, , '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.
SSS'·) /n Fig . .sss, o nel';ble belf rvn& A over the. ec:mpciund pullCij 13 ~ baol<-. over P to o 200- lb we(9ht . The coe ffi o;et>t o{' fri ot rol"I. _ic; Yr bdween Ih~ beH ~ /he compovrid pt..1lle!:J P . f1ncl thei rnoicrmulY\ wei9t-.t W tho1 oan be s uppor ted w/ot4f r0tot ln9 -/he> pun~ · P or G/ipp;ng the- be/Jt on the, pu lie'::! P. -
fa
~
e,
(O.s1&)(e0 xTl/r• ;-.
..
,) = 1 . 6+9
T2 fa
a
1.6t'3 Te
but Ti ·W/e
T3
•
~
7o+.7
t H .9 ( w/z.)
T3 " 0.0H3W '
109
600
lb ·
,, l
~'.
60U Dc:termino t~ mogniiude of the ~ultont • il~ pdinting. ~
I
I
its; direction c.osines for the following systom of non -coplonor-. C
Q)M~na;ms
forces P. Q.. ti...., F hove o resultan t · of ..s lb. forward S.., up to right ot -e->1 = 60., -e-.., = 60•. -frz • 4.S P e directed p01nt 01 .1,4 ). The QtJOls 2.0 lb~ pa10ses through .the ar.1gin ~ the
F z • f x2 t fy 2. t fz 2. · - (:-22.-%)2 t(- B.3'0)
:. Pit =8.73; Py " 4-.37; P2 " 17°
-&.. •a>S-
=Reos&..
F•/F
-e-,.=4e.2·
£.y • Rcasc:7y ~ s(cosGo·),,, :i.a
f!ty •
£.. • Rcos-&z
-e,.
5
1
1
., .5 (cos;60·) =2.s e
t(-23.67)"'
• cos- 2Vl6/33.68
:. ~'i ~· 16.23; ~Y •6.+g; ~z • g.74
~x
!t
F = 33.60 lb .
~ s _fu_~• 20 y ~ 6.16 )(
(cos""'"-) • 3.S4
4
1
cos- a.::lG/
0
~z =cos- 23.'17/33.60 :..e-z. ,; .....5.3.
F.,. • 2 ..5 - 4.:37 - 6.'49 fy
·12£_.,' fu'._=~· -400 ,l( y z 9~7 I • ; Bx-= -273.66; Bj • 102.44; 0z,. -2~05 >(
1 2
1
.·. 'A,. "'113.15 ;Ay = '141;.~4 ; Az.•-&1-.97
c,. • ..fL • Ci!.
2
where : R ".!'>lb.;-&,. • 6o"; &,, - 60•; ftz" -+s•
-e....01
t oge't Dvse: 0::1,,xtt -ty"-t ~ e ~ = 6:.-0~ = 200 , '
)C.Q?t.l\P •. YCOMP. xCOMP.
(0)
x
y
~
F• ?
. ~.). Dotcrmino the mognitude the rosvltanf, il.s poin11ng, ~ ifc; direction cosines for the followlng sysiem Of non -eoplonor, ·concur-rent forces. 200 lb( 4, s,-3); 400 lb(-6,4 ;!IJ ;aoo lb ( 4~ -:2, FMCE
FORGES
DIST.
CDotR)t15'11TS Of 0
FoACE p = 20lb .
a
-6.'36 lb .
... pointing
F" · :2.5- 16.:23-8.73
bock wards. ~ down......ord
to the left.
F,. • -22°46 lb.
f2. - 3 .s+- g,74 - 17.+7 fz ' - ~ .'17 lb.
.300
.5 •.39
/\ force of 100 lb is dlrecied from A toward B in ~he cube shown in fig . P-607. Oden-nine the m
.•. c,. = 222. 63; Cy" -111.32; Cz. • - 166.98 .%,. • 113.15 - 273 .GG t 222.6.3 = 6:2.12 lb . ~i"
1'11 . ++t 1e2.+.ot -1:11.32-: ~1'l-·!:l6
of the coorchna+e axes .
ro .
·-
.£z • -B+.S7- ~28.0.S - 166.96" -
R'.
£." ",..~ y 2. t.-.;.z. L 2. "" . ( G2.1'2 ) "
t(212.s~) 2 t(-47g.g)
f./
cos&,. ~ 62 .12/s2e.s.3
cos&y
c
= 0.110
212.56/s2e.sEJ. Q.402
Cos &z. " 4-79.9/.s20.S3 " O· 908
: . po;nt ing bockwords, ~upwards to ihe right
/
y
/ /
//,;./
2
R ".528 . .5.3 lb.
-.
J,"
{'
A
I
,~
r,
-
.~
d
141 -. b...... / /
I/"''
;
I/
d2
/
II
v
c
•' a'
,
/ /
/
a,..,. "'
1.1
Ji'
= 4~ t 3 2 t'4 2
= G.403
.ft.
113 112
I l
610~
f'~
• Xz:.:_ = fL,, 100 lb
• >=
·6~2.!ilb • Fy~~ . g lb ; fz•62 ..51b.
~ , •• r>t
-+
a
A force of'~ lb i~ d irected from C towan:I E. in the cube
~ in fig . P-610 . Determine ,eoch of the coordi note axes .
~-....,~
the
moment of lhe force about ~111.,.
'
~"1,. "fy(+) - fr:.(+)
.. 326.5(+)-163.3(4)
\ . 4'i.9(....) - 62.5(+) ~Mr:
~M .. = - G2.4 lb-.fl.
c
~Mic..,. ~2 . 8
-Fi(4)
~My ~ t=w (~.) • 62.!i (+)
" - 62 . .5 { +)
n.
ztv\z ~ - 2:!>0 lb.
.t.'My ~ 26'0 lb .
d · ....g n.
.. 163 .3(4) t 1tJ3.3(2) ~My" 919.e lb- H .
:t.Mz.
6 11.) A force P. d irected frorn F . toward B in the cube .shOV'ln ·,,, Fig.P-
~l<\y : -Fr:(+)
6079
= -1"l0.6(4)
..
~Mz • f,.(1) t F\-: (+)
dl!. +'t!I• t2.'
m .+(+)
d .. .s.~89ft.
zMz •.510 .Q ib-fl .
Fr "111 ...... 1b.; fz • 1+e. ~ lb.
oog.) /\ · for ce of ~·6o 'ib. ic;directed from B toword 0
in
i
Py~~ .. ~
~Mt= Px(2)
the cube
p)I
Olle& .
~
= - 100(4-) t (>00(2)
:. p .. 1077 lb .
= -000 (+)
I
Mi! • 612.) A force P is directed from
£ My = -Fz(+)
d .. .5 ff,
.. 0
. 3
...
.5
I
(-- --+..-.::-.
- 160(+)
; I
.£Ml!. = 120 lb- ft.
~
,'
/
.,
o'
~- ~·-· a
~::_ __P. _ __ __ _ ,.,/'
..
:lli!M>1 • Fz (-1) • 24C (1)
:£Mi1 • ~ lb-
o point A ( ""• 1.;4) towond o point B
,,~· .: ---~~---::;! . , I , ,'/ : : ,/' :
-~40(-4-)
z~y ~ -960 lb-ft .. 1'.Mz - fv (+)
' £!__ • ..fr_• _fi_" ~
- 3200 lb-fl .
(-3,4,-1) . If it cou~es o moment Mz· 1000 lb- 0, Oeterrnine -the men~ of' P obout the x ~ Y oxes .
df. 0•13• .... •
"
,
;
d
.,, ~
~- "-a-
= 0 .11 ft .
P • Px • Pv • P.t
Q.11 7
3
p" 911 lb .
114
115 '•
T
: . P. ~ 1iy'g.11 ; P1 £3o/9.11; Pi!: =-!:!o/g.11 £Mr• P,.(1) t Py(+) 1900 • 7/g,1' (P) ~ 3(+)%.11
f-1.
·1
Ml\= - 400 lb -·fl . £tlli! • - p.,. (+) .
0001b.
..&.~ _£____ ... 5,399
y
I
:. Py =60011;>. ; P.11 • 4001b.
£ M,. ~-Pi!. (-t) t Py ('2.)
z
1600 .. p,. (2.)
shown in fig . P-601. Oeterrfl1.n e the rnornent o f the force obout
the coordina te . :· ;'
<.?Ouc;es o moment My ... .1600 lb-rt . Oelerrrlme P ~ '1ls moment
A force of 2oolb . 1s directed from B toword C in the cube 0608~ ~hown ·1n fig . P-607. Determine ttie rnornent of the forc.e obout
each of
.. Fy(,...) - fz(<4)
'liO
,.
••. p,. • 700lb
61~ -) The fromoY110rk shown in fig . P- 615 consists of three members /\fl. /\C," /\0 whose lo....-er ends .or-e in the .some hori:cont ol plone . 't-. horizontol forc;6 of 1000 lb octing poro ll~I t6 the ')( o~is i~ oppl ied oi A Oe\erm·1ne -the in each m~mber.
; P1 =3001b; Pz • ..sOOlb
~M" ~ -Pd1) - Py(.,.)
= - .!?OQ(1) -:- 700(+)
force
M11 " -17tx:l lb-fl .
llr .
.Z.My - P;? (.+) - flll ( +}
d"1 •(6-:a}'+(6·0)'t(o-o)•
- ,!500(+)- 700(+)
dMI •6.706 1 dN:. •(a-o)• +(6-ot' t(o-(-3))"
]v\y - - BOO lb-11. 61+.) The sheor-1~ derricl-\ .shown in flg.P - 61+ suppod" a ....ed ical lood of 2000 lb . owliec! al /\ . Points B.C. "'- D ore in ihe some hof'iz.onta 1 plone ~ f\,O,~ 0 o~ in lhe XY pl<;ine· Determine the force ·in eoch member of \he derrick . . 'Y
i . - - - 20,
10'
df.c - 7.35/
d,.J •«~-0)'.1-t (6-0)i +(s-O)'z d.-.o. 8,367 ,
--l A (10.1s.o) d~· • (10-0)2 +(1s-0)2+( o-(·s>)t
r
d.-.e .. 2
d~
16.708
l ~ote
fl .
·Right Side View,
2 =(10- 0) t(1!ro) -t (10-0) 2
d"c •
2
20.616 2
d....; ~ [10 - (-20))
dAI). 33,~
lsolote Front Vio_w,
fl. t(1s - 0 )
2
.fl. £Mc. •O
£Mo•O
f.soloVing To p View,
-1ooo(B) tCy(9)'"0
rsoiatin 9 · Fro~t View, '
Cy"62Slb*MP.AC i) •O: By(6)-1000(6)•0
I
1.se·
By• 1000lb. .t:Me~ o
~ ..
·
6
o,.. (.s) - c,. (1s) .• o
Ct.
=[2000(•1J/15
- Di<(1o)t B>< frs)
C• ~ 666 .1 lb.
a .. "' [2000(1o)J/ 1s Bx • 1333-'3. lb.
£M.,.c.8 ~ 0: WX>(10)- Dy (w) ~o Dy e 10()0
L:£J_~ ~~ ~=
.33-f>+!'
15
.30
lb.
-2.L
ao
~ a fL ' ~·· C ii! - · 20 · "'6
()
10
AC"
1!! 1374 .~
6'
l\C "765.(o lb.
Jh_• Bi<
6.1tl&'
AC -~ ·
·o
. Dy• 37.S lb .
~-~ 6
8.~ '
l'\O •
.5~:2.9.+
lb.
3 ~ .. 5001b.
""8"1118 lb. ;
616.) Referring to Fig.P-615. replace. the 1000 JO force by a vertical downward lood of 'l.00() lb. Detemiioo the force in eoch member under \his revised looding .
61&.) 1he unGytnmetr·acol cont"llever fromework .shewn in fig.P-618, suppor-tli a ved icol load of 17~\b ol /\ . PO.nt~ C ~ 0 ore in the wrne verticol p lane while 0 iG 3 in fr-ont of th'1s. plane . G:lmpu{c '
Isolate Righi Side Viffl'N,
·Isolate fr-on\ -View,
y
n:
"
the force in eoch msmhsr . d,.,e•rC0-3) 1 -1(0 -(-,))1 t(+--0) 2 0 C4i:1)
:. By • ,6001b; f /\B• 877.!!J lb-·-T ; ,Bi>-400lb.
Cy (1+.120) ~-~;...;ca
'
••
/\C ·1201b ---C
10.7703
~Mc•O
0,.(10) .. -.500(+)+....oo(a)t 13600
D" "12901b. ••• /\D '\ 1~..+ lb ,
621.) /\
9 90.7'+!1 10
~. ~ - 25
;
10·7703
2000lb (0,10,0)
d"5 •
: ~- ~ 1"·"1+ 10
AP·~6 lb-C
vertical lood P" 8001b oppf1ed to the tripod sho"M'l in
fig. P- 621 cous;&.s a compress1Ve. force of 2.56 lb in \eg t\8 ~ o ccmpr.e~ive force of 293 lb 189 /\C. Determine the foree in leg NJ '°"' 'the ooordinates 'Xo ~ ~o of its lower end 0.
620.) The fromewJrk shown in Fig. P- 620 supports o verf1col ·lood of 2000 lb. Points B,C,~0 ore in ihe same horiz.ontal p\one . Oete~mine 1he for-co in each member. A
623.) Determine the mox1mum safe vertical load w that con be suppor\ecl by the tripod shown in fig . P-623 w1lhout exce.ed'1ng o compressive lood of' :Z"'
w dOI'\ ' /(0-(:2))1 t(6-0)'l t (4-0)'l.
'c:J,.,c = 1+.1+£ n.
f (o-(-~))
2
t(10-0)2-
dAO = 11.3S6
t
lb-C
/\D "..501 .5 lb ·C
d ......c .. /(0-0) 2. t (10-0) 2 + (G-o),_
dAO ..
~ 452. s
.S12 10
do,.,• 7. ~ ...
f
('2-0) 2
~I
H.
2
I - .; , ' e_/~-;r' ,.<+~
,'•
"(-~ •.o,+)
.
ei.20 -.s) doe " /
I
doe . a.062 ... /.
~~-'f~--._
doc -.t(+-0)2.•(6-0) 2 t (0-0)1
""--._c (+,o.o}
.
11
dco = 1.211'
·l'• 122
123
Llv1& .. o
lc;olote Top View,
.,. if. [)/\ • 2..fOO b. l' 2.WO .. ~"' k 7."'Hl.3
Solve Prob. 62~ ·i r; in 01do1hon io the -1000 lb, the. pla~e we1CjhG 1200lb Cen\r01d: '#,(1/JCgH - 1{i(3)(,,6)(10/3) +ft(,)(~(~~.3)
fi x : .D (10/3) ~ : 10/3
if exc.ecds 2"'1-colb
'/{(jd}(ld)
fl .
r • 'fe (?>){,¥> {_Gt Vs(,)) g-y .
:. DB I ~lb-
t
•fe(,)(~('z./g ·fi> )
1'. !J'
1 .. a ft .
125
124
=O
-400(~)-t 1000(6)
6!17,)
'
OA ·= 270-+.s lb oo ro1' occep1 Hi1s
r
/\ = 'l.00 lb . T
01\ ; 22:32. 7
volue ~
fC 3',
z MA~·o ,
t c~o) ~o
C ' 4
7.~
W
+
Z:M,....,, - o
•·,r oa • 2..+
·Ot
The; plate shown in fig . P-626 oorf'i~ o load of 1ooolb opplied ol E it; is .suPP'f"fod in o hor'1~ntol pa61fon by ~hree ver\ icol eobles ottoched ol A,.!3 .~ C. Compute the tenG10'1 ir'I eoch ooble .
~(i530·"W16 • 192+..36)
Ct •
·,r
:. oe • 2068..56 lb . :' ../ ·,1
'1 <; 2<1-00 lb ·
ma><. vQlue,
62.6·)
check DC ;:>2400lb _QQ_ . '1731.Q24 lb :. DC • 2061·MI lb 7.211 6 /'it doeS not~
= 1s3g.488
OB
~
1539-~
checl<.. it oe 72400\b
a.oa
w '1 ll
E&UILIBRIUM FCR NON -CONCURRENT SP!\CE FORCES
.. .;-,...------ c.,.
.£Mc;,"'O
~ich
·,t
Right
tol
toke front View,
l:
c
ZM0·0 JJ
Vfs.w,
1.-
Oz(s) -0.(•)- ~(a)
, r I
A
J
~ids
l
•
3'
(o./+)(§) -
•' tUtM
12CX>(10/1) t -tOOO (+) -c(10) =()
eco lb · T
.B ..
Ch·
)'
zF1 • O, -A+12aot1oa::>-000- aoo
·o
t\ • 600 lb -T ,29.) Refer. to unsymmetr1'col .conHlever framewo~k described Prob . 610 an pa96 W6 ~ repea-ted here as F«J. P-620 .. If the '-'l"'tiool ,\ood of 17001b iG sti1f\ed lo o\ the rr11dpo1nt.ormerfl ber /\f?; cornpu\s the cornponen\s \he reocf1on ci\ l3 ~the for·.
in
oot
or
oes in \he.
bor'S .AC ...., /\0.
i
.
e ,,., ""
,,,,.,,,. ; ""
dNa··~ /(e·3)1 t (o-(-f.))1 t~ ·O)._ • 8.77.S fl. d....c. /(0-0)2 t (<1--0) 2 +(e-o)" • 12 n. d...o - la-o)'- t (+-o)' t (o-(·.t))2 "'g,166f).
The boom BC of the i;diff · leg def'ricl\ .shown in Fl(JP~ ; contained in the '/.Y plone . The mo~t f\!3 \G vertical. ~ rosfs in 6 .sool<.B-t ot I\ . Points I\~ 0 ore in t~ same horizon to I plone ·. 0 Point.G . D ~ E. in the .same vertical plane-· Determine theforc,e,s In the legG l3E ~ 130 ~the comp'.ments ·or the- beai 10' c
(4}1\y - Oy • 7000
6:31.)
lb.
. ·.BE c.51S.4 lb .
~ .37.5
lb
• "t-0/:?.o (31s)
-·-@
/\y ( +) • 1000-t..SOO
Ar ,. 1soo/ + :. At
-'-9y, " 70()() )
UGin.g E,
4
c
1P,75 lb .
10/15Ey .. 14<(5(37S).::: :zso lb .
05!. • [60'.)(101Jho =- 2SO lb .
ring 1eoct'1on o1 /\.
reter'f'ir'lg
to l'.l.iBh+s;de, View :
A.,,, • Dz - l=11t
Az. •
, I
129
Z.F1 • O,.
~o
I:~ - Dz ~ Z-50- 2'50
: . Ai';•
128
=.WAo
Dt • «>/.lo Ey
~ tAx (:z) -~· 0000
- (Ay.!_~
c1000 :. El'~ 2!50
Dr = .soo lb. BD/~ • 5Ct>/20 :. BD • 612.4 lb. U~ng@ Arl"') - Dy • 1000
... ,.._, ...5
ore
·IOOO
2
t.J.sing ~y·1..s~ < 1 . ~(2.56) Using
'1000
~ 1000
~/30(H,Ex)
BEj.J1Ts
/\y(2)-1 A .. - Oy/2 • 400010
Ay(2) ~
t
-t b
Ay(+) t A.,(:2) -0y •00()'.) lb-~
15
7\Z • 166.7
Dy (1) - 40/1.s Ey .. 0 [)y. • .ote/3'> lfy sub&. fo @ @ 2E1 t Oy ~1000
cenko1d' of the shaded area shown in fig · which '"' bounded by the "'/. o·i1G', the line -x •a ~ the po-
10s.) Delerrnine t he
p- xis,
707.) Determine
.. r
shown in y
robOlo yt/• KX .
J\ •
f:
ydl< •
a
0
fo° b ~ dx
"·r
• r-!Q r~~~J:
Ay • ( Y~ydA
~ r~ ':j~dl< -g,a•b -~ ~ s ~ ' y: =:30/.s
Ay. -=
c
ydx
L&t )( u
=
dt "
..
hr b::
f bfoJo"-y;~ cJx
b% f ~]~ ·
= X/o
a
/\·%f
- ob
r
- ~b
1 • Y2 [1 t
% [-Yi_·"% (o"-l<") o/o [-Y$(o). y, o~]
9h1:
M .. f:}'i !l(lJdi<) • ~r y'dx c
COG20<]~
r Kt (i''<::-x.))
- bho"
[O( t ~i:h
.aj?-[Th +(~ - 0 -~ ] 0
a
" r)((%J~·-l<')dx
Ax .. -sac." ~x 3 + y: - 40/glr
OCOS"<~
- b/t11.fo1'/c o.. cos'o< do<
clx
.,.
Uo4- ;-:') Ax "kl<(l_:ldx)
s onO("
1.
)(~a'l
tJ "%
• ••
c
b•
.
1-
ocosoc·~
Y~f: ~(ydic)
c
O"
dx
0&1no<
£.. +Ji..
b 2 (o• -)1 12) o«
b
A= ~3ob
=
~%~
rr
y - t>JX,4a
- o/-ro: r~a&~a - ·oJ
A'1.
equotion of' !he ellipse ·,s
1./ "
• b%
y'Z" b(2;y{:, -
of the ell ipse
.
'(._.,
~
K-~%
0
Fig. P-707. The }(
!lr• KJC
'/
f he centroid of lhe quodron t
dx
ro.•,. - )(~1:
• bfao" [o.9 - o.%] - o.b,h
0
~ii q~
-
y- ~hlr
Ac ob~ 100.) Compute the oreo
of lhe sponclrel
in Fig . P -7oe bounded by
the )( oic1G, the line x ·b, ~ the curve y .. KXn where n ~O. Whof is lhe loc.otion of ·,t s centroid from I~ line " • b? Prepare the toble of r
area~ ,.., Jocotlon of centr<,.ld for -.olues
o
r
f
~
AX· xdt\ ~
"
(t>,h)
S: x(yo><)
or
n t 0 , 1, '2, &so., 3.
~ · l'-x"
""·"
~/1-n
= h/bn
I
~.,.hx%n
I
·I· V
--::;:--
j( •
~
yg r'* 7j"' b(nt1) bu! !hisx nt2
= +%.-
ie
reffered to(o,o)
132
133
x=
b
_ b(nt1}
71, .) Determine the coordinates or the centroid of the q r-eo .shown in Fig . p-71s w "1ih respect to the given 011es . ' Aij = ~Adlt
- b(nu) - b ( nt1)
nt2
n t2
f:-r(6X9) + •
~T(-a)!j .Y ·[~l')(9)('/3·9il ._[1'2lf32](~ t9]
4u+y = ·307.23
!J
x Oj
c
in .
7.+7
[v2 (6)(9)('f3.6)]
-41.1+Ji •
+1.1+ )i ·j
t [ '121(3) 2(35]
" 96.41
=2.3+
in .
716-) A slender homogeneous wire of uniform cross .seqtion i~ bent info the shope shown in fig. P - 716 . Oet. the coordir\Otes of' ils cen -
trc)1d . no.) Locate the ceni roid of !he Of'eo bounded b~ the I< oxi6' the sine curve y = os1n T">{_ from x " 0
26.56'6
r.,
=66
!j
y • 2.+e in
to :x =L- ·
26.soGx ~ -6(-+) te(+msao·t-4) 1cs.~i
•
· 'f: • 0
::·
;.. ".C
.!Jdi<
;\ij :
f>
12
~ ~o
f
~ o sm'!"~ dx
l.J LJdi<
'ft (
"J-i. ch
Cs 1n!Z.Tii. dlt
- ~a« Yr [11<
., a 'if (- caGT~)~ ,. -ol(, [ cosT9b_ ~ co~o]
~
~L
717.) Locate the c.en+ro1d of the beot wire shown in rig. P-717 . The wir-e of uniform croSG Geetion .
'is homogeneous~
..
••nso· • !Z.065 in.
;: " rs•rict ..; a ~ :aoic T/t90
- s1.,211xlL '1>
L
35,7129
1.3-4- In.
•
s1nao·~ ~~in [ 2(2J1.3J1. (.30°.><
~
•
1.+a~•1n .
8.293 y = 9 .996
- -.91_(-1 -1) l
.g = 1.086in ~~ .. aC.J,.
T
-'ij",,.lo
a
t
7i-+.) The dimensions of the T- section of o
~
cosot -.iro~
beom are.
shown in f ig . P - 714' . How for K; the oentro"1d of the areo obove -!he boea. ~f~
-2.
J'/ieo•)) t t] ~ · [2(tx3l<(adJ1.•l,W>))(1.+a2~
T
/\ij=~A~
[1 (9)t 1 ~~1g~ .:· ~(B)(4t1)]+[1 (,)(o.s)J
4
~ .. 3 . 07 i11
i
x• o
718) Loc.o•e the centroid of the shoded oreo snown in. Fig. P - 11e . Y [c6)(12)(Ys)• Y:z(<.)(<.) • Yt(<.)(')1 ij = [~M(11)(~·,· .,)] ,,: i ['12 (6)(,)(%·')] + ,. . . ['It (<.)(<.)("h · '")l -~,...+-=-'"-
,..
72E° .. 43~
~
s
oil"I.
721. =[12 (<.)(t2)(Y-'·125]
72 'i ~
t['l2('X')(r/,·i;)J +[~~)(,)(9/a ·6 to)]
= 360 .. sin. ·1
.1
135 134
I
no~ Determined
the centroid of the lines t hat f'orrn t he boundo-
12.a.) Loco~e the centroid
ry of the shoded o r eo in Fig. P - ne. [ 1H ,f18o t6· t
ffe +.J'7iJ x "' t
1:2(6) t
J100 (-lift?>h)(o/~) t .Jn (.f%)(V.fT)
.rnt(. ~)( Y-l2) +6)1
[Y!2(1.s)(G)
.
_
s
..
. 40.30~
5-26
g = 12(G)t6Cahf19oK~)(Y.rs) t 6] t ffe("!%')(Y.ff) i ffe("%)( Y-11)
24.43142
of the shaded arooo in Ffq · P-120 '1s required to
lie .on t he Y o.11is. Determine the dist once b thot will fullnll thi s requiremE.nt.
rec tong le jc; . divided into t wo padG by the cur-ve y • l'-x"
o& sliown ii' fag.;P-729. Using #"le ~n°"'n locoiion of the centroo ~ the lower part /\ oG g"1ven in toble v 11-1, show. thot the cen troid of the. upper por-t B ·,c; loco1ed by 7. B - Vii iI/d.•> ~ e • 2 gA . Y _ 61:...-en th:lt y,.. -I n_t1 ) h .. j..,. • b - ..IL ~1
\,.40t2
b(nt2)-b
nt2
~" = (nriLb · [.
~
(y(7n1).\ ~·n1-~..} n t«} )
nt1J
to~1hel"' os ;shown In
ore welded
f1rid the moment of Ql"'SO
or the upper ohonne-1 obout
the hor-lzonto l centroidol O>tlC 'k ·of the. e.nf1re, Geof.ori. oreo of 1oin - 15.s lb - + .47 in7.
·r
2 (4A7)g - +.+7 ( 0
1 10) t
+."f1(s)
e .91 g ~ i;7. .s0'+
3- 7.s& in
nt2
"I .
=
find "Xe'ii_iYe b(n) - bhl~e"' bh(ttt)-lbh
_ib chonnsl~
7.a.+ "1n~
73fl.)
A
in F.'9 ·
bridge
0
P - 7~ .
16 ccmpo.ed
fnJGS
or
the element Ghcwt"\
Refer lo Tobie. vu - g for +he properl1-es of the angles
~ locote !he centroid of' the built -up ~eotion. 19•1·
\
[ 9(1)i l'(ll..)tJ,7S ti; .-n;J g" ~"J06WS) t '1-('lz.'i..e)
• ..,s("-o-78) t ~;7s (1 .&a) ~.s
5 = 2_79.3+7.s B • 10.s-+ :ri·.
7~) U:x:ate
the centroiol of the. bu'dt -up Gection «.Chon 'inown ii"\ fig. P -73a. Refu lo fable v11-2 for- the properh-es of the elements .
-,.o) A beam hos the cross; GeOfion shown in Flq. P-730 , Compvte 1he momeni of ar-eo of the shoided portion obovt the hori:roniol cen\ro100\ o·i1G ")(., of tre enhre se,o\ion . ( t\ote : il lQ ~ in G"!~ .!J'ft of "'o\er.olG ~t thic; recooulh; is ~ in compulin<3 •he rriox1ini..1m ""heor11"19 st~~ .
I 138
139
:j
I
1M.)
/>. right trion9le of
;\~ "' (
s·1des b'--,h ·,~ rototed obout 011 o)( iS
,/~ ,,,,... .. .. ~
• b%« [
Yit bh
- ; ··
- .o.·
~ 4 dx
Vff'!.Ov.n. • •
0
2
x-
J:
~T. 'ij • I\ ~ -
• Tab
~
V " 1-Tab~
x%
• b%"-[o'-o% ta 1 - a%l
= ViJ Tb.rh
Y
,' h .
r:
-): b%·.Co2-,a)dx
coincid··ng with side h lo genef'ote o right circular ccne . y : !2tr . ii . AA • ~i . ~ b .
~(yd~) .
A~
.
De~ive the e11pressiorlG fur the s ...rfooloreo ~ volume 9enero· tod by rotoling o semicircle of ,...od ius r obovt it<> d ;om&ter .
737.)
=
b~ [
40_%]
ll!fh ~ : \bf.b.. ~:~
.!IT
V " alT. ~. Area of holf' c1ivle -~'f · ~. T r 2 ;81;.
" .. ~ bllyh De ter~med the volume of' the e llipsoid of revolu tion gene.rated by rototinq on ellipi:e abovt o) ·,t~ m0.Ja- dxiQ (pr'Olote elllp,oid) ~ b) ·,js m"1nor 011is (oblate. ~llipsoid . Tot--e the larger .sem~o,.~ oG o
A· 2J . o. 21Tr A= 4lf
7SP.) .
o.)~ · r%-.. +~81 I>
i
···--- . • __a .• ,'
A i=u.iRS;i;
140
M1.) /I 60. pipe elP<'w hos an irifernol d ;o meter ot + il"l . th d of curvo t.L)re o /he p•pe ' .~ oen~er l1"r1e. ie 6 in . f ind f he i0 ternol e ('()vol. ot ths e lbow · V ~Areal\ centroid ~d1stonoe teverGed • lr(~yz x "x (6<:>' x T/,Ba') V -= 79.96 in:'
r
.ii,., t~ Smoller ~em; - 0><1-s os b.
•
lrob
141
•ii~·
or
7~.) find fhe volume the. .sphel'icol wedge formed by rototi119 \hf"OlJ9h Ori on9le of +~· O &emicircle of 1"0d.1us r obout ·it~ 0068
diometef'. y. 'f,4 ' lk'Ai:: • T~
n .~ oreo
7'45.)
of rodi i 1.s
contoined between two c.oncer'ltric Qerr11circle. 6 in . ~ 3 in . iG ro~oted obout on ox is +in . owoy ~
porallel- . to the booo d iometers of the .semic1roles. Compuh3 fhe surfooe oreo ":> volume generoted by o comple~e revolu -
..surface oreo ~ volume 9ereroted by rota~ in'3 in Fi~. P-7+3 through one revolu t;on obovt the. X·oi<"
the
1\1~. ~~ t [~ t
!l ' I +' !!
l i(), 'O?!
c 1.485 Y'" 2T.(1.+8St+). 10.Go:3
v
· .365.1- in~
Oofet.mirie fhc surfooe oreo ~volume gene.roted by o complete revolubon obou• OlC i s of the .shaded .area Prob 719.
'
~~Gin
TMt H
2J~·H.. ~
.
= 2r. '". 7fl
Y
,..
V • 1667 in~ = T(-+)(
1{7.'
1j-;l ti;) 1 2 (') t 2 (~s·,,• .s1ni;~·+J
7"4-7.)
Aa
('C.volutio.n obouf the )( a~i~
v
7.++) The r.·101 of o pu lley hos the cross ssction -shown in Fiq. P-7#. It the rim '1G 11\0de of stee,J we:1t3ht'ri'3 4-90 lb per cu. rle.term1ne. H1C the
A= 2r. ~- le~lh of c·1 rcorrcribina,Areo A • 2lT . .3.91-[-., (11)1 + 6 t .s ~ ~,•nscJ A "" +92. s7 ·1n.2 .
n.
of'
= 2714-:!S 1n ~
]" •.s.oe 1n . v c !Zf. c..os . [ +.s ((,) t '/11(,){1.S) - T(9Y%J
1lf .s.7. 27.99
i
weight
/\ •
A " 16+
.
9- .s.7in.
,.
..Lll<..- - , - - --___::.L-.::"
2(21.%")
~ =.!!'l in
v • .2 J • !3 . Areo
:<>1. 6"'" _g - 1ai. 1:i2
/•
or
•he )(
t
yf 1l1 . 6 -1.5
in.
7-46.)
1.(;l] E • ~ F'~· ][-~<;) + ~ ~[~][ 2~l ~" 6.15 in .
g • 15. 75
~
143
[-.;,, (6)•(1,) - T(4~..)j"~
7 s1 .) Determine the centf'Oid of o hemisphere of radius r, taKing the ox is of symmetry as the z O)(is.
v-,. • (x("t':l')d1t =Tr x (r"-J<•) dx ~(~Tr~)~ c'[~ -~I nr' x - l
.3
r
~
1.sa)
A
%r
v-\ ,. " ~
"1"(-+)(~)t e(s++cos:ao')
:26.S6'°"j = 1+1.919
[
nn.) Repeat Prob. 756 if the cyl1.ndricol portion of the body in fig . P - w6 is replocod by o right conieo l port.on with 0 2 fl rodivs . base~ olf d ude h . .
ffi {J)fh(hf,-)
I
.'. )( " 0
%
lr (1)" t
Q.
in,
of o steel rivet having o cy-
.()f,9'
WsfoeJ
,.j, of
T{0.9)~(t5] ~_ •[(o/1)1f(1)ry[1 f %~1il I T(M'f(«)(1) ~
"
1 -~~ in.
·
·
1' so(16 - 11) ... ~ ..-s in from base-;from 1'>-ob.755 't
7~.) /'.. bod~ consi.s+s of o r1qht circular cone whose base Is 12 in. ~ whoi;e oltiiuc:le is 16 in . A hole a ·,n. ,in d•'ometer.~ '.'" in. deep hos 0een drilled from the boee . The o)(is of the hole co1nc1de,s w/ the 0
4-
Wtimber
"[r(sin11 1~io\) (Mm xJP"Hl)(100Jb;h') 1
Wr ~ . 109.09 \~
3.66.5 ~ = 6 -545 •
. . . . .&·
= '1/11 {#(?'!~(2))
hllp ..
heod of 1 in , radius . Use ihe result of Prob 761. fJ'2•
I.
7.se.) /\ steel ball is moun~od on top of o timber cylinder os .shown in fig . P - 7s9. Stool woighs -+90 lb porcu 0 ~ timber woigh~ 100 lb per cu ft. Ooter-rninc tho position of tho eontCI' of' qrovity.
lii:idrico I body 1 in . In diomeler ~ 2 in . long with o hemispher10ol I
2(1)
h - 3 -'f-64-ft . .
as+
Locot e the center of grovi!y
..
h11 ~ 12
e
z • 1.ZS
iri. from ope,.
h .. 1.-+1+0 .
- - -~ff
-=
;l6. 664iz
h1
-~
x •.s.:s+ in. T(.+) (~) . 26'.-56, y ... :!32 :y. u04 in. ' ~6".'566 z 6 (3) t ·a ( 't son 3¢")
_,,~ ~66"6~
1
= 11
,:r'(ef°h(h/1) - VJ1:r'(~>•C.S'1S(I))
uniform wire "1s bent info fhe .shope .shown in f19 . P-762.
( 6 t11(•)tg)x;
11/9(,;)Y16)[34(1,)] - [T(+1(+X16-25]
or
The .skaig h~ segment~ Jic in •h~ }.-z ~Ions, ~ the_ 9.i n . length mo~ on angles of :so' wifh the X ox1s. The -sem1c1rculor -s~. rnents i s 1n the x - Y plane. Locate the oenl er of gro-'•+y of' the w ore
7 5+.)
j"
•
.. ++~3 .:;If;
)I
Determine thc.-height h of the cylinder mounted on the hetriispherical l:loGe show" in Fag . P- 756 so tho\ lhe com~ite body will be in sioble equi librivm on ils bose . tfinl : J\s Jong os the cenler grovity docs riol lie obove the 'j.-'1- piano there will exist o restoring oovplo whon the bod~ is tipped.
2:11- ·'l[~l y;
402.11
796 -)
:-tj
r2.. -
•
ne-t
( 7(;.02 t
109.oe) 9 18S.1
;
76.02 (2e
Y,r) f
109.0B (12
g " 286.46 ~"' 1.ss
ft. from bose
x ~e)
ii I
I
I
I
volume . UGe the
7-f.G .
145
144
x
·.
SO-..) Oe~ermine tb moment of iner\ia of o tr"longle- of base b ......, oltducle h wi\h rcsped \oon Ol
-~· . ~~
--
I .. bh8A
•/»h
b
~) Delord•he the moment of in&-tio of the guorler 61rde ~ 10 f(Q . P-805 with res;pecl to the 91von OKe&.
f p'dl\
:I
·(ff"df
.I
J•
J •
J; p•(Tf/i·df)
o•r
.J " .L & I
Chopto: 8 Moments of Inert io
J
o
I
~ J..r-+ 6
• ix tly • lr;1 • I" tl,. '
I,. • 1r~6 l y .. 11r/{6 806.) Oetermif'lB the moment ci' inertio of the .sem·1circle shown
607.) Show the moment of inertia of o Gernicircle of rodius r is o.11r• with respect to a cenfroidal OJ(i~ parollel to the c!io~ I,. ,,. ifr4
\he radius of' g~ ion,w/ respec\ to \he Y axiG , of the Ol'\90 CU\ lr0111 the flrs't quodt•ont by ihe curve y ~ 4 - ')( 11 where ')( "*., y ore in inches.
I,.
c·i"
tAd
o.3927r• •
1,.
~
J,. +·lfr:(~)~ 2 3'1f
T(
+ o.2e2..94r
t,. "' 0 .10976 r 4 eoe.) Oeierrnine the rnomen\ of
I"'
0 .11
inel"'ha for \he quoder c;rcle
~
- (o-~). '" 4
)( ; !
~
'k
•(l'f/A)
11.y
A•r yd>< =_1'(4-l(~)dl( A
I,.v]C
")(%
wheny=O
r~
shO'N<\ in Fig. P-sos w ith respect too c.efl\roidol X- ox1s .
L
- (y-"") ')( 2 •
4
-+-/
y
0 1
1'
a10.) 'Determine the moment of inertia
J:
-[+ll -'ll•/~
A = [...{2-0) - («-o)~~J
1G
••• Ky .. ~~V(1,/3)
:.~ '4-/.s of
011.) Determine the moment
of the shaded porobolo oreo
'1nerf10
w/ respect
to +he X
oic 1s
sho'f'm in Fig . P - a11.
=f y dA
(a-i<)dy j )( •K/ tc.. z X/y''f. •a/t:J&"'( a • afb• y ~(a - o/p"y .r) dy . =_1~[0y"- ~a(t 4)] dt 11 ~ [ o/bt(Y% )]: • [ )e] 2
I)
dA
·S:
c
oyx -
I,. •
ab3 - ob% · - ob_%.s (s-a)
~
2ab){s
91~.) Oeform•ne the ly for ~he ,shaded pGll"\?boltc area
Iy ""fx•dA dA ~ yd'!-
11 -
r
Jt~ b.p/Q d)l
:s: b/w;
1.!!Jh. d;t
,. [b/,ro .,.11~h.
J:
hX0/ro) ( a - o) h. 7
-(2
~ ~
b/2.
148
1
'2./1 (o/ra) ( o ~)
"Ir ' (2/7){o-'b) .. 2ct'ly;
149
of
Fig. p -811 .
i 'I
I
e1g,) Determine the moment c£ inerlia of \he T - 6ecfion in F~ . P-820 w/ ~~ to ·,tso cen\ro1dal o*. •"
l'ITY •£Ay Y • 2(0)(2+...)
Sfl
_!So
?
t-
ShQWrl
tt(9)(1)
"~.s·1n
ix • .!:M. t z(eX1t.5)'+,S~i t2(e)(ll.5)f.
•• ••
11l
k • 2(s)(2) -32·10~ 820)
J·l~ +ly
•~1' +~ .. 270in~
"'• -[i7A .. ~21%ffl ..
1• .:: 290-67 in:+ Determine the moment of inertia of the oreo show'n
fig .P- 020 wiih respect
in . . 617.) Determine. the. rnomern of inei'~IO ~ rodn..IS of gyrohon with resopec~ to 0 polar (;.e(lh-0100\ OXIS of the Ol'OSS 9$0'\ion of0 ho 1\0W tub.9 wnOSe ou'\.$ide dlan'IC-tsr- iG 6 in· "" ·1nGide d°1ometer ie-t~ .3.07.3
to ·· ts
cenh~idol
•*
Ar " 6(1) H2(1) H2(1)
.
30y .. 12(1)(0.s) t 12(1)(1) + b(1)(13.~)
12
1r.i
oi
Atr • SfAy
y :: ~ .7in . -
lit
.3 : 12.(1) +12(1)(s.2)~H(1't +12(1)(1.3i 12
••
1
+ ~ + 6(1)(1.st 12
• T'/tJ.(r/-rt) t\"1&.11 ·1n 2
J •TAz(5...-2"'') • : K
"~J/,.: ... ~~11~11
102.11n.~ -= 2.ss in . ~
021.) Find the
rncment of inerlio
about
the indicated X
r., • e~;ol3 + e(1o)(s)~ . ~
r.. 1
"
Q')(iS
\
for the shoded oroo .shO-Nn In f 19. P- 821 . I" = f. t i'\d~ 2666°67 in ~
26~.<07 -1u;o ..S:3 :
g°",14 in~
02.2.) F1·n d the cenfro1dol moments o( inerh~ of !he ~rop<'.to'1d ~hown in Fig . P- 622. Ar = 60t(>)(6/~) .. 72 m 2
ix • [ ~3 + i2 (~)(<> )(o.sl] 2 + r;>Cf,i .
+ r;,(r;,)(o.st K .:: 6 ·7.3 in.
r~.: lu tl~ : IJ71.-a3 +211..33 =.5S"1'. G(;°1n~ JT •J1-J<1. ~2730.~C-..!!54.~<0 "' .21 7~·int
150
I,. = 190 in•
161
12
I 1
i'\c; base b hon:wn~ol. ShoH t hat ~he c.en\r-01dol rnornen\ d" inerha w1\h respect h hori-