For Combined Statics and Dynamics courses.This edition of the highly respected and well-known book for Engineering Mechanics focuses on developing a solid understanding of basic principles rather t...Descripción completa
Descripción: Lectures and notes in Statics (Handwritten).
Descrição: For Combined Statics and Dynamics courses.This edition of the highly respected and well-known book for Engineering Mechanics focuses on developing a solid understanding of basic principles rather t...
For Combined Statics and Dynamics courses.This edition of the highly respected and well-known book for Engineering Mechanics focuses on developing a solid understanding of basic principles rather t...
For Combined Statics and Dynamics courses.This edition of the highly respected and well-known book for Engineering Mechanics focuses on developing a solid understanding of basic principles rather t...
Solutions to Engineering Mechanics "RESULTANT OF ANY FORCE SYSTEM" 3rd Edition by Ferdinand SingerFull description
SOLUTIONS Engineering Mechanics
ENG MECHFull description
Dyanaamics 2nd
Link full download : http://testbankcollection.com/download/solution-manual-for-engineering-mechanics-statics-and-dynamics-2nd-edition-by-plesha-gray-costanzo/ Language: English ISBN-10: 00733803...
, , '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 .
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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.