Engineering Mechanics Statics and Dynamics by Ferdinand Singer Solutions

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. Tet.'~.' 741-49· 16 • 741·49·20 1977 C.M. Rec:to Awn~ Tel. Nos. 741-49· 66 • 741 -49-67 Menil., Philippinff- '

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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. . . "~ an ·. 1 earn e . s'!bJect . matters that ""JII ' d P.ro~essu .of " thinking . . e.~.e. lop .theJr . orde-rly . .. .. . ... . ... . , .· " " " " f'.'J. t.h .. the . present e 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 · ' . . . . ... ' 1 " Extra - eliort ", • " ' ... b'e pr-es e nted .Jn~ ::: exerted so . that - e~eryth,Jng "ill ". and . . i .. . . . y ,p~rle c tly . unders l-ood bu' t Jn Clear 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 0 t e c onc:ept wJth out u.nderstandihg·. .. · •

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- The autho_r s wJsh · to ac~n~ 1 d .. t . " . " .. JUcba·eI Si .. " e i .e. hf:dr Jndebtedne s·s . ongco, J e ttrey Borl . ~ tin"lg, ... a nd Romeo Adr·Ja l . . on~an, , ·:~.onaldo ·contrJbutJons Jn p~.eparJng an;o . ~~ their valuable· wr ng tbe manuscript.

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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 .

P~ted

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N · c-: De 1 aRaaa A · G. Mendo:za ·

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by REX 84-Bfi P-. Fiorentino St.. Sta. ~es;. H~ts. Quemn.City. Tels. 71241-08. 71241-01;

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ACKNOWLEDGEMENT .

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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. '

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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
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"' .Ir 'its

mdmen h. about . A ,is zero, .delef'tm·lne

refecv,.t~ the f'igure· 231 . when .we resolved ot /\ Mo == 100 • Fic (a) Fx =;. 60 )bs.

fy

=/F,_

2

==

;+s lbs.

-+ fy ~

F "'

c

7S

30}b. 60}b.~lb.

40lb.

J;_;_._l ;3' ! t . ~·· ·' 2'

'

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

-&·

-&

dfr.A ., .6,6p/110

.

~~7.)

., 6ft. '

\

re~ulton)

.

R • ' so +
from A .

-~

o( }he four-

orm of fi'g . P - 237 .

~·

~orallel forces be~

·

-~o· • cso

10:

~Mo ·." -.so(6) t "f.Q(~)- 20(3 ) t 60(a)

1

""..:r20.0

fl - I b (cw)

~Mo= R (d fr.

4

~ -36,67 "

(11) +o

MA"'R(dfr.A)

foree ·

= o/6o = f-on- 1 ..-~/60

ton·-eT

'

. R • -30-60 HO...:.W ? -1·1 0 lbs.

c;Jfr.O ...

w hen we resolved of .. 8 . Mo= 90 " - F~ (a)t· f~(6)

. f

llie

o

.236.) ;\ parolle~. force sxslem_ o.d s. ~D }he le~er shown in f(g: P-2P6. Delerm1ne, Jhe ma .., . ...1_11.:. ·1· .. . ·: · ' ,· · · · , ., · _9 n11uuo 11on. of l he ·rea u Iiion ·l . •. ,:<) pbs1 .

Determine }he ing . on }he rocker

.300 fl

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

.

,L·

ol "·'

Mo= P)l.(3). "'?0<:>.fl-lb , . . px:, ": 300/a =:;1'00 lb. to )he right.' when res0)v.;a dl-·13 ' • .

0

. p}'. .. 30)b.

200~o

"'

-4-rl . -right of 0

'

:zaa.~ The beam J\ 13 in fig. P- 238 suppods o I

varies from on 1nferi.s-ity ~f· .so IQ. per Calculate the mo ni lude ~ ·1 ·

· 11 · t · R

..· .~i"

.

o)

d

n .~o 20:~b· ·

·

•

.

h. h . w. ~ ·.

P~

1

•

9 , "') posi JIOn of the re6ultonl food. eplo~ lhe given loading by 0 un iform! d . 1 'b ·t·--' load of .solb . n . .. y 1s r 1 u c;Y· 1 o - triongulor load vory1n9 per', . pus . c:-, rrorn

•r ·

----

/60~ t 4-9._

: '

lb.

'?

10

''

.11

',,

'I

~.I.

'l

,.

'

''

·;;• '1' ; . . ·•

'

·"

.

.

'·

-

.

;

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

:•'

, 241.) .Locate

· : · loode

· '·

. . -;~ .~·· . ·:·{~

·~0., :,i{ I ·'II

.l

R= 2oot·300t_m•ao0t2oot 2000 R .. .a-+00 .lb ( djreqfe.d downwqrd} £Mc= R'(s)

8

~· ''\ .200t aoo t

'

"'fOO

f~ t

2oo

R' "'· 1...00 Jb ( d 1reofe? do-Nnword) Q- lb . , Rdc = :i:Mc - de -= 7000/_,;400 "' 2.06 fl r-ight of

200011:1

~M.c • 1"!00(5) "' 7ood .

d .. 12.'06

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.

d: ~ 3686+ ·~... . g,60 ·fl .

i".;•· ·.. .

!

24 2.),

. aa:10. .

#

\_.i'

,.2-te-)

The shoded oreo in fig . P- 2-.0 represents a

~feel ploto

~: . ~~E!~. !;'~:~~~:~:E~~~::~~~!0~~· ~[.

~' r,.~· ''

the weight of fhe moterrdl cut a·,,/alj . Rep~'Sent the original weight of ·the plole .by Cl downW'Ord force oc~inq of the center of the 1o x1+ in . r.ep.tongle . P-.epre sent thewelght ~f the maleriol cut owoy by on .upward foree 'a cting · 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

F

~lb

::_.! ~:t

~Mp= -F(~) ~ zoo(s) 1' 100(~) • ·aoo(£)

or191nol · plate minus

12 -

C or

o. right or!,\ . .

f ~ •• .r.

200

lb ( do1Nnvvord)

. ~MF"' aoo(1) ~

P ..

-tOO

P(3) - ·1oo(s) - 2oo(e) lb

(upward)

;·~;1!;~

"

·': •' ~ .··.

"

'·

,J

; .·

'

i.·.

'

t-dle · puHey lhQl ' ore req~ire~ fo. Pci10~0e0 Jheg,l'ven ·.sys}~:; ~c

sol'n':

.. ·ao(1z1.-+t9(16).,.6d(e) . ' '

= '~760' itf :ri> (.;.im~ns .eou11fi::t-.cw} , ·. .·

·f=:~ 76%~ . =·. 6a.a lb:(Cdupl~of 'thls . m~c.n .. 0

.:.;,,;.::'--::::::_ _. ·· : ~I

.. ''"' requ1f-ed) ... , .

.

.\

'

.

'.

\ ' '

,'

.

"

~'··: ~: '." .2~8.) To dose

·:.

.~;

' 4011:1 :

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 4 ff. ou} rr-6.~ lhe cenle~:: ; ._Delerrrune ·}he_for(je ·r:-equJ~d ~ cir.ow,, a f~· d.r~grom ·

(Oree

booj

·o f the· b.or'.

·

·

6()1b

f'.'-tlqnq

.

'

1

.

.

,4 P"

w·~~I ·a ~t. in di~.melei. ·~ .

60(a)

P="48 lb:..

. ·t''·'\. 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 )he resu I tonl cou p)e. /\)so deJ~inine lhe forces ocHng ot lhe rim of them~

or.

"

.l

....... :

\

'.··. · ,·\

249.) f'.9 · ~-21:9 re.pr.esen ls }he bp vi.ew of d spe.00· · ~u­ . Ger v.:h101J II>; .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 reod rQn is caused by the unbci "lonced . torque~ whioh is o ; coi:;ple. ·"

,.;

I·'

15\ ,.

..-.

..

.•

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 ils line of ocJ ion wilh .lhe x ~ Y oxes .. ·~ · .tM.) A kTc.e ..y.;fein oonsisls of

C"'30R

(-100-100) ft-lb -(30/i2)R

B"' 120 lb d fr·eded

verJ ica lly up ot /\ ~ down .o t B.

c

a

R" = F,.

~ 240 cos~·

= 207.. 85 I>..

Ry= 'Y

(to lh6 righl)

= 2-ta .sin 30.

- 120 (u~)

2so.) The con ti lever truss shown in Fig. P- 250 corries o vertical load of 2400lb . The }rusg '1s (;uppor)ed by' bea-

rings at /\ ~ 13 which. exerl the fOrces ;\ v

, /\ h. ~

or

or

S

.

Bh

2400(6) - Bh (4)

Bh .. Ah "' 3600 lb

~ d;·reo-tion of forces

Pol /\ ~

4'

'at

(2oolb-H

IP

3 '..

+'

A

3'

J R•1oo lb

R-1001b

ix =~ -

= 300

Fiq- P- 253 o G)IGfem of fbrceG roducos loo downward fOroes or -t«> I> lhrough A plus o oounlercJoc)(wise downward I • l

I

-~

Mo:-MR .

u x

.4oo(-4-) - 800 = 100X

0

x =2n right of o . 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 ...

ffB ·

R=F

+ 2.VO....P(~)

= .300 lb MR "' Wlo = -

a.s fl

)he.

left

('"'> meonG belovv o)

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:., 256..)

') 11{)

16

lo

300y =.300(2.) - 750

lb. ( downward)

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

p

o

2Sa-? In

y

l=l-):

.... rl . lell of

120

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

p

fl . obcNe 0

254.)

P ~ F.

· rt. G ~lb tr "'I ~~ :::r'

2.31

12

ono}her vedi~I ·F f3 111 Fig. P- 2s1 pr;'Odvce .o resu liont of' 100 lb d own a t D ") a Geufl•oPolOC,..,;.,ise couple c of 200 lb-0 . Find the magnitude· 2-51.) /\ vert.1'c 6) force

=~ 2>Z85

Av. 1

i:Y

verli'Col

1n order to const itute o couple Av ., 2'foo lb (upword)

24001b

i.y

Bh . The

forces shown con1 1tu~e two couples which must hove opp:>s'ite momen~ effects }o prevenf movement lhe truss . Determine }he mo9n·1tude the .supporti ng forces.

four

M -C"' F.,,

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 :ioo

~·

~

coupl.e .

P ~ wo)b.

. w '1lh lhe oddi\ion of o pair a \ opposHe

axial )oods eoch aqvo) lo 200 1b ) w e ge)

200

f ,. 200 lb

( dOWf\V"Ord) 400 lb-in

c - ~oo(~)

9

cw

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,..,. 141.-+(1/a) t 22+(2/.JB)-a61(o/Ji3) :141.4 lb 2!fy. 100.()fj !bOo lhe righl) I .' .ZFy = 1:+1 .+ (1/a) - 22-+(1/-19) - .361 (o/$6) I • ~Fy = - 300. 56 (<-> meons downword) G

/

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 C ·, ~ opp\it:d force P - 1BO lb. Forc;e P is perpePdioulor lo lhe plor'le con0 )ain;n9 poinlS /\' 13, ~ c . Delermine lhe lwisl inq ef'fed or p on )he -G'haft AB ~ lhe bending effe/;l al point !>. .

/

2s6.) /\ ver )icol ·short /\B

R

r--;-+--4c---1--J 8 1' 1' ........_ ......._

o

361 lb

•

rhe system ~r forces

rosullon) R o) /\

~

p

p

F

lhrovgh /3 ~ C. 2011>

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)

R

E ""33.37 l b b) 4P = 1001 .1 P"' ·2S.03 lb .

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

.1.;

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

~

I

f '-, I\

0)

w/c

=zf,. ly . i. y = 3601'90 ..

zMo

f

l

R "' ~F,.

ao- 20- 60

.R = - .so lb ~Ma

(~) meons downward)

=C

f ( 2) F "' 110 lb thru J3 ~ C os shown

60(4)-20(1),.

fl

-l

below 0

lersecls }he X o"Xis. ~Mo ~ zFy

Lx "°

Lx

+00/160 ~

19 18

4

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 -

~

I \

0

1- I 3'

the right)

3f c 100.1

0

ading on ·u-ie frorne in fig . o c ouple' od;ng hor•'z ontolly

lo

-e-,. = 71. 58.

o.)

f

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

by o

316.79 lb (down

;n4ton-&,. .. C.Fy/.:tF><

/\

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

p-257

·/(100.01) 2 + (-.aoo.~6).:z

ei<· ton' 300.s6/joo.og

or

2s1.) Repioce

0

r ighl

or Origin

l ~

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

26+.) Compleldy dderrnine

,i·;l!J ' '

.262~ Determine cornplelely lhe resuHon) of lhe forces oc1ing on lhe

sl~p

pu11ey .shOwn in fig.

1

'1

·1

R.. =

750

·

~-.262.'

::fr>' ·.. 7!50 cosso· ·

i

1

2so

• 899.SZ lb(lo )'tie .right)

,

£.fy ~f,.11

t

G

1so s1n30· -12so

.

R " ./:~.F,. 1

..,,_975 lb (- means down'/llO

,.

~Fya.

(099.s2)a • (-a1s)~

,B =.125+- SQ

lb <)OWn 87.S/899.152

·

B =

0

ton ex

-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.

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

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

>

265.) 266 ,

361 lb

.tfy " 300 ~ao +~~( /./6) - a61 (a/..Jiit) = .sg.61 ( upward)

R"' ./~F'/.'2 t~Fy"" .. ./(11·9,9)'1 t (.s9.61)"... 161 . 9161!?... (up to }he r ight)

-(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 ~.Y

~f7'., 390 (1~3) t 722(o/-1i3) ~ (sinao) - 810.47 lo lhe righl £.fy"' aoo(-7(a) - 1~(iz/..fi§)~aoo~·

I

1

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

!.)( -

5+4.68 0

Jy

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

224 lb

~

Cornpule l~e resuHon) of lhe three forces shown in fig PLocale ils 1nlerseclion w'1lh lhe X 'rs Y oxe.s .

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

..A.+----'f-4---1-'---'

ft. right of O

.. 100.6/1+9·9 = o.67 fl below o

/:E Fx

,. '

o

t' .__.._-l._.L.,~'-1.-.J.- _ lL.

1n lb

tor,&J< ,• .:Fy/~Fx -e-.,,. = lon-1 510.3/810.1'7 -Er)l = 3!2.19°

I • -.s10..s \-moons downward).

R .. ./;EF.,,.a + ~Fy¢ •/(810.1-7)~+ (-s1o.3 )2

R • 957. Q7 down lo right

.£Mo= 390(12Aa)(2)-3go(•Aa)(s)+ 122(2/,ffe)(-+)- 300sin 30(a) ~Mo= 1121. 97 0-lb CW . ·

,~ l.,c = 1121.97

Ly

o

.

.

-&,,

..

9 10·3

•

tic

Ly .:; 1121 .97

R

~.2 n .

righl o) 0

-

-= 1.3 8Q. obove Q

910.47

21 20

}on-1 2s1.0.3/419.7g

-&x ... 2B.2!5 • .

oxle

1'

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

=· :i.fy

-&)(,.,

0

.~ l:y

~Fy¢

t

·l(+1g.7g)f2 t (257·0:3)'2

Jo righl

-&.,. " lon- 1 =.t;t.12.1° Rd .. ~Mo• 1so(1.25)-12.so(o.s) - 2so(1.!Zs) Rd .. o :. d 0 j so R po~Ges lh~gh the

....+--l-'',..._,,_-1---''""=- __ _x_

· .ifi<" 14-1.4(Y.JI) t 300 s in6Q 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.

1+1:4Jb

I

111;1;·

r '~.1;

1..

I

I

rh,1''1i

[' 1jI 1u•:;'

-(}-JC "'

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 ·,~ ?

ton·1 Z.fy/zf5c - · ion· 1 10001.os/-t00a:e2

266)

h•I'": ji;,1'.

l~;l I

~MA "' t

-e-,. ..

68.2·

++BO ( 1/~)(.s) t +4-90 ( o/.f6)(10) t '1000 (s)

t

30006~)

'10o0('l0) t 1ooo(ao) .. 160087. 7'1 o.- lb.

x "' 160097· 7'1/10007. 0.3 "' 16:0

n.

~ight of"

.•

iiii·· .,,

.lfi< " 10.000 - 6000 cos 30.

11>: 1

= 4803 .9{1 lb(lo lhe right) ,

!!;hi

,i,.I

--=-==-- --=--=--

•! ~.

\l~'i

I

'1~~·

6'

·'·1iW1 }

a·

R ..

1

·1111·•' I

;j '

il1 1~ ~I

l:l·ij

11

~f,.

t~Fy

a

!Y

1101b

180

~Fx - R>t

b

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

I

I

£F.,. ·Ry F7 - HW t 1so(-t/s) = o

R = 2742+.02 Jb.(down · to the

"$-~ = ton· Z.Fy/.%.f-x ... lpn-1 27000/-400.3.85 -e->C ~ 79.91. i.~ = -24000(11) + 10000(6)- booo(+) ....:: -22~000 n -: lb (- means ccw) 1

I' ;:1~ I'

""ord)

comple t ely .

=/(:+B03.65)a. •(-rnooo)a.

l \ i - - --

11•l j• •

I~

~fy• -24000 -6000 sin30· ::: -27000(- meons clown-

p ... 10,ooolb.

1111·!:

268.) The resu Iton 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

~~-:=:u ---.-..-rz- ~

r!ght)

~

/

_L.

0 /

therefore

r . f)( -

1~0tb

'100 lb

to the left

c~~Mo

+so· -M,. + 110(-i.) + 120(2) Mr. • 200 0-lb (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)

MF• Fd

d ~ 200/~oo • 1 ff . obove 0 ·i1!J 11

111 1

I

:!j~ I:~

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.

.if,. - 2000

~Fy

t

4490(1/..s&J

- "'t()03 •.S2

.. -

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 . 269.)

110

•"/

I

15()

,'

lb

~ ~ 2000 -

3000lb. '1000)b.1000lb

Rll • zF-x2. + ~F1
=./(4003.92)'! .. (-10007 . 03)~

~D ..J077~_.17--1Q.(clown lo the right)

390(1ajia) -·r,. .t 110 t 1s o(3&)

1, >/~~ IT•""'

10QO - 4490(2A!f)

_:.:,12Q07_.03JQ.(- meons down -

""'orcl)

~h •zfl'

I

I

l ho

ri'gh~.)

Ry · zfy

x li)390 -:390('5/ia) • Fy -120

+1so(+ls)

~ f):' • -1so (downword)

M - - 1

R .. F 2.

F = 2::FJC

..

+ ~ fy~

F

23

22

Uo

.:Efx • 160lb

c

- (160)~ + (-150)~

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

-Oj

= -S.15-

Mtt · ~~

390{.Yia){3) + 3'J!0(1~)(2) =110(+) t 120(2) . t MF Mr~ 1'"10~

3.~l7 n. "'90/460 - . :a.067 fl .

b. .,. ....,oJ{eo Ly>

ON

f"i9h• or o ()bOV'S

0

The t~ forc:;es -shown · in fig. P- no ore- roqoired to a ho"'::iioofol ~tont ocfing fhrouqh point A- If F = . 316 lb. doterm'ine the volues of T. Hii1t = /\pply MR""'~ io defer~ R .. then MR - ~IW\c. to find P. ~ f.ilolly e;lhcr Mrt210.)

c;x:uJGe

P,,

.:Oto or Ry - ~Y to c.orr-pufe

1:IF !

T.

£a.fe - ~

-316(1/.t;o)(1)t316(•1/.,it0)(2) s R.(1) R .. 4t"-6+ lb.(to the right.)

a

M1t ·

''n \P

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

MR. = £til'le ..,00.
3

E.quilibrit.nl of force Systems

~Mc

4f'J").6+(3)

Chopter

r(1410¥1)

~l ="1~

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

-f00.+9

(+)

25

D

304:)

Th& cylinder C ir" fig . P - .ao2 ""ei9hs 1ooolb. Orow

of cyl1'nder

C'

Cl

fBO

of rod /\B.

Cir

~

+

Ev

3'

Wc;•1oOO lb

Ah

· Av

~h 303)

1he un; forrn rod

Gel'ler

or _grovity

ih101<.ne5S

II"\.

J\~13 .· Eooh

wei911G 4!20 lb

pull--~ "' 11.. -,~ 1.. "">

~

hos

bor.

Its

FBO of '\he rod . lie_glec1 the. o65ume all COr"toot ,s:urfoces to be,srroo\h.

o t 6 . Orow a

of the rod ~

ao+) The fr'Ome

.Fi9 . P- aoa

of

D ._.J o f th e bor /\ D shown .'" Fj9 P- 306. /\ss.urn6 oll hin9es to be smooth ~ ne_gleot the 'NC1ght of

goo.) Drow a FBD

-4-

~hown i" fi9 . P- .304- Is supported '" pivots at w~1ghs .so .lb per fl Drow.. a ' FBD ofeoch

member

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

Method I

(u~ing

l'l'&mber ·. /\

horizontol 'vertical Axes)

~Fh • o

Tcosao· • Ccoe +s~ 6('.)0lb

CD

£f., - 0 Tson'.30· t Cs1M·s· - 600

'

r I '

\

l.'

©

svbst. eq 1 io 2 (~uote 1 in ~enns of" T) T ~i~30• t- (i'cos.ao/c.os4"')(sin4s') = 600 8v

I=

lood is suppo""ted by a cable whi ch 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 hinges to be .smooth ~ n e_gleoi t he ws1c;ilh~ eool\. BOS .) ;\ 600 lb

or

lb. (cos3o•Vcos-+S·

4.39 . Q.3

C - 439.2.3

cc 5 37 ,.2,!.S JP;. 1< Method

T

I(using rdotion Ol'es) c ~fv-O : C.sin7$> • 6~.s; noo·

- c - .S37.9't.5

bor.

lb J

l

lI

I

26

27

lI

''· 'I

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

FBO of the block

.ifh =O: T"' 600 c:Os 60'

C GOG 75• .. 600 cot> 60 t .S.37· 94.5 (cos 76.) t

~f.,aO

p

300 • Nsin60• - Psin+s'

T "' ~Cbg.3 lb .

©

zfh·O

•Meihod JL (uSing force triongle)

Pcos 11i - Hco560. P = H cos60;/cos1s"

®

.subst. 2 in 1

Method C(Uoing Roioled I'. ,.

.,

"

~.)

/\ 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 ~-

N ~. +1s . 60S tb

ti'" Ol'~)

.4.

zFh"' 0: Pcos·~·

B

c

>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 Tr.ongle)

~~

,Hc,osl91i

p: N co:--ar~~· ©

v

P.sin2S• ~ Ii sin3,. • • 4'00 @ _,bc;t . , to i [N c;or..3s• (r.1n lls·j/~a,.· i Hr.in as•

p=.378-35

- ...00 N -11s.~9s lb p ·= H co~-as/co.s2&

lb..

.ti = "'1-18. 60.S lb·

0

P

~fh 0 0 :

p:

P = 'lo9.eo1(C064!1>·)- 3ooc.oco"7s" p "' 1212. 1.312 lb .

311.) If the value

or

P in Fig . P-a10

1he plane '

"f 28

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::.,

= -HB·60S(cosaGY006llt0'

0

2 12.132 lb .

t1cos45· - 300COS7s'

_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 c:let. the volue of P "'> the normol pres:isu~ H exerted by

N"' 409.007 lb .

p=

H • -to
-tOo & 1""5"

Force Trion9 le)

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

.:Efv =0 : H &in:+s• • :SOO sin 7s •

~fv •O : · /1~n60· •

FBO of' Cyhhder

Me1°hod I ( using

30C?lb -4001b

Method l(v.:;lng

Hcos ~/co~a· N,; ~.907 lb p .. 212,1.32 lb

300 - tfsin60· -

p

8~

cried on the.cyhdcr:

A¥."5)

rJ

<><. : 56.+4·

:. -e- "

'.3.3.ss ·

29

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

:s1 &.) The · 3':'° lb f ~r~ ~ the 4()() Ib force shown in fi'g. p-315 ore to be held 1n equil1bnurn by a third fo.....,..~ · o t on un,..no1. ·...... F oc+ang wn a119le -e- with the horizontal. Determine the values off~&; •

• .O

p-

•

-133.24!1 lb

)

~fhTQ

.


·~oolb 3d ·o.. i&

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

:E.Fv •O

Q.

400sin:ao' = fs1ne- (D

F

~ -o

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

300c
f • :aoocos6o' -13a,2"1-.s (cosao~- :zoo cos1os' F

~

.subst .

86..s7 lb.

©

CD to eq. ®

3QO .. 4

cos ao'

+(.oteo s1nao' ) /cos&-) sinel.

B1a.) · f 1g . .:i1a l"epresents tte concur"rent force sysi em 00Hr"1C3 ot Cl joint . of a bridge tru ss , De termine the voluos P ~ E . to mo·1n-

or

- e- '"

toin eq1iilibr'1um of the forces.

-76.935° ~ 76,935 belew x-axis

F "' 400.sinao" - 400-•n ~· ·

us;ng Rotated ·A -,.es 'M ethod

sine-

F=

.;:Efv • O

20S.31S

lb

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

· F =·· 412.435

lb

316.)

pt

400 COG 4'!;;

0

~ 200CCG .,~· i

300ccs

is' t

~

Wlb~lb

by cosine law w« ~ :ao.2 t-fe"- 2 ( ao)(40) C06&cos -&- "' c>.a1s

ao

-e-

f = - 126 . 33 lb F

"28·96° "' Q9· •by sin low

~ fh "' O 0

t tlCOG6<>' -

.s;u b~t;t ...

te.

f cos:so' i aoo o:>sao' ·

the vol'-'e. of' F

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

P "'"

- .sa.129 IV .

20

-

lh

30

.30

~-~

.s1no<. '"[3o (sin 29°))/ao Sino<.

<=

o( "

,I~

cos-1 o.07s

e

-(7

p t AK>O cos 7s

• ,l i~tl

·

..

pute the volues of P ?», F · zFv

that the for-

..Olb

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

p

60

FCDSJS

p - 165-4++ lb . 314)

Determine the values of fhe ongles O\" ~fr F1'g . P - 316 will be in e.quilibrium .

ces shown in

.:£Fh • O

0. 7272. 46.6.S 0

II

I

31

,/

£Fv c Q

/ 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 ·

Csm1s · "Pco6

.P

c

.,.. 5 •

(lfl3 ,07. & In 76 °

cos ...s·

p"

30A-. 719 lb

y using Force. Tl"'1on9 1e

~ .·/

Problem ~19 so•..ition .

\

by .sin~ law ~srn60'

~Fh

C

iiio- -~·

,.

"0 400 COG (;)-

A .etn.ote'

ff "60°

c c '2'23-07 /b

~ fv

y

£Fti-=0

0G1n-t6·, a oo1'. 'C61n60·

e•

3()()t-

100(s1n60•)

G•n+s·

a =-

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

br s ine law _P __ " C

I\

61n 7s • .

k

946 . .f-1 lb-

"C

"°' '->

p-

.stn60~

!l!l3. 07 ( s 1r175 -)

3£>+. 719

lb

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
of .p t hat w i II prevent rnotion . usi~ ~1ot~ 1v
0

- I

_Q__

s 1n45•

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

319.)

&

sin,+s•

P."

914.162 1b.

~-·

ders

~

o~le

porrno P~'Sc.Are ,. ._ 11onzon o 1 sur roce . "">

the smooth ,__ .

t

~fy~O

/\S'llUJ• = ~ san4S• /\ • 163. 1.!"9 lb

.£Fh = 0

sut>sJ. A

c •'UXJ ~16- t

/\C066(>.

c = ~23-07 lb.

J 33

•

'

't1 1

32

=o

H t "I"()() Sth $- = soo H ~ S00 - 400 &1n60• · H = 45.:1.6 ~ 4-s4 /b .

c

~fv •O

~ 200

N bei th ·

~

·

e cylr'1 -

ler of /\

incl 1'ned

~ a

hinge

.at 13: The

rner:n ber . Oetcrrrline

jc;; ,Gupported by

a rol-

given loads ore riormo l to tne the reoctlonG of /I~ B. 1(1nf : Re-

~

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

32.!l)

.aita:) The Fink' truss shown in f ig . P-32£

at

/I.

~

Q

a.

A~

D

pla~ ihe ioodG by thei_r re~ultont. fOOOlb

• in-.o'

~M" • 30

0

t ~000(101n3dX
Re • 65900. 76 30

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

o

Re• £000 (cos30' )(1s)

~Me •O

2000 lb

Re"' 2199. 36 lb ~ 2200 lb

0

~fh. 0

Rev .t R>.

Rev •

=

R.... h

BODO sin60•

BIXl061fl60 - "t618.8

eos .ao•

= 2000

f(Ah - 1732.05 lb

Rev • ~.309 ..+ lb ~Fv

4fn•O Reh· a00o

R>.v eo&60'

I

•O

= Re -

" (~309,+):z

t

R,._ ·• R.Ah t JV.v Q. R-- ·./ (1 73£.os) 4 t ( 1199, 35)2 R"' = 2106 lb

(40oo')2

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

1

~

Taf'l& • R,..,v t an-
-e- '"'

~309. 4 4-000

ton- 1

I

ao•

RAv = 1199.3e /b 4

Reh .. 400o lb . Re4 " Rev" +Reh"

'l.000 su1

R"'h

·-e- "

fa.,_,

4000

34

11a~'. os

11~.

35

1732. Os

Qa09A·

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

· 1199.35

-e- "' .R.....

c

~106 lb

34.70 •

down t o the lef'+

-& ".3+.7.

·35

at

.. I 1 •

11 ':

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) the force P

.32.+) A

3~.s.)' Determ1'ne

the amount ~ direct1'o n 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.?

·,r

..,. I. ii i 't"

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.

fi " +1.+1 •

.sin~ • s in 71"t1"

£OOO(a)-Px(b)-Py(o)•O

2000(1.a9) - Pws<><.(0.6t}-Ps1n<><(1.89)ro

,.'.'·

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

1"

s in&-=

(o.lrt) P(-sin.._) t o.6+

a/n ."

pz

o)

u.o

b. o.6tf1.

\-&- •tl:'11t "'

0 • 1.&9 fl.

-&~

71."/1•

~ 1.2.9 Pcos<>(. i 1.99 .9f. sin"< • o

(0,,6+.COG<>
.s1no<

p .:= 1732.:651' lb zf)I no p •Ro cos::10° " 1732.o S1

0 .6t ~

p i6 m'1n irnurn

• tan - 1.99 1

0°64o(

b.)

0

•

= lone<_ ~ ~

COS"(

Ra .. 2000\b.

= 11.29 ·

"'"' 71 . .s·

if i~ w i ll be ..L. to Ro

hence,

2000 s1n9o'

-e-" "o'

p s-,-n71-.!2-~-·

R .sin 19.71°

~Ma ~o

10 P min

= 1000(10 COs3o')

Prnin " 866 lb. ~Fi<

~a

=: O

cos 30° = Pmin COS60. Ro e [Prnin(c.os60·fl/cos 30' = [B6G(COS60°)]/cos 30' Ro=

.soo

lb ·

36 ' 1•

%

1,99 ¢cos°' "'0.61-y/s1 n<><,.

.sP " 1000(10(;0S30')

I,·11

r

.3790

~ (o.6tPs1n<>< - 1.99Poos0( ) _

dp

do<.

Ra zMo

tJf< co~"<

2

';

Co,;71.+t

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

-e-- 30•

I

Cos13" 1.;4 ~Mo "' O

l1.,11

,1;

p, •COG<><. p

FY

p"" 1994 0

Rsm9D

..

R ..

lb ot 71,3° with the horizonto l 12ooo(sin1s'-n•)

641.6 lo ~ 64'~ lb

37

I I I I

,I '

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.

326)

FBD of tho big cylinder,

Two weightless bors pinned together os s;;hown in Fi9 P329 support o lood of 3SO lb . Determine the force P ~ F octing respect ively olong bors AB~ AC tho! mointoins eq1iilibr;um

329.)

of p;n

}---•'••~~

~·

~

11,il, I I

I

fon O< •...@... ; o< • 39.66 • 10 ~Me·O

~Fv=O

I

Ro &111 ao• .. Re -4<>0t "400 Ginao' • Re

'400 t

FeD

of .small cylinder,

___L_

Re- 600lb . ~lb

Pc =4-00lb

"-> -&-

~Fh- 0

that defined the pos1t10n of equilibr1.um.

cor;.30•

--&- tO( I r 9().

Ro • 100cosao·

e- .. 9()-«..

a6+.41 lb.

-tool'cos<>< • :Joorcos&-

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 6

1oocosot: aoo cos(9o-O()

100CoG"\•2oo(ctJ~·ts1n~s1noc)

P61t1'-~·

,

c

100 OOG<>< • :'ZOOSl•~'tO &•nO\

~···.::f.M(;.•O

l&.(f) i;insa.1 - 100(6)

I

1,

t

~-

~(12)

'

-e-- 90-C>(.

1s(P)s1n6a·+ ... 100(9) - .300 (12) P = - 1-+7. 61.I lb (- meon6 compres-

- 90 -26•.s3 ',s..i. '

-6- • 6.:1°U'.!t,e" .

sion)

lo oheOk :

Fcoso sa.1 - Pco406l5..+ - :aoo aO S90•15 (cosse.1) - ( - 147.6.r)(c:.os6a..+) - -aoo ., o

38

~&•n«Jo0

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

~Me"'O

1801b

fOt"I"'( • ..:.:1;....00.;;__-'-

~0(

F"' 390.1s lb

tt ',

. 'I,:

©

~Mo•O

327.)

r41;1n...1·, 9 ·.

3&0(10)

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

200- P.c .s1nao" .:

Po·

-+Pco&,B •

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

2'.fv • O

Ro = Re

2.Ps1n~ •

2P(e1ns6.a1•).+ ~P(cosS
RA :; Re coc;ao• 'Rit. ., 400 COSl50" Rio. • 346.41 lb.

Ro

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

-

~Fh•O

f~lb

(+)fcosO<. t (2)fs1n<>< =.ssO(s) -+ f~os 39.66)t 2 f(sm.!!&.66).. :5!!10(8)

e'

39

_,.~32.)

Oeterrnine

:as? 'The roof fruGS in

the reoctions for the beom .shown in f1~. P-~2. ..t'.M~,

p

0

c

~(4)-100(1,..)(9) t R1 (10)-300(16) •

a hinge al

fig. P- ~ is supported by 0 roller ot B . find the values of:' the reactions .

A~

.

o

R1 = 1.seo lbs. ~fy:O

. R1 • R: " ~ t 100~-4-) t 400 R2 • .s20 l:Js.

or fhe

beom . in f ig. P- 333 looded with a concentroled· load of 1600 lb~ a lood varying of 400 lb per fl . from z:.ero

'133·) Determine

the reachol"\S

R, ""'-. R2

~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

t'

1

1

16Ra " 16 F.2 t +F1 t 1600(a)

Ra •

-o

Re

:.

Re

e

935.33 lbs

aoooJb .

•

2100

~M,...

1

10

+100(112)(6) Re". 1s600 lbs.

1cclb/rf .._,

lbs .

...,, 8

6'

R1

t

I<...

Rii ~ 1600 t goo t 1600

R,

,1600 i>'.

-o

2 Re • !looo(tz)

~FvcO

r, -fr-4<>0(12)=1/2)- soo Fe

1~'

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

F1 • eoo Jbs . . F1t Fa ~ !4oo(12 )]/2

r'

~M11-t-O

- ...ao(12) (4) • ·o 2

~f\,

a'

P.,

~MFc•o(t! 12 f1

·o

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. 33 6 )

= Rs - 12000 - 100( 12)

RA ~ 15600 - !2000 - 1200

= 1900 lbs.

R,..

e

12400 lbs .

The upper beorn in fig . P-337 is S1Jpported by at Rs \.., o roller o1 I\ whion separates the upper ' bt90":15. Dete.:niine _ the volueG o( the reactions . 337.)

334\.) Determine the

f i9ure

reoci ions for !he beam loaded o6 she.vi"\ in

P - 33"1'.

-£Mitt =0

'4
--r

.sss lbs.

R1 1 r<'a • 120

R1 - 930-sss

t 60(6) t

101<,..

-t'

1so(6l !l

-6C0(14) - 1900(+)

R"' ~ 160? lbs ~Fy=O

10'

R, - sn lbs .

40

ro'

~

£F.,.•0

R~

fONer

.:l!MRs'O

1sRa •12o(3)t16¢(6)(1a) -t 60(6)(6)

Rz -

reoctlo 11

0

Ri 600/b

190('.)fb 10'

f +'

600 t 19CO

l +'

Ra

r~

41

E

=R,., t Rs

9oolbs .

"0

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

...ooolb 4'

l

R2 ... 3840 lb. %.MRiz •O . .10R t 1600(4) - -4000(6 ) . o 1 P.1. 1760 lb. .

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 m0x1rnum lood W that c.tln juGt be rc)1~eq by~ pu'll Popplied as shown. > &Mo-0

he two 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 .

wfa(O/~) •Wfa(d#) t P(0/12) ·

n

be from DP

w0/4 - w44 = Po/fl w/1' (o-d) .. Po/2 W-4PD !l(D-0)

P•960lb

l

11•-·

W •

..

lc~====Y===~c~l§.;~,~~lR=2~~====;1~

;HO.) for the

aR • 960 R • 3!ld-= R, • R:r =Ra

of pulley1;1. ~ho-Nn in rig. p - 3"f0.

Gy6 fem

the rotio of w

t1on

to p to .mointoil'l

t,

zfv • O Re "'R1tRe

• 9P Wjp :: 9

~Mit:s

"' o

p(i<) ....

Re (y) =R2(12)

6ft.

l I

Re (12)

!

P(i<) ,. &t0(1£)

960 )( '>'

~M~ · O

y" 320(1;1 6'40

rnc-

w 12-y

• 320 t320 Re ~ 6'4-0 lb

~uilibrium . Me,g\eot O'llle

W "'3P t 3P t 3P

Re

'(

determ'1 ne

~ the weights of the pulleys .

~eriri(j ihe Jower .beam.

y ..

!2PO (o-d)

;>(

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

&tt)

""en.

find P to rnaintoin the equilibrium .

1. If eooh pulley 6~n in Fi9 . P-340 .....eighs 36 lb ~ w 3W1

. . R 2 ~Ra ·, s 5f1 ~

p ·,s

opari ·

oleo

en.

f'rom

1

3P - a6 t W.,

36 t 7!10

•

W

•

252Jb

.

· p.,,(~6H252.)/a

p. 961b.

o.

I

I I

I 43

42

• 12olb,

I

" 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 3-16~

3.,.11) The wheel Joods

mine t he distonce

twice

o'

'JI

Ot"\

o jeep ore_given in Fia· P-a ....!2 . Deter -

so that the reaction of the beam a1 A ·,s

.9reot o.s \he reootion ol

r1zonta l , ~ verhco I oornponenls Ieot th e size of the pulley ol D.

i3.

or the

reoc t.ion

0

I

Neg:.

/\

.

RA·2Re (speo1f1ed condition) £MA •O

ton-e- • BA

T

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

-- ·
+ aoo

1s Re • 900,,.

e- -

3Re

2oo(d.) c

= 600 t ~oo

aoo :.

T

---©

Re= 800/.a

in 1, . 1s (e
a'

of

0

t ro"elii;g or?ne

'1c;

20

Rh

s~n r'~ht wh-

t ons aof1ng as

or

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 .

~Fv

.z~1 cQ

(

..,.,he,n

~(><) = '20(6) -6{1" ~ 120Tons

~ 12e.02

ibs. ·

• O

Rv " Rv -=

eJ0.06 lbs.

1!20 t

~()(t5)

in !2,

s1nao• .

)(/+

ton&' • 6/+eo&.ao•

)( • !l'

fr " 60°

i R1 • o P =O

-· - CD

.s6? • !2'20

~ = 20

=

20(1) t 20(10)

~x t~~') "" !l!ZO Tons--- ©

~fh"O ~M,.,, 0 0

Rh • Tcos-&

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

./ans

Rh • 216.s co.soc• Rh .. 100 . 2s lb .

T ~ !Z.16 . .so Iba. ~fvso

From eq. <1),

Rv + TGm~ = ~00+ 1 00

' Qx = 1!20 x = 110/Qo

= 6

44

n.

.

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 t J. ISI inc lined ot ::io•· a vvve ...-. +he . . hor' :zontal . The . Ioo d s · remain ver 1co .

~Mi:t2=0 (whe-nP-20) wbsF1 \ule .

=1too t 100

3 00 - (a7 9.01Z)( .sin 6a.4:::i")

i R2=0

when pc20

T C06'63 .43°

Rv t Tein 6a.+a·

Lirni Hng Conef1+ions, P=O

lbs.

• (1219.!!>£)(cos63.4'!3•)

ii' f ig · P-340 . To pt"event the cran e f('()m fippin~ to the en .corryi\19 0 lood p '20 tons' Cl covn·ter w-'1.,g ht ~ is used . Det.

when

100.(6) - T (.sm 63.+3 · ) (+)

Rh •

100/ti

. )( :: +fl.

w

1

= 279.afl.

~Fh·O

.eu~tltute '2

s.+3) The weigh+

IZ

~M" •O

0'

~r,.cO

RA t Re

c

63.+.3"

Rv • 300- 216. s (sin 6 d) Rv = 1H!.SO lbs .

45

349.) · The

frame

s~

~Fh"O

in fig .

a. fooh m~bor,weighs

oc.f10!"\ oi

A. ~

P-3"t0

Rah= soo lbG.

·,s i;upported in pi>10/G at A 1,.,.

.so lb perf1· c.ompute theho~1zontol ,-e.at the hor'rzonlol ~ verfico l oomponerih• the

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

or

RA=

reoction ot B ·

~Fv

Leoglh o\

h

Fo

e

. ro:

.f8c t 6 ~ 1on.

Re .. /(.soo)11. t (1z1+.12e)"

i.z...a .61

I..

361.)

The

beam -sho wn in. ·

,i

B.,i

I

." . &

c

76". 12 •

715,12·

f igure ' p - ss1 .as suppo"ted by a hi~ of /\

~ a roller on a 1 to 2 slcpe ot /3. Oeterrn'1ne the re 11· t . at !-. ~ 8 . su on reochon;S

cO

11z'

&, • 600 t .SOO t 600 t a.oOO

]I!

., 12ao.79 lbs.

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

0

~fv

tan-e- .; 112M.aejaoo

lbs.

£fh - o Ah 13h =12-t66.67 lbs.

"

11200 + soQ

Rev "' 1121+.~e lbs.

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

Ah ..

soo(120)- soo(l2o)

ibs.

•O

Rev t R.-.. • ooo t

~Me, " 0

Ah(12)

1oss. 71

=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.

16(Rev) • "t00(1z)

Rev::

6001b ~fh

~fheO

Reh - 2-t0 lbs .

.:EMe." O 2"'!-f
2-tRev • 600(1£) t R1-0(16) - -400(12)

Rev" 260 lbt;.

•1!5o lbs. 0

:EMe • O

1fJO. 21 l bs.

J6(~v)"' 400(4)

RA .. 7-+0 lbs. ~JAt. • O

300lb6.

.. O

R-.h·R~n

~(36)

By Ratio ~ Propor-tion , Reh • _1_ Reh· 300(1/2) 2 P.av "' 160 JbG . Ra -..f1so)::i t (soo)2 .. 335.41-1 1bs.

~M..,•O

R....v " :asz) '

!'I

100 lbs. pu IIey -trn ~ . 1.n d.1ameter ~ supporting a load

or

200 lb ie mo..inted 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

rr11ne the reac hons ot A

~ C.

the weight

or the beom, doter·

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

l I

I I

p - 3SQ .

.I

I 47 46

,. 1· ~\

f,

or

F00

F80 of the pulley,-

.w. /\

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

the beom,

P~· RJ8(1t6~/5) ..

lRe·:,~

+'

.

c

'11000[1 t 6(0..SG)l • 1780 lbs. 1g1B ·J

· 3S"f'.) Campl.lte the foto l reactions

Pl~v

fRc

in Fig.

ot fl i.., 8 on the truss shown

P-~.

~M,+.-0

Rc(Q) " 100(+)

A!Ma•O

T(a) .• !200(2) . T = ~(){)Ins. .

Re" go lbs. ~fv"O

!<,Av

~fv•O

R.e • JZoo-Tsin3o·

' Ii'

a

·Ra·

gO()- QOO(Ginao•)

Re

t

Ji..v

a

a

20 1

100 lbs.

100 - SO

~fh cO : R.Ah • I cos BO•

RM •

100 Ibo.

!200

·ton.a-' 1%0 . "

cos:ea·

R>. •

(so) 2 t(17.3.W)~

ton¢

BO Rev

P.ev

16.10 •

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

~Fh - O

. right. :ss!!.) The forces oct"1n9 on o

1-n \enght or o

t

"' 190.27 lb6.

= .50ji7:a.~o "'

dam ore shown

Y1 ) -e- = ~6.56°

i

ac2+o(s1nu.si>)(10) 1200(10) = ::z::zi) 1 :a()()0(6o) • eooa(4<>) i 1000 (~) ~ 46:27.49

lbs.

RAh "' 20CO t ~!240 sin 26. e;cS

1

..

3001 .18 lb&-.

BO RA" = Q000(10) t 1000 (6o)t12ooo(~)t aooo(21j)

so the hor1z.onla)' resistance to .slid 1.n g. ~o s lidir19 ,

+2240 (co526 ,66)(21>)

· t 2::z40(s1n ::Z6•.s6)(10) RAV .. 3376.09 lbs .

R,., • ./(?oo1 . 7s)a. t(:a:a76.09)2 tan~

.. -t.517.57

tbS.

" 3376. 09

%fh • O

3001 .75

f.: 10000 - 60CX) cos 30• f = ...,_go3.9S lbS. ~ 4900 lbs. p.10.000 1b

-t%: tC1n ~i

,0

R "' 606o .srn :.o• t

,a"'l{)OO

3376, 09

I

3001. 75

~ = ~F"

0

Z.Me-.o

if\

fig . P-353. The upword ground reaction varies unif'ormly from on inte;1siiy of p lb/rt . ot /\ to p2 lb/11 ol 8 . Delerrnin~ P1,.,,p2
Resistance

::!
.::fM""O

• 11:a .20 lbs.

1:1

!looolb

10001b

SO lbG .

<>

+e.3s

0

• • R.-_ "' -+.517.57 lbs up to the ~i9 h~ at

-e;, .,,. 4e. 36 •

:I l ; I

: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 -

- 27000 lbl;. ~t.'1a=O

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

x .. e.++ e., 9 -

n.

9 .++ = 0,56

lor to A E ""-..BE

oi their rPidpoinls .

n. 49

48

I

357.) The uniform rod in . grov'1ty ot (j. .

o

F.1 P

9· ·357 we 'g h D~ J • ' s +20 lb~ hos 'ds ceflt ,he tens10 · th ~moothermine r n in e coble ~ the reoc• sur1aces of /\CJ... 8

·f

i1onG ot the

W" ac:Jlb HQt'>45

"'\

•

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

9Nsln4a• t 6tlCOG+e'

i.--- - - --60'

ton & :

1~0

AC "BG

=

j {)- • 26!:16•

1!:;/cos~6. s6·

- - -;..o·-----! :o- BF .? 16.1vcos A D'" BF,. 19176

but, sin~· =co.& 45•

2T t 2520 • 16 . Ns1n.+&•

26•.s6

ft .

2'

R,.. c.os:ao· (60)

8

Ncos-+s· -1260

" B ti s1n-+s" - 1260

AC c BG : 16.77. fl . ~Ml'>cO

T•

.o!Fh"'O : T ~ H C05-"t5.

= 1200(~) t aocx:>(+1.2.S)+ +DOD (16.n)

©

subsi . eq :2. in 1, N cos+s· - 8 Ncos-t-s·

:. T • 2s+.s6 (cos-46·) T

= 190 lb

©

-1:260

- 7Nco.s."'te• • -1260

H .. 254.56 lbs. ~MAtiO

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

60 Rev

Rsv • 6 1.!\
=>1

6130 lbs.

~fh·O

Rah • c

'Rah

~6) ged

=

-tOOO s1n26.G6° - R" s1nso• 4000.s1n26·~6~ - .5361·27 s '1n:ao' 359.) ' fl

/A'

bor

/AE I

~'

~up~ded by 0 hin-

The canti lever I truss shOV'Jn in fig. p-3.56 i s ot /\ o strut 8.C . Determine the reoction ot 'A

~

1()()()1b

• . IS In

orces shown in f ig p

892 .o96. fos .

eCj u1·j'b ' um uncle ' r1

r

fh

~ B. .

or ihe ri:_,e

e ocf1on

De.term ine P., R '

- 359 '

T.

=£MA'"0

Tcoe.~(e)-1 T~1n~(s)

R (16) t

-400lb By Resolving the forces to its equ1vo -

t

lenl force triongle,

16R t 8Tcose t 6T.sine-

v

subci SCj. 2 in 1,

R

16(600 - Tcos&) teTcos&t6T . / . e 1n&

..&.ic;1n30'

T =1ffiS .71 lbs .

~ ~

s 1n90'

p_,._ "" 2000 lbs up 'lo ihe r '1 9 ht ol

a

From 6().

eq.2 ,

R ~ 600 - (128.s,71) cos .36.97•

R • 4~8.57 lbs . (upward) Re ... .346+.1 lbs.

51

50

. 6000

16~600 - TqOG.36.S7 •) t 81 COS.36·07" t 6Ts I n

By s ine low,

R = 600

~ soo - Tcose-

,36. 87 •

"' 6000

600(4)

6000 .

~Fv'"O TcoG~ t

'o.

400(9)

@

©

351.)

Rer~·ng

angle

-e- at

· i

.

·

u is 1n a

..£fh-O

1~'1.

r-rvv. ~ 399 ~ It

·

.po6•t;

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

/ 3tW.) A

t0

. T ~ 30:1lb · 11 be . . ,,.. )( -

which the bar

of

00

w• . .b . 1nol 1ned equ1 1 . num.

1

/

( 't2(00&&tsone-)

O

1 f ion60

.,

3

n • d~tennioe ~l-.e honzon~ol wroc-u ._ __

-to fL• oe> -

t 1. Ion

/

.

60!,_CD60-t s 10&)\ 1 t for)6l:> 'J

bar of negtigibte weighf n?GlG in o hon:z:onto \ pa-

sitian 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

F~p--a59,

~ h()ri%Ql'liol . C

p-360 ...361

By Sine low,

~= ___&!-- ,, .s.nao·

904:s-

100x t-f1J0(9) • R,..ro$:a0•(t2) 100)( ==

Re

219..61(~"')(1~)- 1800

"'155-!Z
x .. -+.92f1 :a60)

300 .son'I06 •

Re ferring ;to Pr-ob· 359 , who\ volue . of I

frorn J3 w i ll

~. the bor nor-tzof"liol ? s.O'IS·

-~

~c

octi~ ot J< ~ 3 0.

~

..Y = 12(C06& +s in&) -x - ·· eq,~. ~ 11~CDS&t&tn&) - .3~1

J

-

1 •ton60 33 COSi}

NJ -

+..39'

60069-

~[
ton60

1 • ton60

~ 60(~ 6 intt). 1-t ton60

I

£.tf1c "' O l ( -4. 61) : ~00(1.39) = 60 . .3 tbS .

53 52

-f

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

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

T

-' : nfoo w (cos~t s ine-) 1 t fon60

ffiC06&-•~

- s.7sn . e,.7S

from sq. It'\ ,

LMo=O·

.

cos «> .. ~ .

12(cos EHG1o
...Y - 126.n& =· - JJ11""'° · ,, ) lx - 1..:GOS&

12 .51(l 7 b

)l :

~}.

lbs -

r 1 +-fan60

;:.G

_y = ~

foe- RA. ~ -o = ton(6d>(x-o) foi-Re.

of ."' .-l •.o

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

~ .. )l fan60- . t'.D RA, .

JV. - ~19.61 ft>G.

~-o

~ the coordinoles

l151°9 Analytic 6eomei ry. ..9:91 - m(x-x,)

]

.

I

-4<:1~.) Joint B of lhe fruss shown 1n fig. P- -+0.1? ·1s euQjected to the fOrces e)lerted by t he ~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

Chapter 4

II'·J

..'A nalysis

of

J

.

_______.:

~C

.•

11

I!

6

I

the flXCe in the fh1'rd z.er-o.

mernb~r muGi' be

K

p

'fQs) Determine fhe forces in ecich bc.ir of the truss shown in Fi9 p:. -403 . Hirif: ffr-st ' determi'n e wh·1ch bqrs c.orry no load us'1n9 . pr 1noiple devebpecl ·11"\ Prob. "402. P ce -cF-o

Structures

£fvc0

P"' co S1n6<:>· t OEs1rnso· G1hce : CO"' DE·

p CO

+

,.11::

c

QCO s 1n60°

00

o .e,77p =OE

-

C

co =ao "'o.!!!J77P

F

I l

. ~Fh~O

BF =- BDO?S 60•

J3F

~ 0.577 P

f3f ...

cos 60 •

0.577 p (0.5)

BF" O. !l.99P

- T

·I \

I .i

'I······ I';::

.

~

·1 54

55

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

@pt. 0 40+) Oeterrroine

the f()l"C96

o. seD+o.~ec,. Qooo - · · -

showr> in fig . P-~-

the- u>hole figure, R,..v + Rov = 100+ 1oosinao•

o;ni1der'1ng £fv =O :

~Fh =o :

~fh··o: Re»1 -1ooca;ao P.oH - S6,6 lb.

0

0°96"6

~C"'866lb---· C

~... --o

RAv " l\BS tn ao• RAv •o.sAe-@ AC " AB

100lb

'1 f

o.seo + OJ5M .._ 100+~ eo+Ae= 1.!JO(~) .. :Mo-© "'fh•O: SOcos-eo• :Aeco&ao• t 1oooos30•

eo ~ "e t 190--©

.. eo - .Ae +100.=

1200 it>

b

@)

- c

eoch member .

+10oo

..

Q0.20 lb

0

·1 s

__·- T

o.e66

-

ci=.

407.) In r. .

the con1ile'<'er truss shown in .Fig · P-...,...,.., ""',

conipute

~

@

pl·"

.

~Fv•O

10001 b

AB1>1nao •1ooos1n 60

The con ti \ever trvss in fig . P- 406 in

966 (o.B66)

1orce 1n memben:; AB, 1;3e, ~OE.

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

96.6 lb· ..-T

@

pt. s.

0

0C = i00 1b·· · - T

force

co•

CE"" 2020(0.<:>) +1732t 8$. (o.!J) = 3 17.?l lb -· . - C

~,AO• o.B66(A0)

~fy

the

CDs1n6'0°• ec81060• t 1000

COcos60· t A.C t BC ()Ot;(it).

. !
:li!fh"'O AC ·CD

"406.)

.

~fh•O

"'0·5AB

@:Cr o:

-c

I!

...c

.,o.e~,C1 o0) "'96.61b-T

~ .s in.+

o. 2!3 eo + o~ ec - 1000 ao - o. +.s~ ., 1.soo

0.7!!J

BD • 2800 lb ~- -- T

A0 = 1o0 lb

RAv

+

CD

cosao·

f':.C = o.966 A0 -

/

. @ pi.C

~fh - 0

100lb

hi~

U61~

<::it D ""--E. Find

Rototio" of A")(•S,

Uv-o .

SE

"'.

@ pt./\

..a.e

ao 1000 lb

£fv•O

.£P..,•0 ASs1nao"

=100 :

Ae-~lb-· · T 0

OE. '"'" 6'0° • 100Q

PE• 116+1b -· -·

lll!rn · O: AC. AP>COS"° AC~ 200ocOS.90. /\C =173Q lb -- .. C ·

I

1cxx>lb

57 56

- --·@

6y Elirnino1ion : 1 ~ r; + (o.s.eo t o .aGGec = ~ooo)'o.s ( 0· 9~ 00 - o.!!)sc ~ 173Q) o .96'6

if' BO• ~SOO lb in 1 BC -eooo- o.e(2000)

@pt./\.

@pt- a.

(!)

BOcosao· w J:>Coos6"0" t ABoos:ao• o.e66BD•o."e l3C t ~000(0,966')

P.Av t RtN • 1!30-...-6)

.

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

= 1000+ 200?(0.~)

o.eBDto.B6'66C

the rriemberG d(' ths l'OOf trusG

1n

c

i "T

@

sho'l'ln in rig . p -~e.

OE:61n60· "'CDs1nro· t 4!000

PE "

20QO(O.B66) t 3000

oe - .5:+e+.2

"" ..!34eo 1b

c

co/ I \oe

~Mi!!•O

.Determine . . fhti iorce 1 l'. · n members /\0, Ac, 130,CD, ~CE of' the conh lever truss .sh
+

411.)

::iOOO(!j)

p_,._v ..

£Fi,,•O

~

coMidering the whole' f1qu"e:

R....v(ro)., 2000(19) t -1
o.

~~

...ae.1 Col'l:'pule the force ir'I eoch member of the Wo"ren tru(;a '3()00 1b.

pl.

or

4-~50 lb.

3o

e

pt .

@

-e-. 33.69. izooolb

@

pl . /\ ,

~~e

~ £fv•O IV.v " ABs1n 6C)

'

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

'

s1noo·

l\6c 4250

o.966

/\£> "' 4007.62."" 4910 lb-C

~h..O :

ec 0

4910 (cos6d)

,AC,- :245.S lb - .. -T

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

SlZ:

~·

eC 6 1r'l60• t

£fv ==0!

C051r'l60°

-e-

"'"\000

2600(0.066) t co (o.066)

C O"' ~01~ .94-

<;:1

=400o

~o201b---T

BC c oo' 2600(0.9)

t

Ct= ..

t

/\C -= CAJ CX:foq:l

Q.+55 "'

t

CE

'20~0 (o.~) t CE

,93.112·

~= ~ · Oe" f3~g· '.20 -'CJ I • ~i=-..·o

C0 61nsa.1!2· - BC

::S:fh •O CJ'O c C O COG .s~. 12

1300 t ~45E> - 1010

Ct= .. !ZH~ lb. - · .- T

(

10

c

I

co- a.90 lb - .. - T

~fti co:

..T

ionO • fa.93

/

/ pt. G

AC- 180lb

Aacosaa.~· "' aOC06118-6'•

BD = A8 C/JS60° t l?>C ocs6o

BO ;.. 375.5 lb - .. - C

/\e"' 1so.2s lb ·

~rh ==o

fl

l?>D • 4910(0.5) t :2500 (0.!'>)

A0cos60· "' AC

AC-a Aacosaa.89

00

- ~ ~Fv ·o ~ · 19·'"' ec • !200 lb

BC "' ~~ lb - .. - 'T

/

:tfh .. 0 /\C "

@ pt. 8,

t

AC

,,. 200 cos s a .12 t 1.so C~

" .300 lb~..C

/

.'

I 59 58

~I

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

409.) Determine ,

"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

an per sec . Who~ .change in ih66B fo;ross, ·jf any. resulh; fr()IO

p\ocing the roller Guppod ot D

~\he hinge support ot

0

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

A?

8

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

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

Hv = 9.50 lb.

~~.

at I\,

:%fy•O, ~Fv=O

Av+ Ov

but

=12.0t ~/s (120)

Av = Ov

•• 20v = 12.l>t

a/s(120)

lnterchonqinq hin¥ ~ roller support will rot chonqe \he forces in each bar ell°'Pt for /\C 1*. CO

2 Ov .. 192

Ov • 96 lb.

,..f>~

ot A :

=·gc;;

t

9/10 (AS)

= 22:4 lb (teos1on) · ,

~~-£96

.BE - ~oo lb(c;vmprcssion)

:. AB• %100 lb (~ion)

/'C • 1820 ib ( tooGion) BO• BEQJG~· t f,()()(et:>s
AC

=Av

co c so(e/10) ·

-4' )

BO"' 1500 lb (eomp~iori)

.

10- Determine 1he forc.e in eoc;h member ci' the PrcrH fI in fig. p.. . roo rvss eho-Nn

410

= 126 lb(icnGi<>O)

o+ 0,~fl'

lb

.,.

,.,c

ec;

~

~-AC ,. Ae 6

AC • 12S

1900(6) = Av (32) Av = 2700 lb .

••• CD

AC

CE. BD • 160 lb

Av=~~

(c;ornpres1>1on)

3

eo(e,fo) - Oh

- -·-C =- FH

£Fy=O /"w +O""' BG

BC • 192. 1b(tension)

61

60

c

o/"13 CO t

= :a600 -

~

O&

Cl=.

Y~ ( 216::1.33)

2400 DE " O

AC= GH ~ 3600 lb-·-T

: • co - 32 lb ( ~c:ns1on)

2163.'33 tb -··-T

00. : t=.6

AC

---+""

• • /\8 • +soo lb

bvt. Oh· 96 lb

c:

s

3

~Fi< = O

M A. ~

at

0C • 1s001bC = FG

• •

%l1/b4•0, 1800(H)t1!IO'l(-t4i)t

o\ 0 :

=


in rne Of'1g1nol o>
lt:>. (tens1~)

A0 " 160 lb..( CO("lprcss1on)

co

.

U6iog Hew Axos

Z:Fy • O,

AB-BO ~ Of' "' +soo lb - ··-C

""10

-ll-

1;00

BE s1n60• - ~(sinoo") •O

c

lb - ··- T

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 down -

@ pt . E,

41f.)

+

word 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?

~r.-

4001b

@ pt.

DF

~fv"O

~rv

_., - T

SQ)lb

I

I

~

-o .

DE = !100 Jb

f

OF

- ~or =

/·

..Je(7!2o)

OF • 1609.97~1610lb- .. -c. .j.ta.) Determine the iorce r ·in each member of Ghown P-+13.

the crone

coi;1ne LOW:

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

@' pt.

(l'.)nsidering the whole figure •

:a;:f... ·o R>.v t ~FV

R.-.v t

'IOOlb

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

P,pv " 1600 lb

+ eoocao) eeo lb i;ubst the volue o f ~ ..."

b ...

..:-©

·" / R.-.v (oo) '"100(-40) t ..+C(a:x>) t

@pl. A,

~ad)

!Ct=h=O

~80 "'~ /\B

G

1,.

in

BO " 98% ~ -49~. 5

~ 1600-BSO .. 720 Ii:>

"""'1'9~ 1b-.:1C

(

@pl~~··:.:::..

.....!..-,.. W +~lie 'W '<5

~ (-t9Q)

R>.v cA8.sin6a.4g· "6 ~ .,993. 9 1 '<:: 99$ lb- ..

c

{s--{9f3s) - 100 t ec lb - .

-:;.I

-.C ~fy "'O

J:>C "{e co +czoo

A0 COS68.4S •

996

700 = ...L cot

COGGa.+s·

IC .. 440.ee ~

440lb -· ·-

"9

T/

~b

czoo - coe 1118.0?>

/

~ 11120 lb-..- C

. o.eeo • 0.966SC -"@ .

12000 "o.s66(o.:!,6 ~ 13 100lb -·'-T

/'

1n

=A8

@k

/\8 • BOcosao' + BC cos 60•

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

62

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

o.s

~Fh•O t o.~ ec

ac) + o.sec

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

AB"' 12000 lb- .. -CI'

1aa:io c o .966 eo

•

EUbsri . 2 1'11 ,

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

BO

~fv aQ

eoGinac) • ecs1r-i60•

J..Cs•n23.a9· .. s200

/\C cos 2a.a9•

1il -08,

6"1•(1a.oa)Lf92 - a(13.oS)(9)cos/\

Ar. ~3 .39··

c.ont. ()t'l 8,

£.fv=o

~h-9

@ pt . c,

~fh-0 c

t

In 1 ,

/\C"' 13()99.64

= 400 tac

SC - 701 .o+ -=::: 700

zfv•O

/\C

2

e,

" ..,.00 t QOO t 2()()t 800

~Mr"o

Rrv

6 = b 2 +9 2 -2b(9)cosA-© ba."' o-.i +ell.- :zoccosa

63

~ 104ootb-!.'.'.c

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

the force in mernbers fH, Of".~ DG truss shown in fig . P-41S.

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

+16.) Solve for

or the

0

p

conGidering the whole, ~Fv "'o R.iw t R11v "' 300

t

~ t 900

R>.v + Rtiv .. 1500 £.M,., "'0

@)

~v

900 lb

Av(3G)"' 300(2.1)

slope ot A'e • 1 slope at BO .. 1/ 3 slope a t CD " "4./3 at A ,

t

300(1e") t 900(9)

R>Jv(:a6) = aoo(9) i

~ rR:-

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

~fv•:

FHV = FH s1n4s•

Av= 600 lb .

fH "' 900/sin-4!1°

FH

,

Y-li A6: GOO

AG

- ·.-r C

@ f,

~FH

~ ~ •o .

Ya

FH

.

Ae" 048.~3 lb

=1127l2.79 'I: 1270lb -·· -C

@ pt. i=,

2Fy•O

= · ::i/,.io

~vco.

Y-ta FH -

DF

OF " 946.'6 "" 947 lb-· ·- C

300 t 4/!EI

i:.Fx =o

(@ pl. 6 ,

141

AB I BO = 632.40 lb -··-C

*

.;i; Fy

~o

/\B ~

BC ~ 400 lb

- ··-T ·

.J

~·

~rv • o

.

1/"10 BD t BC

co = ec · lb -··-C

CD= 125

+foDG t F6 .. 900

../.e 06

)

t 6CO ., 900

1::?0 "'::ioo ( s) 't 06 = 37.!' 1b - ·· ~r

64

Y..rro

OF

J4w(941)

f6r:: .59S . .B6 ~ 600lb

('3)..W

~fy "' O

f6 t

Y-12(1210) • t=G .t

OF• -1Wfr210)

ot C,

3/..fto BO "

:aoo6e) t

Rttv .. 900 tb.

65

900(21

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.

411'.'>

. 1@.lb

£MJ!·O

1f..; oO

or(ao) - RGV(2o)

o/.Ji BC .. -+00

ac • + 47. 21 ""4+S 1b --·-C

1lz'

Of" BOolb -· ·-C LMoao

360lb

CE(~o) t 1000(10)-t RGH (!lo}. RGv(3o) ce~ecd.:ao) - 1000(10)- 600(:20)

/'ll!l,

BO

.

!20

ce ..

i

·.!c E

to determine .+1Q.) Use the me thocl orr- secfion6' . ao,co,. ~CE of the Worren Trow.

,, .

1001b -··-T

~t.'lc•O

FV,v (12)

go "'

GOOOlb

=BO ( 9)

com:idering ihe whole f'igur-e ,

120(1&

~M..,•o

9

~o .. 160 10 -· .-c ~i:v .

Rv•(ao) •'1
.-

!2000(1!1)

0

J:l.vE

w

•t-7!50 lb .

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

co ..

b/?J (HlO)

I

c e e-e.66 n.

, I

sin60•(•)

B ~- -··· -""B;:;...0_--1

CO"' !ZOO lb- ··- C

'

...:;Mo ' O

P-Av ( !2.4) GE

::

er; (9)

= 11ZO(~~

C.

- 9

CE ~ 31ZOlb - · ·-T

41EJ-) The warren . iruss looded os show" in by

a roller ol C

pu-\e \he

force

CE 10'

'f
~

a ri1n9e a* 6.

in \he

~ember-G

Sy

F•9· P~+1a is

suPPof"'-!ed

the rn&thod of seolions,oom-

BCr OF,

~Cl'- ·

BO "

cons·1derin9 the ~ho\& flgure., .:tfh:O : ~0t1

-

%Mc=0

t;,00('10) t 900(10) t

~Me.•0

~Mc ·O

Bo(e.66) =-3ooo(s)t •nso(1o) 6P • .32soo/e.65

600 lb .

100o(~O) .c -406(~+

.37.stZ .9

CO • IZ1~S l b ....:. . · -

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

co ..

c

.

co " 1Zo20 . ~ lb:::: ~o:zo tb -··-T

RG" - 900 lb . 100Dlb

67

66

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

27+:2.!5 lb - ·· -T

JtGv (-40) -'fOOlb

l'

::!!,Mi; aO

=+750(5)

Ci: "' 257.SO/ei.66 Cc

3000(10)~"!000(6)

co· _a12s,9"'1o

lb- ·· -C

~Mo=-0

( 9 -66)CI:

Rvt=(1s)- co(1o)a

using Method of Secfl()r"I • @

CtfECKEO:

.+20:) l)etennm 1he fOCce. in th& mernberG Of' 06. ,.,..1:6 of U\8 ttowe in>S:~ GhoNO in fig. P-.ov.IO·

::E.Ftt'"C

3/.sao=ae ; BO.· ~a(1roo)

!f.l•!ltCJOI>

BO~

£MG•O: Of(9) • 11.100<12) Pf =!l900 \\:> - · · -C

!2000 lb .

.2:f.... •0 ~ 12£>0 t OIAJ 06 • a100 D6•16001b - · · -C ~f'h·O ~ Df t -4{900•1:6

:zsOo t

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

-f'/:5 BF t

f:6 "' .+oOOlb - --T

~ 'in Fig. P -
<4!M-)

rnettioO

'(()loe .of

ae

by incopection.

cons"ic:Jen°ng the whole figure.

1~ • l:F

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

for 1he truGS

by the

EF .. 3!lOOlb

-=N-0

-4/9(1.500~ .. ~6

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-

..e'11•0

Rott .. 1!/.00 lb ~F,,.=0

RJ.,v t Rav

=~.it<)0112J:X>

RAv t Rov "3600

~I

- ·©

~o·O

-Wt>t>

~v (19)

= 1,,,.00(9)

I

+ 1!100 (12)

~v-2000lb

Method

@l pt.

of Joints .:

A~ . , /'13 "

1'f,,~O

"

N:-

~v

in 1, Rev

c

£FV•O

3600 - 2000 =: 1600 lb .

itFA.v·-Jf8"

BE. •1200 lb

- 0C = 12-400 lb . Z:ft1 -o

Rotl ~ a/9 eo ao • s~c1200)

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

ao ... woot> -

/\B=[s(~+

- g~

£ ..~ 10

x

x c ..s.&9' ..liiiO .. y 00~ Y"'1t.S'

~~

.. o

~-S9)t2(2.s)t 1(12.s) .. ,.6 "(n.10) OF •.s.9139 :w.s.92 ~pG - · · ,SVa(): 4A;oe t '1- • weoF 't !2 i 1

OE .. fl

69

tt.ipc; -· . - T

I

I

Ii

...,..... - fc;¥ .... tc.

J\B= !ZSOO lb

68

-1t2t~Hl+1

c

I ,I

L,.:;ht •O

R/\v (ao) - EK(i7.a2) + 1oo(a<>) t '.ZOO(ao-1•.s) + !200 ( :ao- ~) t ~(aa-2'.1 !1) E:K,. 6912.94-

fij.i•O CE t ::l/s PE ...

-i/ra· PF

431.) Def-ermine

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

+

·m~tnod

f~

Kip~ -· · - ·T

+26.) Show !hot the

the

~ 693 lb-· ·-T

.

,...__~ in the rnem berco

T
tru1:is shown i'n Fag p-+31

or' 06 I '!,., E? fOI" the Po...i..-r

.

' ""'

of joinls connot determ"ine the 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.

If

?£)01b

e panels

R.i.v

~fv -o, conGider the whole f .gure, ~1nce : p.1 • R>.v ·

· 1

.

2R1 • 7(3C>)

R1

9

I

ot l!s' • 2o0'

'

R/\v =1os " 0

ton~·· o/ao a"" 11.:a:2'

cos -ao' = 3o/c

c - :3-4.64'/

ton60·

EG=

17.a2

c

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

-'-t)

u= 10'

1::6 - 110.7 -:::! 1"10 1<.·a ~ -·-T

..C:MG • O

· RAv(1s):-ac(ao)tao(~)t 125 DF(a.e)

coosidering the wrde r.·gure, EK

R.-,v • l
2 FV.v

c

1600

i:t-.v = eoolb "'Rov

R.>.v

30 • Ae t b ·, b•10'

TI "'° "

""&

10

FH

f\ E

oe. ='20(.s1noo·)

=

1100 lb -· ·-

" ,co l"l.fl\

,

~I I

c

R>.v • 30taot $-OF+~ OG D6•

~,,,£11 fos - :ao -ao- ~(wi5]

[)6c 3.12.61

<:::..92.7

f
-~

c

H

I!

In Flg._-a sanao· =DE

~Mi:~o

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

. AE • 30-10 • 20'

·

.£.~·o .

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

In fig :2,

~

t~ or(~) OF• 161 , 87 ~ 162 Kips - .. -

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

~80

to c.ompuf the c . e . P-+a2 rorce -shown .in fig . rn members

"faa.) Use the method of sec t'torw 1£, ;..o, BC, of tn t e

ru GS

GH. '1.00 ( 7 •!!>)\ t ~fr;) t 200 (22.~)

Ij

17.:aa

Gt-I • .519-61'

""~!20 lb

-· · - T

I 71

. II

70

I

,. ,.

~Mc-o

R.-.v(110)t•~M(16)= "'8 =

¥

o/..JS9

A8(20)

[ 30(~)]

[s(0o)-e(1t0]

·

AB~ 70.7S ""' 70.9 KipG

::!EMB •O RAv (l&J_ •

f /;.,·,.\ ~M ..

or

~·

•o

ii...v(s11)• ~400(:M) +a~16)

_I

zfv•O o/t
ao= %[%97(!S9~-~J

.:i!Mt;.=O 30(a<>)'" 1!Z(2o)t

OF .. ~.+s

BG - ~ AJ!J + ""f~ BO BC- ~ (s91 o) t +fe (.sooo)

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

Aa • 5909.31 c:::,.!591011;>-· ·-C

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

·

~Me"O

R->.v(e) t a,.S A0(9)

L~ .AO(Hl)

/\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. ~M1·0

K

RAv (60) •

,

.

1~(sD)tn(..o) t 111(30)+1~(.u>)

+ 1!l(10) JV.v ..

, ~v

J

aoK

~:.

'48'

"

P.Jv "

12

-.0

LJ 10

~

~4 "

5

~:

A

...e Kips; - · · - i

t!2(10)t~ OF(a·2 -~-f)

+ ~ 0~(10)

~ni-o

Af'Mo•o RAv(16) t

AC~

/\C = #}4· ( ao)

BD "'5o01.o+ ~ .!5000 1b-· ·- T ~

-+,.f+';

AC - 4€1-02 ;:,

RAv • 36001b

.i....:..... c

.

.5

73

-c

~ ""1l.5 ~ips- ..

i. 1b

.+21.) Use the method

members Of, EF, ~Eo Prob . +11

for:-e

of ~dions

to compute the in ~he .of lhe cantilever '\'russ described in

Z.M,...•o

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

'*'

page 92. ·

Fv = 720 lb . tan-& = '2q/.,0 "

o. 6

-& "'26.57 °

tori20.s1· " a/w

OE" 40/.9

o~ ~Mo

0

10 '

0

CE(10) = 720('20)

f,..·-nolb

. CE "

1~

lb -··-T

~Meo c~1...:in9

F)

BO ot

7'20(-1-0) "':z.00(20) t 800(20)+

y.fij" 80(i<>)

BO = 491.93 lb - ··-C ~M p ~o (Re.solV.ng CD ~tC) 200(20)

• %ME • 0

11..rs co(.ia) co'" 1118.03 lb ..,.....-c

soo('2o) ,

the force acting in rnember-s Of, EF. ~ EG of the tioYVe truss descr'1bed in fig .

4Z3.) Use the method of sections to deterr,..iine

to

(R~solving

t

OF ot

o)

3/..f13 Of (40/3) .. 100('20) t '200(10)

P-409

on poge 91.

OF, ,. 360.SG lb -··-T 0

£MA•

2

:£MA• O

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

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

{5"" 0

EF .. 335.41

lb. - · - T .:lf.Me.·" 0 (Resolving OF o+ H)

zMr •O

Hv(U>) ~ 4t00(10)

E6 (W) • 100(30) t 2.00('1.0) t 200(10) ''=""-

EG "'450 lb - ··- C %MF

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 I 11

h.-=::.L..::.,H G 10' -4«> tlv

~o

Hv (10) • E!G (10 tonao·)

OF

c

t

OF(s1n.30')(20)

1500 lb:-·· - C

~MH • 0 (l
ol

E)

400(to) - EF s •nao· ( 20) ~F " +oo lb -··-C

E& "' 1G4.S.4.5 lb-··-T -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 : 1o oppI'I th e mefhod o f 6edions

75 .

t

!

•*'

'

-427.) Determine t~ f~

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

-r,

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

1200lb

lb

81"

"'

·~::~

Cv "' 1600 lb.

J_

N;

~fy

1/$>

1e<>0 lb - ·-T

. '

•o, DE t 1~ • 1~ t £,()()

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

r&(; g

~n. .. o ~

otF,~~f Cf

ot c,

=0,

OE" 632.
AC .. 3/~ AB • i200~

~Mf

8'

~so

l\v

of the nooelle

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

"'" .. 4-/~ A0 AB.,. 2500 lb -··-C .q'

1t'

+

Av "' 2000 lb -··

~y•O

' ,; e ,.-+.

~lb

.i:Mo~o

A,

bor-G 60,CD, '1Ao.. Of

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

As o~\e.

ot

in

£Fy,.O,

,._C .. Cf "1500 lb-·· - T

4/5

1to0

CO= .5()()1b-'·- T

co = 34ts BO t Y..i;o DE BO= 360.55 lb-··-T

~F.,. •o, CF • 3/e> BF '· "

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 .

1~ (5/.a) ~ ef=

BF "' ~50011:>-· ·-C

By section, •.,

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 ,

..:tMA•v 2..ok)O(q) t 1200 (u.)

""""~~....,..,..--61

"' ~/5 BF ( 18)

BF"' 2~ · ., -··-C

Av('30) = 2(CO!Pu.e7•{1ot 2o)t (s•n2c;.s1•) (2)(.St10) +GI (15)

£~ ~o : 2.«i~(.30) • -2(sm26.1S1')(s)

. GI ..

0.-1-s.9 Ki~

-··-T

- 2(smU.57°)(10)t ~(610 2~.51)(30) + ~MA • O: fG (30) • 2(cos26..57)(10120)

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

2(C06 2t0.s7•)(20)t 2. (cosu.~)(10)

OF • 2.3 ~ip& -··-C

77 76

)

;

'.:i

440.) For tho fromo loodocl

429.) For t he contilever truss .showro in F1'g . P- 429, determine

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

us

shown in Fig. P- 440,,c:Wtor -

mire lho hor izo~tol t+._ vC?rlicol companonts of' t~ pin pros · svm ot '_ 13· S'poofy dirootions (up or down; left or right) tho forco OG ·• t od~ upon rnombor CO .

H

Len of A-A

c

~MG=O ,

. s/.JZQ Of(z4) ='J.oo(60)t 2.00(40) ~

3001b ~MA

2'

400(20)

%M1 •

.£MF ~o ( ot A-A)

.

2001b

[).... = .5SO lb ~Mo

a-a

o,

'JdJ(80) . FH =1664. 81 lb _ .. -T

o£Fi< c O: GI t7$i'FI =~ .F'°l'I

· Fl

Lofl

Av= 350 lb e.;, Bil

:f'.Me"O

c i

=~oa . 2.7 lb - ..-c

BH

or c-c

3001b

e

0

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,

:E'.Mo .. 0 811 ( 4) " 30(){.6)

BH" 450 1b

0v

~Fy"O

-t'

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

. 8t:.CE, 'ii... OE .

•0

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

.s/~ FH(..w) =-400('1.0) t '.'l-00(40) t 200(~) .,.

UGI = 400(20)t 100(40) t2oo(6o) . 61.., 1166.67 lb-.. - c

=o

CN(4) • 200(2.) t 300(6)

Of-> 1256 •.s+ lb - .. -1

Lef\ of

of

Ott 0.,'5.50lb '

A.H(4)c 350(4) M

t ~(Z-)

2'

AH .. "I-SO lb

""'

~f1< *'0

s.so -ev =o

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

8v "' .5SOlb ( do.vn)

£fy=O

8.., - UJ0 - 3SO = O Bv "S.SOlb

,

H

•I

,

.£Fv=O

A" tJv = 30 ( 7)

: . Av

Svl Av •Jv

=105 !'ipG

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

tho horizo11tol

~vertica l compone nts

of' the hingo force ot B, 1001b

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

27/Jiii'+ 8E(160)

t 30(135) =

Ati

110(10~)

BE = 63. 88 K'.ips - .. -T

i..en

CE "' 97. 22 Kipc;-.. - T

%:t.'\E "0 1o!'l (eio) " 30(2~)+

-'/$

§ 8';e.11001b A Av~ :0,,1.., ~o

ol ti-b

80w • 1oo(s)-+ ioo( 6)

~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

BH . 17S lb .:e:Fy u

BD(32J

.,ei=,. '"0; BH "AH"' 17.5 lb O i Av - 100 - 100 "'o

Av "' 200 lb CH " Bti • 17.5 lb

78

79

Isolating bar CE ...+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

:f;F.,. =o

Cv

Ev -Dv

t

C v = 360 - '2.64 Cv = 96 lb ( downwor-d wiih rospec! to.

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

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

~

le..

.s' (~loting

c:

G i

At-1 '" 120 lb

leO c.tic -4-0 lb { io wthe ifh rei:pect

BH ~ 40 lb to tho right

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~~~

The frame shown 1n F19 . P-+
god ot E i..., o rollor at D. Computo the col components Of the h ingo force ot C

Dv

o~

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

lev

~,

°" =

12' 1'2'\0lb

. .0

Ev"' 1so lb lc:oclating bar DB

Dv

lc;olohng bor AB 2..0

%Maco

B!-+.s

. Ev (0) = 240(s)

BO

Isolat ing bar BO

360\b .

""'

"

-+'

~"

l

c

~Ma•O

w/ ~pee! +o oo.- N:)

Bv " 120 lb ( upword

µ ..

·rev

D ~Fy"O

240(4)

Bv c 160 lb

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

2'

:EM.-. .. o Bv(6)

DH

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

Av=aolb

=

Cv 16o-go

Cv"' 70 lb

ZMe "' O CH (3)

c

70(.3) t 90(7)

CH"' 260 lb.

81

80

= 9o lb

:::E:.Mo c.O

E:v >= 264 1b

0

hiri-

vorti OE it ociG .upon 60.

Dv(S)-= 240(3)

Ev -Av - 240.= o

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

horizon~ol ~

.Z:.ME=O

A

:f.Fv .,0:

.

·'" 445.)

oc(a) -e0(4/.s)(a)=o

Cv(s) - CH (1o)- Bv (~) co 4-

- 120 "'0

l=:v = 120 -(0 l:v ,.,· 60 lb .

Att =O

80 "'(5h)AH BO= 200 lb -··-'-C .%'."M.-1.·0: 120(+)+

+ev

J

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

Con~ering iho wholo f'romo : ~fv"'O: Av

~F11 •0: BD(3/s) -

ZM.-.•O : &i ( 4 ) t

. Av= 60 lb .

....

4'

to AC

~

lsolo hng bor

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

120tb

AC }

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

at iGOIQting bar AB

Bor AB

Mo

~Mo •o: CH(6) +Cv(3)-E.,(~)

X:ft-t .. O. : E~>-At-t .. O Et-1

=Q

I

.

or

~FH

.:iMe •0

~.) A three - hinged orch iG composed two trusses hi'n<;ied. together ot 0 in Fig. P- +'l-0 · Compute the. compoi:onts of the reaction ol /\ ~ then find the fon:;c,s acting 1n ban; AB'&.._ AC . Hinf: f irst 1·Gol0Je each kuG~ ow o freo. body .

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

AH ·-BH •O

/\v-At-1=120 -@

.I ~

@.,

a:i. © !..._

::E:Fv "'0

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

Av :- Bv - 240c o

- (Av - AH ""120)

.%Ma "'0

&,, = +40- 240

sAv "' 1320

Av(eo)-aoot.G0)-600(20) =o

. Av c +40 1b

/\v "' 420 lb

8

2

AH cAv -120

=+'K:l-120 bor

i\15,

I

Ai-1

I

2o' : 'lo':

: . Ai-1

c

= -3:20 lb

320

-to tho r ight

20

3"'

3'

.......

H.r.go

AB • 700 lb _ .. - C AC= 320

/'

R2

•

@member CO

36·87 =O -7oo(cos.36.a7•)

;c, . ,._ - 240 lb.,,,

CH

Two truss;o~ ore joined os shown in Fig . P-4-4~ to form three - hinqod orch . Comput.e the horizontal vodicol com0 pol'"\cnts the hingo forco ot B ~ thcr'I determine tho typa

'*"

or

~ rnogn'1 tudo of force ·,n bors BD ~BE.

I

10'

:!:Mc =o

I 200

&

120(10) t 240(30) t AH(10) -Av (40~ a 0 Av(4)- A..i = 1440

---' ©

0

2001~/fl ~

t.olO

10'

B

I

R2

ro'

1:~, 1

R4 " 300 lb ... Cv

~

lw2

7 '

R4(6) ... 600(.3)

""'

I

• ~aool b 300

. :
lsoloiing loft tr;;ss ~

R~

e

2120 lb .

.zM ~2 ° 0

R:i (10) " 200'.:>(3) + 300( 14) - 1..00( 4-)

R:-l

82

100 l'o/fl

::f::Mcco

@member BC

-t'

m>lb

10'

c

2.40 lb - .. -T

447.)

1

Fl9. P- .+40

10'

.:EF'~ ~o : .320- AC - AB COS

I 10

111

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

'*2o - !\8G1n .36.87• ~o

/

·"

'l-40ib

. . 377.+ 'fb.

A %Fy=O

10'

(300)2

400lb

@ Joint A,

A

(~)2 t

iG

comf)()!;od of throe GC
Av

l'C

8,, 2 t BH 2

c

A boom carrying the . loads shown

446.)

2':N\o: O

Av(40) -36CJC20) t AH(30) cO At-1(30) =- ¥Q(W) - 4 2.0(40) I

8

=

AH - 32.0lb

0

)

=0

= 11 ao

lb.

83

c

.:::EMA,.0

449,) The bridge shown in Fig. P-449 rons'1sts of tvvo end i;ecf1onG, each weighing 200 tons with ccntcr of grovlty ot G, hingod '\o a 1,.1niforrn center. span y.1cighing 120 tons. COm-

3000(9) +1000(6) t Fi1 (4')= Fv(12) . fv(3)-fH "'0250 -@

puto ~he rcoctionG ot A, B, E, .~ F .

· ~MF

4'

=O

3000(s) - 1000(6}t Ai-I (4)

= Av (12)

4' 1 -- - '' - -

@member 0v

@member CO Gil

E(ft I 20

I

c

-40

0

Chi

CH

Dv "' 100 tons

~Fx

'/fl

.;

E~·

• .

30

•

0

BH

. I

Ql member CO; CH

C

~C>v DH 0

'

Fv (so) =- z.oo(ro)-100(1.0)

Fv

01-1 " CH=1$00 lb

~I

@ membor OF,

fv"' 40 ~onG

I

D

~

.2'.Mi. =O: OH(+)= F~ (4}

.Z:Me "'O

Av (SD)

FH ~ 01-1 "1li00 lb

20o(20) - BO(W) Av " 48 tons· "

EH

EH =3000\b

F

Bv (so) ~ · 200(30) t 60(70) 8v = 232 -tons

F.;

1000 lb is .subjecfocl 1o o 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. 300 lb

p rcss;u..-e = 300 l_b/H. = 300 1b/f1,

l(

::EFx •O:

F~ t

?>ODO - AH

=o

AH= 4.500 lb

-450~ A billboard BC weighir19

Wind

.:!F.c =o: E1-1 -1500 -1500 =o .+'

.::€MA = O

vv'1ncl pres.sure of

=1500 lb.

8v

Ev :::: 260 tonb . F-Fw•() %tv' is =

CH -BH "0

6H "'3000-1500

B

Ev(.50) ., wo(30) t 100(10)

1

=-1soolb.

=o : 3
.:i::Mo ='O

..:!!M p"O

D

•

~Me"O

C11(10) .. 3000(5)

C.-(6o) "' 120(30) t 60(W) Cv = 00 t onG

Ot1

CH

-®

ce.

Dv(oo)"' 1~(30) t 60(4<>)

Ov

Cv

C

..::tMc • O

Av(3) - A H .. 2.Q.SO

·

fv(::i) = 0250 t. 1500

Fv

= 32.SO

lb

Av(3)"' 2QSO t 4500 Av == 2250 lb.

10 'fl

- 3000\b. i

84

85

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>. )

300tb

:a!!!.F>e "'0

:l!M.A. •O

EH - 2-40

Ev(12.) "'300(16) - 24-0(10)

i::.,, =::EMi=- =-0

.EH

2~ lb

c:

.,:;Q

2.W lb .

Choptor .5

•'

Av(12) = 2~(10) - 300(6)

friction

Av F SO lb ,I

@member CE

c

1I

G."

+'

D1-1 = 480lb

0

2'..Mo =O

°"+' Ell

:t:Mc=o DH (4) "'2.40(0)

Ct1

c.. (.+) = 24-0(4')

E

CH = 24-0 lb

Ev

.2!"'1e "'0 . Dv(•)=.so(6)+aoo(12)t...eo<'12)

t'

. .°" T

= 010 lb .

.:i!fx =O: D.i "'8H c.400 lb

6'

~Fy"'-0

Bv

.ll!Fy -=O

s.... .. w

c:

·

.SO t e10 - 300 Bv = 580 lb.

Cv =s
86

87

block weighing W lb is plo~d upon o plane in on a ngle , -&- with the. horizontol. Discuss wha t 'f"ill hoppen ·it the angle of f6ctlon ¢ IE> (o) greater than & , (.l:>) ~quo l .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. 'its 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. -'l).5.} /\

clined at

,·

__ P_· _

.,

.s1n,sg,o+• P

=

~OO l b sin 11io.g6°

60.:>-o l b .

R sinoo«96·

.

ooolb s in

~g.o+·

P=300fb .

o rolJ9h horiz.ontol . '.surface, for .whioh the COCJfficietnt Of friction jG Q,40. Def. the force. P reql) ire.ct 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;~ .506·) /\ -tOO- lb block Is rest1'ng on

c.)

.to

to start the motiO-,;.

.Zf;oc:::O :

P= 160lb.

400(COSi5') .t f-'(.8.oO)(cos45")=0 F = 70 .71 lb

sotl) The 200 lb block s hown 1'n f'19. P-008 hos 1mpe,nd1 ' n9 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

p - "IOQ/b .sin 2-1,e · .- sin. 68,2'

O·

F

•1 '

contoot Gurfacei; .

I

.t::f.1,3 .=0: /'I"" 40o(s1nzio') t •.•·

i

- - .sin. zie.z• · p --- 400lb

Glr'\.u~·

400

2oo(ros30')

/b .. ~F-x. =O: F c 4oo{cos13d) - .zoo(simio') F . . 246·11 lb : /'/ "'.373.21

,,«. •.yr,

:

246.41, '373 · 2.1

0

0.66

'The: blockG s hown in fig. P--50.9 ore conne-otcd b~ f'Je"Xiblc ine-xfen~1b/e, cords passln9 ove r fr1Ct1onlcss pul/;ys. At A fhe coofflotenf of' (rict ion ore f. ~ o.oo ~ fk · o. 20 while at /3

-:09,)

400lb .sin go •

Pmin -= 1-+8.5 lb.

88

, 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
89

)

:-.·.

I .l

I. I

fio1'e nt of frich.on I

1s 0 .10, ~

the

· to

pulle!jS are, oGSumed

•

be fr1o t 1onlesG' . sol.

or

blf.) find fhe Jeos ~ volue p required fo c.ause fhe Sl.dS.fem ·of blocl
fBO

.!'og .

TA/\

W

;.#"'

A

.

~~~

TA·

I~·

JO()lb

I

I

~ f.9•0 : N" =.?JOO(cosa6.0•) HA. 240 lb.

F&>of f3locl<..J : .

· ·. f,... • f"- N11 • (0.2)(140) f,.. • +8 lb . ~f'H·O: H 11°2oo(c.o.s.s~.1)

c

a6

~

~

'""

1

N

T ~lt co: T tf-Pcos~ cO

100

sin.,1.~1·

67.Mt(M.)(900-Psin<><,)-Pc.osO(•O

His= 1~0 lb . ·.Fe • f,_ Ne • (o.?.)(110)

rs

·,g

block. .

.

fA

.

or

of Block /\ :

r . 67:3Z lb.

127.32-o.2Ps1nC\ - Pcos<>{ "o F'(0·~SlflO(tOOSC() • 127.32 Jb.--+

Jb.

dP • P(o.a,OD&0( - '5'tt10() c 0 d& o.~GQS<>c - s rn"'< cO p.2 =~ ~ ton<><

·~ . .510·

. i.

.sin 79.1,9 •

p: · 127. .'J~ o.11s1no< tCOS"<

'i

P .. 12.4 . 6 lb .

COS<>(

c

O(' • 11-31 •

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. .!Jf~.)

r... -

601b

~fl:! ·O : Ne.• -400 cos ao·

ti 8

•

.·

v.,:> \·

~ r

P

3-+6·-t lb

£f-,. c O: Te "" Fe t T" t.of00£iri all

'fl

I

~

.

Reaolvlng wot Pf.O

w =·Te

.,.

p ..

F tW~na6.e7 ·

P • (o.s)(wcosa6.87\ t '} W {t;m a6.e1•)

p • O.&-+W

Ts 2 (0.1)l3'K·~ tl.OtlOO Te"' .294 .bt lb . ~F~=O

.e:~-o: N•wcosa6.&7• · ~f~co:

~~-=O

P(h) =Wcoo.36.&1'(1.)tWtm 86.97•(-.) o.6fV\l.(n) • 2.~ t 1.ow..

h -..±.. o.et

w • 20-+ .6+ lb .

q· I

I II

i 90

91

I

.I

~n ·O

fn f,'g . P-.912, ihe horno,genoolJS Ploc~ wel,ghG' ,.?>oo lb~ 1h coeAT1.oient fr1'c ho" i~ 0.40 . If h :s 1n., determine the

513 )

or

force

motion.

to 1'mperd ·

~...!.1-o

.:E:P~·D: 11•300COS36.e7' " ~-40 lb · ·

F "(o.+)( ~40) • 96 lb · ..c!MA ~o : ..SP .. wGin a6.87(4) t wc,os36.e7•(g_) SP • (.3oo)(?in 36,97.(+) t C.Of>.36.87'(2))

p

Rt ~1+.ot-'

i

R2sin1+.o+ •

from1 : R~ "a2+. - R1 - R.j "2.06. 2 COS..9-

~.

.sP • 1wo 111 p ... .z+olb .

: Ws1ne- • Rt .s11"J1+0t

~oos1n&- •(R1 tR~) G1n1+.01-· . . (R1 t R~) ~ B!.l-f..f .sin-e- (f) . : R12.. cas1+.o+ - R1cos1+.o+ • w oose (R:t. - R1) cos 1+.o+ - 200 COlX!>R12. - R1 ., :206 ,!J. coi;e:- @

~

82-f.+.sin&- -£ R1

:J06.2 COGe-

: but R, '1Qo.7G.06& ... 82f.+sine- - 2{123.7c.o&e-) • zo6. 2cose-

lhe .100-lb cylinder .shown In f19 P-814 i6' held o t res+ the ..3o" incli'ne by o weig ht P suspended from o cord .

s1+.) 00

wropped around the ~linde.r " If stipplrg lm~G' dc::.termine P'*-. the c.oofTioienf ol frd1on . . • ~Fx ·o : Hcos60 "'Fcoeao

p,

W<;!li .• 100IP ·

(,,

.

'f:>lCOS 60

a

f- 'tJ•COS.30 •

)I- ~ cosoo: - 0.677'0 00630 ~Ms•O

p1(1) "' r(1)

,. p

p =JA-H ~

·' I

•

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 .

s1s.) Bloc.I<. ./\

I p

Ir\

P )IAl'l=(0.5773)(173,~2) : 100 1b:

Fe>D of t3:

p~ ·to~, 30·96

block B weighs ~':°lb, inoline. ·. Ir the. coerfic~Clf"\t

t:. ·

cosao.96 • R~ " .S~.3 . 17

A'

Fi9. P- si.s woighs 120\b,

~the cord IG porolbl to . ~_he of fr 1otlof\. fol"' 0 1! .sur.f'oc:.eG' 1n. '?°ni ovf -' ~ o.w , d~ierm1ne. the angle -e- of the 1nol;ne cit wh1oh rnohon. or f3 impends .

~f~ aQ: R¢.COGa0.06 . 2«>C06ao· + R1

fl~

p- 128.6 lb .

s19.) Jn

fie . fl,.si0 ,

two blocks

sf rut otfoohed. to ooch

If

the c.oemcront ~ /j wo'9hs '.z701b .,

~

or

f ind

-

-R1 -- ----,,

s111(00-~)

12.3,7 COSf)--

93 92

with

fr'1dionless

p1n£>.

und~r eoch block. i s o.~s the min. we.i,ght of .A to pr:evenf

rr-;ct •on

R1 • 12J.7 .sin (~o -&) • R 1 ~12a.1 (singo·rosa-- eos9o'Gin-e-)

R1

ore connected ~ a solid

b~ock.

mot ion.. 120 s1n7s.~M·

lb .

~Fx "'O; P • R~.s;nao.96 t R1 Gm ao.06 .;ioo s 1n ao·

sn) Aepeo~ illu6'. · Prob .s11, 055um1ng thot the .siNJi j6' 0 uniform rod weighln9 .300 lb Hint: rt'nsf /solofe fho sfruf

:ao

61t'\

WI\

-=-

e

as a Freeboqy dt'qgrom, resolving :fs end fbrc~ /nfo compononfs oofi'ng along ~ perpend/cu/or to the G-fru f.

1+·<>+·

.590.3 lb

c -

.324. 0 lb

r1 c ssota2+.a(s1nao·) H • 712,4 lb . P w F .t C =sao· a ;"-rj +¢>2t .a)(o.e 66) p = (0.2)(712.+) t 2."1.29 pe+tz4jb

force of. 4-00 !b Is oppl ie.d to the pulley shovvn ;f\. The pulley JS preve.nted F'rorn rotaflng bu Cl fo~ .P ?~led to the erd cJf the bro~ le-ve-r. Jf coef offr1cf1on. lot the brak.e surroce ls 'o.20,de-f . thevolue-of'P. 5 ~.)

1

A

F 19· P -523.

the

J

•

·,r

,

s:z.1) In ·ff9 P- s •9 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.

j•

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

95

526 J ;4i ladder ~on long we;ghs iO lb ~ '•ts c enter of grova~ '1s 8f1 from. the bottom. The. !odder ~s placed ogo;ns+ o vertic;ol wall GO traf i ~ mo~ on orgle of f:-0° fhe. 9ro(;,lnd . HO..., for l)p . the !odder con Cl 160 lb man o l1'mb befOre the !odder

w/

.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

i6' on -the venae of slipp ng . The ore le of fn;oi ion of o 11 r--

A

o.-+L-12 = O

i.F,y~O ; 'It•

f1 t F1. £F>i =o ! N1 c t-l!l. · , F1 •t-li1 • O.QH1 = o.2N,,

r. =b o.f

R:z.

(10,11.at)

~ lY"'/1,()'

L~.e>m

@!:I
is• H

I ~lb

o;ub,, the volue of.!:! :

bef!

I c

A(S1!1)

~;lb,

i- I

f1 = f!J.

w-= F1 t F1.

con -

tod Gur foce J{; 15 •

Tan n;'l( -17,32 ~ - Tofl1.S.X t 10Ton 15 •

is·Jt 0.0)

o.2N1 ( 2)

(Ton 1s• +. Ton 1S·)

Ri[lt!

W • 0.4N1

@ ~M,., ·o

+o(& - acos60°)•160{x-s)

:it - 20

x --s ft .

"'fO (1) ,. 160)( - eoO

.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~. J o l f Th~ angle of' friofion. o t oil contoot surfaces I~ t 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~.

· ,•

.

.

I

w

iS-

,'

,I' I'

~I

I , !-f~W"-':cr--- ------1 @ (~-.!h)~-m(X-'X~ . . (~- Li;ine-) 0 - TonW'{x -L
0

or

!

.

of ~il ibr•um

~IJ) 5 )TonuiL?&tLS.•~ \ Ton70 Tonl!O t

0) ~=m?.tb _y =Ton 70 •.X

'

I

.

~(Tan10' t 1onzd) = To nzo"L0%'9- i LGin~

(Ton7o'

Ton 1ZO.L0%9-t l~1ne- - L co~&- •O y Ton ~o· t Ton 10•

-4::

f3 - 1'47.6 lb- ft

• o: ti2 cos1s· c N, c.os 1s· N'L · ~ N1 cosu;•

coio1,·

l'l!l

~f!:j"O:

-

N1&1n75' ·1 H2 s1ri1s' .. .300

~m'" .ltj, ax7~·) sin 11;' ~ 300 '\ cosw· N1(s1n 1.r't cos1s'ta'1'1s') = aoo li1

N1 • 289.78 lb .'.Nz•77. 6slb

:. F2 .. jt1,,

·

~(tonisJ(~76) · -: To"!1s ' (n6s) 11"77.6/b. Fii=zo.0 1b.

2.3935 2

s ze.) In stead of o covple; d eterm;n'e fhe. minimum horlzonfa:I force p oppli'e.d tonge.n~io11!:1 to the Jefl of the, top the qy li'nder described In. Prob . .s27 to stort the c,1:1llnder rotallng

or

Tofl -&- ., 1.1g11s -ft- e ,.50 o

counterolock.w'1se,.

Ton w· Lc.oG~ t J..G1ne- • ..!::.coseTon 70~ Ton 20·

~f~

··· f1 =,.M.N.

21on20•) (f:s9-

@~Ms·Ow('l(- ~-~a-) .o

tTonzo •)

2 L~1n &- 'LG1n&{ron]0° tTan20 -

si'n & •

.

Tan 10• t Ton 20•

•

.J!M" "13 13 " f.:o (1.s) t F, (1.s) 13 .. (20.s)(1.s)t(200.7e)(1.s)

1l Ton 1.0' LcoG& t 2 Ls 1n& • Leo~(>

7an70X- L&ine" .:_Ton~o·.x-tfonw'LCOG&­

x =1on2o·Lcoce1 l{;in&-

~llnder -3 fl . In d1'omek.r ~ weighing /6 restin g On two 1'n ofined p lane as Ghow n in. fig .P.;f7 If the ar-gle of ff1dlon °1G' 1s fo r o il contact surfaces compute the rnognitvde the eovple reciuired to stort ' the cyl'1 der rotating counf~ock.wlse . 527.) /\_ homoge.A""leouG"

14

I

f./ .

.3001b

(t.cos.e 1 u ant - · -- 2'>11

w

ioos.,.

160 X ~ 8-to ~ .X - 5. 2S

2

96

97 '

<

I

; :.

.slipping imperida.

~M,t.·O

p6-.ft). • f;(1,~ t F, (t:Sl_

Pl.14 ~

£M•"'0

20.&(14 + 77.6(1-st

.. WJ.. 21-"'°'"
p • 20.s t n.6

T0t11G"t 'btl60 '

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 o ..30 •·. e- ~16-7'

t 2.tCO!P.Ton60' -12.J{"n"( =)!.c:cis..(ronH•+ra,,¢)

GUI¥.~:.·.

Ton1,• flonoo·

p " 103. 021b

'."

..ss.69 lb

1.06b

In.

plank wff. long is place? .ir;

In f(g · .P-.S.32, two, b loc.N; eoch we19· h;na · ~ f.50 Jb o· ~connected Qy a l)fl• for'ITI hor1:z.onfo I bar which weighs 1ooJb. If th . -of'.\ '11> 1.s· undo..- c:ooh block. find p d; ted e- onII_ 9 le, o f nnof1 ith · o e. <+.s •·1nol1·ne. tho~ will oOl.u;·e ;mpend;ng rroho"recto the-para le-f1. .S.32.)

.

p(H o.B66) ~ 120 (o.U6)

1

J(Tan11~·t fQt"160°) = 2LC06"lTon60°-12~no<. ")( • IZL. ~O(Tan60•- 12l'ino<

Pl 1 n) ; w('!:)

c lose- the lood

1i-npe.nds. f11.

o)

A

',..

',..td·'X

p' con be pl oced • st. ) ,..!- ,.. \ ~9.£t.!:! ,,.,,,,,. ' ., ·

' p

')(

© Isolating, t he. bor

P

horiwntol posi its endg restin9 on two me.lined p lane os ..shown 0

w/ fig. p-530. The, o~le- of fr1d;on i~

t 0n.

1. 461' .. 0 • nu. :J. e>{ 0 36.2.

Tancx. .,

c~-~;)--m(x-~ .)

~1 · iZOlb

£M11 ..0

5.30.) /\

.

y tilt.''~: -ton60'(x-2t.eosC<)

J-

p

~

>

~·m';\tb

20· .

"'

Oe-fc-rminc:. ~v;

tttl

\

fso

~~

"-.

,/

p

r?

":';,,, "" 10 ~~',

aof

@ 010CI<,.,,

.-,J (:.ei,'6)

I .u'

I!!

---.. . ffi•:)

to eooh end befOre .slipP1 :.

lf«)

'·

200

~· C

~~ fonJis·~

y -ton6s"JI (!:J-O) "-TorHo· ( x-10) Ton25.,X .. - TaneO'x t ~on"(10) Ton6s·11; -Ton·ll t10Tan..a'

')( (Ton£.s• t Tan.~o') x c 9.2+ zMs =o:

.:£f,y • o: N· ?,()(}Ginti;' tllOO•rn+s•

(1c1-o) s -1on86'(11-10)

= 10 Tan60°

~=2.&1

p(g.2+ :. (10-11))~0

zMi;cO:

P(2.01 - !:1)~0

2.01-~ · o

._9.2+-10,x-O

,...

)( =

:X(Ton66°tTon-t
.Y

o. 76 !.!.+

= ~.81

~r,. ~o: re tc~~ - ~i;· .. 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 rl mohon lo fh< nght.

I

ff\ total Length 2L 1s ploc.e.d oG shown in fig . ~-5'1 with ·,~ ends Ir\ co"toct the,.· inci•"noo po~ . lhc. orig le of f r lcti'on '1i; 1-5°. Defen"nlne.

,;Jf

sa1.) /\ uniform . ploril\ of weighl W

w/

98

2.00 lb

(o. 268) ( 2u t-t) ~ o • p

H " 282.84' lb.

#

ft.

e

p

i?1:J 99

zfi. ·O: R1 cos-4'5' = R:z cos 1s •

"'+ ·. ~

%.F!:I •

.

:EF.!1 ·~ : We • R2 sin"s ·. -

• 200 t Av H = 101.6 lb

P" AH

+-

We• 2.73.Q./b

2IZ'i.,OOS+s' t(O·U&)t\6'eo&tG. -Hs'i..,s1n+6' -o

In Figui;c- ,determine the volu~ of P. just surficieni io Gt~rt_ the.10 wedQe under the -+oo-.lb plock. The ongle. oF' n:- ion 1s 20' for all contact .surfaces.

537.)

f

p = 2.86-'t t (o.~)(101.6) p . +27lb

4ti(-f-OOG-t&') tfs Lea&i-6·-lis siri•s'L •O

fi 1

R,A400 .

1.tg.13 "'0.-'18tte

-~

Ne= 289. 4 lb

AH • He• 289.'ff6 :,wte = (o.ua)(zs~.+) Fe· n.6 lb .

~<

fl.it

. ·.p~·o:

.

{!,¥.'le)·

/

'

R

.

·

R

.

£f;, • O'

2R&•n (~q•) ·P

.

.2 R[S•fl?JC.OS~ +51noYac.os¢}• P

.2R[o•nt(·Gin"f'L)(i) -+(cos3X~)~~·o

dP

P

d"<

.

.

-.s1n~s1nc
.

-t

2

.,( coso/2 co.sl" • o 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:Jlock /I up the, incline surface, of f3 . The- we ight /I ;~ 100 lb ~ the onglo of fric~ ion for oll s urfoces in coniooi is 13..

or

p

~1. •. ~1.:9 . 2 lb -. P • 41g.'Z. &1n~·

Av - +i'f'tfe =-,s01 .6 lb

WE06ES 53s .) t4i wod' . ongle1.o< of thewe00e. so that · ii- '1rn re-main embedded on the.- log· p

R1 s in .+s.0

Ws •(as6.-t)s1n7s'-(141,+a)sin4r.'

o:N

i..fx uO :

.. · o

COS7&"

H /\v tOO

.

~

frn+2)cOG+~· ~ Re - 386.4 lb

r,

f600~fi00 p

•

"°

'-f

F1 ... N~'

100

tOO ~R1

~

P.1 ~ 141 . 42 lb.

'

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 wvuge -"' · t firom under the 400- Jb """' ou Ploc/< .

~-.......----.-.'

p~:

of~ .

FBD

~Ht

,..

~

J

- ~ Fa

~~

.

Ll . •

P

r.

Fe

11.

R.

-13.1__ • 400 sl'l

.cm 100•

xr

R2 • -"01.7 lb

_ P_ smac•

""

391 · 7 -"'m7o•

.._

P•

203.1

lb.

-

sag.) Jf t he 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~.

p

.

101

.

F

I()()

II

~ !7~· -

.

.i.

11.t-

"t ..

fl)

Ii+ R.t

ongie of friohon. .' 1~ •

Rt • 601 .1 lb .

~"°"*'

R3 •

=·

10~.01 lb £F~-o:

.

P • (109,&1)sinao• 1-l13.z) fNl~ Pc =; 91.sli>

P,.. " Pt.-P1 " g1,5 -73.~

p,.. t:: 18-'' lb .s+1) De_for~inc:. the for·ce P required lo starf the wedge ~hown 1n f1~ure io

p.

contact

P·&tl .

reo of B

P

+76 .7 lb

R1 • Rt cos1~,. - sob

R. coi; &•

--+--

~fy ..o: R, cos1s' - R~ sinr& • 2000 , .su~titu+e- R.1 : "· 111 . fr.1.~1.s·- i;oo) cosu;· - R2sin15·2000

CO' 1i:•

t Rt.

.Sff'\ 19•

:

R~ - -Rq&>~·

~F.!1 •0 :

+

Go111;• •

R,CO'~· - R, .s1ni:' (r,•ni,).-w

.eol:)

..S11"11~ ·

c0&1.s•

R, (cos,•-s1ns'tonH:

0 )

..

c

Rt • 11ss."f lb. p QO+s•

s-fO~ As shown i11 Fig P-sw, t wo blooks we1gh1ng ~oolb ~ resflfl9 on a hon2.ontol .surfbce- aret to be- pushed opar1 by a 30• ~t Now

'1 '" 1s

0

for a ll confoot .surfbcos.

value of P jf; required_. ta Gtort rnovemCl'lf or ihe blook.s ~

wo4fd th'1i; .

onswe-r be chC11"<3ed 'if the woight of

one

bl~ were increa&'ed to 3oolb f

- 11.55.-+

~

p - .Q.of-3 . + lb

.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/\

of the

.

4' 100

J3!_ .. ..!E2""1 1&°

.

R1

s

""'~

1a.~

P.1 ·~ ·

~Fa • O : z~ ..sin30• • P

p-

lb

.

1Z.(13.~).s1n110•

P. •

715.~

lb

103 102

.

R2 ( a.~-+6) ~ :it1!166

.+ 11.1 lb ,

-"? p .. 7~.g lli.

wed9e.. The angle- of f,..jot1.on

1

.

Rir (cos16· co1-t6•) - 1509cot j5• - Re &1n1e"c ~ R2 (c.o&1s·cohs"-sfm5•) = ~ooo t soo cotu~·

~

R, (o.97a) "-too R1

Gurfaces

.e"-,. "0: Rtc.oG1e· -Rt&in1e•., ~

1ooolb

~f).·o : ~.sln1>'1e-tc=is··o .

The ongle or friction for oil

16 f5~ ·

To adjui;t the 'v6rt 1eol 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 . J>+3.)

(.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· co.s llO •

...

- o.4-66"'

005 70 ' t

ODS"<

Ton o< = o. +663 o<

2S "

><

SSIW'IRE - THR~~ .sCRrew

s-fSJ

~

Rsln6o• =

p

R

20 00

= 23og.4 lb

_ _.:_P_ . _R_ s1n&S

..sin66'

;\

fhrooded jook.screw hos 0 pitch Ot O.S II'\

.single

or 1.7.S.in. The, coer(i'o icni or Gtafrc rriof1on or Q.JS, ~or f\1net;c , ( r icl1.ori of 0.10 (o) Oeforrnln~ force P applied of the end of o levor 2 r1 , Jong which will .sfor+ .'.if'fln9 Q we!9ht 2 tons (b) whoi volue- of p wlll Keep fhc jOOlt!t) · -&- ~-/on-• (o.0'166) 0 meon roel;L>S

the

or

4:

p = 208 il • .3 Jb

-fT- :- :2.6·

h • 0·15

o • .2('j. •£-+in

ton1c0.1 .S

W•

1c8.53'

.

2ton' • 4000/b'

P = -t.2.6

Jbs.

ton

P • .Yi£... (-e t ¢>) a P • (+000)(1 .1s) fa n(:z.6t8·53')

12+

.

p • -57. ...,. Jb.s;. ~.) The- d 1'-tonoc::. be.fwei:-n o~oceni ·1hreo~~ jocl
R1 (6irr'IO -~) ~ R1 ..

1000 Gin100 •

9134 . s (.sin10"-«.)

_P _ _ • ·_R_1__ ; s~bf1 tule- R.1 .•. -. P · .-oneo• . s1n(<1oto<}

P

=

1000 s1n(10--<)

-+ p(s1n(10--<)) son(wto<.)

(S1n1o'c0s -< - r;ono'•nnO() 1000 ( s in 20 •cos-1· - oosw·&•n-<)

~

~

1000

1000

oquolu~ P • 1poo ~ 151('1

or rri"ot1on

{7-

.271"

70- -<

o .10.

tone- • ~:(z) •

:z1r

c

0. 15.9

f - b.JO -

tan1

9,oa •

x.....ill,_ • o. 1667 rt • 1.21-A.._

f

•0.10 ·-5.71'

•·n~o t-<

f "Wr

to" (cp t&) 12000

104

threods on a /r;ple,r-oo;us 1.;: zi11 , the lood 0011 be raised b~

mean

whof ~e.-t1r19 0 momeot or 2000 · ib - n ~ Pifch e % in ( srn~ ·,~ ·,s lnple lhreodcd muli_iplj t~ pih::l-d~~ · ·. Pitoh : '2 in

~ll

61n(110t"\)

.s1~t<><)

coetT'.01e,f"\t

i6' % ;n. The

QOOOlb-n • W{o.1667rt} 1-an (.s.11°t9 ,0.3') lb·-n : w(o.o+30-s1aan)

105

,.~ -"'lb

w=

o.o+.3&5733'fl_

.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 .

Pillih. =- 0.2.sin ( 6inoo d ouble. ihreoc:le.d

j-o.1s

rnultiplij b'.:j 2)

fanf : o.1s

to" fr

f> = a.53" tons•. -teOO Jbs ·

·

T • Wr ton (tp t B-)

-& •

o.s it')

-

- 0.0.531

£n b ·~ 1r\)

r: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"

, P .. o.'f' in.

I~

.:n(1 ..f)

:ar

. ton.& .. o.o+GS -fr~ 2-6

T~ - Wr ~Of' ( tp t~) . " ..-000? lb ( 1-"f' ll

T~

z

a.+ ton& • .Q±_

&-

:t

·

•

.·. ""' • 1..+

,

JI"\ -

f ..0.1

c

T~ ~

tonlfi. 0·1 T

eELT

-t-&)

"f0000(1.+1'.t") ton.f.s.71-~.6·) 253 .6

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.+ what wC.9ht 00'1 bo -suppar-lod h9 4

106

olhcr

end

or th~

"'

2

.&s~) /\ rope vvrapped

t1.S7 lb·

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" . "°'b on9Je or ocntocl : no·

q

~~

Ln

-

f~ -

Ln 80 •

f

l.n

~o · - f (rw·J( Io·)

J (1~·~n)

= ·~

~

f • o .a-1<3

12..s7

ss 2 .) /\ boot c:J1erts o . pull of "'1'000 lb on its haweer w/c Is wrapped obovt o capGion on the dock. Jr the o.ao, how rnon~ Ii.Ams must the hawser mo~ around 1he. oopstari rotrot 1he pull al the other docs rct 01
f•

Ln ]!._ .. j {J" fa /

" Ln 1VT1Z • l.n~ • ln eo • 1+.61 rod-

f

/+.67 x

0.30

1~

0 .30

= 8217.1 •

~~7. ( ~ .:360• = ~.33

-

furne .

553) /\ torque of 2-tO lb -11 ~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 fnOt iol'\.. 1·,s oao. Oe-t~ine the (of"G<:/ p Of tne- encJ O( the broke le.Yer. e~ "'· 2"° 1b · fi -T1 · c(o.'!>) ('60'• V~) 12 .Ji_· -3.701. .·. T2 ,, _ T _1 _ ~MA

FRICTION

~el"'Hng a .so - lb ro~ at k

T1· .J!>O (23·1+) .·. W • 115 f lb ,

buf T1 • W

Ti

Jb-n-

.sso. /'\ rope- mok.'1'"'9

eJ~ ·(.so)e0.+(~"Teo)

•

tp •.s.71•

1 ~~ ton {,s.71• t 2.6·)

11 =le

1

= 681.82 lb- ff

. T~ • Wr ton (
w

tan- • (o. 0531) " a.o'37 °

0

:z.:;;inD

1

2

-tOOOlb

• ~(1.s.,,.Y< ;.P~) ton(a.sa 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'" · 1he. c.oo£ri'Oie'.nt ·or rrd•ol'\'•:~,"r o.10. De-fc:.rniine the momel"\I ~Ga­

TPI ·

·0.4

..IL· e-f" -

l

..S"t7.)

'1

angle. of conlool ·1~ (a6o t 90) = "hS0°

. f

woolb-~

rope-?

c

3 .70Z

M

(T, -Tz) (8 iA)l.-fP"") •

240 lb- 11

To - T2 = .360 lb

T, - T,

'360 lb

3.102

T1 (o.7~gq) " 360 -

107

T1 ~ +93.23lb

' ·· ~ · )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 dovvnp lalle mohol"I. ol the , '.

.flV!p 0 (.::s)T' t T2.(2.) - (2)fa -(a Jr1 cO ST t , - 2Ta · 311 c CJ .:, (""/~) - 2 (o.az~)w - .:3(~0()) co a

0 · 06 1tW

)

w=

1~ l b bod~ ·

vi .

1Fy •O

lt

rl,

'

.

~

T.e -f,

c

wS..na6.e1·

T~ -(o. -i) (WC
~.,.

1,

w cos 36-67 •

~fx•O :

l\11.

T 2 = 0.16 W

rlc

...!!~

. o .: H2. • 1000 COGa6.e1· +rt 1 rh "' eoo

/ ' '

bi.it T1 • 1.B7h fi,= 0·76W

H1

t

rl1

.£1',..~o: T, 1 f1t f ._• 1000&i1'36-87 °

,

11 t

JA-111

+JA-rt.. -

600

-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 1·7'f-W · ·.,."f'Olb

I

w

~

f 5.2. g 112 .

r.-om

SSS'·) /n Fig . .sss, o nel';ble belf rvn& A over the. ec:mpciund pullCij 13 ~ baol<-. over P to o 200- lb we(9ht . The coe ffi o;et>t o{' fri ot rol"I. _ic; Yr bdween Ih~ 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

~

e,

(O.s1&)(e0 xTl/r• ;-.

..

,) = 1 . 6+9

T2 fa

a

1.6t'3 Te

but Ti ·W/e

T3

•

~

7o+.7

t H .9 ( w/z.)

T3 " 0.0H3W '

109

600

lb ·

,, l

~'.

60U Dc:termino t~ mogniiude of the ~ultont • il~ pdinting. ~

I

I

its; direction c.osines for the following systom of non -coplonor-. C
Q)M~na;ms

f()RCE

y -4

.IC

aoc>.lb. ii B • ..OOlb. -2 c•20011'1 -+

A•

....

(D)

6

7.81

-5

6°7i

-.3

7,07

!!S

r'ORCE.S

olST.

1

llOOMP. {15.2..f.

-1~.23 ~-4!> - 298.<>6 -11.iMS .1'41. - -EH.R7

-117.H

1DTAl..f~1

I

f t .. ~- .Aie ._6_ y

1l

300/7,91 ""

Av/a

=(-117.H)

.". J\,. " 115.24 lb.

30Q/Mn • l'w/--4- .•• Av

a<:io/7.e1 " A¥6 .·.

n

·z

Y

o

Fon:i

.

"

Systemb

in

Space

~· ..fL 9~·

x

Y

a 117-~Qt;.~ .. o.::ig.s .

6·71

..s;_. D

l!.

Cos-"y"' 226·2o/.f!96.00 .. Q.762.

eos-ez ..

WO 7,07

1s2.-t0/~96·90

. ·.·Cx ,. -11.:L1S; Cy= 141 -~; Cr."' -&\-.87

= o.&13 :. poi'n\109 bock"Mln::ls,

~ .. ... 115.2+-110-23-113.15 - -117.14 lb .

\..up-Nord to ihe ldl.

.;Ey

n

- 153 . ~ t :Z38.45t

1l

(;o66;. •h/R

••• Bll .. -119.23 ; B,. = 230.16 ; 8%. c -1!190 .QS

Chapter 6

2

t 226.2.f. +(-1~.-tij

R" 296.90 lb.

-1.5::1.65 lb .

l\z." 230..+7 lb.

_fu_• ..fu:_- ~. JL· ~ ')14

I

R.2·~11."t:t:I'"+&•

0

L.

~26.2-fo -1.52.~

o•· ll2+y•n:.•

to get OiGfor'lGe(O) use )(

ltalMP.

Y~P.

-153.65 23().47

141·++ • 226.:Mlb.

~z·

I

- 230.+7 - 298.06 - &t. 87 == - 1.52 - ~ lb . 603-) Determine -the mognitude of the resuHont, its poinh119 , ond

'

ils d 1recf;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 ')(

2 lb· -3

A•1QClb. s~300

C=-

!ZOO lb.

I y

li!.

DIST. (0)

3

+

.6·:305

-+

0

0

TOTAL{<1

to got {D) u.sc o 2

5

7.07

+

4

,....'(

D

37.1+ ;

Bx/le .. By/y ~

- 127,-?QI -169 .7.;I

Ay

c

Bz/l:

5s.71 :

'·~ I i

0 0 ~00 - 90.156 -114.()2 "tt36.44

.. , r 1'

:. c)I. -. o ;cy =o; Cz

I

.. 2.00

Ar.= 1-+.:za

" 300/7. 07

•• Bx .. -127.296; Br " -169· 73; Bz - 21 ~2.'t6

110

2H'.16

= x~ t y•o~; 2

..b_" ~ ~ -6.z._ - _jQQ_ x y z 5.385 :.

FoRC.E.S ')( o::>MP. YCOMP•. !:.c.oMP. 74 :28 37-1"1- .55.71

111

I

I I

\

, ;.

.-

forces P. Q.. ti...., F hove o resultan t · of ..s lb. forward S.., up to right ot -e->1 = 60., -e-.., = 60•. -frz • 4.S P e directed p01nt 01 .1,4 ). The QtJOls 2.0 lb~ pa10ses through .the ar.1gin ~ the

6().5.) Three concun-ent £><

0

37.14 -127· 298 .. 0 - -oo.1se lb.

a

•

£.7 "' -1H-.02 lq; 1+. ~0 1" 212.16 " '2/Jo ... 4a6.441b.

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.

:z.21 -

..,olue cf:~ is ol.so 20lb ~

p.2 =~?<)2t(~y)ll•(~~)2

~ (-90._1 se) 12 t (-114.02) ~ t (-..e6.+4)

2

· R "'-507,7 lb . po;nbnq forworcl \...clO'Nfl \o· the lefl . Cos~ · .. ~:1
£:.z/R ~ 4e6.-M-/.507,7 .. ().958

•

or

- 3,): COMPOHENT!O OF D y z:. X·

DIST. (0}

FORCES llCOMP. YCOMP. .Z.OOMP .

.s +

-.3

7.07

11.!3·15

-5

0.77

- .3

...,,39

-:l13.66 182 .44 -228.05 '21/2,63 -H1.a2 - 166·98

A .~~lb.

4

B • 4«> lb. -6 C= 300 lb. TOTAL(.;)

+

-2

62.12

141.+4

,l(

7-0~'l·

i!

y

212 .56 -479'.9

f

y

i!

•

4

4.se

e.73

Q.2 20 10 .

.s

.3

6.16

16.23

4,37 6·"'1-9

2.6

? 2.5

_fr_ .. _fr_ ".EL~ E£.. y

)(

4,se

i!

-,

17.47 g .74 ?' ,3 • .!'>4

F z • f x2 t fy 2. t fz 2. · - (:-22.-%)2 t(- B.3'0)

:. Pit =8.73; Py " 4-.37; P2 " 17°
-&.. •a>S-

=Reos&..

F•/F

-e-,.=4e.2·

£.y • Rcasc:7y ~ s(cosGo·),,, :i.a

f!ty •

£..
-e,.

5

1

1

., .5 (cos;60·) =2.s e

t(-23.67)"'

• cos- 2Vl6/33.68

:. ~'i ~· 16.23; ~Y •6.+g; ~z • g.74

~x

!t

F = 33.60 lb .

~ s _fu_~• 20 y ~ 6.16 )(

(cos""'"-) • 3.S4

4

1

cos- a.::lG/
0

~z =cos- 23.'17/33.60 :..e-z. ,; .....5.3.

F.,. • 2 ..5 - 4.:37 - 6.'49 fy

·12£_.,' fu'._=~· -400 ,l( y z 9~7 I • ; Bx-= -273.66; Bj • 102.44; 0z,. -2~05 >(

1 2

1

.·. 'A,. "'113.15 ;Ay = '141;.~4 ; Az.•-&1-.97

c,. • ..fL • Ci!.

2

where : R ".!'>lb.;-&,. • 6o"; &,, - 60•; ftz" -+s•

-e....01

t oge't Dvse: 0::1,,xtt -ty"-t ~ e ~ = 6:.-0~ = 200 , '

)C.Q?t.l\P •. YCOMP. xCOMP.

(0)

x

y

~

F• ?

. ~.). Dotcrmino the mognitude the rosvltanf, il.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, FMCE

FORGES

DIST.

CDotR)t15'11TS Of 0

FoACE p = 20lb .

a

-6.'36 lb .

... pointing

F" · :2.5- 16.:23-8.73

bock wards. ~ down......ord

to the left.

F,. • -22°46 lb.

f2. - 3 .s+- g,74 - 17.+7 fz ' - ~ .'17 lb.

.300

.5 •.39

/\ force of 100 lb is dlrecied from A toward B in ~he cube shown in fig . P-607. Oden-nine the m
.•. 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

of the coorchna+e axes .

ro .

·-

.£z • -B+.S7- ~28.0.S - 166.96" -
R'.

£." ",..~ y 2. t.-.;.z. L 2. "" . ( G2.1'2 ) "

t(212.s~) 2 t(-47g.g)

f./

cos&,. ~ 62 .12/s2e.s.3

cos&y

c

= 0.110

212.56/s2e.sEJ. Q.402

Cos &z. " 4-79.9/.s20.S3 " O· 908

: . po;nt ing bockwords, ~upwards to ihe right

/

y

/ /

//,;./

2

R ".528 . .5.3 lb.

-.

J,"

{'

A

I

,~

r,

-

.~

d

141 -. b...... / /

I/"''

;

I/

d2

/

II

v

c

•' a'

,

/ /

/

a,..,. "'

1.1

Ji'

= 4~ t 3 2 t'4 2

= G.403

.ft.

113 112

I l

610~

f'~

• Xz:.:_ = fL,, 100 lb

• >=

·6~2.!ilb • Fy~~ . g lb ; fz•62 ..51b.

~ , •• r>t

-+

a

A force of'~ lb i~ d irected from C towan:I E. in the cube

~ in fig . P-610 . Determine ,eoch of the coordi note axes .

~-....,~

the

moment of lhe force about ~111.,.

'

~"1,. "fy(+) - fr:.(+)

.. 326.5(+)-163.3(4)

\ . 4'i.9(....) - 62.5(+) ~Mr:

~M .. = - G2.4 lb-.fl.

c

~Mic..,. ~2 . 8

-Fi(4)

~My ~ t=w (~.) • 62.!i (+)

" - 62 . .5 { +)

n.

ztv\z ~ - 2:!>0 lb.

.t.'My ~ 26'0 lb .

d · ....g n.

.. 163 .3(4) t 1tJ3.3(2) ~My" 919.e lb- H .

:t.Mz.

6 11.) A force P. d irected frorn F . toward B in the cube .shOV'ln ·,,, Fig.P-

~l<\y : -Fr:(+)

6079

= -1"l0.6(4)

..

~Mz • f,.(1) t F\-: (+)

dl!. +'t!I• t2.'

m .+(+)

d .. .s.~89ft.

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

i

Py~~ .. ~

~Mt= Px(2)

the cube

p)I

Olle& .

~

= - 100(4-) t (>00(2)

:. p .. 1077 lb .

= -000 (+)

I

Mi! • 612.) A force P is directed from

£ My = -Fz(+)

d .. .5 ff,

.. 0

. 3

...

.5

I

(-- --+..-.::-.

- 160(+)

; I

.£Ml!. = 120 lb- ft.

~

,'

/

.,

o'

~- ~·-· a

~::_ __P. _ __ __ _ ,.,/'

..

:lli!M>1 • Fz (-1) • 24C (1)

:£Mi1 • ~ lb-

o point A ( ""• 1.;4) towond o point B

,,~· .: ---~~---::;! . , I , ,'/ : : ,/' :

-~40(-4-)

z~y ~ -960 lb-ft .. 1'.Mz - fv (+)

' £!__ • ..fr_• _fi_" ~

- 3200 lb-fl .

(-3,4,-1) . If it cou~es o moment Mz· 1000 lb- 0, Oeterrnine -the men~ of' P obout the x ~ Y oxes .

df. 0•13• .... •

"

,

;

d

.,, ~

~- "-a-

= 0 .11 ft .

P • Px • Pv • P.t

Q.11 7

3

p" 911 lb .

114

115 '•

T

: . P. ~ 1iy'g.11 ; P1 £3o/9.11; Pi!: =-!:!o/g.11 £Mr• P,.(1) t Py(+) 1900 • 7/g,1' (P) ~ 3(+)%.11

f-1.

·1

Ml\= - 400 lb -·fl . £tlli! • - p.,. (+) .

0001b.

..&.~ _£____ ... 5,399

y

I

:. Py =60011;>. ; P.11 • 4001b.

£ M,. ~-Pi!. (-t) t Py ('2.)

z

1600 .. p,. (2.)

shown in fig . P-601. Oeterrfl1.n e the rnornent o f the force obout

the coordina te . :· ;'

<.?Ouc;es o moment My ... .1600 lb-rt . Oelerrrlme P ~ '1ls moment

obout the X ""z aJ(es .

zMy • -.594."'f lb - fl . - 1+.3 (1) t

:. r)l .• 7.....,, 1b.;

'f:

-F~(,.) - Fy (2)

~le" -1.306·2 lb-ft .

~M>1 " Fi!!.(1)~1~.6 lb -f\.

d ".s.3BS fl .

s

=-16a.3(4)-3:26.S(2)

: • .f,." 163.3 lb . ; Fy • 326.slb; F.w;•163.3 IP .

' eoch of the coond1no1e axes.

d'J.· :z"+3•t+2

lb-ft.

%My • Fx(+) t Fz~2)

rt .

A force of 2oolb . 1s directed from B toword C in the cube 0608~ ~hown ·1n fig . P-607. Determine ttie rnornent of the forc.e obout

each of

.. Fy(,...) - fz(<4)

'liO

,.

••. p,. • 700lb

61~ -) The fromoY110rk shown in fig . P- 615 consists of three members /\fl. /\C," /\0 whose lo....-er ends .or-e in the .some hori:cont ol plone . 't-. horizontol forc;6 of 1000 lb octing poro ll~I t6 the ')( o~is i~ oppl ied oi A Oe\erm·1ne -the in each m~mber.

; P1 =3001b; Pz • ..sOOlb

~M" ~ -Pd1) - Py(.,.)

= - .!?OQ(1) -:- 700(+)

force

M11 " -17tx:l lb-fl .

llr .

.Z.My - P;? (.+) - flll ( +}

d"1 •(6-:a}'+(6·0)'t(o-o)•

- ,!500(+)- 700(+)

dMI •6.706 1 dN:. •(a-o)• +(6-ot' t(o-(-3))"

]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 . . 'Y

i . - - - 20,

10'

df.c - 7.35/

d,.J •«~-0)'.1-t (6-0)i +(s-O)'z d.-.o. 8,367 ,

--l A (10.1s.o) d~· • (10-0)2 +(1s-0)2+( o-(·s>)t

r

d.-.e .. 2

d~

16.708

l ~ote

fl .

·Right Side View,

2 =(10- 0) t(1!ro) -t (10-0) 2

d"c •

2

20.616 2

d....; ~ [10 - (-20))

dAI). 33,~

lsolote Front Vio_w,

fl. t(1s - 0 )

2

.fl. £Mc. •O

£Mo•O

f.soloVing To p View,

-1ooo(B) tCy(9)'"0

rsoiatin 9 · Fro~t View, '

Cy"62Slb*MP.AC i) •O: By(6)-1000(6)•0

I

1.se·

By• 1000lb. .t:Me~ o

~ ..

·

6

o,.. (.s) - c,. (1s) .• o

Ct.

=[2000(•1J/15

- Di<(1o)t B>< frs)

C• ~ 666 .1 lb.

a .. "' [2000(1o)J/ 1s Bx • 1333-'3. lb.

£M.,.c.8 ~ 0: WX>(10)- Dy (w) ~o Dy e 10()0

L:£J_~ ~~ ~=

.33-f>+!'

15

.30

lb.

-2.L

ao

~ a fL ' ~·· C ii! - · 20 · "'6

()

10

AC"

1!! 1374 .~

6'

l\C "765.(o lb.

Jh_• Bi<

6.1tl&'

AC -~ ·

·o

. Dy• 37.S lb .

~-~ 6

8.~ '

l'\O •

.5~:2.9.+

lb.

3 ~ .. 5001b.

""8"1118 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 .

10

I>' '' ' '''

lb ..

1-..(.M>.0)

I

·'.· /\D • 12236.1 lb.;

'•

D>< • 2£XX> lb ; 01:. • O

d111a. 6.708 fl .

.__ J____

"&,1

_L_ ,._J,~:

.!1-',-'0 3' AB = 2494-. .3 lb.

D ,,/~

/

I! J'

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.

,. (O.o,&) ~-

116

1000(3)-0.,(e>

117

..

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 in fr-ont of th'1s. plane . G:lmpu{c '

Isolate Righi Side Viffl'N,

·Isolate fr-on\ -View,

y

n:

"

the force in eoch msmhsr . d,.,e•rC0-3) 1 -1(0 -(-,))1 t(+--0) 2 0 C4i:1)

I ,

..tlllo •a (~c00-1ooo)(s)-C,,(8)

£Moii.c" 0 By(') - 2otxh) ·o

'

.

AO ... .Qx_

M

6'

8.1'9., ...

Dy• 37S lb .

...L&..· ~ 7.3$

·~

•

1

d.-.c • /(s-o)*-.(+<>) t (s-o) • 1a' 1 d>-o • Jte-o)2 -+ (~) 2 t(o-(-.t)) • 9.16!1'

O

Oy(S) -(WcXHooo)(3) a O

Cy:: 62$

. By • 1000 lb. ___J,k_

~ 6.70&'

~Mc ·

•o

1

e.367

AC • 76!5,6 lb .

6

r.o .. .522.9 lb.

/\B • 1116 lb -·- C

· The point(; B, C, II., D of the con Ii lever frorneworl'- Ghown in fig . P"-617 ore ottoched too vedicol wall . The 4
Isolate l op View,

617.)

!Y

d~ "v't-10--0)2. t(o-o)~

I

0 (c),,,q)

,,,I

dN:- .. /(10-or

7

f

Isolafo

(o~(-4-))2 .; 10.n '

t

co-o)" + C-+- o) 2

c;~

• 10:11'

C"(10)-1
.tMc.o • " 1100(0) -ex (10) - 8y(~) eo C ·":._--;-

;_,../(o,o ,'1-)

s ~ (10) t Br,(a)

1~1b

1.sola+e Top View, ~Mc-o

.001b

a,.(s)- ~(.+) -100(10). o s~

AB

£.Mc.o,e •

o

2'Me • O

1200(10) - 0.(6)• 0 o~

- 2ooolb.

...t:Q_ . _Qz_ < 11.66

10

~ 10

l\D • 2332 lb - ·-T

•

"1eoo lb .

1~00

-10

10-77

. c~

.AS " 1615.S lb-·- C

.2000(<1-) - -'f-00(10)-C~(il) •

o

c." .soo lb

118

e

:. ey

~ 6/s s,.

a,. (10)

f

-

AC ·1260 lb-·-,.

+

6

·ZM c •O 1000(-4-)- ~(3)-o,. (10) •6

sub,, lo 1

°" . 160 lb

6/.s 8JC (S) • 1S6ai

13,. • 1ooolb. :. By " 1:zoo lb. ,'. ;\e, .: 17.S.S lb-·-C B.i ·· BOO lb-

"

1\0 • 160 ~----S-1\0 ... 193,3 lb -· - T

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 /\.

61g.)

o"'e

10

AC • .538 . .!5 lb

s.77~

8-ia

----s-

from Prob. "71.. 610

_&__ · ~

10· 77

/\C • ----:j"2

£13600 - ··-(j}

~~...fu_ · ~·k._

.I.

lsolot .e Front V iew,

~

.=::Mo=O

-·-c

=

e.17s

n.,

d.-v

o"'o = 9.168 0.

119

= 12

n.

;

lsolote

lsolote front Viow,

Isolate Front View,

Top View,

~t.1c•O, -Oy(&.66) t By<'.-+), •o.

90t1Dlb

"

By = 6 .66/+ Oy

-· -©

Z.Ma•O, ~(...)-Cy(...)-()y(12.66) •o Cy(4)t Oy(12.'6) - 8000

•• ' I I

·,

- ~-

.

. •'

I

: a'

lsolote Right Side View,

..... :

! Cir

.iMc:,o • 0

0,. (10h By(3) - 1100(.+)

c

~"10

Qr -C..(10) t

"'0

s

P,.71!5

~

.+

;,/~ Bit - sutw. to© 6/s a,. (a) • 6000

/'C

J3.G 811 "6800

e,.- .500 lb.

:. By • ,6001b; f /\B• 877.!!J lb-·-T ; ,Bi>-400lb.

Cy (1+.120) ~-~;...;ca

'

••

/\C ·1201b ---C

10.7703

~Mc•O

0,.(10) .. -.500(+)+....oo(a)t 13600

D" "12901b. ••• /\D '\ 1~..+ lb ,

621.) /\

9 90.7'+!1 10

~. ~ - 25

;

10·7703

2000lb (0,10,0)

d"5 •

: ~- ~ 1"·"1+ 10

AP·~6 lb-C

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.

:I

d.-..e • J.o-o)• t 60-0)~ • (o-(-8))2

·1 jf

d-'6 = 12· 806, d/\C - J..&-o)at (10·0)2. ~ (6-o)•

8

•

. d>.& ~ (o++))'l tb0-0)11 t (o--O)e

10

/\C=009.07 1b -C ;

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. A

eooo

= &.66/,... ( ...53) ,;. 980.74.5 lb.

AB •10.56.a lb- ·-C

-0~(10) -B..(+) •Bz(.3)+ 1100(0) co

. /\0 • 1200 . g.165 0

~~

&

Cy =.!5 66.25 lb in@, Oy = -+/!!> (s".2.5) • ""'.!53 lb.

in©, By

1r- -6.-

:. By "

109't t

©

Cy<'.<4-) t +/.,Cy(12.G6) • 6000

B~(l.)tB.!(3)-1100(t)= O

00

•

+Is Cy - · -@ t;u~. to

in©,

c,.- 80 lb.

/\B .. __fu_-~ · ~

()y •

-o

c,..(lo)·~)t~(:l) - 3'400 .

Bit (10) i-By(:l) " 6000 -·-©

£Ms:O, Dy(s)-Cy(.ot) •O

llOOO

&z

....:..- ®

(0,0;6)

d.-G"' t+-11'2 ,

10.7703 ft.


o,....s ..

dAO

("""·o,o)

fl.

(S.&6-0)1 +(10-o)• t(o-(-s)) c

1+,1.....

2

!

fi.

./>z

'/

,/'

(9,0,6)

120

121

~

:[.

l&alote P.ight Side View,

Isolate Front View,

lsolote front View, eQOlb

I

Vt

Of

Ito Dy

zMc •o .J> -01 (,-zo) t OOO(f>) - Sy(1") --+00(6 -~o)t

:. By· 19g,g lb.

LMo•O

• Oy

gy

.£Mo •O !)

6/

C1 (-.) t 1200(2)- By(10) •

By(S) t C,...(13)-1200(9); 0

•O

.!I

.subs.
- 2400. -4-0C>Zo . -2001 .4

/\C .. _k.~ 1+.1+.t 10 1•M'ta

{?QQO ~Cy(13)(1o)-Cy(-+) • 2400

Zo • 0.997 '

: . C1 , 100.1 lb

12a:x? - Cy(JZc) - Cy(.+) = 2400

~Fy•O

Oy

t 199.'~.

-Cy(ao)' .. -9600

Dz • [-4(o.gg1)]/10 '· Dz = 39.66 tb

200.1-000·0

Dy .. -4{)0 lb %Me.•0

Cy~

:. Br

.

/\0 • J..t60) 2

t

~t. t(39.06)f.

1<0 • [200.1(el]/4<:lqf°

/\0

a

432.7 lb-·-C

0y (){o)- Cy(S) "0

~o • +1 0

14-1.+2

a

s

I

12.00<0

/\C 1+-i+!Z

10

/\0 - 471 .3\b -C

/\C

_6!L..

120.1 lb.

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

Dy; .!!512.lb . 32()

----:;o--

6

!

c~ 160.~ lb. . ~

~"' ~

tCz ·

320lb.

6000 -(a20)(13) =-· 360 lb . .5

Cr +DY t 0y -12
•

_!1;._.~- ~ 1·+.1~

o

0y(w)-Cy(4) • 2100 - · -©

St • 6000 -Cy(13) - ·- ©

-+soo- 2190.e .. o

...

_!'_4__ ~:--~ .. __

o a s' i;- - -- -

p

d~ ~ /co-o) t(10-o)'2

d.....a

t(o -l-a))2

= 12 .006

n.

11.358

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"'
w dOI'\ ' /(0-(:2))1 t(6-0)'l t (4-0)'l.

'c:J,.,c = 1+.1+£ n.

f (o-(-~))

2

t(10-0)2-

dAO = 11.3S6

t

lb-C

/\D "..501 .5 lb ·C

d ......c .. /(0-0) 2. t (10-0) 2 + (G-o),_

dAO ..

~ 452. s

.S12 10

do,.,• 7. ~ ...

f

('2-0) 2

~I

H.

2

I - .; , ' e_/~-;r' ,.<+~

,'•

"(-~ •.o,+)

.

ei.20 -.s) doe " /
I

doe . a.062 ... /.

~~-'f~--._

doc -.t(+-0)2.•(6-0) 2 t (0-0)1

""--._c (+,o.o}

.

11

dco = 1.211'

·l'• 122

123

Llv1& .. o

lc;olote Top View,

.,. if. [)/\ • 2..fOO b. l' 2.WO .. ~"' k 7."'Hl.3

1l

1331,3(.5) -

. .1YL +

'

Ax (g) = O

/\-;, ""' 7.39.61 lb. '

:. /\x • &+1.4·.5 lb ; /\y· 1QM.36 lb ;.A.z •1202.91 lb. IGOlote

lsolote· R1gh l' Side View,

checl<.

it OA 7 24QO lb

PA

7.39.61 ~

'7 •.o>tf>8

Front VieN,

CV- , 2 767. 2.5 lb. : . -this conno! be acaipl beoause

the hrri1f

.~

.,

""'" ,..• •'

: • .so1e c c.

,,.'

LMr ·O (By•!w)t

l\"f(-4-) - l3y(.e) - O ~. I

192+.~6(+) •By(!>)

By•

I

l

f l

Gt ·.

6

doe&

noi

ei<~ ..,

~.()61

-c1 (+) •o .

j g .5195.8

lb .

2

6

B

W •(17ai.9H)(:9) • 5191;.g lb.

7.~11

.!I

~

6

'°°' 1 ,6'

4

C>

11000

~ 1

zM...,e • o, - 100!>(..}

·,r DC· t"KJO • ~-~·2t_·.Q:_

Bz

P,(gh.·G"1de \/t'eW I

Fr-on! View .

"I

w(lf) - c.,c;) =o .

t

.az_-fu:'.....·

1'131.92+ lb.

6'

a.,.,sgs.30 lb; Sy=110b.1" ;ez·1+08-~ lb .

- C(4")1 100o(6) - J3(g)

B

.ZF;r eO, A

C1 • 1006.9 lb .

ot the r~h l side view,

c~ ~

.tMc · O, /\y(-t) - By(/;) •O

/\y ·• ..9/-t(17061G) diecl<. ·,f DA "'7 2-400

: .. C1 • 1331 .:3 lb

~ '223:Z.1 lb

o

L

2

9 B ~ 400 1b T

1000 - 400 - -400

Solve Prob. 62~ ·i r; 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

if exc.ecds 2"'1-colb

'/{(jd}(ld)

fl .

r • 'fe (?>){,¥> {_Gt Vs(,)) g-y .

:. DB I ~lb-

t

•fe(,)(~('z./g ·fi> )

1'. !J'

1 .. a ft .

125

124

=O

-400(~)-t 1000(6)

6!17,)

'

OA ·= 270-+.s lb oo ro1' occep1 Hi1s

r

/\ = 'l.00 lb . T

01\ ; 22:32. 7

volue ~

fC 3',

z MA~·o ,

t c~o) ~o

C ' 4
7.~

W

+

Z:M,....,, - o

•·,r oa • 2..+
·Ot

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 .

~(i530·"W16 • 192+..36)

Ct •

·,r

:. oe • 2068..56 lb . :' ../ ·,1

'1 <; 2<1-00 lb ·

ma><. vQlue,

62.6·)

check DC ;:>2400lb _QQ_ . '1731.Q24 lb :. DC • 2061·MI lb 7.211 6 /'it doeS not~

= 1s3g.488

OB

~

1539-~

checl<.. it oe 72400\b

a.oa

w '1 ll

E&UILIBRIUM FCR NON -CONCURRENT SP!\CE FORCES

.. .;-,...------ c.,.

.£Mc;,"'O

~ich

·,t

Right

tol
toke front View,

l:

c

ZM0·0 JJ

Vfs.w,

1.-

Oz(s) -0.(•)- ~(a)

, r I

A

J

~ids

l

•

3'

(o./+)(§) -

•' tUtM

12CX>(10/1) t -tOOO (+) -c(10) =()

eco lb · T

.B ..

Ch·

)'

zF1 • O, -A+12aot1oa::>-000- aoo

·o

t\ • 600 lb -T ,29.) Refer. to unsymmetr1'col .conHlever framewo~k described Prob . 610 an pa96 W6 ~ repea-ted here as F«J. P-620 .. If the '-'l"'tiool ,\ood of 17001b iG sti1f\ed lo o\ the rr11dpo1nt.ormerfl ber /\f?; cornpu\s the cornponen\s \he reocf1on ci\ l3 ~the for·.

in

oot

or

oes in \he.

bor'S .AC ...., /\0.

i

.

e ,,., ""

,,,,.,,,. ; ""

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).

0

,,r~-

:•' I

•

c ~~:'"""'° '. e·

1: ------

From@,

AC. [4-M(1eU/.ot

AD ~ 91.66 lb

IC • 12GO lb.

c. · ~·2s(eo).

Retell"•r'lg lo Front Yiew ,

in CY , S()()(-t) - 13;!. (3)

Z:Mo 0 6 Bi. (6) t 62:(3) - C..(10) ·, 0 s.. (6) t Bz (a) · '!-200 -- - @ 0

0x(+)- 8i'.(.!!J)

r

0CO

c

800

Sl!.c 400 lb.

.£Mc.• 0 ';J B,. (+) - Bz (3) - 0,.(10) ·CJ

in ' ©, -!100(10) t Sy (a) : 9350 0y = 14-SO lb .

- ·-@

1'00.@~®

. B,. (+) - ~ ~ 000 B,. (4 ) ~uG) ~ 4'200

.630.) The Boom BE

I ,,

of lhe .st.IT· leg

derr1c 1<.. shown in Fig. P - 630

IS rotal
h"TO'

f

@ Cr · G / 2 :

- B

9 .1(.5

@ 0,. • D~h.

t\B ·,, vedicol "'- '1s suppor\ed in o .socl
Ci _-c,.

.6IL.

f ront View,

.!2!_ • 6

EL •..Tu+

ec

2

: Oz " o.:/+

subso.@t....© lo © (o.,.f~· t C../2)::. -(c,..+O•}IO + +2.!50

o. +C•

.£1Ac,o • 0,

O

I

• .soo' -·-@

deo ·dee

·o

e~ (10)-1 By~) · g::iso - ·-©

.t.Me •O, • c 1)(3) - (c. t0.)(10) ~ 1700(2.5) co

(0

1

(o.,. • c 1)(3) - (c. i o.)(~ o) • 42.&0 ~ o- · :.@

126

·1

ij c

a.5 o,.-+ e.-' c.· = +2~0 g. (1o) +By l~) -11oo(e-M)

-+-~o 1b

ax'· .so0 lb.

1I I 1· •; ' "'~ I

eolb

;\O. [eo(0.1~6 )1/s

81< (10) = .sooo

.,' , ,

/ -~.

c.(3) ' c,. (+) . 0

subs. @ to@

1200(5) ~ 1000 (f,) • 800(') - .13l9) •O

C ,:;, 600 lb - T

o,.(G) -

c,. ... ·.s.2.!!J o,. -' · - @

~M,..·O

:='
.c,. (•) ·o

0u+: 0.a • 0•/+ ~ C• ~ c,.

11

127'

= ...[i1s·

11

l~olofe

lsolote Front

' t - •-

I'

~·

dei: s ~lf • (:20· !5)• , (10-0)• = ~4:Zs

R19ht ~ide View,

deo

Ill

2 t(20-o).,. t(o -(;-16)) 2

= ,Ao-(-10))

lsolote front View I

Koo

c

l sol ot.~ Righi s ide. View ;

•

I

0

·I.

0

1;•;-

c;;. I?< c..O

I

10

0,

c.,

~o·O, C,(1lJ)- 1~(s) =O

±"'c;.o~o,

Cy

/\y(!i)-~(!!H &.~) cO

/\y .i
= 1~ lb .

{fi5

9

G83 lb .

s

. ~

:. K

/\y(-i.)t Ex

l

Oy • 1~ lb

Oy ~ 866 1-·-© £Me·(), .

:. AO

-AJ.(1s)-t (c,.-. o~)(+s) t .soo(s.~)-(CytD,Xs)-o

"(c,. • 0,.)(1$) -o -· -@

. {77-5 subc. i.

o"

~

""M ~ (:z17.7t.~;~

~ 183

~ ®1

~ 103

-:;a-

• • o~

= 200.1 ,1b .

:Z27.7

1.sE,.

:a;,.+ 4C/30Er

10<;tJ(2)- (1 .~ !:,.)(~) - {)y(t)- E,x '0

Bu•: Ey = 1.s E,.

Bu~:

<4E•

e,,;;~

t

• e,.f0

Dy/:20; D•/10

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

•••

20,. · ·2()0() - ·

:. r.,. "6000- 0r)/a.

lb .

>.

Ey-=

-@

21::"

.s1.1bs. fo©

0)1 • Otl:z -.s1.1bs . ~o@

(160C>:z-!?.l..) ~

€y

A~(+) I 1ocv.l - Oy ~ 8000

lb .

The boom BC of the i;diff · leg def'ricl\ .shown in Fl(JP~ ; contained in the '/.Y plone . The mo~t f\!3 \G vertical. ~ rosfs in 6 .sool<.B-t ot I\ . Points I\~ 0 ore in t~ same horizon to I plone ·. 0 Point.G . D ~ E. in the .same vertical plane-· Determine theforc,e,s In the legG l3E ~ 130 ~the comp'.ments ·or the- beai 10' c

(4}1\y - Oy • 7000

6:31.)

lb.

. ·.BE c.51S.4 lb .

~ .37.5

lb

• "t-0/:?.o (31s)

-·-@

/\y ( +) • 1000-t..SOO

Ar ,. 1soo/ + :. At

-'-9y, " 70()() )

UGin.g E,

4

c

1P,75 lb .

10/15Ey .. 14<(5(37S).::: :zso lb .

05!. • [60'.)(101Jho =- 2SO lb .

ring 1eoct'1on o1 /\.

reter'f'ir'lg

to l'.l.iBh+s;de, View :

A.,,, • Dz - l=11t

Az. •

, I

129

Z.F1 • O,.

~o

I:~ - Dz ~ Z-50- 2'50

: . Ai';•

128

=.WAo

Dt • «>/.lo Ey

~ tAx (:z) -~· 0000

- (Ay.!_~

c1000 :. El'~ 2!50

Dr = .soo lb. BD/~ • 5Ct>/20 :. BD • 612.4 lb. U~ng@ Arl"') - Dy • 1000

... ,.._, ...5

ore

·IOOO

2

t.J.sing ~y·1..s~ < 1 . ~(2.56) Using

'1000

~ 1000

~/30(H,Ex)

BEj.J1Ts

/\y(2)-1 A .. - Oy/2 • 400010

Ay(2) ~

t

-t b

Ay(+) t A.,(:2) -0y •00()'.) lb-~

15

7\Z • 166.7

Dy (1) - 40/1.s Ey .. 0 [)y. • .ote/3'> lfy sub&. fo @ @ 2E1 t Oy ~1000

100?(10)- Ey(10)- O.,(~o)- El' (-') •.o

:. c,. ·:Z27.7 1b .

"u'@ to

[)y(1.)-1o/1S Ey - Ey (£) "0

A.l( -Ox • "\-000 - · -®

.ZM,.. -o,·

, 61 iv .

•o -@

Ei!j1o ~ey/1s : . f:'i! • 10/ts Ey

·o

Ay(lo)t /\>1(s) - OJ1(s) • 20000

t-.1 (1.) t

AO• 202. .3lb.

D~ .; ' 61 W/5-.. - 61 lb. , I ~ ~ 227.7~/~~

0y(2.) - E~ - Ey(2.)

0,.(10)t1ooo(w)- l\y(10) - Ax (s)

'83 t Oy = 06G

Ct t

A•

.tM... •O, Oy(10) - Ez(.s):- Ey(10) ~o

-·- ©

~s·O;

C,.(5 ).t Oy (.!1) " .+?130 lb

-/\.. (1s)

~<1000

10'

/'
---~r~y-----

~Mo · 0;1oao<_ro)-A.,..60)- Ex(.s)•o

.·. ec " 1~ 10.

15

Cy(s)d)y(s) -soo(s.~)·o

'

•

.h.:.

... "

o

=o

The boom of the c}erricl'- shown IP Pig P-f>32.

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

c;32.)

&(•o,.,.o)

·w;

.

ir.(~,ll,·!>) ~z • smao"(to) • ,!;f} . I

0

1- " 10 COS 'a()

fltl(X)Jb

i~.+s· -1tJ/;;:D "'n'°"• -p()/t.o

I

:

..

,,, 1 __ ••Jfli

-~

'.I.. - o·--71,. " ~·

0 ·-

- - ----

(10,0.0)

(0. 0,11• .,..)

ton fiO"

8.66.fi .

.·. 00"17.32

n.

i5A • 10 n. 10/CA.. :. l5A,. :zo ft .

..17.3t/6A :.

C.CSE()' ,.

·

•

:. N5 .. 200.

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.~<;

dee

·,/60-o't+Gi:>'o)'it(o-(;17.!Jt)) 2 •

26.~& {1 .

~ 1s.~ fi . -oc .. 1otoo oo· . 17;~2 n.

0C = S77.3~6

160iote fron\ View,

t.&.'26

:.

2a

ec "816.3 lb. Chop tcr

7

&, CentorG

or

'" Eq.©

Cy tOy•1132 ~ • 1732- 577.316 " 11.s+-'&+ lb

UCO I

: I

co9

I

s·

°' -__1~~-- ~--- : Cy

Ay

,·

f&,26

Refern~ to F~n1 View: ~f,.,, 0

A. - c.. -o~ "'0

• 0.

-A1 (10)•2¢00(11uc}~ o

"268,i:i?i8 lb.

Cit.• .&77.31G (17.~2)

1·

• sn. 3-'v.l lb.

I

Oz· 11~.66+(17.:32)

Q .. ,.99.96 lb

- ·-@

Oy • 1132

o~

'LO

- 2000(a.1;~) • o

I

0, • 11s+.~(10) w

A,..-~ lb

;t"tv1.-. ~ 0 (£,

Gt t

c. • sn.3~(10/i~)

Ax .. ue."5a +.s11."M1Z

Ay • 3732 lb

(Ct• Dy) 1 0

'}.()

.

8.'40'

Or ~Mc.o.o

G rovity

••• 00~1632 .7 lb .

BO • 1154.60-+

.~ -

Centroid<>

I

w

loolo+e Righ~ Sida View,

Dz

,, ~-:

=99.9 _qr; lb .

Rel. \o P.'ogh t ~ide View : £:Fe ·O

"

Dz - A~ .-cz ' 0 Ail. ~

0 De Oy

__.. _- ---------- c,. "•

i?-32 .

A

l\y

G1

a

11

999-96 - 4:99 .96

Al-

a

.SOO

lb

C.

17.31,

~tvlo s9; ·3732.(17.32)+Cy(~.~) 1icn?(na2)=0 Cy(3"!-.c.-+) • 1999S·Hll'.J

130

131

cenko1d' of the shaded area shown in fig · which '"' bounded by the "'/. o·i1G', the line -x •a ~ the po-

10s.) Delerrnine t he

p- xis,

707.) Determine

.. r

shown in y

robOlo yt/• KX .

J\ •

f:

ydl< •

a

0

fo° b ~ dx

"·r

• r-!Q r~~~J:

Ay • ( Y~ydA

~ r~ ':j~dl< -g,a•b -~ ~ s ~ ' y: =:30/.s

Ay. -=

c

ydx

L&t )( u

=

dt "

..

hr b::

f bfoJo"-y;~ cJx

b% f ~]~ ·

= X/o

a

/\·%f
- ob

r

- ~b

1 • Y2 [1 t

% [-Yi_·"% (o"-l<") o/o [-Y$(o). y, o~]

9h1:

M .. f:}'i !l(lJdi<) • ~r y'dx c

COG20<]~

r Kt (i''<::-x.))

- bho"

[O( t ~i:h

.aj?-[Th +(~ - 0 -~ ] 0

a

" r)((%J~·-l<')dx

Ax .. -sac." ~x 3 + y: - 40/glr

OCOS"<~

- b/t11.fo1'/c o.. cos'o< do<

clx

.,.

Uo4- ;-:') Ax "kl<(l_:ldx)

s onO("

1.

)(~a'l

tJ "%

• ••

c

b•

.

1-

ocosoc·~

Y~f: ~(ydic)

c

O"

dx

0&1no<

£.. +Ji..

b 2 (o• -)1 12) o«

b

A= ~3ob

=

~%~

rr

y - t>JX,4a

- o/-ro: r~a&~a - ·oJ

A'1.

equotion of' !he ellipse ·,s

1./ "

• b%

y'Z" b(2;y{:, -

of the ell ipse

.

'(._.,

~

K-~%

0

Fig. P-707. The }(

!lr• KJC

'/

f he centroid of lhe quodron t

dx

ro.•,. - )(~1:

• bfao" [o.9 - o.%] - o.b,h

0

~ii q~

-

y- ~hlr

Ac ob~ 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 r

area~ ,.., Jocotlon of centr<,.ld for -.olues

o

r

f

~

AX· xdt\ ~

"

(t>,h)

S: x(yo><)

or

n t 0 , 1, '2, &so., 3.

~ · l'-x"

""·"

~/1-n

= h/bn

I

~.,.hx%n

I

·I· V

--::;:--

j( •

~

yg r'* 7j"' b(nt1) bu! !hisx nt2

= +%.-

ie

reffered to(o,o)

132

133

x=

b

_ b(nt1}

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

- b(nu) - b ( nt1)

nt2

n t2

f:-r(6X9) + •

~T(-a)!j .Y ·[~l')(9)('/3·9il ._[1'2lf32](~ t9]

4u+y = ·307.23

!J

x Oj

c

in .

7.+7

[v2 (6)(9)('f3.6)]

-41.1+Ji •

+1.1+ )i ·j

t [ '121(3) 2(35]

" 96.41

=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 . no.) Locate the ceni roid of !he Of'eo bounded b~ the I< oxi6' the sine curve y = os1n T">{_ from x " 0

26.56'6

r.,

=66

!j

y • 2.+e in

to :x =L- ·

26.soGx ~ -6(-+) te(+msao·t-4) 1cs.~i

•

· 'f: • 0

::·

;.. ".C

.!Jdi<

;\ij :

f>

12

~ ~o

f

~ o sm'!"~ dx

l.J LJdi<

'ft (

"J-i. ch

Cs 1n!Z.Tii. dlt

- ~a« Yr [11<

., a 'if (- caGT~)~ ,. -ol(, [ cosT9b_ ~ co~o]

~

~L

717.) Locate the c.en+ro1d of the beot wire shown in rig. P-717 . The wir-e of uniform croSG Geetion .

'is homogeneous~

..

••nso· • !Z.065 in.

;: " rs•rict ..; a ~ :aoic T/t90

- s1.,211xlL '1>

L

35,7129

1.3-4- In.

•

s1nao·~ ~~in [ 2(2J1.3J1. (.30°.><

~

•

1.+a~•1n .

8.293 y = 9 .996

- -.91_(-1 -1) l

.g = 1.086in ~~ .. aC.J,.

T

-'ij",,.lo

a

t

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~

-2.

J'/ieo•)) t t] ~ · [2(tx3l<(adJ1.•l,W>))(1.+a2~

T

/\ij=~A~

[1 (9)t 1 ~~1g~ .:· ~(B)(4t1)]+[1 (,)(o.s)J

4

~ .. 3 . 07 i11

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]

72 'i ~

t['l2('X')(r/,·i;)J +[~~)(,)(9/a ·6 to)]

= 360 .. sin. ·1

.1

135 134

I

no~ Determined

the centroid of the lines t hat f'orrn t he boundo-

12.a.) Loco~e the centroid

ry of the shoded o r eo in Fig. P - ne. [ 1H ,f18o t6· t

ffe +.J'7iJ x "' t

1:2(6) t

J100 (-lift?>h)(o/~) t .Jn (.f%)(V.fT)

.rnt(. ~)( Y-l2) +6)1

[Y!2(1.s)(G)

.

_

s

..

. 40.30~

5-26

g = 12(G)t6Cahf19oK~)(Y.rs) t 6] t ffe("!%')(Y.ff) i ffe("%)( Y-11)

24.43142

of the shaded arooo in Ffq · P-120 '1s required to

lie .on t he Y o.11is. Determine the dist once b thot will fullnll thi s requiremE.nt.

r'ig. P-n-4.

value shou ld ·the

1+4,!593)( 1f ,

~.666s ,,,~

or the f\onge

width

t 2. 5

b "

40 t 0·8 b

T;t_l.)(

•(¥~£)(~~ - (~<2 )(%)

.IC.~

_:rv«~ 8 !l

~!J B

..

T~ 3°

" .9~ 1

r% - V-'# c

6.57S

r~

>t$..6 ~ •

relative to the Y

OJCiS ·,~

of the '6Clme fbrm

OG ~

e

20 y~ • "' f-2 <1~M _'f11(t>)~2~j "f%(111)(~)(3J~-1'1~-r~(cx11x%-1~J ,

12-,. -

o

111'

. )<

12

·

-~ -~ 12T

-!57.6

x • +.at-1. ~ =[%(12)(c;)cafa ·"~ : [!111 (4>)(H1)(Y~ .,5) 1-ig •

r,% - rX2 .

e(v~)( -t~-,) - (irr,Ya )(rlz)

vr,% ij

01<

t<-x 11 with respeot to the X 011 is .

by cutting o semicicle of d iorneler- r ft'Om o quo.-+er c irc le of' r.

B

Jr • s .oa in .

the curve y2 •

b-1oin.

_

:g • 7.;Ja,76!l

nts.) Locai'e t he centro'1cl of the shoded oreo enolosed b~ t he curve y 11 •ox'!+... the stroigh~ l•ne .shown in Fig . P-726'. Hint : Observe thof

72~.) Locate fhe centroid o f the shoded oroo in Fig. P - 72'1 crcate::I

[ J~C

't/11-)

~(12 - 1jc;>)

1-t.+.593

be chonged so tho+ the

2b '"20

,.

1118.531 iri .

1+4-.s9ag; 12(1e)(c;)- '111(c;)(c)(o/s) - v(+)"(12 -

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 ,

e

j( ~ 7 . 7+

".

T- section of Prob.n+ shown in Fig . P - 7•4. To ...mot 6 in .

T~)~ i< • 111(1e~)-~~c;) (2/~(G') t H)- T~•t(
..

b "6.53 in . 121.) Refer to the

in.

- - 12{E1) - Yt.(c)(c;)- lT/.f (a)«(va) -

--tr-"'-.-TTT77'77i'-+-

in • .. 2 in b 2

ba "

2.69

~

n-+.) find the coordinores of the centroid or the shoded or-ea

shown'" ,

[ ~ (-.)(eiJ['lf -aJ c[MbX6U[1/!tb] 95 . 333

-y : 6'-9.599

y•

in.

·

g ~ +.s(s)(a) t ('h)(1.s)(c;)(2/5.c;) - Tr(,)""(r; - ~)

24.4314-2

49,397 if " 261,659

no.) The cen t roid

~ + .s(G")- T(~)~V:., Y11(t.s)(<;){Yt(1.s)t+.s) · +(+.s)(c;)(;.~s)-: v~L)(W)

x • 3.o+ in.

in .

g - .s.+1

the shoded area in Fig. P-723.

)( 24."431.+2. 714.15


Y: "'

or

y

. ·1

I

3c;

g : .::ifl.

" o.637 r

121.) Loco~e

•he cen~roid of" the culvert ..shown in Fi'g .

tegrot1on '1s unnecesso r~

·,r !he oreo

P-727 .

H1'nf : In-

·,s subd'1v;ded 'i nto 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)(~)

- (~Is) (-+ )(1a) (g/a ·1>)]

-j 6'

'59~

c

204

.H ·a.sf'!.

6.57~C

(ff

136

·

II

137

I ·1

[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 •

731 .) Two 10 in.-15.3

fig. P - 731. 1119.) /\

rec tong le jc; . divided into t wo 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 gA . Y _ 61:...-en th:lt y,.. -I n_t1 ) h .. j..,. • b - ..IL ~1

\,.40t2

b(nt2)-b

nt2

~" = (nriLb · [.

~

(y(7n1).\ ~·n1-~..} n t«} )

nt1J

to~1hel"' os ;shown In

ore welded

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.

·r

2 (4A7)g - +.+7 ( 0

1 10) t

+."f1(s)

e .91 g ~ i;7. .s0'+

3- 7.s& in

nt2

"I .

=

find "Xe'ii_iYe b(n) - bhl~e"' bh(ttt)-lbh

_ib chonnsl~

7.a.+ "1n~

73fl.)

A

in F.'9 ·

bridge

0

P - 7~ .

16 ccmpo.ed

fnJGS

or

the element Ghcwt"\

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 .

. .. we - ~ - b ~)

2

2nt1 2

- bh%.

.:hruLg& -~bh'-00

~&

:

h(n•1¥nt1)

• 6.,Xo.r.)(.a.2a)•(~"f..o.7) 27-7GS

~(ent1)

(n+1)

[111t3.7.5 tll;(o.e)1(~)] ~ .: l!l("to.211 - 1.,5)I3:1!l(o.-,s~o.211)

ant-i

~e • 2~,;

g = !ZG!l,
E .. 9:~

in .

-,.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 rriox1ini..1m ""heor11"19 st~~ .

I 138

139

:j

I

1M.)

/>. right trion9le of

;\~ "' (

s·1des b'--,h ·,~ rototed obout 011 o)( iS

,/~ ,,,,... .. .. ~

• b%« [

Yit bh

- ; ··

- .o.·

~ 4 dx

Vff'!.Ov.n. • •

0

2

x-

J:

~T. 'ij • I\ ~ -


• Tab

~

V " 1-Tab~

x%

• b%"-[o'-o% ta 1 - a%l

= ViJ Tb.rh

Y

,' h .

r:

-): b%·.Co2-,a)dx

coincid··ng with side h lo genef'ote o right circular ccne . y : !2tr . ii . AA • ~i . ~ b .

~(yd~) .

A~

.

De~ive the e11pressiorlG fur the s ...rfooloreo ~ volume 9enero· tod by rotoling o semicircle of ,...od ius r obovt it<> d ;om&ter .

737.)

=

b~ [

40_%]

ll!fh ~ : \bf.b.. ~:~

.!IT

V " alT. ~. Area of holf' c1ivle -~'f · ~. T r 2 ;81;.

-

b)

g._

v • 'Vs Tr• A • 2T. ~ . Le.n9fh of oul'Ve

to one-holf'

Yor """"bol.,;d , 20. ~ . " .c'eT ..::!J2. • .a..bh 'Q.2

Y=~f.a .Tr~

v ~ 21f·r~a

'5.

" .. ~ bllyh De ter~med the volume of' the e llipsoid of revolu tion gene.rated by rototinq on ellipi:e abovt o) ·,t~ m0.Ja- dxiQ (pr'Olote elllp,oid) ~ b) ·,js m"1nor 011is (oblate. ~llipsoid . Tot--e the larger .sem~o,.~ oG o

A· 2J . o. 21Tr A= 4lf
7SP.) .

o.)~ · r%-.. +~81 I>

i

···--- . • __a .• ,'

A i=u.iRS;i;

140

M1.) /I 60. pipe elP<'w hos an irifernol d ;o meter ot + il"l . th d of curvo t.L)re o /he p•pe ' .~ oen~er l1"r1e. ie 6 in . f ind f he i0 ternol e ('()vol. ot ths e lbow · V ~Areal\ centroid ~d1stonoe teverGed • lr(~yz x "x (6<:>' x T/,Ba') V -= 79.96 in:'

r

.ii,., t~ Smoller ~em; - 0><1-s os b.

•

lrob

141

•ii~·

or

7~.) find fhe volume 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~

n .~ oreo

7'45.)

of rodi i 1.s

contoined between two c.oncer'ltric Qerr11circle. 6 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 .

y • .J.r!._

[T/j! (~•- 1.s•~ '1 = ~ (•)' fW?J - %{t·!S)2f4,.(~e)J

6

A = l~

7.of
Totol Gurfoce Area • A.1c.a.- T3 .T r 2

¥

t

lrit

..surface oreo ~ volume 9ereroted by rota~ in'3 in Fi~. P-7+3 through one revolu t;on obovt the. X·oi<"

the

1\1~. ~~ t [~ t

!l ' I +' !!

l i(), 'O?!

c 1.485 Y'" 2T.(1.+8St+). 10.Go:3

v

· .365.1- in~

Oofet.mirie fhc surfooe oreo ~volume gene.roted by o complete revolubon obou• OlC i s of the .shaded .area Prob 719.

'

~~Gin

TMt H

2J~·H.. ~

.

= 2r. '". 7fl

Y

,..

V • 1667 in~ = T(-+)(

1{7.'
1j-;l ti;) 1 2 (') t 2 (~s·,,• .s1ni;~·+J

7"4-7.)

Aa

('C.volutio.n obouf the )( a~i~

v

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. rle.term1ne. H1C 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~

V e o.+ess rl

1

!2f. S.+1 .

cp~plete or the GI-laded oreo of' Prob. 72!5.

779.92 in.?>

m

] for length

r3

i

.

lit.!~ t f•s,H~ ~ ~ ~6.s)t +.S(•)>6(') • ~1.st.,.,,,(J1.i:1&' )(~) .

19.997

t

T(%

_g =3·9+ in.

i '

·1

238 lb .

142

(6-~)

g ~ 78 :a2g

w • o.4BS.S ( 490) w"

(1H~tJ4•t1fi ti'~)

A= 2r. ~- le~lh of c·1 rcorrcribina,Areo A • 2lT . .3.91-[-., (11)1 + 6 t .s ~ ~,•nscJ A "" +92. s7 ·1n.2 .

n.

of'

= 2714-:!S 1n ~

]" •.s.oe 1n . v c !Zf. c..os . [ +.s ((,) t '/11(,){1.S) - T(9Y%J

1lf .s.7. 27.99

i

weight

/\ •

A " 16+
.

9- .s.7in.

,.

..Lll<..- - , - - --___::.L-.::"

2(21.%")

~ =.!!'l in

v • .2 J • !3 . Areo

:<>1. 6"'" _g - 1ai. 1:i2

/•

or

•he )(

t

yf 1l1 . 6 -1.5

in.

7-46.)

1.(;l] E • ~ F'~· ][-~<;) + ~ ~[~][ 2~l ~" 6.15 in .

g • 15. 75

~

143

[-.;,, (6)•(1,) - T(4~..)j"~

7 s1 .) 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

.3

r

~

1.sa)

A

%r

v-\ ,. " ~

"1"(-+)(~)t e(s++cos:ao')

:26.S6'°"j = 1+1.919

[

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~ olf d ude h . .

ffi {J)fh(hf,-)

I

.'. )( " 0

%

lr (1)" t

Q.

in,

of o steel rivet having o cy-

.()f,9'

WsfoeJ

,.j, of

T{0.9)~(t5] ~_ •[(o/1)1f(1)ry[1 f %~1il I T(M'f(«)(1) ~

"

1 -~~ in.

·

·

1' so(16 - 11) ... ~ ..-s in from base-;from 1'>-ob.755 't

fhe- cone. . Loco•e fhe centroid of the

result of Prob .

~II: [413f(4~11 ~}~] (-.9<) lb/ff 5) Wss. ~ 76 . 02 lb.

.

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

4-

Wtimber

"[r(sin11 1~io\) (Mm xJP"Hl)(100Jb;h') 1

Wr ~ . 109.09 \~

3.66.5 ~ = 6 -545 •

. . . . .&·

= '1/11 {#(?'!~(2))

hllp ..

heod of 1 in , radius . Use ihe result of Prob 761. fJ'2•

I.

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.

lii:idrico I body 1 in . In diomeler ~ 2 in . long with o hemispher10ol I

2(1)

h - 3 -'f-64-ft . .

as+

Locot e the center of grovi!y

..

h11 ~ 12

e

z • 1.ZS

iri. from ope,.

h .. 1.-+1+0 .

- - -~ff

-=

;l6. 664iz

h1

-~

x •.s.:s+ in. T(.+) (~) . 26'.-56, y ... :!32 :y. u04 in. ' ~6".'566 z 6 (3) t ·a ( 't son 3¢")

_,,~ ~66"6~

1

= 11

,:r'(ef°h(h/1) - VJ1:r'(~>•C.S'1S(I))

uniform wire "1s bent info fhe .shope .shown in f19 . P-762.

( 6 t11(•)tg)x;

11/9(,;)Y16)[34(1,)] - [T(+1(+X16-25]

or

The .skaig h~ segment~ Jic in •h~ }.-z ~Ions, ~ the_ 9.i n . length mo~ on angles of :so' wifh the X ox1s. The -sem1c1rculor -s~. rnents i s 1n the x - Y plane. Locate the oenl er of gro-'•+y of' the w ore

7 5+.)

j"

•

.. ++~3 .:;If;

)I

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 grovity docs riol lie obove the 'j.-'1- piano there will exist o restoring oovplo whon the bod~ is tipped.

2:11- ·'l[~l y;

402.11

796 -)

:-tj

r2.. -

•

ne-t

( 7(;.02 t

109.oe) 9 18S.1

;

76.02 (2e

Y,r) f

109.0B (12

g " 286.46 ~"' 1.ss

ft. from bose

x ~e)

ii I

I

I

I

volume . UGe the

7-f.G .

145

144

x

·.

SO-..) Oe~ermine tb moment of iner\ia of o tr"longle- of base b ......, oltducle h wi\h rcsped \oon Ol
-~· . ~~

--

I .. bh8A

•/»h

b

~) Delord•he the moment of in&-tio of the guorler 61rde ~ 10 f(Q . P-805 with res;pecl to the 91von OKe&.

f p'dl\

:I

·(ff"df

.I

J•

J •

J; p•(Tf/i·df)

o•r

.J " .L & I

Chopto: 8 Moments of Inert io

J

o

I

~ J..r-+ 6

• 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

f/ &1n'~de­ • r4/.+ J; Yt(1-cc.s.i~)de-

I~ • r~

J~

z

Iy •

Iy • (

f

0"

• r'Ys [-e- - s11·1 2&/tl: • r•/11 [(1-o) - (4>in2lr-s1no)] '2 Tr"';/e

f ~ dA 1

f'cas~ fd
. ri"er~ - ~J

it

0

~ r'"/e [cr-o) - (casalf -coso)]

Iy ::

146

lfr./e

147

.

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

\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.

I,.

c·i"

tAd

o.3927r• •

1,.

~

J,. +·lfr:(~)~ 2 3'1f

T(

+ o.2e2..94r

t,. "' 0 .10976 r 4 eoe.) Oeierrnine the rnomen\ of

I"'

0 .11

inel"'ha for \he quoder c;rcle

~

- (o-~). '" 4

)( ; !

~

'k

•(l'f/A)

11.y

A•r yd>< =_1'(4-l(~)dl( A

I,.v]C

")(%

wheny=O

r~

shO'N<\ in Fig. P-sos w ith respect too c.efl\roidol X- ox1s .

L

- (y-"") ')( 2 •

4

-+-/

y

0 1

1'

a10.) 'Determine the moment of inertia

J:

-[+ll -'ll•/~

A = [...{2-0) - («-o)~~J

1G

••• Ky .. ~~V(1,/3)

:.~ '4-/.s of

011.) Determine the moment

of the shaded porobolo oreo

'1nerf10

w/ respect

to +he X

oic 1s

sho'f'm in Fig . P - a11.

=f y dA

(a-i<)dy j )( •K/ tc.. z X/y''f. •a/t:J&"'( a • afb• y ~(a - o/p"y .r) dy . =_1~[0y"- ~a(t 4)] dt 11 ~ [ o/bt(Y% )]: • [ )e] 2

I)

dA

·S:

c

oyx -

I,. •

ab3 - ob% · - ob_%.s (s-a)

~

2ab){s

91~.) Oeform•ne the ly for ~he ,shaded pGll"\?boltc area

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:

hX0/ro) ( a - o) h. 7

-(2

~ ~

b/2.

148

1

'2./1 (o/ra) ( o ~)

"Ir ' (2/7){o-'b) .. 2ct'ly;

149

of

Fig. p -811 .

i 'I

I

e1g,) Determine the moment c£ inerlia of \he T - 6ecfion in F~ . P-820 w/ ~~ to ·,tso cen\ro1dal o*. •"

l'ITY •£Ay Y • 2(0)(2+...)

Sfl

_!So

?

t-

ShQWrl

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~ 820)

J·l~ +ly

•~1' +~ .. 270in~

"'• -[i7A .. ~21%ffl ..

1• .:: 290-67 in:+ Determine the moment of inertia of the oreo show'n

fig .P- 020 wiih respect

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$0'\ion of0 ho 1\0W tub.9 wnOSe ou'\.$ide dlan'IC-tsr- iG 6 in· "" ·1nGide d°1ometer ie-t~ .3.07.3

to ·· ts

cenh~idol

•*

Ar " 6(1) H2(1) H2(1)

.

30y .. 12(1)(0.s) t 12(1)(1) + b(1)(13.~)

12

1r.i

oi
Atr • SfAy

y :: ~ .7in . -

lit

.3 : 12.(1) +12(1)(s.2)~H(1't +12(1)(1.3i 12

••

1

+ ~ + 6(1)(1.st 12

• T'/tJ.(r/-rt) t\"1&.11 ·1n 2

J •TAz(5...-2"'') • : K

"~J/,.: ... ~~11~11

102.11n.~ -= 2.ss in . ~

021.) Find the

rncment of inerlio

about

the indicated X

r., • e~;ol3 + e(1o)(s)~ . ~

r.. 1

"

Q')(iS

\

for the shoded oroo .shO-Nn In f 19. P- 821 . I" = f. t i'\d~ 2666°67 in ~

26~.<07 -1u;o ..S:3 :

g°",14 in~

02.2.) F1·n d 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 .

+ r;,(r;,)(o.st K .:: 6 ·7.3 in.

r~.: lu tl~ : IJ71.-a3 +211..33 =.5S"1'. G(;°1n~ JT •J1-J<1. ~2730.~C-..!!54.~<0 "' .21 7~·int

150

I,. = 190 in•

161

12

I 1

i'\c; base b hon:wn~ol. ShoH t hat ~he c.en\r-01dol rnornen\ d" inerha w1\h respect h hori-

923.) /\n oquilo\ero\ triong\e hos

:Wl"l~ol ,._ vo
:I1

b

12.5(6)3

t

~(HZ.!i)(<;)(
r)t

~

il
t

Ille ..

t ..c

o.Wb(ti/~t .

1iZ

t

of 1,, 1-,. 11 .

cf side o .

= 1.9o+it10' b+

ton ao

Y-n

•WLYh =t>Mlh

b

c

c

I•

5

0[2(0~/t)]~ -t ( 1h(a)(o.JY,4) ]4 1~

AT

tonqle shown ip fig . P-82S 1 .Oclined ol on ' onqle-fJ- "' sin - ¥~ . Hinf : ReE>ONe the fiqure


.38Y •

Y

b•7•.5

in.

d • 1!1 -7.5 = 7.5 in.

.n.

h = c.osa6.87(7.s) = c.oG .s3.1a· =

a

fx.; -

12.!! (•) 3

t

'12

2

a/10

=~in.

(12..5)(<> )('15 .,) = 225 in~

~

152

in.

~

2 • [1Cs)3 t 1 (s)(o.-t7) ] 2

n

lir ~ 37G.~ io ~

u,;og Tobie Vlll- 2 Proper-fie, of Gtrvcturol Section.2 9•,.'"•,,i •: j,. - 38.8 1n~ ; y -1.G."? in . ; A~ 1~ 1n.

!,,

~ 36.97• = h/d

12°(1)(G) t[8(1)(11.s)'t .S(-1)(9.5)] 2

= 9.g7

I,, .. ~(f,, •Ad•) = 1{12~ H2.ft)(t.97f +[~ (1)(2.53Y] 2 u

ton 53.13 =10,tb

~[e(1) t !S(1)]2i-12{1) " :39 in~

A,y · ~ ~Ay

the 1oin . by~ in. reoobout the X ax1~ to which if is

K. • 1ojs1n53:t1\ " 12.s

tf

-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~.

toe moment <:A inedia of

ifl\o parls A. B,

>

A,. .+.s(:a/.+) +[3.is(3.4)]2 ~ e .62!> ·.n~ The built - up S«:fion shown in Fig. P-627 is composed of

,.•' 825.) Cornpvte

4

i 1 =iis."M in:4

1/2 (o) 827.)

12

iri

= +.s(
b/o

h "' a-1!/2

o a

.~~

8'0/ei -= 60•

:3GOO

i .... +2.11 in~ ly = ~(fy t Ad 2 ) 12

-& •

0

a

.. 3/+(+.r.)3 t (3.5(3/4l t 3,5(3/•)(2.'925)2] 2 12 12

2

3

o .M§b(l:Vt) 12

momenl

Compute lhe paei;ing through

900

i,. ~ zO~ • /\d~) .

cf 1iierlia with reGJ)CCt toan ax~G two oppo&ifo opeJ!eG cf o mgu\or hB"
ez+.)

24'75 t 22.!i ,i-

0

36 ;

2-.1-' in~

3li

826.) The. cross GBCtion shown in . Fig. P-026 ·,s t hot of o sfructurol member "'1\oWn OS 0 Z SeC~ion. Determine the values

~

.

f,.,.. •

= 1 (12} 3 t12(1 )(2.
z

12 4&

in.

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 c honnels 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

________ i IYo __________ _

from Table Ytl l -2 Prop. of Shvchm:il A!°'Olys'1s

t. ly x

Olcn"lel Size /\reo 12'20.7

Iir

~

l,.

-(1:2a.1) 2

ly ~

__ J_ . -

(,.03 128.1 3,9 ().1 0

!y • 3019-63 1n.+

r.

2 12e.1~ -~3-9 +6.os(d/2 to.1) ] d = 1.~a in 929.) Oefermine the d 1s4once d ot which the two 3 in. by s in. rectangles shown in f i9. P- 829 should be spaced so ihal f,. • ly .

•

h =

...;·· .

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~

f1

<

.

7601:99 ·.n.+

2

[ 14(•) 3],A2 I [G9.G t 11(3.!!S) ]4

!y

c

3021 -1 ·, n~

f.-+

Iy =1.

3

j,.

; [ 8(3) + 6(3)(Glh t 1•!>)2] 2 •

Ad?.

;.[3(S)3 +o;~~[e(a)3 12

+S(3Xcl/2

12

tu)~

plotes riveted to two 12 in 'J.0.7 lb channels. s---

cl"' 1:20

Strvdural Geetionc;

11 =3. 9 in.4 ;

~

!,.

2C&

in.

too web plote 2.3~ in by S/16'1n . to fQNT'l the ~ ·in fig. P- 030. Compute tho volue of i •.

-

U~ing

in~

t

.

1~

Tobie v111-2 Pl"Qpertie'? of Giruci.,rol sections

•h : Arco - 4.7.s in 2 = 6.a ·in 4 ~ I 1 •11.+ in 4 ;

6>\4>'

!,. I. " [ 6.3 t 4,75(11.01)~] 4

t

y· o.99·..,_ ~

ii... or:igle column

·1c; composed

1.99 in

Prob. 826, ore riveted

1o o

12 bt1 in. plate to form the seoi ion

12

12

.

[ +.i;(a;..,.)9 t +.!>(::i/+)(s-~:25)2 ]

' 3 a/+(3.5) t 12

+tr

3,4(3.5)(7)~]-!-

3

[.sA6 (2a.s) );12

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

in:+

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

2

f,. .. 2G
2

833-) Four Z bo~. eoch hovin9 the Gize ~ properties determined ·,n

2

'~~·~t t

6(0.5)(11.1s) ] .... [o.s(a.!:>)'5 t o.6(:a.s)(g ,;g)~]+

r .... :2667.5.5

11 e(3,9t~.oa(+:i) ]i +[1C~~)3] 2

Iy " gs6.87

y • 12 in. 3 f!JA61''1.3.s) .. [

1

J. • 1610. 9 ° 111~

12 •. !O.Tlb

ore connected plate~ ongle girder-

I.. ..

/\reo •G.os 1 n~~lx = 120.1in y • o it._ il • o.1·1n . " 128.1(2) +[%~) 3 + 1 6(1)(6.~)'2] 2

12"- 20.1 lb :

or four 6 by +by 1h ir'l. ongles

930.) The ~hod legs

From Tobie Vlll-2 Propert ie!O of'

--<1>" lb

I

I

+f[ 2..2S(10)"412} (2)

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·

""

L •

• 3(1)(.ot.7?1)i) +

1

Jy • 1i1)3 t (1.26(18)3 ]2. +(1(~)3+1(0)(+.5)2] "t 1 12 1 t [~fi)'& + 3(1)(•)~] 4

Jy "[:i.9 I ~03(d/iz +0·7)t] 2 =iy

r

]+ t [-i(i)3

I. - 7~&+.65 in~

------- ---~---------

+Ad ~

12 12 f,6(1)-a t 8(1)(,,1s)1 ~

-

L '

i.c t A~"

h-

j,. • 1(1+)3 +(1e(2.2st + 10(2.25)(0.375)~]2 •

'

'i

1,. "1291.31s in:t 11 ~ ~ t [ :a.s (3/±l3 t 3.s(3/+)(o.01st' J + 12 1~ t[-"1+\+.s)3 1

t

s/-t(+.s)(s.s)e.]+ +[3.!5(3/-+t 12

155

t 3/4(a.s)(6.m;)

2 ]+

iy

I= MK'

t Md2. = Yt Mr'" t M(r)1.

.591.6 in .~

&

I\ 10 in. 1!1.3-lb chonnel is welded to the fop of' o 1+ WF 34 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-. 1 339. 2 in ~ C:Ompu-le 9 "". the moment of inertia obovt lhe centroidal ;>( all.i6'. From Tobie v111-2 ProperheG of 5trvchJral ~ ·

I = s/2 Mr'Z

834.)

beort'l

)l

or

10·- 1s.s lb Area • +..ot1 ·,n :l.

Area• 10 '1n~ 1,. =~.01n~ !I... ly 2_.310:4' 1"=339.2 in~ S.,. t•o.2
A1y £Z.Ay 14.'4-7y '-10(7) t 4.47(13.6)

i.3

t

4 .47("t.!l6)Z t

f 1' = 476.01

I .. MK 2 t Md 2 2 2 ~ = % Mr t Mr I • 7/s Mr11. 866.) 9y using the transfer formula 'lili., the result Prob. 003, determine the moment of' inedio of' o rod w ·dh re~pect to on oxii;; through the center gravity . perperx:llculor to the rod.

~

Y"' g.o+in.

I,. "

(Tr\

or

'-14.47in 2

/\T • 10t4.'47

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 .

3~.2 t

10(2.04)

_L:~ ---1

2

in~

83S.) Two 10in fS.3lb chonMIS Ol"C weldeO toqether as sho-Nn in FK).

of Le

P-835 : Compute t\;:le,volues

AT y

Q

Ya ML2

I"

t

t

'I

Md 2

Ya ML : 1 + I .. 1 ML2

tor orrongemen+s (o) ~(b). Eoch

M(1-12)

2

~

From Tobie Ylll-2 10 '-15.3 lb

Areo

I"

2

chonnel web '1io .'0 24 in. thick.. o.)

or

Ty"' 2.31n:+ i( .. o.C04 in. ""' y e 0 4..+1 111 2 ;

.. zAy

&...l• =66.9 in~

- 4.41(2) y "'4.47(!5)t + .-t1(9.6)

Y"

using the tronsfer formula~ the result of' Proh.861,determine the moment of inedio of the rectangular porollelepiped shown in Fiq. 8 - 29. with re~ct to a medlon 1i·n e orthe z face. Tak.e the median line porollel to the )( axis.

860.) By

1.3.1n

'

I~ =".9t4.47(2.'3)z ~ 2.3 t 4.47(2.3)2 lJ< = 116.5 '1n.-+ b.)

" MK.2 2t_Md2 . = M(b +c 2)(1.42) + M(C/:z)2 " 1A2 M(b 2 t c 2 ) t 114 Mc I : 1,42 M(b 2+.+c2)

b

2

2(4.47)y :. .ot.47(5) t 4,47(10.6+)

y • 1.82 ·1 n.

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,de-

termine the moment of inertia of'o homogeneous r ight circu lor cylinder Obovt on O~iG through on element on surface. The cylinder hoc; o rnass M 'b,, o rodiuG r .

as

869.).

0

Determine the morneot

or inerlio o\ the rectongulo r porellel-

ep1ped !Ohown in Fi9 .P-8GG or f ig 0-29 v./1th re.s·pect to o n ox•G" t hrough

one edge porollel \o the Y o.xis . I c MK 2 +Md 2 =1.l.;2 M( o 2 +cz)+M(o12.)2+(c/1f .. 1/t~ M(o~+c 2 ) + M [./(Ola)• 1(c;,d• ~ 1A2 M (o~1c 2 ) + M(o}'+ t c)l..f.) I r % M(a 2 +c 2 )

157

156

J

f

·l l

I

Y• 4% [1011- 8 3]

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~ .

--3£'~· 2

Ill.

V•1f(1.9•-1 2)(1)" 3:Q27 n~

w= 4SO_lbj,r((M83fi'} ""S32.3Slb

-

W" ~Olb/f\1 (v)

v21 mt for fue hole

Determine the momeni of inertia of \he c~t -iron flywheel 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

Ws "(.+.90)] (1.5)(2)(23) 1728

.. ~ (1138.03/32.2) [::10

or 20 in

outside diameter ~ 16 in· inside diorneter, compute thcS mo.men-\ of inert ia~ -the

lo

o d a·ometer.

w

=- 450 lb/ft 3

= .50.45 lb .

Ir= Y2 M(R1 tr~)

I= 1.l(;z ML:i :. 1.42(~2.~)(6') +{4-0/32.2)(1)' 1 = +.97 fl -lb-.seci

158

Ws = globL i for one spol\es Wr •(+SO)lf ("30 1 -28 2 ) 12 ,. 1138.83 lb.

1729

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 .

wei9h~ 450 lb per cu. fl . Given·. Oo # 2c/':.. R=-1oin .

6m

Wh = 'lST(R.•- r 1 ) L i hub .

1

'"1/t M(R2 +r 1 )

rodiu<0 of' gyration w '1t h res;-pect

w, • i•(R' - r')L ;

·~lfR h \12•1fr'h-rTl1.""Tr'h

1 V1"lR h -m1

:""*

87.S.) for' 0 hollow cos\-iron sphere

7.35 in.

shew:

r Is. ~ NIR2 M~gv I': Y2 [ m1 R2 - mt r 1] 2 M ·'f,irh(~~-:· 2 ) .::. 1/~ [ ~TR1 h(~)-llrr h(r~)1 4 4 lSlfh [ R -r ] 1 "' 1h glfh (R'- r')(R +r•) 1 i

=-

878.)

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 .

v-t\. m, for the cyhnder

..532 -35

K

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 -

R•

.....

1,...~_,,.2 (1_+4_)....,(.-3-2....... 2)

l : 6.i fl-lb-.SCJC( .,.

=~O lb/;:r' (3.927~ 32.:2

h

~ -

K"

.

w = 192.+.23 lb 1 2 I . Y2 M(R' tr') = 4ft(19i-+.2"'J [(1.5) +(1) ]

1fl.

"' 1.183 ff.3

(12)•

Cod ·,ron

Ir "'

2~. s

1

+2si)A+t1

fl -lb-sec(

Ih • 11.l M(R« t r 1 )

- r~(1:::i7.++/32.i)[(s~+2 2.Vi
In "o.+290 fHb-.sec2 Is ... n(i t Md 2 ) 1 c n (1A2 ML'+ Md )

=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 e.

t Ih •Is 206.e + 0°4298 t 23.11

1 ... 230.34

ft - lb-sec~

159

.~

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 ap2 plied at once ~ retard ihc lroin· ol the uniform role; of 2n por soc -

1002.)

.

(}ivon: Vo~

60mph

0

-2fl/so=-"

Q

•·

..

I.

Vo .. 60 x s 2so ~ eeft/~ec

f-vo' • 20s

3600

-SI/ = -2(.2)S

.s .. 1936

n or

-

o.36G7 mi .

1003·) /\ sfono· is thrown xcdicolfy Upwor-d ~ returns f0 Oorth 1osoc. Whof was i~s initial volocity 'I.._ how high did

Given t=1osoc .

I(\

if go?

y(vo col -Vo

c

-9.Q1 (10)

= 99.1 mis

or -vo" -;32.11+(10) "' 321.7+ fl/s

Choptcr 10 S

Roctilineor lronsloiion

s

U

Viz gt~= ~(.9:g1)(10)'1 ¥129+~

D

c

~ (3.2,174)( 10) 2

4-90 .S C

I

m.

or

1609-7 ft,

n

dropped From the top of o tower oo 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 ? .

100-+.) .~

co

/\ boll is

I

! l

1 !I !

~ ' vo.l t ~gt2

•

"!Ot - ~(M.:i)t.

2

-

..

-©

ii

80-h ~ y~ (3.2.2)l ~

h ~ eo-16.1t' - ·· -@ substitute 2 to 1

80-~ .. 4-0t - Y12~ t:

12SOC .

Qo - 16 .1(2) 2 ~ . 15. 6 from

hs

tho

botlom

s' = 80-15.6 - G-+.-4- from the top

Vf1 -Yot Yf1

2

-.0-32. 2(2)

Yf, • 312. ~ OZ)

11 I

: ot e

'=

- 24-4fl/GOG

64.+

ft/soc

~II !

160

161

" j\.

r·p·

ar

what ic; tho
a

~

i~ fOoe.) ~

1llfOI "" .5SOC iofor Ibo 60~ 1r lhG «>loc.ily of sound ~ mo n ~GO(; ,

1CtJ5,) /\ sfono Ir; droppOd ~ splat;tl -. . heOl"d-

O

the~?

v~

h"40.3t -

tl ..5600- ttmt, ~,,._1 tl

'IHI0(-5-l,..) = 16-1

. Y'· .

fhc souod d .. fl2I0(41-~)"---@ &ubsfilule 1 lo~~- .

fq..t_=

0"' '12~'

d"'

·1'-1(3.793)

=

~(3£2)t,.r --- @

•

-b :!:J_b_*~-40--C

231.'3

fl_

..9 =3~ Sl/scc

'IOCO-n = h tr

1

t - tt.Ar.c ~ lhG 1!s[ ~ l~-L~

I

fer t1hG. f'._.j~

.

I • I

t £

I'\

h

fl .

~ 386. 4

.second ball

'1-He(l-2)- '1~ (::n)l(l - 2)~ h ~ 2-tel:-~ -116!.~ tEftt - 64 --·-

2nd boll

Vo " 1!09. 69

n /soc .

t2

~10.)

t • :2.S36 -sco

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 /\

or

1-sl ·s tono

©

.S '" Vot • I

15400 =312 l t".S SCC- :.h: f()0() - 14i(li)~ : 600fl.

162

386'. 4 • Vo (9.~+ - 4) - 16.1 ( 9."64-4) 2

s1onc f rorn •he bp of 0 600 n low~r ·.( fhc two ~fones o ro co ch othor ~co fl fri:Jm the fop the fow er .'7

(~)l L

tooo-~~- 21-St -~ -.u-V t *I.: - G+

11!JGQJ~·

fl frorn tho ground f'

ft I9GC

Bt ~uodroHc E~ t ~ 9.41-G+soc s,.....

1ooo.-16 l~ - · - - @

SJ1il6f'itiuille 1 fo 2

-~bi n

386: 4

Vo, • 193. :2

396.+· 193.:2 ~ - 16.1

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 tho 9'0 nd w ith ...,doolly of He fl per .«c. If 9 ·l'i 3'2 R per scx;Z. ~ 11>:-~ 'lllflll the 6blGs pciis& ecdl ofhcr ~ =

Is shof vcrtic.olly ·,nfo tho o;f' ot a vo loci t y of 193.2

t-+ " limo tet' the ..s .. vot - Yi.>q!~

1007..)

"'s flefl/soc, h: 1~ fl_

16.1

t • l'1mo for the 1st boll

t!O

2

1000.) /\ boll

fir~t bo ll

I

16.1lll - .. -
0 per sec-_ After :- sec, arothc,r ball is shat vort1cally ·,nto tho air. ~hot ind1al vcloody muGt t ho 2nd ball havo il'I ordor to moot tho

for- tho .stone

IOr

P, ~i°G

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 h .. 120.et - 1.:;.H ll - 112.1 _ .. - @

"fSX:-

16-1 ~ ~

112. 7 "' 80.!! t

h~ oo.6(t.-1)- 16.1 ( t - 1):l

fq- you gi;t t, = +..a45 liOCd= t(S..1 (+.QH5)S - 3.53 -'31 ff0)6) Rapoot Prob - tooS if tho sound of tho splosh i& ~ ofter

+t80- tt20fp ~ 16... t.

subst. 1 to !l

4S:st -~." 129.~ l ·- ~ -112.7

for lhc 2nd stone

By~

c

• 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'

subsfifulc; t lo ~

1120(+-t..) '

with

Vo, • +B ..3 n/soc

d • ~(M_t')t.,S --·@

' l.

th~ ground

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 ?

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)

·I

stone is thrown vorlicolly upwon:l from

•o pos~

2nd Gtorio

t Y~gl:z

200 .. )[z@a.<2) (11.100 -

t')"

-400 ~ 300t - 1i;.11,?

:200 • 1 G.1 [(t1.1ee)'J - ~(17.1 BB. l') _ t'

By~uodrotic · t ~ 1 7. 1ss soc.

200"' 'l-7.S6.30 - 5.S3.4's t ' -+16.1i,'z

i..__

'JJ

0-= .;sSG, 38- SS3.45t'11 6 .11.•!l

t __ < 1.44 &cc ~Y_ '"" *":__u_a_d ro11c • · you 90 \ • _ _ __ _ _ __ _

163

t :

13,1;~ i;cc.

8y ~uodrotic fonnulo you 90\ vf "' 2-40 ,H/i;

~40 " Yf ( t1 th)

~

t 1 ~i'.1.tts:-+1

1011.) / , ship boing lounohcd GlidoG down the ways v./1lh o consiont

41 - tg

occ.olorotion . She takos B soc to .slido tho first foot . How IMq .,...·,11 ~al
Q'40 •

,Sc G!l~f'\

~61-0 = vf(41-:!j-)

t1 d2

sho

s • 1 ft when t a soc . ..s • V12~t:l • Y2a~:l 1 c Y2 a (e)Q

E

vf= ee H/s

tJso or 90+ 00 fU, Bfl fi {s x a~e. 'x

vf (-41 - ta)

1~

n

mile

mph .

60

-!'>2&0R

$

Cl - o.031~.5

i

Qoo.sec or .3min , 2osoc.

c

2G-40

B

n/sec,•

?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 o t th'1s Gpcod for • 2

time, k. finally c.omcc: to rest w'1\h a dccelorotian or ~n por sc
£0\'n :

6 ivon: Vo=O

.

a•-tf'Vi;'.I.

y s '\ONs.

o•- sfU

..

~

Yf = -+efl/G

~

0=4-r+/s" a -~n/s•

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". outorn0bile leoves the 9o"°line stotion 1*.._ oooolcratcs at the c:Onstanl retie of 6fl porsoc 2 • How .soon wi ll the sc.c.ond o utomobile overto~c the first' ? Given : V1 • +.s fi/s

o ~eq'd :

40 "'4 t1

.G. •

Req'd : total Time

-

.,

Y.i (-+X 10)

2

•

I

~ooH .

t ' = 1s. 79 - e

when t • 2600.

e10CI'.)

ta

~

top Gpcocl ? • o • &fl/," Soln: vr -v/,·ot.

6iven: d· 'le m 'i le t1. 41 soc Req<:l : tnO.)C. ¥pcocl in mph

,.,__ disti'nce trove. I ot

thl6 top .speed

Yf • !;;11/i11

- vr - st,

1~0) /\ porhclo mo-.-es In

= ta -

40t whero

Q

cc.loro Hon ? 61von : s - t

12 1') ts2 +Vft1 -

24(1l)

-

tip- t

3-

+

=

98 n;s~

6froighl- line occordln9 to the low

ihc s Ii; in f1 g,.._ t in GC.COnd6 . (o) Whon

+ot

ago in comes to

dur:n9

re~;t, what

l = '51SOCS, the 4th

is 'its oc-.

Rcq'd : a 1 ve-1. b ~ avo. vel. !!o... c~ occclc.-ot ion .

~ 165

164

2

(X)rflpu~c fhc veloc'.ity .(b) find rhc ave. veloo.•ty G"OCOnds . (s;) 'NrY:ri the pci0t iole

-

~ + Yfh

'o

Ql~

Q -

G

S1 tS.itSa • o,e mi le.

26+o •

ei"' - t.}ti; t

3

'

Roqa : vlr..,a

-'If= - 6t& G3 .; Yf ta -

sea .

t• fGOC.

.3-+soc .

th'1G

=1s.79

.

$.von: .s .

=-+ot2 -"" t2: 16 Gee . 10~ 16

~ ~uodrotic

1he motion of o pariicle- is g ivon by the equation s • ~t+ - t. Qt~ whores Is iri foot 1t..., t inGCXX>nd6 . Compute tho values of v

~tSo.tS3

t

'Jl>OtSt+1"0s~OOO --s2=6+on

-

4

t

t • 10.79 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

ot

1st · t 2 - +t-t+ t 12 -1 9 t ++ ~ o

1019)

Tt ~ t 1 tt'.I. +t,,

-©

~t c ' Viz (
time

t!I - is :: 8-sec .s~ =-tole) - 'Y2 (J!,)(s)'.I. = 1GoH.

b'"IO

trovol

d - Y2 .o (t - 2) 2

19t "! 12

- -1-0 ~ - s

·,t

6H/s

4 i; t .. d ~©

-

t1 • 106CC

·' V" S~ 2 - S2 • "!Ct" Vf-Vo • -ota

doc:>G

~

Q>l'n : . Y=d/t

2

Vf·Vo • at1 ·

d = ~ooon .

/

= 41 Vf - n_•

,,I

' I

>~~

' .p

solh: .a§ dl

"'v

..,.. t • c

2

=

= fq / .sa;e- •

ton-sf

l -~

- Bl\ I~., ~ ~ .

.+o ;; (s) 2 - +o .. ss ft/s

" 3l

-

c~

- b)

>

et; "" J(v"-'9)

t =3. 6.S1 sec Wr• ~e on C:>J '!,..__ ae. = J(0.11Ah11tl<~ t hllY"/~. Compute Ve 'II,.,, o a from these relations.

£,'/J>I

Tan z

e- = -Ye_/...;" s ince Ve is Ya = -v.... tan&

cli tfcronf1a~c O"'

'

o~f°" Oe

~=VA

+l
~ :.pz+n;z --.!X•th 2

-j.

-x'1.VA'2.

~~"'Ya

h

-Y~xYA..(x 2 +'ri~)-;f(~y.) ~t.

1

1

2

-tYA (x th )-11 YA ( -..G"'th•}s

2

Req~

f:.ol 'n'-

: s-t, v- t, ~o-t

St9=v2- -

__Q§__""' dt

..fGt"

~

a

""'~ "Q£.

i;ec, 2.&

,;e-

~ 9+9ton~

s= aton'l7"' :::ise0"-e- ded<> ~ aGeG1-e- de3 sec &-

dt

diGf'm;r;cji b y -jhG

~ - ·u

&Von :

t-1

'l-.2

Reqa : v-t, .i;-t, v-~

3S'='£(%)3 t11

Solh : d'l'/dl: 4t

35 "" ;r,(.:(v)3 't 1! ~...rr

'/-=~ i-C =.2i~t-C .

2

3s

c=o;;f i.-1

""°

:ztL - t -=-.fo/.w dt&/dt = d~ 3 -S = ~t -t C 1

a-

= 2 (1)"

=4..F'f

"'"

t 1· .

(3s- 1)« "'~-k

tC

3

3"' !l

~ ilho~

i

!I

o~"'i-t

+ ac

c" 1/a

167 166

ft/80C~

'
2

;vc3

Tho moi1on of O llJllllfiiCb j 5 o is in ft per GCG/z ..._ t in

8

iroHaliiorii a •-4.t Is .s=1fl ""' v= ~{per soc -11:ion. t - 1 -soc- De\onrilfl(; ·llOO rc.DofiorlS ~ 102.5.)

*here

VA2

The roctilinoar motion of o particle> Is g1:Vcri by s~v -9 whcro 5 i.s 1n foct ~ v in foci- por GCCOnd. Vv'.hcn t "0, S"O~ v•3

s~o

I

d"'/ci'1- =~ .

o-t relo t iol'\-5·

·l

'lcN = odht

0 = 4(2}=

1M.3.)

l"O;

o

I

I

'ti= •ff/sco

·

v'a [9 (10'?]/.[9•-1 1? -= 6 f!,4 3 2 . as= [9(•0}(]9<-1.,e 4 tr1rlll(10~:µ1~9•t12 2) .; (;,<:,7 f'f/s

Gi...on" .s-= v g_-9

"~ £2'f' - '4{1l)2- .. 1>( 2)

d'YdJt "'"3~~-8Jt +6

~(x 2 th~}'!>

the s -t,v -t, 'ii<.,

S:>Bn.7

is ~

a: Vdo/
3

ft per sec· find

)t•- -f-)l2 i- 4'X where v

ctt

2

= ')(CV,

Ux'-111"-)3 '1~'"'+h2 2

feiof por sc.ca;d "-. ;ic; l.s ill Cocnpuio fho vol.Jic of tho occobrotm ~ ~=· dt _ by Y =

'%? ( xo. +h 2 ) -'12 (2x dx) - _Qy dt

cJt;

Tho volooii'Y of o porhGlc; ~ olonq tho -X ollits is do-

Retfd :

l

Qe"" XOA .

3

f.ne.d fGOt-

1

. VB = XVA(x 1 + h e)~YQ . . dVa = Oe =d;c VA(xo.th 2tl'2 + xcJVA (x-zth 2fYll cH Cit I at Qe=V.AQ.

·• ---

t 9

dt

~=2

xll +hci-= z.2 i L ~~ + h-y @

=(v'--g)2 v 2- 9

IOU.)

6iwon: ..., ... -r.3 -~...,.s;x

downwo!"d

=

di;

'fhod:

=Va j

1

3

1022.) Checl-. tho answers tc Tilus. Prob. 1018 by t he followlng mo-

1

c

Y£•S-t-9 ..SL

·- 4-0=:3t.'2.

;~A

11-

..tsa-n t ~J .t = In . 1-a-3 J

Y =a.l 2 - -to o = atll-40

c .)

t

]l

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 "-._

10£6.) The mOtion of a particle

v=2ft perecc. Pet ermino tho relation bot woon v..._t,.s1to..,t , v,s. Given :

0/5 2

Q • -

~

V •

4dt

.solh :

.+t

ods "VdV YdV -%~ ds· - 1

+c

1

V Q = 1' S- tC

"if

S ~ "'f 'ro..V•2 .; C =O

Yi c - 1ct+ 60 S2" - 1ot/t t 60t

ds ~Sa/IZ tC

12t

s = 16,NQ

£

=

(t%e) 3 /z

aH6"

G 1Ve'1 '.

S .. Ya (')(10) = 600.

126 - 4 2

t•O, 9='1 , V·•O

Roqo : v-t , s- t, v-s ' sotn: d%1; - Gv"e

. d'0;.(v ~ dt

I

d%~ ~ 9t.~

0 • GvVe.

,,.

' ,'

dV(v)·Y: • Gdt. I '.2.Vve = 6t +c '

s = 93 1;

.,.

6

= 3t

+6

5 ".:s("o/3)a +~ "'~ v ~
s

i f t . 0 '-. v - 0 ••• c .. o . 012 'l V • Gt

v"e ".3t v = 9te

t:

·.stj1a

i

governed by the r elation a "4!2, where a ~,G ~ t '1s in GOC · When t 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 3 t•O, V• ~,5".f V"' 4-}.$ (a) 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 '1G

0

168

-.0

:

: I I

I I

I I

t.

.S1 •..st•/t

10a+.) Thc ' rnotion of o porticlc .Gtorfinq from rest' jc; 9ovcrnod by the o-t curve shown in fig. P-1034. S t
o parf 1olc

~~

51•2.&t.

C1

-+-----'.,__~:..;;. tel;)

in· 0/s 2

.sc...• ::'.!.2.!.~t-120

Vi •!It

-iv"~

1028.) The motion of

- -

S,•01;...l•O:.c, · O

"if"s .. c;'k..,.t"" o ••• C " 6 3

#I)

ll " 'TI ·~

8 ---a- t C

G" 9t

3

•-40ff

al 6.seG

-4

yr

'10

Ca " - 120 2 S1 = -5t t60t.-i.u>

S• 'h(+)(t O)

givon by a " G v " 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 .

1027.) The rnot ;on of .o roriiclc

t C2

S1 •.G.,• 4IO t·+ 40 " -.s(4) 1 +60(+) +c~

- 3 --

c

to ..scolc.

. Y· -'o/a(t-4)

'l(t1'')

~ .rs e

112' t

Vil."" 1~/s

.curves approximately

dG dt

+ ,. 2/3 ( +) 3/11. tC c = -+/~ +t ~ 2/3 .s ;lfiz t (- 4./3) 12t = ~GWct _ +

3

v/f. "' -as-1

..

$"

t=1, G·+,-v:ct R~q'd: v-t, s-t, v-!;,

"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

v, .

6

.9

V2

·"'°His V(fl/')

" --

2/li't t t Vt

if V1 •O Jo...t•o :. C, • O v, • t :r. = ' 2 • 36 nIs

3' -

- :--it~ -:-'(«~

- «-- - -

" .t

\ "\

....0

~ ~

I I

I

Ot = -4./3t t 2.0 11

Vt • -+/, t t !lot tC.t.

Vt· t"

H' v-ll. -v, =9

Si;

""- t • 6 36 • -+/,(<.)tt20(6)tC2

t.'/a tC,;lf s, •o Jrr. t·o:. c,·o ttls =(6°th - 7Z. n.

&1::

169

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?

Yt .. -f/,gf1 t ~ot - '° .!;1. =

2

-1./g t3 t z({'d

-

'7Gt t C1.

I!.-. t ~ 6 2 ~. :Z/g (') 3 ~ 10 (6) - 60(.~) t C1.

.G1. • G1 _- 72. 72. •

c~_;

w= 3220

120

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

,45 "' v(tsd1) .. AQrOO.H • t,. "' -'1>("·-o) t(&l')/~]('~·')

8 0

1::.$

=D

2

6

tC. V* 16 \.., t =b

·:·

v = Skzt.. - s

A mon weighi ng 1b1 \b is in on olevafot"' 'rnoving upward w/ on occolcrotio n of 8 pcrsc.C~- (0) Whof prossuro docs he O)(ori on tho fl oor the ol<5volol"' ?(b) What will tho pros.suro bo ·,f lho c lovolol"' ·," dcsoonding w ith the some accclorol ion ? 104.s.)

or

n

Soln :

Gi-.en:

lb a~ 'T- W .. W/g (a) o • 8f!/s 2 T - 161 • 16%2.~ (a) Rcqo : a -) prossuro ho 6i
6iven:

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 !>CGin\orvol . fulvo by ui;'1nq rno\ion cuNCh ~ ~ by ca\culvs -

. o~ -ei6lt.·6)

8

·

v = ~ Wl'".tet - 14

Goin:

if v ·'4

"- t • o

A~

' 0. - 41.3t t0

=v(t1.- l 1) t =

(,

-)4(" - o)

/:;.f, " 12fl -

i fv~10S...tc6

Aroa A.-t • h t 6/i. (• )(1/.5 • '

)

P.eq<:J: Coeff1c'1eol of l'-ine-tiC

O.Ll\lh

"°

2

=

a-= 8

frloi1on p - F =mo .

11 ·

Qo(100)

ff/s~

60-161 (f-) "1'~2 .2 (e)

/"- :0,12422

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

7''1'~

ye. • ~as -

thorc fore ,C· -1~

170

~

~

~ -100 1! .

'\047.)

Yo • ;+fl/f'

t

2/a t :2 t et t

52!2%.v2 ( 2.-5)

"

T = .34-70 lb

'1f \I cYo \.., t =O

" =-

T - 3220

1oofl,.siorlinq from t"'CGi- Compufc tho oooffiOient of 1<-1noiic fricA 1on between the blook. ~ the ground .

-8

0

wh (a)

T-W ~

acoolcl"Ofion T "' 121 lb. 10<1e) The blovk In f ig - p - 1 0~ roaohos o vclocdy of 40 f{ per sco In

16 ., eM.(r,) 2 t C

c"

w

= 2CJ(00)

a • 2.5 ft/s<1

w19

t 01 v ·"'11d

I '/

Y -= 600 ft/min 5· 200 . RoqCl:

~.2. = -o/0 t • t 1ot~ - "ot t 120 ~ = -1/g l9) 3 t 10(9) 2 - l>O(~) t120

( ~60)2 1

Given:

I I

I.

l II t

)

o+s.) A rnognclic porllclc .•.•~;ighing a., grtOTl.

1

,subef if ufo 3 lo 1

or

Gotn :

6ivon :

F·

W•:a,cr;g • 7; t;1Z%J( I0-3 lb

a - 6 m/s ~.3,.~ irl(s Flcqo : f()((Xl

2

4 •

•

ma

=

200 - 1,37(100 t .3.110) = 6.Q1 a

a:M!

19·"9 ti/sit

200 - 131 - 4,2,1CI "G.210

time until tho blocl
0 • (Q,OQ

>

f!/ scoll

10.5.+.) Two bodies / \!1;, B in f ig . P- 1os + are s eparated 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 :

"-fa= o.+

P•2001b

Roq'd: Timo to elapGcxl

5,... o.ao

unt'tl .solution : 8Ginao• -

tho blool<-6 \ouch

OA-8<:.06.90.

0

~.2

J8. 0.10

as

Qa =4.% fi/s ~

j\Ginao· -

a,..

o-Q /\ c,$'30. 3

10..sa

')/g~~ a,.,

O/s 2

.S()

t /\ .,

- 200 t % (..oo)- 4h(4<)())(o.a) ts t 4'?l£Ma

'° 1 2~ a --Q) ot B, S/a(600) - s • +ls(60::))(0.1) .. 600,.32.20

·. cs -~aal aXz ("l-9.S)ill / · ~ts ,= Y2aA tll .. ~ (10.~ate) I

0

- .s6 l .5

2

f 2,"f'7,5t~ • ,5.2'5t£

from

t · = +. 2.3 .sec. os2.) Detcr m1nc the occcbrotron of the bodies In Fig. p-1o.s2 1 1 tho focccl drum ·, ~:/.smooth """ A ls hoovicr thon 13.

w,_- T.. w... ( 0 /9) -© 1- We ·We ( o/g) -@ l " We t Ws(~/~)

W>. - We

G -= 12. 420 i'56

1

·,r

( WA-We)g ~a

~

-s

1

,subi;f tiutc

-W13 ( 0 /9) :w'.o. (Gl/g)

~ 18,b.3 0

31Q

to~

/ 312 - 12420 - ,sc; : 18·6a CA Q.56 • .g1.asa .S ~ 1!2.42(8-24) t.$6 ° 1S8.4 lb.

or

1os1.) The cooffici onl !
o\ ..c, :300 -12

10:53.) Rorcrring to f\g . f> - 1os2, aGSumc A woighs QOO lb ~ 13 ~el9hs

t\t

. +he acx::clorotion of the bx:lies ·,r tho cocffidtont of 1<-i'notic frsCtion ;, 0.10 bot the cablo i.... tho fl~od dn.im . w,., ~ 200 lb ) • 0.10 T~fie c e ro- 0 •1 (r)

Te - 10 0 • 1ooh.e.~

-=-

ai A,

~~.t.a

= ~1.11a 11 "1.s.ga +".11a

- ·

to@, 11'!-&-s ·99~~ .1101) -1.31-.<::+ • c.!.11 a 1~ -210.6~ c 9,o/i'OI - ·

172

a

©

T, -1a::>G1f'l&.'90(o.s) • 1~,q a

1, - 7.S·98

.subSt. 11

(o j - .. @

a - ··© .

1 1.!-11 - 1 ~. ,"I- - , , 210 _ ...

T~/r 0 • e i,. s 1.z.11e - · ·@

200-T" .,. 2ooh~.12(a) - ·{j)

"' a(X)4~~a

t>, h-11 -!ZOO.s•nao• - am C
100 \b. poi

we" 100 lb ..so1n :

i

a " s.24 f'-l/.s:z

.sub!;t. 2 lo 1

f1

+

from 2 , Jfl " 100 t 1ooh 2 .2 a · GUb6f1tuto. 2 lo +

7.~9sx10-"'(19.69)

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

~ = O.'J..

200-1;a7lj .. 200/3fM! 0 - ..

173

·· ©

·®

~

T~

" 210.G2 t g,32 a

Sob!.i1luto to 1 300- 210.1;2 t ,9 . s~a

ag.3e

:

.soo~,a

a

= 1a64a

n4e

a,. ·4.a 11 = 1s.ge t ::1.11(.of..a)

d

go.01 lb.

1£ ,. 210.bf t 9.3~ (.;.,a)"

2"5. :3G ft) .

> . i·

Choptcr

11

C urvi linc;or Tronsloi ion ,. ·' .

I I

174

175

1102

.) /\

s tone is thrown frorn

o

hill ol on angle of 60'

to the

t

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? .

1 = 1003.;Yx t 447914.9/)(

- 100~.9>< - 447.914.<3

xe

7(

/ er/ - 257.~

·o

e

19:1.2 S1n3o't -

'/ll la2.2) ! 2

- 1<0.1 l ~

~ t ..

'!-=Yo cos&t • 193 ~

11os) Rcpoot Prob . 11o+

\

fl. or

743. 97

to the

n.

o modor w/ a muzzle veloci~y of .sooft/.soc; ot ro· w·1th the hof' i.u:>nt al. PcAcrrri1no tho positl0t1

of tho .sholl "" i+G rosulion\ volooty wi ll

'lO .soc offor flr'1ng .

·

v,,: soo fVs t . 20.SOC ·

= 2:220. 2~

1

.

;

V-= j_V_x_«_t_Vl-,,- •

J

'2!>0• t

2 10.99~

·• :327.13

:zg

n/s .

1'1L

....

- .. . . ' '

-e-·so·

_g CoGf>

''

fofl~ :S1/><

'

Cc6 ~

R~Js.iz t;ic.!l

S " ~s7.G

=

x/fl._

1.60C.

fl .

O"Vos1nG-,.t - Y:z9t.t

(x:o)

1/ll

Yo.sin&~

R = '2.Vo~(CG~~(tori~ t

~

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 .o d by the projc.ct ilc . _(ya!~) h .. Vos1n9-t - Y1Z9tt Y. • Vo O'.).Se-t, .-' h . •.. \ at (x,o) i h =o

1

/I. pr'Qioctile ·,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

)(

t

fl.. (32.~

.so\n:

ay ~uoorotlo fof'mulo

)( "' 334. 63

(0,0)

ft .

1104.)

~cqa :

__

,.- .- l .· ..

fl;$ .

h= V~ s 1n 1fJ- • sooes1n2-ro• "" 2911 • .5

.

n.

- Yx -·soo~oo· a Q - Vx ~ 1260f1/s. Yy - .soo.s1n<00' • - 32.2 (,.20) Vy = 210, 99

Vo· 193.~ H/s

'J.57.' - 96· ' I:. t 1'1·1 ~ ll

x

)( ~vocose-l "' 1 9:1. 2cos-30·(~)

x .. sooo ft . y =Vo sint:tt - 1n9i 2 2 '/ =.soos1nfi0' (20)-Yq(a2.~)(20)

-fr = 60°

0ivon :

/l57,,

How high

s o\n : y=voGiriet 1 Vi29i ll 2.su, = 1Q'3.2s•nsot t Y2 (,9:v:z)t 2

.so\'n : x =Yo c.os~ t x "' soo ~ f,(/('20)

6iven :

projectile ·1s f.r"Od dowl"lwon::i at 30•

ao•

.._

·at ·,.:1so?

ir the

Ssec. c.osao•(s) = 1338.5:3. fl .

hori zen ta I .

1103.) /\ shell lcovcs

d ;rcc,tcd upward

2

• 1338.S3 fi .

- 2 5 7,, ~ 9G.~ t

y · -743. 97

~) x .

ton-e-)

'/I. "' Vocose-(2vo s1rie-)

01(x/a,h) ; t= t ./1Z

9

'l:zgt -- t,

• v0 l2s1n2-&

9

~ 2Vos·1~

9

.

.

h= Vo.sin& [2.Y~~ne- -.V£g(v.o;•nerzJ h "'

Voe.s,n 2-G- - Vo 2 s1ri!eg ~9 2 h =_Voit sin e-

-1.9

1"

176

177

lj

shOwn 111 f:g. P-1107 is ju'°* to cleor tho wotorfilled gop· find tho to\<.e -off vclociiy Vo.

1107.) The

c;:Jf"

/"°

solh:

~ ·Y/x

;gy '•J'. - .. ©

-tf,f c.Vo .s1n:adt - YL(~fl·,2H'-··©

.sub6i ituto l' fo 1,

-u.!l - y{, srn30 ( 2o/~)

- 16·1 ( 20;V0 )

-y

t .

-'t.t Vo • 6440

~'-'""

Vo

c

14114

x .. 77. 8' (.s.51) :. 429. ft.

. r·.,fl.

1< .. 6oft -tr ~6Cl

~"

= " •• 7i0t_- 1~.1t 2 - ·

~.SIZ.~

n.

_n.

Retfd:

Vo

'Aro %s+:s -

..._t'

ft.

cos-I== x/5

3/w =

x4st..s =VoS111-&t

Y.

j

= €41 ...+4.f1 .

l

-Y4!!g.f.2 -· · -(j)

2

~. '

~ 4 • 4 = Yo £9Glt'I~ --VoG1 oe-"'G4·4~ . <,gJ

SUb6fitufo .3 s.,_ 1 /

'°·'

1

·I I

-eo.s =64.4t -

16.H «

t • .S 6oC.

'· t

60 ,. Yo COG~/>

241-++ = Vo case. (s)

t • 1£0/vo h • Vo 6111-6-t - Y~ 9t q .. 10-.s - Vo s1nGo(12D/y0 ) - 1,.1 (1t0;(,0 )

Yo ~

cose- •
~ 2.318'40 Yo 6 1r'l .!13.137

4e.4ff/soc.

the distoncc s a1 whion o ba ll thrown w/ o 1 velocity Yo of 100 fl. par soc, of on angle ft-' ton- 3/4 will .strike

1109.) Deter mine

ihc lricilnc shown in

S •

241."M = Vo ca;&t . - .. - @

: . Yo. . I solh: x cVoros&t / ..

Yo=

-s=J(41Z9l +(14t.
1110.~ In Fi'g. P-1109, o ball thrown dOV'ln tt'lo 1nolino .stn1
!'

~oqct

ga.9 v/

fl.

·©

'/. =77.86t -··@ 1"':S, 3f. • 77.8H ; y •fS.%t - ·8)

J · __%. /'

· ·

~ L.s.g.s (.s.61)" 14i'.·9S

y.

-80.s

Given" h =1ofl.

= 64-·4

0ivon :

-e"

~e:=-·

Vo

p

6

100

/ .

-

:...--fr

c

.sa 1::17'

~

'

ao,s N/s couso the proj cetilo to

!I

.Vo "

.

t,·, t

I! l[ I

ft/s

tan - • '3/+

-fJ : :38.67°

RcqCI : distonco .s )(

•

178

'4...,. /.,.a.teB

1111.) Rof~ to flg.P-1111 ""'- find 0( to p61nt B in v1-0Ctly 4 sec. What i s the distance x (' .

f.q. P-1109.

,...-1---\

- I

.

I

ft/Gee .

1100.) /\ boll is thrOWfl so thot ·, t .just clearG a 1oft forico "oft owoy. If il 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?

.

t "" .s.s1 Goe.

)( "'~00 C.O!i.38.&7 t

- · ·-@

~·

~6t.7i0;(-16.1 t~

-ts,g.s,.t'

-y"' 1oo~l11 3e. &7t -1,.1 t 2

Vo oosao·t

17.~£ =VoGOSO!O·t t • 20/vo

t~-t,

Y. =-Vo GOS O-t

h "Vo61n&t - Y~glt ~-

-y =Vo,s111&t - 1/Lgt~

179

5.) /\ pof'f1cl~ hm; such a cur-vi linear mo t 1.on t h0 t I·+f; )'. COQr . f ' d r 3 1no c 1~ c11nod bv , inches~ t i'n so d , r x =..s t - 1ost w h ere x ·'" in 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, dcf crrri1no the t ofol velooty when t •.4scc. · · 111

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 t be wotcr ?

111ll.) Boot /\ moves w ith

d~ v t - 100 : Q.O(s)

,,rt/sec

Given: )( , -sf3- 1ost

.Soln : x

t=2soc, a - 7-S in/&:

t

0

0.

8' t

t •2

=4soo.

-sP- 1os t

= 1si 2

0)(

;

. Rcci'd : V

e

No. g ·1rol'1 irdined at .::12 • . . I Given: V0 60 fVs .Solh : 2

)( ~ 100

.

h/

RcqCI: rio. of Jron

.

'f.

= Yo~ s 1n2e-

= .31 · 7~ •

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 --6-

Use No. 9

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/s 12 oy "-~o

fl/s

2

,

2

Rcqa: horimnto l di.stance Sbln : y = Vos1n&t - '/12 gt 2 ..'oo = 100.sin30t - 'h. (QO) i -60 ~.so t

- 10t

By ~uod(l)tic

valu•c

of t

~

-;.="\-47.GQfl .

Vy e Oyt 2 V = Vy. t t Yy2

.

..

.

:2.25

th/sec .

Reqa : .Spcc;p i.n rpm.

&:irn: 011

=~ r ·

-

8"~ s

Y =2ooff/s V = lfO H

t1 "6-+rps (6~)

c

.382 rprn

1119.) . .At fhe boHorri of a Ioop th c .sp eod of 01 olrplane Is 4-00 mph Thi~ COUSC-G 0 norma l occclcrotion g .9 fl pcr .gx2. Detorm1no the ro~1us of the loop.

or

V c ~rnph

Ro9'd : RodluG

You get the

135 in/G

1110.). The. normal acce lorotion of o po"tic lc on tho rim of'o ulley 10 ft 111 d 1amctcr '1Gconstant ot 8 f l 2 · p ..,,..,,.,., · f th . pcr
011 =ggftj~>!

IZ

=

= "l-5 (4) = 180 in/G

~ /.?!S e t180::z

V =

Given:

2

= ~ in/so:;"

200 = 1 (10) t1

x =Vocos&t t Y2ot . y. = 1oo cosao(6) t V~(-4)(f) IZ

y = 60fl .

= f;O"to:r"

0'1':'en : d" 10f1 On = 8000 fl/sz

g

a!Z.~(100)= GOiz s 1n!Z&

I

2

=

~75"

V-x = 1St IZ- 10S Vx = 1S(4) - 10S

V

= 60

= Ja.,.-:r.toy~ Oy

t e-t€cc.; 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 .striking the green, which is·on tho some lc;;bel as the point · from which tho boll IG struck . ,.Assumo t he 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

- .1os

30l

O;ic = 3o(q,)

a

li'

'l

=

Yx

Sa in :

Or1

= v"/r

.9('32.e)r- [4oox.s~ox 1;8600Je. r - 1187. 64 ft .

I

l l

:1

II I

.1

6 sec .

181

l80

I :] I 11

1120.)

A padiclc

mo'«iG on o cif'COlor path of 20 fl rodius;;

coo thot 'of ..s ..

"its arc dl
•f'

occ.elcrot ion at tho end of 2 sc.c. 61~on: rc2ofl Seth: d6 •12tll - 10 • V

s•+t~-1ot

-at""'

t• ~GeG · Req8 :

1!Z (£)

41

path. .

-10 •3e fl/GOC

.sfi._=. 24t rOt dt

a

2

Ot. " 2+(«-) .. 48 n/s

•

ao .. vo/r • (;ge))/w 2

1121.) /\ porlidc .'1s lllOVinq

a

H • \l&Ls1n «4 • fl.Oy

tan- 1 o."1$

c

,'I'" ~6.~7

0

2

(WO)

Q:::

~a./· ta/ . ~" c t

c

(4/e)

•

a,.

=11. " 2 rt;,,.•

10"

~.l.-.

~-

l!}' · "4rAid -1''-"-w-tO ft.

-

Vy

"T

, '~~)<;

-

. \ ',,

~

\.

- --

.

~(20)

2C0(1t/s) t - Yt.(izo)t z '40 "1rot - 1ol 2 - · -1ot f60t i 640 ~ o tL - 16t t64 -0 - (t-&)(t- 6) " 0

olonq a curvod po\h. /\t a ooriain

*-

•

.•. ta8G(IC. V~ -

'1,()()(:t./.s) • -HZ(i)

v. -~4ff/,

Ori " Oy COS-&- - a,. .sine• 20
Vy " wo(+M - ~(e) Vy • O ; Y·24ff/!;

0'1 .. QO f t/.s e

an vy,.. - r " 247to r· 2&. 8 fl a

0 41 •Ota t On11. (11 .c;~2)c. Ot2 t (+.+)( 41 ,! CH " 10· 8 H/s 1 1122 .) A s tone 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-

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 HlO partiolo is 48-IT ·ft por 6!Xi!7 . Compute tho radius of the; oirclo.

..si \ion. Given : Yo; 1oofl/s

a•48~ft/sa Rcx:iCI : rod iuc;; of tho

Roq'd: Ct \.,,On . ,r

I'

\ ·

H ·Vo.sin ~l - Y~yi 1 " b«> •

86.7 fl/Gt:e;t.

inGtont whon -the .slope of the poth Is o.1s. 0)1 c Gft per S!Ctjz ~ ay - 10 fl per soo41 • Compute -the voluos of ab t,.., On at this · instant~ st
~-

.

61'vcn : Yo· ux:>ft/s o.. • -112. ff/.st1, Or· -120 fl/st1 m• -+/a RcqCI : rodi'ui;i of curvc..duro Soln :

On ~ .72·2 fl/s 2 a« • at« t Ont. .. (46) 2 t. (72.2)

0

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

p-60•. x ··. sofl

Req'.d : ~ 1u<0 of Curvofurc

Vy - 100Slfl60 • - 32. 2(1)

Vy "..s+.4 ft/s y ... ~ 50 ct(..s-t.4)l ..

ton &

".Vy/y,.

"

73 - 89

ft/s .

S
b'olh: Oy "9 c 32.:2 ft/s' x "Vo cos~t .so :. 100 cos roo• t

t

~

On· v'lr

1.soc.

Vx -100 GOS60• • 0

v~ •$0

On ~ Oy cose - a., sin-eOn • :92. ~ Q:>s +7."11 • .. 21.79

ft/s

r -c-n~.6~)~1.79 r - 2so.6(; fl -

or

Given : .s- t+- et ; l • !ZGCG . .

C1i-Glo

&>tn: .s -t+-at

3 Vt = -"t(t)a-8 .. 4(a) - 6 ~ 2+ Q,, •

>./'/r

41'> r

=

Vt" 4t -6

ai." 1!l!*

fl /re. f

Ot; 1q,(a)«" 46.fYs• a 41 '" aL 41 tOn-i (1-9~)1' .. (48/ t On fl On• 48 0/6<

183 182'

(H)~

r - Hlf! .

3

fl/re.

•I

1127.) £olvo llluG. Prob.

fl pcr-.scc ;

L

11~ . uG1nq the ff : dota :

11~9.) A wci9ht

W "' 100 lb i v" M 3

vcloody Y of the weigh t the vcrt ical .

=18 in . 0ivon : L:1ein • t.511 w ~ 1001b ~ v .. 5 .(Ja fi/s So1h

0 •

.Sinec.ose-

£

t=

COG -z& ,. V-z(;()Gl!T -1 • 0

W " TCOG-6- 0. ·100

=

t · nJrJ9t~Mt 1120.) /\ rod 4fl.

c

COSIT" o.s~B -IT ".S7. 67.

V ~ t.9G ft/s

~ ~TJ1.2.,;%~.dans 7,,7•

!onl~totcs in a

-

o. 9<13GCC·

horizontal piano obout

;

the 120-lb ball

0.901 m/s

Is tcrccd to rotate

o vorii°'I

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 H - s0.2s lb T =IZOto.s(.s!M!S) n • 1 re..Y,;-t ~ "'1-/t rps

a round

+he

v~

for

160/s

rl ::.o /

Tcos:a0- T1sinso• - W "' 0-© ~f11

o.sT

~. 1

)(= 1.Sft.

r " 1.s tlZ • .:a.s ft .

To"& =v~/9 ,. tM :30. ..

40ubGtitutc to "' 8·07 (60)

t

~·,,_ 2ov/12(a2)

=o.::i125V~ v - 6.079 f-1 /s

v

= ~trn

6.019 • [-.?ir (2)

n ..

from 1, T • (20 t o.s li)/o.g6t;

"/'32.11. (~.s)

o.s,, T - ~: 20

o.s(2s.1)

·0

Tslnso t rleo&:ac>'- wv/9 .... •O--@ -- o.U6T - o.st\. = 20 i:;q.2 - o.sr + o.S6';t1 = 20 (16)/32 (fl)

.Sin :30• = X/3

lb.

Ts 2s .1 lb.

.z!f'v-0

Goin:

.57. 3

0.06G

" = i>tf'n = 2 1((2)(.Yt)

QT (3,5J n

':::!

iho

1, 2'1s fl

-e- .. 31l"

Y • 2Trn

= 0.756 2 1

rorcl 'ii!o.... iho force on the conica l Gholl . At whof ,speccl in rprn viii\

Reqa : s peod

v = 8.07 n/s

Y2

.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

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
v ~r

J~.2 ton30'

1130.) In Fl9· P-113",

lCOG.S7.67•

T = 187 lb r = L.s1"& = 1.s (.sin &7. ,1•)

-f :30°w/

r =o.471ft .

g!. 1 • O _. By~uo d rotic.

0

ton.:3o' " v y 32.1(0:+71)

zirj ~tan~o·

1 • 2 11'

,9 L

-

ton-&:

Rcqo : vcioci ty f>olh :

.G•n~~ " v c.ose-/..9 L 1-cos 2 e- - v "cos& .:: 0

0 cor"d form 0 copor1'cx:I ic; 1 sc;c . Dotcrmmc tho

if the cord rotates Inclined

:.o•

-& •

v/g Ls •~fr

2

C0611& t s.re~co&& ,3!I.'l (.1.~)

the

6ivori ·. t , 1 sec .

=

-to"~ "' v2/ qr

concontratod al the end of

n 1col pendulum for whicn

n],/Go

29 rprY'l.

2

o.s(20to.6tt) .. 0.86(; rl 0·8"

=

80 'I

I

n = 22 rprn.

i I 184

185

Ii :1

1t31.) /\ body of weigh\ W ros\s on the. smooth inclinod surfoco 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)

~r""o;

r .. w

•

,.o·;

Ts;t'l:;o· t Hs in6o•-=- W

Tcas:30'- ti cos6o·

.

•

tt:ss.) Wh~t counter:--c'1ght W will mointoin tho Corla's$ en
or

V°

:S2.2 O•.S) .3g.+e lb .81o •O

"''!:!.£:. © gr

wsin30 t rt s 1n 60 cw · N = w- o.sw ~ o.577+ w

(6)AB.S1n1s•

A•LJAti CEJ

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 aftor bolnq · rolcased fn::rn the 91~d' p001\ion 1 (.~)What Is the beorinq rooc\ion just before impoot ot e ii thG vdooi~ of the haromcr is thon -s.g ft por6CG?

. ·'

6iv6n : I

.

~ '

T cos~o

1. =4f1 .

T

Reqa : (o)Roocfon

,

le ..'

, ._.,,.

also

at (b}

.c

c ""'"""

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 M.1s1n14>q"t'a f;oln : zFy=o [!70(4-0x 68/&o)iJ/a2.2r = !lo.1.sin14,q+3'

1133.) . To chock the

0

ft .

in fig. P-:11:% iG 80 long 1'woighs

of' tho onqine

rpm. Detonn1i1e tbo mox 1.mum bcndinq momeni M In the rod if M • WL/s I where w ic; tho totol cliGtnliutod lood "'L jc; the lenght -the rod . · Given : W •100 lb

or

n"aoorpm r-181n

1.._L~sft

. M " 100 lb-ff.

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 prc6SUrc on tho wheels of the cor- when it is at tho crest of the hill wh4 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]~

e>-c 'I

T = 81.8\b . == R .

r .. 400 . 5

co-s+s• (2) • W W • G3.781 lb .

1137~ The segment of rood poss1n9 ovor the crest ·of o hi II iG definod

b~ T > w ~wv71r - , ........ +[,41-.-t(.s.9)~]/322(4)

Tcos& :W

"1-S.1

~~ : M

7+.:%10 • R.

0

l\8GOS4'0• i/\BC0$415• •W

100 lb· lhe cranl
.

Soln: a) Tcasaa •W

W~6•1-.+lb

1136.) Tho side rod

30.~s.smcoo"(10) - 16,1 .sinae~~

M • +s.1 lb.

·~

o.866

subsotitll\e to•. 'f/cosa-0· - 0,5774-.,f cos6o• = .)l(vY3~.~ (s) cosao

0

a

Given : r+rt. ; v • ao O/.s

.

We •..3220lb

X =20

" '10 ±.. - 100 2x ~

Reqa: pref.Suro R

'l

IJ.._ Gpeod whon r-ood

y" • - 2/100

possurc ;,

ZOl"O

Goin: .of.

£--L 10 100

;l(«-+o)(

~ :

12Aoo

4Ao - 1.(1l1J)/,.A ;.>:tc.v .....

I'. son.

R+ wv1l/,,.r :::.w /':.I

R c aa20 -[aQ2.0(ao)~/a2'.2 (.so)

t+oo-o .·.(x-2o)(x-U>)·O

R., 14~0 lb .

·I 186

187

'if R• o

y;

W'_v 2

e

,gr

Given: Wp • 100 lb

-.. vtJ. gr ; 32.Q (so) "'1610

'J4

s

Y

0

Soth :

tar"& .. -v~r

v· 460 mph

40.1~ ff/G

A boy ronning o foot race rounds o f1at curvo of .soft.rod. · fho ronG ol the ro.ic. of1s rnph, ot -NhDt onglc v4 the v~riicoJ :'ii~

1141.)

tone-

ReqGI : prossuro

fon
V •1Smph

Rcgl:i:

-e-

1142.) /\ darcaevil drives

=

o. 3006

o rnotOl"'Cyolc around

o 01rcular vorti-

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 2 Rc;qCJ: Ym'in ton'& - q r/v • 32.~(so) .. o.f.V V"" 51 . 0 fi/s • 51.& [::1~00/s.280] c::is.a mph.

,-a-

ot eo rnph? @ivcn : r:: "2000

n.

-tft a.s i'r) - .sG.s ,;, Req'd : the .supcrolcvotion e. for o t rocl<- ·

or

tontr

0.1102.

=

-e-,.

~.s,

0

". 60 mph

r -o.

9r

90

3872 lb

32.2(2.00)

to.nf • o~ g

RoqCI : friction Force

9..

R

~jtj3220) 2

4-Q .

+(3872) 2

-

..s03s.gs 1b .

.Sino/" F/iz. ~.S1n42°"' 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;

Rcqli: roto9~pced

ton(1+6-)=v'l9r

fon (ao.96 Hr)

cfioo )( .s2eof3600J

2

3!Ml (300)

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 presOl
:z

~ .. 3Q~O [ 60 x 88A;o] •

r = ~oon ;-e- • 30·

fan-& : eh e/s6-'S e = 6.79 in

he

Solh:

<01vcn: W 3220 lb

3<"B6 t ir "' ~S.82 .. -& ... 34.86°

0. 1202 -

clOGG

= 462 ..97 lb

0

v=60mph.

sure

1soll + 438 2

when the car ilO froyo\lfng ot 60 mph. ihc c.oofll"oiont of frivt10n botweon tno tire' ' t he rood iG o.<~o.

or

a roilrood trock is the number cl' inoi"i 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'
~r

= 71. 07 °

A car ..,..,c1(Jhinq .3220 lb rounds o curvo of 200 f1 rQdlu.s bQni
114.s)

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

11""'3-) The GupOl"OICvofion

N ~~

_:1:l.'L = !l.q2(1so) .. 43" lb.

= ~.92

"fr

=~5l<.S280j3600]faM(so)

2

32.~ (s~ao)

r .. 1 mile

1

ho inotino his bod'( ? Given: r~.sofl ·.sotn : tan& -v%r

·&aox,9280@6oo]

on his

-scat f' 188

I .1

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

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 tire~ ~ 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

·-(:] c

1'). 7.3 0

ton 1 o., c

ao.g<;

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 overfuf"n ? +57Q9.48 '•fl30(~·'H7) t 5799,+8 (t){;.3C(a) -322ot»in30•(a)

row

Sotn:

- '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 .3024,G1 lb F1 ~ O.G (38H,61) • 2294.B lb at outor wheels r

~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 l s

~8 ~ o, rit.(-+.8333) =3200 ~30°(2.+11) t~799 · 1'8 G1030(2A17) t 3'2'20 s 1ri ao· (5) ·

ion 4- o.G l/J "'.30.96°

- 5799.+s cosao·cs)

ion(~ 1~) • v2./9r

H~ • n".4" lb

ton (ao.96" 1 ao') vo/32,2 (soc) v - 110,129 fl/6 W'i" • 3'220 (110.,eq)
gr

f'+/, ..

t11 - 4963. lb F1 •JAN1 • o.G(49G3) '" ~977.8 lb at tho ouier -Nhoel~

p. "0 · 6 lk.qo : f rnohon "-hi~h

.

3e,2 (s~)

V • 84.11 ( 60/86) = ..S7. 3S mph .

w-3220 lb

• o.6

~" ~0.96

yf. c 19853--49

c

'

ton~

WV/gr • a2ro(HJ&58,19)

fan (ao.q~·+ 1,,73•) " vV32.2 (1200) v 04.11 fl/s

r esooft

~h

~In'.

v%2 2 esOO)

0

ton ( 4t&) -= vo/gr

Given :

r=.sooft

tan(~-ttJ) ~v 2/9r tao (30.9' t 20) =

#

~

orao

G1v0r1 : J_A "0·'

RcqCJ: FfnGf.;o.,

ton-&

RcqCI : rnoximum GpOXJ

Is bon~d ot !lo· 1nfoad

¥(. 3220

•[3ox.S~80/3600]J3!2,:2(200J ton fr = 0.301

Vr· aomph

·,r t ho rood

11-49.) Ropoat Prob.114& OG 'hown In Fig. P- 1M·&.

32,Q (soo)

l Ii

'I

f:z. "fH:z. • ~' (12
outOf' .....-tioo IG

!i

2.41'''Vh

tM .gc.g6" h .. 4,0278

ft .

I

.I

~l\"0

N, (4.6333) "afl20C"5acl(~.+11)

li 190

191

II ~I

flywheel 6fl. in diomefor occelorotes from rosl oi i he 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 . Rcql:J : the norma l ~ tonqcnhol 1202.) /\.

0( • 4

rpm/sec .

tho oc.cc.lcrotion

components of

t ~ 1oscc . Goin : o< ~ 4 ~ :L

/l}l'ln~

o(

4

o(,,

x .1.!.CQQ. x JP'f

1.P'lfn 60sec

o. ~ rodAcc~

urA cur

o.4~(10)

uT "'4·2 rod/s '·

Qn cref2 ~

"' 3 (4.Q)

C h optcr 12

P.ototion I

~­

I

'

i

c

.52.92

. n/s

11

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 . Tt comes fo

rest ofter the r im hos trovolcd a rincor distonco of 600 n. v.kol

f

ore the c.o:'\Gfont ongulor occ.elcratlon ~ tho whc.ol makes in comlnq to rest? 61vcn: d "..SO in Roq~ : o< is...._'17 v =60mph

I

I

.s "600 fl. .SOlh: r = 25/i!l

2 .oe.s ft .

o<. ~

.S"'r'&

600 .. (2,083) (Ji}- = 26B rod fr = :?S0 rod '1- 1 rc~n rod -$- =- 4.5. 64 rev . V =r ur.

(~ xf>?.>/Go)

w

2.oe5 lli.

'll( ~ 42.21' rod/s

192

number

~"lJJ/t2~& -(42•24)t

=

the

193

e

~o( (~88)

- 3 .10

rod/sc.ct

or revolut i

\ '

12as.) Whon the angular vclooHy of o -+fl diomolor pulley 12 ·," 3 rod/s. tho totol oc.colorofion of a p:i111i on it!O rim '1s 30 fl pcr-soc • Ootorrnine the O~lar ao::.olorat 1on \he pulley OI th\G °instant '. 0iven ~ d = 4f\. RcqCi : o<

or

ur=:arod/s a=sonN1 Soln

1209·) The .stop pu lloys shown

cro&G~ tx:H . If t~ onqulor who+ is for A to trove! 1BO flfrorn 1"'<%f? who~ chstancc will D rn
ti~c

·~c 180{l.

s. : ? "."hen s-.._

=

On • ru/ 2

-

c

ot'1+"1e

0<

SA ~

~ 12 rocJ/sz..

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 t ho givon po51tion,ur•+radpcr..soo. ~0<. =1!l rod pcr-.scc1:, both

o\oc1<.wiooG1:.0cn: uf· 4rod/G o<." 1'2 rad/s 2

~rou

h

'3

~

L"-.3ft

Rcqa : hori-z:onto l ~ vcn•ool compononh:;· of the accdcroHon

Ot"ro<.. "' 3 (12) ..

a"· rur 0

=

2

•

36

f*/s

2

3 ("1-)2 " 48 fl /G

~ at" t On 2

a= 6on/s

or "- 2 roo/s . a ·1o fl/G 2 cliomcfer

2

Ot

•

J~!L

t 46

2

10

2

=

Cit 2

~3

Ori ~

t

"(.:ir) 2 t (4r} ~

100 = ~r 2 t1Gl

r d

=

4(1~

=

-tn .

2

194

Q

2

2n

r

'/£ (z.) (7. 1.s) 2

-

60

rod .

t ~.

o

1n.

b.) f)A • S.../r " 24<>/(aoM) -{)A~ 96 rad 't1A " 1'€ O(" t !J

- 2

t . 0.48 6CC

<>
B-o • Y2~ct~ .. Y2(2)(0.48/

o<.. _ .. z. 67 rod/s,,_ . SA ~

ro<.. • r (3)

a

a2

rodk:z.

1, SO<"' "2(2)

2

soc.

~ • 1/1(2.61)t1.

H1

12ce.) A pulley ha-s a consbnt onqular occc\ercrllOI'\ of ..3 rod pcrs~:.2. When tho onqulor vcloci~y iG 2 rod por 6C.C, tho+ total ocoolcral1on 2 of o point on the rim of \he pullcy is 10-!1 pcrGC.0 - Com~~c +119 cl16rriotvl" of the pulloy . Sol'n: On = r~ = r (1.)2 Given: o< • .3rod/s

~ 18 1n.

a~ ..11l0<,..

2

V1<.(4)J:'1

~ .. r&o ~ .3 (60) ~ 100 f'I .

so1n :

~

= 7,75

Y2 <;::l(c. t 2

120 =

if' the rodii of pulley a ore chonqcd

RcgCl: tA .~st> whcl"l S.,:2<10'

~

I

Y20( t 2 ,._

-ffo ..

~epc.of Prob. 1209 ~-a

Qp =120rod t

--&v :

t,. • 6 ' 71 &CC . 1210 ~

0

.,.180 • 2&A

s..._ =1oon .

~

I

't:t

go · Y2 (-1-)t...2

6ivcn :

.solh:

Y

re-.._

itA • gorad

1200.)

:

0

r.so::;~

n. b) -t}A -

2

Ol·rcx.

R~(;I

!.140

o<,...' 2o!..c °
2

Ot .. 24fi/s '24 • a<><. -

4

4'~

by

!l (3)e..

.an"18fVs 2 o~ • 0t 2 t On 2 302

~u1rccl

Rcq'd : l,.. = :'

)

1n fl9. P- 120g ore c.onncctcd occ.olcrotion of c i's Q rod

re-..._

=P- 160 &

~ = 2

-&..._ = Y.e <>< t,.,

7'2

6-o ~ 71.
39'-12 -e,.,

rad ,

.

7rJ. ~ Y2 ( 2 .i;7)

1

t,..

t.... • 7. 34 sec.

.

of o pu lley is dcfl.f'lcd by Lh 1 · r c· r·o 1011ori & " 2t +_ 6' whc;ro-tf fs rYIOOGurod In radius 'l.t. t · ...1_ "') •n sccon(.A.!>. COmputc ctvoluhcs ot f onqulor· voloc:ify ' ongular accclerot 1on at the inston w en ~ 4GCC. 121.3.) The rotai 1on

.30 ~

-th i

t

0 '1 ven ·. & ~

t

zl 4

Re<:ja : ongulor vcloofy - 30t'116

Ol"Jgulor acc.oloroti'on

~ 4-GCC-

195

~

sol'n : d&/dl :

ur = 8(4)3 -60(4)

rod/.s

:212

uf " Sl3 -6ot

c1ur/ot : o<. "24l.~ - 60 o< w 24(4) 2 - i;;o "324 rod/s . 121"!'.) Tho ,r ota tion a \lrN°h col is governed by the cquo i ion iJf• .[[ ;w is fn rod ;ans per Gccond ~ t i..l in seconds . -e ~ 2 rad whcr 4 t • 1 sec. G:imputc ihe values of -6' ~ 0( at the inslant when t ~ a sec.

or

6i...cn :

we -+-lr

Solh :

~ = 2 rod

Rcqo : -e~cx

di%t ,/ r

·'

o<.

= Bfa(~,. - o.6,7

.:3,

c ;,

i t "'/?t

c

;f or. + rod/-s (0)

ur . 2 '

d& -&

-, r -e ~ o

= =

I)) •

fl. t

4

~

t · o :. c

oz

,, _ _

" - .

a S-t.

J2t

.!---'t

'

of

ur~ ~ · ;

o Ovwhee l decreases un i -

4'2

£

(6 )'~ -6" • HIO rad c

4-('

t ·6 S(<;)- 6~ tC

2

)

-

t

1 ~(')

I ·l

ca 1~

23/3

f} : =

t 4t

if i

1 -t(SJ

10. ,1 rod .

"2.GCC .

uf·uf. t ·D uJ • et - t &/1

ul -

Hl

11

t 12

roo/s

!I

-o 196

rev .

dB'/dt "' Bt -{ 2/:2 t H! 2 -fi- = 4{ -t'/6 t 12t 1 C it fT •O 1o.._ t • o .·. c • 0 8- · f j t·6

c

iT

23.~

A

i7

or • st - t'lR t c

·e- ~ t%

t 1 1+ C)'t t%1 t .+t t c

2Tr~

or

duf'/c::lt "8-t

vo

t t (U;cc. = e ract/.soc.

l t.H

-&e. -146.
-f7 • 1%.~ rad )(~ 1n-v •

Glope. ~ e-a - 1 • <><-& c - 1(t -a)

a

f

c

60·'l-4
2

--f- - " --

t

t

~7

fo~ly, from 8 rod per scc in 6scc of wh ic h fonc ifs on9ulor vc lov1ty 1.s 42 rod per.sec.' Compute the ini.fio l angular velocity ~ ih~o( number revoluf1onb mode during lhe 6 sec inforvo l .

,('"'«

c ·1' cl&/cft -

't

-9/!e i" - "f6. 3.9 l

-&t =6B/6(1) 2-efie(1f - -+s.SE(7) t60.443

The angular accclorotion

1211.)

t c. I

2

'& .. f'

-1'.S. ~8

0

when

2

W ;:: t jl t "!' \

=66/6 t

ufi • (;6/3t -·8/6i 1 - -t.s.33

= 1.164 7 rod / st.

4- -

"frt

G9/..s ("r) - 8/,(+)'-

4 )".:

2 ( 3 ) _ •It

ur

... c -o

C=60,443

6~ t - 8/, t 2. t c 11'2 · W11!.... t·f

\S/fl.

Q,(;{;1

t31t - o.GC.7

c

t •o

d fhjd t = 69fo t - 8/6 {2- 'tS.3-3

.3

•J-

(ont u>hcn t - 2 Goe. Ga°ln : o<.- gt 6 1ven: 0<, - 2t .dvi / dt - ~ ,. 2t 4rod/.r. duT = 1t cit .

or" t7

::..ill.§.§.

W,· 3,.1zt2. w;• f '~ % t

/\ 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-

Rcqa :

;r-e, •o ,

- ett'8

o( •

5 -3/<;(1) 3 • 68/6 (4t - 8f!g (+) - "'5.33(4) tC

1216)

or-

~ O( ..

B;idl-' - +.s.33t i c if -B-1 =t:T,. "" t ·+

-&- "' 1a.2 rod • o<. "' 'h. (4) t - •It o( .

-&, ·-~i

.3"( • 3 ' - St HI£

\

..

C• -

0/S

=

=:at:

1

-~ • 69/6 f;" -

2 " 4(1) 31t('l./8) t C

-e-

ir

•

0&/dt " W"4.[t 06' • + l w dt 31 TJ · +t. 9.(#3) tc if fr• !l. roo u>hcn t • 1 scc.

uf" 4Jt

tho number of revoluhons throuqh w/c 0 pu l lc~ will rototo from rcsi ifs angular occclcrof1 on is incrcoscd urn formly ffl?rn :zero to 12 rad por sa/· durinq 4 sec ~ i hcn uniformly decreased lo
ur • el~ -6ot

I 197

ii

..•

I.

I" Who! force P will g i-10 the sys\om of bodios .shown 1n fj'9 P0 vcloci'\y of 30 fl per soc a fl er rnovin9 20 rorn rosi f'

1406.)

t

nr

1406

t

~ '4fJ'

100

,,,,,,

>

y•,, 205

· a 301/2(20) 4

=2:z.sfl/s~

~ f~1

p = .agg,34 -

ll/az.2

Zf)l / T1 -100 (0.2) • 600(22.~

T1

•

so(o.2) • .so(22.sl a2,2

69.SS lb.

Jc ...

1

£f.1S/"O

Ti ..

Choptcr 14 Worl'\ .~

a~u,6 t

0

2ocicas-.s (0.t.) t eoor.in.+.s

-[~(2t.oil/at.d Tc - 399.3+ lb. 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.

Energy

1 i....2

2 Tt • 300 t 9,32 o,... - · - @ ..... a,_3 2(~-12.412a.._) .. ~.g.320,...

-4a>- 24. MC:V- .. .3CO t g,32 o,...

l .

100 " ..:l-4 ·165CIA .

°" - s,..

2.
H/G2

v,..~ !lQ v,.. ~ ~---2(~.93)(10) .

,.

v,.. " 7.65 ft/s

:I

I '. I

T2 • 200 - 6.21 Qa T~ • W0 - 6 , L1 (20,..)

i l.

Ti - .eoo - u ..;20,.. -· ·-@

! 198

199

11

:I

"

1408·) Through who! distance w·,11 body A

. chonginq its velocdy frorn 6

n par.soc

in fig.

or

1-+10.) In what d1'.. .fonco w·dt block. A

P-1400 m&1e in

lo 12 n per'6eG·

.,;.

City of 12

Ft'g. P - 104<3 attolfl ~t ?

fl por sec, storflnq from

1 "'-.ll

.!I ""-- 2

1ZT1

300 t

=

+·6' QA

_ ..

1ZT~ ·OD+ t 1fl.. 4~ CIA - · · - $

-@

4 &._3

4 ~1

£(200 - ~.210A) d 300t4.66QA

L { W4 - 18 .G
400 - 12 .~IZQA = 300 t 4.G6 DA

400- 37.2.8 0A

a.-. .;, s.es ft/s 2 7 - YA1 - 2as,...

VA2

12

'.i

1 -

£,...

c

304- t12.42

°"

0..., •

IZ. Og

fl/s ~

·Yt...i " 2as... 12

z(s.es)s...,

= g.IZ.3

G

1()4 • 4g,7QA

100 - 17.• 00 OA %

velo-

0

IL c

fl.

2 (fl,
GA

£A" 3"\..41fi.

•;I I

.:j1

Gubsfdule 4

s,... 3

Zo=~l5Lt6L

£(£o+-'0·6+o ... ) ~ so4 +12.+~ a" 4-00 - 37. 20('.M =::1D4t11Z.4{?0A

104

=~.7

°"'

o"' • .e.09 n/.svA' • !ZOG y,., ·~ .~1Z(.t.oq)(1~)

VA

2

7.08

f'i/.s

Ve

to 1,

1400 , a .11 (0) 1 t

Ille • 304 t 12.~ O" - ·· : ~ +

:,' ;_•

4,6,

v,./'·

v,., . . 11.33 n/s.

II

"17

i

zr~Jost6" -8' i!.~ ~ ei+xL 2~dt/ot = llxdx/dt

. J'

11

~Vt? • XVA

. Rw - ~ uJ~g (v 2-v,, L) 100 (11. -8) = 2oo;t+.4 v8 "'

l t

aa:1/6 4, 4 v,.,Z -·- -(!)

14 ~ 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 g iver) paGitlon, for 9 flalonq fhe fr1cf 1 onl es~ surface shown xz10~ ~ z:~

2/\d:11/at • u.ot/ot

2ot- 9,32 as m £04-- g,~£ (La..,)

• TL.

T~

-

·

~ ~~

Zf

=

)
2

15 10

2

~ '~ t 0 ~ -

-

·17 10

I f.

i I I i

J

I 11

200

201

l

';

800 = 1,5.SVA

2

t t.33Vs

2

- -@

subsiduto 1 fo f, eioo • 1.sB (rA-Ve''/ a,) t £.33 Ve£

t1/s 1417.) I'.. W:cigh ~ iGdropped f ~mo p<:>&ition Jus;f obovo, but ~al- to!.1chlng, 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, Ve .. 1Lo.S
~1<.b1 • w'1 b, • 2W/K 1

W • l<.bt.

6~ - W/1<..

~lb•

•(tYJ/ty64'J • R &

IZ

SOb" 1• ~v"' «.t 1.si> Ys t 0;11' Vis

= 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

f.

.soo • 1,ssVA" t ~. a3Ve~ - · ·-© ~ituto 1to1l

soo: u;s(a6YA)"~) +£.3:3Ve £

h ·-tft ..

Ve. c 1£.S fi/~-

K• 100 lb/iri •

x1 t y
6a+in

I

•10~

,,

6t+Yt '.' 1o
RcqCJ ! Velocity

I

xVe • -7VA VA ( ~ (<;Vs/~) t

-· .

-(j}

.z. u.1,49 (v'-v., ")

IZ

100 (10 -9) t .so(6) • 100/
1414.) Repeat Proh tt1a when i< c

t'·

10

1

f

~

. v • 1.s ..32 fl/s.

2yd"1at =o

6Ve = -SVA

RW c

"wv/:zq .

7s an ~xclx/at t

w(htb)-12 K.b"

gM; (4t4A~) - Y2(100)(12)(4A2) • q6.6/6+.+V

1419.) A 600 lb bk~;k s li'dcs down on inc.line having o .slope of .3 vertic.al to
~f\ ~ ... • _,....,.-~"b ~

W=60olb· ·

en·

- ef. - y = 6H

Glope; .3/4 K • 100 lb/fl

/).l<., Q.Q d " 4fl · Rcqo : mo)( . velocity

)(2ty1.<101.

2i
xVa : -yv" 8Ye = - 6VA

202

......_

Y2 K6.r = wvj'2q . 600(4 tb) [3/_s - 4/s(o.t.J] - 100/.2 bL~(600/64.f}v f [w(.a/s) - W(4/s)(o.ti] (dtli) 2G+o

t

1os6

-soob..

203

£

g.32v ..

d iffcrentiolo vokx:;•tY. with ~pc.ct.

~6+-1oob · -· g,3~

Neglc.cl fricl ion of hie 40-lb collar ogo·1ns l ii!> vertica l gu·1c1c 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 . L " ~ .3 2 t 3~ , -f,243 ft .

1-423.)

tob,

·'°" compute

6 .. 2.~+fl ((or mo><. velociiy) G00(4u.6+) [ a/s- 4/5(9;2)] - so(e.6+)' " g,32v' Y • 12-26

what

fl/s.

&•

or

muGt the 600 lb blocli:. Prob. 1+1Q s lido from re"t boforc louchin9 the spnnq ·,f Its -velocity i" 10 fVG at the in~o nt the .sprinq i" doforYr'IOd 3 fl f Assume the spring c.ons-

I

tont jc; changed to .30 lb/in .

f·

1420.) lhrough

cliGfoncc

Given :

Solh:

w • 6001b

[600(3/s)

1ofl/, 6 • 3 fl

.Y·

Rc:qCJ :

d

=6.67 fl.

distonco

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; velo1421.) /\weight of W lb

'1G

I

gh'!

·,c 'rtG

of

"'

vt =¥ v -sJ "-o/w

ft .

332 .21 ft 'ls 2

142-4.)

fl/s .

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)y 2 ·

v • ~0. 88s fl/s 3 fl -to(+)- 1/2(sx12)(3)'J. =("f<);i+.+)(v/ 02

=~HZ •

Yt'J ~ 259.1

Y1.. = 16. 1

·

20

.aest)

.

fi/s

142.s.) Tho ~r i~ Fig. P - 1+1s i& movlnq toward 1hc bumper .spr'1ng

1'4212.) Tho rigid hor"1x.on tal .shown in fig- P1+~22 is supp::irtcd .by 2 Gpring" . ,A. 300 lb weigh t iG the plocod upon tho bar ot 8 /\ .wclder1 blow prQ.fccts the weight toward/\ with an initia l velocity of M\jt;.

Whot

• .!!

·.Qn.

Y2 = 18. 23

I

,·

.s-.:9

Vz 2

'*K8" "'wvyrg "'-

=(-10fa+.4)Y 2

"f()(4)- !/2(.sx12)(2)2 .. (40fi;+A)(v,ii- 1 ".~s")

city when if re\or,iis to the oquilibr1urn position .

w

6t ..

a'

(600(10)''_1/6..,..4

I\.· 30 lb/i'n

n.

v • 16 . 36.5 fl/s. L • ~ 3• t"1'1

-ooo(~~)(o.z~ (d t3) {30(1~)(~)~/1

1~+ (d t3) • ~s.s1 .7

4.243 -3 ,. 1.243

40(3) t V2(sx12)(1.24~) 2

veloerty when it rcochcs t...? tleglccl fr iction Yt._, tho wei-

Inc bar .

"

V1 - 19.9

Yt - -t .~1 fi/s

204

~ 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 modu lus of soon por 1n . \IVl'\o1 the CQr iG' b'r ought 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 ~ .!'()() 1 b/ 1n

P.cqCt : .s I!.., /.

205

J[ I• I

I

/2(1000)(12t6) 2

1

t

'/2(2)(~)6

.500 ( 12 £ t 2.46 t 6

2

1

e .. output /1·n put

=100, 000

0.9

t EOObt • 1001 000

)

hp(~) - 3.5. 7S

1ooob2 t 120006 - ~aooo ,, o b'- t12b - 28 ·o

6

-12 t J12• - .ot(1)(:-2&) ~ (1)

0.9+ hp/3Sas hp " 33.G1 hp .

~jn .

•

Powor Output = 33.61(0.1+61<..w) ~ 25.1 KW. e~fcr5 o hydraulic react.on turb'1no w/ 0 vcloeify

1431.) Wofor

nper GCC·

G

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 fh e fra in iG . inorroGod frorri ro fl p<]r .sec to 40 ft pc;r i;cc. In a distan co of 1axfl. find the rnox1inurn ~rGC fOV'
w-1oo ~onG 1a>o

ful'ri : ZJ.. s

·

R• 1010Aon$ c

hp.

A

a

S ~ 6 i 12 = 2 +12 = 14 in . 2 KE~ Y21<.(12tb) 2 /<.Es " Y2 (1000)(12-t2) = 90000

s

hp /3C,J,73

=

-100(10)-100(2000)(.2;100~ x1000=100<'20002(402-.eJ}

[F

n.

6+.+

RCCJO : Po-Hen in +.Ip- ,, r f a, 726 . 71 lb . . Hp .., fV {EVZ6.n ( 40l}/sso Hp ~ ~::i4-
in 1<-i lowoHs ? G i'ven : d • 1 il"I, .Rc'1Cl : Power G\Jtput

·

vf/zg •

J/.f ][
4-00

y = [2(:32,l1)(400)

p

160.S

~

136 .136

Vt a tf!h;

e • 001. h .. 3ff. Req~

: Hp Oufput

Se in: V..j49 " V1ftg (12)~(su) • 4l

i

bt

l'f

µ (3it,:2) t 3 + H

ti ~ ~·9876 fl. Hp1U-~& .. u)H •

tip :

e

=

3 100x10 3 498.7G 1<'10

(f.907'1) = +98.76 x103 n-lb

fNQ. )( 1 hp

sso f.+..-)£

=

725.S

E

906. B4 hp.

.

I

l

l

Pcutpul /Pinput

hp

I

I

lb.

·1

11

rJ /s

II

hp• fV • [l3 M3G(160.sTI/sso hp = 3~;L 73

206

Q

V1 • fZ fl/5

Poutpu/

PONc" "' ~Qh • fv . f'J/ ~ lS AYh 2 f ; "12.-tgv(1M)

lower w'd h

O. 8 -= Poutput / 906.64 hp

~oln:

of 12

or n

volocity 4 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 : ::irt

%ter()) ... 1ooe.:1 fb/s

wfa~ (v -v0 2 ) 2

=

·,t

~ loavc.G

207

!.

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 veloc°ity of 40.a f1 . per sec, stor ting from roGt ? If the ..SO-lb force·,~ thon removed, how much

o

\onger wil l tho block conilnuo 0 ivon : > .Goin :

to move?

·

W • ~lb, ' f &.90 1t:> ;(.C • 0,1, V"' 48.3 fl/G ~Fi< =o [so - 300(0.1~ t • 300/32.~ (48.3)

Rcqo: time

t • ~.s sec . .300(0.1) t

300/a2.~ (40.3)

c

t : 1s sec .

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.,

1504.)

Chapter

lmpulGc

~

15

Momentum

determine the I imc required to incrcoso f ho velocity of t he block frorn 10 to so por sec . . Given: f>ol'n: f

n

f".300lb~W-= ~OJ lb. rn=%. i )A... " o'.2

v1 ~1ofl/G,Y~~soft/.s

Rc.q(:j: t -1mo

,

~

o/.s(~(4ls) · [300-1!20-160(02)] t ,, 2D'.lp~2 (so-10) 148t

t

c

240,4$

= 1° 68

soc.

1543.) 0 1 ·root central ·1mpocf occurs bet. a 100 - lb body moving

or

to tho right of .s n per 6cx:i ~ 0 body 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 ofhc.r. body . G ivcn : W,.= 100 1b ,

Sol'n :

e c o.s

V"= ~ n/s , v,; ~ 2(1/.s Ws =W, Ya c.=ift/G

Rcq'd : Wc(ght

w,..v" tWeVa

c

WAv"' 1

wave

1oo(s) - .3W • 100(:-2) t wve'

e

700 "sW+Wve' -.·-GJ = va' -v..:/-tA - V~

o..5" Va'H/.s-C-.a)

Ye= Z(I/~

208

209

·!

·i

''~l . t

subsfi-luto Ve

'o

154-6.) The S!:-JS\ern shown in Fig. P - 1s4.s is used to defer' -

equation 1 ,,

mine tho c.o::ffic1onl of resii lut ion . If ball A iG refeo.s~ from

700 = 3W • W(IJ..)

rest ~ball B GWings through -&= 53.1• oflor ooing struck, determino & . Soln: eos·6o .. h/10 C.0S Sl!l,1• = h/e h -8ft. ti'="'t.eff b=10-.5 .. C" 9-4.e = 3.2{1 . v" =.[29h · ~1-2-(3-2.2.......)(,,....s Ve' " J2(~:Z:2)(:3.Z) = 14.35 ff/s = 17-9+ M Wl'..V/\ =w"v,: t WaVfJ' .

·W"HOlb

f\ golf boll iG droppod . f rorn () height Of 2~. fl ,upoh Q hordonod. .stcol plate; . 1ne cooffi·c ienf of rOGti\ufion IS 0.894; find the ho1gh~ tC> which the boll rebounds on the ffrs~, ind, v...._ third bounces. > 1544.)

h • .20 ft , c= o.s94

Re.q'c:l: h, , h2 ~ha

e =~ h1fii

•

.5n.

(o.eg4l(h1) "h2 h~ - 1~.70 ft .

Given;

.soln !

0

(o.eg4)2(hz)

=

h2

30(17,g4) • 36V,...' t 20(14.35)

h:3 = 10.21 fl .

f %, •Jh3/h2

1

Vl'.,

0.094 = f1/~o _,.. h1 = 1.s.gs.fl . . . 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 1545.)

to Giril<-0 a. If tho cocffic1ant of rc'1.1tution le o.6 , dci~rrn1ne the; onglo tT 1hrou9h w hich ba II B wi II .9W1ng· If tho lo.Gts for 0,01 sex;,.: dlGO find the aveirago impact force . / r~1' . 6 1ven : ! · '3<./ n ·

1mpoct

"

8.37

n/s

e = f/a'-yl'..' )/(:./A -va) • (14,35 - S.37)/(17.94-0) e - o.333 1547.) A . ball is t hrown 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

orrcsfi tul'ion is

& wdh the normal. Show thot !he coclTi'c1'eni oxpres.sc.d by o • tan 6-/tan-e-'. ·

·

Q,

IH

i

' ,i7 ! -e',,../ "(jY ,,,,,.,,,,,,,»;;,,,,,»,,,,

I.

.soln :

,D

c

C().&60 = h/10

vv..v,...

c = 4,61 h·0-4.61 ; 3,'.S.9 ft . COG tr -=. 3.3.9/a • O· 42375

2ova' -' ·-(!)

o.G .(v& -y,..,'. )/@1. 0'f- O) to,17 = Ye'-Vp..' - - ·

-@

-tr =6.S"

from 2, Y,...'• Ye/ -10.77 su ~titutc fo 1, .s39,2. ~ e,o(ve'- 10.77) -t QOVe" '> •.30,Q "

ao ve' - 3 .' Z3.112ova'

86P~

=..sova' -- vs'"'

11. 1226

e =(-)t'(s111-e-)(-cos-e-')

vs' · f29Z ; h· a-c

+ MVs'

©

cose-

from 1, Y/ ~ - V1.S111G-/s 1ne-'

I

17,2Q6 • ~~a2.2)c

~ w,..,vA' tWsYa'

.s:;e.2 • 3ov,..· f

,'

~ -v/~e-/y1

sin&'

h =.Sf\ · j b • 10-S = .S fl YA - ~ = j2(a2.2)(s .. 11.g4f\/s 30(17,94) " 3oV,..'

'~ ...

1

mY1 cos& • mVi''cose-' V1 sin &'· \I.I .Srn9- - · ·-

e I

~ (vi'' -v,')/(v -w.)

?o'·'

=

tone cot -e'

V{QJ!.e

v " ton-&/tan&'

a

or .so

1548.) /\ is thrown w·dh velocity fl per .sec d1·r cctcd oi 60 w ith lhc horizontal agoinst o .smooth vcrhcal woll . · 1hc ball i~ role.a.God from o -~·1 tion . 40A the woll ond 6 (lobovo

the level ground 'i frovcls in o verhcol plono. The coefficien t rcstiluhon between ball ~ woll iGo.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

or

fl/.s ;'

210

2ll

......

"

.ti t

.. '

1~.) As .shown in f(q. P-1549, o 4
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>

sol'n:

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 bp ll d1'rodly aflcr .

t1

-L --.. o·---

· lmpoot.

34,07 = 11.1s1nfb·"

Fl~Yocose- l

16.1 !2 ~ 0 .~t

"!<> = sac.osoo t

t: 1,6SGG· ycYof;1n&l - Y2Ql

1 y-

~.so

H~

t•-

1

2'.

07

-o

"

)

o< ~ ton-1 e,11./2s

,'

• 10.16

}:

ti

I

c

{

=

from rz, ~ -1otv1; sub.st. to© t

.30·(~o•v1;)

'°'l • '40V1le t ~ t -2'39,8" 70V1~ "1.,: ::-~..+a

ac>V1 ~

n;,

-fJ ~

Y4<' " 3,.+3 fi/~

to tho loft

I

'

•

I I I

fan· 1·s,4,57

0

371 27.

~G)e~

16 .. t g.Q:z.

ft/s

~.. " Vit MN;' t '/1 M1Vi 11

e • Ve'-v.,' v1 -v'J.

t.1,\4 t-Ma'ff • M.V.1' +MtV/ l
~ .ee. 66

•

1

',iM,V1't'/tM2V1

1

1

.~

~

V30 •a.~~ fl/& up to right ot 37,27• w/ the ~zonfal

a. .2 nJs

tor1·1 s.2:1.s

f

Vty -v,; •s rt/~

"'..;id

& • fan-1 V•)' /,..,;

0

n/s .

y 62 ~ 11.1+

1

• nljV1y t msV2 y

~

26, 3 1sin18.16' .. Ve2.Y

Vs~

-©

w.'

· e ~ ve2 "/. /vs1~

•

..

&

L,,,,. . . . ~ y82 y

-

from a. , v.; 10 t (-3....3) - 6 •.s7 ft/.s v,' = J~'-+v{yi. 1 Y1' •Y30 ·~'·51 +51 • 8, f6 nJs

.

Vf!ZY.- " 15

m,'°"1 t!ni.V2y

Y/ •

/<~:].,,·

O." = Y&!l'j./~6.31~10.1"

f

-&,; - v,x')/8 t 10GO!i30•

:.a6 (10&rn3(/)

,l

H/s •

2 :r. . Ile 26.31

,

60.t • 40V1'Jt

c

V1a1o('ffr;

10 ::v1~ -v1 /

x. 10.3 n.

~, -(sosin6o).2 c.·~(32.'1)(2.a.01) 1 Vy e..e ft/s' . f v•

_

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(,. · - - · -© 0.6

y,2 _ y0y 1. •.20yS '.,

"Y81· -~ \JQS I £

~-~·

e ..(vt.; -v,.,.'Vv1,. -v•~

ScG ·

x -Vocos~t · • 1 ~, 1 Q)S £8,66 (1.222)

V:,..' - Yox ii • 2 a~s It v./ -(socos6o) .. o Yx • 250/c

.,

J(8.£.); - +(161'1)(-34-.07)

t • 1.222

fl

6 t28.01 = 3+.010-

8.2_+

--a:.

.t 116·1 t z

2(16.1)

,-A(1 •6) - 1/.2 .sin.,.,. , , (32-2)(1.6)

Q0.07

w,·10lt> ....-ao11>

.l-- -" 1 1 7 • Vosin~t t hgt

212

213

Chapter 2: bor CABO supported os shown ·.n Fig . B is octed

1.) f\ r '1gid

· J ?l=, . '

upon by two equal hor·1zon-\ol fore.es P applied a t 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

h><. Rb= -Ro • MSP

~·

I

PROBLEMS .

.·

)

.,

:.:-: r

t

~

2.)

\

Owing to the weight W of the locomotive

GhO"
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. A St.. B, determine the magn.dudes of vertical ·reootions Ro 1b.. Rb. Ans. Ra " w - Pb J Rb= W t Pb 'T ~ T 2Gi""

+he

or contact

214

215

/\ 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. • roo (Fig . A). Deierrnine th<9 rnogri1tude of the resl)l\ont pul\' on \he boo\ ~-\he ongle f.> 'b...,"'< oGlelhown in the figure . /\no . P.-= .:3S2 lb ~ fd 33• ; -Y =· 2( .

.a.)

0

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

5 .)


..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 .

.

\

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 a vediool lood P applied ot 0 . Ans. ~· " QP (tene1on) 9

217 216

of

T of roclil)
i

2) /\ roller

roller

O'ler

the cvrb .

l\C &...

. p\ted

the a>
r. r 1 ~ .:::11 !Ir.... S« 1"nduc--' . th b ....... \(\ e OrG in he F19vre oue to the acti~ of the hori :i:.on.tol o

ec .

t

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.)

Ans. Pa 2-00 1b .

10001:>

~

,.

1,

,t,

.i!J.) !In electric -. light f1")(t1..1re of weight '1':1. ~ 40 lb .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 .

eorry

1r


Al60 f ind the

~ssure

/\ns. S1 " 2g.;s lb ; S2

R bdween

the bol\ 1*.. the plar,le.

,r

'I

'

3 .) A bOll of wc19ht w reGt upon 0 .smanh hor-izonfol plane oncl hos aHochecl to · con tor two Gfring~ AB ~ AC .,...\iich poss oYer ft"iationleGG pulley~ ot B ~C !!:... toads P &i-..\S!, t"e1>pecti--ely, oG Ghown . the strong AB ·,c; hon%.Ol"\tol , f"ir'd ~heanqle o<. thoi the string /\C,.~ rt;ia~ w ith tho. hor-i :t:ont o l when the ball i s

·,n a poGdlon of

4) Determine

Alis. cos« - P/6( ; R .. w -/~Lp'J.

218

'\".

219

" 2C>.7 lb .

(

Chopter

.+: .

1.) Determine the reoction o\ /\

I' · i bo OEcl !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 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

or

"

+.) A plone figure - four frorne ABO~ i~ ~upporfed on on inc li ned plane ~ looded os shown . Colcu la~e the oxi~t force induced in the member BO. Ans . Sbd .. 106 .7 lb, tenG"ion

e

the oxiol force• ·,no bon; 1.2.3,4, ~5 of the pl~ne trus~ s;;upporied ~ looded os GhCJwn. Ans. S, • -P ; S2 "+P ; .

2.) Oelermine

0

Sa• -o.,P ; 5.t"'

o.....+2 P ; Se • - 0.33.3 P.

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 produced ·,n eoch of the1:1e bore clue to the ochan of' t he 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

3 .) Oeiermine the for-ce S ·1n the bor /\S of \he plane truGG looded 1..,, supported 06 shown . An{; · S .. o.-+3 P. Chapter S',

Ki the necessary coefficie nt of friction between ~i res &.,, roadway to enable the four wheel drive ouiomob'1le to clirnb 0 30 'l. grode? A ni;. p ~ o.3 w 1.) Whot

220

221

s~

. ht"' W, t.. Wt. ore connected by 2 ) Two block. hav .ing weig fi .o 1Gtnn9 · r ~ . on hori:r.on\ol planes OS s how n . If lhe .ongle rest . of r1cr1on lhe I ior-t ch bloc!<. ·1~ f F1.nd the mog0-Yiude g..._ d1~tron o . . eas ea tied 'to the upper bloc!<- that will induce s liding . force p opp Ans. Pmin • (w1.tW2) s in'('.

r

' ".

I

, I

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 di 6tance h/+ c 3 in . o~ t~ base . This cone rests on on incl ined · plane AB, which makes ·on 01191e of 30° with the hori zon 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 the cone &..obts in the 'vertical plane of the f 1gu"' os shQl.vn . find the mo.ximurn 8tc:._ O') inimurn values of ' P consistent with equ'dibriurT) of the cone the weight w = 10 lb . Ans. Pma.>< ,,. 4.61 lb ; Prn1n" 0 ..590 lb.

or

·,r

3

l'

t

I

. Ior ) A .smooth cir-cu

CY

r nder of weight 1Q

&. rod.1us r is suppor• the .some werq. . .r en tre

1

rodiu~ ~

· · ....1- ooch of ted by two oomlcirculor cylinUCl ht ""'!2 OS Ghown . If the coomoent of' stoi1c fr-d1~ betwe •· of the sem1c . 1rculor . plane cyl1.nders ~the hor1:wntol . t h on fl at '\'faces . r bet cen the eyl1nderG em fi which \hey rest is»-= 1/1 &.-. r-19 ion . w . .-.· t b . between •--t-...1 GelveG .1<:> neglOU =- / chtermine the rnox1mum · <.111s ance b l ../ \ th nter.s B ~c for which equilibrium w ·111 be p0(;SI e w, ou e ce . · t I Plone . the middle cylinder,. touching the honi:on · / ' /\nG. bmox 2.03 I

°

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 whic h they rest. To prevent

or

~

I'

s lipping 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 Pis o vertical load . Ans . S = 1/g P

vertical~

p

I

jl 0

1/

Referring Lo the figure the coefT1.cients or fr1'dion ore OS follow: 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 .2 1b . b lock. thot will hold the {;)'Sfem in equ1hbnurn . ns. m1n· 1 .+.)

I

II

:l 2.) De ler rnine the forces pr oduced 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:>
222

223

\,•·•

' \

,·

~

distonoe between hori zontol plones ACBO A'B 'c'o' is 2a. f\ns. S -== S =- 0.2.5 P; S3 =s,.. • -1.06P; Ss=Sio "0. 1

p

2

p

111£---..10

.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 a s shown. A fl e"ible cord ·1G fostef1(jld to the wolf at E,pas:

~AF

a'

~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 .

995

3) Find the tension 1 i.n each of \he guy wires BO

~BE of the

______ --=-::~··----· · ·-·-

/\n
crof\e loaded o s shown .

e p

f

..

Chapter 7:

1~

/\

ght +)

A s trut AB o\\oched to the foce of o verhcol wollo\ A by

y

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 wail , two fof>'".es P os shown. -E< be'1ng hor1"z.on1ol bi,_P; ver-

1iool . Usin9

duced in the mernoors ·,f p " soo lb ~ \Q

the

oxiol forces pro-

""' . . 2.0 1 A~G. ".n ; Ye= Zc = 0:67 in.

ter of gravity .

o

~~ the rn~ihod of (l1()!Ylent~, flixl

homogeneous slender wire 12 in . long is bent in lwo "r ion9I~ as shown . Determine the coordinates of '1\G cen-

A

+" 0 :i

+·

...

e·

"

7

4-00 lb . Ans . S1 = 103 lb ; S2 = 162.5 lb; G3"' - 1 , 600 lb.

225 224

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 \he short ~o the center or ·g rovity c of the composite body. Ans. k;,r 6°2.8 in.

or

,.

! "·::

.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

/\ homogeneous body consists of o c'1rculor- ey\indricol porhon 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 3 .)

·I

Ans. h

A

i

·I

Ans. Ye• 1.1s in.

"'/

Ye

or

c of ~the cir-culor ,yiore fore or the hemi sphero.

.

f

I

r/{2

T

41·' I

Chapter 8: j

1.) f 1'r.d . J

4 .) A homogeneous body cos'u;ts

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

of o righ ~ c irculor coni col por-

=bh3/4 t hbj'46

or

. hon oHoched too hem·1spher'1ca l porhon rodiu ~ r os shown . Determ ine the qltihx:le h \he cone the .~nter of 9rovdy of the, composite bocly coincides w ith the centerC ot the arculor base the cone : Ans. h ,. % r.

or

·,r

or

., 226

227

i I

2) find the p0\or moment cf ined io of

&~ ~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~

ihe shaded oreo

sho-.vn ·,n figure v-i1\h respeot to poin~ 0 . Ans . J 0 = 0.27.+ ,. ....

0

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

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.

.

;

or

-...) Calculate the moment inert ia of the shaded /\rca in fig. 0 w·1th r espect to the -x - a'X 1S. Ans. I,. c :26.83 in.""'

.

228

229

' , .

. i:!ll . }~,

>·

~n:»1 n is moving dow n o slope of Q.CX)B with o veloc'dy of 30 rnph . /\t o certoin ins1ont the engineer oppli"e s the brokes 'lit.._ proch..iccs o to to~ resistance to rnohon equal to one - tent h 2.) A

o( the weigM of the troin . Whoi distonce x w'1ll the troin · trove I before stopping? /\ns. 'J. = 327 ft . .

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 . 4 .)

Ans . s ~ 60. 76 1<..m/hr. ;

t

=

7.6.39 S .

.I

.,

1

I

3.) The depth of the' crater of the tool Volcano was colclJloted

in the

following ~·o hner : From a helicopter flying vertically up-

won:J ol 6 m/s, ·a ,smoll bomb

'NOS

rcleosed ot the inston~ the

9r

.5.) ."The motion of o portio le \s 9iven by the eqlJofion s= 4t -'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.

helicopter was zorn. obove the crater surfoce. The oound eitplosion wos ~ord 9 ~econds later. If thc•·speed of sound ·,s 335 m/s' what ·, ~ the depth !he croter ? I

or

Ans. v=11 O/s ;

Ans . depth" 240.2 m.

o =16 fl/s 2

3

l !

.\'

I•

! 230

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.

~) In figure ~low, o hammer of we'19ht W - 2 lb s 1orls from rest ot /\ &... shdes down 0 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 .t c poin. w here ·,t ht .s the ground . Ans. :>< = 14-.4 ft.

roor

4 .) In whot proportion will the maximum range of o pf'Qjec tile be increased ·, f the initial velocity ·,G inoreosed by 10 per cent ?

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 .

Ans. 21 per cent .

. 2 .)

ihe pilot of on oirplone f1y1ng hon':zontol ~ ly with constant speed V"' ~ rnph ot on elevation h =20000 obove a level ploin wishes to bo;nb a forget B on the grotJnd . .At who~ ongle-e>- below the hori:c.ontol should he .see

.s) In. the figure,

the torget ot the instant of rel.easing to score

1 1

,..

o

hit? Neglect oir resistonce . v

9-=(~--- -

' -'"' .

;',,,_ ',

B

/flT7777777ij/)7/777777/,IT:ll/; /)7

232

233

the

bomb in order Ans.

-e- .. 22• 12 •

;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 s peed '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 .

Chapter 12: lhc · ormoture o f on ·eloct r 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 dcceler o tion o< assuming -\hot ·,t ·,s consiont . (n) How rnon'y complete rcvohAions docs the ormoture rnoke dur·1ng 2 tn1s ponod? /\ns .(o) ex= 101T scc- ; ( b) go rev.

1.)

For tho figure s hown w,...,. 8 rocl/G ; o<,... "' -1.s rod/st ; t ,. 2socs .. find the veloe1\y ~ occclcrotion of blocK C .

4:)

Ans. Ve= 12s rnm/s ; Oc: 37.5 mm/s2 ¢" 100mm

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. A ns. fT., 30•

1-...e

~I

-P--t 0 .!!.) For

~he

234

I

the figurn shown , o<. =

60

rod/s2.

~

Ans.

blocK C is rising .

235

S'" 6 m · How fosl Ve

c

14.7 m/s.

'i' I

'

,1 ·

'I

1,

·1·

•'

or

3.) When o boll wo'1 ght W roGt On 0 spring of constant K. producos o static doncction of 1 in. How mvch will the some boll cornpross the .Gprin9 ·,f ·,+ i~ droppc.d fromo height

·,t

h,. 1 ft? Hcg\oct tho mac;;G of tho spring

Ans.


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 . Octor mine the coefficient of f rictiori, oGcouminq thal -.t ·,s independent of vcloe1\y. 2 /. Ans. p. •Yo/ 29x

11

~

I

+.)I\ smoll b lock. of weight w=10 1b is given on 'i n itial veloc ity Yo•10 fps down tho inclin~ plonc shown in figure . If the cocrfici"cnt of frichon botwccn the pl one ~ .tho blocK· ·,s µ. == o.a, f ind tho velocity v of -the block. al B oflcr ·,+ hos \reveled o distoncod ~=.soft . Aru;. v=29.Sfps . 2.) Determine

the dynamica l deflection

o that ·will

,·ii

'I ·1

bo pro-

duce<::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 tho boorri . Ans. c5 • 1.00 in .

I

or

I 237

236

.~r

If tho syst~rn in figurv is re1eoscd from roc;t '1n tho eonflgurotion shown · by solid lineG, fi'nd tho rno1
!I.)

h -\hot tho weight P will foll . Neglect frichon "*..assume ·thot the pulleys A '*-. B aro very smo ll . Ans. h • 4P~t/(4~~-p2) .

3) A mon wc.1ghing 1so lb runs~ jumpi; from o p'1cr into

boo~ wi~h

o

o horiz.ontol velocit y Y1-= 10 fps. A~uming .tha t the impoct i'7 en t ire ly ploshc, fi'nd ~he velocit 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 .

,.:

~

'/

Chapter 1.5: ' , A locomotive weighing 60 tons hos o volocity of' 10 mph~ txlcks into o froight o:J.r weighing 10 tons thot•1s ot rest on o level trocK · Aftif' 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.

. 1.)

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 . Ans. e "'o.9.

rc:Gtit ution .

i·

ll "

I•

l

2.) A ""'ood block wc'1ghing. .5 1b 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 the

I

or

Ir 11

bullot?

Ans . v

s1010

fps.

l

I.

.1 239 238

ll

II

I

..

)

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 b lock. 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 fric tion between th~ .b locK. ~piano :1c; ,µ.. ... 1/tZ. , colcl) lo te the cliGfo nce 'X to point£ where tho bodicG c:Omo to rost . · An.s. x"" 14.2 ft .

or

.I

'-!

J;i 1tli

*~jl;

A'i

;

~i

';

.I

~!'i

..

1,1

/1 ,.-·

j

ii

/

~

I

i

lI

ti

/'

!.

I

I

1!!

..,

~

f

., I

.1· I

240 1.