, , '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·. .. · •
~
•
) ~oj,'• B~ for"th~~ ·
- 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 th<>Sb who bring the grCaicst.joy into Niel' s lU:ti. Nico, Nikki M
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Choptcr2: ~ltonl~ of~ Sydom& Choptor. 3: Equilibrium of foroo ~
;/
Chopicr .... ; /\no~is;:
-ChoPt« 6 ·="
\
Ctioi>ki- -i: .s
OloPto- a:
'"
of Strvcturos:
ChoptQr s: Friction
:I,
furn:; SysterJWi!
frl
SpOoc '
Controidif I\ Cenlc:iO; of ' Grovity Moments cl" lnortio
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' "'· 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
~\;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..)
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 !>. .
: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
-&.,. " 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
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
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.)
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)
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
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$> • 6~.s; noo·
- c - .S37.9't.5
bor.
lb J
l
lI
I
26
27
lI
''· 'I
Method I . (us;ng tt'"-.V oKe~)
FBO of the block
.ifh =O: T"' 600 c:Os 60'
C GOG 75• .. 600 cot> 60 t .S.37· 94.5 (cos 76.) t
P.sin2S• ~ Ii sin3,. • • 4'00 @ _,bc;t . , to i [N c;or..3s• (r.1n lls·j/~a,.· i Hr.in as•
p=.378-35
- ...00 N -11s.~9s lb p ·= H co~-as/co.s2&
lb..
.ti = "'1-18. 60.S lb·
0
P
~fh 0 0 :
p:
P = 'lo9.eo1(C064!1>·)- 3ooc.oco"7s" p "' 1212. 1.312 lb .
311.) If the value
or
P in Fig . P-a10
1he plane '
"f 28
i6 180 lb, determine fho an9 1e
-e- ot w/c .... ~ mu"t . be 1nchned with. the. smooth plone to hold the 300 lb t)())( in e<:juilibriu111 .
Resolving the fon:-os to 'dG equivolenf fOrce t::.,
= -HB·60S(cosaGY006llt0'
0
2 12.132 lb .
t1cos45· - 300COS7s'
_E - 370.36 lb.,
a10~ /\ :soo lb b°" is helc:I of rest on a smooih plone by a force P anclinod o~ on angle f7 w/ the plone as .shown in f iq. P - 310. If tr"' -ts c:let. the volue of P "'> the normol pres:isu~ H exerted by
N"' 409.007 lb .
p=
H • -to
-tOo & 1""5"
Force Trion9 le)
~ _aoo z _ r t . _ P_, !J()D ~.s1n-os" ,.,nio.i 6IO;;id
.:Efv =0 : H &in:+s• • :SOO sin 7s •
~fv •O : · /1~n60· •
FBO of' Cyhhder
Me1°hod I ( using
30C?lb -4001b
Method l(v.:;lng
Hcos ~/co~a· N,; ~.907 lb p .. 212,1.32 lb
300 - tfsin60· -
p
8~
cried on the.cyhdcr:
A¥."5)
rJ
<><. : 56.+4·
:. -e- "
'.3.3.ss ·
29
aHi.). Ootermi ne the magnitudes of P ~ f nece$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
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
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.)
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 .
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.
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 .
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.