GH. '1.00 7 •!!>) t ~fr;) t 200(22.~ 17.:aa

Gt-I • .519-61' ""~!20 lb -· · - T

70

L,.:;ht •O 00 :ao ~ t ~(aa-2'.1 !1) R/\v (ao) - EK(i7.a2) + 1oo(a<>) t '.ZOO(ao-1 • .s) + !2 ( - )

E:K,. 6912.94- ~ 693 lb-· ·-T .

T<A "-"" rnem berco or' 06 I '!,., E? fOI" the Po...i..-r 431.) Def-ermine the ,...__~ in the tru1:is shown i'n Fag p-+31 . ' ""'

If

R.i.v e panels ot l!s' • 2o0'

~fv -o, conGider the whole f .gure,

~1nce : p.1 • R>.v ·

2R1 • 7(3C>)

R1 9 R/\v = 1os " 0

..C:MG • O

· RAv(1s):-ac(ao)tao(~)t 125 DF(a.e) ~ ·

t~ or(~) OF• 161 ,87 ~ 162 Kips - .. - c .£.~·o .

EG= ~~(162) - 2~(ag.1)

1::6 - 110.7 -:::! 1"10 1<.·a~-·-T

R>.v • 30taot $-OF+~ OG

D6• ~,,,£11 fos - :ao -ao- ~(wi5] [)6c 3.12.61 <::: . .92.7 f<jps _ .. -~

"faa.) Use the method of t' sec torw to c.ompuf the c . 1£, ;..o, BC, ~80 of tn t . e rorce rn members

e ru GS -shown in fig . P-+a2 .

71

. '

I · 1

I j

I

I

f /;.,·,. \

~M .. •o ii...v(s11)• ~400(:M) + a~16)

RAv • 36001b

~

Af'Mo•o

RAv(16) t V..,;; ,.o..e(-ia)-o/,m Ae(16)

Aa • 5909.31 c:::,.!591011;>-· ·-C

~Me"O · R->.v(e) t a,.S A0(9) L~ .AO(Hl)

~· zfv•O

o/t<X)0

t %PP• ~AB

ao= %[%97(!S9~-~J BD "'5o01.o+ ~ .!5000 1b-· ·- T

~ni-o

BG - ~ AJ!J + ""f~ BO

BC-~ (s91 o) t +fe (.sooo)

Be"' 6:4001b - .. -c .

/\0 • 3000 It:> - .'·-T

-t.aa.) Cornp:ite the fOl"G6~ ir. bars- AB, N;, OF, "OE of the sciss~ frur;s .eho.vr. in fig, P-.+99.

K ~M1·0 , .

RAv (60) • 1~(sD)tn(..o) t 111 (30)+1~(.u>) + 1!l(10)

JV.v .. aoK

'48'

~:. J ~

~4 " -.0

" 5

LJ ~: ,

~v P.Jv

" 10 A .5

12

or

,. ,.

_I

~Mc-o

R.-.v(110)t•~M(16)= o/..JS9 A8(20)

"'8 = ¥ [ 30(~)] [s(0o)-e(1t0] ·

AB~ 70.7S ""' 70.9 KipG .i.. .. :..... c ::!EMB •O .

RAv (l&J_ • -+,.f+'; AC~ /\C = #}4· ( ao) AC - 4€1-02 ;:, ...e Kips; - ··-i

.:i!Mt;.=O

30(a<>)'" 1!Z(2o)t t!2(10)t~ OF(a·2 -~-f) + ~ 0~(10)

OF .. ~.+s ~ ""1l.5 ~ips-.. -c

73

i .

I 11

.+21.) Use the method of ~dions to compute the for:-e in ~he members Of, EF, ~Eo .of lhe cantilever '\'russ described in

Prob . +11 '*' page 92. ·

OE" 40/.9

• to

%ME • 0 (R~solving OF ot o) 3/..f13 Of (40/3) .. 100('20) t '200(10)

OF, ,. 360.SG lb -··-T

£MA• o / (Resolving EF ot E) 2 EF (2.0) =- 200 (10) t 2.00( 20)

{5""

EF .. 0

335.41 lb. -·-T

zMr •O E6 (W) • 100(30) t 2.00('1.0) t 200(10)

EG "'450 lb - ··- C

422) Reier •o the descr1~od in Prob. 41'2. on poge 92 , ~ com­pute the force in f'(lemben; 80, CO. it.-, CE by '\'he method of

seot;on6.

. 74

1b Z.M,...•o

. Fv(so) • ... 00(10) t 200(10) t 000(30)

t 200(30)

Fv = 720 lb .

tan-& = '2q/.,0 " o. 6

-& "'26.57 °

tori20.s1· " a/w o~ 10 '

~Mo 0 0

CE(10) = 720('20)

f,..·-nolb . CE " 1~ lb -··-T

~Meo c~1...:in9 BO ot F) 7'20(-1-0) "':z.00(20) t 800(20)+ y.fij" 80(i<>)

BO = 491.93 lb - ··-C

~Mp ~o (Re.solV.ng CD ~tC) 200(20) t soo('2o) , 11..rs co( .ia)

co'" 1118.03 lb ..,... .. -c 4Z3.) Use the method of sections to deterr,..iine the force acting

in rnember-s Of, EF. ~ EG of the tioYVe truss descr'1bed in fig . P-409 on poge 91.

''=""- h.-=::.L..::.,H G 10'

-4«> tlv

%MF ~ o

0

Hv (10) • E!G (10 tonao·)

E& "' 1G4.S.4.5 lb-··-T

:£MA• O

Hv (-4-0) =400(30)t 1000(W)t ~(1~) Hv "' 900 lb

.:lf.Me.·" 0 (Resolving OF o+ H)

Hv(U>) ~ 4t00(10) t OF(s1n.30')(20)

OF c 1500 lb:-·· - C

~MH • 0 (l<eGolving l=F ol E)

400(to) - EF s •nao· ( 20)

~F " +oo lb -··-C

-4-2-1-.) For the truss shown '1n Fig. P- 424, determ·1ne the forGe in BF . by the rnethod of joints ~ then check· th·1s resul1 using the me~hod of .secf1on<0 . Hint : 1 I th f o opp 'I e mefhod o 6edions

75 .

t !

fif"St obto1n the vo\uc of 0E by ins~tion. As o~\e.

.i:Mo~o

Av(1&) -1~(12) t 2400(9)

Av "' 2000 lb -··

·~::~ ot A,

~y•O

"'" .. 4-/~ A0 AB.,. 2500 lb -··-C

~so

.q' AC .. 3/~ AB • 1e<>0 lb - ·-T

l\v 81"

otF,~~f

lb i200~ ot c, N; r&(; g

~n. .. o ~ Cf

1to0

~F.,. •o, CF • 3/e> BF '· " 1~ (5/.a) ~ ef=

BF "' ~50011:>-··-C

By section,

..:tMA•v

,._C .. Cf "1500 lb-·· - T

2..ok)O(q) t 1200 (u.) "' ~/5 BF ( 18)

BF"' 2~· ., -··-C

76

•*' '

. '

•.,

-427.) Determine t~ f~ in bor-G 60,CD, '1Ao.. Of of the nooelle

troGS shO"m in Fio. P-.+f.7.

~lb 1200lb ' ,; ,. -+- 1t'

-r, +

"' 8' J_

e . ~Mf =0,

c.,(1&)H200(~)" ~24)t 1200(1e)

Cv "' 1600 lb.

~fy •o, 1/$> DE t 1~ • 1~ t £,()()

OE" 632.<t-6 lb -·· -T

ot Jdtot 0, £Me •o (Re501ving coot c) 1/-11o DE (18) " ·aA; (12)(co)

£Fy,.O, CO= .5()()1b-'·- T

4/5 co = 34ts BO t Y..i;o DE

BO= 360.55 lb-··-T

428.) U~ the mc:rlhod of GCC{iOl'lS fo deier...,...ine the force in mCJm -bers Of. F6, 1t..,61 ofthe triangular t10V'le trvGS shown in Fiq. P-~ . Hint : fir'Gi. d6icrmine by inGpection . the forces in the web members of the right s·ide of the tru!:>s .

by inspection web members JK, IJ, HI,

Ila.. Hc9 car-ries no lood

-&- • ton-1 1/~ - ~ .. !!J7°

~~·k=O, /\v(«J) = 2cos2,.57•(!50) t 2

(cos 26s1•)('4o) - 2(s1t"126.57.)(!S) - 2.(1o)(sin

26.!!17) --- Av ." 2.46 ~ps .-!MF•O ,

""""~~....,..,..--61 Av('30) = 2(CO!Pu.e7•{1ot 2o)t (s•n2c;.s1•)

£~ ~o : 2.«i~(.30) • -2(sm26.1S1')(s)

- 2(smU.57°)(10)t ~(610 2~.51)(30) + 2(C06 2t0.s7•)(20)t 2. (cosu.~)(10)

OF • 2.3 ~ip& -··-C

(2)(.St10) +GI (15)

. GI .. 0.-1-s.9 Ki~ -··-T ~MA •O: fG (30) • 2(cos26 .. 57)(10120)

+(2)(~n~57•)(1!St10) FG .. 2. 24 ~ipi; -··-T

77

) ;

429.) For the contilever truss .showro in F1'g . P- 429, determine

the forces ·,n members OF, FH , Fl, GI ,~ fG . @. H

Len of A-A ~MG=O, . s/.JZQ Of(z4) ='J.oo(60)t 2.00(40) ~

400(20)

Of-> 1256 • .s+ lb - .. -1

Lef\ of a-a %M1 • o,

.£MF ~o ( ot A-A) . .s/~ FH(..w) =-400('1.0) t '.'l-00(40) t 200(~) .,. UGI = 400(20)t 100(40) t2oo(6o) . 'JdJ(80) .

61.., 1166.67 lb-.. - c FH = 1664.81 lb _ .. -T

o£Fi< c O: GI t7$i'FI =~ .F'°l'I ·Fl = ~oa . 2.7 lb - .. -c

Lofl or c-c zM .... cO: 60FG = -1-00(60) t"'\00(40) t 200(21>)

F6 "',,7f3.33 lb -·: - T 430.) The loods on tlie· 'parker tru6S shown in fig . P-+30 ore '1n

~ip.~ . One l'-i p equolG . 1000 lb . Oeterm'1ne \he forees in members 00,

. 8t:.CE, 'ii... OE . , •I

H , .£Fv=O

CE "' 97.22 Kipc;-.. - T

%:t.'\E "0

A" tJv = 30 (7) Svl Av •Jv

: . Av = 105 !'ipG

Z.M at the inleri;eciion o f 60 k, CE • 0

27/Jiii'+ 8E(160) t 30(135) = 110(10~) BE = 63. 88 K'.ips - .. -T

i..en ol ti-b ~Mot lhe inierc;ection of BO~CE•D

160(30) t 160 Of:. t 30 (135) "10!'l(110)

OE "' 16 . 87.S ><1pG - .. -c

1o!'l (eio) " 30(2~)+ -'/$ BD(32J

78

'.:i

440.) For tho fromo loodocl us shown in Fig. P-440,,c:Wtor -

mire lho horizo~tol t+._ vC?rlicol companonts of' t~ pin pros -· svm ot '_ 13 · S'poofy dirootions (up or down; left or right) of

tho forco OG ·• t od~ upon rnombor CO . c

2001b

3001b

2' ~MA =o CN(4) • 200(2.) t 300(6)

[).... = .5SO lb

~Mo •0

Av ( 4-) .. 300(6) - 2.00(2)

Av= 350 lb e.;,

Bil :f'.Me"O

c 3001b :E'.Mo .. 0 A.H(4)c 350(4) i

e BH

0v

-t'

0 Ott 0.,'5.50lb '

811 ( 4) " 30(){.6) M

BH" 4501b ""' ~Fy"O

s.so -ev =o 8v "' .5SOlb ( do.vn)

2' t ~(Z-)

AH .. "I-SO lb

~f1< *'0

(left) ... AH "' BH "+60 lb

£fy=O

8.., - UJ0 - 3SO = O

Bv "S.SOlb

44'1·) The struetvro shown in Fig . P-441 is hinqoo ot /\ tii.,c. find

tho horizo11tol ~vertical components of' the hingo force ot B,

1001b

Ati § 8';e.11001b A ~

Av :0,,1.., ~o

80w • 1oo(s)-+ ioo(6)

BH . 17S lb

.,ei=,. '"0; BH "AH"' 17.5 lb

.:e:Fy u O i Av - 100 - 100 "'o Av "' 200 lb

CH " Bti • 17.5 lb

79

...+3.) The frome shOwn in fig. P-.++3. is hinged to rigid sup-. i:;orts ot /\ ~ E . find iho Q?fTipo;\ont~ of tho hingos forcos ot

/'\ "'- E '6i... tho fon:;os in rnembors . BC !,... 80. 1e:>lb

~ .s' le..

(~loting Bor AB 120tb

4' ....

~F11 •0: BD(3/s) - Att =O

80 "'(5h)AH BO= 200 lb -··-'-C

zt.V.=O: 1'20(4) - EH (4) "'0

e:1-1 ... 1w lb

X:ft-t .. O. : E~>-At-t .. O

Et-1 c: At-1 '" 120 lb

at iGOIQting bar AB

:.t:Me"O: Av(B) - 120(4) -::o .Av= 60 lb .

Con~ering iho wholo f'romo : ~fv"'O: Av +ev - 120 "'0

l=:v = 120 -(0

l:v ,.,· 60 lb .

.%'."M.-1.·0: 120(+)+ oc(a) -e0(4/.s)(a)=o l3C ., 100 1,b -· - T

+++) The frornc ~fuwn in fig . P- +'14 is suppodod by o hin­god ot /\ ·&. o rol1er ot E . Compute tho. horizontal ~ vorti­

. ~I componen\.s o.fi the hingo furc.os o t B ~ C os -they od upon rnombor AC~~- ~

A

Isolat ing bar BO B!-+.s ~, . 12' 1'2'\0lb

lev .0 DH

~Mo "'O. : 240(3) - Bv(6) "'0

E:v >= 264 1b

:f.Fv .,0: Ev -Av - 240.= o

Av " - 24() "t t=v Av = .2+1b .

o~ BO

%Maco

240(9) - 0v (~) - 0

°" = 360\b.

Bv " 120 lb ( upword w/

~pee! +o oo.- N:)

80

·'"

i

Isolating bar CE

lsolo hng bor ~

:f;F.,. =o Cv t Ev -Dv = Q

Cv = 360 - '2.64

Cv = 96 lb ( downwor-d wiih rospec! to.

AC } ~Mo•o: CH(6) +Cv(3)-E.,(~) Mo

C..1 -= 264(2) - 96'(3) G

c. { io the leO tic -4-0 lb w ifh rei:pect to AC J

ZM.-.•O : &i( 4) t Cv(s) - CH (1o)- Bv (~) co

s,..: -to(10) +120( 2) - (96)(s) 4-

BH ~ 40 lb to tho right .

445.) The frame shown 1n F19 . P-+<Ts is supported by 0 hiri­

god ot E i..., o rollor at D. Computo the horizon~ol ~ vorti -

col components Of the h ingo force ot C OE it ociG .upon 60.

Dv

lc;olohng bor AB

2..0

""' " -+' l 2' µ .. ~" ·rev

:EM.-. .. o Bv(6) c 240(4)

Bv c 160 lb

~Ma•O

(6)Av = 2"1<>(2.)

Av=aolb

.Z:.ME=O

Dv(S)-= 240(3)

Dv = 9o lb

:::E:.Mo c.O

. Ev (0) = 240(s)

Ev"' 1so lb

lc:oc lating bar DB

D

~Fy"O

Cv = 16o-go

Cv"' 70 lb

ZMe "'O CH (3) c 70(.3) t 90(7)

CH"' 260 lb.

81

)

I .

~.) A three - hinged orch iG composed or two trusses hi'n<;ied. together ot 0 in Fig. P-+'l-0 · Compute the. compoi:onts of the reaction ol /\ ~ then f ind the fon:;c,s acting 1n ban;

AB'&.._ AC . Hinf: f irst 1·Gol0Je each kuG~ ow o freo. body .

bor i\15, 0

I

I I 2o' : 'lo':

Av

@ Joint A,

.%Ma "'0

Av(eo)-aoot.G0)-600(20) =o /\v "' 420 lb

2':N\o: O Av(40) -36CJC20) t AH(30) cO

At-1(30) =- ¥Q(W) - 4 2.0(40)

Ai-1 = -3:20 lb

: . Ai-1 c 320 -to tho r ight

%Fy=O

l'C 20 / '*2o - !\8G1n .36.87• ~o ·" AB • 700 lb _ .. - C •

A .:EF'~ ~o : .320- AC - AB COS 36·87 =O AC= 320 -7oo(cos.36.a7•)

' / ;c, ..,._ - 240 lb.,,, 2.40 lb - .. -T

447.) Two truss;o~ ore joined os shown in Fig . P-4-4~ to form

0 three - hinqod orch. Comput.e the horizontal '*" vodicol com­

pol'"\cnts or the hingo forco ot B ~ thcr'I determine tho typa

~ rnogn'1tudo of force ·,n bors BD ~BE. 'l-40ib m>lb

10' I 101 I 10' 10' :!:Mc =o &

120(10) t 240(30) t AH(10) -Av (40~ a 0

Av(4)- A..i = 1440 ---' © lsoloiing loft tr;;ss

t.olO 10' B ~

82

.:iMe •0

Av(20)-A>.(oo)-240(10) =0

/\v-At-1=120 -@ a:i. © !..._ @.,

A,,(-+.)-AH "'-1440

- (Av - AH ""120)

sAv "' 1320

. Av c +40 1b

AH cAv -120

= +'K:l-120

AH - 32.0lb

~FH =0

AH ·-BH •O

::E:Fv "'0

Av :- Bv - 240c o

&,, = +40- 240

8 2 = 8,, 2 t BH 2

c (~)2 t (300)2

8 ... 377.+ 'fb.

446.) A boom carrying the . loads shown 111 Fl9. P- .+40 iG

comf)()!;od of throe GC<Jrnont~ . It is supporlod by four vor­ti~I roodions ~ joinod. by two frictionlot;i; hinges- Ootc­rm1ne tho values of ihc reodionG' .

400lb

A f=;;:::~;:;::=;.::::;=200:;:::1=bA;n;:::::!:=:::;:c:4::~100=:'%:'.::Jo 3"' 3' ....... 10'

H.r.go R2

@member CO

C H c 100 l'o/fl 0

@member BC

I 2001~/fl ~ -t' I ro' ""'

200 R2 lw2 ~ 300 1:~, 1 7 ' I

• ~aoolb

. :<E'M R:i "'o R2 (10) .. 200(14) t 2000(7) - 300(4-)

R~ e 2120 lb .

.zM~2 ° 0

R:i (10) " 200'.:>(3) + 300( 14) - 1..00( 4-)

R:-l = 11ao lb.

83

::f::Mcco

R4(6) ... 600(.3)

R4 " 300 lb ... Cv

.I ~

449,) The bridge shown in Fig. P-449 rons'1sts of tvvo end i;ecf

1onG, each weighing 200 tons with ccntcr of grovlty ot

G, hingod '\o a 1,.1niforrn center. span y.1cighing 120 tons. COm-puto ~he rcoctionG ot A, B, E, .~ F .

@member CO I E(ft I

Gil c -40 20 0 Chi

..::tMc • O

Dv(oo)"' 1~(30) t 60(4<>)

Dv "' 100 tons

.:i::Mo ='O C.-(6o) "' 120(30) t 60(W)

Cv Ov

Cv = 00 tonG

Ot1 D ..:!!Mp"O

Ev(.50) ., wo(30) t 100(10)

Ev :::: 260 tonb

• 1 .F-Fw•() %tv' is = 0 '

E~· 30 • Fv Fv (so) =- z.oo(ro)-100(1.0) '/fl .; • .

I fv"' 40 ~onG

.Z:Me "'O Av (SD) " 20o(20) - BO(W)

Av " 48 tons·

.::€MA = O Bv (so) ~ · 200(30) t 60(70)

8v = 232 -tons

-450~ A billboard BC weighir19 1000 lb is .subjecfocl 1o o vv'1ncl pres.sure of 300 lb per 0 ·as Gho~n ·,n fig . P · 450 . Ncg­lecf1ng the weights . of the suppor \'Ing members, dotorrnine

the components of the hinge forces o~ /\ &,., F. Wind prcss;u..-e = 300 l_b/H. ·

= 300 1b/f1, l( 10 'fl - 3000\b.

84

EH

c

4'

4'

.:::EMA,.0

3000(9) +1000(6) t Fi1 (4')=

Fv(12) .

fv(3)-fH "'0250 -@

·~MF =O

3000(s) - 1000(6}t Ai-I (4)

= Av (12)

1---''-- Av(3) - A H .. 2.Q.SO -®

@member ce. 0v ~Me"O

C CH

B BH 8v

C11(10) .. 3000(5)

CH =-1soolb.

~Fx =o : 3<XJO - CH -BH "0

6H "'3000-1500 = 1500 lb.

Ql member CO;

C CH

~C>v DH 0

01-1 " CH=1$00 lb

D

.+'

@ membor OF,

~

F

F.;

.2'.Mi. =O: OH(+)= F~ (4}

FH ~ 01-1 "1li00 lb

.:!F.c =o: E1-1 -1500 -1500 =o EH =3000\b

::EFx •O: F~ t ?>ODO - AH =o AH= 4.500 lb

fv(::i) = 0250 t. 1500

Fv = 32.SO lb

Av(3)"' 2QSO t 4500

Av == 2250 lb.

85

. I

~ I

i

J l'

~ ti'

451.) 1'hc lrome stiown in fig . P- -4-.51 1$ hinged o\ E. !t.._ rollor svppork0 o\ A. Oetor~1nb horizon\ol ~ vorticol c.omponenti; of tho hinge fon:;C;s ot B, C, """D .· l'leg\cct tho woightli of tho

mcmbon>. )

:l!M.A. •O Ev(12.) "'300(16) - 24-0(10)

i::.,, =- 2~ lb

::EMi=- =-0 •'

Av(12) = 2~(10) - 300(6)

Av F SO lb

@member CE G." I

c

Ell

Ct1 +' 0

°" +'

E

Ev

1 :t:Mc=o DH (4) "'2.40(0)

D1-1 = 480lb

2'..Mo =O

c .. (.+) = 24-0(4')

CH = 24-0 lb

300tb

:a!!!.F>e "'0

EH - 2-40 .,:;Q

.EH c: 2.W lb .

.2!"'1e "'0 . Dv(•)=.so(6)+aoo(12)t...eo<'12)

T.. . °" = 010 lb .

.:i!fx =O: D.i "'8H c.400 lb 6'

~Fy"'-0 ·

t'

.ll!Fy -=O

s.... .. w Cv =s<eib.

86

Bv c: .SO t e10 - 300

Bv = 580 lb.

Choptor .5

friction

,I

87

-'l).5.} /\ block weighing W lb is plo~d upon o plane in -clined at on angle , -&- with the. horizontol. Discuss what 'f"ill hoppen ·it the angle of f6ctlon ¢ IE> (o) greater than & , (.l:>) ~quol .to -e-, (c) less than .e- . (o) If ¢ is greater thon ~ th& plock will not .slide ~)own in.stood it will re-to;n '1h; poGi ti'o0- becolJGe the frlct1onol force iG so ,rnuoh that it w·i11 . hold · th~ blociK ·

, (IJ) If¢ iG c.<:Juol to fr- lhe ·blool<. w ill sfill not .sh'de, down b~- having -e- vqlJol to¢ th~ sldstern will 1.Sfill be Inc­quil;br1.um ~· here the fr1ot ioriol f~ree IG In. 'i ts minimum. (c.) If ·¢ Is less thon tr. t hen sli'pping ocot..irs tJeoa;'.se the fr1'ctioriol force. '1~ not enou.9h to hold t he blooK.

.506·) /\ -tOO- lb block Is rest1'ng on o rolJ9h horiz.ontol . '.surface, for .whioh the COCJfficietnt Of friction jG Q,40. Def.

the force. P reql) ire.ct .to cause ~otion to ' lmpc-nd '1f op­ph'od the block. ( oY hor1'z.onto 11 ':1 , or lb) downward o t 00 • with 'the horizonta l : :·(~) w hot minimum force '1s n::qu;~ to start the motiO-,;.

O·

400

88

p - "IOQ/b .sin 2-1,e · .- sin. 68,2'

P= 160lb.

p --- 400lb --Glr'\.u~· .sin. zie.z• ·

400lb .sin go•

Pmin -= 1-+8.5 lb.

,·

•1 ' I

•.•· i

c.) F

__ P_· _ ., ~OO lb .s1n,sg,o+• sin 11io.g6°

P = 60.:>-o lb .

R ooolb sinoo«96· s in ~g.o+·

P=300fb .

.Zf;oc:::O : 400(COSi5') .t f-'(.8.oO)(cos45")=0

F = 70 .71 lb

sotl) The 200 lb block h ' f' ' s own 1n 19. P-008 hos 1mpe,nd1n9 motion up the plane caused by the f-ic?ri:wrrfa/ (Orce of "l-00-lb . Pd· fhe c.oe(ficient . of .sfcd lo frlo~·ion between · the contoot Gurfacei; .

.t::f.1,3 .=0: /'I"" 40o(s1nzio') t 2oo(ros30') /'/ "'.373.21 /b ..

~F-x. =O: F c 4oo{cos13d) - .zoo(simio') F ... 246·11 lb :

,,«. •. yr, :: 246.41, 0 0.66 '373 · 2.1

-:09,) 'The: blockG s hown in fig. P--50.9 ore conne-otcd b~ f'Je"Xiblc ine-xfen~1b/e, cords passln9 over fr1Ct1onlcss pul/;ys. At A fhe coofflotenf of' (rict ion ore f. ~ o.oo ~ fk · o. 20 while at /3

, f~ey ore fs • O...,.~ ~ [1<.. • O,i30 · Compute. the rnagnifude- ~ direc -f1on of fhe rriot ron force odin.9 Of'\ eoch plOoK.. .

l.!!10 .) What wei9hj W is necc-ssary to stod the sysiem o f' bloof<G Ghow n In. Fig , P -..510 rnov1'ng to the r ight ? ·The. ooef-

89

. :-. ·.

)

.l

fio1'ent of frich.on 1s 0 .10, ~ the pulle!jS are, oGSumed · to I •

be fr1o t1onlesG' . sol. .!'og .

·~ . .510·

~F~=O

w =·Te w • 20-+ .6+ lb .

fBO of Block /\ :

W TA ;.#"' /\ .

fA TA·

A JO()lb I~· . . I I

~ f.9•0 : N" =.?JOO(cosa6.0•) HA. 240 lb.

· ·. f,... • f"- N11 • (0.2)(140) f,.. • +8 lb .

~f'H·O: H 11°2oo(c.o.s.s~.1) His= 1~0 lb

. ·.Fe • f,_ Ne • (o.?.)(110) rs c a6 Jb.

c

r. .. - 601b

~fl:! ·O : Ne.• -400 cos ao· ti 8 • 3-+6·-t lb

£f-,. c O: Te "" Fe t T" t.of00£iri all

Ts 2(0.1)l3'K·~ tl.OtlOO

Te"' .294.bt lb .

90

. i.

.·

blf.) find fhe Jeos~ volue or p required fo c.ause fhe Sl.dS.­

fem ·of blocl<s .show In . Fig. P -s11 to have. •'mpe,r;d c'nq rrot 1~11. to the left . The. coefficien t or fr1.chon. ·,g o.i under ~ooh block. .

~~~ ~ ~ '""

F&>of f3locl<..J : . N 1

T 100

~lt co: T tf-Pcos~ cO sin.,1.~1· .sin 79.1,9•

67.Mt(M.)(900-Psin<><,)-Pc.osO(•O r . 67:3Z lb. 127.32-o.2Ps1nC\ - Pcos<>{ "o

F'(0·~SlflO(tOOSC() • 127.32 Jb.--+ p: · 127 . .'J~ dP • P(o.a,OD&0( - '5'tt10() c 0 o.11s1no< tCOS"< d&

o.~GQS<>c - s rn"'< cO P .. 12.4 . 6 lb .

p.2 =~ ~ ton<>< COS<>(

O(' • 11-31 •

.!Jf~.) A homogeneous J:>Jook of weight w rest upon the ln­ol1ne ·shown in Fi9. P-.s12 · If thet .. c.oemcionl of frlcf ion ;~ o.\!IO, d etermine. the groatoet he-1,ghf h. af which o F'orc:;.e, P porollol to tho incl1'ne maybe applied so thot the block. wills­l~e vp the inclrne w/oof f•pp;ns over.

v.,:> \· ~ .e:~-o: N•wcosa6.&7• · ~f~co: p .. FtW~na6.e7·

r P P • (o.s)(wcosa6.87\ t

~ .. ,. '} 'fl I W {t;ma6.e1•)

p • O.&-+W

. ~~-=O

P(h) =Wcoo.36.&1'(1.)tWtm 86.97•(-.)

Reaolvlng wot Pf.O o.6fV\l.(n) • 2.~ t 1.ow.. h -..±..

o.et

91

I I I .

' i

q· I

I I I

i I . I

I

513 ) fn f,'g . P-.912, ihe horno,genoolJS Ploc~ wel,ghG' ,.?>oo lb~ 1h coeAT1.oient or fr1'cho" i~ 0.40 . If h :s 1n., determine the force motion. to 1'mperd ·

.:E:P~·D: 11•300COS36.e7' " ~-40 lb · ·

p

F "(o.+ )( ~40) • 96 lb · ..c!MA ~o : ..SP .. w Gin a6.87(4) t wc,os36.e7•(g_)

SP • (.3oo)(?in 36,97.(+) t C.Of>.36.87'(2))

.sP • 1wo111

p ... .z+olb .

s1+.) lhe .100-lb cylinder .shown In f19 P-814 i6' held o t res+

00 the ..3o" incli'ne by o weight P suspended from o cord . wropped around the ~linde.r " If stipplrg lm~G' dc::.termi-

ne P'*-. the c.oofTioienf ol frd1on . . • W<;!li .• 100IP ·

(,, p, ,. p .

·'

~Fx ·o : Hcos60 "'Fcoeao • tJ •

'f:>lCOS 60 a f- ' •COS.30

)I- ~ cosoo: - 0.677'0 00630

~Ms•O p1 (1) "' r(1) p = JA-H ~ F

~F~ • D: ttsin60' t rs1n30' -P • 100

N (s1n60°t 0.5773s1n30';0.sn'3)""100

H • 100/o.s773 c 17,:,. 2 z lb . I p P )IAl'l=(0.5773)(173,~2) : 100 1b:

s1s.) Bloc.I<. ./\ Ir\ Fi9. P-si.s woighs 120\b, block B weighs ~':°lb, ~the cord IG porolbl to . ~_he inoline. ·. Ir the. coerfic~Clf"\t of fr1otlof\. fol"' 0 1! .sur.f'oc:.eG' 1n. '?°niovf -' ~ o.w , d~ierm1ne. the angle -e- of the 1nol;ne cit wh1oh rnohon. or f3 impends .

t:. · ~

120 R1 --- ----,, s1n7s.~M· s111(00-~)

R1 • 12J.7 .sin (~o-&) • R1 ~12a.1 (singo·rosa-- eos9o'Gin-e-)

R1 - 12.3,7 COSf)--

92

~n ·O : Ws1ne- • Rt .s11"J1+0t i R2sin1+.o+ •

~oos1n&- •(R1 tR~) G1n1+.01-· .

. (R1 t R~) ~ B!.l-f..f .sin-e- (f) . ~...!.1-o : R12.. cas1+.o+ - R1 cos1+.o+ • w oose

(R:t. - R1) cos 1+.o+ - 200 COlX!>-

R12. - R1 ., :206 ,!J. coi;e:- @ from1 : R~ "a2+.<f ..s1n&- - R1 ;,. Rt ~1+.ot-'

~. 82-f.+.sin&- -£R1 ~ :J06.2 COGe-

8:H.+sinf> - R 1 - R.j "2.06. 2 COS..9-

: but R, '1Qo.7G.06& ... 82f.+sine-- 2{123.7c.o&e-) • zo6. 2cose-

Fe>D of t3:

p~·to~, 30·96

A' fl~

~f~ aQ: R¢.COGa0.06 . 2«>C06ao· + R1

cosao.96 •

R~ " .S~.3 . 17 lb .

~Fx "'O; P • R~.s;nao.96 t R1 Gm ao.06 -

.;ioo s1n ao·

p- 128.6 lb .

s19.) Jn fie . fl,.si0 , two blocks ore connected ~ a solid

sf rut otfoohed. to ooch b~ock. with fr'1dionless p1n£>. If the c.oemcront or rr-;ct •on und~r eoch block. i s o.~s ~ /j wo'9hs '.z701b ., f ind the min. we.i,ght of .A to pr:evenf mot ion..

93

:ao

61t'\ 1+·<>+·

WI\ e .590.3 lb -=-

J • ,

s:z.1) In ·ff9 P-s •9 ·,r J•0.3 vndor both blool<-6 ~A we\ghs 400

lb. Find the- rnox . wei,ght ol 13 thot con be storted up the fnohne qy oppl~1n9 to A a ri,ght word ror1zoniol force P of .soolb.

cf.,y"o : H = .+OO t C sln.ao· ~~o: soo = Coor;ao· t F

..soo •ccoi;ao· t '(o, ;,o) ( 4Q'.>t csinao·) .:3eo ... c (1 .016) - cc .:i74 .02 Jb .

We - .!)74.o.e ( s 1n+l:)·a·) .sin 76'7'

We = .26.3.7 lb.

94

j•

sn) Aepeo~ illu6'. · Prob .s11, 055um1ng thot the .siNJi j6' 0

uniform rod weighln9 .300 lb Hint: rt'nsf /solofe fho sfruf as a Freeboqy dt'qgrom, resolving :fs end fbrc~ /nfo com­pononfs oofi'ng along ~ perpend/cu/or to the G-fru f.

c - .324. 0 lb

r1 c ssota2+.a(s1nao·) H • 712,4 lb . P w F .t C =sao· a ;"-rj +¢>2t .. a)(o.e66)

p = (0.2)(712.+) t 2."1.29

pe+tz4jb

51~.) A force of. 4-00 !b Is oppl ie.d to the pulley shovvn ;f\. F19· P-523. The pulley JS preve.nted F'rorn rotaflng bu Cl

fo~ .P ?~led to the erd cJf the bro~ le-ve-r. Jf the coef offr1cf1on. lot the brak.e surroce ls 'o.20,de-f . thevolue-of'P.

95

i.F,y~O ; 'It• f1 t F1. £F>i = o ! N1 c t-l!l. ·

, F1 •t-li1 • O.QH1 = o.2N,,

f1 = f!J. w-= F1 t F1. c o.2N1 ( 2)

W • 0.4N1

.t.MI\• 0 W(L-1) tfll(~)-Hi(fl.) c o o.+yiY1.-1) + o.Q(~)~ - 2 pf .. o

· o.-tL -oA t Q.4-2 co o.-+L-12 = O

r. =b o.f

L~.e>m

.s2.5.) /\ uniror rn lodde-r t6ff. long~ weigh;ng W I~ is plo~ : w/ one end on the ground ~ the- o ther ho_nd oga1ns+ a Y~-

t . J o l f Th~ angle of' friofion. o t oil contoot surfaces I~ ICO w . ' t ,/, th I ct

2d. Find the:. minimum volue. of ~he ongle -e- o w;~ e. 0

dcraY\ be dcSin~a · w/ tho horiz.onial be.fore s), ltpp 1ngocou~. · , • iS- (t.cos.e1u ant . . I - · -- - 2'>11

w ,' w 14

I ,I' 5)TonuiL?&tLS.•~ ~IJ) ~ I

'I \ Ton70 t Tonl!O

I , !-f~W"-':cr---------1 0) ~=m?.tb

ioos.,. ! . _y = Ton 70 • .X

@ (~-.!h)~-m(X-'X~ . . (~- Li;ine-) 0 - TonW'{x -L<XJ~~) .

7an70X- L&ine" .:_Ton~o·.x-tfonw'LCOG&­~(Tan10' t 1onzd) = Tonzo"L0%'9- i LGin~

x = 1on2o·Lcoce1 l{;in&- . Tan 10• t Ton 20•

@~Ms·Ow('l(- ~-~a-) .o Ton 1ZO.L0%9-t l~1ne- - L co~&- •O

y Ton ~o· t Ton 10•

Ton w· Lc.oG~ t J..G1ne- • ..!::.cose-Ton 70~ Ton 20· 2

96

1l Ton 1.0' LcoG& t 2 Ls1n& • Leo~(>

(Ton7o' tTonzo •)

2 L~1n &- 'LG1n&{ron]0° tTan20 -

21on20•)

si'n& • 2.3935 (f:s9- 2

Tofl -&- ., 1.1g11s

-ft- e ,.50 o

I •

' < I

526J ;4i ladder ~on long we;ghs iO lb ~ '•ts center of grova~ '1s 8f1 from. the bottom. The. !odder ~s placed ogo;ns+ o verti­

c;ol wall GO traf i ~ mo~ on orgle of f:-0° w/ fhe. 9ro(;,lnd . HO..., for l)p . the !odder con Cl 160 lb man o l1'mb befOre the !odder

i6' on -the venae of slipp ng . The ore le of fn;oi ion of o 11 con -tod Gur foce J{; 15 •

r-- A(S1!1) A ~;lb, is• H

I ~lb R:z. i- I (10,11.at)

@!:I <m'l( -tb - ~·Tern ?s·x ®(~·0.) c-m{x-;ic .)

{._y -11.n} • - fon1.s· (x-10) ~ lY"'/1,()'

I bef! o;ub,, the volue of.!:! :

is·Jt 0.0) Ri[lt!

@~M,., ·o +o(&- acos60°)•160{x-s) "'fO (1) ,. 160)( - eoO

Tan n;'l( -17,32 ~ - Tofl1.S.X t 10Ton 15 •

(Ton 1s• +. Ton 1S·) :it - 20

x --s ft .

160 X ~ 8-to ~ .X - 5. 2S f./ .

527.) /\_ homoge.A""leouG" ~llnder -3 fl . In d1'omek.r ~ weighing .3001b /6 resting On two 1'nofined p lane as Ghow n in. fig .P.;f7

If the ar-gle of ff1dlon °1G' 1s0

fo r o il contact surfaces com­pute the rnognitvde or the eovple reciuired to stort ' the cyl'1 der rotating counf~ock.wlse .

of ~ilibr•um ~f~ •o: ti2 cos1s· c N, c.os 1s· N'L · ~ N 1 cosu;•

coio1,· -4:: l'l!l '

-~f!:j"O: N1&1n75' ·1 H2 s1ri1s' .. .300

.J!M" "13 13 " f.:o (1.s) t F, (1.s) 13 .. (20.s)(1.s)t(200.7e)(1.s)

f3 - 1'47.6 lb- ft

li1 ~m'" .ltj, ax7~· ) sin 11;' ~ 300 '\ cosw·

N1 (s1n 1.r't cos1s'ta'1'1s') = aoo

N1 • 289.78 lb

.'.Nz•77. 6slb

··· f1 =,.M.N. :. F2 .. jt1,, · ~(tonisJ(~76) · -: To"!1s ' (n6s)

11"77.6/b. Fii=zo.0 1b.

s ze.) Instead of o covple; determ;n'e fhe. minimum horlzonfa:I

force p oppli'e.d tonge.n~io11!:1 to the Jefl of the, top or the qy li'nder described In. Prob . .s27 to stort the c,1:1 llnder rotallng counterolock.w'1se,.

97

~M,t.·O

p6-.ft). • f;(1,~ t F, (t:Sl_

Pl.14 ~ 20.&(14 + 77.6(1-st

p • 20.s t n.6

pc 9,0.+ Jb

sw.) ;\s shown in Fig p - 259, a homogeneol)s o,:il incler 2fi In diameter~ we-1ghing 120lb is acted l)pon b~ a ver\ieol forc.e p. Oe,,\errnine ihe mo_gn ;fude.. of P necc.ssory to ~;tart . the. qyl;nde.r tl)m'1ng . /\G~vme that J-o . .30 •·. e- ~16-7'

p ~1 · iZOlb

£M11 .. 0

Pl 1 n) ; w('!:) p(H o.B66) ~ 120 (o.U6)

p " 103. 021b '." ..ss.69 lb 1.06b

5.30.) /\ plank wff. long is place? .ir; 0 horiwntol posi -t

10n. w/ its endg restin9 on two me.lined p lane os ..shown

In. fig. p-530. The, o~le- of fr1d;on i~ 20· . Oe-fc-rminc:. ~v; c lose- the lood p' con be ploced to eooh end befOre .slipP1

• st. ) .-,J (:.ei,'6) 1i-npe.nds. ,..!- ,.. \ ~9.£t.!:! : . "-.

,,.,,,,,. ' ., ,/ ":';,,, "" f11. o) ',.. ' p \ I .u' p 10 ~~',

A ',..td·'X · ')( r? '·

~~ fonJis·~ (1c1-o) s -1on86'(11-10) y -ton6s"JI (!:J-O) "-TorHo· ( x-10) Ton25.,X .. -TaneO'x t ~on"(10) Ton6s·11; -Ton<K>·ll t10Tan..a'

')( (Ton£.s• t Tan.~o') = 10 Tan60° :X(Ton66°tTon-t<f)c 10 Ton"\O'

x c 9.2+ ~=2.&1

zMs =o: p(g.2+ :.. (10-11))~0 zMi;cO: P(2.01 - !:1)~0 ._9.2+-10,x-O 2.01-~ ·o

,... .Y = ~.81 ft. )( = o. 76 !.!.+ I

sa1.) /\ uniform. ploril\ of weighl W ff\ total Length 2L 1s

ploc.e.d oG shown in fig . ~-5'1 with ·,~ ends Ir\ co"toct w/ the,.· inci•"noo po~ . lhc. orig le of f r lcti'on '1i; 1-5°. Defen"nlne.

98

;

: .

> ~·m';\tb . c~-~;)--m(x-~ .)

.slipping imperida.

£M•"'0 .. WJ.. 21-"'°'"<Ton60·-2ism _ L<:oe"<) • 0

T0t11G"t 'btl60'

t 2.tCO!P.Ton60' -12.J{"n"( =)!.c:cis..(ronH•+ra,,¢)

~ Tancx. ., 1. 461' .. 0• nu. :J.

e>{ 0 36.2.

y tilt.''~: -ton60'(x-2t.eosC<) GUI¥.~:.·. J(Tan11~·t fQt"160°) = 2LC06"lTon60°-12~no<.

")( • IZL. ~O(Tan60•- 12l'ino<

Ton1,• flonoo·

.S.32.) In f(g · P-.S.32, two b loc.N; eoch we19· h;na f.50 Jb o-t

. , · ~ · ~connec-

. ed Qy a l)fl• for'ITI hor1:z.onfo I bar which weighs 1ooJb. If th _ le, f n . - ' e- on 9 o nof1of'.\ 11> 1.s· undo..- c:ooh block. find p d; ted II

ith •· · ~ · rec para o e. <+.s 1nol1ne. tho will oOl.u;·e ;mpend;ng rroho" to the- le-f1. P © Isolating, t he. bor

"' lf«) I!!

fso aof tttl @ 010CI<,.,,

---..... ffi•:) ~~

200

~· C e 2.00 lb

.:£f,y • o: N· ?,()(}Ginti;' tllOO•rn+s• ~r,. ~o: re tc~~ - ~i;· .. p

H " 282.84' lb. (o. 268) ( 2u t-t) ~ o • p

pc 7..s.g lb

53~.) ".A unifo~ bar /\B, weigh;~g 4.24- lb , is fo?tencd b~ a fr1~ f 1onle66" pin to o bloc-k. we.9 hin3 200 lb 06· shown . Al the

v~hcol wolf' f # o.268 wh'de under the- blook, f•o.~o . Oeter­rn•n~ th~ for= p neeaaJ ,;Jf rl mohon lo fh< nght.

p i?1:J 99

. ~ .... · o

"'+·. H ~ . /\v tOO

%.F!:I • o : N • 200 t Av

H = 101.6 lb

i..fx uO : P" AH +- f p = 2.86-'t t (o.~)(101.6)

4ti(-f-OOG-t&') tfs Lea&i-6·-lis siri•s'L •O

2IZ'i.,OOS+s' t(O·U& )t\6'eo&tG. -Hs'i..,s1n+6' -o p . +27lb

1.tg.13 "'0.-'18tte

Ne= 289. 4 lb

fl.it AH • He• 289.'f-f6 :,wte = (o.ua)(zs~.+)

Fe· n.6 lb .

. ·.p~·o: Av - +i'f'tfe =-,s01 .6 lb

WE06ES 53s.) t4i wod<ae- is used to 10plit logs . ff ¢ is ihe angle of fhc-

. tion between th-e wt;4~.~ ~ fhe- log, delt . the mo;>' . ongle1. o< of the-we00e. so that · ii- '1rn re-main embedded on the.- log·

p . ' R R . {!,¥.'le)· £f;, • O' 2R&•n (~q•) ·P . . · .2 R[S•fl?JC.OS~ + 51noYac.os¢}• P

/ P dP .2R[o•nt(·Gin"f'L)(i) -+(cos3X~)~~·o d"< . . .,(

-.s1n~s1nc<J:z -t coso/2 co.sl" • o . 2 2

cos~ cos¢ ·= s 1n¢ s,ir1~a cos~ _ .sin¢ .s1noV2 cof:¢

Tan ¢ " col °"'~

sao.) · /n Figur:e, determine the minimum weight ol l'locl\ B thoi w·i11 Keep ·it ot tc.bl o force P Giorfs t:J lock /I up the, incline

surface, of f3 . The- weight or /I ;~ 100 lb ~ the onglo of fric­

~ ion for oll s urfoces in coniooi is 13 ..

p f600~fi00 p tOO ~R1

F1 ~ • ... p

"° '-f N~' .

P.1 ~ 141 . 42 lb.

100

r, zfi. ·O: R1 cos-4'5' = R:z cos 1s •

frn+2)cOG+~· ~ Re - 386.4 lb COS7&"

:EF.!1 ·~ : We • R2 sin"s ·. - R1 s in .+s.0

Ws •(as6.-t)s1n7s'-(141,+a)sin4r.'

We• 2.73.Q./b

537.) In Figui;c- ,determine the volu~ of P. just surficieni io

fiGt~r1t_ the.10 wedQe under the -+oo-.lb plock. The ongle. oF' n:- ion 1s 20' for all contact .surfaces.

R,A400 . -~

~<

~1. •. ~1.:9 . 2 lb -. P • 41g.'Z. &1n~· ' 6 in 70•

p~.341 .? lb

S38) In Fi9 · 537, dofermine, the volue of P act ing to fhe left thot is r,..,.,uired to pull the -"' · t fi """' wvuge ou rom under the 400- Jb

Ploc/< .

~-.......----.-.' p~ : . ~~ Ll

r. Fe -13.1__ • 400

FBD of~ . . • 11. R. sl'l xr .cm 100•

,.. ~Ht ~ R2 • -"01.7 lb

-~ P _ P_ - 391·7 .._ P • 203.1 lb. Fa J smac• -"'m7o• -

"" sag.) Jf the wedge described in lltuG-tration had o weighl of 4001b' whot value of p would be required (o.) to start the- wedge under the block , ~ (b} to pull the wedge. out from under the: bl~.

101

. . fl) i

ongie of friohon. .' 1~ • Rt • 601 .1 lb .

p. +76 .7 lb

~f).·o : ~.sln1>'1e-tc=is··o R~ - -Rq&>~·

. CO' 1i:•

~F.!1 •0 : R. coi; &• t Rt. Go111;• • +<X>

R,CO'~· - R, .s1ni:' (r,•ni,).-w c0&1.s•

R, (cos,•-s1ns'tonH:0) .. ~

R, (o.97a) "-too

R1 c .+ 11.1 lb ,

-"? p .. 7~.g lli.

s-fO~ As shown i11 Fig P-sw, t wo blooks we1gh1ng ~oolb ~ resf­

lfl9 on a hon2.ontol .surfbce- aret to be- pushed opar1 by a 30•

wed9e.. The angle- of f,..jot1.on '1 '" 1s0

for a ll confoot .surfbcos. ~t value of P jf; required_. ta Gtort rnovemCl'lf or ihe blook.s ~ Now wo4fd th'1i; . onswe-r be chC11"<3ed 'if the woight of one of the

bl~ were increa&'ed to 3oolb f .

4' J3!_ .. ..!E2-

100 ""11&° . ""'~ R1 s 1a.~ lb

P.1 ·~ · . ~Fa • O : z~ ..sin30• • P

p - 1Z.(13.~).s1n110•

P. • 715.~ lb

102

II

F I()() . . ~ !7~· -=·

11.t-~"°"*' R3 • 10~.01 lb "t ..

Ii+ R.t P,.. " Pt.-P1

" g1,5 -73.~

p,.. t:: 18-'' lb

£F~-o: .

P • (109,&1)sinao• 1-l13.z) fNl~ Pc =; 91.sli>

.s+1) De_for~inc:. the for·ce P required lo starf the wedge ~hown 1n f1~ure P·&tl . The ongle or friction for oil Gurfaces io contact 16 f5~ ·

P reo of B .e"-,. "0: Rtc.oG1e· -Rt&in1e•., ~

R1 • Rt cos1~,. - sob 1ooolb

: .Sff'\ 19•

--+-- .eol:) ~fy .. o: R, cos1s' - R~ sinr& • 2000

, .su~titu+e- R.1 :

"· 111 . fr.1.~1.s·-i;oo) cosu;· - R2sin15·2000 ..S11"11~ · 1 . .

Rir (cos16· co1-t6• ) - 1509cot j5• - Re &1n1e"c~ R2 (c.o&1s·cohs"-sfm5•) = ~ooo t soo cotu~·

R2 ( a.~-+6) ~ :it1!166

Rt • 11ss."f lb.

p - 11.55.-+ ~ p - .Q.of-3 . + lb QO+s• .sin oo·

s+a.) Whoi force P mus• be appliOd to the wedgc.G' shown ;I"\ figlJ('e P- .s+.£ to start them .unc:k.r the block ? the angle ol

fr0tion for oil c.ontoof surf~ is 10·.

FDD of/\

103

J>+3.) To adjui;t the 'v6rt1eol position. of o pos.1t1on of o column supporting a ')J)O() lb lood • two-!!! ' wedges are usod os shown .in f igure P-s+a. Detorm1ne the fbrc.o P necessary to 1Siort the wodge-s if the angle:, of friction a.I oil surfaces is · 20~ Negleot the fr1clion · of lhe· rollers .

p Rsln6o• = 20 00

R = 23og.4 lb

_ _.:_P_ . _R_ s1n&S ..sin66'

p = 208il • .3 Jb

R1 (6irr'IO -~) ~ 1000 Gin100 •

R1 .. 9134 . s (.sin10"-«.)

_P __ • ·_R_1 __ ; s~bf1 tule- R.1 .•. -. P · 61n(110t"\) .-oneo• . s1n(<1oto<}

P = 1000 -+ p(s1n(10--<)) ~ .s1~t<><) s1n(10--<) son(wto<.)

(S1n1o'c0s -< - r;ono'•nnO() 1000 ~ 1000

1000 oquolu~ P • 1poo ~ 151('1 70- -<

•·n~o t-<

(s in 20 • cos-1· - oosw· &•n-<)

104

(.son7o'oos"<. - cos1o's•""') (~1nW'c.o.So\ - COS1Z0'61'l"'<)

(.sorno' - s rn20· ) COS"<. • (oos10' t ooa20°).s1n"<

.sin o< ... SJn70' - s rn 20· - o.4-66"' ODS"< 005 70 ' t co.s llO •

Ton o< = o. +663

o< >< 2S "

SSIW'IRE - THR~~ .sCRrew

s-fSJ ;\ .single fhrooded jook.screw hos 0 pitch Ot O.S II'\

~ 0 meon roel;L>S or 1.7.S.in. The, coer(i'oicni or Gtafrc rriof1on or Q.JS, ~or f\1net;c , ( r icl1.ori of 0.10 (o) Oeforrnln~ the force P applied of the end of o levor 2 r1 , Jong which will .sfor+ .'.if'fln9 Q we!9ht or 2 tons (b) whoi volue- of p wlll Keep fhc jOOl<sCf"'~-N f urnln9 ?

(o/ ~ tan-Cc~ (b) ~o.~ 2lr

'&n r --$-·ton~(~) · -&- ~-/on-• (o.0'166)

-fT- :- :2.6· h • 0·15 o • .2('j. •£-+in

ton1c0.1 .S W• 2ton' • 4000/b'

1c8.53' .

P • .Yi£... ton (-e t ¢>) a

P • (+000)(1 .1s) fan(:z.6t8·53') 12+ .

p • -57 . ...,. Jb.s;.

f1<. • 0 ·10 t on 4 : 0 .10

¢ c5.71 0

P · V-:,,~ ian(l/>t!t)

P = -t.2.6 Jbs.

~.) The- d 1'-tonoc::. be.fwei:-n o~oceni threods on a /r;ple,­·1hreo~~ jocl<Gcrew. i6' % ;n. The mean r-oo;us 1.;: zi11 , the coetT'.01e,f"\t or rri"ot1on o .10. whof lood 0011 be raised b~ ~e.-t1r19 0 momeot or 2000 · ib - n ~

Pifch e % in ( srn~ ·,~ ·,s lnple lhreodcd muli_iplj t~ pih::l-d~~ · ·. Pit oh : '2 in

~ll tone- • ~:(z) • 0. 15.9

:z1r {7- c 9,oa •

.271" x.....ill,_ • o. 1667 rt 1.21-A.._ •

f - b.JO - tan1 •0.10

f ·-5.71'

f "Wr to" (cp t&) - QOOOlb-n • W{o.1667rt} 1-an (.s.11°t9 ,0.3')

12000 lb·-n : w(o.o+30-s1aan)

105

w= woolb-~ o.o+.3&5733'fl_

..S"t7.) .iis .shown ;r. fig p - s+7, o squore threadw 5cre,w 'is

used ;n o vi Ge to O"Jler'f a pres&ure cf' i tons. Jf H-.e. 69~'N l1; daJblo f/)re,ado:I ~ ho6 pitch of.o.125 In.. 'o mea(\ d oomekr. cf 1.6 in . dotermt'ne fhe- torque tN:iit l'nUb~ be oppl;ed ot E3 to create. fhi(.' presl;ure . AsslA'l'le- the coefr•o;eni of' rnclion.. cf fr1d10" to be. 0.1.s .

j-o.1s

fanf : o.1s

Pillih. =- 0.2.sin ( 6inoo d ouble. ihreoc:le.d

rnultiplij b'.:j 2) f> = a.53" to" fr - o.s it')

£n b·~ 1r\) - 0.0.531

'1 tons•. -teOO Jbs · · -& • tan- • (o. 0531) " a.o'37 °

T • Wr ton (tp t B-) • ~(1.s.,,.Y< ;.P~) ton(a.sa0 t.J .~3'7')

T = 102.a lb - rf o+&) A .single- fhre,aded s~are. screw hos a !Z. Y.z ftv-eode /1rch. ihe. root d1omefc:r Is 2.6 i" \, fhe, 01.1t.sldo d1amo-ter ;~ a'" · 1 he. c.oo£ri'Oie'.nt ·or rrd•ol'\'•:~," r o.10. De-fc:.rniine the momel"\I ~Ga­r:y fo Sfarf fi(11n9 0 vertiool O'Xial loocl of' 4-0, 000 Jb. what

moment is n~ory ; to 6fari. iowerinq ihe load ;' , P. _1 _ =-1.... ·· ," · Dm • Oo - .!!... - a - ~ = e.e I"

TPI :z.:;;inD z :t , · I~ a.+ ton& • .Q±_ .·. ""' • 1 . .+ JI"\ -

P , &- .:n(1 . .f) ·

.. o.'f' in. :ar . ton.& .. o.o+GS f .. 0.1 • tonlfi. 0·1

-fr~ 2-6 T T~ - Wr ~Of' ( tp t~) . tp •.s.71•

" ..-000? lb (1-"f' ll 1~~ ton { ,s.71• t 2.6·) T~ = 681.82 lb- ff

. T~ • Wr ton (</J -t-&) c "f0000(1.+1'.t") ton.f.s.71-~.6·)

T~ ~ 253.6 Jb-n-

eEL T FRICTION

.sso. /'\ rope- mok.'1'"'9 1 '.4- tvt"'nS around o stoi'1ono"1 hon:zon­tal drum ji; UfPOC/ fO '5Upp:ll"'1 0 hGavy wei9hi • If lhe coern' -

olent of r...:;ct;on ic; o.+ 4 what wC.9ht 00'1 bo -suppar-lod h9 ~el"'Hng a .so - lb ro~ at k olhcr end or th~ rope-?

106

,.~ -"'lb angle. of conlool ·1~ (a6o t 90) = "hS0°

. f ·0.4

l ..IL· e-f" - 11 =le eJ~ ·(.so)e0.+(~"Teo) w 12 T1·

2 .J!>O (23·1+) "' t1.S7 lb· buf T1 • W .·. W • 115 f lb ,

.&s~) /\ rope vvrapped t ~(o:;, oround a post wi II .support a

wo19ht of 41<700 lb when o force of' .so lb is exerted of the o ­ther of"'d · CbJ&-rn1T-e H-.e.- coofTi.0tent of' (ricf10" .

q"°'b on9Je or ocntocl : no· Ln ~~ - f ~ - l.n ~o · - f (rw·J( Io·)

-tOOOlb Ln 80 • J (1~·~n) f = ·~ ~ f • o.a-1<3

12 . .s7

ss2 .) /\ boot c:J1erts o . pull of "'1'000 lb on its haweer w/c Is wrapped obovt o capGion on the dock. Jr the f • o.ao, how rnon~ Ii.Ams must the hawser mo~ around 1he. oopstari rotrot 1he pull al the other docs rct 01<cecd -50 lb.

Ln ]!._ .. j {J" fa

/ " Ln 1VT1Z • l.n~ • ln eo • 1+.61 rod-f 0.30 0 .30

/+.67 x 1~ = 8217.1 • - ~~7. ( ~ .:360• = ~.33 furne .

553) /\ torque of 2-tO lb -111

• ~c~s on ~he. b.ro~rum ahown in·fl9 P-ssa . If !he- brake bond 1s 1r.. con loot ..,..,,H, the. brokedrvm thro1.19h. zso· ' the ~m-o•.ent ol fnOtiol'\.. 1·,s oao. Oe-t~ine the (of"G<:/ p Of tne- encJ O( the broke le.Yer. e~ "'· 2"° 1b · fi T1 (o.'!>) ('60'• V~) -·c

12 .Ji_· -3.701. .·. T2 ,, _ T _1 _

Ti 3 .70Z

~MA c M

(T, -Tz) (8 iA)l.-fP"") • 240 lb- 11 To - T 2 = .360 lb

T, - T, '360 lb 3.102

T1 (o.7~gq) " 360 - T1 ~ +93.23lb

107

, '.

' ·· ~ )j~.· .::!M.11=0

p(16~ ... ~) : 11 (2~11.!f:l) HZ!f\.

p(p1a3fl);.. +9a.~3 (0.1667)

p"' 61-7 lb ·

.ss+) /n fig p-ss+, Hit:, c0efl:-1(i1e,r)t of f Nolt01"1. tG 0.20 be.tween -\he rapo ~ the- fhed drum ~ be.-lween. oll surf'occs in. con foe/. Cbtet" mine the, rninimum weight W to pre-vent dovvnplalle mohol"I. ol the

1~ lb bod~ ·

vi . lt

1Fy •O

. ' rl, ~ w cos 36-67 •

1,

bi.it T1 • 1.B7h

fi,= 0·76W

l\11. ~.,.

rlc

~fx•O : T.e -f, c wS..na6.e1·

T~ -(o.-i) (WC<X036·t'1) • Wson36.&/

T 2 = 0.16 W

...!!~ . o .: H2. • 1000 COGa6.e1· + rt1

rh "' eoo t rl1 , .£1',..~o: T, 1 f1t f ._• 1000&i1'36-87 °

/ ' ' 11 t JA-111 +JA-rt.. - 600

H1 -w~,j6.r,7• j 1.a7(0·76W) t (o.z) wc,os -;i6.&7 ' t o. ~(600.tWC»!.~•lil N2•.soqttt1 j · = 600

· ' 1.4·!lWto:16w+o·16w=600-16b

I 1·7'f-W · ·.,."f'Olb

w ~ f 5.2. g 112 .

SSS'·) /n Fig . .sss, o nel';ble belf rvn& r.-om A over the. ec:mpciund

pullCij 13 ~ baol<-. over P to o 200- lb we(9ht . The coeffi o;et>t o{' friot rol"I. _ic; Yr bdween I h~ beH ~ /he compovrid pt..1lle!:J P . f1ncl thei rnoicrmulY\ wei9t-.t W tho1 oan be s uppor ted w/ot4f r0tot ln9

-/he> pun~ · P or G/ipp;ng the- be/Jt on the, pu lie'::! P. fa (O.s1&)(e0 xTl/r• ;-. - ~ e, .. , ) = 1 . 6+9 T2

fa a 1.6t'3 Te

but Ti ·W/e T 3 • t H .9 ( w/z.) T3 " 0.0H3W '

.flV!p a 0

(.::s)T' t T2.(2.) - (2)fa -(a Jr1 cO

STt , - 2Ta · 311 c CJ

.:, (""/~) - 2 (o.az~)w - .:3(~0()) co 0 ·061tW ~ 600

) w = 7o+.7 lb ·

109

Chapter 6

. " Fon:i Systemb in Space

' I

110

60U Dc:termino t~ mogniiude of the ~ultont • il~ pdinting. ~ its; direction c.osines for the following systom of non -coplonor-. C<JOcurn:m• forces. 300 lb ( 3, -'4. 6) ~ '400 lb (-2.-+, -s); 2001b(-'4-. !I

-3).

f()RCE Q)M~na;ms or= D

.IC y 1

A• aoc>.lb. ii - 4 6 B • ..OOlb. -2 .... -5 c•20011'1 -+ !!S -.3

1DTAl..f~1

to get OiGfor'lGe(O) use o•· ll2+y•n:.•

ft .. ~- .Aie ._6_ )( y L. 0

olST. (D)

7.81 6°7i 7,07

300/7,91 "" Av/a .". J\,. " 115.24 lb.

30Q/Mn • l'w/--4- .•• Av n -1.5::1.65 lb .

a<:io/7.e1 " A¥6 . ·. l\z." 230..+7 lb. _fu_ • ..fu:_- ~. JL· ~

')14 Y ·z o 6·71

••• Bll .. -119.23 ; B,. = 230.16 ; 8%. c -1!190 .QS

~· ..fL 9~· ..s;_. WO x Y l!. D 7,07

. ·.·Cx ,. -11.:L1S; Cy= 141 -~; Cr."' -&\-.87

~ .. ... 115.2+-110-23-113.15 - -117.14 lb .

r'ORCE.S llOOMP. Y~P. ltalMP. {15.2..f. -153.65 23().47

-1~.23 ~-4!> - 298.<>6 -11.iMS .1'41.- -EH.R7 -117.H ~26.2-fo -1.52.~

R.2·~11."t:t:I'"+&• 1l 2 1l = (-117.H) t 226.2.f. +(-1~.-tij

R" 296.90 lb.

(;o66;. •h/R a 117-~Qt;.~ .. o.::ig.s .

Cos-"y"' 226·2o/.f!96.00

.. Q.762.

eos-ez .. 1s2.-t0/~96·90 = o.&13

:. poi'n\109 bock"Mln::ls,

\..up-Nord to ihe ldl.

.;Ey n -153 .~ t :Z38.45t 141·++ • 226.:Mlb.

~z· - 230.+7 - 298.06 - &t. 87 == - 1.52 -~ lb .

603-) Determine -the mognitude of the resuHont, its poinh119 , ond ils d1recf;o" cosines for the follo..,.,,ing system of non -coplonor. concurrent forces. 100 1b(2.3,4); .300 1b(-3,-4~s); QOOlb(o.0.-1-).

F~CE COMl'\:)r'li;rtrsoet' o

')( I y li!. A•1QClb. 2 3 + s~300 lb· -3 - + 5 C=- !ZOO lb. 0 0 + TOTAL{<1

to got {D) u.sc o2 = x~ t y•o~;2

..b_" ~ ~ -6.z._ - _jQQ_ x y z 5.385

DIST. (0)

.6·:305 7.07

4

:. ,....'( D 37.1+ ; Ay c 5s.71 : Ar.= 1-+.:za

Bx/le .. By/y ~ Bz/l: " 300/7.07

FoRC.E.S ')( o::>MP. YCOMP •. !:.c.oMP.

37-1"1- .55.71 74:28 - 127,-?QI -169 .7.;I 2H'.16

0 0 ~00

- 90.156 -114.()2 "tt36.44

:. c)I. -. o ;cy =o; Cz .. 2.00

•• Bx .. -127.296; Br " -169· 73; Bz - 21 ~2.'t6

111

,, l

~'. I

I

I I

'·~ I i

.. , r 1'

I I I

I

\ , ;. . -

£>< a 37.14 -127· 298 .. 0 - -oo.1se lb.

£.7 "' -1H-.02 lq;

:z.21 - 1+. ~0 1" 212.16 " '2/Jo ... 4a6.441b.

p.2 =~?<)2t(~y)ll•(~~)2 ~ (-90._1se) 12

t (-114.02) ~ t (-..e6.+4)2

· R "'-507,7 lb . po;nbnq forworcl \...clO'Nfl \o· the lefl .

Cos~ · .. ~:1</f'\ = ,go.1ss/.so7.7 ,. o.17s Cor;.tty "'~,../~ = 1H-.02/.!001.1 • o.~2s .

Q)s-&'i!. • £:.z/R ~ 4e6.-M-/.507,7 .. ().958

. ~.). Dotcrmino the mognitude or the rosvltanf, i l.s poin11ng, ~ ifc; direction cosines for the followlng sysiem Of non -eoplonor, ·concur-rent forces. 200 lb( 4, s,-3); 400 lb(-6,4 ;!IJ ;aoo lb ( 4~ -:2,

- 3,):

FMCE COMPOHENT!O OF D

X· y z:. A.~~lb. 4 .s -.3

B • 4«> lb. -6 + -5 C= 300 lb. + -2 - .3

TOTAL(.;)

t oge't Dvse: 0::1,,xtt -ty"-t ~ e

~= 6:.-0~= 200 , ' f

,l( y i! 7-0~'l·

DIST. (0}

7.07 0.77 ..., ,39

.·. 'A,. "'113.15 ;Ay = '141;.~4 ; Az.•-&1-.97

·12£_., ' fu'._=~· -400 ,l( y z 9~7

I

llCOMP.

11.!3·15 -:l13.66 '21/2,63 62.12

• ; Bx-= -273.66; Bj • 102.44; 0z,. -2~05

c,. • ..fL • Ci!. • .300 >( y i! .5 • .39

.•. c,. = 222.63; Cy" -111.32; Cz. • - 166.98

.%,. • 113.15 - 273 .GG t 222.6.3 = 6:2.12 lb .

~i" 1'11 .++t 1e2.+.ot -1:11.32-: ~1'l-·!:l6 ro . .£z • -B+.S7- ~28.0.S - 166.96" - <t7.9.9 lb.

FORCES YCOMP . 141.+4 182 .44 -H1.a2 212 .56

R'. £. ",..~ 2. L 2. ( )" 2 " y t.-.;.z. "" . G2.1'2 t(212.s~) t(-47g.g)2

R ".528 . .5.3 lb.

cos&,. ~ 62.12/s2e.s.3 = 0.110

cos&y c 212.56/s2e.sEJ. Q.402

Cos &z. " 4-79.9/ .s20.S3 " O· 908

.Z.OOMP. -e .... 01 -228.05 - 166·98 -479'.9

: . po;nt ing bockwords, ~upwards to ihe right

112

6().5.) Three concun-ent forces P. Q.. ti...., F hove o resultant · of ..s lb.

directed forward S.., up to right ot -e->1 = 60., -e-.., = 60•. -frz • 4.S 0

• P e -

QtJOls 2.0 lb~ pa10ses through .the ar.1gin ~ the p01nt 01 .1,4 ). The

..,olue cf:~ is ol.so 20lb ~ it passes ih rough the p0in~ (s , 2~3).

l)ete.m11ne ·1he mogn'1\ude of i~ third force f ti...., the angles it makes witn the reference oxes.

FoACE CDotR)t15'11TS Of 0 DIST. FORGES

~ y x (0) )C.Q?t.l\P •. YCOMP. xCOMP.

p = 20lb. 2 1 4 4.se e.73 4,37 17.47

Q.2 20 10 . .s 2 .3 6.16 16.23 6·"'1-9 g .74

F• ? -, ? ?'

2.6 2.5 ,3 • .!'>4

where : R ".!'>lb.;-&,. • 6o"; &,, - 60•; ftz" -+s•

_fr_ .. _fr_ ".EL~ E£.. )( y i! 4,se F z • f x 2 t fy 2. t fz 2.

:. Pit =8.73; Py " 4-.37; P2 " 17°<t7 !t "' · - (:-22.-%)2 t(- B.3'0) t(-23.67)

~s _fu_~• 20 F = 33.60 lb . )( y ~ 6.16 -& .. •a>S-

1 F•/F :. ~'i ~· 16.23; ~Y •6.+g; ~z • g.74 • cos-

1 2Vl6/33.68

~x = Reos&.. ., .5 (cos;60·) = 2.s -e-,.=4e.2· £.y • Rcasc:7y ~ s(cosGo·),,, :i.a f!ty • cos-1 a.::lG/<J3.G8

£..<? • Rcos-&z e 5 (cos""'"-) • 3.S4 -e,. 4 7:S. 6 0

F.,. • 2 . .5 - 4.:37 - 6.'49 ~z = cos-1 23.'17/33.60

fy a -6.'36 lb . :..e-z. ,; ..... 5.3.

F" · :2.5- 16.:23-8.73 . pointing bock wards. ~ down......ord ..

F,. • -22°46 lb. to the left.

f2. - 3 .s+- g,74 - 17.+7

fz ' - ~.'17 lb.

6()7.) /\ force of 100 lb is dlrecied from A toward B in ~he cube

shown in fig . P-607. Oden-nine the m<irnent of the force oboo.1+ each

of the coorchna+e axes . J,"

- .

·- / / / / f./ y / / II

b ...... //,;./ /

A I {' , 141 -.

c v a / ,..,.

•' a' ,~ - "' /

r, ; I/"'' I/ 1.1

.~ Ji'

d 2 = 4~ t 3 2 t'4 2

d = G.403 .ft.

113

I l f'~ • Xz:.:_ = fL,, 100 lb

~ a -+ ~-....,~

• >= ·6~2.!ilb • Fy~~.g lb ; fz•62 . .51b. , •• r>t '

~"1,. "fy(+) - fr:.(+)

\ . 4'i.9( .... ) - 62.5(+)

~M .. = - G2.4 lb-.fl.

~My ~ t=w (~.) • 62.!i (+)

~Mr: c -Fi(4)

" - 62 . .5 { +)

.t.'My ~ 26'0 lb . n. ztv\z ~ - 2:!>0 lb. rt .

0608~ A force of 2oolb . 1s directed from B toword C in the cube

~hown ·1n fig . P-607. Determine ttie rnornent of the forc.e obout

' eoch of the coond1no1e axes.

d'J.· :z"+3•t+2

d ".s.3BS fl .

: . r)l .• 7 ..... ,, 1b.;

..

~M>1 " Fi!!.(1)~1~.6 lb -f\.

~l<\y : -Fr:(+)

= -1"l0.6(4)

zMy • -.594."'f lb - fl . ~Mz • f,.(1) t F\-: (+)

- 1+.3 (1) t m .+(+)

zMz •.510 .Q ib-fl .

Fr "111 ...... 1b.; fz • 1+e.~ lb.

oog.) /\ · for ce of ~·6o 'ib. ic;directed from B toword 0 in the cube

shown in fig . P-601. Oeterrfl1.ne the rnornent o f the force obout

each of the coordina te Olle& . . :· ;'

df. 0•13• .... •

d .. .5 ff,

'

I i

£!__ • ..fr_• _fi_" ~ 0 . 3 ... .5

:lli!M>1 • Fz (-1) • 24C (1)

:£Mi1 • ~ lb- f-1.

y

114

£ My = -Fz(+)

.. -~40(-4-)

z~y ~ -960 lb-ft ..

1'.Mz - fv (+) - 160(+)

.£Ml!. = 120 lb- ft.

'•

610~ A force of'~ lb i~ d irected from C towan:I E. in the cube

~ in fig . P-610 . Determine the moment of lhe force about , eoch of the coordinote axes .

~111.,. .. Fy(,...) - fz(<4)

d · .... g n.

: • . f,." 163.3 lb . ; Fy • 326.slb; F.w;•163.3 IP .

.. 326.5(+)-163.3(4)

~Mic..,. ~2 .8 lb-ft.

%My • Fx(+) t Fz~2) .. 163 .3(4) t 1tJ3.3(2)

~My" 919.e lb- H . :t.Mz. s -F~(,.) - Fy (2)

= -16a.3(4)-3:26.S(2)

~le" -1.306·2 lb-ft .

611.) A force P. d irected frorn F . toward B in the cube .shOV'ln ·,,, Fig.P-

6079 <.?Ouc;es o moment My ... . 1600 lb-rt . Oelerrrlme P ~ '1ls moment

obout the X ""z aJ(es .

dl!. +'t!I• t2.'

d .. .s.~89ft. Py~~ .. ~ :. Py =60011;>. ; P.11 • 4001b.

~Mt= Px(2)

1600 .. p,. (2.) z p)I ~ 0001b.

..&.~ _£____ :. p .. 1077 lb . ... 5,399

£ M,. ~-Pi!. (-t) t Py ('2.)

= - 100(4-) t (>00(2)

Ml\= - 400 lb -·fl . £tlli! • - p.,. (+) .

= -000 (+)

Mi! • - 3200 lb-fl .

612.) A force P is directed from o point A ( ""• 1.;4) towond o point B (-3,4,-1) . If it cou~es o moment Mz· 1000 lb-0, Oeterrnine -the 'liO

men~ of' P obout the x ~ Y oxes .

,,~· : -- -~~---::;! . , . I , I

,'/ : : ,/' : d = 0.11 ft . (-- --+..-.::-. ~- ~·-· .,, ~ P • Px • Pv • P.t ; / a ~- "-a- Q.11 7 3 T I ~ o' ; ,' ., , ~::_ __ P. _ __ __ _ ,.,/' .. "

: . P. ~ 1iy'g.11 ; P1 £3o/9.11; Pi!: =-!:!o/g.11

£Mr• P,.(1) t Py(+)

1900 • 7/g,1' (P) ~ 3(+)%.11

p" 911 lb .

115

'f:

I · 1

ll r .

••. p,. • 700lb ; P1 =3001b; Pz • ..sOOlb

~M" ~ -Pd1) - Py(.,.) = - .!?OQ(1) -:- 700(+)

M11 " -17tx:l lb-fl .

.Z.My - P;? (.+) - flll ( +}

- ,!500(+)- 700(+)

]v\y - - BOO lb-11.

61+.) The sheor-1~ derricl-\ .shown in flg.P - 61+ suppod" a ....ed ical lood of 2000 lb . owliec! al /\ . Points B.C. "'- D ore in ihe some hof'i­z.onta 1 plone ~ f\,O,~ 0 o~ in lhe XY pl<;ine· Determine the force

·in eoch member of \he derrick . . i.---20, 'Y 10' --l A (10.1s.o) d~· • (10-0)2 +(1s-0)2 + ( o-(·s>)t

rsoiatin9 · Fro~t View, ' I

£M.,.c.8 ~ 0: WX>(10)- Dy (w)~o

Dy e 10()0 lb.

L:£J_~ ~~ ~= -2.L .33-f>+!' 15 .30 ()

·'.· /\D • 12236.1 lb.;

D>< • 2£XX> lb ; 01:. • O

r d.-.e .. 16.708 fl . d~2

=(10-0)2 t(1!ro)

2 -t (10-0)

2

d"c • 20.616 fl. d....; ~ [10 - (-20))2

t(1s -0 )2

dAI). 33,~ .fl.

f.soloVing Top View,

.t:Me~o ·

o,.. (.s) - c,. (1s) .•o

Ct. =[2000(•1J/15 - Di<(1o)t B>< frs) ao C• ~ 666 .1 lb. a .. "' [2000(1o)J/ 1s

Bx • 1333-'3. lb.

~a fL '~·· Cii! - · 20 · "'6 10 1 !! 10

A C " 1374.~ lb ..

AB = 2494-. .3 lb.

116

,.

61~ -) The fromoY110rk shown in fig . P- 615 consists of three mem­

bers /\fl. /\C," /\0 whose lo....-er ends .or-e in the .some hori:con­tol plone . 't-. horizontol forc;6 of 1000 lb octing poroll~I t6 the ')( o~is i~ oppl ied oi A Oe\erm·1ne -the force in each m~mber.

lsolote Front Vio_w,

*MP.AC i) •O: By(6)-1000(6)•0

By• 1000lb.

~ .. Jh_• Bi< 6.1tl&' 6 3

""8"1118 lb. ; ~ .. 5001b.

d"1 •(6-:a}'+(6·0)'t(o-o)•

dMI •6.706

dN:.1 •(a-o)• +(6-ot' t(o-(-3))"

df.c - 7.35/

d,.J •«~-0)'.1-t (6-0)i +(s-O)'z

d.-.o. 8,367 ,

l~ote ·Right Side View,

£Mo•O

-1ooo(B) tCy(9)'"0

Cy"62Slb­

AC -~ · 1.se· 6'

l\C "765.(o lb.

£Mc. •O

1000(3)-0.,(e> ·o . Dy• 37.S lb .

~-~ 8.~' 6

l'\O • .5~:2.9.+ lb.

616.) Referring to Fig.P-615. replace. the 1000 JO force by a verti­cal downward lood of 'l.00() lb. Detemiioo the force in eoch member under \his revised looding .

I>' 1-..(.M>.0)

' ' I ' ' ' ' ' . • "&,1

_L_ ,._J,~: __ J_ __ _ .!1-',-'0 3' I! J'

/ D ,,/~

,. (O.o,&) ~-

d111a. 6.708 fl . d"'c ·vk~-0)2 t (6-o)'' • (o-(-3))1

d>.c ~ 7,35 ft . d.a.o .. vf.3.-o)2 -+(f,-o)1 + (s-o)1

·

ct,.., .. e.361 n.

117

·Isolate fr-on\ -View,

£Moii.c" 0 By(') - 2otxh) ·o

. By • 1000 lb.

~ M ___J,k_ 6.70&' 6'

Isolate Righi Side Viffl'N, y

I ,

..tlllo •a (~c00-1ooo)(s)-C,,(8) •o

Cy:: 62$

...L&..· ~ 7.3$ '

"

~Mc · O

Oy(S) -(WcXHooo)(3) a O

Dy• 37S lb .

AO ... .Qx_

AC • 76!5,6 lb . e.367 6

/\B • 1116 lb -·- C r.o .. .522.9 lb.

617.) · The point(; B, C, II., D of the con Ii lever frorneworl'- Ghown in

fig . P"-617 ore ottoched too vedicol wall . The 4<Xrlb lood ii; poro­llel to the z. pxis , "&., the 1200 lb lpoCl is vertical·. Compute the fon:::e in eoch member.

!Y I 0 (c),,,q)

I 7 ,,,

C ·":._--;-;_,../(o,o,'1-) .I.

lsolot .e Front V iew,

£.Mc.o,e • o 1200(10) - 0.(6)• 0

o~ - 2ooolb.

...t:Q_ . _Qz_ < ~ 11.66 10 10

l\D • 2332 lb - ·-T

d~ "v't-10--0)2. t(o-o)~ t (o~(-4-))2 .; 10.n '

dN:- .. /(10-or f co-o)" + C-+- o) 2 • 10:11'

1~1b

1.sola+e Top View,

~Mc-o

. c~

a,.(s)-~(.+) -100(10). o .001b

s~ "1eoo lb .

AB • 1~00 10-77 -10

2'Me •O .AS " 1615.S lb-·- C

.2000(<1-) - -'f-00(10)-C~(il) • o

c." .soo lb

_&__ ·~ 10·77 10

AC • .538 . .!5 lb -·-c

118

61&.) 1he unGytnmetr·acol cont"llever fromework .shewn in fig.P-618,

suppor-tli a ved icol load of 17~\b ol /\ . PO.nt~ C ~ 0 ore in the wrne verticol p lane while 0 iG 3 n: in fr-ont of th'1s. plane . G:lmpu{c '

the force in eoch msmhsr .

·~ .

Isolafo

.tMc.o • " 1100(0) -ex (10) - 8y(~) eo s~(10) t Br,(a) £13600 - ··-(j}

d,.,e•rC0-3)1 -1(0 -(-,))1 t(+--0)2 • 8.1'9., ... 0C4i:1) d.-.c • /(s-o)*-.(+<>)1 t (s-o)1

• 1a'

d>-o • Jte-o)2 -+ (~)2 t(o-(-.t))1

• 9.16!1'

Isolate l op View,

c;~ ~

.=::Mo=O

C"(10)-1<l00(")-800(3) =o

Ct-= g-+o lb .

/\C • 8-ia ----:j"2 ----s-

~~...fu_· ~·k._ s.77~ e 6 +

AC ·1260 lb-·-,.

·ZMc •O

:. ey ~ 6/s s,. - sub,, lo 1

a,. (10) f 6/.s 8JC (S) • 1S6ai

13,. • 1ooolb. :. By " 1:zoo lb.

,'. ;\e, .: 17.S.S lb-·-C "

1000(-4-)- ~(3)-o,. (10) •6

°" . 160 lb

1\0 • 160 ~----S--

B.i ·· BOO lb- 1\0 ... 193,3 lb -· - T

61g.) Golve Prob. 618 ·,f the 11oolb lood in fa'g.610 acts horiz.on­

\olly Ou~ward from/\ in the direotion from E to-Nord /\. from Prob. "71.. 610

o"'e = e.17s n., d.-v = 12 n. o"'o = 9.168 0.

119

.. ;

I

lsolote front Viow,

.iMc:,o • 0

0,. ( 10 h By(3) - 1100(.+) "'0

Bit (10) i-By(:l) " 6000 -·-© /\B .. __fu_-~ ·~ P,.71!5 s ~ .+

:. By " ;,/~ Bit - sutw. to© 109't t 6/s a,. (a) • 6000

J3.G 811 "6800

e,.- .500 lb. :. By • 6001b; , f

/\B• 877.!!J lb-·-T ; ,Bi>-400lb.

lsolote Top View,

' I • • I ·, . •' - ~- . : a'

..... : &z

! Cir c ~"10 -o

Qr -C..(10) t B~(l.)tB.!(3)-1100(t)=O c,..(lo)·~)t~(:l) - 3'400 .

c,.- 80 lb.

/'C • 00 1r- -6-. -

' /\C ·1201b ---C

~Mc•O

-0~(10) -B..(+) •Bz(.3)+ 1100(0) co 0,.(10) .. -.500(+)+....oo(a)t 13600

D" "12901b.

. /\0 • 1200 . ••• /\D '\ 1~ . .+ lb , g.165 0

620.) The fromewJrk shown in Fig. P- 620 supports o verf1col

·lood of 2000 lb. Points B,C,~0 ore in ihe same horiz.ontal p\one . Oete~mine 1he for-co in each member.

("""·o,o)

2000lb A (0,10,0) • . d>.& ~ (o++))'l tb0-0)11 t (o--O)e

d"5 • 10.7703 ft. <JN;• " (o- o)• +(10-o)e ~ (o-(-+))'

d1'C • 10:1103 fl. o,....s .. (S.&6-0)1 +(10-o)• t(o-(-s))2

dAO c 1+,1 ..... fi.

120

Isolate Front View, 90t1Dlb

" ~t.1c•O, -Oy(&.66) t By<'.-+), •o.

By = 6 .66/+ Oy -· -© Z.Ma•O, ~( ... )-Cy( ... )-()y(12.66) •o

Cy(4)t Oy(12.'6) - 8000 ....:..- ®

lsolote Right Side View, £Ms:O, Dy(s)-Cy(.ot) •O

llOOO ()y • +Is Cy - · -@ t;u~. to © in©,

Cy<'.<4-) t +/.,Cy(12.G6) • 6000

Cy (1+.120) & eooo

~-~;...;ca Cy = .!566.25 lb in@, Oy = -+/!!> (s".2.5) • ""'.!53 lb.

in©, By = &.66/,... ( ... 53) ,;. 980.74.5 lb.

• • ~~ 9 90.7'+!1 ; ~. ~-25 : ~ -~ 10.7703 10 10·7703 10 1"·"1+ 10

AB •10.56.a lb- ·-C /\C=009.071b-C ; AP·~6 lb-C

621.) /\ vertical lood P" 8001b oppf1ed to the tripod sho"M'l in fig. P- 621 cous;&.s a compress1Ve. force of 2.56 lb in \eg t\8 ~ o

ccmpr.e~ive force of 293 lb 189 /\C. Determine the foree in leg NJ '°"' 'the ooordinates 'Xo ~ ~o of its lower end 0.

d.-..e • J.o-o)• t 60-0)~ • (o-(-8))2

d-'6 = 12· 806, 8 d/\C - J..&-o)at (10·0)2. ~ (6-o)•

(0,0;6) d.-G"' t+-11'2 ,

./>z ' /

,/'

(9,0,6)

121

:I ·1 jf

!

~ :[.

lsolote front View, eQOlb

:. By· 19g,g lb.

/\C .. _k.~ 1+.1+.t 10 1•M'ta

: . C1 , 100.1 lb

~Fy•O

Oy t 199.'~. 200.1-000·0

Dy .. -4{)0 lb

%Me.•0

0y (){o)- Cy(S) "0 .

1<0 • [200.1(el]/4<:lqf° ~o • +1

•

_!1;._.~-~ tCz · 1·+.1~ 0 14-1.+2 I 6

c~ a 160.~ lb. . ! ~ s 120.1 lb.

Ito Vt Dy

zMc •o .J> -01(,-zo) t OOO(f>) - Sy(1") •O

--+00(6 -~o)t -+soo- 2190.e .. o - 2400. -4-0C>Zo . -2001 .4

Zo • 0.997 '

Dz • [-4<X:>(o.gg1)]/10 '· Dz = 39.66 tb

/\0 • J..t60)2 t ~t. t(39.06)f.

/\0 a 432.7 lb-·-C

csu.) In fig .P-62.1, ·,r Ps1irolb. ~ tne coorcl'tnates ci Dare 'f.o .. S'

~ Zo - 2 fl , compute lht:i force in each leg of the tripod . I

p

d~ ~ /co-o)2 t(10-o)'- t(o -l-a))2 d.....a = 12 .006 n.

d ...... c .. /(0-0) 2. t (10-0) 2 + (G-o),_

' c:J,.,c = 1+.1+£ n. dAO .. f (o-(-~))2 t(10-0)2- t ('2-0)2

dAO = 11.3S6 H .

122

Isolate Front View, l&alote P.ight Side View,

I ...

o a s' _!'_4 __ ~:--~ .. __ i;- --- -

Of gy • Oy

LMo•O 6/ .£Mo •O !)

By(S) t C,...(13)-1200(9); 0 C1 (-.) t 1200(2)- By(10) • o 0y(w)-Cy(4) • 2100 - · -©

.subs. <Y lo@.

St • 6000 -Cy(13) - ·- © .!I

{?QQO ~Cy(13)(1o)-Cy(-+) • 2400

12a:x? - Cy(JZc) - Cy(.+) = 2400

-Cy(ao)' .. -9600

Cy~ 320lb.

:. Br 6000 -(a20)(13) =-· 360 lb . .5

Cr + DY t 0y -12<l? .. 0

320t Dy •368-1200-0 - Dy; .!!512.lb .

~"' ~ /\C 32() 12.00<0 10 1+-i+!Z ----:;o--

/\0 - 471 .3\b -C /\C ~ 452. s lb-C

_6!L.. .S12 11.358 10

/\D "..501 .5 lb ·C

623.) Determine the mox1mum safe vertical load w that con be suppor\ecl by the tripod shown in fig . P-623 w1lhout exce.ed­'1ng o compressive lood of' :Z"'<l? lb ifl any member .

w

dOI'\ ' /(0-(:2))1 t(6-0)'l t (4-0)'l.

f do,.,• 7.~... .

~I ei.20-.s) doe " /<o-(;2))2

t(G-0)1i(O-(-e))1

I - .; , ' . e_/~-;r' doe . a.062 ...

,'• I /. ,.<+~ ~~-'f~--._ doc -.t(+-0)2.•(6-0)2 t (0-0)

1

"(-~ •. o,+) ""--._c (+,o.o} dco = 1.211'

123

11

·l '•

~. I I

l

f l

.,. if. [)/\ • 2..fOO b. l' 2.WO .. ~"' k . .1YL

7."'Hl.3 1l ' +

:. /\x • &+1.4·.5 lb ; /\y· 1QM.36 lb ;.A.z •1202.91 lb.

lsolote· R1ghl' Side View, IGOlote Front VieN,

.£Mc;,"'O l\"f(-4-) - l3y(.e) - O

192+.~6(+) •By(!>)

By• 1539-~

checl<.. it oe 72400\b

OB = 1s3g.488 a.oa 6

:. oe • 2068 . .56 lb . :'

../ ·,1 doe& noi ei<~ .., .

• ·,r oa • 2..+<lO ,

~- .az_-fu:'.....· Bz ~.()61 2 6 .!I

. ""'" ,..• •' ~ ,,.' c c.

.. .;-,...------ c.,. ., LMr ·O

(By•!w)t -c1 (+) •o Gt ·. ~(i530·"W16 • 192+ . .36)

. +

Ct • 1'131.92+ lb.

check ·,r DC ;:>2400lb

_QQ_ . '1731.Q24 lb :. DC • 2061·MI lb

7.211 6 /'it doeS not~

Z:M,....,, - o w(lf) - c.,c;) =o

W •(17ai.9H)(:9) • 5191;.g lb.

t ·,r DC· t"KJO •

~-~·2t_·.Q:_ 7.~11 ~ 6 C>

a.,.,sgs.30 lb; Sy=110b.1" ;ez·1+08-~ lb .

ot the r~hl side view,

: .. C1 • 1331 .:3 lb

C1 • 1006.9 lb .

c~ ~ o .tMc· O, /\y(-t) - By(/;) •O

/\y · • ..9/-t(17061G) ~ '223:Z.1 lb

diecl<. ·,f DA "'7 2-400

01\ ; 22:32. 7 7.~ '

OA ·= 270-+. s lb oo ro1' occep1 Hi1s

volue ~ if exc.ecds 2"'1-colb

:. DB I ~lb-

124

lc;olote Top View, Llv1& .. o 1331,3(.5) - Ax (g) = O

/\-;, ""' 7.39.61 lb. '

checl<. it OA 7 24QO lb

PA 7.39.61 '7 • .o>tf>8 ~

CV- , 2 767. 2.5 lb.

: . -this conno! be acaipl beoause ·,t w '1 ll ~

the hrri1f ~ich '1<; 2<1-00 lb ·

: • .so1e ma><. vQlue, ·Ot W j g .5195.8 lb .

E&UILIBRIUM FCR NON -CONCURRENT SP!\CE FORCES 62.6·) The; plate shown in fig . P-626 oorf'i~ o load of 1ooolb opplied ol E it; is .suPP'f"fod in o hor'1~ntol pa61fon by ~hree ver\ icol eobles ottoched ol A,. !3 .~ C. Compute the tenG10'1 ir'I eoch ooble .

Fr-on! View .

"I '°°' ~ B 4 1 , 6' 1 zM...,e • o, - 100!>( .. } t c~o) ~o

C' 4<lO lbT .ZF;r eO,

A tB t C - 1000 • O

A L 1000 - 400 - -400

/\ = 'l.00 lb . T

P,(gh.·G"1de \/t'eW I 11000

6' fC 3', r zMA~·o ,

- C(4")1 100o(6) - J3(g) = O

B 2 -400(~)-t 1000(6) 9

B ~ 400 1b T

6!17,) Solve Prob. 62~ ·ir; in 01do1hon io the -1000 lb, the. pla~e we1CjhG 1200lb Cen\r01d: '#,(1/JCgH - 1{i(3)(,,6)(10/3) + ft(,)(~(~~.3)

fi x : .D (10/3) ~ : 10/3 fl .

'/{(jd}(ld) r • 'fe (?>){,¥> {_Gt Vs(,)) t •fe(,)(~('z./g ·fi> )

g-y . 1'.!J'

1 .. a ft .

125

toke front View, c

J

:='<M.-..v"'O

tol<e Right ~ids Vfs.w, 1.-

lA • l 11 : , r 3 ' I

•' t UtM

~M,..·O

1200(5) ~ 1000 (f,) • 800(') - .13l9) •O 12CX>(10/1) t -tOOO (+) -c(10) =()

C ,:;, 600 lb -T . B .. eco lb · T )'

zF1 •O,

-A+12aot1oa::>-000- aoo ·o t\ • 600 lb -T

,29.) Refer. to unsymmetr1'col .conHlever f ramewo~k described in Prob. 610 an pa96 W6 ~ repea-ted here as F«J. P-620 .. If the '-'l"'­tiool ,\ood of 17001b iG sti1f\ed lo oot o\ the rr11dpo1nt.ormerfl ­ber /\f?; cornpu\s the cornponen\s or \he reocf1on ci\ l3 ~the for· . oes in \he. bor'S .AC ...., /\0.

i 0 .. ,,r~-e ,,., "" '

,,,,., ,,. :• ;"" I •

c - '"""'° .. ~~:- ' e· 1: ------' , , ., 1 1· I I I ,,

• "'~ h"TO' ; ' I f

/ . -~ - B

f ront View,

.£1Ac,o • 0, g. (1o) +Byl~)-11oo(e-M) ·o e~ (10 )-1 By~) · g::iso - ·-©

.t.Me •O,

dNa··~ /(e·3)1 t (o-(-f.))1 t~·O)._ • 8.77.S fl. d....c. /(0-0)2 t (<1--0) 2 + (e-o)" • 12 n. d...o - la-o)'- t (+-o)' t (o-(·.t))

2 "'g,166f).

@ Cr · G / 2 : Ci _-c,. .6IL. .!2!_ • EL • ..Tu-9 .1(.5 6 + 2

@ 0,. • D~h. : Oz " o.:/+ subso.@t....© lo ©

(o.,.f~· t C../2)::. -(c,..+O•}IO + +2.!50 c O

a.5 o,.-+ e.-' c.· = +2~0

o. +C• • .soo' -·-@

(01

• c 1)(3) - (c. t0.)(10) ~ 1700(2.5) co (o.,. • c1)(3) - (c. i o.)(~o) • 42.&0 ~o-· :.@

126

ZM0·0 JJ Oz(s) -0.(•)- ~(a) .c,. (•) ·o 0u+ : 0.a • 0•/+ ~ C• ~ c,.

(o./+)(§) - o,.(G) - c.(3) ' c,. (+) . 0 c,. ... ·.s.2.!!J o,. -' · - @

subs. @ to@

Ch· eolb From@, c. · ~·2s(eo). -+-~o 1b

;\O. [eo(0.1~6 )1/s AC. [4-M(1 eU/.ot

AD ~ 91.66 lb IC • 12GO lb.

Retell"•r'lg lo Front Yiew ,

Z:Mo 0 0 6 Bi. (6) t 62:(3) - C..(10) ·, 0

s .. (6) t Bz (a) · '!-200 -- - @

.£Mc.• 0 ';J B,. (+) - Bz (3) - 0,.(10) ·CJ

0x(+)-8i'.(.!!J) r 0CO - ·-@

1'00.@~® . B,. (+) -~ ~ 000

B,. (4 ) ~uG) ~ 4'200

81< (10) = .sooo

ax'· .so0 lb.

in CY , S()()(-t) - 13;!. (3) c 800

Sl!.c 400 lb .

in ' ©, -!100(10) t Sy (a) : 9350

0y = 14-SO lb .

.630.) The Boom BE of lhe .st.IT· leg derr1c 1<.. shown in Fig. P - 630

IS rotal<!ld fOrwol"tl :.!!Jo' measured In o horrzon\ol pkme- . The mast

t\B ·,, vedicol "'- '1s suppor\ed in o .socl<et ot /\ . The points I\,

C. """- 0 ore In the .some horizonta l plane . Oe.!emi;ne ~he forces in ihe legG ec ~ 130 \..._ th9 components of the bsarinq reocrion d A

deo ·dee = ...[i1s·

127'

·1

ij I

11

I'

~· I

·I.

' l

lsolote Front l~olofe R19ht ~ide View, ' t - • - Ill

0 1;•;-c;;. I?< c..O

0, c.,

±"'c;.o~o, ~o·O, C,(1lJ)- 1~(s) =O

/\y(!i)-~(!!H &.~) cO

/\y = 1~ lb .

.i<f.M,..~O 9 Cy(s)d)y(s) -soo(s.~)·o

C,.(5 ) .t Oy (.!1) " .+?130 lb

Ct t Oy ~ 866 1-·-© £Me·(), .

Cy s G83 lb .

:. K . ~ .·. ec " 1~ 10. {fi5 15

'83 t Oy = 06G

Oy • 1~ lb :. AO ~ 183 AO• 202 . .3lb.

-AJ.(1s)-t (c,.-. o~)(+s) t .soo(s.~)-(CytD,Xs)-o -/\ .. (1s) "(c,. • 0,.)(1$) -o -· -@

. {77-5 ~ subc. i. ®1 o" ~ 103 • • o~ , 61 iv . ~ -:;a-

""M ~ (:z17.7t.~;~ A• = 200.1 ,1b .

D~ ; ' 61 W/5-.. - 61 lb. . , I

~ ~ 227.7~/~~ :Z27.7 lb .

:. c,. ·:Z27.7 1b .

Refe1'1"'ing -\o ~~igh~ Side View, .£Me •O: /\z6s)+(DL tCi!.) (1s) +Cy(,;)- Oy(s) - .soo ~s) • O

Az ; -(221.1 t61)(1s) -: 683(.s) t 103(1!1) t 2500 15

7\Z • 166.7 lb .

6:31.) The boom BC of the i;diff · leg def'ricl\ .shown in Fl(JP~ ;6

contained in the '/.Y plone . The mo~t f\!3 \G vertical. ~ rosfs in

0 .sool<.B-t ot I\ . Points I\~ 0 ore in t~ same horizon to I plone ·. Point.G . D ~ E. ore in the .same vertical plane-· Determine the­forc,e,s In the legG l3E ~ 130 ~the comp'.ments ·or the- bea-

i 10' c ring 1eoct'1on o1 /\.

128

, I

dei: s ~lf • (:20· !5)• , (10-0)• = ~4:Zs deo = ,Ao-(-10))2 t(20-o).,. t(o -(;-16))

2 c Koo

lsolote front View I • l sol ot.~ Righi s ide. View ;

I 10

.h.:. " ...

~Mo · 0;1oao<_ro)-A.,..60)- Ex(.s)•o

/\y(-i.)t Ex ~<1000 -·- © ~s·O;

0,.(10)t1ooo(w)- l\y(10) - Ax (s) ·o Ay(lo)t /\>1(s) - OJ1(s) • 20000

t-.1(1.) t A.l( -Ox • "\-000 - · -® .ZM,.. -o,·

0

• /'<fl. 10' ---~r~y-----

.tM ... •O,

Oy(10) - Ez(.s):- Ey(10) ~o

0y(2.) - E~ - Ey(2.) •o -@

Ei!j1o ~ey/1s : . f:'i! • 10/ts Ey "u'@ to [)y(1.)-1o/1S Ey - Ey (£) "0

Dy (1) - 40/1.s Ey .. 0

100?(10)- Ey(10)- O.,(~o)- El' (-') •.o [)y. • .ote/3'> lfy sub&. fo @

@ 2E1 t Oy ~1000

Bu~: e,,;;~ • e,.f0 ••• Ey-= 1.sE,. :a;,.+ 4C/30Er ~ 1000

10<;tJ(2)- (1 .~ !:,.)(~) - {)y(t)- E,x '0 Bu•: Ey = 1.s E,.

<4E• t 20,. · ·2()0() -· -@ 21::" t ~/30(H,Ex) 2 ·IOOO

:. r.,. "6000- 0r)/a. .s1.1bs. fo© -t b c1000 :. El'~ 2!50 lb. Dy/:20; D•/10 >. 0)1 • Otl:z -.s1.1bs .~o@ BEj.J1Ts =.WAo . ·.BE c.51S.4 lb .

Ay(+) t A.,(:2) -0y •00()'.) lb-~ t.J.sing ~y·1..s~ < 1 . ~(2.56)

/\y(2)-1 A .. - Oy/2 • 400010

Ay(2) ~ (160C>:z-!?.l..) ~ '1000

A~(+) I 1ocv.l - Oy ~ 8000

(4}1\y - Oy • 7000 -·-@

... ,.._, ...5

€y ~ .37.5 lb

Using Dt •«>/.lo Ey • "t-0/:?.o (31s)

Dr = .soo lb. BD/~ • 5Ct>/20 :. BD • 612.4 lb.

U~ng@ Arl"') - Dy • 1000 /\y ( +) • 1000-t..SOO

Ar ,. 1soo/+ :. At c 1P,75 lb . ~tAx (:z) -~· 0000

- (Ay.!_~ -'-9y, " 70()() ) UGin.g E, 4 10/15Ey .. 14<(5(37S).::: :zso lb .

05!. • [60'.)(101Jho =- 2SO lb .

reter'f'ir'lg to l'.l.iBh+s;de, View : Z.F1 • O,.

A.,,, • Dz - l=11t ~o

Az. • I:~ - Dz ~ Z-50-2'50 = o : . Ai';• o

129

c;32.) The boom of the c}erricl'- shown IP Pig P-f>32. ·w; rotated bock. ~· from the Y.Y plone . Oe.terrri1ne the forces in BC ~ BD ~ the component~ of 'lh6 beoring reoc\iOf'\ a\ \he ~\- /\ of the~ AB

&(•o,.,.o) .

ir.(~,ll,·!>) ~z • smao"(to) • ,!;f} .

I fltl(X)Jb 1- " 10 COS 'a()0

• 8.66.fi . I : i~.+s· -1tJ/;;:D :. N5 .. 200.

,,, 1 -.. __ •• Jfli "'n'°"• p()/t.o .·. 00"17.32 n. '.I.. - o·--71,. " -~ ton fiO" ..17.3t/6A :. i5A • 10 n.

0 ·- - - ---- (10,0.0) ~· (0.0,11 • .,..) · C.CSE()' ,. 10/ CA.. :. l5A,. :zo ft .

dBA .. .)jo-w)l+ (26-o)•+ (o-o)' " 2Cfi. Ell • <5A t x

dso • 16o-c)" t (w-o'f +67.32·0"'f : 28-280. b • 10~ s.~<; ~ 1s.~ fi . dee ·,/60-o't+Gi:>'o)'it(o-(;17.!Jt)) 2 • 26.~& {1 . -oc .. 1otoo oo· . 17;~2 n. 160iote fron\ View, 0C = S77.3~6 :. ec "816.3 lb.

UCO I I : ,·

co9 s· I .~ -°' -__ 1~~-- ~--- : .

t.&.'26 2a

'" Eq.© Cy tOy•1132

~ • 1732- 577.316 " 11.s+-'&+ lb

BO • 1154.60-+ ••• 00~1632.7 lb . f&,26 '}.()

Cy Ay 8.'40' Or

Ref ern~ to F~n1 View: ~f,.,, 0

A. - c .. -o~ "'0 c. • sn.3~(10/i~) ~Mc.o.o • 0. -A1 (10)•2¢00(11uc}~ o

Ay • 3732 lb

;t"tv1.-. ~ 0 (£,

(Ct• Dy) 10 - 2000(a.1;~) • o

Gt t Oy • 1132 - ·-@

loolo+e Righ~ Sida View, ,, ~-:

Ax .. ue."5a +.s11."M1Z

A,..-~ lb

Cit.• .&77.31G (17.~2) 'LO

"268,i:i?i8 lb.

0, • 11s+.~(10) w

o~ • sn. 3-'v.l lb.

Q .. ,.99.96 lb Oz· 11~.66+(17.:32) w

Dz = 99.9 _qr; lb .

Rel. \o P.'ogh t ~ide View : £:Fe ·O

Dz - A~.-cz ' 0

Ail. a 999-96 - 4:99 .96

0 ~ Al- a .SOO lb De "• A G1 C. __ .. _ - ----------

Oy i?-32 . l\y 17.31, c,.

~tvlo s9; ·3732.(17.32)+Cy(~.~) 1icn?(na2)=0

Cy(3"!-.c.-+) • 1999S·Hll'.J

130

Chop tcr 7

Centroid<> &, CentorG or G rovity

131

I

1· I I

" 11

10s.) Delerrnine t he cenko1d' of the shaded area shown in fig · p- xis, which '"' bounded by the "'/. o·i1G', the line -x •a ~ the po-

robOlo yt/• KX .

'/

0

A'1. = -g,a•b -~~ s ~

' y: = :30/.s

r

o r ~ "

AX· f xdt\ ~ S: x(yo><)

V j( • yg r'* --::;:--

~ = +%.-

!lr• KJC

K-~% • b% y'Z" b(2;y{:, - y - t>JX,4a

J\ • f: ydl< • fo° b ~ dx

• r-!Q r~~~J: - o/-ro: r~a&~a - ·o J

A= ~3ob

Ay • ( Y~ydA c Y~f: ~(ydic)

~r~ ':j~dl< .. hr b:: clx

Ay. -= b% f ~]~ ·

132

707.) Determine fhe centroid of lhe quodron t of the ell ipse

shown in Fig. P-707. The equotion of' !he ellipse ·,s £.. +Ji.. c 1 . O" b•

}( .. '(._., . .

r ~ rr ~%~ c 1 - )(~a'l y

b 1./ " b2 (o• -)1 12)

o«

tJ "% Uo4- ;-: ') .,. 0 a dx

"·r ydx = f bfoJo"-y;~ cJx L&t )( u 0&1no< • •• sonO(" X/o

dt " OCOS"<~

ocosoc·~

/\·%f<oc.oso()(ocos0<d0()

- b/t11.fo1'/c o .. cos'o< do<

- ob r1• Y2 [1 t COG20<]~ - ~b [O( t ~i:h

Ax "kl<(l_:ldx) " r)((%J~·-l<')dx = % [-Yi_·"% (o"-l<") 9h1: a o/o [-Y$(o). y, o~]

Ax .. -sac." ~x 3 + y: - 40/glr

M .. f:}'i !l(lJdi<) • ~r y'dx

c r Kt (i''<::-x.)) dx

- bho" r o.•,. - )(~1: • bfao" [o.9

- o.%] - o.b,h a .aj?-[Th +(~

0

- 0 -~0

] Ac ob~

~ii q~ - y- ~hlr

100.) Compute the oreo of lhe sponclrel in Fig . P-7oe bounded by

the )( oic1G, the line x ·b, ~ the curve y .. KXn where n ~O. Whof is lhe

loc.otion of ·,t s centroid from I~ line " • b? Prepare the toble of area~ ,.., Jocotlon of centr<,.ld for -.olues or n t 0, 1, '2, &so., 3.

(t>,h) ~ · l'-x"

""·" ~/1-n = h/bn

~.,.hx%n

133

7j"' b(nt1) bu! !hisx ie nt2

reffered to(o,o)

I I

·I·

x = b _ b(nt1} - b(nu) - b ( nt1) nt2 n t2

no.) Locate the ceni roid of !he Of'eo bounded b~ the I< oxi6' r., the sine curve y = os1n T">{_ from x "0 to :x = L- ·

0

::· ;\ij : f>12 l.J LJdi< ~ 'ft ( "J-i. ch

;.. ".C .!Jdi< ~ f o sm'!"~ dx

., a 'if (- caGT~)~ ,. -ol(, [ cosT9b_ ~ co~o]

- -.91_(-1 -1) l

~ ~o C s 1n!Z.Tii. dlt ..

- ~a« Yr [11< - s1.,211xlL ~L '1> L •

~~ .. aC.J,. -'ij",,.lo T t a ~

7i-+.) The dimensions of the T- section of o cosot -. iro~ beom are.

shown in f ig . P -714' . How for K; the oentro"1d of the areo obove -!he boea.

~f~ /\ij=~A~ T [1 (9)t 1 ~~1g~ .:·4~(B)(4t1)]+[1 (,)(o.s)J

~ .. 3 .07 i11

134

71, .) Determine the coordinates or the centroid of the q r-eo

.shown in Fig . p-71s w"1ih respect to the given 011es .

' Aij = ~Adlt f:-r(6X9) + ~T(-a)!j .Y ·[~l')(9)('/3 ·9il ._[1'2lf32](~ t9]

• 4u+y = ·307.23

x !J c 7.+7 in .

Oj -41.1+Ji • [v2(6)(9)('f3.6)] t [ '121(3) 2(35]

+1.1+ )i " 96.41

· j = 2.3+ in .

716-) A slender homogeneous wire of uniform cross .seqtion i~ bent info the shope shown in fig. P - 716 . Oet. the coordir\Otes of' ils cen ­

trc)1d .

26.56'6 !j = 66

y • 2.+e in 26.soGx ~ -6(-+) te(+msao·t-4)

1cs.~i • 35,7129

· 'f: • 1.3-4- In.

717.) Locate the c.en+ro1d of the beot wire shown in rig. P-717 . The wir-e

'is homogeneous~ of uniform croSG Geetion .

;: " rs•rict ..; a ••nso· • !Z.065 in. ~ :aoic T/t90

s1nao·~ ~~in -2. ~ • 1.+a~•1n . [ 2(2J1.3J1. (.30°.>< J'/ieo•)) t t] ~ · [2(tx3l<(adJ1.•l,W>))(1.+a2~

8.293 y = 9 .996

.g = 1.086in i x • o

718) Loc.o•e the centroid of the shoded oreo snown in. Fig. P - 11e .

Y [c6)(12)(Ys)• Y:z(<.)(<.) • Yt(<.)(')1 ij = [~M(11)(~·,· .,)] ,,: i ['12 (6)(,)(%·')] +

,. . . ['It (<.)(<.)("h · '")l -~,...+-=-'"- ,..

72E° .. 43~

~ s oil"I. 721. = [12 (<.)(t2)(Y-'·125] t['l2('X')(r/,·i;)J +[~~)(,)(9/a ·6 to)]

72 'i = 360

~ .. sin.

135

·1

.1

I

. .

no~ Determined the centroid of the lines t hat f'orrn the boundo­

ry of the shoded o reo in Fig. P - ne. [ 1H ,f18o t6· t ffe + .J'7iJ x "' 1:2(6) t J100 (-lift?>h)( o/~) t .Jn (.f%)(V.fT)

t .rnt(. ~)( Y-l2) + 6)1 <fS.387 j' ... Q.5'4.322

Y: "' 5-26 in . . 40.30~ g = 12(G)t6Cahf19oK~)(Y.rs) t 6] t ffe("!%')(Y.ff) i ffe("%)( Y-11)

49,397 if " 261,659

g - .s.+1 in.

no.) The centroid of the shaded arooo in Ffq · P-120 '1s required to

lie . on t he Y o.11is. Determine the distonce b thot will fullnll thi s

requiremE.nt. [ ~ (-.)(eiJ['lf -aJ c[MbX6U[ 1/!tb]

95.333 in • .. 2 in b 2

ba " ~.666s ,,,~

b "6.53 in .

121.) Refer to the T- section of Prob.n+ shown in Fig. P - 7•4. To ...mot value shou ld ·the 6 in . width or the f\onge be chonged so tho+ the

centroid of the ON:!O Is 1.s in . obove the bage ?

[ 1(9) t b(1i2.~) .. 1(0)(•1-11) • b(1)(0.~)

'10 t 2. 5 b " 40 t 0·8 b .. , 2b '"20

b-1oin. ,. 72~.) Locate fhe centroid of the shoded oroo in Fig. P - 72'1 crcate::I

by cutting o semicicle of d iorneler- r ft'Om o quo.-+er c ircle of'

r .

[ J~C _ T;t_l. )( •(¥~£)(~~ -(~<2 )(%)

.IC.~ r,% - rX2 .

!l ~!J B

B

_:rv«~ " .9~ · -~ -~ " o.637 r 8 1 12T

e(v~)( -t~-,) - (irr,Ya )(rlz)

vr,% ij .. r% - V-'# T~ 3° c 6.57S r~

>t$..6

~ • 6.57~C (ff

136

12.a.) Loco~e the centroid or the shoded area in Fig. P-723. y

. s

[Y!2(1 .s)(G) ~ + .s(G")- T(~)~V:., Y11(t .s)(<;){Yt(1.s)t+.s) ·

_ +(+.s)(c;)(;.~s)-: v~L)(W) )( 24."431.+2. 714.15

x • 3.o+ in. ·

24.43142 g ~ +.s(s)(a) t ('h)(1.s)(c;)(2/5.c;) - Tr(,)""( r; - ~) - ~

24.4314-2 y : 6'-9.599

y • 2.69 in.

n-+.) find the coordinores of the centroid or the shoded or-ea

shown'" r'ig. P-n-4. --tr-"'-.-TTT77'77i'-+- -- 12{E1) - Yt.(c)(c;)- lT/.f (a)«(va) - T~)~ i< • 111(1e~)-~~c;)

(2/~(G') t H)- T~•t(<f-) - T~t (1s-~) , .. 1+4,!593)( e 1118.531

1f , " . j( ~ 7 . 7+ iri . 1+4-.s9ag; 12(1e)(c;)- '111(c;)(c)(o/s) - v(+)"(12 - 't/11-)

- ~(12- 1jc;>) 1-t.+.593 :g • 7.;Ja,76!l

Jr • s .oa in .

nts.) Locai'e t he centro'1cl of the shoded oreo enolosed b~ t he curve

y11 •ox'!+... the stroigh~ l•ne .shown in Fig . P-726'. Hint : Observe thof the curve y2 • 01< relat ive to the Y OJCiS ·,~ of the '6Clme fbrm OG ~ e

t<-x 11 with respeot to the X 011 is.

y~2•0"' , f-2<1~M _'f11(t>)~2~j "f%(111)(~)(3J~-1'1~ -r~(cx11x%-1~J 12-,. - -!57.6

o 111' . )< x • +.at-1. · 12 ~ =[%(12)(c;)cafa ·"~ : [!111 (4>)(H1)(Y~ .,5)

1-ig • 3c;

g : .::ifl.

121.) Loco~e •he cen~roid of" the culvert ..shown in Fi'g . P-727 . H1'nf : In­

tegrot1on '1s unnecessor~ ·,r !he oreo ·,s subd'1v;ded 'into the e lement

to be- found In Table v 11-1 -l---1 , --t [1(Y11 )(,X&)rn2(')- o/gf ... x•~1lr ,r(reWt~)('J9) . 111(6)(~)

-j - (~Is) (-+ )(1a) (g/a ·1>)] 6'

'59~ c 204

.H ·a.sf'!.

137

I I

. ·1

I

I ·1

1119.) /\ rec tong le jc; . divided into two padG by the cur-ve y • l'-x"

o& sliown ii' fag.;P-729. Using #"le ~n°"'n locoiion of the centroo ~ the lower part /\ oG g"1ven in toble v 11-1, show. thot the cen ­

troid of the. upper por-t B ·,c; loco1ed by 7. B - Vii iI/d.•> ~ e • 2 g A . Y _

1 61:...-en th:lt y,.. -I n_t1 ) h .. j..,. • b - ..IL

~ \,.40t2 "I . nt2

= b(nt2)-b nt2

~" = (nriLb ~

find "Xe'ii_iYe

[b(n) - bhl~e"' bh(ttt)-lbh \ (y(7n1).\

· . nt1J ~·n1-~ . .} n t«} )

. .. we -~ - b 2nt1 - bh%. 2 ~) 2 ant-i

.:hruLg& -~bh'-00 (n+1) ~(ent1)

~& : h(n•1¥nt1) ~e • 2~,; -,.o) A beam hos the cross; GeOfion shown in Flq. P-730 ,Compvte

1he momeni of ar-eo of the shoided portion obovt the hori:roniol

cen\ro100\ o·i1G ")(., of tre enhre se,o\ion . ( t\ote : il lQ ~ in G"!~­.!J'ft of "'o\er.olG ~t thic; recooulh; is ~ in compulin<3 •he rrio-

x1ini . .1m ""heor11"19 st~~ .

138

[1(,) 1 12 (1 ) 1 ~12)(t)]g • t(,)~s) t12(t)(7)112(1)(M)

:ao~-111

~ - .s.1 in .

.ZM:ico • 1 (<.)(J,s) t ( 1)(7.!!)(-'·6.s)

· L.M~o • 7.a.+ "1n~

731 .) Two 10 in.-15.3 _ib chonnsl~ ore welded to~1hel"' os ;shown In

fig. P - 731. f1rid the moment of Ql"'SO or the upper ohonne-1 obout

the hor-lzonto l centroidol O>tlC 'k ·of the. e.nf1re, Geof.ori.

oreo of 1oin - 15.s lb - + .47 in7.

2 (4A7)g - +.+7 ( 0·r 1 10) t +."f1(s)

e .91 g ~ i;7 . .s0'+

3- 7.s& in

73fl.) A 0 bridge fnJGS 16 ccmpo.ed or the element Ghcwt"\

in F.'9 · P - 7~ . Refer lo Tobie. vu -g for +he properl1-es of the angles

~ locote !he centroid of' the built -up ~eotion. 19•1·

[ 9(1)i l'(ll..)tJ,7S ti; .-n;J g "~"J06WS) t '1-('lz.'i..e)

• ..,s("-o-78) t ~;7s (1.&a)

~.s 5 = 2_79.3+7.s

B • 10.s-+ :ri·.

7~) U:x:ate the centroiol of the. bu'dt -up Gection «.Chon 'inown ii"\

fig. P -73a. Refu lo fable v11-2 for- the properh-es of the elements .

[111t3.7.5 tll;(o.e)1(~)] ~ .: l!l("to.211 - 1.,5)I3:1!l(o.-,s~o.211)

• 6.,Xo.r.)(.a.2a)•(~"f..o.7)

27-7GS g = !ZG!l,<t-03

E .. 9:~ in .

139

I :j

I

1M.) />. right trion9le of s·1des b'--,h ·,~ rototed obout 011 o)( iS

coincid··ng with side h lo genef'ote o right circular ccne .

,/~ ,' h . . .. ,,,, .. ~ .. - ; ··

y : !2tr . ii . AA • ~i . ~ b . Yit bh

Y = ViJTb.rh

- .o.· . 737.) De~ive the e11pressiorlG fur the s ... rfool oreo ~ volume 9enero·

tod by rotoling o semicircle of ,...od ius r obovt it<> d ;om&ter .

V " alT. ~. Area of holf' c1ivle -~'f · ~. Tr 2

;81;. - g._

v • 'Vs Tr•

A • 2T. ~ . Le.n9fh of oul'Ve

Yor """"bol.,;d , 20. ~ . " .c'eT . .::!J2. • .a..bh

'Q.2 '5.

" .. ~ bllyh

to one-holf'

7SP.) . Deter~med the volume of' the e llipsoid of revolu t ion gene.ra­ted by rototinq on ellipi:e abovt o) ·,t~ m0.Ja- dxiQ (pr'Olote elllp,oid)

~ b) ·,js m"1nor 011is (oblate. ~ll ipsoid . Tot--e the larger .sem~o,.~ oG o

. ii,., t~ Smoller ~em; - 0><1-s os b.

o .)~· r%-.. +~81 I> i

···--- . • __ a .• ,' A i=u.iRS;i; • lr ob

140

;\~ "' ( ~(yd~) . r: ~4dx -): b%·.Co2-,a)dx

• b%« [ 02x - x% J:

• b%"-[o'-o% ta1- a%l

A~ = b~ [ 40_%]

Vff'!.Ov.n. • ~T. 'ij • I\ • ~ - <t-b • Tab

.!!l; ~

V " 1-Tab~

ll!fh ~ : \bf.b ..

~:~ .!IT

b)

Y=~f.a .Tr~

v ~ 21f·r~a A· 2J . o. 21Tr A= 4lf<ra

M1.) /I 60. pipe elP<'w hos an irifernol d ;o meter ot + il"l . th d of t r /h ' . e ('() . curvo .L)re o e p•pe ~ oen~er l1"r1e. ie 6 in . f ind fhe i0 ternol vol ot ths elbow ·

V ~Areal\ centroid ~d1stonoe teverGed • lr(~yz x "x (6<:>' x T/,Ba')

V -= 79.96 in:'

141

~· •ii

7~.) find fhe volume or the. .sphel'icol wedge formed by rototi119 \hf"OlJ9h Ori on9le of +~· O &emicircle of 1"0d.1us r obout ·it~ 0068

diometef'. y. 'f,4 ' lk'Ai:: • T~ y • .J.r!._

6

A = l~ Totol Gurfoce Area • ¥ t lrit

A. - 3 .T 2 1c.a. T r

7.of<!! .) Cornp~e. the ..surface oreo ~ volume 9ereroted by rota~ in'3 in Fi~. P-7+3 through one revolut;on obovt the. X·oi<"

1\1 ~. ~~ t [~ t 1.(;l] E • ~ F'~· ][-~<;) + ~ t ~[ ~][ 2~l

:<>1. 6"'" _g - 1ai. 1:i2

~" 6.15 in .

yf 1l1 . 6 -1.5 . 2(21.%") /• V • 1667 in~

TMt H 2J~·H .. ~ = T(-+)( 1j-;l ti;) 1 2 (') t 2 (~s·,,• .s1ni;~·+J 1{7.'<J' ~ - 15(/. 66 .

9- .s.7in. Aa 1lf .s.7. 27.99 i

7.++) The r.·101 of o pu lley hos the cross ssction -shown in Fiq. P-7#.

It the rim '1G 11\0de of stee,J we:1t3ht'ri'3 4-90 lb per cu. n. rle.term1ne. H1C

weight of' the rirn .

¥ [4(4) -2 (1:'&21(!1)] g . 4(4)(-2-) - 2(1·~"~)(+-%)

11.5 'j = 1 s.s g- = 1.61 in.

v. 2T .(1. 61 •10) ,11.~

_ :__ 10• V = 939.9 in~ - _ Ve o.+ess rl 1

w • o.4BS.S ( 490)

w" 238 lb .

142

7'45.) n.~ oreo contoined between two c.oncer'ltric Qerr11circle.6

of rodi i 1.s in . ~ 3 in . iG ro~oted obout on ox is +in . owoy ~ porallel- . to the booo d iometers of the .semic1roles. Compuh3 fhe surfooe oreo ":> volume generoted by o comple~e revolu -

t1on . [T/j! (~•- 1 .s•~ '1 = ~ (•)' f W?J -%{t·!S)2f 4,.(~e)J

l

! l ' I +' ! ! i(), 'O?! g • 15. 75

~ c 1.485 in. Y'" 2T.(1.+8St+). 10.Go:3

v · .365.1- in~

7-46.) Oofet.mirie fhc surfooe oreo ~volume gene.roted by o complete

revolubon obou• •he )( OlC i s of the .shaded. area or Prob 719.

' ~~Gin ~ =.!!'l in

v • .2 J • !3 . Areo = 2r. '". 7fl

Y = 2714-:!S 1n ~

..Lll<..--,----___::.L-.::" /\ • !2f. S.+1 . (1H~tJ4•t1fi ti'~) ,.. ,. A " 16+<f.77 in.t

7"4-7.) Compute. the. .surfoce. oreo ~ volumS generated by o cp~plete ('C.volutio.n obouf the )( a~i~ or the GI-laded oreo of' Prob. 72!5.

]" •.s.oe 1n .

v c !Zf. c..os . [ +.s ((,) t '/11(,){1.S) - T(9Y%J

v m 779.92 in.?>

A= 2r. ~- le~lh of c ·1rcorrcribina,Areo

A • 2lT . .3.91-[-., (11)1 + 6 t .s ~ ~,•nscJ A "" +92.s7 ·1n.2 .

] for length . r3 i lit.!~ t f•s,H~ ~ ~ ~6.s)t +.S(•)>6(') • ~ 1.st.,.,,,(J 1.i:1&' )(~)

. t T(% (6-~) 19.997 g ~ 78 :a2g

_g = 3·9+ in.

143

ii '

·1

7s1.) Determine the centf'Oid of o hemisphere of radius r, ta­King the ox is of symmetry as the z O)(is.

v-,. • ( x("t':l')d1t =Tr x (r"-J<•) dx

~(~Tr~)~ c'[~ -~I nr' x - l r r2 .. - :-tj .3

2:11- ·'l[~l y; ~ % r

1.sa) A uniform wire "1s bent info fhe .shope .shown in f19 . P-762.

The .skaigh~ segment~ Jic in •h~ }.-z ~Ions, ~ the_ 9.i n . length mo~ on angles of :so' wifh the X ox1s. The -sem1c1rculor -s~-

. rnents is 1n the x- Y plane. Locate the oenler of gro-'•+y of' the w ore

v-\ " ,. ~ 1

( 6 t11(•)tg)x; "1"(-+)(~)t e(s++cos:ao') :26.S6'°"j = 1+1.919

x •.s.:s+ in.

_,,~~66"6~ -= T(.+)(~) . 26'.-56, y ... :!32

:y. u04 in.

' ~6".'566 z e 6 (3) t ·a ( 't son 3¢")

;l6.664iz as+

z • 1.ZS in,

75+.) Locot e the center of grovi!y of o steel rivet having o cy­lii:idrico I body 1 in . In diomeler ~ 2 in . long with o hemispher10ol

heod of 1 in , radius . Use ihe result of Prob 761.

I fJ'2• .'. )( " 0 .

I [ % lr (1)" t T{0.9)~(t5] ~ _ •[(o/1)1f(1)ry[1 f %~1il I T(M'f(«)(1) 3.66.5 ~ = 6 -545

Q. • ~ " 1 -~~ in. · ·

..... &· 1' so(16 - 11) ... ~ .. -s in from base-;from 1'>-ob.755 't

7~.) /'.. bod~ consi.s+s of o r1qht circular cone whose base Is 12 in.

~ whoi;e oltiiuc:le is 16 in . A hole a ·,n. , in d•'ometer.~ '.'" in. deep hos

0een drilled from the boee . The o)(is of the hole co1nc1de,s w/ the 0 ,.j, of fhe- cone. . Loco•e fhe centroid of the ne-t volume . UGe the

result of Prob . 7-f.G .

144

[-.;,, (6)•(1,)-T(4~ .. )j"~ • 11/9(,;)Y16)[34(1,)] - [T(+1(+X16-25]

• 402.11 )I .. ++~3 .:;If;

j" = 11 iri. from ope,.

796-) Determine thc.-height h of the cylinder mounted on the hetriispherical l:loGe show" in Fag . P-756 so tho\ lhe com~ite body will be in sioble equi librivm on ils bose . tfinl : J\s Jong os the cenler or grovity docs riol lie obove the 'j.-'1- piano there will exist o restoring oovplo whon the bod~ is tipped.

,:r'(ef°h(h/1) - VJ1:r'(~>•C.S'1S(I)) h1 .. 2(1)

h .. 1.-+1+0 .

nn.) Repeat Prob. 756 if the cyl1.ndricol portion of the body in fig . P - w6 is replocod by o right conieo l port.on with 0

2 fl rodivs . base~ olfd ude h . .

- ~ - - -~ ff

ffi {J)fh(hf,-) = '1/11 {#(?'!~(2)) hllp .. 4-

h11 ~ 12

h - 3 -'f-64-ft . .

7.se.) /\ steel ball is moun~od on top of o timber cylinder os .shown in fig . P- 7s9. Stool woighs -+90 lb porcu 0 ~ timber woigh~ 100 lb per cu ft. Ooter-rninc tho position of tho eontCI' of' qrovity.

. ()f,9' WsfoeJ ~II: [413f(4~11 ~}~] (-.9<) lb/ff5)

Wss. ~ 76.02 lb.

Wtimber "[r(sin11 1~io\)1(MmxJP"Hl)(100Jb;h') Wr ~ .109.09 \~

( 7(;.02 t 109.oe) 9 ; 76.02 (2e x Y,r) f 109.0B (12 x ~e) 18S.1 g " 286.46

~"' 1.ss ft. from bose

145

I.

ii I

I I

I

·.

Chopto: 8

Moments of Inert io

146

SO-..) Oe~ermine tb moment of iner\ia of o tr"longle- of base b ......, oltducle h wi\h rcsped \oon Ol<is; through ihe opex porollel to the base · U~ the trons;fer formula ~ ihe- r&:ult- Ot illuG . Prob· 602. .

-~· -- . ~~

•/»h b

I,.= bh°Y.96

I"'!,. t Ads • bhf36 t bh/2 (2/3h)2

• bh3/36 t a.bh}'g I .. bh8A

~) Delord•he the moment of in&-tio of the guorler 61rde ~ 10 f(Q . P-805 with res;pecl to the 91von OKe&.

J• f p'dl\ J • J; p•(Tf/i·df)

·(ff"df .J " .L o•r ~ J..r-+

& I o 6

J • ix tly • lr;1 • I" tl,. ' I,. • 1r~6 l y .. 11r/{6

806.) Oetermif'lB the moment ci' inertio of the .sem·1circle shown

I~ • r~ f/ &1n'~de­• r4/.+ J; Yt(1-cc.s.i~)de- • r'Ys [-e- - s11·12&/tl:

• r•/11 [(1-o) - (4>in2lr-s1no)] '2

J~ z Tr"';/e

Iy • f ~ 1dA Iy • ( f0" f'cas~ fd<Jdf • {,,.for p ' of 00$

2-6-dfr • _!," f~ (cos~de)]; • fo" r~ (cos'-~dtr) • r'/+ foT (1t(;•nlle-2 d&

JT it . ri"er~ -~ 0 .

~ r'"/e [cr-o) - (casalf -coso)]

Iy :: lfr./e

147

:I

I .I

607.) Show the moment of inertia of o Gernicircle of rodius r is o.11r• with respect to a cenfroidal OJ( i~ parollel to the c!io~

I,. ,,. ifr4

0

I,. c·i" tAd1 ~ T( J,. + ·lfr:(~)~ 2 3'1f

o.3927r• • 1,. + o.2e2..94r4

t,. "' 0 .10976 r 4 I"' 0 .11 r~ eoe.) Oeierrnine the rnomen\ of inel"'ha for \he quoder c;rcle

shO'N<\ in Fig. P-sos w ith respect too c.efl\roidol X-ox1s .

L I,.v]C ~ 1G

~ b/2.

148

a10.) 'Determine the moment of inertia 1' \he radius of' g~ion,w/ respec\ to \he Y axiG , of the Ol'\90 CU\ lr0111 the flrs't quodt•ont by

ihe curve y ~ 4 - ')( 11 where ')( "*., y ore in inches.

y -+-/ - (y-"") ")(%

wheny=O

')( 2 • - (o-~). '" 4

)( ; ! ~

'k 11.y •(l'f/A) A•r yd>< = _1'(4-l(~)dl(

A -[+ll -'ll•/~ J: A = [...{2-0) - («-o)~~J

••• Ky .. ~~V(1,/3) :.~ '4-/.s 011.) Determine the moment of '1nerf10 w/ respect to +he X oic 1s

of the shaded porobolo oreo sho'f'm in Fig . P - a11.

I) =f y2dA dA c (a-i<)dy j )( •K/ tc.. z X/y'-

'f. •a/t:J&"'( a • afb• ·S: y ~(a - o/p"y .r) dy .

= _1~[0y"-~a(t4)] dt 11 ~ [ oyx -o/bt(Y% )]: • [ <y,(ti -0):1 ;- [O/b"<~)J{»-<>)e]

I,. • ab3 - ob% · - ob_%.s (s-a) ~ 2ab){s 91~.) Oeform•ne the ly for ~he ,shaded pGll"\?boltc area of Fig. p -811 .

Iy ""fx•dA dA ~ yd'!-

11 -r Jt~ b.p/Q d)l

: s: b/w; 1.!!Jh. d;t

,. [b/,ro .,.11~h. J: -(2hX0/ro) ( a - o)

7h.

~ '2./1 (o/ra) ( o 1~)

"Ir ' (2/7){ o-'b) .. 2ct'ly;

149

i 'I I

J·l~ +ly

• ~1' +~ .. 270in~ "'• -[i7A .. ~21%ffl .. .3.07.3 in . .

617.) Determine. the. rnomern of inei'~IO ~ rodn . .IS of gyrohon with resopec~ to 0 polar (;.e(lh-0100 \ OXIS of the Ol'OSS 9$0'\ion of 0 ho 1-\0W tub.9 wnOSe ou'\.$ide dlan'IC-tsr- iG 6 in· "" ·1nGide d°1ometer ie-t~

• T'/tJ.(r/-rt) t\"1&.11 ·1n 2 J •TAz(5 ... -2"'') • 102.11n.~

: K "~J/,.: ... ~~11~11 -= 2.ss in . ~

K .:: 6 ·7.3 in.

r~.: lu tl~ : IJ71.-a3 +211..33 =.5S"1'. G(;°1n~ JT •J1-J<1. ~2730.~C-..!!54.~<0 "' .217~·int

150

e1g,) Determine the moment c£ inerlia of \he T - 6ecfion in F~ . P-820 w/ ~~ to ·,tso cen\ro1dal o*.

ShQWrl

•" l'ITY •£Ay

_!So Sfl Y • 2(0)(2+ ... ) t- tt(9)(1)

? "~.s·1n

•• ix • .!:M. t z(eX1t.5)'+,S~i t2(e)(ll.5)f. •• 11l

k • 2(s)(2) -32·10~ 1 • .:: 290-67 in:+

820) Determine the moment of inertia of the oreo show'n 1r.i

fig .P- 020 wiih respect to ··ts cenh~idol oi<es. •* Ar " 6(1) H2(1) H2(1) = 30in?

. 12

••

Atr • SfAy 30y .. 12(1)(0.s) t 12(1)(1) + b(1)(13.~)

y :: ~ .7in . - .3 lit : 12.(1) +12(1)(s.2)~H(1't +12(1)(1.3i

12 1

+ ~ + 6(1)(1.st 12

021.) Find the rncment of inerlio about the indicated for the shoded oroo .shO-Nn In f 19. P- 821 .

X Q')(iS \

I" = f. t i'\d~ r., • e~;ol3 + e(1o)(s)~ . ~ 2666°67 in ~

r .. 1 " 26~.<07 -1u;o . .S:3 : g°",14 in~

02.2.) F1·nd the cenfro1dol moments o( inerh~ of !he ~rop<'.to'1d ~hown in Fig . P- 622.

Ar = 60t(>)(6/~) .. 72 m 2

ix • [ ~3 + i2 (~)(<> )(o.sl] 2 + r;>Cf,i . 12

+ r;,(r;,)(o.st

I,. = 190 in•

161

I

1

923.) /\n oquilo\ero\ triong\e hos i'\c; base b hon:wn~ol. ShoH

t hat ~he c.en\r-01dol rnornen\ d" inerha w1\h respect h hori­

:Wl"l~ol ,._ vo<i1eal oxe'" ore equal . ~

h - Jb1 - (~)1 - o.8"b i,, " b(o.9"bf = 1.80+)1.1()-tb+ . 36

b :I1

; o.Wb(ti/~t t o .M§b(l:Vt)3 = 1.9o+it10'

2b+

. 1iZ 12

ez+.) Compute lhe momenl cf 1iierlia with reGJ)CCt toan ax~G paei;ing through two oppo&ifo opeJ!eG cf o mgu\or hB"<l90""\

cf side o . 0

a

•' ,.

-& • 8'0/ei -= 60• .~~ c b/o

ton ao •WLYh b c 1/2 (o)

Y-n =t>Mlh o h "' a-1!/2 I• 0[2(0~/t)]~ -t ( 1h(a)(o.JY,4)

5]4

12 1~

825.) Cornpvte toe moment <:A inedia of the 1oin . by~ in. reo­tonqle shown ip fig . P-82S obout the X ax1~ to which if is .Oclined ol on ' onqle-fJ- "' sin -

1 ¥~ . Hinf : ReE>ONe the fiqure

ifl\o parls A. B, <it. C .

ton 53.13 = 10,tb b•7 • .5 in.

d • 1!1 -7.5 = 7.5 in.

K. • 1ojs1n53:t1\ " 12.s .n. ~ 36.97• = h/d

h = c.osa6.87(7.s) = 4& in.

c.oG .s3.1a· = a/10 a =~in.

fx.; - 12.!! (•)3 t '12 (12..5)(<> )('15 .,)2 = 225 in~

~

152

f,.,.. • 12.5(6)3 t ~(HZ.!i)(<;)(<O t y3.,)2

3li 2-.1-' in~

r)t ~ il<A t Ille .. t .. c 0 24'75 t 22.!i ,i- 900 a :3GOO iri4

826.) The. cross GBCtion shown in . Fig. P-026 ·,s thot of o sfruc­

turol member "'1\oWn OS 0 Z SeC~ion. Determine the values

of 1,, 1-,. 11 .

i,. ~ zO~ • /\d~) . > .. 3/+(+.r.)3 t (3.5(3/4l t 3,5(3/•)(2.'925)2] 2

12 12

i .... +2.11 in~

ly = ~(fy t Ad 2 )

= +.s(<V+f t [¥t(a&f t ¥+(3.s)(1.37?S)2 ] z 12 tf

i 1 = iis."M in:4 A,. .+.s(:a/.+) +[3.is(3.4)]2 ~ e .62!> ·.n~

827.) The built - up S«:fion shown in Fig. P-627 is composed of -two 8 x ~ x 1 \n. angles riveted to o 12><1 in. web plate . Determine

the moment of inerlia w'1\h re~pect to the cenh'Oidol )( Olli~.

AT ~[e(1) t !S(1)]2i-12{1) " :39 in~ A,y · ~ ~Ay

.38Y • 12°(1)(G) t[8(1)(11.s)'t .S(-1)(9.5)] 2

Y = 9.g7 in. I,, .. ~(f,, •Ad•)

= 1{12~ H2.ft)(t.97f +[~ (1)(2.53Y] 2 • [1Cs)3 t 1 (s)(o.-t7)2

] 2 u ~ n

lir ~ 37G.~ io ~

u,;og Tobie Vlll - 2 Proper-fie, of Gtrvcturol Section.-

9•,.'"•,,i •: j,. - 38.8 1n~ ; y -1.G."? in . ; A~ 1~ 1n.2

!,, = 1 (12}3 t12(1 )(2.<n)2 • [3S.6 t 13 ( 1.se)2] z 12

j,. = 37<..97 itl :+ s20.) T""o 12 in. 20.7 lb chonnelc; ore Iott iced toge+ her to form the Geet.Ori Ghown in Fig. P- 626. Oeten-nine how for oport the chonnels sha..ild be placed so oso to moke I. equol to f y for the Gechoo. (tieqlecf the loHice bars which ore 1i-idlcotecl by

the doGhed linec; .)

163

from Table Ytl l -2

Prop. of Shvchm:il A!°'Olys'1s

Olcn"lel Size /\reo t. ly x 'i 12'20.7 (,.03 128.1 3,9 ().1 0

Iir ~ i.c t A~" l,. -(1:2a.1) 2

- ~ ly ~ h +Ad

Jy "[:i.9 I ~03(d/iz + 0·7 )t] 2

r. = iy 12e.1~ -~3-9 +6.os(d/2 to.1)2]

d = 1.~a in

I Yo ________ i __________ _

'

__ J_. -

L '

----------~---------

929.) Oefermine the d 1s4once d ot which the two 3 in. by s in. rec­

tangles shown in f i9. P- 829 should be spaced so ihal f,. • ly .

• r "" h = f.-+ Ad?. ; [ 8(3)3 + 6(3)(Glh t 1•!>)2] 2 • j,.

L • ... ;·· Iy =1. ;. [3(S)3 +o;~~[e(a)3 +S(3Xcl/2 tu)~ 12 12

. I

cl"' 1:20 in. I

930.) The ~hod legs or four 6 by +by 1h ir'l. ongles ore connected too web plote 2.3~ in by S/16'1n. to fQNT'l the plate~ ongle girder­

~·in fig. P- 030. Compute tho volue of i •. y • 12 in.

I. . .. f!JA61''1.3.s)3

.. [ '~~·~t t 6(0.5)(11.1s)2] .... t .

[o.s(a.!:>)'5 t o.6(:a.s)(g ,;g)~]+ - 1~ r .... :2667.5.5 in~

U~ing Tobie v111-2 Pl"Qpertie'? of Giruci.,rol sections

6>\4>' •h : Arco - 4.7.s in2

!,. = 6.a ·in 4 ~ I1 •11.+ in 4 ; y· o.99·..,_ ~ 2 1.99 in

I. " [ 6.3 t 4,75(11.01)~] 4 t [.sA6 (2a.s)3);12

f,. .. 2G<OG. 3.S in:"'

031.) /\ plo te ii... or:igle column ·1c; composed of' four B by 4 by 1 in .

angles w1\h the short legs connected to o web plote 14 in by 1 in

Plvc; t wo flongs plotes eoch 16 in by 2 •;_.. in a6 .Ghown in fig . P-0.31.

154

j,. • 1(1+)3 +(1e(2.2st + 10(2 .25)(0.375)~]2 • 12 12

f,6(1)-a t 8(1)(,,1s)1]+ t [-i(i)3 • 3(1)(.ot.7?1)i) + - ~ 1 I. - 7~&+.65 in~

Jy • 1i1)3 t (1.26(18)3 ]2. + (1(~)3+1(0)(+ .5)2] "t 1 1 2 1

- t [~fi)'& + 3(1)(•)~] 4

!y • 3019-63 1n.+

Using Tobie v111 -2 Properfiec; of strvcturol secl•ons a·)(+·><1" ; Areo • 111"~ ; 1, .. 11:, 1n•;Jy • 69.6 in"°j i c .a.os. in ~ 'j'• 1.0.s in

1. ·[1(1<4YJ,A2 ~ [1t.6t11(6.2)~1 + + [(1e(2.2sl~~ -t 1S(MS)(B.37.6)'l.] 2

r~ . 7601:99 ·.n.+

f 1 < [14(•)3],A2 I [G9.G t 11(3.!!S)2]4 +f[ 2. .2S( 10)"412} (2)

!y c 3021 -1 ·,n~

832.) Determi~ the controidol momen•s of inertio of /he built_: up calumn section shown in Fig. P-832.. It is camposed of two 1o"x1·

plotes riveted to two 12 in 'J.0.7 lb channels.

s--- --<1>" lb From Tobie Vlll-2 Propert ie!O of' Strvdural Geetionc;

12"- 20.1 lb : /\reo •G.os 1n~~lx = 120.1in

11 =3.9 in.4 ; y •o it._ il • o.1·1n .

~ !,. " 128.1(2) +[%~)3 + 16(1)(6.~)'2] 2 2C& 1

12•. !O.Tlb J. • 1610. 9 ° 111~

11 e(3,9t~.oa(+:i)2]i +[1C~~)3] 2

Iy " gs6.87 in:+

833-) Four Z bo~. eoch hovin9 the Gize ~ properties determined ·,n

Prob. 826, ore riveted 1o o 12 bt1 in. plate to form the seoi ion shown in Fig . P - 033. Oe~ermine the cen koidol rnomen ls of inertia .

i. = i(12)3 t[S/4(3.!>? t at+(a.s)(-t.25)~] + t 12 12 . '

[ +.i;(a;..,.)9 t +.!>(::i/+)(s- ~:25)2] +tr a/+(3.5)3

t 12 12 3,4(3.5)(7)~]-!-

1,. "1291.31s in:t

11 ~ ~ t [ :a.s (3/±l3 t 3.s(3/+)( o.01st' J + 12 1~

t[-"1+\+.s)3 t s/-t(+.s)(s.s)e.]+ +[3.!5(3/-+t 1 12

155

t 3/4(a.s)(6.m;)2]+ iy & .591.6 in .~

834.) I\ 10 in. 1!1.3-lb chonnel is welded to the fop of' o 1+ WF 34

beort'l oio sho-Hn 1n fiQ. P-03+. The wide non9e beam has an over­oll height of 1+ in. on oreq of 10.00 in~. 'lt-. 1 )l or 339. 2 in ~ C:Ompu-le

9 "". the moment of inertia obovt lhe centroidal ;>( all.i6'. From Tobie v111-2 ProperheG of 5trvchJral ~·

10·-1s.s lb

Area • + . .ot1 ·,n :l.

1,. =~.01n~ !I... ly 9 2_.310:4'

x ·0.6411"1,.., y · 0

/\T • 10t4.'47 '-14.47in2

A1y £Z.Ay 14.'4-7y '-10(7) t 4.47(13.6)

Y"' g.o+in.

Area• 10 '1n~

1"=339.2 in~ S.,. t•o.2<t in.

I,. " i.3 t 4 .47("t.!l6)Z t 3~.2 t 10(2.04)2

f 1' = 476.01 in~

83S.) Two 10in fS.3lb chonMIS Ol"C weldeO toqether as sho-Nn in FK).

P-835 : Compute t\;:le ,volues of Le tor orrongemen+s (o) ~(b). Eoch

chonnel web '1io .'024 in. thick.. o.) From Tobie Yll l-2 10 '-15.3 lb

b.)

Areo Q 4..+1 1112 ; Ty"' 2.31n:+ &...l• = 66.9 in~

i( .. o.C04 in. ""' y e 0 AT y .. zAy - 4.41(2) y "'4.47(!5)t + .-t1(9.6)

Y" 1.3.1n

I~ =".9t4.47(2.'3)z ~ 2.3 t 4.47(2.3)2

lJ< = 116.5 '1n.-+

2(4.47)y :. .ot.47(5) t 4,47(10.6+)

y • 1.82 ·1n.

I, ~ COG.9 t 't.47(2.8~)2 t 2.3 t 4.-n(2.e2) 2

Ill " 140.3 in:"

8'04.) By using the tronsf'er formula ~ the result of Prob. 060,de­termine the moment of inertia of' o homogeneous r ight circu -

lor cylinder Obovt on O~iG through on element on as surface. The cylinder hoc; o rnass M 'b,, o rodiuG r .

156

I= MK' t Md2. = Yt Mr'" t M(r)1.

I = s/2 Mr'Z

eoo.) By using the transfer forrl'lulci ~the re!OuH of Prob·. B62, determine the momen~ of inertia of o homogenea><0 sphere of rno~ M ~ rad'u; s r w1\h respect ~o a tangent .

(Tr\ I .. MK 2 t Md 2

~ = % Mr2t Mr 2

I • 7/s Mr11.

866.) 9y using the transfer formula 'lili., the result or Prob. 003, de­termine the moment of' inedio of' o rod w ·dh re~pect to on oxii;;

through the center or gravity . perperx:llculor to the rod.

~ I _L:~ ' ---1 I" Ya ML2

I " t t Md 2

Ya ML2 : 1 + M(1-12)2

I .. 1 ML2

~

860.) By using the tronsfer formula~ the result of' Proh.861,de­

termine the moment of inedio of the rectangular porollelepiped shown in Fiq. 8 -29. with re~ct to a medlon 1i·ne orthe z face.

Tak.e the median line porollel to the )( axis.

b

' " MK.2 t_Md2 . = M(b2+c2)(1.42) + M(C/:z)2

" 1A2 M(b2 t c2) t 114 Mc2

I : 1,42 M(b2+.+c2)

0 869.). Determine the morneot or inerlio o\ the rectongulor porellel-ep1ped !Ohown in Fi9 .P-8GG or f ig 0-29 v./1th re.s·pect to on ox•G" t hrough

one edge porollel \o the Y o.xis . I c MK2 +Md2 =1.l.;2 M(o 2 +cz)+M(o12.)2+(c/1f

.. 1/t~ M(o~+c 2) + M [./(Ola)• 1 (c;,d• J f ~ 1A2 M(o~1c2) + M(o}'+ t c)l..f.)

I r % M(a2+c2 )

157

012.) Determine the moment of inedio of o hollow steel cylinder w'1th respect to its geomeiric o)(i6. Tho cylinder iG 1 fl long ~ hos Of" ov\sicle diorneter of af\ ~on inside diorneter of 2 fl. Steel

wei9hs 4QJ lb per n~ . V•1f(1.9•-12)(1)" 3:Q27 n~ .

W" ~Olb/f\1 (v)

--3£'~· =~O lb/;:r' (3.927~ w = 192.+.23 lb

2 Ill. 1fl. I. Y2 M(R' tr') = 4ft(19i-+.2"'J [(1.5)2

+(1)1]

32.:2

I· 97.11 ft-lb -sec' 973,) By using the rnethocl d1scus&ed in Prob . s10.determ.1ne tha moment of ir-edia.w1th respect to the9eomefr1c axico.of'o cylin ­der of' rodiuc; R from which is drilled o .concentric hole of n:r

·diuc; r . Denote the mass of ihe resulting hollow cylinder by M, "'-, the moSG" per unH volume by 't .

h v-t\. m, for the cyhnder R• v21 mt for fue hole V1"lR

1h -m1 ·~lfR1

h r Is. ~ NIR2 \12•1fr'h-rTl1.""Tr'h

M~gv I': Y2 [ m1 R2 - mt r 1

]

M ·'f,irh(~~-:·2) .::. 1/~ [ ~TR1 h(~)-llrr2h(r~)1 : ""* lSlfh [ R

4 -r4]

1 "' 1h glfh (R'- r')(R1 +r•)

i '"1/t M(R2+r1)

874·) A solender rod 6 f\ \of\9 rotates about on o)tiS perpendicul

io it ot 0 p01nt Qf) . from one end. The rod we19hG . 40\b. Compute

the moment of ·,nertia obout the 011'1s of roto\ion .

I= 1.l(;z ML:i :. 1.42(~2.~)(6') +{4-0/32.2)(1)'

1 = +.97 fl -lb-.seci

87.S.) for' 0 hollow cos\-iron sphere or 20 in outside diameter ~ 16 in· inside diorneter, compute thcS mo.men-\ of inert ia~ -the rodiu<0 of' gyration w '1t h res;-pect lo o d a·ometer. Cod ·,ron wei9h~ 450 lb per cu. fl .

Given·. Oo # 2c/': .. R=-1oin . w =- 450 lb/ft 3

158

Y• 4% [1011- 83] "' 1.183 ff.3 (12)•

w= 4SO_lbj,r((M83fi'} ""S32.3Slb

l : 6.i fl-lb-.SCJC( .,.

K" ~ -- 1,...~_,,.2 ..... (1_+4_)....,(.-3-2 ....... 2)

..532-35

K =- 7.35 in.

878.) Determine the momeni of inertia of \he c~t -iron flywheel shew: in F'.Q· :-078 with res!)C'ct to the o,.ic; of rotoliOl"I. The flywr.i,,I hot;; 61>' ell1pt1col spok~. 3x+ in. 1n cross section which may be a:m­s1dered os slender roc1~ . O:lGt iron wO.ghs 400 lb per-cu.fl.

+I w, • i•(R' - r')L ; 6m Wh = 'lST(R.•- r 1

) L i hub .

Ws = globL i for one spol\es Wr •(+SO)lf ("301-282

) 12 ,. 1138.83 lb. 1729

Ir= Y2 M(R1 tr~)

Ws "(.+.90)] (1.5)(2)(23) = .50.45 lb . 1728

.. ~ (1138.03/32.2) [::101 +2si)A+t1 Ir "' 2~. s fl -lb-sec(

Ih • 11.l M(R« t r 1 )

- r~(1:::i7.++/32.i)[(s~+22.Vi<KJ In "o.+290 fHb-.sec2

Is ... n(i t Md 2)

c n (1A2 ML'+ Md1 )

= 6 [~2(.sG . .+s_h2.2)(23A~)1. +(.!56.1-o/32.2)(16.5A2)2

]

Is= 2.3.11 t1-lb-.sec 2 ·

I"' Ir t Ih •Is e . 206.e + 0°4298 t 23.11

1 ... 230.34 ft - lb-sec~

159

· l l I

. ~

Choptcr 10

Roctilineor lronsloiion

.~

160

1002.) 0.i a ccr~oin .sfrdch of track, tr01ns run oi 60mph. Ho.,. for boc\<. of' o stopped lroin should o warning torpodo bo plocod

to i;ignol on on coming troin? .A.ssumo thdt tho brokes oro ap­plied at once ~ retard ihc lroin· ol the uniform role; of 2n por soc

2-

(}ivon:

Vo~ 60mph

0 Q -2fl/so=-"

. f-vo' • 20s

-SI/ = -2(.2)S

Vo .. 60 x s2so ~ eeft/~ec 3600

.s .. 1936 n or o.36G7 mi . -1003·) /\ sfono· is thrown xcdicolfy Upwor-d ~ returns f0 Oorth I(\

1osoc. Whof was i~s initial volocity 'I.._ how high did if go?

Given t=1osoc . y(vo col

-Vo c -9.Q1 (10) = 99.1 mis or

-vo" -;32.11+(10) "' 321.7+ fl/s

S co Viz gt~= ~(.9:g1)(10)'1 c 4-90 .S m . or

s U ¥129+~ D ~ (3.2,174)( 10) 2 C 1609-7 ft,

100-+.) /\ boll is dropped From the top of o tower oo n high of the some instant thot o second boll ·,s thrown upward frorn tho ground

with on inihol vo\oci~ of -40 fl/soc . Whon t+.._ Whore cb H~y pass, 'ii<... wi\h what ro lotivo veloc1~y ? .

~ ' vo.l t ~gt2 • "!Ot - ~(M.:i)t.2 - .. -© 80-h ~ y~ (3.2.2)l ~

h ~ eo-16.1t' - ·· -@

substitute 2 to 1

80-~ .. 4-0t - Y12~ t: 12SOC.

h s Qo - 16.1(2)2 ~ . 15. 6 from tho botlom

s' = 80-15.6 - G-+.-4- from the top

Vf1 -Yot : ot Yf1 2 -.0-32. 2(2) '= - 24-4fl/GOG

Yf, • 312. ~ OZ) e 64.+ ft/soc

161

•· .. I.

I I

! l 1 !I !

ii

11 I

~ II

!

j\.

r· p·

·I ' l.

1CtJ5,) /\ sfono Ir; droppOd ~ O 1llfOI "" .5SOC iofor Ibo 60~ ar ~ splat;tl -.. heOl"d- 1r lhG «>loc.ily of sound ~ mo n ~GO(; , what ic; tho <k;plh a the~?

f....csmc '115 c t'100/soc

t. "" time b-hslorio lo drCJfPOd f.... - '1 • i'ilne IOr-i1hc;5GQOd iobG hGotvJ:;(Or-60Urld " d c"sl

tor ihc.stono d-= -'i~ ct• 1t20(.s-t,..)-6) d • ~(M_t')t.,S --·@

subsfifulc; t lo ~ 'IHI0(-5-l,..) = 16-1 tl ..5600- ttmt, ~,,._1 tl

By~ fq- you gi;t t, = +..a45 liOC­

d= t(S .. 1 (+.QH5)S - 3.53-'31 ff-0)6) Rapoot Prob - tooS if tho sound of tho splosh i& ~ ofter "fSX:-

IOr fhc souod d .. fl2I0(41-~)"---@

&ubsfilule 1 lo~~- . 1120(+-t .. ) ' c 16-1 ~ ~

+t80- tt20fp ~ 16. .. t.

for- tho .stone 0"' '12~' = ~(3£2)t,.r ---@

• P, ~i°G fq .. t_= -b :!:J_b_*~--40--C

I t!O

d"' ·1'-1(3.793)2 231.'3 fl_ 1007 .. ) A Sto.nci. droppc;di from o cvpflvG balloon of on elevotioo of 100QI a ~ ~ ~ llailcr of'Olho:-.,sbio CG ~oafad vcri;colDy up­'f<Ofd fioro> tho 9'0 nd with ...,doolly of He fl per .«c. If 9 ·l'i 3'2

R per scx;Z. ~ 11>:-~ 'lllflll the 6blGs pciis& ecdl ofhcr ~

"'s = flefl/soc, h: 1~ fl_ 'IOCO-n = h (~)l L

tr 1ooo.-16 l~ - · - - @

'1-He(l-2)- '1~ (::n)l(l-2)~ ..9 = 3~ Sl/scc1

t - tt.Ar.c ~ lhG 1!s[ ~

l~-L~ fer t1hG. f'._.j~ h~ 2-tel:-~ -116!.~ tEftt - 64 --·- ©

II n .. -~bi 11!JGQJ~· I • t

I £

SJ1il6f'itiuille 1 fo 2

tooo-~~- 21-St -~ -.u-V t *I.: - G+

15400 =312 l

t".SSCC- :.h: f()0()- 14i(li)~ : 600fl.

162

"

• I

. Y'· .

i~ fOoe.) ~ stone is thrown vorlicolly upwon:l from th~ ground with

o velocity o.f +e .::1 fl por sec. Ono soc.end lotor orothcr .stono ·,9

thrown 'V'ortic.olly upword wif.h o volocity of 96.6 fl per soc_ How for obovo from tho ground w i ll .stono bo ot tho .some lovol ?

Vo, • +B .. 3 n/soc v~ • 96.6 ft/soc

t " t ime for tho 1st -stono t -1 ~ timo ta- lhc 2nd Gtono

for tho 1"t .sfono

h ~ 46.3 t - Yll (.3::<>.2) t' h"40.3t - 16.1lll - .. -<!)

for lhc 2nd stone h~ oo.6(t.-1)- 16.1 ( t - 1):l

subst. 1 to !l

4S:st-~." 129.~ l · -~ -112.7

112. 7 "' 80.!! t

t • 1. -+ sec .

h" +9·3 (1.+) - 16.1 (1.4):2

h = 36.064 n .

h" 96.(; t -96'.6 - 16.1 t. ~ t 32.:2t - 16.1

h .. 120.et - 1.:;.H ll - 112.1 _ .. - @

1000.) /\ boll Is shof vcrtic.olly ·,nfo tho o;f' ot a vo loci t y of 193.2

0 per sec-_ After :- sec, arothc,r ball is shat vort1cally ·,nto tho air. ~hot ind1al vcloody muGt tho 2nd ball havo il'I ordor to moot tho fir~t bo ll 386:4 fl frorn tho ground f'

Vo, • 193. :2 ft I 9GC

h ~ 386. 4 fl . t • l'1mo for the 1st boll

t-+ " limo tet' the 2nd boll ..s .. vot - Yi.>q!~

396.+· 193.:2 ~ - 16.1 t2

.second ball

386'. 4 • Vo (9.~+ - 4) - 16.1 ( 9."64-4) 2

Vo " 1!09. 69 n/soc .

Bt ~uodroHc E~ t ~ 9.41-G+soc s,..... t • :2.S36 -sco

I'\ ~10.) /\ storic 'rs thrown vc;rlicolly up frol'l'l fho ground wifh 0 velo ­

city of 300 fl/soc . How long mus~ one wo'it boforc dry;>pplng 0 Gee

s1onc f rorn •he bp of 0 600 n low~r ·.( fhc two ~fones oro •o pos~ co ch othor ~co fl fri:Jm the fop or the f ow er .'7

1-sl ·stono

.S '" Vot t Y~gl:z

-400 ~ 300t - 1i;.11,?

By~uodrotic · t ~ 1 7. 1ss soc.

2nd Gtorio

200 .. )[z@a.<2) (11.100 - t')" :200 • 1 G.1 [(t1.1ee)'J - ~(17.1BB. l') _ t' 'JJ 200"' 'l-7.S6.30 - 5.S3.4'st ' -+16.1i,'z

i..__ 0-= .;sSG,38- SS3.45t'116 .11.•!l

t < 1.44 &cc ~ '*" d • · \ ___________ Y_ "":__u_a_ ro11c you 90 • t : 13,1;~ i;cc.

163

1011.) / , ship boing lounohcd GlidoG down the ways v./1lh o consiont

occ.olorotion . She takos B soc to .slido tho first foot . How IMq .,...·,11

sho ~al<o to s lide down the woyio 'if thC.r length ·1s 62.5 f\ . '

,Sc G!l~f'\

s • 1 ft when t $ a soc .

..s • V12~t:l • Y2a~:l 1 c Y2 a (e)Q

Cl - o.031~.5 n/sec,•

i c Qoo.sec or .3min , 2osoc.

?J'v\013.) /\n automobi le starting from rest s peeds up to -40 fl persa:.

w1\h o constant occolcratlon of -+ft per soc-2, runs ot th'1s Gpcod for • 2

0 time, k. finally c.omcc: to rest w'1\h a dccelorotian or ~n por sc<? ·

If the total disfonco f rovclc.d is 1000 fl , find the ~o\ol time r0qd ·

6 ivon: Vo=O

Yf = -+efl/G

0=4-r+/s" a -~n/s•

d = ~ooon .

Req'd : total Time

£0\'n : . .. a•-tf'Vi;'.I. ys '\ONs. o•- sfU

~ ~ Vf·Vo • at1 ·

40 "'4 t1 - t1 • 106CC

.G. • Y.i (-+ X 10)2 • ~ooH .

·' V" S~2 - S2 • "!Ct"

Vf-Vo • -ota - -1-0 ~ - s t!I - is :: 8-sec

. , .s~ =-tole) - 'Y2 (J!,)(s)'.I. = 1GoH.

I ~tSo.tS3 e10CI'.) /

'Jl>OtSt+1"0s~OOO --s2=6+on b'"IO =-+ot2 -"" t2: 16 Gee .

Tt ~ t 1 tt'.I. +t,, - 10~ 16 ta ~ .3-+soc .

101+.) A train trovolc; bctwoon fwo etotions Y2 mile apor~ in. o

rri1n'1mum timo o(4160C · If the troin acoe\o,.,atos ~ doocloratosot

8 fl per J;oc/l, stort from rest at the 1st Gtation ~ coming to o s top at tho Qnd .stat;on , who! iG ·,fa mo11i~urn -spoed ;n mph 7 How long

doc:>G ·,t trovol ot th'1G top Gpcocl ? • 6iven: d· 'le m'ile o •&fl/,"

t1. 41 soc Req<:l : tnO.)C. ¥pcocl in mph

,.,__ disti'nce trove. I ot

thl6 top .speed

Soln: vr -v/,·ot. - vr - st, Yf • !;;11/i11 -

-'If= - 6t& G3 .; Yf ta - 12 1') ts 2

S1 tS.itSa • o,e mi le.

26+o • ~ + Yfh +Vft1 - ~

164

~40 " Yf ( t1 th)

t1 ~i'.1.tts:-+1

t1 d2 E 41 - tg Q'40 • vf (-41 - ta)

~61-0 = vf(41-:!j-)

2G-40 = 41 Vf - n_• B

8y ~uodrotic fonnulo you 90\ vf "' 2-40 ,H/i;

~ vf= ee H/s tJso or 90+ 00 fU,

Bfl fi {s x a~e. 'x mile 1~ n -!'>2&0R

60 mph .

VJ..' 1016.) An oufomobi le. mo¥1ng at a conG\ant '<'elocity of -+5 ft por

s;ec .paGGos o gaool1ne -station . Two GOCCnciS IC\ter, onoH·lO". outo­

rn0bile leoves the 9o"°line stotion 1*.._ oooolcratcs at the c:Onstanl retie of 6fl porsoc2 • How .soon wi ll the sc.c.ond o utomobile over-

to~c the first' ? Given : V1 • +.s fi/s

o ~ 6H/s2

~eq'd : time

Q>l'n : .Y=d/t - 4 i; t .. d ~©

d - Y2 .o (t -2)2 -© ~t c ' Viz (<O) (tl1-+tt+)

1st · t 2 - +t-t+

19t "! 12 t 4

t 12 -1 9 t ++ ~ o ~ ~uodrotic t • 10.79 soc

t ' = 1s. 79 - e = 1s. 79 sea .

1019) 1he motion of o pariicle- is g ivon by the equation s • ~t+ - t.3

. ' t Qt~ whores Is iri foot 1t..., t inGCXX>nd6 . Compute tho values of v 'o when t • 2600.

$.von: .s . ei"' - t.}ti; t Ql~

t• fGOC.

Roqa : vlr..,a

Q - 24(1l)2

- tip- t + = 98 n;s~

1~0) /\ porhclo mo-.-es In Q 6froighl- line occordln9 to the low

G = ta - 40t whero ihc s Ii; in f1 g,.._ t in GC.COnd6 . (o) Whon l = '51SOCS,

(X)rflpu~c fhc veloc'.ity .(b) find rhc ave. veloo.•ty dur:n9 the 4th G"OCOnds . (s;) 'NrY:ri the pci0t iole ago in comes to re~;t, what is ' its oc-.

cc. loro Hon ? 61von : s - t 3 - +ot Rcq'd : a 1 ve-1. b~ avo. vel. !!o...

c~ occclc.-ot ion .

165

I ,,

solh: .a§ "'v " 3 l2

- .+o dl = ;; (s)2 - +o .. ss ft/s

- b)

c .) Y = a.l 2 - -to o = atll-40

> ·-4-0=:3t.'2.

t = 3. 6.S1 sec 1022.) Checl-. tho answers tc Tilus. Prob. 1018 by t he followlng mo­

'fhod: Wr• ~e on C:><prcssion rototinq -x ~ y, ~ by· -succossivo d ifforontiotion -Ghow tha,t Ye= -XY/\/JTi<ttr->J '!,..__ ae. = J(0.11Ah11tl<~ t

hllY"/~. Compute Ve 'II,.,, o a from these relations.

Tan e- = -Ye_/...;" s ince Ve is downwo!"d

z Ya = -v .... tan& xll +hci-= z.2 i L~~ + h-y @

cli tfcronf1a~c 1

O"' '%? ( xo. +h 2 )-'12 (2x dx) - _Qy

' dt ctt ;~A

1 =Va j o~f°" Oe ~=VA ~~"'Ya ·

. VB = XVA(x 1+ he)~YQ . . . dVa = Oe =d;c VA(xo.th2tl'2 + xcJVA (x-zth2fYll -Y~xYA .. (x2+'ri~)-;f(~y.) cH Cit I at ~t.

l

Qe=V.AQ. +l<OJ\ -x'1.VA'2. = ')(CV, -tYA2 (x1 th1)-11

2YA

2

~ :.pz+n;z Ux'-111"-)3 '1~'"'+h2 ( -..G"'th•}s

Qe"" XOA . -j. h2 VA2

--.!X•th2 ~(x2th~}'!> v'a 3 [9 (10'?]/.[9•-11? -= 6 f!,4

. as= [9(•0}(]9<-1.,e4 tr1rlll(10~:µ1~9•t122) 3 .; (;,<:,7 f'f/s2

1M.3.) The roctilinoar motion of o particle> Is g1:Vcri by s~v 2 -9 whcro 5 i.s 1n foct ~ v in foci- por GCCOnd. Vv'.hcn t "0, S"O~ v•3

ft per sec· find the s -t,v -t, 'ii<., o-t relot iol'\-5·

Gi ... on" .s-= v g_-9 l"O; s~o ;vc3

Req~ : s-t, v- t, ~o-t

f:.ol 'n'- St9=v2- - ""'~ "Q£. __Q§__""' dt ~ a i;ec, 2.& ,;e- dt

..fGt" ~ 9+9ton~ s= aton'l7- "' :::ise0"-e- de-

d<> ~ aGeG1-e- de- 3 sec &-

166

' I >~~ ' . p

..,.. t • c = fq / .sa;e- • ton-sf - Bl\ I~., ~ ~ l -~ t 11- c

c~ . t = In 1..tsa-n t ~J . . -a- -3J

et; "" J(v"-'9) 1 t 9

3

£,'/J>I = cJt; di; dt

Y£•S-t-9

..SL =(v'--g)2

v 2 - 9 ·•---3

IOU.) Tho volooii'Y of o porhGlc; ~ olonq tho -X ollits is do­f.ne.d by Y = )t•- -f-)l 2 i- 4'X where v is ~ feiof por sc.ca;d "-. ;ic; l.s ill fGOt- Cocnpuio fho vol.Jic of tho occobrotm ~ ~=· dt _ 6iwon: ..., ... -r.3-~ .. .,.s;x S:>Bn .7 "~£2'f' - '4{1l)2- .. 1>( 2)

~=2 'ti= •ff/sco Retfd : o 'lcN = odht

a: Vdo/<JJI'.

d'YdJt "'"3~~-8Jt +6

d"'/ci'1- =~ 0 = 4(2}= 8 ft/80C~

102.5.) Tho moi1on of O llJllllfiiCb j5 diGf'm;r;cji by -jhG iroHaliiorii a •-4.t

*here o is in ft per GCG/z ..._ t in ~ - ·u Is ~ ilho~ .s=1fl ""' v= ~{per soc -11:ion. t - 1 -soc- De\onrilfl(; ·llOO rc.DofiorlS ~ '<l"'-.t • .s1...t.,., ~6-

&Von : o~"'i-t

s~ 1 'l-.2 t-1

Reqa : v-t, .i;-t, v-~

Solh : d'l'/dl: 4t '/-=~ i-C =.2i~t-C

. 2 c=o;;f i.-1

""° :ztL - t -=-.fo/.w dt&/dt = d~

-S = ~t3 -t C a-

1 = 2 (1)" tC 3

3"' !l + ac c" 1/a

3S'='£(%)3 t11 35 "" ;r,(.:(v)3 't 1!

~...rr

3s =4..F'f t 1 · .

"'" (3s-1)« "'~-k

167

I I

· l I

i

!I ]l

10£6.) The mOtion of a particle i4" governed by the equation o = -~, -tihcro o ·;s in f eyt per .soc~ 1J.., s le In feet · 'Nhon t "1.scc., s • + fi "-._ v=2ft perecc. Petermino tho relation bot woon v..._t,.s1to..,t , v,s.

Given : Q • - 0/5 2 V • ~ .. dG $" dt

t=1, G·+,-v:ct

R~q'd: v-t, s-t, v-!;,

.solh : ods "VdV

YdV 3 -%~ ds· v/f. "' -as-1 +c

- 1

V Q = 1' S-1 tC

"if S ~ "'f 'ro..V•2 .; C =O

Vil."" 1~/s s = 16,NQ

4dt ~ .rs ds .+t e ~Sa/IZ tC

- 3--+ ,. 2/3 ( +)3/11. tC

c = -+/~ +t ~ 2/3 .s ;lfiz t (- 4./3)

12t = ~GWct _ + 112' t c £ (t%e) 3/z -4

12t = 126 -4 yr

1027.) The rnot ;on of .o roriiclc '10 givon by a " G v "2 I where a le; Inn

per- &'OV;z ""v ic; in fl per "'OC· When t Is zero, ~"6 t..., v •O. find thb

relot s"Qn.s botwccn v ~ t , s 'I.., t, v ~ s . G 1Ve'1 '. 0 • GvVe. d%~ ~ 9t.~

t•O, 9='1 , V·•O G" 9t8 t C ,,. ---a-

Roqo : v-t , s- t, v-s ' "if"s .. c;'k..,.t"" o ••• C " 6

sotn: d%1; - Gv"e s = 93

1;3

.,. 6 = 3t3

+6

. d'0;.(v ~ dt ' ,' dV(v)·Y: • Gdt. I

'.2.Vve = 6t +c '

i f t . 0 '-. v - 0 • •• c .. o . 'l V

012 • Gt

v"e ".3t

v = 9te

t: -iv"~

5 ".:s("o/3)a + ~ s "'~ v ~<Z t b

-SI (9)

1028.) The motion of o parf1olc '1G governed by the r elation a "4!2,

where a ~, G in· 0/s2 ~ t '1s in GOC· When t 0

11> :zero, v • 2 ft/s· """s • +foot .

fsnd the valuoio of v li.._S whon t = 2 sec . G1von : o -+t £ {l. ~ ~ + c ~ c • £ .•. v " +ta t 2

3 3 t•O, V• ~,5".f V"' 4-}.$ (a) 3

H' = 1~.67 f"-1/B Reqa : v ~s when t .. ~ dS/dt "+t.,% t 2 ••. s = 4t.~1Z t .et t4

Solh : d%t • +t 2 s = +t).(1Z +et t c £ = +(2)/'.12 t 2(tZ) + + v"' +t~ +c -,r t·o~s =+ ; c=4 s = 1:a.:a3 rL

168

"1033.) From the v-t curvo in Fig. P-1033, determine the distanco tr-ovelcd in 4 Gee~ olso in 6scc. /\lso sketch the o-t ~ s.-i .curves approximately to . .scolc.

'l(t1'') . Y· - 'o/a(t-4)

aH6" S• 'h(+)(t O) •-40ff al 6.seG

S .. Ya (')(10) = 600.

ll " 'TI Vi •!It ·~ ·.stj1a i C1

S,•01;...l•O:.c, · O

.S1 •..st•/t

Yi c - 1ct+ 60

S2" - 1ot/t t 60t t C2

S1 •.G.,• 4IO t·+ 40 " -.s(4) 1 + 60(+) + c~

Ca " - 120

S1 = -5t2 t60t.-i.u>

#I) - - .sc...• ::'.!.2.!.~t-120

~~ : 51•2.&t. :

-.0 I I I I I I

t.

10a+.) Thc ' rnotion of o porticlc .Gtorfinq from rest' jc; 9ovcrnod by the o-t curve shown in fig. P-1034. S t<ofch the v-t &:. s-t CUfVcd. Dofcrmlno the cl1splacemont ot t • g soc.

o (ft/s•) Os·B • ·+/3(t -o)

12 /.O'l · -4,(,t. t to 8

I I I

-+-----'.,__~:..;;.tel;) ' .9

v, .6 2/li't t t Vt

if V1 •O Jo...t•o :. C, • O

v, • t :r. = '2

• 36 n Is Ot = -4./3t t 2.0

Vt • -+/, t 11 t !lot tC.t.

H' v-ll. -v, = 9 ""- t • 6

36 • -+/,(<.)tt20(6)tC2

C12 ~ -60

Y2 = -2./a t 2 t wt - 60

• - o/..3(0)2 t 2o(g) -60

V2 ·"'°His .... 0 V(fl/') ~ ~

169

" - - -:--it~ -:--'(«~ I

3' - - «-- - - I " .t \ "\ I

Vt· t" Si; t.'/a tC,;lf s, •o Jrr. t·o:. c,·o

&1:: ttls =(6°th - 7Z. n.

Yt .. -f/,gf1 t ~ot - '°

.!;1. = -1./g t3 t z({'d 2

- '7Gt t C1.

.G1. • G1 _- 72. I!.-. t ~ 6

72. • ~. :Z/g (') 3 ~ 10 (6)2

- 60(.~) t C1.

c~_; 120

~.2. = -o/0 t • t 1ot~ - "ot t 120

~ = -1/g l9) 3 t 10(9)2 - l>O(~) t120

s;,. - 220 n 1040.) /\n objoct attains o veloc"ity of 16 fl por ~ by mo"lnq in a siroi9h\ N'ie wi\h oh occc!Gro\ion 'Hhich vo..;oi; un-1fonnly from :zero to 6 fl por GCG2 In 65CC- Cornputo its ·101\ial "°'ocity"' tho c;hat'lgo In dlGplocomcnt durinq the 6 sex; imCN"ol . Solve

t>y uGlng rnot1on curve<: "' ohcck. by calcµlw;.

q 8

0

t

0 1 2 6 v ·"'11d tC. ·:·

V* 16 \.., t =b

16 ., eM.(r,)2 t C

c" -8 v = Skzt .. - s

'1f \I cYo \.., t =O

,45 "' v(tsd1) .. AQrOO.H • t,. "' -'1>("·-o) t(&l')/~]('~·')

1::.$ = D

Vo· - efl/i; 11>'t1· Tho oc.c- of on ob'p:;f dcO'GOGCS uniformly lll)Trl iHl W scc;.7.

to 7.0r'O In 6 ~ ot w/v h"rnc a, ... ~ty i{; 10 fl pei-~- Bnd thv in­itiol vclx1~y k._ tho chongo it'l d:splocoi:ncnl during tho 6 !>CG­

in\orvol . fulvo by ui;'1nq rno\ion cuNCh ~ ~ by ca\culvs -0 · v= ~Wl'".tet - 14

8 . o ~ -ei6lt.·6) if v ·'4 "- t • o

t

' 0. - 41.3t t0

" = - 2/a t :2 t et t (, ifv~10S...tc6

thorcfore ,C· -1~

Yo • ;+fl/f'

A~ = v(t1.- l1) t Aroa A.-t • h

= -)4(" -o) t 6/i. (• )(1/.5 • ' )

/:;.f, " 12fl -

170

104-4.) f.lfl elevofor weighing 3220 lb st~ls from resl t..,. ocquircs an upward velo01ty of 6000 per min in o distonco of 2of1. lf 1ho

oc.cclerot1on ·,s conslont . Whol ii;; \ho \cns1on in lhe olcvotor ooble?

Given: w= 3220

Y -= 600 ft/min 5· 200.

RoqCl: I (

1

~60)2 = 2CJ(00)

'/ a • 2.5 ft/s<1 T-W ~ wh (a)

w T - 3220 " 52!2%.v2 ( 2.-5) T = .34-70 lb

104.s.) A mon weighing 1b1 \b is in on olevafot"' 'rnoving upward w/ on occolcrotion of 8 n pcrsc.C~- (0) Whof prossuro docs he O)(ori

on tho floor or the ol<5volol"' ?(b) What will tho pros.suro bo ·,f lho c lovolol"' ·," dcsoonding with the some accclorol ion ? Gi-.en: Soln :

Wm= 161 lb a~ 'T- W .. W/g (a)

o • 8f!/s 2 T - 161 • 16%2.~ (a) Rcqo : a-) prossuro ho 6i<Ori T c 201 lb .

b~ pros&urc ·,f' elcvofor b.) T -w "' - w19 (o)

iG doscc.ndin9· v'<f lho somo T - 161 "-(1'Ya2.~(e) acoolcl"Ofion T "' 121 lb.

10<1e) The blovk In f ig - p-10~ roaohos o vclocdy of 40 f{ per sco In 1oofl,.siorlinq from t"'CGi- Compufc tho oooffiOient of 1<-1noiic fricA 1on between the blook. ~ the ground .

Goin: ~ O.Ll\lh 6iven:

~ -100 1! .

P.eq<:J: Coeff1c'1eol of l'-ine-tiC frloi1on

7''1'~ ~

ye. • ~as - "° 2 = Qo(100)

a-= 8 ff/s~

p - F =mo - 60-161 (f-) "1'~2.2 (e) . /"- :0,12422

'\047.) Delcrm1no . tho force P thOi w i ll gi-lc iho body in fig- P - 1647 on

occoler-of1011 of' 6fi pcr ooc,<. The coofflciont of Nriohc friclion 1s o2.

0 ;ven : o "' c0n/6e. f"M .sol'n= F• (:a22-Py)o.2 RcqCI : force P "(3Q2 -% P) oQ

+~P -(:3Q~ - % P)o.t. = a~2.~ (6)

4/~P - 64.+ t O·~/.s P ~ 60

p .: 135,22. lb -

171

I q I

11 ·

I I I .

l II t

) 1o+s.) A rnognclic porllclc .•.• ~;ighing a., gr<irns is pullod through a

.OOloroid with an accc\crof1on or 6 moto~ per GCGe · Compufo the fO~ in ~c acting on tho particlo . Nole : 1 lb ·"'\&t9rom' ""-1.n .=2,f>tOTl.

6ivon : W•:a,cr;g • 7; t;1Z%J( I0-3 lb

a - 6 m/s4 • ~.3,.~ irl(s2 • 19·"9 ti/sit

Gotn : F· ma = 7.~9sx10-"'(19.69)

a:M!

Flcqo : f()((Xl f " o. Oo+85 lb .

1051,) Two blockb A '-8 arc roloosocl frow rost on a '30° 1nohno w

fhoy orol?!Ofl afXlri. Tho oocffioic.n ~ ot friclion urdor the uppor blodlt\

Is o,Q ~that under tho lower blool<.. 8 '6 O·f. Cornpuf o tho olapGc.d

time until tho blocl<s; touoh . , Gr~en : -fl- ~ 30•

~ = O.'J .. "-fa= o.+ Roq'd: Timo to elapGcxl unt'tl tho blool<-6 \ouch .solution : 8Ginao• - OA-8<:.06.90. 0 ~.2 as

Qa =4.% fi/s~ j\Ginao· - o-Q /\ c,$'30. - ')/g~~ a,.,

a,.. 3 10 . .sa O/s2

·. cs -~aal2 aXz ("l-9.S)ill / · ~ts ,= Y2aA tll .. ~ (10.~ate)

I .S() f 2,"f'7,5t~ • ,5.2'5t£

t · = +. 2.3 .sec.

1os2.) Detcr m1nc 1the occcbrotron of the bodies In Fig. p-1o.s2 ·,r tho focccl drum ·, ~:/.smooth """ A ls hoovicr thon 13.

f1 w,_- T .. w ... (0/9) -© 1- We ·We ( o/g) -@

l " We t Ws(~/~)

.sub!;t. 2 lo 1

W>. - We -W13 ( 0/9) :w'.o. (Gl/g) (WA-We)g ~a

10:53.) Rorcrring to f\g . f> - 1os2, aGSumc A woighs QOO lb ~ 13 ~el9hs 100 \b. poi . +he acx::clorotion of the bx:lies ·,r tho cocffidtont of 1<-i'notic f rsCtion ;, 0.10 bot the cablo i.... tho fl~od dn.im . w,., ~ 200 lb ) • 0.10 T~fie c e ro-

0•1 (r)

we" 100 lb T~/r0 • e ..so1n : i,. s 1.z.11e - · ·@ 200-T" .,. 2ooh~.12(a) - ·{j) Te - 10 0 • 1ooh.e.~ (o j - .. @

172

,subef ifufo 3 lo 1

200-1;a7lj .. 200/3fM! 0 - .. +

from 2 , Jfl " 100 t 1ooh2.2 a ·

GUb6f1tuto. 2 lo + 200 - 1,37(100 t .3.110) = 6.Q1 a 200 - 131 - 4,2,1CI "G.210

0 • (Q,OQ f! / scoll > 10.5.+.) Two bodies / \!1;, B in f ig . P- 1os + are separated by a opn n9 . Tholr motion down tho ·1nolino is resiG'foci by a force P " 200 lb. the rooff1c1ent of t<.inoh·c frlcf1on '1s o.:so under A If...,, 0.10 undor- 8 . De -tcrm·mc the fof'CO In the &pr1ng . G1¥on : P•2001b

5,... o.ao J8 . 0.10

0 t /\ - 200 t % (..oo)- 4h(4<)())(o.a) ts t 4'?l£Ma ., - .s6 l .5 '° 12~ a --Q)

ot B, S/a(600) - s • +ls(60::))(0.1) .. 600,.32.20

31Q - s ~ 18,b.3 0 ~

from 1 1 G -= 12. 420 i'56

,subi;f tiutc to~ / 312 - 12420 - ,sc; : 18·6a CA

Q.56 • .g1.asa

a " s.24 f'-l/.s:z .S ~ 1!2.42(8-24) t.$6 ° 1S8.4 lb.

1os1.) The cooffici onl or !<Jnclro frictior\ under blcx;k. A in Fi<J _P-1os7

'1s o.-ao ~under blod<-. B ii: o."20 · f ind ~ho ocx:.olcratiorr of 1hc Gystorn .._ ~he fcnG'1on if\ covh c.horci .

o\ .. c, :300 -12 "' a(X)4~~a a - ··© . t\t t>, h-11 -!ZOO.s•nao• - am C<J&.30° (o.!l)

-=- ~~.t.a a 1 1.!-11 - 1~.,"I- - , , 210 _ ... ©

ai A, T, -1a::>G1f'l<J0·- 100Q:>&.'90(o.s) • 1~,q a 1, - 7.S·98 = ~1.11a - · ·® 11 "1.s.ga +".11a

.subSt. 11 to@, 11'!-&-s ·99~~ .1101) -1.31-.<::+ • c.!.11 a 1~ -210.6~ c 9,o/i'OI -· ·· ©

173

i

~

T~ " 210.G2 t g,32 a Sob!.i1luto to 1

300- 210.1;2 t ,9 .s~a : .soo~,a a ag.3e = 1a64a

a,. ·4.a n4e 11 = 1s.ge t ::1.11(.of..a) d go.01 lb. 1£ ,. 210.bf t 9.3~ (.;.,a)" 2"5. :3G ft) .

,. ·'

. I I

174

>

. i·

Choptcr 11

C urvi linc;or T ronsloi ion

175

\

1102.) /\ s tone is thrown frorn o hill ol on angle of 60' to the

horizonlal ,,/1\h on inif1ol velocity of 100f1 per soc . Afler hiHlng level ground ot l hc base of the hill \he slonc hos covorcd a

. horiwnto l cl'istonco of soo ft . t\ow hiqh IG the hill? .

y · -743.97 fl. or 743. 97 n. 1103.) /\ shell lcovcs o modor w/ a muzzle veloci~y of .sooft/.soc; d;rcc,tcd upward ot ro· w·1th the hof' i.u:>ntal. PcAcrrri1no tho positl0t1 of tho .sholl "" i+G rosulion\ volooty 'lO .soc offor flr'1ng . How high

wi ll ·at ·,.:1so? 6iven :

v,,: soo fVs -fr = 60°

t . 20.SOC ·

. ;

.so\'n : x =Yo c.os~ t x "' soo ~ f,(/( '20)

x .. sooo ft .

y =Vo sint:tt - 1n9i2

'/ =.soos1nfi0' (20)-Yq(a2.~)(20)2

1 = 2:220. 2~ n . - Yx - ·soo~oo· a Q - Vx ~ 1260f1/s.

Yy - .soo.s1n<00' • - 32.2 (,.20)

Vy = 210,99 fl;$ . V-= j_V_x_«_t_Vl-,,- • J '2!>0• t 210.99~ ·• :327.13 n/s ..

h= V~s1n 1fJ- • sooes1n2-ro• "" 2911 • .5 ft . . :zg fl.. (32.~

1104.) /I. pr'Qiocti le ·,s f ired with on indiol voloc'ity of 19a.2 O/soc · upward ot on ongic or 30' to lhc hor-i:zont.o l from 0 po·1nt 257.6 n. above o level ·plain . Whot hor1zon1al dic;tonco wi ll ·.t e<:Nor boforo

0ivon :

Vo· 193.~ H/s .... - .. ...

'

-e-·so· S " ~s7.G

~cqa : )( 1"

' ' ' '

176

.so\n: R = '2.Vo~(CG~~(tori~ t ton-e-) _g CoGf>

fofl~ :S1/>< Cc6 ~ = x/fl._

R~Js.iz t;ic.!l

t ~) x .

1 = 1003.;Yx t 447914.9/)( 2

xe - 100~.9>< - 447.914.<3 · o 7( • 1338.S3 fi .

/ er/ - 257.~ e 19:1.2 S1n3o't - '/ll la2.2) ! 2

- 2 5 7,, ~ 9G.~ t - 1<0.1 l ~ ~ t .. Ssec. '!-=Yo cos&t • 193~ c.osao•(s) = 1338.5:3. fl .

11os) Rcpoot Prob . 11o+ ir the projectile ·1s f.r"Od dowl"lwon::i at 30•

to the hori zen ta I . · ao• s o\n : y=voGiriet 1 Vi29i ll

2.su, = 1Q'3.2s•nsot t Y2 (,9:v:z)t 2

/l57,, 'J.57.' - 96·' I:. t 1'1·1 ~ ll

x ay ~uoorotlo fof'mulo .._ __ t ~ 1.60C.

)( ~vocose-l "' 1 9:1.2cos-30·(~)

)( "' 334.63 fl . 1106.) /\ projecti le is fired w·1~h an ·.nitiol volocity of'h ft. per sec. upward at on angle of ~ w·1th the horizontal . Find tho horizontal d istance c.o-tcrcd before the projectile returns to

·11<; on9.1na1 level . Also dcicrrnine the mox'1mum height oHoin ­.od by the projc.ct ilc .

_(ya!~) h .. Vos1n9-t - Y1Z9tt ,.- .- l . ··.. Y. • Vo O'.).Se-t,

1.-' h . • .. \ at (x,o) i h =o

(0,0) 1'1L 1/ll (x:o) O"Vos1nG-,.t - Y:z9t.t Yo.sin&~ 'l:zgt -- t, ~ 2Vos·1~

'/I. "' Vocose-(2vo s1rie-) • v0 l2s1n2-& 9

.

01(x/a,h) ; t= t ./1Z 9 9 .

h = Vo.sin& [2.Y~~ne- -. V£g(v.o;•nerzJ

h "' Voe.s,n 2-G- - Vo2 s1ri!e-g ~9

h =_Voit sin 2e--1.9

177 lj

1107.) The c;:Jf" shOwn 111 f:g. P-1107 is ju'°* to cleor tho wotor­filled gop· find tho to\<.e -off vclociiy Vo.

/"°

.sub6i ituto l' fo 1,

h "Vo61n&t - Y~glt -tf,f c.Vo .s1n:adt - YL(~fl·,2H'-··©

~ - Vo oosao·t 17.~£ =VoGOSO!O·t - · ·-@

t • 20/vo

-u.!l - y{, srn30 ( 2o/~) - 16·1 ( 20;V0 ) ~ ·

t . -'t.t Vo • 6440

~'-'"" Vo c 14114 ft/Gee . 1100.) /\ boll is thrOWfl so thot ·, t .just clearG a 1oft forico "oft owoy. If i l loft tho hon.d -'fl otx>Vo the ground 'Iii..... of o on9le of 60' to. the hor.1%on"tol, whot wo.s the iriii io l veloe1ty of the boll?

Given" h = 1ofl.

__%. ~" !'

1< .. 6oft -tr ~6Cl J · . /'

r·.,fl. · · ~. '

. ~oqct : . Yo. . I

solh: x cVoros&t / .. '· 60 ,. Yo COG~/> t

'°·'

t • 1£0/vo

h • Vo 6111-6-t - Y~ 9t q .. ~

10-.s - Vo s1nGo(12D/y0 ) - 1,.1 (1t0;(,0 )

ga.9 v/ ~ 2.318'40

Yo= 4e.4ff/soc. 1109.) Deter mine the distoncc s a1 whion o ba ll thrown w/ o velocity Yo of 100 fl. par soc, of on angle ft-' ton-1

3/4 will .strike

ihc lricilnc shown in f.q. P-1109. ,...-1---\ ~e:=-· "

- I )(

•

178

0ivon : Vo p 100 ft/s -e- 6 tan - • '3/+

-fJ : :38.67°

RcqCI : distonco .s

solh: -y =Vo,s111&t - 1/Lgt~

~ ·Y/x ;gy '•J'. - .. © .

Y. =-Vo GOS O-t -y"' 1oo~l11 3e.&7t -1,.1 t 2

-y = " •• 7i0t_- 1~.1t 2 -· ·©

t~-t, -ts,g.s,.t' ~6t.7i0;(-16.1 t~

t "" .s.s1 Goe.

x .. 77. 8' (.s.51) :. 429. ft. y. ~ L.s.g.s (.s.61)" 14i'.·9S fl.

)( "'~00 C.O!i.38.&7 t '/. =77.86t -··@ -s=J(41Z9l +(14t.<JB)!l.

1"':S, 3f. • 77.8H ; y •fS.%t - ·8) S • ~.SIZ.~ n . 1110.~ In Fi'g. P-1109, o ball thrown dOV'ln tt'lo 1nolino .stn1<os 'it at a _drsfaflce s • f.s-t.s _n. If the ball rlsoc. io a moximurn heiqht h 6+.4fi · 000¥0 the point of release, CO'Tlputc '1t.s ·1niiia l velocity Yo 1f., inolination -fr.

I

Given: h ·<0+.+f1. Solh : sin~= H/s s = t.s+:.s rt. 'Aro = %s+:s - H - go.s ft .

cos-I== x/5 Retfd: Vo ..._t' 3/w = x4st..s - Y. = €41 ...+4.f1 .

-80.s = VoS111-&t -Y4!!g.f.2 -· · -(j) 241."M = Vo ca;&t . - .. - @

~4•4 = Yo 2 Glt'I~ --VoG1oe-"'G4·4~ SUb6fitufo .3 s.,_ 1 /

£9 . <,gJ

-eo.s = 64.4t - 16.H « t • .S 6oC.

241-++ = Vo case. (s) Yo cose- • <tS .~ee

vq 6 111 e- : Tone- " '4 . ..,. /.,.a.teB :...--fr c .sa1::17' Vo~ / . '

Yo 6 1r'l .!13.137 = 64-·4 - .Vo " ao,s N/s 1111.) Rof~ to flg.P-1111 ""'- find 0( to couso the projcetilo to t,·,t p61nt B in v1-0Ctly 4 sec. What is the distance x (' . .

179

j

l 1 ·I

I

~ !I

I! l[ I

li'

'l

111ll.) Boot /\ moves w ith o coni>tor'lt volocity of w fl pcrsoc,stor· h'""3 from the posrt ion shown in Flq. P- 1111. foiel -& in order for t.ho projectile to ha · ~he boat .s .sec. ofter .starting. under tho q:indition qivcn . How h'1gh ii; the h'ill obovc tbe wotcr ?

,,rt/sec d~ v t - 100 : Q.O(s)

111s.) H Is desired to pitch o golf ball across a trop to o grocn 1ooft owoy. What is the;; best c\l..lb to use H' the initial velocity of tho ball i" 60ft per soc? .Assume that tho boll stops deod of­ter .strik ing the green, which is ·on tho some lc;;bel as the point

· from which tho boll IG struck . ,.Assumo the clubs hove slopes groduafod ot ir'ltcl"vals of 6. Q) thot a No.1 ·iron hos a foce "1n -

chnc.d ot BO· to ]hd ground, the tio. 2 iron ot 74 ·, c1c ., clown too No. g ·1rol'1 irdined at .::12 • .

. I Given: V0 2 60 fVs .Solh : 'f. = Yo~ s 1n2e-

. )( ~ 100 h/ . g

RcqCI: rio. of Jron a!Z. ~(100)= GOiz s 1n!Z& I

--6- = .31 ·7~ • Use No. 9 iron. 1114.) /\ stono hos on initie1l velooity of 100 ft per.sec up to the right at .::io· w·rth the horizontal . The components of o~lcrof oro canstont Qt Ox = -4 fl per sccfl ~ O~ = - QO f1 Per'"'-seoll. C.Om ­

put o the ho.-:1:z.onto l 9h;toncc covered unti l the ..stone reaches

o point 60 fl below 'its or'1ginol clovotion .

Given: Vo• 100 ft/i;; -e--·~ 30 • , ·a", - -+ ft/s12

oy "-~o fl/s 2, y = 60fl .

Rcqa: horimnto l di.stance Sbln : y = Vos1n&t - '/12 gt 2

..'oo = 100.sin30t - 'h. (QO) i 2

-60 ~.sot - 10t IZ

By ~uod(l)tic You get the valu•c of t ~ 6 sec .

x =Vocos&t t Y2ot2

.

y. = 1oo cosao(6) t V~(-4)(f) IZ

-;.="\-47.GQfl .

l80

1115.) /\ pof'f1cl~ hm; such a cur-vi linear mo t . t h t ·+ d

. f ' d r 1on 0 I f; )'. COQr -1no c 1~ c11nod bv x =..s t 3 - 1ost h · , , r w ere x '" in inches~ t i'nso-

condi;; . Whon t ~ QG'cc, the total accclcrohon is 75 ,·n per. soc~ . If tho Y component of accclcroh on '1s constan t t._ the parti cle storf~ from rest at the orig in whon t ~o, dcfcrrri1no the tofol velooty when t •. 4scc. · ·

Given: )( , -sf3- 1ost t=2soc, a - 7-S in/&: t 0 0. 8' t =4soo.

. Rcci'd : V

.Soln : x = -sP- 1os t Yx = 1si

2- .1os

t • 2 ; 0)( e 30l

O;ic = 3o(q,) = 60 a = Ja.,.-:r.toy~ ~75" = f;O"to:r"

Oy = ~ in/so:;" t e-t€cc.; V-x = 1St IZ- 10S

Vx = 1S(4) 2 - 10S = 135 in/G

2 Vy e Oyt = "l-5 (4) = 180 in/G

V = Vy. t t Yy2 . .. .

V = ~ /.?!S e t180::z

V = :2.25 th/sec .

1110.). The. normal accelorotion of o po"tic lc on tho rim of'o ulley 10 ft 111 d 1amctcr '1G constant ot 8<XX> f l 2 · p ..,,..,,.,., · f th

1 . pcr<SCC · Octcrm1nc ihc-s-

1"""-"-' 0 c pul cy 1n rpm . 0'1':'en : d" 10f1

On = 8000 fl/sz Reqa : .Spcc;p i.n rpm. &:irn: 011 = ~ - 8<XX>"~ r · s

Y =2ooff/s V = lfOH

200 = 1 (10) t1

t1 "6-+rps (6~) c .382 rprn

1119.) .At fhe boHorri of I th . a oop c .speod of 01olrplane Is 4-00 mph Thi~ COUSC-G 0 normal occclcrotion or g .9 fl pcr .gx2. Detorm1no the ro~1us of the loop. Given: V c~rnph

011 =ggftj~>! Sain : Or1 = v"/r .9('32.e)r- [4oox.s~ox 1;8600Je.

Ro9'd : RodluG r - 1187. 64 ft .

181

I l l

:1

I I I

.1

I :] I

11

I'

1120.) A padiclc mo'«iG on o cif'COlor path of 20 fl rodius;; coo thot

"its arc dl<donco f rorn o fi,lcd po'1nt on the path i10 gi..,en 'of ..s .. •f' -1ot whcro .G ·,~ in tho foot s,..._ t . In sccondG. O?mputo -tho total

occ.elcrot ion at tho end of 2 sc.c. 61~on: rc2ofl Seth: d6 •12tll - 10 • V

s•+t~-1ot -at""' 1!Z (£)41 -10 •3e fl/GOC

t• ~GeG · .sfi._=. 24t rOt dt Req8 : a Ot. " 2+(«-) .. 48 n/s2

•

ao .. vo/r • (;ge))/w On ~ .72·2 fl/s 2

a«• at« t Ont. .. (46)2 t. (72.2)2

0 a 86.7 fl/Gt:e;t.

1121.) /\ porlidc .'1s lllOVinq olonq a curvod po\h. /\t a ooriain inGtont whon -the .slope of the poth Is o.1s. 0)1 c G ft per S!Ctjz ~ ay - 10 fl per soo41 • Compute -the voluos of ab t,.., On at this · instant~ st<Otoh how the path curves . 61.ven: Ox = 6 fi/s< Soln: On " Oy Cos& - a,. G1""°

On ,. 10 cosa~.67° - ~ so'l 36.07° 01 . 10(1/5

2

slopo - 0 .75 /· r - 1 ,'I'" 0 ~ - tan o."1$ c ~6.~7 Q::: ~a./· ta/ . ~" c t 10" = 11. "2 rt;,,.•

Roq'd: Ct \.,,On 0 41 •Ota t On11.

Qn ,. ,....+ fl/c; 2

. ,r (11 .c;~2)c. Ot2 t (+.+)(

,! CH " 10· 8 H/s 41

1122 .) A s tone 1 is throWri w"ith on ·1nit'ial vctoo'1ty of 100 rt pcr60C .

upwardo.t w· to the rori:wntal. Compvtc the rodius of cur'w'Oturc of i\s pclh at the point where ' it is .soft horiwntally frOfl"I it' initial po-

..si \ion. Given : Yo; 1oofl/s

p -60•. x ··.sofl Req'.d : ~1u<0 of

Curvofurc b'olh: Oy "9 c 32.:2 ft/s'

x "Vo cos~t .so :. 100 cos roo• t

t ~ 1.soc.

Vx -100 GOS60• • 0

v~ •$0 ft/s

Vy - 100Slfl60 • - 32.2(1)

Vy "..s+.4 ft/s y ... ~50ct(..s-t.4)l .. 73 -89 ft/s . ton & ".Vy /y,. " S<t-.4/so • 1.088

On ~ Oy cose - a., sin-e-

On • :92. ~ Q:>s +7."11 • .. 21. 79 fl /re. f On· v'lr r -c-n~.6~)~1.79

r - 2so.6(; fl -

182'

1tu.) A stono I 1os on init iol v61ocify of ~oofl per soc. uP to· 1ho ri9ht ot o slopo of + to .3. Tho c.onpononl-s of occelerotion ore constont a\ . Oic. • -1~ f1 J?OI"' -scc/z.11.., Oy " - :zo ft pci:-.soc ~ C.om­pt.1le the rodius of curvatur-e ot tho stort 81<, ot ihc top of fho path. . . \ Vy "T

61'vcn : Yo· ux:>ft/s · a,. , ~~)<; ~.l.-. - ' o.. • -112. ff/.st1, Or· -120 fl/st1 ~- . \ ',,

m• -+/a l!}' · ~ \. RcqCI : rodi'ui;i of curvc .. duro "4rAid -1''-"-- - ---

Soln : 2 c .

H • \l&Ls1n «4 • (WO) (4/e) • w-tO ft. fl.Oy ~(20)

H ·Vo.sin ~l - Y~yi 1 " b«> • 2C0(1t/s) t - Yt.(izo)t z '40 "1rot - 1ol2 - · -1ot * - f60t i 640 ~ o • tL - 16t t64 -0 - (t-&)(t- 6) " 0

V~ - '1,()()(:t./.s) • -HZ(i)

v. -~4ff/, Vy " wo(+M - ~(e) Vy • O ; Y·24ff/!;

.•. ta8G(IC.

Ori " Oy COS-&- - a,. .sine­• 20<PGO -1a~1nO

0'1 .. QO f t/.s e

an a vy,.. - r " 247to r· 2&. 8 fl

Ms.) /\ podiclc moves on o circlo in occorclonco w'1th tho cquo ­tion G ~ t•-~t whcro .sis tho diGploccrnent in foct mOOGurod

along the ciroulor path ~ t it0 in sooond.s . Two .seconds of· for start1nq frorn rest the total ocoelorotion or HlO partiolo is 48-IT ·ft por 6!Xi!7 . Compute tho radius of the; oirclo. Given : .s- t+- et ; l • !ZGCG . .

a•48~ft/sa Vt = -"t(t)a-8 .. 4(a)3

- 6 ~ 2+ fl/re. Rcx:iCI : rod iuc;; of tho

C1i-Glo

&>tn: .s -t+-at Vt" 4t3

-6

ai." 1!l!*

Ot; 1q,(a)«" 46.fYs• a 41

'" aL41 tOn-i

(1-9~)1' .. (48/ t On fl On• 48 0/6<

Q,, • >./'/r 41'> r = (H)~

r - Hlf! .

183

1127.) £olvo llluG. Prob. 11~ . uG1nq the ff : dota : W "' 100 lb i v" M 3

fl pcr-.scc ; L = 18 in . 0ivon : L:1ein • t.511

w ~ 1001b ~ v .. 5.(Ja fi/s

So1h =

-to"~ "' v2/ qr 0 • .Sine- £ v/g Ls •~fr

c.ose-

.G•n~~ " v2 c.ose-/..9L 1-cos2e- - v"cos& .:: 0

,9L

COG -z& ,. V-z(;()Gl!T -1 • 0 g!.

C0611& t s.re~co&& - 1 • O _. By~uod rotic. COSIT" o.s~B ,3!I.'l (.1.~) -IT ".S7.67.

W " TCOG-6- 0. ·100 = lCOG.S7.67•

T = 187 lb

r = L.s1"& = 1.s (.sin &7. ,1•) c 1, 2'1s fl t · nJrJ9t~Mt ~ ~TJ1.2.,;%~.dans7,,7 • - o. 9<13GCC·

1120.) /\ rod 4fl. !onl~totcs in a horizontal piano obout o vorii°'I axic: throuqh · ,t~ eontor· Ai ooch encl of tho rod iG fOGtcnod o rord .::ift . long. Eoth' ~ supportG o weight W . Compute tho G_. pa<X1 of ro\ofion n .1i" rpm to 1noline eao h corcl at :30• w1'th the vortiool . ;

Given : 1.. • :af\. -e- .. 31l"

Reqa : s peod Goin:

.Sin :30• = X/3

)(= 1.Sft.

r " 1.s tlZ • .:a.s ft . To"& = v~/9,. tM :30 . .. "/'32.11. (~.s)

v = 8.07 n/s Y • 2Trn QT (3,5J n "' 8·07 (60)

n = 22 rprn.

184

• I

11~9.) A wci9ht concontratod al the end of 0 cor"d form 0 co­n1col pendulum for whicn the por1'cx:I ic; 1 sc;c . Dotcrmmc tho vcloody Y of the weigh t if the cord rotates Inclined 0 -f :30°w/

the vcrt ical . 6ivori ·. t , 1 sec.

-& • :.o•

Rcqo : vcioci ty

f>olh :

t = zirj ~tan~o· 1 • 211' J ~.2 ton30'

r =o.471ft .

ton-&: v~r ton.:3o' " vy 32.1(0:+71)

Y 2 = 0.75621

V ~ t.9G ft/s ':::! 0.901 m/s

1130.) In Fl9· P-113", the 120-lb ball Is tcrccd to rotate a round +he .smooth inside -Gul"focc of o c.on1ool -shell ai Hio rote of ono rovolu­

hon in T/4 Gee . Asi;urninq thot 9 • :a2 fl/Gc;c~ find fhc tons.ion in tho rorcl 'ii!o.... iho force on the conica l Gholl . At whof ,speccl in rprn viii\

iho fO"Go on tho &holl bccomo zoro? <3ivon ; w-20 lb

tcn/-tSGGi 9 =:a2fi/stl Rccid : toriGion ~force

- t.......spoed when F8 •0

Gl:>ln: r .. Lsine-= -i sin30· = 12 n

n • 1 re..Y,;-t ~ "'1-/t rps " = i>tf'n = 21(( 2)( .Yt)

v~ 160/s .z!f'v-0

Tcos:a0- T1sinso• -W "'0-© ~f11 ·0

H - s0.2s lb T = IZOto.s(.s!M!S) - .57. 3 lb.

0.06G

for rl ::.o / o.s,, T -~: 20

Ts 2s.1 lb.

o.sT t ~·,,_ 2ov/12(a2)

o.s(2s.1) = o.::i125V~

v - 6.079 f-1 /s v = ~trn

Tslnso t rleo&:ac>'- wv/9 .... •O--@ ~. 1 -- o.U6T - o.st\. = 20

i:;q.2 - o.sr + o.S6';t1 = 20 (16)/32 (fl)

from 1, T • (20 t o.s li)/o.g6t;

40ubGtitutc to 2

6.019 • [-.?ir (2) n],/Go n .. 29 rprY'l.

o.s(20to.6tt) .. 0.86(; rl = 80 0·8"

185

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1t31.) /\ body of weigh\ W ros\s on the. smooth inclinod sur­

foco of the fromo shown in fl9. P-1131. A poq attached to the frorno forces t ho body to rotate w/ ·.~ about thv VCl"tlcol a­xis . Ootcrrri1nc the speed in rpm ot which tho tension in the

cord ·,i; CGluol to the wc1ghl of the body.

r =GOS3o·(s) ~ ~ ,.o·; Ts;t'l:;o· t Hs in6o•-=- W

r .. w ~r""o; Tcas:30'- ticos6o· "''!:!.£:. © • . • gr

wsin30 t r t s 1n 60 cw· N = w- o.sw ~ o.577+ w

o.866 subsotitll\e to•. 'f/cosa-0· - 0,5774-.,f cos6o• = .)l(vY3~.~ (s) cosao

v • s.91:z ff /s V~~lfrn = 8.~rn! cH(-!5co.sao·)ri/(;()

n .. 19.79 rym ~ ~o rpm.

1132) The hammer of on impaot · tes;tinq machine weighs "..-.+lb. /\s shown ,;, flq P-1132, 'it is attached to the oriel of o li~ht rod

,,.. A br-q whioh Is p~vo\od too honzontol oxis ot A . {o) What is tho

bcor-(ng roootion <», :Hio pivo~ on instat'lf aft or bolnq · rolcased fn::rn the 91~d' p001\ion 1 (.~)What Is the beorinq rooc\ion just before im­

poot ot e ii thG vdooi~ of the haromcr is thon -s.g ft por6CG? . ·' . ~ ' .

6iv6n : W~6•1-.+lb I

1. =4f1 . , Reqa : (o)Roocfon

le ..' also at (b} , ._.,,.

Soln: a) Tcasaa •W

T cos~o c ""'"""

T . c 7+.:%10 • R.

b~ T > w ~wv71r - , .... .... +[,41-.-t(.s.9)~]/322(4)

T = 81.8\b . == R .

1133.) . To chock the radius of o roi lroad curve, tho oITocl of o ~o lb wo19ht Is observed to be 20.7 on o .sprin9 svole .su.sporidGd from the roof of on expenmenta l c.ar rounding the ~NC at· +omph .

'Htlot i.s the rod1us of the ourve i' Given : v • 40 mpl'i - 66.6 ft/s cos &- = 2ofao. 7

1 = 20.1 lb : w ~ !20 lb -t:r - 1-t.943° RpqCI : rod;us -'l'!f,. =- o i TG1n& "WtT'~r 0 M.1s1n14>q"t'a

0

f;oln : zFy=o [!70(4-0x 68/&o)iJ/a2.2r = !lo.1.sin14,q+3'

Tcos& :W r .. 400.5 ft .

186

tt:ss.) Wh~t counter:--c'1ght W will mointoin tho Corla's$ en<Ji governor 1n the poGihon show in Fig. P-mis at o ro•ional spood n • 120 rpm·. Eooh plyboll wo·1ghs 16·1 lb. Neglect iho weight or the other links. ·

.SOl'n : s1nzio· • dAo - d "s" r•.st1 ·~· ·-o.s;· 'J·2irrn•2v(o.1'~)

wv' .1,.1 ,.2&)~ v•<a.28 fl/s V° :S2.2 O•.S)

.3g.+e lb .81o •O

(6)AB.S1n1s • 0 30.~s.smcoo"(10) - 16,1 .sinae~~

·~ M • +s.1 lb. A•LJAti l\8GOS4'0• i/\BC0$415• •W

CEJ "1-S.1 co-s+s• (2) • W

W • G3.781 lb .

1136.) Tho side rod of' tho onqine in fig. P-:11:% iG 80 long 1' woighs 100 lb· lhe cranl<S AD \....BC aro of longht r.,. 18 in. 11.... rotoic at :aoo rpm. Detonn1i1e tbo mox1.mum bcndinq momeni M In the rod i f M • WL/s I where w ic; tho totol cliGtnliutod lood "'L jc; the lenght or -the rod . ·

Given : W •100 lb

n"aoorpm r-181n 1.._L~sft

~~ : M . M " 100 lb-ff.

1137~ The segment of rood poss1n9 ovor the crest ·of o hi II iG definod

by tho parabolic cuNe y c ~ -" A car weighir1<3 :a2ro lb trovels a:­, long tho rood at o oonstont t;pcod of 30 0 per soc. What iG the prc-6SUrc on tho wheels of the cor- when it is at tho crest of the hill wh­e>-c 'I a 4 fl ? .Ai the what 6peod will the road eressul"O be xero? lf'.nl : tho rodius of cur-voturo by t1' c(d'Y /cix 11y[ 1 t(d#J)t)11]~ Given : r+rt. ; v • ao O/.s . We •..3220lb X =20

Reqa: pref.Suro R 'l " ±.. - 2x ~ 4Ao - 1.(1l1J)/,.A '10 100 ;.>:tc.v

IJ.._ Gpeod whon r-ood y" • - 2/100

possurc ;, ZOl"O ~ : 12Aoo ..... I'. son. Goin: .of. £--L R+ wv1l/,,.r :::.w

10 100 /':.I R c aa20 -[aQ2.0(ao)~/a2'.2 (.so)

;l(«-+o)( t+oo-o .·.(x-2o)(x-U>)·O R., 14~0 lb .

187 ·I

'if R• o y; W'_v 2

e 'J4 ,gr

-.. vtJ. s gr ; 32.Q (so) "'1610

Y 0 40.1~ ff/G

1141.) A boy ronning o foot race rounds o f1at curvo of .soft. rod. ·1fho ronG ol the ro.ic. of1s rnph, ot -NhDt onglc v4 the v~riicoJ :'ii~ ho inotino his bod'( ? Given: r~.sofl ·.sotn : tan& -v%r =~5l<.S280j3600]faM(so)

fon<t = o. 3006 V •1Smph

Rcgl:i: -e-1142.) /\ darcaevil drives o rnotOl"'Cyolc around o 01rcular vorti-col woll 100 {\in diamoio"" Tho f bot. fire~~ wall ls o.<'10· Who~ i<> the m'1n'1mum .£Opccd that will prcvonf plG Gliding down the wall? .At whot onqlc 'tl·.it the motor cyole bo inollnod to the hori:i:ontoll What r~ the offoot of frovcli1"1<3 a1 o greafor spcccl?

Given -. c:l • 1ooft. Solh: f .. o.6 17- " tan-10.0 .. ::11· z

Rc;qCJ: Ym'in ,-a- ton'& - q r/v 2 • 32.~(so) .. o.f.V

V"" 51 . 0 fi/s • 51.& [::1~00/s.280] c::is.a mph.

11""'3-) The GupOl"OICvofion or a roilrood trock is the number cl' inoi"i<x ihot tho atft~tdC roi I (1> ro1·Ged to provCf\t side ihroGt On tho

wheel flongCG of con; ro1.>nd1'n9 the curve ot rote speod · Dctcrrri1no iho Gt.Jpor elo'<af10'n c, for a traok hov'1ng a gauge o{' 4 ft 8 Y~· ··ri ·Of

2000 fl. radius '*1 a rate speed of 60 mph . Whof is tho flan9c proG­GUre p on the ~hoolG of o 100,000 lb. cor that rounds t~e curvo

ot eo rnph? @ivcn : r:: "2000 n.

v=60mph.

-tft a.s i'r) - .sG.s ,;, Req'd : the .supcrolcvotion e. for o t rocl<- ·

tontr = 0.1102.

-e-,. ~.s,

fan-& : eh 0.1202 - e/s6-'S

e = 6.79 in 11+'1') An oirplano rnakc o iurn in o horiwntol plonc w'1thout

.Sides.l ip ot 4ao rnph . At what onqlo must the plane be bon~ed irthc rad;uG of the turn is .1 rn't\c,? Jr~ho pilot weighs 1.00\b,whcrl pres­sure clOGG he Ol<Ort on his -scat f'

188

Given: Wp • 100 lb

v· 460 mph

r .. 1 mile

ReqGI : prossuro

Soth :

tar"& .. -v~r ·&aox,9280@6oo]2

32.~ (s~ao)

tone- = ~.92

"fr = 71. 07 ° _:1:l.'L = !l.q2(1so) .. 43" lb. ~r

N ~ ~ 1soll + 438 2 = 462 . .97 lb

114.s) A car ..,..,c1(Jhinq .3220 lb rounds o curvo of 200 f1 rQdlu.s bQ­ni<od at on on9lc or ao·. find tho fr1'clion foroe ootinq on the t/r-es when the car ilO froyo\lfng ot 60 mph. ihc c.oofll"oiont of frivt10n botweon tno tire' ' the rood iG o.<~o . <01vcn: W 0 3220 lb

r = ~oon ;-e-• 30· ". 60 mph r -o. 90

RoqCI : friction Force

Solh: :z ~ .. 3Q~O [ 60 x 88A;o] • 3872 lb 9r 32.2(2.00)

to.nf • o~ g 9 .. 4-Q .

R ~jtj3220)2 +(3872) 2 - ..s03s.gs 1b .

.Sino/" F/iz. ~.S 1n42°"' f/s~.95 f =3368.37 lb.

1f'+6.) find fhc, on9lc of bonk:in9 fOr' a hf9hwoy curve ot :300 fl . · rodiu5 ~csignod to aa:omodotc cars troveling of 100 mph, if' H'O rooff101cnt of friction betwoon the t ires ~ the r0ocl IG o.t;o.

Whot Is the rated Gpccd of tho curvo ( 6 1vcri: Gain : r=~ft . µ_=0.60 ton¢=o.G V = 100 mph. ¢ .. .30,9c;

0

Rcqli: roto9~pced ton(1+6-)=v'l9r fon (ao.96 Hr) cfioo )( .s2eof3600J 2

3!Ml (300)

3<"B6 t ir "' ~S.82 .. -& ... 34.86°

tone= Yr~r fun.34-·U'c Vr~!l.2(300J

Yr Ii! ,,,. .;,2,2(300) +on 34-86.

Yr= gQ,024-H/s

Vr = a~.024' x 6%~ =; .s6 mph .

189

I .1

I :1

11-47.) The roted spoed of ·o highwoy curvo of '200 f1 ro~ius is 30mph. If the coofflciont of fr;cf ion bct~con the t ire~ ~ tho rood :,G O,(,(), whot is tno maximum spocd at wh'ich a car co n

round tho curve wi~hout .i;k.1.dc::ling? @iven : Solh : r~ 200 n. /}- a0,6 torrfr . vo/gr Vr· aomph ton-& •[3ox.S~80/3600]J3!2,:2(200J RcqCI : rnoximum GpOXJ ton fr = 0.301

·-(:] c 1'). 7.3 0

ton 1 # o., ~ c ao.g<;

0

ton ( 4 t&) -= vo/gr fan (ao.q~·+ 1,,73•) " vV32.2 (1200)

v c 04.11 fl/s V • 84.11 ( 60/86) = ..S7. 3S mph .

11~) The coc(ffoicnt or friction bOI . thy rood ~ tho f'1""s. of the CXJf' shown in f l9 P-11~ is Q.<;O. Tho oor w.01(jis 3'2{'0 lb . H Is rou -ndlr'9 the CUr'VC of ~-ft rod.uJS at maximum speed- What is tho voluo of tho fr1chofi foreo ooHng und~ coch whoo! ? how high o~o tho rood must/ tho contet\ of qrovity be to limit +his; ma ­'ltlinum ~~ by th~ tOndcncy to overf uf"n ? Given : ' r esooft w-3220 lb

p. "0·6

lk.qo : f rnohon "-­row hi~h Sotn: ion 4- o.G

l/J "'.30.96°

ion(~ 1~) • v2./9r ton (ao.96" 1 ao') » vo/32,2 (soc)

+ 57Q9.48 '•fl30(~·'H7) t 5799,+8

(t){;.3C(a) -322ot»in30•(a)

t11 - 4963. lb F1 •JAN1 • o.G(49G3) '" ~977.8 lb

at tho ouier -Nhoel~ ~8 ~ o, rit.(-+.8333) = 3200 ~30°(2.+11) t~799 · 1'8 G1030(2A17) t 3'2'20 s 1ri ao· (5) ·

- 5799.+s cosao·cs) H~ • n".4" lb

v - 110,129 fl/ 6 W'i" • 3'220 (110.,eq) <l. = .s 799, <113 lb

f:z. "fH:z. • ~' (12<i, 4,;) •435.96 lb at the out Of' .....-tioo IG

tM .gc.g6" 2.41'''Vh gr 32,Q (soo) h .. 4,0278 ft . ~l\"0

N, (4.6333) "afl20C"5acl(~.+11)

190

11-49.) Ropoat Prob.114& ·,r tho rood Is bon~d ot !lo· 1nfoad orao OG 'hown In Fig. P- 1M·&. G1v0r1 : J_A "0·'

r=.sooft ¥(. 3220

RcqCJ: FfnGf.;o., ~h ~In'.

ton~ • o.6 ~" ~0.96

tan(~-ttJ) ~v2/9r . tao (30.9' t 20) = v%2 2 esOO) yf. c 19853--49 f'+/ , ..

WV/gr • a2ro(HJ&58,19) 3e,2(s~)

- 'i3970.7 lb . ~~-o ... t11 (+.9333) =.~22oco~20(2.411) t a0w.1.s1n20(2."f11)

' t a010,1 a:is20(3) -a220~1n:z.0(3)

H1 r .3024,G1 lb F1 ~ O.G (38H,61) • 2294.B lb at outor wheels

~Me·o

rtr(4.&33.3) ~322oc.as~(~.411) tag10.1s1n20 (1Z,4"1) t 3nos1ri2o(s) - 3970,7 COG ~·(3)

H'l. c S59.87 lb

f~ ".S5~.e1(0.t;) - .335.02 lb 1nnc,.., whoo ls

191

l Ii

'I !i

I .I

li

II ~I

~­I I

' I

I i

f

' ·

C h optcr 12

P.ototion

192

1202.) /\. flywheel 6fl. in diomefor occelorotes from rosl oi ihe constant rote or 4 rprn por sec. Compute the normal ~ tonqcn­tlol c.ompononts of ihe occelcrotion of o particle on ihc r im of 1hc flywhee l. oftor 10 sec. 61vefl : d " 6 fi .

0( • 4 rpm/sec . t ~ 1oscc .

Goin :

Rcql:J : the normal ~ tonqcnhol components of tho oc.cc.lcrotion

o< ~ 4 ~ :L x .1.!.CQQ. x 1.P'lfn /l}l'ln~ JP'f 60sec

o( 4 o.~ rodAcc~ o(,, urA o.4~(10) cur uT "'4·2 rod/s

Qncref2

~ . "' 3 (4.Q) c .52.92 n/s11

Ot =-~ = :a(o.42)

Ot = 1.26 fl/s 2 ·

1~03.) Tho rim of c .so-it1. wheel on o broke.shoe testing machine hos o .gpc.ccl of EIJ mph when the broke Is dropped . T t comes f o rest ofter the r im hos trovolcd a rincor distonco of 600 n. v.kol ore the c.o:'\Gfont ongulor occ.elcratlon ~ the number or revolut i tho whc.ol makes in comlnq to rest? 61vcn: d "..SO in Roq~ : o< is...._ '17

.SOlh:

v = 60mph

.s "600 fl.

r = 25/i!l = 2 .oe.s ft . .S"'r'& 600 .. (2,083) (J-

i}- = 26B rod

fr = :?S0 rod '1- 1 rc~n rod -$- =- 4.5. 64 rev . V=r ur.

( ~ x f>?.>/Go) w 2.oe5 lli. 'll( ~ 42.21' rod/s

~"lJJ/t2~& -(42•24)t e ~o( (~88)

o<. ~ - 3 .10 rod/sc.ct

193

12as.) Whon the angular vclooHy of o -+fl diomolor pulley ·," 3 rod/s. tho totol oc.colorofion of a p:i111i on it!O rim '1s 30 fl pcr-soc

12• Ootorrni­

ne the O~lar ao::.olorat1on or \he pulley OI th\G °instant '.

0iven ~ d = 4f\. RcqCi : o< ur=:arod/s a=sonN1

) Soln =

On • ru/2 - !l (3)e..

.an"18fVs2

o~ • 0t2 t On2

302 c ot'1+"1e2

Ot .. 24fi/s2

Ol·rcx. '24 • a<><. - 0< ~ 12 rocJ/sz..

1200.) Octcrm1no tho hor1ionk1I ~ vertical components of' tho oca:i­lorot ion of paint B on tho r im of' the f)ywhoel ~hown in F~- P-im> Ai tho givon po51tion,ur•+radpcr..soo . ~0<. =1!l rod pcr-.scc1:, both

o\oc1<.wioo-

G1:.0cn: uf· 4rod/G Y o<." 1'2 rad/s2

I

L"-.3ft I

Rcqa : hori-z:onto l ~ vcn•ool compononh:;· of

the accdcroHon

.solh: ~ ~

~

Ot"ro<.. "' 3 (12) .. 36 f*/s2

a"· rur2 • 3 ("1-)2 " 48 fl /G 2

0 = ~ at" t On 2 • J ~!L t 46

2

a= 6on/s 2

12ce.) A pulley ha-s a consbnt onqular occc\ercrllOI'\ of ..3 rod pcrs~:.2. When tho onqulor vcloci~y iG 2 rod por6C.C, tho+ total ocoolcral1on of o point on the rim of \he pullcy is 10-!1 pcrGC.0

2- Com~~c +119

cl16rriotvl" of the pulloy .

Given: o< • .3rod/s 2

or "- 2 roo/s . a ·1o fl/G 2

R~(;I : cliomcfer

Sol'n: On = r~ = r (1.)2 Ot a ro<.. • r (3) ~ 3 r a 2 = Cit 2 t Ori ~

10 2 "(.:ir)2 t (4r} ~ 100 = ~r 2 t1Gl2

r 2= 4(1~ Q 2n

d = -tn .

194

\ '

1209·) The .stop pu lloys shown 1n fl9. P-120g ore c.onncctcd by 0 cro&G~ tx:H . If t~ onqulor occ.olcrotion of c i's Q rod r.so::;~ who+ ti~c is ~u1rccl for A to trove! 1BO flfrorn 1"'<%f? ~rou h who~ chstancc will D rn<NC while t\ Is mavinq 240 fl~ '3 6"1von ·. o<c • Qrad/s2 Solh : ;

·~c 180{l. 4'~ 4 ~ Rcq'd : l,.. = :'

s. : ? "."hen s-.._ 4 !.140 n .

o<,...' 2o!..c

°<A. 2(2) . 4roc1/5 2

SA ~ re-.._ .,.180 • 2&A

itA • gorad go · Y2 (-1-)t...2

b) -t}A - Qp =120rod

't:t 0 Y20( t 2 ,._ 120 = V1<.(4)J:'1

t = 7,75 soc.

--&v : Y2 <;::l(c. t 2 - '/£ (z.) (7. 1.s)2

-ffo .. 60 rod. t,. • 6'71 &CC . ~ .. r&o ~ .3 (60) ~ 100 f'I .

1210~ ~epc.of Prob. 1209 if' the rodii of pulley a ore chonqcd t ~ . ~ 18 1n. o 1n.

6ivcn : ~-a rodk:z.

s..._ =1oon . RcgCl: tA .~st> whcl"l S.,:2<10'

so1n : a~ ..11l 0<,.. - 2 <><c

H1 1,SO<"' "2(2)

o< ..._ .. z. 67 rod/s,,_ .

SA ~ re-..._ =P- 160 & 39'-12 -e,., ~ = 7'2 rad ,

-&..._ = Y.e <>< t,.,2 .

7rJ. ~ Y2 ( 2 .i;7) t,..1

t.... • 7. 34 sec. .

b.) f)A • S.../r " 24<>/(aoM) -{)A~ 96 rad

't1A " 1'€ O(" t !J

~ • 1/1(2.61)t1.

t . 0.48 6CC

B-o • Y2~ct~ .. Y2(2)(0.48/ 6-o ~ 71.<Jl rod.

So • rfJ0

So • .3(n.91)

121.3.) The rotai 1on of o pu lley is dcfl.f'lcd by Lh · 1 1 · + .30 ~ 6

r c r·o 011ori & " 2t _ i t ' whc;ro-tf fs rYIOOGurod In radius 'l.t. t · ...1_ -th "') •n sccon(.A.!>. COmputc

t ctvoluhcs ot f onqulor· voloc:ify ' ongular accclerot1on at the ins­

on w en ~ 4GCC.

0 '1ven ·.

& ~ zl4 - 30t'116

t ~ 4-GCC-

Re<:ja : ongulor vcloofy ~

Ol"Jgulor acc.oloroti'on

195

sol'n : d&/dl : ur • el~ -6ot ur= 8(4)3 -60(4) :212 rod/.s

uf " Sl3-6ot c1ur/ot : o<. "24l.~ - 60

o< w 24(4) 2 - i;;o "324 rod/s . 121"!'.) Tho ,rotation or a \lrN°hcol is governed by the cquo i ion iJf •

4.[[ ;w is fn rod ;ans per Gccond ~ t i..l in seconds . -e ~ 2 rad whcr

t • 1 sec. G:imputc ihe values of -6' ~ 0( at the inslant when t ~ a sec.

6i...cn : Solh : we -+-lr • uf" 4Jt 0&/dt " W"4.[t ~ = 2 rod 06' • + l w dt Rcqo : -e~cx TJ · +t.319.(#3) tc

if fr• !l. roo u>hcn t • 1 scc. 2 " 4(1) 31t('l./8) t C

C • - Q,(;{;1 \S/fl.

-e- = 0/S t31t - o.GC.7 = Bfa(~, . - o.6,7

-&- "' 1a.2 rod di%t • o<. "' 'h. (4) t - •It

,/ r o( . 2 ( 3) _ •It

·' o<. = 1.164 7 rod/ st. 1216) /\ body rot<('k;s occorclir:i9 to tho rolotion <><. -~2t , whoro 0(

is In rodians pc;' ,socOnd ~ t '1~ in .scconclG'. w = ~ rod per .soc ~ tJ is zero when {1G :zero . Q::mpvto tho val ~us of or' ~ & o\ the ins-

\ '

(ont u>hcn t - 2 Goe. 6 1ven: 0<, - 2t Ga°ln : o<.- gt

.dvi/ dt - ~ ,. 2t duT = 1t cit .

or-4rod/.r.

Rcqa : or" t7 ur c i t "'/?t t c ;f or. + rod/-s when vo

4- - (0) 2

t c. c ·1'

W ;:: t jl t "!' I t t c (U;cc.

ur . 2 fl. t 4 = e ract/.soc. cl&/cft - I)) • l t.H ·e- ~ t% t 4t if i "2.GCC .

d& = t 1 1+ C)'t f} : 23/3 1 -t(SJ

-& = t%1 t .+t t c iT = 10. ,1 rod .

-, r -e ~ o ~ t · o :. c -o

196

1216.) 0olerminc tho number of revoluhons throuqh w/c 0 pu l ­lc~ will rototo from rcsi ir ifs angular occclcrof1on is incrcoscd urn formly f fl?rn :zero to 12 rad por sa/· durinq 4 sec ~ i hcn unifor­mly decreased lo <f rod per" sec/' dlJr inq Jhe ne:i1. i -3 sec .

f ~ • 1{ slope,· +- 12 • -iv3 t °'. at 7· '1' l ·

1e;if~ .. --; ---- o< - 12 • - 6/s(t-1) \ .3"( • 3 ' - St HI£

de/0t =:at: /{ - 2

-&, ·-~i c ~O( .. - ett'8 ; r -e, • o , t • o . .. c -o o( • ::..ill.§.§.

.3 dfhjd t = 69fo t - 8/6{2- 'tS.3-3

-~ • 69/6 f;" - B;idl-' - +.s.33t i c if -B-1 =t:T,. "" t ·+

-3/<;(1)3 • 68/6 (4t - 8f!g (+)5 - "'5.33(4) tC

W,· 3,.1zt2. C=60,443

w; • f '~ •J- % t "frt = 66/6 t 2 -9/!e i" - "f6. 3.9 l t 60·'l-4<S

.. 6~ t - 8/, t 2. t c '& .. f' 't ~ 7

11'2 · W11!.... t·f -&t =6B/6(1)2-efie(1f - -+s.SE(7) t60.443

.:3,4)".: G9/..s ("r) - 8/,(+)'- -&e. -146.<n rad c ;, -1'.S. ~8 -f7 • 1%.~ rad )( 1 n-v ~ • 23.~ rev .

ufi • (;6/3t -·8/6i 1 - -t.s.33 2Tr~ 1211.) The angular accclorotion of o Ovwhee l decreases uni ­

fo~ly, from 8 rod per scc2

in 6scc of wh ich fonc ifs on9ulor vc lov1ty 1.s 42 rod per.sec.' Compute the ini.fio l angular velocity ~ ih~o( number or revoluf1onb mode during lhe 6 sec inforvol .

a Glope. ~ e-a A -1 0 • oz

<><-& c -1(t -a) ,('"'« a S-t.

f ,, _ _ " - . J2t

--f--"-- .!---'t

' duf'/c::lt "8-t

or • st - t'lR t c ur~~ · ; t · 6

4'2 £ S(<;)- 6~ tC

ca 1~

uf·uf. t ·D

uJ • et - t &/1 t 12

ul - Hl roo/s

dB'/dt "' Bt -{ 2/:2 t H!

-fi- = 4{2-t'/6 t 12t 1 C

it fT •O 1o.._ t • o . ·. c • 0

8- · f j t·6 i7 c 4-('

2) - (6 )'~ t 1 ~(')

-6" • HIO rad

197

I ·l 11

!I

I

ii

I. ..•

I"

t t

Choptcr 14

Worl'\ .~ Energy

l .

, .

198

1406.) Who! force P will g i-10 the sys\om of bodios .shown 1n fj'9 P-1406 0 vcloci'\y of 30 fl per soc a fl er rnovin9 20 n r rorn rosi f'

~ '4fJ' 100

y•,, 205 > · a 4 301/2(20) = 2:z.sfl/s~

~ f~1

Zf)l / T1 -100 (0.2) • 600(22.~ ll/az. 2

T1 • 69.SS lb. Jc

... 1

£f.1S/"O Ti .. a~u,6 t 2ocicas-.s0(0.t.) t eoor.in.+.s

-[~(2t.oil/at.d Tc - 399.3+ lb.

,,,,,,

p = .agg,34 - so(o.2) • .so(22.sl a2,2

Ho1.) fuid fhc veloci ty of body A in ffq P- 1.+01 ofter ·,t hos moved

10 fl from r-est . N:.Gumc the pu lley~ to be weightless ~ H--ictionless.

T2 • 200 - 6.21 Qa

T~ • W0 - 6, L1 (20,..)

Ti - .eoo - u . .;20,.. -··-@

1 i....2

2Tt • 300 t 9,32 o,... - · - @ ..... a,_3

2(~-12.412a.._) .. ~.g.320,...

-4a>- 24. MC:V- .. .3CO t g,32 o,...

100 " ..:l-4 ·165CIA .

°" -2.<J3 H/G2

v ,..~ - !lQ s,.. ----v,.. ~ ~ 2(~.93)(10) .

v,.. " 7.65 ft/s

199

:I

I '.I

ii l .

! 11

:I

" 1408·) Through who! distance w·,11 body A in fig. P-1400 m&1e in

. chonginq its velocdy frorn 6 n par.soc lo 12 n per'6eG·

.!I ""--2

T~ • 2ot- 9,32 as TL. m £04-- g,~£ (La..,)

1ZT1 = 300 t +·6' QA _ .. -@

4 ~1

£(200 - ~.210A) d 300t4.66QA

400 - 12 .~IZQA = 300 t 4.G6 DA

100 - 17 .• 00 OA

a.-. .;, s.es ft/s 2

% 7 VA2 - YA1 - 2as,... 12 1 - '.i c z(s.es)s...,

£,... = g.IZ.3 fl .

Ille • 304 t 12.~ O" - ·· : ~ + 4 s,... 3

£(£o+-'0·6+o ... ) ~ so4 + 12.+~ a" 4-00 - 37. 20('.M =::1D4t11Z.4{?0A

104 = ~.7 °"' o"' • .e.09 n/.s-

vA' • !ZOG

y,., ·~ .~1Z(.t.oq)(1~)

VA 2 7.08 f'i/.s

200

,; . ..

•;I

I .:j1

' :, ;_•

. J '

1-+10.) In what d1'...fonco w·dt block. A or Ft'g. P - 104<3 attolfl 0 velo­City of 12 fl por sec, storflnq from ~t ?

Zo=~l5Lt6L "17

zr~Jost6" -8' i!.~ ~ ei+xL

2~dt/ot = llxdx/dt

~Vt? • XVA

. Rw - ~ uJ~g (v2-v,, L)

1 "'-.ll 1ZT~ ·OD+ t 1fl..4~ CIA - · · - $

4 &._3

L { W4 - 18 .G<l-O,Y • -ao+ t 11'. 44' QA

400- 37.2.80A G 304- t12.42 °" 1()4 • 4g,7QA

0..., • IZ.Og fl/s ~ ·Yt...i " 2as ...

12 IL c 2 (fl,<XJ) GA

£A" 3"\..41fi.

Gubsfdule Ve to 1,

1400 , a .11 (0) 1 t 4,6, v,./'· v,., ... 11.33 n/s.

100 (11. -8) = 2oo;t+.4 v8"' t aa:1/6 4,4 v,.,Z 14<X> ~ 3: ~1Va~ t . 4..66VAa -·- -(!) .

1-41~. ) Fr'nd th.c veloci ty o( body A, in ft'q . P-1411 oner I~ hos

m~v~, Gtarting from rest ai tho giver) paGitlon, for 9 flalonq fhe fr1cf 1 onles~ surface shown ·

xz10~ ~ z:~ -

2/\d:11/at • u.ot/ot )<VA• ZVs

~ ~ ~ 15 2 102 - ·17

Zf = ~ '~ t 0 ~ - 10

201

II

i 11

l

I f.

i I

I i J

I

11

l

x1 t y<! •10~

6t+Yt '.' 1o<r

7s an

• I

~xclx/at t 2yd"1at =o xVe • -7VA

6Ve = -SVA VA ( ~ (<;Vs/~) t -· . -(j}

IZ f. SOb" 1• ~v"' «.t 1.si>Ys t 0;11' Vis

.soo • 1,ssVA" t ~. a3Ve~ - · ·-© ~ituto 1to1l

soo: u;s(a6YA)"~) + £.3:3Ve £

Ve. c 1£.S fi/~ -

RW c .z. u.1,49 (v'-v., ") IZ

100 (10 -9) t .so(6) • 100/<A-·tvA• HaJ/h'!'.+Vsc: t'5o/'~AVc

Ye •Ve

1414.) Repeat Proh tt1a when i< c en· t'· 10 1- ef. - y = 6H )(2ty1.<101.

2i<cl:it/dt t tydy /dt =O

xVa : -yv" 8Ye = - 6VA

202

';

,, I

800 = 1,5.SVA2 t t.33Vs 2 - -@

subsiduto 1 fo f,

eioo • 1.sB (rA-Ve''/a,) t £.33 Ve£ Ve .. 1Lo.S<f t1/s

1417.) I'.. W:cigh ~ iG dropped f ~mo p<:>&ition Jus;f obovo, but ~al- to!.1-chlng, a $prlng . Show thc rno:><•mum dcf0rrnot1on produood wi II be

· twiec that if the samo wo"19ht IG gnxluo lly low~ up:m the ~prlrig,

~1<.b11 • w'1 b, • 2W/K

W • l<.bt.

6~ - W/1<..

~lb• • (tYJ/ty64'J • R

& = 1Zf>e

1+16.) A block wc"ighing Q6,6 lb is droPPCct from o height of 4ft . upon o spring whose modulus is 100 lb per in . Whot voloci~wlll tho block hove ot tho instant the spring is dcformod 4 In.? 61ven: so1n : W= 96.6 lb

h ·-tft .. K• 100 lb/iri

6a+in RcqCJ ! Velocity w(htb)-12 K.b" "wv/:zq

. f ~ gM;(4t4A~) - Y2(100)(12)(4A2) • q6.6/6+.+V

. v • 1.s . .32 fl/s. 1419.) A 600 lb bk~;k s li'dcs down on inc.line having o .slope of .3

vertic.al to<r horizonto/. It .s.1ort.s frorn rest~ oflcr moving 4fecL ~lril<o<.: a spring whose modulus i& 100 lb po-n. If fhe cocf'fic1cnt . of 1<.;neiic frlclion j1; o,w , find the rno;ic imum vcloc"1fy of the blocK. Given : Gol'n:

W=60olb· · ~f\ ~ ... • _,....,.--Glope; .3/4 ~"b ~ K • 100 lb/fl ...... _

/).l<., Q.Q [w(.a/s) - W(4/s)(o.ti] (dtli) - Y2 K6.r = wvj'2q

d " 4fl · . 600(4 tb) [3/_s - 4/s(o.t.J] - 100/.2 bL ~(600/64.f}v f Rcqo : 2G+o t 1os6 -soob .. £ g.32v ..

mo)( . velocity

203

d iffcrentiolo vokx:;•tY. with ~pc.ct. tob, ~6+-1oob · - · g ,3~

6 .. 2.~+fl ((or mo><. velociiy) G00(4u.6+) [ a/s- 4/5(9;2)] - so(e.6+)' " g,32v'

Y • 12-26 fl/s. 1420.) lhrough what cliGfoncc muGt the 600 lb blocli:. or Prob. 1+1Q

s lido from re"t boforc louchin9 the spnnq ·,f Its -velocity i" 10 fVG at the in~ont the .sprinq i" doforYr'IOd 3 fl f Assume the spring c.ons-

tont jc; changed to .30 lb/in . Given : Solh: w • 6001b [600(3/s) -ooo(~~)(o.z~ (d t3) {30(1~)(~)~/1

. Y · 1ofl/, (600(10)''_1/6..,. .4

6 • 3 fl 1~+ (d t3) • ~s.s1 .7

I\.· 30 lb/i'n d = 6.67 fl. Rc:qCJ : distonco 1421.) /\weight of W lb '1G suspc;ndcd fron1 o verticol s pring (Fig P- ·

1-t!Z1.) whose modulus ic; K.. lb por fl . The weight iG pv llcd dD'Nn s f1. frorn ifo c:quilit:riu"1 P,OG°it1011 ~ thon rc lcoGOd . Determine '1h; velo-

city when if re\or,iis to the oquilibr1urn position .

'*K8" "'wvyrg II

w "'- , · vt =¥ v -sJ "-o/w

1'4212.) Tho rigid hor"1x.ontal .shown in fig- P1+~22 is supp::irtcd .by 2

Gpring" . ,A. 300 lb weight iG the plocod upon tho bar ot 8 /\ .wclder1 blow prQ.fccts the weight toward/\ with an initia l velocity of M\jt;. Whot ·,c 'rtG veloerty when it rcochcs t...? tleglccl fr iction Yt._, tho wei-

gh'! of Inc bar .

" V1 - 19.9

Yt - -t .~1 fi/s

204

1-423.) Neglc.cl fricl ion of hie 40-lb collar ogo·1ns l ii!> vertica l gu·1c1c

·'°" compute tho velocity of the collar offer 'ii ho" fallen j fl s tart­ing frol't'l rest in the position ~hown in Fig P- 1423 . 1hc unsfrotched lcn9lh of l he i;pr;nq is 3f't .

I f· a'

L " ~ .3 2 t 3~ , -f ,243 ft . & • 4.243 -3 ,. 1.243 n. 40(3) t V2(sx12)(1.24~)2 =(-10fa+.4)Y 2

v • 16 . 36.5 fl/s. L • ~ 3• t"1'1

• .!! ft . 6t .. .s-.:9 ·.Qn.

"f()(4)- !/2(.sx12)(2)2 .. (40fi;+A)(v,ii- 1".~s") Vz 2

"' 332 .21 ft 'ls 2

Y2 = 18. 23 fl/s . 142-4.) Repeat Prob. '\423 if ihe unstrefched lon9ht. of \ho' sprin9 is 2ft.

0 1vcn : Soln:

W·"IOlb L~~3c1!3"-= 4 . 241-3ft.

Rc:qCI : Velocity b, • 4 . 24.::i - 2 • 2.2 .... 3 ft. "'I0(:3) t Y2(.sx12)(2. ~3)2 =(40/6'f-.4)y2 ·

v • ~0.88s fl/s 02 =~HZ • 3 fl

-to(+)- 1/2(sx12)(3)'J. =("f<);i+.+)(v/ - 20.aest) Yt'J ~ 259.1 · .

Y1.. = 16.1 fi/s 142.s.) Tho ~r i~ Fig. P - 1+1s i& movlnq toward 1hc bumper .spr'1ng

~ h~ o 1<1nctic e nergy of 1~cm in-lb. The rn:::iin bumper shield (?a) i_s c.onnccfod to the l'flClin ~pr1hg, which hos a modulus of1000 lb per 1n. lhc two oux1'liary bumper Ghlelds (b) orc 12 in . tehind 00 ~

ore atl'och~ to secondary springs, coch of which has o modulus of soon por 1n . \IVl'\o1 the CQr iG' b'rought to rest, whal will hove been ' fuc: grcat&ci mO'<'cmont aa ? What pcrconiago of the erorgy hos ~n . obGOr:-bcd by the rnoin Gprlng ? . · Gr~on: &>In :

KE = 100,DOO in- lb l<a ~ 1000 lb/.n

1<-1> ~ .!'()() 1b/ 1n

P.cq Ct : .s I!.., /.

205

J[

I• I

I

1/2(1000)(12t6)2 t '/2(2)(~)61 = 100, 000

.500 (12 £ t 2.46 t 6 2) t EOObt • 1001 000

1ooob 2 t 120006 - ~aooo ,, o b'- t12b - 28 ·o

6 a -12 t J12• - .ot(1)(:-2&) • ~jn . ~ (1)

S ~ 6 i 12 = 2 +12 = 14 in . KE~ G Y21<.(12tb)

2

/<.Es " Y2 (1000)(12-t2)2

= 90000

1· • K~ /Kt:: = gaooo;1oaooo .. ga Y,- · 1429.) /\ troin wc.1gh1.nq 100 tons iG being pulled up to 2t. 9rado. lhc troin res~fonCC '1G constant ot 10 lb per ton . 1hc Spead of fhe frain

iG . inorroGod f rorri ro fl p<]r .sec to 40 ft pc;r i;cc. In a dist an co of 1axfl. find the rnox1inurn ~rGC fOV'<Cr dc-<c lopcd ·

6rvon : ful'ri : w-1oo ~onG · ZJ.. s = wfa~ (v

2-v0

2)

R • 1010Aon$ [F -100(10)-100(2000)(.2;100~ x1000=100<'20002(402-.eJ} s c 1a>o n. 6+.+

RCCJO : Po-Hen in +.Ip- ,, r f G a, 726 . 71 lb . . Hp .., fV {EVZ6. n ( 40l}/ sso

Hp ~ ~::i4-<o7 ' I . . .

1+.30.) Wcrlor nowG . through 0 nozz le 1 if). in d1omotcr under 0 heocl of' 4CO fl ~o dr'1"1'.: 0 ,/urb'1ne . The turbiM '1G o/J/, . offi(:;'ienf ~ ji; conna:;­tcd fo Cl generator w/c ·, ~ 941. emcien~. w1·ia1 '1s the power ovlput

in 1<-i lowoHs ? · G i'ven : d • 1 il"I, .Rc'1Cl : Power G\Jtput ~oln:

PONc" "' ~Qh • fv . f'J/ ~ lS AYh f ; "12.-tgv(1M)

2J/.f ][<too] ~ 136.136 lb.

vf/zg • 4-00

y = [2(:32,l1)(400) p 160.S rJ /s hp• fV • [l3M3G (160.sTI/sso

hp = 3~;L73

206

e .. output /1·nput

0.9 = hp /3C,J,73

hp(~) - 3.5. 7S hp.

0.9+ A hp/3Sas hp " 33.G1 hp .

Powor Output = 33.61(0.1+61<..w) ~ 25.1 KW.

1431.) Wofor e~fcr5 o hydraul ic react.on turb'1no w/ 0 vcloeify of 12

n per GCC· ~ loavc.G ·,t ::irt lower w'dh Q volocity or 4 n pci-~cc . If

100,000 lb of woicr flow through fhc turbine e,och .sc:;coricl, compute ihc. h:>rscpowc,... output. A'-Sumc the ~urbino jc; 80 f. e.ffic1ent .

6 1vcn :

V1 • fZ fl/5

Vt a tf!h; e • 001.

%ter()) ... 1ooe.:1 fb/s

h .. 3ff. Req~ : Hp Oufput Se in:

V..j49 " V1ftg i b t l'f (12)~(su) • 4l µ (3it,:2) t 3 + H

ti ~ ~·9876 fl. Hp1U-~& .. u)H • 100x10

3 (f.907'1) = +98.76 x103 n-lb

tip : 498.7G 1<'103 fNQ. )( 1 hp E 906. B4 hp. sso f.+..-)£ .

e = Pcutpul /Pinput

O. 8 -= Pout put / 906.64 hp

Poutpu/ = 725.S hp

207

I l l I

I ·1

11

II

!.

Chapter 15

lmpulGc ~ Momentum

208

1s'o.3.) /\ 300 lb block is 1·n confo ..... t w ith o level plane who so coefficient of 1<.1nef 1C. friction is 0.10. If the block is octcd upon by o hori zontal . force of ..so lb, what time will c lopso be­fore the block re.aches o veloc°ity of 40.a f1 . per sec, stor -ting from roGt ? If the . .SO-lb force·,~ thon removed, how much \onger wil l tho block conilnuo to move? · 0 ivon : > .Goin : W • ~lb, ' f &.90 1t:>

;(.C • 0,1, V"' 48.3 fl/G

Rcqo: time ~Fi< =o [so - 300(0.1~ t • 300/32.~ (48.3)

t • ~.s sec . .300(0.1) t c 300/a2.~ (40.3)

t : 1s sec . 1504.) A hor1".i:.ontol forco of 300 lb pu.shes o ~oO-lb block up on incl ino whose s lopo is· 3 ver!i'cal to -hor·1z.ontol. If f ... "' o.2., determine the I imc required to incrcoso fho velocity of the block f rorn 10 to so n por sec . . Given: f>ol'n: f

f".300lb~W-= ~OJ lb . , ~ rn=%. i )A ... " o'.2 o/.s(~(4ls) v1 ~1ofl/G,Y~~soft/.s · Rc.q(:j: t -1mo [300-1!20-160(02)] t ,, 2D'.lp~2 (so-10)

148t c 240,4$

t = 1°68 soc. 1543.) 01·root central ·1mpocf occurs bet. a 100- lb body moving to tho right of .s n per 6cx:i ~ 0 body or weight w moving to tho loP ot .3 ft por .GOC· The cooff1·c·1cnt of rest ituilon e='o,5. Aflcr 1inpact. the 100 lb body rebounds to t'1o loft ai 2 ft/soc. Petcrrri1no the weight w of the of hc.r. body . G ivcn : Sol'n :

W,.= 100 1b , e c o.s w,..v" tWeVa c WAv"' 1 wave V"=~ n/s , v,; ~ 2(1/.s 1oo(s)- .3W • 100(:-2) t wve'

Ws =W, Ya c.=ift/G 700 "sW+Wve' -.·-GJ Rcq'd : Wc(ght e = va' -v..:/-tA - V~

o . .5" Va'H/.s-C-.a) Ye= Z(I/~

209

·! ·i

'l . '~ t

I.

subsfi-luto Ve 'o equation 1 ,, 700 = 3W • W(IJ..)

·W"HOlb 1544.) f\ golf boll iG droppod . f rorn () height Of 2~. fl ,upoh Q hordonod. .stcol plate; . 1ne cooffi·cienf of rOGti\ufion IS 0.894; find the ho1gh~ tC> which the boll rebounds on the ffrs~, ind,

v...._ third bounces. > Given; (o.eg4l(h1) "h2

h • .20 ft , c= o.s94 h~ - 1~.70 ft . Re.q'c:l: h, , h2 ~ha (o.eg4)2(hz) = h2

.soln ! h:3 =10.21 fl .

e = ~ h1fii • f %, • J h3/h2

0.094 = f1/~o _,.. h1 = 1.s.gs.fl . . . 1545.) The balls A'*' B in f (g. P-1545 ore oHochod to sh ff rods of ncgligiblo - weight. Ball /\ 'i.G roloosccl from rci;t ~ a llo:"'od to Giril<-0 a. If tho cocffic1ant of rc'1.1tution le o.6 , dci~rrn1ne the; onglo tT 1hrou9h which ba II B wi II .9W1ng· If tho 1mpoct lo.Gts for 0,01 sex;,.: d lGO find the aveirago impact force ·

. / r~1' . 6 1ven : ! · '3<./ n ·

C().&60 = h/10 h =.Sf\ · j b • 10-S = .S fl

YA -~ =j2(a2.2)(s .. 11.g4f\/s

vv..v,... ~ w,..,vA' tWsYa'

30(17,94) " 3oV,..' + MVs' .s:;e.2 • 3ov,..· f 2ova' -' ·-(!)

o.G .(v& -y,.., '.)/@1. 0'f- O)

to,17 = Ye'-Vp..' -- · -@ from 2, Y,...'• Ye/ -10.77

su~titutc fo 1,

.s39,2. ~ e,o(ve'- 10.77) -t QOVe"

'> • .30,Q " aove' - 3 .'Z3.112ova'

I I

'~ ... , '

vs' · f29Z ; h· a-c 17,2Q6 • ~~a2.2)c

c = 4,61

h·0-4.61 ; 3,'.S.9 ft .

COG tr -=. 3.3.9/a • O· 42375

-tr = 6.S"

86P~ =..sova' -- vs'"' 11. 1226 fl/.s

210

154-6.) The S!:-JS\ern shown in Fig. P - 1s4.s is used to defer' -mine tho c.o::ffic1onl of resii lut ion . If ball A iG refeo.s~ from rest ~ball B GWings through -&= 53.1• oflor ooing struck, de­termino & .

Soln: eos·6o .. h/10 h0 -8ft.

b=10-.5 .. .5n. v" = .[29h · ~1-2-(3-2.2 ....... )(,,....s

= 17-9+ M Wl'..V/\ =w"v,: t WaVfJ' . 30(17,g4) • 36V,...' t 20(14.35)

Vl'.,1

" 8.37 n/s

C.0S Sl!l,1• = h/e ti'="'t.eff

C" 9-4.e = 3.2{1 .

Ve' " J2(~:Z:2)(:3.Z) = 14.35 ff/s

e = f/a'-yl'..' )/(:./A -va) • (14,35 - S.37)/ (17.94-0)

e - o.333 1547.) A . ball is thrown at on ongle e- w ith the normal ~o 0 s­m?'t~ wall, os shown in f ig- P- H.57 . It rebounds ot on onglo & wdh the normal. Show thot !he coclTi'c1'eni or rcsfi tul'ion is oxpres.sc.d by o • tan 6-/tan-e-'. ·

IH .soln :

Q, i ,D c ~ (vi' ' -v,')/(v1 -w.)

' ,i7 ! -e',,../ mY1 cos& • mVi''cose-' "(jY ,,,,,.,,,,,,,»;;,,,,,»,,,, V1 sin &'· \I.I .Srn9- -· ·- ©

e ~ -v/~e-/y1 cose­from 1, Y/ ~ - V1.S111G-/s1ne-'

e =(-)t'(s111-e-)(-cos-e-') = tone cot -e' sin&' V{QJ!.e

v " ton-&/tan&' 1548.) /\ ?o'·' is thrown w·dh a velocity or .so fl per .sec d1·rcc­tcd oi 60 w ith lhc horizontal agoinst o .smooth vcrhcal woll . · 1hc ball i~ role.a.God from o -~·1tion . 40A the woll ond 6 (lobovo the level ground 'i frovcls in o verhcol plono. The coefficien t

or rcstiluhon between ball ~ woll iG o.6. How for rrom /he; wolf doeG the ball Glr'rkc fhc ground ? Given: Re<:jo: d1's~oncc frorn the wolf

Yo " so ft /s when /he ba II .sf~i kc the ground .

R"' 40ff

2ll

;'

.. '

}: ti

11

.~

sol'n:

t1

--L --... o·---

Fl~Yocose-l

"!<> = sac.osoo t t: 1,6SGG·

ycYof;1n&l - Y2Ql1

2'.

,-A(1 6) - 1/.2 (32-2)(1.6) 1 ~.so .sin.,.,. • , ,

y - Q0.07 fl H ~ 6 t28.01 = 3+.010-

V: ' - Yox ii • 2 a~s ,.. It

v./ -(socos6o) .. o

Yx • 250/c y,2 _ y0y 1. •.20yS '.,

~, -(sosin6o).2 c.·~(32.'1)(2.a.01) 1

Vy c e . .e ft/s' . " · ~ I £ " 2 :r. . le 26.31 H/s Y81 - \JQS f v• I •

o< ~ ton-1 e,11./2s • 10.16

/<~:].-,' ,, · .

L,,,, ........ ~ · e ~ ve2 "/. /vs1~

O." = Y&!l'j./~6.31~10.1"0

Vf!ZY.- " 15 n/s . 26,31sin18.16' .. Ve2.Y

y82y • a . .2 nJs Vs~ c { 16 .. t g.Q:z.

y62 ~ 11.1+ ft/s I = tor1 ·1 s.2:1.s

f ~ .ee.66

.l-- -" 7 • Vosin~t t

1hgt 1

34,07 = 11.1s1nfb·" .t 116·1 t z

16.1 !2 ~ 0.~t --a: .. 07 -o

t • - 8.2_+ J (8.£.); - +(161'1)(-34-.07)

2(16.1)

t • 1.222 ScG ·

x -Vocos~t · • 1 ~,1 Q)S £8,66 (1.222)

x. 10.3 n.

212

......

" .ti t

.,

)

,l

1~.) As .shown in f(q. P-1549, o 4<Hb ball moving horizonta lly

to iho right with a velocity of e fl parsec, collldos obtlquoly w/ . a 30-lb ball rnovinq up to the left at 30• to tho horiwntol · ai 1l> fl pcrsc.c-. If tho cooffo:::iont of rcGi°ifution i~ o.6 , oo~erm'1nc .the a""unt \.,. d1'rochon <£. the Yeloc'1ty of each bpll d1'rodly aflcr .

· lmpoot. w,·10lt> ....-ao11>

~-~· _ V1a1o('ffr;

, f

in. V1,. t ma '4 )I .. n'1j Y1x t IT1% Vix

'4<'(&) - 30(10 ~"311J • -4<1 Vix + 3(JV;.,_ ~0,2 --wv,~ t 3tJV(,. · -- · -© e .. (vt.; -v,.,.'Vv1,. -v•~

0.6 -&,; -v,x')/ 8 t 10GO!i30•

10 ::v1~ -v1 / - .. -© m,'°"1 t!ni.V2y • nljV1y t msV2

1y

:.a6 (10&rn3(/) "'..;id Vty -v,; •s rt/~ from rz, ~ -1otv1; sub.st. to© 60.t • 40V1'Jt t .30·(~o•v1;) '°'l • '40V1 le t ~ t ac>V1 ~ -2'39,8" 70V1~

"1.,: ::-~ . .+a n;, Y/ • Y4<' " 3,.+3 fi/~ to tho loft

from a. , v.; & 10 t (-3 .... 3) - 6 • .s7 ft/.s v,' = J~'-+v{yi.

Y1' •Y30 ·~'·511+51 • 8, f6 nJs w.'

~ & • fan-1

V•)' /,..,; • fan· 1·s,4,57 -fJ ~ 371 27.0

V30 •a.~~ fl/& up to right ot

37,27• w/ the ~zonfal

~G)e~ ~ .. " Vit MN;' t '/1 M1Vi11

e • Ve'-v.,' t.1,\4 t-Ma'ff • M.V.1' +MtV/

l<E1 • ',iM,V1't'/tM2V11

213

v1 -v'J.

~ '

I I I

I

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t

., :.:-: r ~

PROBLEMS .

\

214

Chapter 2: 1.) f\ r '1gid bor CABO supported os shown ·.n Fig . B is octed upon by two equal hor·1zon-\ol fore.es P applied at c t.._D. G::il­Cl)lo\e the reoctionG that will be induced ot the points of.su­pport. /\ssume l · -+ft , a .. afl, b'" 2H · J h><. Rb= -Ro • MSP

?l=, . '

2.) Owing to the weight W of the locomotive GhO"<n in Fig· e:. t~ reootionG at ihe two points of support A 'i;.,.B w'dl eoch be e­qool to W/2 . When the locomotive ·,s pulling o iro'1n !t.,, t~e drowbor pl)ll P i<0 juGt equal · to the total friction ot the pt.s. or contact A St.. B, determine the magn.dudes of +he vertical ·reootions Ro 1b.. Rb. Ans. Ra " w - Pb J Rb= W t Pb

'T ~ T 2Gi""

215

.· )

.a.) /\ boci\ i6 mm1ed uniformly oloog o conol by \V't'O hor­ses pulling with forces P• 200 lb ~ ~-= 24<'.>lb octing under

on angle e1. • roo0 (Fig . A). Deierrnine th<9 rnogri1tude of the resl)l\ont pul\' on \he boo\ ~-\he ongle f.> 'b..., "'< oG lelhown in the figure . /\no . P.-= .:3S2 lb ~ fd <r 33• ; -Y =· 2( .

.. q For the particular position shown ·,n Fig .c the connec­ting rod BA of on engine e~ds o force P= !500 lb on the cronk.p1n ot /\. R~t09lve this force iflto two rectonQvlor corn­ponen\s Ph """Pv17 ~9tiog horizontolly ~ ver\icol\y, o\ /\ .

/\ns . Ph = "'tOS ID P'V' "' 11T lb .

. \

216

5 .) f'. vedicol lood P is c;upportecl by o t r ior'\qulor tirock.et os shown in Fig.G · Flncl the forcdS +rorismitled to the bolts /\ Zt._B. f\ssume thot the bolt B f'Hs loosely in o vertical slot in the plote. /\ns. Ro"' 1.2s .P : Rb = o . ..,s p

Chapier 3; 1.) /\ weightle66 bor /\B ·,c; 6Vpported ·,n o verticot plone bf a h'1nge ot /\ ~ o tie bor DC, os shown. Oeterrri1ne, graphi­

colly . the 0"'1al forc.e S induced in the tie bar by the action of a vediool lood P applied ot 0 .

Ans. ~· " QP (tene1on) 9

217

2) /\ roller of roclil)<i r ; 12 ·1n . ~weight ~ s ~oo lb ·,G to be pul-

, \eo C/'ler o cvrb of he"19ht h • 6 in. by o hor .1zontal force P op ­

p lied to the ~ncl of o strinq wound oround the circumference of the roller . f 1hd the mognitucle of P ~1..i"1"'cl to Giort the-roller O'ler the cvrb . Ans. Pa 2-00 1b .

~

3 .) A bOll of wc19ht w reGt upon 0 .smanh hor-izonfol plane oncl

hos aHochecl to · ,t, con tor two Gfring~ AB ~ AC .,...\iich poss o­Yer ft"iationleGG pulley~ ot B ~C !!:... eorry toads P &i-..\S!, t"e1>pecti--e­

ly, oG Ghown . 1r the strong AB ·,c; hon%.Ol"\tol , f"ir'd ~heanqle o<.

thoi the string /\C,.~ rt;ia~ w ith tho. hor-i :t:onto l when the ball i s

·,n a poGdlon of <if~il.ibr-1 u n'I . Al60 f ind the ~ssure R bdween

the bol\ 1*.. the plar,le. ,r Alis. cos« - P/6( ; R .. w -/~Lp'J.

218

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4) Determine the a><iol r. r . 1~ .:::11 !Ir.... S« 1"nduc--' th b l\C &... ec . t ....... \(\ e OrG

. in he F19vre oue to the acti~ of the hori :i:.on.tol o p\ted l~oo at C. Th6 baro ore hinged together at C o. to ti...._:. fovndotion at A "'-.. B . "" •c

Ans. S1" 182 lb.(tentoion); S!t . 6'40lb,(comp.)

10001:>

.i!J.) !In electric - light f1")(t1..1re of weight '1':1. lb . . ~ 40 l'il SUpporied OS &hewn· DeteMT11ne the tensile forces' c t.' c_ · ii...._ ·

. r- h . . . ~ ""-"'-':! rn nic: wires BA ~ 8C Ir terr ongles 0f1nclinofion Ore OG&hown .

/\ns. S1 " 2g.;s lb ; S2 " 2C>. 7 lb .

219

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' 1,

Chopter .+: . I' · i bo OEcl 1.) Determine the reoction o\ /\ !ii... the 1orce S 1n 1he r ue

. to the oction of the loods P ft.,~ opp\ied to the crone. os shown. Neglec\ the weight or the crone !..._ ossume ideal hinges ot A. D, tt.,_ E . Assume tho\ P'>.!lOOlb, ~ • 300lb, o .. eft .

. . . I I Ans. Ro = 1140 lb: s "1~5 lb comprCGSlOn 1-o ..,-f--0 --1 .. . c

e

2.) Oelermine the oxiol force• ·,no bon; 1.2.3,4, ~5 of the pl~ne trus~ s;;upporied ~ looded os

0

GhCJwn. Ans. S, • -P ; S2 "+P ; .

Sa• -o.,P ; 5.t"' o . ....+2 P ; Se • - 0.33.3 P.

3.) Oeiermine the for-ce S ·1n the bor /\S of \he plane truGG lood-ed 1..,, supported 06 shown . An{; · S .. o.-+3 P.

220

( "

+.) A plone figure - four frorne ABO~ i~ ~upporfed on on inc lined

plane ~ looded os shown . Colcula~e the oxi~t force induced in the member BO. Ans . Sbd .. 106 .7 lb, tenG"ion

a) Two beoms AB 'l.t..,BC. joined toqeiher by o hinqe B, aro supf)O("­

ted by four bol"G, hinged at their ends . Determ.me the force produ­

ced ·,n eoch of the1:1e bore clue to the ochan of' the load p,. 500 lb. The dimension of' the structure are as shown. Ans. So= 186 lb, Comp: ;

Sd " Se : 2Gs lb. Compress'1on ~ Sc '° G2 lb • tens.ion . p

Chapter S',

1.) Whot Ki the necessary coefficient of friction between ~ires &.,,

roadway to enable the four wheel drive ouiomob'1le to clirnb 0 30 'l. grode? A w ni;. p ~ o.3

221

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I ~

. . ht W t.. Wt. ore connected by o Gtnn9 ~ 2 ) Two block. hav ing weig "' , fi . 1 · r . ho n If lhe ongle of r1cr1on ior-rest on hori:r.on\ol planes OS s w . . . r lhe I t ch bloc!<. ·1~ f F1.nd the mog0-Yiude g..._ d1~tron o . . eas

ea tied 'to the upper bloc!<- that will induce sliding . force p opp Ans. Pmin • (w1.tW2) s in'('.

. h . I r nder of weight 1Q &. rod.1us r is suppor-) A smoot cir-cu or CY 1 • 3

· · ....1- h f the some rodiu~ r ~ werq-ted by two oomlcirculor cylinUCl ooc o . . . . en tre ht ""'!2 OS Ghown . If the coomoent of' stoi1c fr-d1~ betwe

'\' •· . . .nd ~the hor1:wntol plane on flat faces of the sem1c 1rculor cyl1 ers . th -fi . r bet cen the eyl1nderG em which \hey rest is»-= 1/1 &.-. r-19 ion . w . .-.· t b . between

. •-- t-...1 cht rmine the rnox1mum · <.111s ance . GelveG 1<:> neglOU =- / e b l ./ \ th nter.s B ~c for which equilibrium w ·111 be p0(;SI e w, ou

e ce . h · t I lone . the middle cylinder,. touching the oni:on ° P · / ' /\nG. bmox 2.03 I

.+.) Referring Lo the figure I the coefT1.cients or fr1'dion ore OS fol­

low: 0.25 a l. the floor. Q.3 ot the woll, ti\.,, 0 ·2 between b locks. f ind u.._ . . ·~lue of'a horirontol force P opplied to the lower rne m1n1mum v.... . . . . A P. . ,. e

1.2 1b

. b lock. thot will hold the {;)'Sfem in equ1hbnurn . ns. m1n·

222

I , I

I'

s~ A solid right clrcula · cone of altitude h = 12 fr1 . &... radius d' base 1=3in . has i ts cen~erof grovity Con ·its geometric oidG at :lhe di6tance h/+ c 3 in . o~ t~ base . This cone rests on on incl ined · plane AB, which makes ·on 01191e of 30° with the horizon ta I &_for

. which_ th~ c.eefflc1en~ of' fr-idion is µ.=o.f,. A hori~ntol force P

is applied to the vertex 0 or the cone &..obts in the 'vertical plane

of the f1gu"' os shQl.vn . find the mo.ximurn 8tc:._ O') inimurn values of ' P consistent with equ'dibriurT) of the cone ·,r the weight w = 10 lb . Ans. Pma.>< ,,. 4.61 lb ; Prn1n" 0 .. 590 lb.

Chapter 6 !

1.) In the case of' lhe tripod shown. there is no f'r ict ion between

the ends o f !he leqs &...., the floor on which they rest. To prevent

s lipping or the legs their ends one connected by strings olon9 the lines AS, BC, '&._AC. Determine then the tensi le force S in ·

eoch ol these strings if each leg makes 30• with the vertical~ Pis o vertical load . Ans. S = 1/g P

p

0

2.) Deler rnine the forces produced in the bors 1 to 6 , inclusive,

ol lhe s poce truss shown , 0¥<";'19 to the action of four vertical load c; P oppli'ed o.s shc:><vn . Aceo ~ A'C' 0 ·o ' ore two squares wi fh paralle l sides ol lengths a ~ 20 , res pec f ive ly, lY.. the

223

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1/

II

:l

distonoe between hori zontol plones ACBO ~ A'B 'c'o' is 2a. f\ns. S

1-== S

2 = - 0.2.5 P; S3 = s,.. • -1.06P; Ss=Sio "0.

p p

111£---..10

a'

3) Find the tension 1 i.n each of \he guy wires BO ~BE of the crof\e loaded o s shown. /\n<o. T .== ..+.3etons , cach-

f ..

e

p

+) A s trut AB o\\oched to the foce of o verhcol wollo\ A by

o sph<'ricol hinge £\ands perpendicular \o the woll &... ·,s sup ­ported by two guy w ires of> shoV'l'll . At B . in o plane porollel to \he wai l , two fof>'".es P ~~ os shown. -E< be'1ng hor1"z.on1ol bi,_P;ver-

1iool . Usin9 the rn~ihod of (l1()!Ylent~, flixl the oxiol forces pro­

duced in the mernoors ·,f p " soo lb ~ \Q 7

4-00 lb . Ans . S1 = 103 lb ; S2 = 162.5 lb; G3"' - 1 , 600 lb.

224

\,•·•

' \ ,·

.5.) ': pulley A of rodius o is Gupporf~d frorn \he foce of o -.ert1col wall by two braces f\8&.._AC ioge-l~r w'tth a tie bar AO as shown. A fle"ible cord ~AF ·1G fostef1(jld to the wolf at E,pas:

995 ~ver -the pulley, &.... can"'1es ot its end F a load ~· find the · tensile for.ve S produood 1·n ihe h e bor AO ·,r ..o • 1oolb a = 6 · b • .+ fl = 'I: l'1 ~ ' in.

, C · 7 :Z ' r · Ans. S = .!53 lb .

______ --=-::~··----· · ·-·-

Chapter 7: 1~ /\ homogeneous slender wire 12 in . long is bent in lwo "r i-ght on9I~ as shown . Determine the coordinates of '1\G cen-

~G. ""' .2.0 1n ; Ye= Zc = 0:67 in. ter of gravity . A "- . .

:i

y

A

+" 0

+· e · " ...

225

·I

2 .) A steel shofl of or-culor crosG section hoG a circulor steel hub presGed onto it OS shown . for tho dimension6 .;;hown in tho

flguf'e; de\er mine the diG\anc.e 1-c f rorn \he \er\ end or \he short ~o the center or ·grovity c of the composite body.

Ans. k;,r 6°2.8 in.

3 .) /\ homogeneous body consists of o c'1rculor- ey\indricol por­

hon or rodius r oHoched too . hemispherical podion of rodius

r OS shQ'Nn . Oeterrnine \he height h of \he cy\inclricol pod ion ·,f' HIO center of gravity ol \he composite body lies o t the center

c of ~the cir-culor ,yiore fore or the hemisphero. Ans. h A r /{2

T 41·'

I

j

4 .) A homogeneous body cos'u;ts of o righ ~ c irculor conicol por­

. hon oHoched too hem·1spher'1ca l porhon or rodiu~ r os sh­own . Determine the qltihx:le h or \he cone ·,r the .~nter of 9rovdy of the, composite bocly coincides w ith the centerC ot the arculor base or the cone : Ans. h ,. % r.

226

,.

" ! ·::

. J

.5.) Deter~ine i~· coordinate Ye of the center of grovi ~y of a steel r1vef having the dimensions shown in Fig. H . Assume the head of the rive~ to be hemispherico l.

ti . 16 "'/ Ans. Ye• 1.1s in.

Ye

Chapter 8:

1.) f 1'r.d the polor moment of inertio of on isosceles t r iongle having bose b ~ oltitude h w ith respei:.t to its ope)( A .

An~. Ja = bh3/4 t hbj' 46

227

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f I

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ii I

2) find the p0\or moment cf ined io of ihe shaded oreo

sho-.vn ·,n figure v-i1\h respeot to poin~ 0 . Ans . J 0 = 0.27.+ ,. ....

3 .) Ft'nd the polar moment of inertia of the oreo or a circ.ulor sector of rodiuG r 'iJ.... centrol angle ex w'dh respect to ds center . Ans. Jo "o<r""/4 ·

., .. .

;

. i

-...) Calculate the moment or inert ia of the shaded /\rca in fig. 0 w·1th r espect to the -x - a'X1S. Ans. I,. c :26.83 in.""'

228

.

&~ ~olculate . the moment d(' inertia oft~ 0~0 of the angle section hoVlng the dimensions shown in fig . /\ with respect 1o o c;entroidal oici'°. porollel to the y. o.xiG

Y 11-/ t\nG. I.,. "' s.~ in~

0

Chapter 10:

1.) A body starts . to move verticol~upword under the in­fluence of gravity with on Initial velocity Xo = 2ofps. Find (o~ lhe m()'J(imum height to which ·.t w'ill riGC. ?.,_ (b) the time requirro for ·,t to return to 'tis initial po,Mion . 1ok.e the slot'\. flng pOint os the origin so thot Xo • o ~ neglect ~ir resist ·

Ans. (o~ /Cmo,. • 6.2 f1: (b) t • 1.24 sec.

229

1

I

2.) A ~n:»1n is moving dow n o slope of Q.CX)B with o veloc'dy of 30 rnph . /\t o certoin ins1ont the engineer oppli"es the brokes 'lit.._ proch..iccs o to to~ resistance to rnohon equal to one - tenth o( the weigM of the troin . Whoi distonce x w'1ll the troin

· trove I before stopping? /\ns. 'J. = 327 ft . .

3.) The depth of the' crater of the tool Volcano was colclJloted

in the following ~·ohner : From a helicopter flying vertically up­

won:J ol 6 m/s, ·a ,smoll bomb 'NOS rcleosed ot the inston~ the helicopter was zorn. obove the crater surfoce. The oound 9r eitplosion wos ~ord 9 ~econds later. If thc•·speed of sound ·,s 335 m/s' what ·, ~ the depth or !he croter ?

I

Ans . depth" 240.2 m.

230

I •

l ' , . }~, .

. i:!l

>·

. I

. ,

.\'

4 .) The troin tr-ovel i1> from stat ion A. to station B which ·,s 1 t<.m apart in o min'1mum time of one minule. If lhe t roin stods frorn red at sta tion A , &... occclerotcs o t 2 . .5 m/sec~ continues ot consfont speed ~ decele r"Otes ot Q.5 m/s 2 un­ti I ·, t sfop a t sto1 ion 5, f1ncl the r:no)( ;mum · speed ·,n Km/hr-.

How long did ·,t trove l at . this top .speed . Ans . s ~ 60. 76 1<..m/hr. ; t = 7.6.39 S .

.5.) ."The motion of o portio le \s 9iven by the eqlJofion s= 4t3

-'3t 2 +s t + 6 where s ·,G in fl . k. t iG in sec . C:Ompute ihe

volues ot V ~ o when t • 1 sec. !

Ans. v=11 O/s ; o = 16 fl/s 2 l

231

!

l!

11 i

· Chapter 11 : 1.) /\ motorcycle ?H.. rider of totol weight W = 50() lb travel ·,I"! 0 vedicol plane with cor.stont speed v = ~mph olo~q the cif"'Culor curve /\B of rQdius r "' 100011 , os shown. f ind the reaction R eicerted ori the motorcycle by the trocl' os

·,1 posses the crest C of the ourve. /\ns. R "' 432 lb.

. 2 .) The coefTlci.e0t of frlcfion ootwcc.n wet Olipholt po..,. ...... -. 'it.., 1he tires of on outo~bile iG found to hove the .volue µ.. • o.w. At wl-Jt constant speed v con the automobile fro -vel around o curve of' rodius r- eoofl . w/out skidding ·1 f ihe rood ·,5 level ? Ans. Ymo, = 49 mph.

232

~) In figure ~low, o hammer of we'19ht W - 2 lb s1orls from rest ot /\ &... shdes down 0 roor for which the r-r:· . 1 of fr . t . - c.oen ICICl'lr

nc ion is µ. = 0.2. F.ind -the distonce x to fh . t D h t

. t c poin. w ere ·, ht .s the ground . Ans. :>< = 14-.4 ft.

4.) In whot proportion will the maximum range of o pf'Qjec­tile be increased ·,f the initial velocity ·,G inoreosed by 10 per cent ? Ans. 21 per cent .

.s) In. the figure, ihe pilot of on oirplone f1y1ng hon':zontol ~ ly with constant speed V"' ~ rnph ot on elevation h =20000 ;I

obove a level ploin wishes to bo;nb a forget B on the gro-tJnd . .At who~ ongle-e>- below the hori:c.ontol should he .see the torget ot the instant of rel.easing the bomb in order '.! to score o hit? Neglect oir resistonce . Ans. -e- .. 22•12 •

,.. v

19-=(~--- -

' -'"' .

1 ;',,,_ ', B

/flT7777777 ij/)7/777777/,IT:ll/; /)7

233

Chapter 12: 1.) lhc · ormoture o f on·eloctr ic motor hos on ongulor speed n"' 1800 rprn ol tho ·1nstont when the power is cut ofT. (o~ If ·,I cornes to rest in 6 sec, colculote the ongulor dccelero tion o< assuming -\hot ·,t ·,s consiont . (n) How rno­

n'y complete rcvohAions docs the ormoture rnoke dur·1ng tn1s ponod? /\ns .(o) ex= 101T scc- 2; ( b) go rev.

I I

2.) Considering the system in the f 1gur-c , dct~rm·1nc tho voluo of fJ- for which the negative onqulor occoloro\ iol\ -& of the bor OA hos 'its rnox imurn voluc. Ans. fT., 30•

1-...e -P--t

0

234 I

3.) f\ slender but rigid semicircular wire Ot roi.Jius r iG

Gupported in its own vertical piano by o hinge ot 0 z.... o smooth peg /\ as shown. It tho p09 Gtorts from O ~ rnoves with c.onstont speed 'lo· olonq the hor1.z.ontol '/.. 0')('1s·, find the onglJlor vcloc"ity .& of tho w~rc ot tho '1nston ~ when fJ-,, 60·. /\ns. -6 = Yo/r .

4:) For tho figure shown w,...,. 8 rocl/G ; o<,... "' -1.s rod/st ; t ,. 2 socs .. find the veloe1\y ~ occclcrotion of blocK C .

Ans. Ve= 12s rnm/s ; Oc: 37.5 mm/s2

¢"100mm

.!!.) For the figurn shown , o<. = 60 rod/s2. ~ S'" 6 m · How fosl ~he blocK C is rising . Ans. Ve c 14.7 m/s.

235

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1, ·1·

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ll

Chapter 1.+ : 'i 1.) I\ bloc!<- of 'f"'Cight W ·,s in given on inif1ol vel<;x:ity Vo 010119 o rovgh ~rizontol p!ono ~ is brought to rost by fric­tion in o distonco x . Oct or mine the coefficient of f rictiori, oGcouminq thal -.t ·,s independent of vcloe1\y.

I 2 /. Ans. p. •Yo/ 29x

2.) Determine the dynamica l deflection o that ·will bo pro­duce<::l ot the center of o simply supportod boom by allo­wing o '4;000 lb woi9ht to drop onto ·,t from o hoight of' 4 in . Whon gradually opplied, the same load produces o static deflection of 0·1 in . NC.(Jlect the mosoG or tho boorri .

Ans. c5 • 1.00 in .

236

3.) When o boll or wo'1ght W roGt On 0 spring of constant K.

·,t producos o static doncction of 1 in. How mvch will the some boll cornpross the .Gprin9 ·,f ·,+ i~ droppc.d fromo height h,. 1 ft? Hcg\oct tho mac;;G of tho spring Ans. <S:: 6 in.

+.)I\ smoll b lock. of weight w=10 1b is given on 'in i tial velo-

11 ~

.~r city Yo•10 fps down tho incl in~ plonc shown in figure . If the cocrfici"cnt of frichon botwccn the pl one ~ .tho blocK· ,·ii

·,s µ. == o.a, f ind tho velocity v of -the block. al B oflcr ·,+ hos 'I \reveled o distoncod ~=.soft . Aru;. v=29.Sfps .

237

·1

I

I

I .

!I.) If tho syst~rn in figurv is re1eoscd from roc;t '1n tho eon­flgurotion shown· by solid l ineG, fi'nd tho rno1<imum d iGtonce h -\hot tho weight P will foll . Neglect frichon "*..assume ·thot the pulleys A '*-. B aro very smoll .

Ans. h • 4P~t/(4~~-p2).

,.: ~ '/

Chapter 1.5: ' ,

. 1.) A locomotive weighing 60 tons hos o volocity of' 10 mph~ txlcks into o f roight o:J.r weighing 10 tons thot•1s ot rest on o level trocK · A ftif' c.ouphng is rnodc, with wbot voloe1ty v will the enfiro .s~,stem c.ontinue to move? Neglect oll friction.

Ans· .v " 8.57 mph.

2.) A ""'ood block wc'1ghing . .51b ros;ts on o ..smooth horizon -tol surfocc . /\ revolver bullet weighing 1h oz i~ .Shot hor·, -

:z:.on\olly into tho s ide of the block . If the b.lock. oltoins a velooity of 10 f p~ whot WOG the muzzle vcloc'1ty V or the bullot?

Ans. v s1010 fps.

238

3) A mon wc.1ghing 1so lb runs~ jumpi; from o p'1cr into o boo~ wi~h o horiz.ontol velocit y Y1-= 10 fps. A~uming .tha t the impoct i'7 en t ire ly ploshc, fi'nd ~he veloci t y w i th whic h. the man ~ boot will move owoy from the pier ·, f the boot weigh 200 lb. Ans. v. "4.3 fps .

4) A golf boll dropped from rest onlo o corncn t Giclewolk rebour:ids eight-tonths of the hciqht through which 'it foll . Heglcding o'1r ros istoncc, determine the cocfT1c icnt of . rc:Gtit ution . Ans. e "'o.9.

239

i· ll " I•

l I

Ir 11

l .1 ll II I

. I '-! J;i 1tli *j ~ l;

A'i ;

~i '; .I

~! 'i

1,1 j

ii ~

I

i

l I 1! !

~ ., I

1.

s.) In +he figure shown, a small oar of we igh~ w starts from ~t 6t A "'·rollG w ithout frict ion ' olorig on inclined plonc to B where ·,t ic; ..str'i k.os.· o block. ol~o of' wo'1ght W

~ in itial ly a t reGf. /\ssvming o p lost ic irnpoor ~. B , tho ·car '/,t,... block w ill movc. frorn B to C o,G one podiclc. If the .

. cooff1cicnt or fric tion between th~ .blocK. ~piano :1c; ,µ.. ... 1/tZ. , colcl) lo te the cliGfonce 'X to point£ where tho bodicG c:Omo to rost . · An.s. x"" 14.2 ft .

. . / 1

,.-· ti /' /

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