Engineering Mechanics Statics and Dynamics by Ferdinand Singer Solutions

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PubliiMd.A Oi11ribut6d by:

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IH N~~ Reye,, Sr."st. '

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

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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;

Fax 00- 711-54-12

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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 thSb who bring the grCaicst.joy into Niel' s lU:ti. Nico, Nikki M_:_1~..

175-m

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Rosultants of · Force Systom'f

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~w)~ . ;, . . . -&

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fOin .

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~fh2

f' fvl c./(-.'2"f())2 f- 360, --&- " ton _, :;,6Qfa4C _ s 6 , 3 1 o

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"~" ; :,':, -;\ '·." :-..".... . .. : ... . . "·i" " ", . ' ·· ,. :· : _ .,:!fy ~ '.f'~,6iil~·~ ,~ -~~:sinaa~- 1000. -·s~i) .

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R. ·•../~Fx7 -t 2l~d~f :'or f0e eohc.u~t

lr:i f.19· P- 21a.

ly

;efx ·".° _--'fOO. COS6P• t 3~ C0S4'5 ~f=>< .· "" -:161 :07 lb '.

ioo lb '

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=

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£1+.) Defei-m.1n-6, the res~Jtqnt' of the concurr;ent system of fprces ~/->own In Ff9 · P·- QH· · ~FY.• -of900C-OG30• -soo6(~) taoqoo0s30• -

- -r(3,66 .03 lb.

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O,mpufe: ' ,

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~1~·81)'"

i,;:· d;;~· to_•the r-~h+·

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p

- - --."--,111c---,..-,--,-.,.....-

x·

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~Fr.1oo{_~) t ~s1n6if:.~61_~;ie;~ -.~ 00 .-i.... +5" ~Fy ~ -~·~.9 lb. .

R ::{1"'2 ':fE.'"""F,....,/•...-:.t-:::i:'""Fy.,...,.2,-_ . -&-ox

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ror-- '-

...... 2.67ft.

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~Mo =~F_y i>< ... "fOO .. a~.f (%.61) t,)i< ii · . 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'.

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6()1b

f'.'-tlqnq

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w·~~I ·a ~t. in di~.melei. ·~ .

60(a)

P="48 lb:..

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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.

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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. ·"

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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..)

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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

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: 1

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

!!;hi

,i,.I

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

•! ~.

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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 lelf: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·)- 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 eS ao'- 2oococo1os· ·

300c.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-- 30•

I

Cos13" 1.;4 ~Mo "' O

l1.,11

,1;

p, •COG -&-

~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(

., 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)

-- ·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. •

(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

+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&-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 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'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 "' 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-

- 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:>s1on)

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 :

=

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.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

tlv

~o

Hv (10) • E!G (10 tonao·)

OF

c

t

OF(s1n.30')(20)

1500 lb:-·· - C

~MH • 0 (l-· ·-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 >= 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.(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

. :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. )

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 (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.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,

WD 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 bloolm

@!:I

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-"'°'"){,¥> {_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/.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)

~

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))(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(-ob.755 't

fhe- cone. . Loco•e fhe centroid of the

result of Prob .

~II: [413f(4~11 ~}~] (-.9•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/+)(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:-~ '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 - .. -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 ( 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}(]9tOTl.

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

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 tG1f'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 . 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.

''

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./'/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'- 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_. paq"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 weighir1lb octing under on angle e1. • roo (Fig . A). Deierrnine th