255232539-Engineering-Mechanics-Statics-and-Dynamics-by-Ferdinand-Singer-Solutions.pdf
January 14, 2018 | Author: Justin Carlo Domingo | Category: N/A
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PubliiMd.A Oi11ribut6d by:
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IH N~~ Reye,, Sr."st. '
. Tet.'~.' 741-49· 16 • 741·49·20 1977 C.M. Rec:to Awn~ Tel. Nos. 741-49· 66 • 741 -49-67 Menil., Philippinff- '
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, , 'The · : book/ ~fitt tled Le'a .rnJng Guide In . . > • Jlec~-~flJcs rra,a ~rJtt~n as . text /re~Je.,er . for · EngJneerJng
.o:l ", P!esen t.J ng .. , t'iie pr Jnc Jplea· and the purpose EnrJneerJnr ·Meehan.lea '-Jn ,. . conc'epts · ol .'~pproac1i · to ' help" the· t ' a -. v ~ ry, bas Jc. an,d ~yate.matJc ·th · · . . . . . a udenta understand d. . . "~ an ·. 1 earn e . s'!bJect . matters that ""JII ' d P.ro~essu .of " thinking . . e.~.e. lop .theJr . orde-rly . .. .. . ... . ... . , .· " " " " f'.'J. t.h .. the . present e d topic th · , ·, . very compr~ hen'SJve s 'fuify.• o·l th s'. .. 't '" e 11~·~.ra wJ ~I ~ave a . ol EngJneering lie'chailJcs h e .. U{!~amental P".Jnc·Jplll'.s· var)ety . ot .·pr'actJ:cal sHua~J! ch are· applJ~able,: to . " .fde . the• , f~ .their day . to 'day· actl ~: t ~:=~ally encountered by · ' . . . . ... ' 1 " Extra - eliort ", • " ' ... b'e pr-es e nted .Jn~ ::: exerted so . that - e~eryth,Jng "ill ". and . . i .. . . . y ,p~rle c tly . unders l-ood bu' t Jn Clear cone se· Ian ua ·· · '" . ': a reduce to ~ ~inf t g ge. This ~Jll el.lmJnat~ or , / memorizing t'h .. . ,mum., h.e .!J. tu~ent 's ·a. tti: ~ude .or ,- haliU 0 t e c onc:ept wJth out u.nderstandihg·. .. · •
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- The autho_r s wJsh · to ac~n~ 1 d .. t . " . " .. JUcba·eI Si .. " e i .e. hf:dr Jndebtedne s·s . ongco, J e ttrey Borl . ~ tin"lg, ... a nd Romeo Adr·Ja l . . on~an, , ·:~.onaldo ·contrJbutJons Jn p~.eparJng an;o . ~~ their valuable· wr ng tbe manuscript.
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PEPA's IntematiOaal Book'ASsoriatians ~'bership: Asian , ..l:'aaf;ic.Pu~ ..\sscw=i•tion'tAPPA);.Association al. South EaSt Asian Publishers (ASEAP); lotem.ational Publishers ~tioa .
P~ted
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N · c-: De 1 aRaaa A · G. Mendo:za ·
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by REX 84-Bfi P-. Fiorentino St.. Sta. ~es;. H~ts. Quemn.City. Tels. 71241-08. 71241-01;
Fax 00- 711-54-12
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ACKNOWLEDGEMENT .
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Jbe authors arc perti~ularly indeb~. in the p~eparation of this book to.. Mr: Romc'Adtjano, Mr. 'Ronalda· Caiindig, . Mr°. Getardo Slimson, and . · Mr. Jeffrey BOrlong11r1. '
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They further wish·io speejally rceogni1.e the exerted effort of . Mr. Aritold·QUetua in writilig the m8i11.1sc'ript
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spcc'inl mention 10 thSb who bring the grCaicst.joy into Niel' s lU:ti. Nico, Nikki M_:_1~..
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Rosultants of · Force Systom'f
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~'JI ., f•dG\?lb.
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~w)~ . ;, . . . -&
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f' fvl c./(-.'2"f())2 f- 360, --&- " ton _, :;,6Qfa4C _ s 6 , 3 1 o
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"~" ; :,':, -;\ '·." :-..".... . .. : ... . . "·i" " ", . ' ·· ,. :· : _ .,:!fy ~ '.f'~,6iil~·~ ,~ -~~:sinaa~- 1000. -·s~i) .
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R. ·•../~Fx7 -t 2l~d~f :'or f0e eohc.u~t
lr:i f.19· P- 21a.
ly
;efx ·".° _--'fOO. COS6P• t 3~ C0S4'5 ~f=>< .· "" -:161 :07 lb '.
ioo lb '
a
0
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=
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£1+.) Defei-m.1n-6, the res~Jtqnt' of the concurr;ent system of fprces ~/->own In Ff9 · P·- QH· · ~FY.• -of900C-OG30• -soo6(~) taoqoo0s30• -
- -r(3,66 .03 lb.
4
O,mpufe: ' ,
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..
-1 (-
~1~·81)'"
i,;:· d;;~· to_•the r-~h+·
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p
- - --."--,111c---,..-,--,-.,.....-
x·
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~Fr.1oo{_~) t ~s1n6if:.~61_~;ie;~ -.~ 00 .-i.... +5" ~Fy ~ -~·~.9 lb. .
R ::{1"'2 ':fE.'"""F,....,/•...-:.t-:::i:'""Fy.,...,.2,-_ . -&-ox
!
ror-- '-
...... 2.67ft.
~
~Mo =~F_y i>< ... "fOO .. a~.f (%.61) t,)i< ii · . fer. Th~ugh '_ on·pccident l.he ·wheel fa pro~ : ~ '1he V.ol~e .;· )'_' .~ must ·b~ closed . by lhrusti'ng .o bar' }hrough ·a s/o} j~· lhe •• :1 yqlve . .S}eni ~· exe.:--h"ng 0 4 ff. ou} rr-6.~ lhe cenle~:: ; ._Delerrrune ·}he_for(je ·r:-equJ~d ~ cir.ow,, a f~· d.r~grom ·
(Oree
booj
·o f the· b.or'.
·
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6()1b
f'.'-tlqnq
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,4 P"
w·~~I ·a ~t. in di~.melei. ·~ .
60(a)
P="48 lb:..
. ·t''·'\. I,.">?.:
"...
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2~7.) The Jh~ -step-pulley . shown in Fig . P- 2+7 is subj~ctecl . to the given' couples . Compute l~e vo)ue )he resu I tonl cou p)e. /\)so deJ~inine lhe forces ocHng ot lhe rim of them~
or.
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....... :
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249.) f'.9 · ~-21:9 re.pr.esen ls }he bp vi.ew of d spe.00· · ~u . Ger v.:h101J II>; .geared for 0 ,. fo.lJI"' fO One rdduoffQf\ in S~ , The ~or:-que mptJ} at the hor1:z.onlol shoH C is· 100 l,b.-N.T forque output of }he /;lorizon}ol shaft O; because of' .the s-· . peed red~d1:on , is "'!CO lb ~ f't . Corr:ipu te }he lorq_ue .reoct1on · o) r~~ mounting bo)t,s /\.~ B h9ldlncrthe redtJCf:3r ro lhe floor. ·Hird: The torque reod rQn is caused by the unbci "lonced . torque~ whioh is o ; coi:;ple. ·"
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o clockwise couple of ....solb-A plus o 2"fO JJ-foroe dircded up to lhe right }hrough lhe orqn of X ~ Y o-xes al -&,. = 30•_ ~ lhe given eys}eni · by on equovolenl single f'orce ~ cvmpute . lhe inlercepls ils line of ocJ ion wilh .lhe x ~ Y oxes .. ·~ · .tM.) A kTc.e ..y.;fein oonsisls of
C"'30R
(-100-100) ft-lb -(30/i2)R
B"' 120 lb d fr·eded
verJ ica lly up ot /\ ~ down .o t B.
c
a
R" = F,.
~ 240 cos~·
= 207.. 85 I>..
Ry= 'Y
(to lh6 righl)
= 2-ta .sin 30.
- 120 (u~)
2so.) The con ti lever truss shown in Fig. P- 250 corries o vertical load of 2400lb . The }rusg '1s (;uppor)ed by' bea-
rings at /\ ~ 13 which. exerl the fOrces ;\ v
, /\ h. ~
or
or
S
.
Bh
2400(6) - Bh (4)
Bh .. Ah "' 3600 lb
~ d;·reo-tion of forces
Pol /\ ~
4'
'at
(2oolb-H
IP
3 '..
+'
A
3'
J R•1oo lb
R-1001b
ix =~ -
= 300
Fiq- P- 253 o G)IGfem of fbrceG roducos loo downward fOroes or -t«> I> lhrough A plus o oounlercJoc)(wise downward I • l
I
-~
Mo:-MR .
u x
.4oo(-4-) - 800 = 100X
0
x =2n right of o . Rework Prob - 253 if the .sysfem reduces to 0 left ward honzanlol fo~ of~ lb )~ pain} I\ plus 0 cfodtwiGe couple of 750 Jb - ff ...
ffB ·
R=F
+ 2.VO....P(~)
= .300 lb MR "' Wlo = -
a.s fl
)he.
left
('"'> meonG belovv o)
A short compreGSion member corr-ies on eccefl tric load P = 200 lb 51)uoled 2 in .. from lhe o'll:is of the member: OG shown in fig_ P-1255 .. In strength of moler1ols a is- ,er:., 256..)
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16
lo
300y =.300(2.) - 750
lb. ( downward)
~MA 3 -200 ""'100 ( +) ~_f (a) .f ~ 200 lb (upward)
p
o
2Sa-? In
y
l=l-):
.... rl . lell of
120
~Me., 100(1) =-200 .t P(3)
p
fl . obcNe 0
254.)
P ~ F.
· rt. G ~lb tr "'I ~~ :::r'
2.31
12
ono}her vedi~I ·F f3 111 Fig. P- 2s1 pr;'Odvce .o resu liont of' 100 lb d own a t D ") a Geufl•oPolOC,..,;.,ise couple c of 200 lb-0 . Find the magnitude· 2-51.) /\ vert.1'c 6) force
=~ 2>Z85
Av. 1
i:Y
verli'Col
1n order to const itute o couple Av ., 2'foo lb (upword)
24001b
i.y
Bh . The
forces shown con1 1tu~e two couples which must hove opp:>s'ite momen~ effects }o prevenf movement lhe truss . Determine }he mo9n·1tude the .supporti ng forces.
four
M -C"' F.,,
17
rnec:I· lho) lhe inlernal slre~es ore d etermined from the equivo\enl oxiol loo
265.) 266 ,
361 lb
.tfy " 300 ~ao +~~( /./6) - a61 (a/..Jiit) = .sg.61 ( upward)
R"' ./~F'/.'2 t~Fy"" .. ./(11·9,9)'1 t (.s9.61)"... 161 . 9161!?... (up to }he r ight)
-(7)< ~
fan-I .!?J9 .61f14g,9
-B-x " 21. 69• ~Mo "' 300 sin .30 (2) - 2~4 (1!~)(2 )-361(2/-113}(1) = - 100.6 Q~ lb (-meons Coun~er CW) 100.6/.sg.61 " 1.67 ~.Y
~f7'., 390 (1~3) t 722(o/-1i3) ~ (sinao) - 810.47 lo lhe righl £.fy"' aoo(-7(a) - 1~(iz/..fi§)~aoo~·
I
1
tone-)< = ~ry/.:EF>< - .s9 .61,/1+9.9
!.)( -
5+4.68 0
Jy
~ 1+g.g g_ (to lhe..right)
224 lb
~
Cornpule l~e resuHon) of lhe three forces shown in fig PLocale ils 1nlerseclion w'1lh lhe X 'rs Y oxe.s .
.£fl< u300.s jnao t361(ll/.J1a)-n+(o/,,f5)
..A.+----'f-4---1-'---'
ft. right of O
.. 100.6/1+9·9 = o.67 fl below o
/:E Fx
,. '
o
t' .__.._-l._.L.,~'-1.-.J.- _ lL.
1n lb
tor,&J< ,• .:Fy/~Fx -e-.,,. = lon-1 510.3/810.1'7 -Er)l = 3!2.19°
I • -.s10..s \-moons downward).
R .. ./;EF.,,.a + ~Fy¢ •/(810.1-7)~+ (-s1o.3 )2
R • 957. Q7 down lo right
.£Mo= 390(12Aa)(2)-3go(•Aa)(s)+ 122(2/,ffe)(-+)- 300sin 30(a) ~Mo= 1121. 97 0-lb CW . ·
,~ l.,c = 1121.97
Ly
o
.
.
-&,,
..
9 10·3
•
tic
Ly .:; 1121 .97
R
~.2 n .
righl o) 0
-
-= 1.3 8Q. obove Q
910.47
21 20
}on-1 2s1.0.3/419.7g
-&x ... 2B.2!5 • .
oxle
1'
s.+4.68 lb(up ~o lhe r19hl)
=· :i.fy
-&)(,.,
0
.~ l:y
~Fy¢
t
·l(+1g.7g)f2 t (257·0:3)'2
Jo righl
-&.,. " lon- 1 =.t;t.12.1° Rd .. ~Mo• 1so(1.25)-12.so(o.s) - 2so(1.!Zs) Rd .. o :. d 0 j so R po~Ges lh~gh the
....+--l-'',..._,,_-1---''""=- __ _x_
· .ifi: 1
= 4803 .9{1 lb(lo lhe right) ,
!!;hi
,i,.I
--=-==-- --=--=--
•! ~.
\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 lelf:f".i'l •Cl :
p • -t00 COG6'f.• +rt c.05 60"
p "'~O C.0566° + "'l-18•60& COG60' p"' 378 .36 lb. Method ]I[ ( tJ&in·)- 3ooc.oco"7s" p "' 1212. 1.312 lb .
311.) If the value
or
P in Fig . P-a10
1he plane '
"f 28
i6 180 lb, determine fho an9 1e
-e- ot w/c .... ~ mu"t . be 1nchned with. the. smooth plone to hold the 300 lb t)())( in eS ao'- 2oococo1os· ·
300c.v eo&60'
I
•O
= Re -
" (~309,+):z
t
R,._ ·• R.Ah t JV.v Q. R-- ·./ (1 73£.os) 4 t ( 1199, 35)2 R"' = 2106 lb
(40oo')2
Re c/( 'l.309.~) e t ( ..coo) a Ra "' 4618 .e Jb ~ 46£0 lb
1
~
Taf'l& • R,..,v t an-(
-e-- 30•
I
Cos13" 1.;4 ~Mo "' O
l1.,11
,1;
p, •COG -&-
~Fh- 0
that defined the pos1t10n of equilibr1.um.
cor;.30•
--&- tO( I r 9().
Ro • 100cosao·
e- .. 9()-«..
a6+.41 lb.
-tool'cos< • :Joorcos&-
f~ P ~ F acting olonq the boq; .Shown in Fig . P- 3'27 rna•ntoin equilibr'iurn of pin /\. Determine the volue of P \.if 6
1oocosot: aoo cos(9o-O()
100CoG"\•2oo(ctJ~·ts1n~s1noc)
P61t1'-~·
,
c
100 OOG< • :'ZOOSl•~'tO &•nO\
~···.::f.M(;.•O
l&.(f) i;insa.1 - 100(6)
I
1,
t
~-
~(12)
'
-e-- 90-C>(.
1s(P)s1n6a·+ ... 100(9) - .300 (12) P = - 1-+7. 61.I lb (- meon6 compres-
- 90 -26•.s3 ',s..i. '
-6- • 6.:1°U'.!t,e" .
sion)
lo oheOk :
Fcoso sa.1 - Pco406l5..+ - :aoo aO S90•15 (cosse.1) - ( - 147.6.r)(c:.os6a..+) - -aoo ., o
38
~&•n«Jo0
< .. 26. 33' .!!+.
~Me"'O
1801b
fOt"I"'( • ..:.:1;....00.;;__-'-
~0(
F"' 390.1s lb
tt ',
. 'I,:
©
~Mo•O
327.)
r41;1n...1·, 9 ·.
3&0(10)
~ ~nnected by a rigid rod curv~ parallel fo Hie smooth cylindricol· .&urfoce .shown in fig . P-329. Deferrn1"ne the angles 0(
200- P.c .s1nao" .:
Po·
-+Pco&,B •
s119.) TwQ cyl1"nders / \ ' B, we83hing 1oolb ~ 2oolt> relilpecfively,
2'.fv • O
Ro = Re
2.Ps1n~ •
2P(e1ns6.a1•).+ ~P(cosS(
., 64-0frZ) = 6'f-0(1a) 960
&tt)
""en.
find P to rnaintoin the equilibrium .
1. If eooh pulley 6~n in Fi9 . P-340 .....eighs 36 lb ~ w 3W1
. . R 2 ~Ra ·, s 5f1 ~
p ·,s
opari ·
oleo
en.
f'rom
1
3P - a6 t W.,
36 t 7!10
•
W
•
252Jb
.
· p.,,(~6H252.)/a
p. 961b.
o.
I
I I
I 43
42
• 12olb,
I
" A boom 115 is supporfed in a hor"1zon+ol posi tlm by oh A \.... a. roble wh - t'i fi 1nge . ic ronG rom C over o smoll pulley I D . Ghown in F • e · P-346 . r~ t h . a as ho . . \..Urnpu e t e tenG•Or\ T in the coble ~ the 3-16~
3.,.11) The wheel Joods
mine t he distonce
twice
o'
'JI
Ot"\
o jeep ore_given in Fia· P-a ....!2 . Deter -
so that the reaction of the beam a1 A ·,s
.9reot o.s \he reootion ol
r1zonta l , ~ verhco I oornponenls Ieot th e size of the pulley ol D.
i3.
or the
reoc t.ion
0
I
Neg:.
/\
.
RA·2Re (speo1f1ed condition) £MA •O
ton-e- • BA
T
ISRe" 60D(x) +1200(,.t-t)
-- ·10/G at A 1,.,.
.so lb perf1· c.ompute theho~1zontol ,-e.at the hor'rzonlol ~ verfico l oomponerih• the
. £Me•O 36 R-'. . . 600(so) ... 1aco(~o) -
or
RA=
reoction ot B ·
~Fv
Leoglh o\
h
Fo
e
. ro:
.f8c t 6 ~ 1on.
Re .. /(.soo)11. t (1z1+.12e)"
i.z...a .61
I..
361.)
The
beam -sho wn in. ·
,i
B.,i
I
." . &
c
76". 12 •
715,12·
f igure ' p - ss1 .as suppo"ted by a hi~ of /\
~ a roller on a 1 to 2 slcpe ot /3. Oeterrn'1ne the re 11· t . at !-. ~ 8 . su on reochon;S
cO
11z'
&, • 600 t .SOO t 600 t a.oOO
]I!
., 12ao.79 lbs.
,', Re ::112so.79 lbG up to the lef'l ot
0
~fv
tan-e- .; 112M.aejaoo
lbs.
£fh - o Ah 13h =12-t66.67 lbs.
"
11200 + soQ
Rev "' 1121+.~e lbs.
=soo(_.) t (l:Jd.o) t 2o::>0(12)
Ah ..
soo(120)- soo(l2o)
ibs.
•O
Rev t R.-.. • ooo t
~Me, " 0
Ah(12)
1oss. 71
=3700 lbs.
3-49.) 1he trus s . shown in f ig. P - 349 is supporied on rollers o\ /\ ~a hinge at 5 .'. ·Solve for the components Ot the recidions.
16(Rev) • "t00(1z)
Rev::
6001b ~fh
~fheO
Reh - 2-t0 lbs .
.:EMe." O 2"'!-f. •
(so) 2 t(17.3.W)~
ton¢
BO Rev
P.ev
16.10 •
RA • 1so. :27 lbs. o~ 16.10° up to t he
~Fh - O
. right. :ss!!.) The forces oct"1n9 on o
1-n \enght or o
t
"' 190.27 lb6.
= .50ji7:a.~o "'
dam ore shown
Y1 ) -e- = ~6.56°
i
ac2+o(s1nu.si>)(10) 1200(10) = ::z::zi) 1 :a()()0(6o) • eooa(4) i 1000 (~) ~ 46:27.49
lbs.
RAh "' 20CO t ~!240 sin 26. e;cS
1
..
3001 .18 lb&-.
BO RA" = Q000(10) t 1000 (6o)t12ooo(~)t aooo(21j)
so the hor1z.onla)' resistance to .slid 1.n g. ~o s lidir19 ,
+2240 (co526 ,66)(21>)
· t 2::z40(s1n ::Z6•.s6)(10) RAV .. 3376.09 lbs .
R,., • ./(?oo1 . 7s)a. t(:a:a76.09)2 tan~
.. -t.517.57
tbS.
" 3376. 09
%fh • O
3001 .75
f.: 10000 - 60CX) cos 30• f = ...,_go3.9S lbS. ~ 4900 lbs. p.10.000 1b
-t%: tC1n ~i
,0
R "' 606o .srn :.o• t
,a"'l{)OO
3376, 09
I
3001. 75
~ = ~F"
0
Z.Me-.o
if\
fig . P-353. The upword ground reaction varies unif'ormly from on inte;1siiy of p lb/rt . ot /\ to p2 lb/11 ol 8 . Delerrnin~ P1,.,,p2
+e.3s
0
• • R.-_ "' -+.517.57 lbs up to the ~i9 h~ at
-e;, .,,. 4e. 36 •
:I l ; I
:ass.~ Determine the reaci1ons at /\ ' B on the fink. +russ .show in F•g . P - 35& . Membe.rG ~ 0 ~ f6 are res ped ively perpend icu -
- 27000 lbl;. ~t.'1a=O
P4 11>/f1 R,., =a-t0:XJ(11) t 6000(+) - 10000 (6)
x .. e.++ e., 9 -
n.
9 .++ = 0,56
lor to A E ""-..BE
oi their rPidpoinls .
n. 49
48
I
357.) The uniform rod in . grov'1ty ot (j. .
o
F.1 P
9· ·357 we 'g h D~ J • ' s +20 lb~ hos 'ds ceflt ,he tens10 · th ~moothermine r n in e coble ~ the reoc• sur1aces of /\CJ... 8
·f
i1onG ot the
W" ac:Jlb HQt'>45
"'\
•
/,_B~-~M~CO..!>'
9Nsln4a• t 6tlCOG+e'
i.--- - - --60'
ton & :
1~0
AC "BG
=
j {)- • 26!:16•
1!:;/cos~6. s6·
- - -;..o·-----! :o- BF .? 16.1vcos A D'" BF,. 19176
but, sin~· =co.& 45•
2T t 2520 • 16 . Ns1n.+&•
26•.s6
ft .
2'
R,.. c.os:ao· (60)
8
Ncos-+s· -1260
" B ti s1n-+s" - 1260
AC c BG : 16.77. fl . ~Ml'>cO
T•
.o!Fh"'O : T ~ H C05-"t5.
= 1200(~) t aocx:>(+1.2.S)+ +DOD (16.n)
©
subsi . eq :2. in 1, N cos+s· - 8 Ncos-t-s·
:. T • 2s+.s6 (cos-46·) T
= 190 lb
©
-1:260
- 7Nco.s."'te• • -1260
H .. 254.56 lbs. ~MAtiO
=1200(1.s) t aooo(1a:1s) • 4000(.....,.2s)• 4000 COG!Z6·sl(tM16)
60 Rev
Rsv • 6 1.!\1
6130 lbs.
~fh·O
Rah • c
'Rah
~6) ged
=
-tOOO s1n26.G6° - R" s1nso• 4000.s1n26·~6~ - .5361·27 s '1n:ao' 359.) ' fl
/A'
bor
/AE I
~'
~up~ded by 0 hin-
The canti lever I truss shOV'Jn in fig. p-3.56 i s ot /\ o strut 8.C . Determine the reoction ot 'A
~
1()()()1b
• . IS In
orces shown in f ig p
892 .o96. fos .
eCj u1·j'b ' um uncle ' r1
r
fh
~ B. .
or ihe ri:_,e
e ocf1on
De.term ine P., R '
- 359 '
T.
=£MA'"0
Tcoe.~(e)-1 T~1n~(s)
R (16) t
-400lb By Resolving the forces to its equ1vo -
t
lenl force triongle,
16R t 8Tcose t 6T.sine-
v
subci SCj. 2 in 1,
R
16(600 - Tcos&) teTcos&t6T . / . e 1n&
..&.ic;1n30'
T =1ffiS .71 lbs .
~ ~
s 1n90'
p_,._ "" 2000 lbs up 'lo ihe r '1 9 ht ol
a
From 6().
eq.2 ,
R ~ 600 - (128.s,71) cos .36.97•
R • 4~8.57 lbs . (upward) Re ... .346+.1 lbs.
51
50
. 6000
16~600 - TqOG.36.S7 •) t 81 COS.36·07" t 6Ts I n
By s ine low,
R = 600
~ soo - Tcose-
,36. 87 •
"' 6000
600(4)
6000 .
~Fv'"O TcoG~ t
'o.
400(9)
@
©
351.)
Rer~·ng
angle
-e- at
· i
.
·
u is 1n a
..£fh-O
1~'1.
r-rvv. ~ 399 ~ It
·
.po6•t;
p + T GI0-0- "' -iOO p = .-f00- (1~·71)s"'::J6. 07• P "'371.1-£ lbs. (to the lef1)
/ 3tW.) A
t0
. T ~ 30:1lb · 11 be . . ,,.. )( -
which the bar
of
00
w• . .b . 1nol 1ned equ1 1 . num.
1
/
( 't2(00&&tsone-)
O
1 f ion60
.,
3
n • d~tennioe ~l-.e honzon~ol wroc-u ._ __
-to fL• oe> -
t 1. Ion
/
.
60!,_CD60-t s 10&)\ 1 t for)6l:> 'J
bar of negtigibte weighf n?GlG in o hon:z:onto \ pa-
sitian on ttie
~
plane s hown in Fig. P - 359. Compute the
cJeonce -x ot w hieh lood T .. 1oolb should be plaoed from pt. 8 to
Keef)
the
F~p--a59,
~ h()ri%Ql'liol . C
p-360 ...361
By Sine low,
~= ___&!-- ,, .s.nao·
904:s-
100x t-f1J0(9) • R,..ro$:a0•(t2) 100)( ==
Re
219..61(~"')(1~)- 1800
"'155-!Z&-t 6lrl6)
Oo = "t.39 - ~ "' 1 . :39f~ m - 6 - ·1. 39 = ... 6 1 fl.
T
-' : nfoo w (cos~t s ine-) 1 t fon60
ffiC06&-•~
- s.7sn . e,.7S
from sq. It'\ ,
LMo=O·
.
cos «> .. ~ .
12(cos EHG1o in fig . P-~-
the- u>hole figure, R,..v + Rov = 100+ 1oosinao•
o;ni1der'1ng £fv =O :
~Fh =o :
~fh··o: Re»1 -1ooca;ao P.oH - S6,6 lb.
0
0°96"6
~C"'866lb---· C
~... --o
RAv " l\BS tn ao• RAv •o.sAe-@ AC " AB
100lb
'1 f
o.seo + OJ5M .._ 100+~ eo+Ae= 1.!JO(~) .. :Mo-© "'fh•O: SOcos-eo• :Aeco&ao• t 1oooos30•
eo ~ "e t 190--©
.. eo - .Ae +100.=
1200 it>
b
@)
- c
eoch member .
+10oo
..
Q0.20 lb
0
·1 s
__·- T
o.e66
-
ci=.
407.) In r. .
the con1ile'lb
57 56
- --·@
6y Elirnino1ion : 1 ~ r; + (o.s.eo t o .aGGec = ~ooo)'o.s ( 0· 9~ 00 - o.!!)sc ~ 173Q) o .96'6
if' BO• ~SOO lb in 1 BC -eooo- o.e(2000)
@pt./\.
@pt- a.
(!)
BOcosao· w J:>Coos6"0" t ABoos:ao• o.e66BD•o."e l3C t ~000(0,966')
P.Av t RtN • 1!30-...-6)
.
ASt 100tAB"' 300 ,..,e =1001b
= 1000+ 200?(0.~)
o.eBDto.B6'66C
the rriemberG d(' ths l'OOf trusG
1n
c
i "T
@
sho'l'ln in rig . p -~e.
OE:61n60· "'CDs1nro· t 4!000
PE "
20QO(O.B66) t 3000
oe - .5:+e+.2
"" ..!34eo 1b
c
co/ I \oe
~Mi!!•O
.Determine . . fhti iorce 1 l'. · n members /\0, Ac, 130,CD, ~CE of' the conh lever truss .sh "' 4007.62."" 4910 lb-C
~h..O :
ec 0
4910 (cos6d)
,AC,- :245.S lb - .. -T
,._a - ao - 1eo. 2 s lb -
SlZ:
~·
eC 6 1r'l60• t
£fv ==0!
C051r'l60°
-e-
"'"\000
2600(0.066) t co (o.066)
C O"' ~01~ .94-
- 1010
Ct= .. !ZH~ lb. - · .- T
(
10
c
I
co- a.90 lb - .. - T
~fti co:
..T
ionO • fa.93
/
/ pt. G
AC- 180lb
Aacosaa.~· "' aOC06118-6'•
BD = A8 C/JS60° t l?>C ocs6o
BO ;.. 375.5 lb - .. - C
/\e"' 1so.2s lb ·
~rh ==o
fl
l?>D • 4910(0.5) t :2500 (0.!'>)
A0cos60· "' AC
AC-a Aacosaa.89
00
- ~ ~Fv ·o ~ · 19·'"' ec • !200 lb
BC "' ~~ lb - .. - 'T
/
:tfh .. 0 /\C "
@ pt. 8,
t
AC
,,. 200 cos s a .12 t 1.so C~
" .300 lb~..C
/
.'
I 59 58
~I
:·' i · rn6'mucrrs L..• .the fr....ro "' """.' in AB, BO, BE ,~ OE of -the Howe roof irvSG shown " fig. P--409. ·
409.) Determine ,
"tos.) Determine the force in each bar ti' the \russ shown in Fio P-"1-0S covood by \if\inq lhe 12.0-\b \ood oi cons\on\ ve\oci\y
of
an per sec . Who~ .change in ih66B fo;ross, ·jf any. resulh; fr()IO
p\ocing the roller Guppod ot D
~\he hinge support ot
0
£MHoO Av("+o)-6Cd..30) - 1ooo(w)--400(1 o)= 0
A?
8
1\-.i :: 1oso rb. 'k'a::=:::-;::1t-~-'?'i~~-d--1~~H.. ~ £MA ;. 0
H..r(..io)--400(30)- 1ooo(:u>)-600(~0) :o
Hv = 9.50 lb.
~~.
at I\,
:%fy•O, ~Fv=O
Av+ Ov
but
=12.0t ~/s (120)
Av = Ov
•• 20v = 12.l>t
a/s(120)
lnterchonqinq hin¥ ~ roller support will rot chonqe \he forces in each bar ell°'Pt for /\C 1*. CO
2 Ov .. 192
Ov • 96 lb.
,..f>~
ot A :
=·gc;;
t
9/10 (AS)
= 22:4 lb (teos1on) · ,
~~-£96
.BE - ~oo lb(c;vmprcssion)
:. AB• %100 lb (~ion)
/'C • 1820 ib ( tooGion) BO• BEQJG~· t f,()()(et:>s1on)
3
eo(e,fo) - Oh
- -·-C =- FH
£Fy=O /"w +O""' BG
BC • 192. 1b(tension)
61
60
c
o/"13 CO t
= :a600 -
~
O&
Cl=.
Y~ ( 216::1.33)
2400 DE " O
AC= GH ~ 3600 lb-·-T
: • co - 32 lb ( ~c:ns1on)
2163.'33 tb -··-T
00. : t=.6
AC
---+""
• • /\8 • +soo lb
bvt. Oh· 96 lb
c:
s
3
~Fi< = O
M A. ~
at
0C • 1s001bC = FG
• •
%l1/b4•0, 1800(H)t1!IO'l(-t4i)t
o\ 0 :
=
t>
~v (19)
= 1,,,.00(9)
I
+ 1!100 (12)
~v-2000lb
Method
@l pt.
of Joints .:
A~ . , /'13 "
1'f,,~O
"
N:-
~v
in 1, Rev
c
£FV•O
3600 - 2000 =: 1600 lb .
itFA.v·-Jf8"
BE. •1200 lb
- 0C = 12-400 lb . Z:ft1 -o
Rotl ~ a/9 eo ao • s~c1200)
-./!J/\S ~ S:V.v
ao ... woot> -
/\B=[s(~+
- g~
£ ..~ 10
x
x c ..s.&9' ..liiiO .. y 00~ Y"'1t.S'
~~
.. o
~-S9)t2(2.s)t 1(12.s) .. ,.6 "(n.10) OF •.s.9139 :w.s.92 ~pG - · · ,SVa(): 4A;oe t '1- • weoF 't !2 i 1
OE .. fl
69
tt.ipc; -· . - T
I
I
Ii
...,..... - fc;¥ .... tc.
J\B= !ZSOO lb
68
-1t2t~Hl+1
c
I ,I
L,.:;ht •O
R/\v (ao) - EK(i7.a2) + 1oo(a) t '.ZOO(ao-1•.s) + !200 ( :ao- ~) t ~(aa-2'.1 !1) E:K,. 6912.94-
fij.i•O CE t ::l/s PE ...
-i/ra· PF
431.) Def-ermine
Ct:."' o/{S(s.02)- '3/5(!2) CE e
+
·m~tnod
f~
Kip~ -· · - ·T
+26.) Show !hot the
the
~ 693 lb-· ·-T
.
,...__~ in the rnem berco
T.v
30 • Ae t b ·, b•10'
TI "'° "
""&
10
FH
f\ E
oe. ='20(.s1noo·)
=
1100 lb -· ·-
" ,co l"l.fl\
,
~I I
c
R>.v • 30taot $-OF+~ OG D6•
~,,,£11 fos - :ao -ao- ~(wi5]
[)6c 3.12.61
tlv
~o
Hv (10) • E!G (10 tonao·)
OF
c
t
OF(s1n.30')(20)
1500 lb:-·· - C
~MH • 0 (l-· ·-C
By section, •.,
by inspection web members JK, IJ, HI,
Ila.. Hc9
car-ries no lood ton-1 1/~ - ~..!!J7° ~~·k=O, /\v(«J) = 2cos2,.57•(!50) -&- •
t
2
(cos 26s1•)('4o) - 2(s1t"126.57.)(!S) - 2.(1o)(sin
26.!!17) --- Av ." 2.46 ~ps .-!MF•O ,
..:tMA•v 2..ok)O(q) t 1200 (u.)
""""~~....,..,..--61
"' ~/5 BF ( 18)
BF"' 2~ · ., -··-C
Av('30) = 2(CO!Pu.e7•{1ot 2o)t (s•n2c;.s1•) (2)(.St10) +GI (15)
£~ ~o : 2.«i~(.30) • -2(sm26.1S1')(s)
. GI ..
0.-1-s.9 Ki~
-··-T
- 2(smU.57°)(10)t ~(610 2~.51)(30) + ~MA • O: fG (30) • 2(cos26..57)(10120)
+(2)(~n~57•)(1!St10) FG .. 2. 24 ~ipi; -··-T
2(C06 2t0.s7•)(20)t 2. (cosu.~)(10)
OF • 2.3 ~ip& -··-C
77 76
)
;
'.:i
440.) For tho fromo loodocl
429.) For t he contilever truss .showro in F1'g . P- 429, determine
the forces ·,n members OF, FH , Fl, GI ,~ fG . @.
us
shown in Fig. P- 440,,c:Wtor -
mire lho hor izo~tol t+._ vC?rlicol companonts of' t~ pin pros · svm ot '_ 13· S'poofy dirootions (up or down; left or right) tho forco OG ·• t od~ upon rnombor CO .
H
Len of A-A
c
~MG=O ,
. s/.JZQ Of(z4) ='J.oo(60)t 2.00(40) ~
3001b ~MA
2'
400(20)
%M1 •
.£MF ~o ( ot A-A)
.
2001b
[).... = .5SO lb ~Mo
a-a
o,
'JdJ(80) . FH =1664. 81 lb _ .. -T
o£Fi< c O: GI t7$i'FI =~ .F'°l'I
· Fl
Lofl
Av= 350 lb e.;, Bil
:f'.Me"O
c i
=~oa . 2.7 lb - ..-c
BH
or c-c
3001b
e
0
F6 "',,7f3.33 lb -·: - T 430.)
The loods on tlie· 'parker tru6S shown in fig . P-+30 ore '1n
~ip.~ .
One l'-i p equolG .1000 lb . Oeterm'1ne \he forees in members 00,
:E'.Mo .. 0 811 ( 4) " 30(){.6)
BH" 450 1b
0v
~Fy"O
-t'
zM .... cO: 60FG =-1-00(60) t"'\00(40) t 200(21>)
. 8t:.CE, 'ii... OE .
•0
Av ( 4-) .. 300(6) - 2.00(2)
.s/~ FH(..w) =-400('1.0) t '.'l-00(40) t 200(~) .,.
UGI = 400(20)t 100(40) t2oo(6o) . 61.., 1166.67 lb-.. - c
=o
CN(4) • 200(2.) t 300(6)
Of-> 1256 •.s+ lb - .. -1
Lef\ of
of
Ott 0.,'5.50lb '
A.H(4)c 350(4) M
t ~(Z-)
2'
AH .. "I-SO lb
""'
~f1< *'0
s.so -ev =o
(left) ... AH "' BH "+60 lb
8v "' .5SOlb ( do.vn)
£fy=O
8.., - UJ0 - 3SO = O Bv "S.SOlb
,
H
•I
,
.£Fv=O
A" tJv = 30 ( 7)
: . Av
Svl Av •Jv
=105 !'ipG
44'1·) The struetvro shown in Fig . P-441 is hinqoo ot /\ tii.,c. find
tho horizo11tol
~vertica l compone nts
of' the hingo force ot B, 1001b
Z.M at the inleri;eciion o f 60 k, CE • 0
27/Jiii'+ 8E(160)
t 30(135) =
Ati
110(10~)
BE = 63. 88 K'.ips - .. -T
i..en
CE "' 97. 22 Kipc;-.. - T
%:t.'\E "0 1o!'l (eio) " 30(2~)+
-'/$
§ 8';e.11001b A Av~ :0,,1.., ~o
ol ti-b
80w • 1oo(s)-+ ioo( 6)
~Mot lhe inierc;ection of BO~CE•D 160(30) t 160 Of:. t 30 (135) "10!'l(110) OE "' 16 . 87.S >= 264 1b
0
hiri-
vorti OE it ociG .upon 60.
Dv(S)-= 240(3)
Ev -Av - 240.= o
Av " - 24() "t t=v Av = .2+1b .
horizon~ol ~
.Z:.ME=O
A
:f.Fv .,0:
.
·'" 445.)
oc(a) -e0(4/.s)(a)=o
Cv(s) - CH (1o)- Bv (~) co 4-
- 120 "'0
l=:v = 120 -(0 l:v ,.,· 60 lb .
Att =O
80 "'(5h)AH BO= 200 lb -··-'-C .%'."M.-1.·0: 120(+)+
+ev
J
s,..: -to(10) +120( 2) - ( 96)(s)
Con~ering iho wholo f'romo : ~fv"'O: Av
~F11 •0: BD(3/s) -
ZM.-.•O : &i ( 4 ) t
. Av= 60 lb .
....
4'
to AC
~
lsolo hng bor
:.t:Me"O: Av(B) - 120(4) -::o
120tb
AC }
C..1 -= 264(2) - 96'(3)
at iGOIQting bar AB
Bor AB
Mo
~Mo •o: CH(6) +Cv(3)-E.,(~)
X:ft-t .. O. : E~>-At-t .. O Et-1
=Q
I
.
or
~FH
.:iMe •0
~.) A three - hinged orch iG composed two trusses hi'n.(oo)-240(10) =0
AH ·-BH •O
/\v-At-1=120 -@
.I ~
@.,
a:i. © !..._
::E:Fv "'0
A,,(-+.)-AH "'-1440
Av :- Bv - 240c o
- (Av - AH ""120)
.%Ma "'0
&,, = +40- 240
sAv "' 1320
Av(eo)-aoot.G0)-600(20) =o
. Av c +40 1b
/\v "' 420 lb
8
2
AH cAv -120
=+'K:l-120 bor
i\15,
I
Ai-1
I
2o' : 'lo':
: . Ai-1
c
= -3:20 lb
320
-to tho r ight
20
3"'
3'
.......
H.r.go
AB • 700 lb _ .. - C AC= 320
/'
R2
•
@member CO
36·87 =O -7oo(cos.36.a7•)
;c, . ,._ - 240 lb.,,,
CH
Two truss;o~ ore joined os shown in Fig . P-4-4~ to form three - hinqod orch . Comput.e the horizontal vodicol com0 pol'"\cnts the hingo forco ot B ~ thcr'I determine tho typa
'*"
or
~ rnogn'1 tudo of force ·,n bors BD ~BE.
I
10'
:!:Mc =o
I 200
&
120(10) t 240(30) t AH(10) -Av (40~ a 0 Av(4)- A..i = 1440
---' ©
0
2001~/fl ~
t.olO
10'
B
I
R2
ro'
1:~, 1
R4 " 300 lb ... Cv
~
lw2
7 '
R4(6) ... 600(.3)
""'
I
• ~aool b 300
. :ODO - AH
=o
AH= 4.500 lb
-450~ A billboard BC weighir19
Wind
.:!F.c =o: E1-1 -1500 -1500 =o .+'
.::€MA = O
vv'1ncl pres.sure of
=1500 lb.
8v
Ev :::: 260 tonb . F-Fw•() %tv' is =
CH -BH "0
6H "'3000-1500
B
Ev(.50) ., wo(30) t 100(10)
1
=-1soolb.
=o : 3. )
300tb
:a!!!.F>e "'0
:l!M.A. •O
EH - 2-40
Ev(12.) "'300(16) - 24-0(10)
i::.,, =::EMi=- =-0
.EH
2~ lb
c:
.,:;Q
2.W lb .
Choptor .5
•'
Av(12) = 2~(10) - 300(6)
friction
Av F SO lb ,I
@member CE
c
1I
G."
+'
D1-1 = 480lb
0
2'..Mo =O
°"+' Ell
:t:Mc=o DH (4) "'2.40(0)
Ct1
c.. (.+) = 24-0(4')
E
CH = 24-0 lb
Ev
.2!"'1e "'0 . Dv(•)=.so(6)+aoo(12)t...eo (o) greater than & , (.l:>) ~quo l .to -e-, (c) less than .e- . (o) If ¢ is greater thon ~ th& plock will not .slide ~)own in.stood it will re-to;n '1h; poGi ti'o0- becolJGe the frlct1onol force iG so ,rnuoh that it w·i11 . hold · th~ blociK · , (IJ) If¢ iG c.Jook of weight w rest upon the lnol1ne ·shown in Fi9. P-.s12 · If thet ..c.oemcionl of frlcf ion ;~ o.\!IO, d etermine. the groatoet he-1,ghf h. af which o F'orc:;.e, P porollol to tho incl1'ne maybe applied so thot the block. willsl~e vp the inclrne w/oof f•pp;ns over. .!Jf~.)
r... -
601b
~fl:! ·O : Ne.• -400 cos ao·
ti 8
•
.·
v.,:> \·
~ r
P
3-+6·-t lb
£f-,. c O: Te "" Fe t T" t.of00£iri all
'fl
I
~
.
Reaolvlng wot Pf.O
w =·Te
.,.
p ..
F tW~na6.e7 ·
P • (o.s)(wcosa6.87\ t '} W {t;m a6.e1•)
p • O.&-+W
Ts 2 (0.1)l3'K·~ tl.OtlOO Te"' .294 .bt lb . ~F~=O
.e:~-o: N•wcosa6.&7• · ~f~co:
~~-=O
P(h) =Wcoo.36.&1'(1.)tWtm 86.97•(-.) o.6fV\l.(n) • 2.~ t 1.ow..
h -..±.. o.et
w • 20-+ .6+ lb .
q· I
I II
i 90
91
I
.I
~n ·O
fn f,'g . P-.912, ihe horno,genoolJS Ploc~ wel,ghG' ,.?>oo lb~ 1h coeAT1.oient fr1'c ho" i~ 0.40 . If h :s 1n., determine the
513 )
or
force
motion.
to 1'mperd ·
~...!.1-o
.:E:P~·D: 11•300COS36.e7' " ~-40 lb · ·
F "(o.+)( ~40) • 96 lb · ..c!MA ~o : ..SP .. wGin a6.87(4) t wc,os36.e7•(g_) SP • (.3oo)(?in 36,97.(+) t C.Of>.36.87'(2))
p
Rt ~1+.ot-'
i
R2sin1+.o+ •
from1 : R~ "a2+. - R1 - R.j "2.06. 2 COS..9-
~.
.sP • 1wo 111 p ... .z+olb .
: Ws1ne- • Rt .s11"J1+0t
~oos1n&- •(R1 tR~) G1n1+.01-· . . (R1 t R~) ~ B!.l-f..f .sin-e- (f) . : R12.. cas1+.o+ - R1cos1+.o+ • w oose (R:t. - R1) cos 1+.o+ - 200 COlX!>R12. - R1 ., :206 ,!J. coi;e:- @
~
82-f.+.sin&- -£ R1
:J06.2 COGe-
: but R, '1Qo.7G.06& ... 82f.+sine- - 2{123.7c.o&e-) • zo6. 2cose-
lhe .100-lb cylinder .shown In f19 P-814 i6' held o t res+ the ..3o" incli'ne by o weig ht P suspended from o cord .
s1+.) 00
wropped around the ~linde.r " If stipplrg lm~G' dc::.termine P'*-. the c.oofTioienf ol frd1on . . • ~Fx ·o : Hcos60 "'Fcoeao
p,
WD of t3:
p~ ·to~, 30·96
block B weighs ~':°lb, inoline. ·. Ir the. coerfic~Clf"\t
t:. ·
cosao.96 • R~ " .S~.3 . 17
A'
Fi9. P- si.s woighs 120\b,
~the cord IG porolbl to . ~_he of fr 1otlof\. fol"' 0 1! .sur.f'oc:.eG' 1n. '?°ni ovf -' ~ o.w , d~ierm1ne. the angle -e- of the 1nol;ne cit wh1oh rnohon. or f3 impends .
~f~ aQ: R¢.COGa0.06 . 2«>C06ao· + R1
fl~
p- 128.6 lb .
s19.) Jn
fie . fl,.si0 ,
two blocks
sf rut otfoohed. to ooch
If
the c.oemcront ~ /j wo'9hs '.z701b .,
~
or
f ind
-
-R1 -- ----,,
s111(00-~)
12.3,7 COSf)--
93 92
with
fr'1dionless
p1n£>.
und~r eoch block. i s o.~s the min. we.i,ght of .A to pr:evenf
rr-;ct •on
R1 • 12J.7 .sin (~o -&) • R 1 ~12a.1 (singo·rosa-- eos9o'Gin-e-)
R1
ore connected ~ a solid
b~ock.
mot ion.. 120 s1n7s.~M·
lb .
~Fx "'O; P • R~.s;nao.96 t R1 Gm ao.06 .;ioo s 1n ao·
sn) Aepeo~ illu6'. · Prob .s11, 055um1ng thot the .siNJi j6' 0 uniform rod weighln9 .300 lb Hint: rt'nsf /solofe fho sfruf
:ao
61t'\
WI\
-=-
e
as a Freeboqy dt'qgrom, resolving :fs end fbrc~ /nfo compononfs oofi'ng along ~ perpend/cu/or to the G-fru f.
1+·+·
.590.3 lb
c -
.324. 0 lb
r1 c ssota2+.a(s1nao·) H • 712,4 lb . P w F .t C =sao· a ;"-rj +¢>2t .a)(o.e 66) p = (0.2)(712.+) t 2."1.29 pe+tz4jb
force of. 4-00 !b Is oppl ie.d to the pulley shovvn ;f\. The pulley JS preve.nted F'rorn rotaflng bu Cl fo~ .P ?~led to the erd cJf the bro~ le-ve-r. Jf coef offr1cf1on. lot the brak.e surroce ls 'o.20,de-f . thevolue-of'P. 5 ~.)
1
A
F 19· P -523.
the
J
•
·,r
,
s:z.1) In ·ff9 P- s •9 J•0.3 vndor both bloolm
@!:I
7an70X- L&ine" .:_Ton~o·.x-tfonw'LCOG&
x =1on2o·Lcoce1 l{;in&-
~llnder -3 fl . In d1'omek.r ~ weighing /6 restin g On two 1'n ofined p lane as Ghow n in. fig .P.;f7 If the ar-gle of ff1dlon °1G' 1s fo r o il contact surfaces compute the rnognitvde the eovple reciuired to stort ' the cyl'1 der rotating counf~ock.wlse . 527.) /\_ homoge.A""leouG"
14
I
f./ .
.3001b
(t.cos.e 1 u ant - · -- 2'>11
w
ioos.,.
160 X ~ 8-to ~ .X - 5. 2S
2
96
97 '
<
I
; :.
.slipping imperida.
~M,t.·O
p6-.ft). • f;(1,~ t F, (t:Sl_
Pl.14 ~
£M•"'0
20.&(14 + 77.6(1-st
.. WJ.. 21-"'°'"){,¥> {_Gt Vs(,)) g-y .
:. DB I ~lb-
t
•fe(,)(~('z./g ·fi> )
1'. !J'
1 .. a ft .
125
124
=O
-400(~)-t 1000(6)
6!17,)
'
OA ·= 270-+.s lb oo ro1' occep1 Hi1s
r
/\ = 'l.00 lb . T
01\ ; 22:32. 7
volue ~
fC 3',
z MA~·o ,
t c~o) ~o
C ' 4/.lo Ey
~ tAx (:z) -~· 0000
- (Ay.!_~
c1000 :. El'~ 2!50
Dr = .soo lb. BD/~ • 5Ct>/20 :. BD • 612.4 lb. U~ng@ Arl"') - Dy • 1000
... ,.._, ...5
ore
·IOOO
2
t.J.sing ~y·1..s~ < 1 . ~(2.56) Using
'1000
~ 1000
~/30(H,Ex)
BEj.J1Ts
/\y(2)-1 A .. - Oy/2 • 400010
Ay(2) ~
t
-t b
Ay(+) t A.,(:2) -0y •00()'.) lb-~
15
7\Z • 166.7
Dy (1) - 40/1.s Ey .. 0 [)y. • .ote/3'> lfy sub&. fo @ @ 2E1 t Oy ~1000
100?(10)- Ey(10)- O.,(~o)- El' (-') •.o
:. c,. ·:Z27.7 1b .
"u'@ to
[)y(1.)-1o/1S Ey - Ey (£) "0
A.l( -Ox • "\-000 - · -®
.ZM,.. -o,·
, 61 iv .
•o -@
Ei!j1o ~ey/1s : . f:'i! • 10/ts Ey
·o
Ay(lo)t /\>1(s) - OJ1(s) • 20000
t-.1 (1.) t
AO• 202. .3lb.
D~ .; ' 61 W/5-.. - 61 lb. , I ~ ~ 227.7~/~~
0y(2.) - E~ - Ey(2.)
0,.(10)t1ooo(w)- l\y(10) - Ax (s)
'83 t Oy = 06G
Ct t
A•
.tM... •O, Oy(10) - Ez(.s):- Ey(10) ~o
-·- ©
~s·O;
C,.(5 ).t Oy (.!1) " .+?130 lb
-/\.. (1s)
~
L
35,7129
1.3-4- In.
•
s1nao·~ ~~in [ 2(2J1.3J1. (.30°.><
~
•
1.+a~•1n .
8.293 y = 9 .996
- -.91_(-1 -1) l
.g = 1.086in ~~ .. aC.J,.
T
-'ij",,.lo
a
t
7i-+.) The dimensions of the T- section of o
~
cosot -.iro~
beom are.
shown in f ig . P - 714' . How for K; the oentro"1d of the areo obove -!he boea. ~f~
-2.
J'/ieo•)) t t] ~ · [2(tx3l))(1.+a2~
T
/\ij=~A~
[1 (9)t 1 ~~1g~ .:· ~(B)(4t1)]+[1 (,)(o.s)J
4
~ .. 3 . 07 i11
i
x• o
718) Loc.o•e the centroid of the shoded oreo snown in. Fig. P - 11e . Y [c6)(12)(Ys)• Y:z(-ob.755 't
fhe- cone. . Loco•e fhe centroid of the
result of Prob .
~II: [413f(4~11 ~}~] (-.9•C.S'1S(I))
uniform wire "1s bent info fhe .shope .shown in f19 . P-762.
( 6 t11(•)tg)x;
11/9(,;)Y16)[34(1,)] - [T(+1(+X16-25]
or
The .skaig h~ segment~ Jic in •h~ }.-z ~Ions, ~ the_ 9.i n . length mo~ on angles of :so' wifh the X ox1s. The -sem1c1rculor -s~. rnents i s 1n the x - Y plane. Locate the oenl er of gro-'•+y of' the w ore
7 5+.)
j"
•
.. ++~3 .:;If;
)I
Determine thc.-height h of the cylinder mounted on the hetriispherical l:loGe show" in Fag . P- 756 so tho\ lhe com~ite body will be in sioble equi librivm on ils bose . tfinl : J\s Jong os the cenler grovity docs riol lie obove the 'j.-'1- piano there will exist o restoring oovplo whon the bod~ is tipped.
2:11- ·'l[~l y;
402.11
796 -)
:-tj
r2.. -
•
ne-t
( 7(;.02 t
109.oe) 9 18S.1
;
76.02 (2e
Y,r) f
109.0B (12
g " 286.46 ~"' 1.ss
ft. from bose
x ~e)
ii I
I
I
I
volume . UGe the
7-f.G .
145
144
x
·.
SO-..) Oe~ermine tb moment of iner\ia of o tr"longle- of base b ......, oltducle h wi\h rcsped \oon Ol(::i/+)(s-~:25)2 ]
' 3 a/+(3.5) t 12
+tr
3,4(3.5)(7)~]-!-
3
[.sA6 (2a.s) );12
of' four B by 4
by 1 in .
angles w1\h the short legs connected to o web plote 14 in by 1 in Plvc; t wo flongs plotes eoch 16 in by 2 •;_.. in a6 .Ghown in fig . P - 0.31.
154
in:+
shown in Fig . P - 033. Oe~ermine the cen koidol rnomen ls of inertia . i. =i(12)3 t[S/4(3.!>? t at+(a.s)(-t.25)~] + t 12
2
f,. .. 2G:-~ 'lllflll the 6blGs pciis& ecdl ofhcr ~ =
Is shof vcrtic.olly ·,nfo tho o;f' ot a vo loci t y of 193.2
t-+ " limo tet' the ..s .. vot - Yi.>q!~
1007..)
"'s flefl/soc, h: 1~ fl_
16.1
t • l'1mo for the 1st boll
t!O
2
1000.) /\ boll
fir~t bo ll
I
16.1lll - .. -OtSt+1"0s~OOO --s2=6+on
-
4
t
t • 10.79 soc
101+.) A train trovolc; bctwoon fwo etotions Y2 mile apor~ in. o rri1n'1mum timo o(4160C · If the troin acoe\o,.,atos ~ doocloratosot 8 fl per J;oc/l, stort from rest at the 1st Gtation ~ coming to o s top at tho Qnd .stat;on , who! iG ·,fa mo11i~urn -spoed ;n mph 7 How long
ot
1st · t 2 - +t-t+ t 12 -1 9 t ++ ~ o
1019)
Tt ~ t 1 tt'.I. +t,,
-©
~t c ' Viz ( Is g1:Vcri by s~v -9 whcro 5 i.s 1n foct ~ v in foci- por GCCOnd. Vv'.hcn t "0, S"O~ v•3
s~o
I
d"'/ci'1- =~ .
o-t relo t iol'\-5·
·l
'lcN = odht
0 = 4(2}=
1M.3.)
l"O;
o
I
I
'ti= •ff/sco
·
v'a [9 (10'?]/.[9•-1 1? -= 6 f!,4 3 2 . as= [9(•0}(]9tOTl.
1
,subef if ufo 3 lo 1
or
Gotn :
6ivon :
F·
W•:a,cr;g • 7; t;1Z%J( I0-3 lb
a - 6 m/s ~.3,.~ irl(s Flcqo : f()((Xl
2
4 •
•
ma
=
200 - 1,37(100 t .3.110) = 6.Q1 a
a:M!
19·"9 ti/sit
200 - 131 - 4,2,1CI "G.210
time until tho blocl
f!/ scoll
10.5.+.) Two bodies / \!1;, B in f ig . P- 1os + are s eparated by a opn n9 .
Tholr motion down tho ·1nolino is resiG'foci by a force P " 200 lb. the rooff1c1ent of tG1f'l&.'90(o.s) • 1~,q a
1, - 7.S·98
.subSt. 11
(o j - .. @
a - ··© .
1 1.!-11 - 1 ~. ,"I- - , , 210 _ ...
T~/r 0 • e i,. s 1.z.11e - · ·@
200-T" .,. 2ooh~.12(a) - ·{j)
"' a(X)4~~a
t>, h-11 -!ZOO.s•nao• - am C . i·
Choptcr
11
C urvi linc;or Tronsloi ion ,. ·' .
I I
174
175
1102
.) /\
s tone is thrown frorn
o
hill ol on angle of 60'
to the
t
horizonlal ,,/1\h on inif1ol velocity of 100f1 per soc . Afler hiHlng level ground ot l hc base of the hill \he slonc hos covorcd a . horiwnto l cl'istonco of soo ft . t\ow hiqh IG the hill? .
1 = 1003.;Yx t 447914.9/)(
- 100~.9>< - 447.914.
''
fofl~ :S1/><
'
Cc6 ~
R~Js.iz t;ic.!l
S " ~s7.G
=
x/fl._
1.60C.
fl .
O"Vos1nG-,.t - Y:z9t.t
(x:o)
1/ll
Yo.sin&~
R = '2.Vo~(CG~~(tori~ t
~
1106.) /\ projecti le is fired w·1 ~h an ·.nitiol volocity of'h ft. per sec. upward at on angle of ~ w·1th the horizontal . Find tho horizontal d istance c.o-tcrcd before the projectile returns to ·11./'/r
41'> r
=
Vt" 4t -6
ai." 1!l!*
fl /re. f
Ot; 1q,(a)«" 46.fYs• a 41 '" aL 41 tOn-i (1-9~)1' .. (48/ t On fl On• 48 0/6<
183 182'
(H)~
r - Hlf! .
3
fl/re.
•I
1127.) £olvo llluG. Prob.
fl pcr-.scc ;
L
11~ . uG1nq the ff : dota :
11~9.) A wci9ht
W "' 100 lb i v" M 3
vcloody Y of the weigh t the vcrt ical .
=18 in . 0ivon : L:1ein • t.511 w ~ 1001b ~ v .. 5 .(Ja fi/s So1h
0 •
.Sinec.ose-
£
t=
COG -z& ,. V-z(;()Gl!T -1 • 0
W " TCOG-6- 0. ·100
=
t · nJrJ9t~Mt 1120.) /\ rod 4fl.
c
COSIT" o.s~B -IT ".S7. 67.
V ~ t.9G ft/s
~ ~TJ1.2.,;%~.dans 7,,7•
!onl~totcs in a
-
o. 9'- wv/9 .... •O--@ -- o.U6T - o.st\. = 20 i:;q.2 - o.sr + o.S6';t1 = 20 (16)/32 (fl)
.Sin :30• = X/3
lb.
Ts 2s .1 lb.
.z!f'v-0
Goin:
.57. 3
0.06G
" = i>tf'n = 2 1((2)(.Yt)
QT (3,5J n
':::!
iho
1, 2'1s fl
-e- .. 31l"
Y • 2Trn
= 0.756 2 1
rorcl 'ii!o.... iho force on the conica l Gholl . At whof ,speccl in rprn viii\
Reqa : s peod
v = 8.07 n/s
Y2
.smooth inside -Gul"focc of o c.on1ool -shell ai Hio rote of ono rovolu hon in T/4 Gee . Asi;urninq thot 9 • :a2 fl/Gc;c~ find fhc tons.ion in tho
axic: throuqh ·,t~ eontor· Ai ooch encl of tho rod iG fOGtcnod o rord .::ift. long . Eoth' ~ supportG o weight W . Compute tho G_. paq"t'a f;oln : zFy=o [!70(4-0x 68/&o)iJ/a2.2r = !lo.1.sin14,q+3'
1133.) . To chock the
0
ft .
in fig. P-:11:% iG 80 long 1'woighs
of' tho onqine
rpm. Detonn1i1e tbo mox 1.mum bcndinq momeni M In the rod if M • WL/s I where w ic; tho totol cliGtnliutod lood "'L jc; the lenght -the rod . · Given : W •100 lb
or
n"aoorpm r-181n
1.._L~sft
. M " 100 lb-ff.
by tho parabolic cuNe y c ~ - " A car weighir1lb octing under on angle e1. • roo (Fig . A). Deierrnine th
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