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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
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, , 'The · : book/ ~fitt tled Le'a .rnJng Guide In . .
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.o:l ", P!esen t.J ng .. , t'iie pr Jnc Jplea·
and
the purpose
EnrJneerJnr ·Meehan.lea '-Jn
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conc'epts · ol
.'~pproac1i · to ' help" the· t ' a -. v ~ ry, bas Jc. an,d ~yate.matJc
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e . s'!bJect . matters that ""JII ' d
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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 ·
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e c onc:ept wJth out u.nderstandihg·. ..
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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 <IPA> .
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<i<Nlkka
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lhe 361 lb force wi)1 hove a · clockwise moment of 400 .f}-/p
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= -+50('4-/.5)(3) - 4,50(.ak)(1) Me· r~sdve a"l' pl.+ I . .
Mo reGOlve al pl 2
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ii<;,. £.Mo/~F/ ~ 1001.. ~/1.53·.31 ,;; 6·~ In. to. 'the )en of'O
·, iy, = ~Mo/.:e.F,. .. 1oo1.2:y6s3,gQ c 1.!;a fn obove O
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~
·~j\:~~ ,' "~6,) In fig. p1~·i~6 . ~~~uml.~ b'1od.,wlse ~menfa os pc;;~ifivo.
~;.:::.1.,,'.... Wn;ip~to.,. fh.e ,inom~~l .of. .force f ...+~. lb~: of force P=3611b.
.::.
~~\
.
-L-•
'
·R -·~6.* + E Fy~ "' .381 ..58 lb.
· :
T
. :- ..., ,. ;: (,700..sin ao~)(-:t).• (~~cosao~)(3).... ' (too sin 60~)(+)-(100 cCs60~)(3)
·
1
I
~
I
•
--=>'-
...... 2.67ft.
~
~Mo =~F_y i>< ... "fOO .. a~.f (%.61) t,)i<
ii<"' 2.oq Jen or o
~Mo = ~Fx ~.!f ' = .3~1 ~%, 61)(ly~ ...
Lii • 1.33 ff obove 0
"f-00'
. I
;
. : '~
I
"
'
?~~·f,
'~-<-.
I
'
'
.2:9.)' I~ fig. p-2~.1, a .~rce.: p inlersecls
lfy~. X o~1s q) -i' ft, to t.l;ie
rig?t ~fa . .If ,1tG ..~t obout Ai~ .t7b fl,- Jb <X't.mlerGl~k.~f ,: . .
e - ~· d· e -i.1erm1ne
.
•
"t
'·.. .· : ···
·~l ils.,,_tno.
l.m
·· enl:· abQU(
· . ' · .J3.· is· 40 ff-lb· clockw1s·
1 6' y , ·
1n ervop ·.
;
'
AT .
. .when_~~olved 0 } ·
c
3'
•
.Ma• Py(.2) .... P.y • Me/2
--;:::+-'---:----="''-:-· ,..:..B
-. P,x ..."f<>'fl-1%8,. ~ 20 lb.
pl.
, Mr." Py.(+) t B-.(3) '"170ft-lb
" 2olb(4fl) t P1< (3) ;·,,, 170fHb
''
de"' 1.!1.180
reGolv·e oi . 0 :'.'
.t.MG • f(3/.J1o)(16) t f .(1/@)(12)
..
:-
•.
••
•
·!·.
.
•
.
•
soi.y ... -t(w) ~ Lx· 2.61 ft below
=Fdg
·$·"· 1~;~7·· rt : '.· . . .
. .
' 2a1:) A force P passing lhrou9h" 'p:bii1h A· ~· B in..,Ffg. P-~3 1
~S Q cloc')f..;vi$e. rr'lOl'Tfo~t Of 300 n-(b obout 0 . 9Jmpute the ·
.
. vOlue of
r.
P·
\::··,·; :-._,
- y
/
j
•
""
.
when
re'solved
Mo· py(6) ;.
,
= 111.Blb
In-, Frq p - 2~1, . lhe rnomer.11 ·oro ceda'•n force: Fis, 1sof1lb.n\ocK:.wise.,.' obo.uf 0 ~ gq n - lb coun}erclock.wlse obQiJ.t /3.
232.)
"' .Ir 'its
mdmen h. about . A ,is zero, .delef'tm·lne
refecv,.t~ the f'igure· 231 .
when .we resolved ot /\
Mo == 100 • Fic (a)
Fx =;. 60 )bs.
fy
=/F,_
2
==
;+s lbs.
-+ fy ~
F "'
c
7S
30}b. 60}b.~lb.
40lb.
J;_;_._l ;3' ! t . ~·· ·'
2'
'
£M" c .30(2) t 60(s)-: 20(1) t
.
:;~ 660 lb- (f.
-&·
-&
dfr.A ., .6,6p/110
.
~~7.)
., 6ft.
'
\
re~ulton)
.
R • ' so +<oho
from A .
-~
o( }he four-
orm of fi'g . P - 237 .
~·
~orallel forces be~
·
-~o· • cso
10:
~Mo ·." -.so(6) t "f.Q(~)- 20(3 ) t 60(a)
1
""..:r20.0
fl - I b (cw)
~Mo= R (d fr.
4
~ -36,67 "
(11) +o
MA"'R(dfr.A)
foree ·
= o/6o
= f-on- 1 ..-~/60
ton·-eT
'
. R • -30-60 HO...:.W ? -1·1 0 lbs.
c;Jfr.O ...
w hen we resolved of .. 8 .
Mo= 90 " - F~ (a)t· f~(6)
.
f
llie
o
.236.) ;\ parolle~. force sxslem_ o.d s. ~D }he le~er shown in f(g:
P-2P6. Delerm1ne, Jhe ma .., . ...1_11.:.
·1·
..
.
·: ·
' ,· ·
· · , ., ·
_9 n11uuo
11on. of l he ·rea u Iiion ·l .
•.
,:<) pbs1
.
Determine }he
ing . on }he rocker
.300 fl
-lb
Py = 300/6 -: so lb.
· P=./P,.2-tPy2 . =/100,•+~o~
.
,L·
ol "·'
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 ·
----
/60~ t 4-9._
: '
lb.
'?
10
''
.11
',,
'I
~.I.
'l
,.
'
''
·;;• '1' ; . . ·•
'
·"
.
.
'·
-
.
;
I~ 'omouril ~- pOGitiori:.·of '~ resuiton'l or.'the
acting '"!OOlb
on ,lh'e fink
., ..Ir.us~ -shO.:....n in •fi9 .• p'- 2-"A
· ·I ·
•
-r.1 •
:•'
, 241.) .Locate
· : · loode
· '·
. . -;~
.~·· . ·:·{~
·~0., :,i{ I
·'II
.l
R= 2oot·300t_m•ao0t2oot 2000
R .. .a-+00 .lb ( djreqfe.d downwqrd}
£Mc= R'(s)
8
~· ''\ .200t aoo t
'
"'fOO
f~ t
2oo
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
F
~lb
::_.! ~:t
~Mp= -F(~) ~ zoo(s) 1' 100(~) • ·aoo(£)
or191nol · plate minus
12 -
C or
o. right or!,\ . .
f ~
•• .r.
200
lb ( do1Nnvvord)
. ~MF"' aoo(1) ~
P ..
-tOO
P(3) - ·1oo(s) - 2oo(e)
lb
(upward)
;·~;1!;~
"
·': •' ~ .··.
"
'·
,J
; .·
'
i.·.
'
t-dle · puHey lhQl ' ore req~ire~ fo. Pci10~0e0 Jheg,l'ven ·.sys}~:;
~c
sol'n':
.. ·ao(1z1.-+t9(16).,.6d(e) . ' '
= '~760' itf :ri> (.;.im~ns .eou11fi::t-.cw} , ·. .·
·f=:~ 76%~ . =·. 6a.a lb:(Cdupl~of 'thls . m~c.n ..
0
.:.;,,;.::'--::::::_ _. ··
: ~I
.. ''"' requ1f-ed) ... , .
.
.\
'
.
'.
\ ' '
,'
.
"
~'··: ~: '." .2~8.) To dose
·:.
.~;
' 4011:1 :
o
ga~e
".'91ve· ·il ·1'.s necessary
R:iex~rl tw~ :for...:
~\;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
<XJUple of im IJ-ll. De~ioe }h6 snJe fON;e, Jhol w~ll pro~ on equiVolenl effecl _ ·
!:----;-- - - -,A
1II
R-F= 100 I> 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..)
') 11{)
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<: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 !>. .
/
2s6.) /\ ver )icol ·short /\B
R
r--;-+--4c---1--J 8
1' 1'
........_ ......._
o
361 lb
•
rhe system ~r forces
rosullon) R o) /\
~
p
p
F
lhrovgh /3 ~ C.
2011>
C = =EMo
3F = 361(Q/.Jj3)(1) t361(sA13)(1) t141 .4
(1/.a)(2)-141.-f (1/-f2)(3}- 224(2/,ra)(1)
- ~2+(V..rs)(1)
R
E ""33.37 l b
b) 4P = 1001 .1
P"' ·2S.03 lb .
: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
f ig. P-~65 ~ '1ls "~ y int ercepts
>
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<" 14-1.4(Y.JI) t 300 s in6Q i 260~3"$9)
- 240 sin.30 • +79.]9 lb.
£F:1" 1-+1.4 ('/.Jl!) t260(M3) -t240 cosoao·
-3ooc.os60° • 2s1 . 03 lb.
1+1:4Jb
I
111;1;·
r
'~.1;
1..
I
I
rh,1''1i
['
1jI 1u•:;'
-(}-JC "'
Derer-mine lhe resuHont of }he lhree fOrces ochnq on )he
dom shown in fig . P-266, ~ locale "1ls intersedron w"1lh lhe
base 1416. for good design , lhis 1nler~ec1ion should occur w/i"n
lhe middle third orlhe base. Does ·,~ ?
ton·1 Z.fy/zf5c - · ion· 1 10001.os/-t00a:e2
266)
h•I'":
ji;,1'.
l~;l
I
~MA "'
t
-e-,. ..
68.2·
++BO ( 1/~)(.s) t +4-90 ( o/.f6)(10) t '1000 (s)
t
30006~)
'10o0('l0) t 1ooo(ao) .. 160087. 7'1 o.- lb.
x "' 160097· 7'1/10007. 0.3 "' 16:0
n.
~ight of"
.•
iiii··
.,,
.lfi< " 10.000 - 6000 cos 30.
11>: 1
= 4803 .9{1 lb(lo lhe
right) ,
!!;hi
,i,.I
--=-==-- --=--=--
•! ~.
\l~'i
I
'1~~·
6'
·'·1iW1
}
a·
R ..
1
·1111·•'
I
;j '
il1 1~
~I
l:l·ij
11
~f,.
t~Fy
a
!Y
1101b
180
~Fx - R>t
b
F11t110 "'1so(3/s) • o
f11 " - '100 lb . (-meonG to the le<l)
I
I
I
£F.,. ·Ry
F7 - HW t 1so(-t/s) = o
R = 2742+.02 Jb.(down · to the
"$-~ = ton· Z.Fy/.%.f-x
... lpn-1 27000/-400.3.85
-e->C ~ 79.91.
i.~ = -24000(11) + 10000(6)- booo(+)
....:: -22~000 n -: lb (- means ccw)
1
I'
;:1~ I'
""ord)
comple t ely .
=/(:+B03.65)a. •(-rnooo)a.
l \ i - - --
11•l j• •
I~
~fy• -24000 -6000 sin30·
::: -27000(- meons clown-
p ... 10,ooolb.
1111·!:
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
~~-:=:u ---.-..-rz- ~
r!ght)
~
/
_L.
0 /
therefore
r . f)( -
1~0tb
'100 lb
to the left
c~~Mo
+so· -M,. + 110(-i.) + 120(2)
Mr. • 200 0-lb (ccw)
~Ma"'
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.)
110
•"/
I
15()
,'
lb
~ ~ 2000 -
3000lb. '1000)b.1000lb
Rll • zF-x2. + ~F1 <Z.
=./(4003.92)'! .. (-10007 . 03)~
~D ..J077~_.17--1Q.(clown lo the
right)
390(1ajia) -·r,. .t 110 t 1s o(3&)
1, >/~~
IT•""'
10QO - 4490(2A!f)
_:.:,12Q07_.03JQ.(- meons down -
""'orcl)
~h •zfl'
I
I
l ho
ri'gh~.)
Ry · zfy
x li)390 -:390('5/ia) • Fy -120
+1so(+ls)
~ f):' • -1so (downword)
M - - 1
R .. F
2.
F = 2::FJC
..
+ ~ fy~
F
23
22
Uo
.:Efx • 160lb
c
- (160)~ + (-150)~
219, 32 lb (dawn ~o l"'ight .)
-Oj
= -S.15-
Mtt · ~~
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<nown to be 3611b~ pom~fe the valu86 of r .,,.p .
.:Et.41; ~MR
-36~ ($)(c)t a6l (~X+) = R (~)
R.,,. .:'f00.-+9 lb (do.Nrlvvord)
MR. = £til'le
..,00.<f.9 fi) = F (3 /9){!z)F· - ~sa. n lb
3
E.quilibrit.nl of force Systems
~Mc
4f'J").6+(3)
Chopter
r(1410¥1)
~l ="1~
P(24s)(-t)• (153.1t9) ("'k.o)6) - 1Z.53 - ~(~)(1) ...
p - 419~ - .s+ lb .
-f00.+9
(+)
25
D
304:)
Th& cylinder C ir" fig . P - .ao2 ""ei9hs 1ooolb. Orow
of cyl1'nder
C'
Cl
fBO
of rod /\B.
Cir
~
+
Ev
3'
Wc;•1oOO lb
Ah
· Av
~h
303)
1he un; forrn rod
Gel'ler
or _grovity
ih101<.ne5S
II"\.
J\~13 .· Eooh
wei911G 4!20 lb
pull--~
"' 11..
-,~ 1..
"">
~
hos
bor.
Its
FBO of '\he rod . lie_glec1 the.
o65ume all COr"toot ,s:urfoces to be,srroo\h.
o t 6 . Orow a
of the rod ~
ao+) The fr'Ome
.Fi9 . P- aoa
of
D ._.J o f th e bor /\ D shown .'" Fj9
P- 306. /\ss.urn6 oll hin9es to be smooth ~ ne_gleot the 'NC1ght of
goo.) Drow a FBD
-4-
~hown i" fi9 . P- .304- Is supported '" pivots at
w~1ghs .so .lb per fl Drow.. a ' FBD ofeoch
member
soa) The coble~ boom shown in fig . P-aoe suppor+ 0 food
of 600lb - Deterrn1'n e the lens.de rorce T ,·n the coble ~ tho
compressive force C in the boom .
t
Method I
(u~ing
l'l'&mber ·.
/\
horizontol 'vertical
Axes)
~Fh • o
Tcosao· • Ccoe +s~
6('.)0lb
CD
£f., - 0
Tson'.30· t Cs1M·s· - 600
'
r
I
'
\
l.'
©
svbst. eq 1 io 2 (~uote 1 in ~enns of" T)
T ~i~30• t- (i'cos.ao/c.os4"')(sin4s') = 600
8v
I=
lood is suppo""ted by a cable whi ch runs o ver o pulley~ iG fos\ened to !he bor OE in f ig . P- ao.s. Oro
o fBD of bors /IC ~OE ~ of the pul ley . Ass~mer oil hinges to be .smooth ~ n e_gleoi t he ws1c;ilh~
eool\.
BOS .) ;\ 600 lb
or
lb.
(cos3o•Vcos-+S·
4.39 . Q.3
C - 439.2.3
cc 5 37 ,.2,!.S JP;.
1< Method
T
I(using rdotion Ol'es)
c
~fv-O : C.sin7$> • 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
~f.,aO
p
300 • Nsin60• - Psin+s'
T "' ~Cbg.3 lb .
©
zfh·O
•Meihod JL (uSing force triongle)
Pcos 11i - Hco560.
P = H cos60;/cos1s"
®
.subst. 2 in 1
Method C(Uoing Roioled
I'.
,.
.,
"
~.)
/\ cylinder. weighing "400 lb. i" held 09a'1nGt a smoot h inc line by mOQnS of' the wo'19hi less rod A0 Jn fi'g. P·309 · Oet · pt, ti ~-
N ~. +1s . 60S tb
ti'" Ol'~)
.4.
zFh"' 0: Pcos·~·
B
c
>lf: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<;! fol"Ce Tr.ongle)
~~
,Hc,osl91i
p: N co:--ar~~· ©
v
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<J, 807 lb
-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
.
<fOOlb
·~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<KJO~ao· ~ FcoG~
f • :aoocos6o' -13a,2"1-.s (cosao~- :zoo cos1os'
F
~
.subst .
86..s7 lb.
©
CD to eq. ®
3QO .. 4<X>
cos ao'
+(.oteo
s1nao' ) /cos&-)
sinel.
B1a.) · f 1g . .:i1a l"epresents tte concur"rent force sysi em 00Hr"1C3 ot Cl
joint . of a bridge tru ss , De termine the voluos
P ~ E . to mo·1n-
or
- e- '"
toin eq1iilibr'1um of the forces.
-76.935° ~ 76,935 belew x-axis
F "' 400.sinao" - 400-•n ~·
·
us;ng Rotated ·A -,.es 'M ethod
sine-
F=
.;:Efv • O
20S.31S
lb
;;ioosin1s• t F s ·1 n7s' = -+00 s1n•5° + 200 sin10!l'
· F =·· 412.435
lb
316.)
pt
400 COG 4'!;;
0
~ 200CCG .,~· i
300ccs
is' t
~
Wlb~lb
by cosine law
w« ~ :ao.2 t-fe"- 2 ( ao)(40) C06&cos -&- "' c>.a1s
ao
-e-
f = - 126 . 33 lb
F
"28·96° "' Q9·
•by sin low
~ fh "' O
0
t tl<X>COG6<>' -
.s;u b~t;t ...
te.
f cos:so' i aoo o:>sao' ·
the vol'-'e. of' F
p+.,ooco&-;is"t aooo:x.60• •-126. 33 (~ao') +::100 oosao •
P "'"
- .sa.129 IV .
20
-
lh
30
.30
~-~
.s1no<. '"[3o (sin 29°))/ao
Sino<.
<=
o( "
,I~
cos-1 o.07s
e
-(7
p t AK>O cos 7s
• ,l i~tl
·
..
pute the volues of P ?», F ·
zFv
that the for-
..Olb
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<n 1n 1gure
1 rrc 1onleGs pu lley5'
to
. ht'
"
Determine the
f} L . Hl8
. I .wet9 s 01 2oolb
.+oolb.
of .p t hat w i II prevent rnotion .
usi~ ~1ot~ 1v<es Method
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<fO)
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-<r •
-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<><f 1 . S~ S lri&)
.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
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<S.31•) • 3500 ·
p e 901,39 lb (tenG1on)
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.)
334\.) Determine the
f i9ure
reoci ions for !he beam loaded o6 she.vi"\ in
P - 33"1'.
-£Mitt =0
'4<lO()lb
--r
.sss lbs.
R1 1 r<'a • 120
R1 - 930-sss
t 60(6) t
101<,..
-t'
1so(6l
!l
-6C0(14) - 1900(+)
R"' ~ 160? lbs
~Fy=O
10'
R, - sn lbs .
40
ro'
~
£F.,.•0
R~
fONer
.:l!MRs'O
1sRa •12o(3)t16¢(6)(1a) -t 60(6)(6)
Rz -
reoctlo 11
0
Ri
600/b
190('.)fb
10'
f
+'
600 t 19CO
l +'
Ra
r~
41
E
=R,., t Rs
9oolbs .
"0
considering the lower beam.
%.Mit1 •O
10P4 - ·1600(1+) --+000(4). 0
...ooolb
4'
l
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
.
wfa(O/~) •Wfa(d#) t P(0/12) ·
n
be from DP
w0/4 - w44 = Po/fl
w/1' (o-d) .. Po/2
W-4PD
!l(D-0)
P•960lb
l
11•-·
W •
..
lc~====Y===~c~l§.;~,~~lR=2~~====;1~
;HO.) for the
aR • 960
R • 3!ld-= R, • R:r =Ra
of pulley1;1. ~ho-Nn in rig. p - 3"f0.
Gy6 fem
the rotio of w
t1on
to p to .mointoil'l
t,
zfv • O
Re "'R1tRe
• 9P
Wjp :: 9
~Mit:s
"' o
p(i<) ....
Re (y) =R2(12)
6ft.
l
I
Re (12)
!
P(i<) ,. &t0(1£)
960 )(
'>'
~M~ · O
y" 320(1;1
6'40
rnc-
w
12-y
• 320 t320
Re ~ 6'4-0 lb
~uilibrium . Me,g\eot O'llle
W "'3P t 3P t 3P
Re
'(
determ'1 ne
~ the weights of the pulleys .
~eriri(j ihe Jower .beam.
y ..
!2PO
(o-d)
;>(
., 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)
-- ·<D
+ aoo
1s Re • 900,,.
e- -
3Re
2oo(d.)
c
= 600 t ~oo
aoo :.
T
---©
Re= 800/.a
in 1, .
1s (e<XV'a) ~ eoo(x) t goo
a'
of
0
t ro"elii;g or?ne
'1c;
20
Rh
s~n
r'~ht wh-
t ons aof1ng as
or
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 .
~Fv
.z~1 cQ
(
..,.,he,n
~(><) = '20(6)
-6{1" ~ 120Tons
~ 12e.02
ibs. ·
• O
Rv "
Rv -=
eJ0.06 lbs.
1!20 t
~()(t5)
in !2,
s1nao• .
)(/+
ton&' • 6/+eo&.ao•
)( • !l'
fr " 60°
i R1 • o
P =O
-· - CD
.s6? • !2'20
~ = 20
=
20(1) t 20(10)
~x t~~') "" !l!ZO Tons--- ©
~fh"O
~M,.,, 0 0
Rh • Tcos-&
.ofT • 1ZDc{_r1..)c,()l;ao~ i 1oo(6)coc;ao'
./ans
Rh • 216.s co.soc•
Rh .. 100 . 2s lb .
T ~ !Z.16 . .so Iba.
~fvso
From eq. <1),
Rv + TGm~ = ~00+ 1 00
' Qx = 1!20
x = 110/Qo
= 6
44
n.
.
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.
16(Rev) • "t00(1z)
Rev::
6001b
~fh
~fheO
Reh - 2-t0 lbs .
.:EMe." O
2"'!-f<A I 2-40(16) .,600('1e) t
2-tRev • 600(1£) t R1-0(16) - -400(12)
Rev" 260 lbt;.
•1!5o lbs.
0
:EMe • O
1fJO. 21 l bs.
J6(~v)"' 400(4)
RA .. 7-+0 lbs.
~JAt. • O
300lb6.
.. O
R-.h·R~n
~(36)
By Ratio ~ Propor-tion ,
Reh • _1_
Reh· 300(1/2)
2
P.av
"' 160 JbG .
Ra -..f1so)::i t (soo)2 .. 335.41-1 1bs.
~M..,•O
R....v "
:asz)
'
!'I
100 lbs.
pu IIey -trn ~ . 1.n d.1ameter ~ supporting a load
or
200 lb ie mo..inted
at
Bon
a
hor12onia
l
beom
(
Fig
P-3!52)
·The
b
.
ti
.
.
eom IG supporled by a
·~at " ~ rol~9"5 al
c . tie.gleoi1n9
rr11ne the reac hons ot A
~ C.
the weight
or the beom, doter·
'350) Compi;ie fhe to\ol reactions a l A"-. B for the truSG sho'Nl"I in Figure
l
I
I
I
p - 3SQ .
.I
I
47
46
,.
1·
~\
f,
or
F00
F80 of the pulley,-
.w. /\
Pi =R/e (1- 68/e) • ~1000 [1 - c;(o.e6)l "' n2o lbs .
19
18 "j
the beom,
P~· RJ8(1t6~/5) ..
lRe·:,~
+'
.
c
'11000[1 t 6(0..SG)l • 1780 lbs.
1g1B ·J
· 3S"f'.) Campl.lte the foto l reactions
Pl~v
fRc
in Fig.
ot fl i.., 8 on the truss shown
P-~.
~M,+.-0
Rc(Q) " 100(+)
A!Ma•O
T(a) .• !200(2)
. T = ~(){)Ins. .
Re" go lbs.
~fv"O
!<,Av
~fv•O
R.e • JZoo-Tsin3o·
'
Ii'
a
·Ra·
gO()- QOO(Ginao•)
Re
t
Ji..v
a
a
20 1
100 lbs.
100 - SO
~fh cO : R.Ah • I cos BO•
RM •
100 Ibo.
!200
·ton.a-' 1%0 . "
cos:ea·
R>. •
(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<J(COG!26·56)(6l>) 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 <!+.._ ol -
Resistance
::!<J()Olb
.::fM""O
• 11:a .20 lbs.
1:1
!looolb
10001b
SO lbG .
<>
+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.!\<l-.67
=>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<J
x .. -+.92f1 :a60)
300 .son'I06 •
Re ferring ;to Pr-ob· 359 , who\ volue . of I
frorn J3 w i ll
~. the bor nor-tzof"liol ?
s.O'IS·
-~
~c
octi~ ot J< ~ 3 0.
~
..Y = 12(C06& +s in&) -x - ·· eq,~.
~ 11~CDS&t&tn&) - .3~1
J
-
1 •ton60
33 COSi}
NJ -
+..39'
60069-
~[<JC06& - n(C06&• s ino-)]
1
ton60
1 • ton60
~ 60(~ 6 intt).
1-t ton60
I
£.tf1c "' O
l ( -4. 61) : ~00(1.39)
= 60 . .3 tbS .
53
52
-f
~ 0?69- - 36(C-OE>&-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<J.)
1 t ton@
...Y - 126.n& =· - JJ11""'°
·
,,
)
lx - 1..:GOS&
12
.51(l 7 b
)l :
~}.
lbs -
r 1 +-fan60
;:.G
_y = ~
foe- RA.
~ -o = ton(6d>(x-o)
foi-Re.
of ."'
.-l
•.o
x fan60 : 12 (rose- isin<>-) - x
~ .. )l fan60- . t'.D RA, .
JV. - ~19.61 ft>G.
~-o
~ the coordinoles
l151°9 Analytic 6eomei ry.
..9:91 - m(x-x,)
]
.
I
-4<:1~.) Joint B of lhe fruss shown 1n fig. P- -+0.1? ·1s euQjected to
the fOrces e)lerted by t he ~hree memberG AB, BC, ~BO . Membe · ·
M~ BO ore. in the some siroighf line, but /3C iG inclined of an angle of-& dE'.9rees with this stroigh\. line. Show that the force in
fjC must be :tero . 6enel"'Olt'ze fhis result ~ then show thol the
for.
ces in m,smber CD,OE, r::F,Fl,tfl,'HI<., ~.JI<..' ·,s also zero.
Ans. Ai o Joint suQjected to ihe. action
of three members but no oiher load,
·
r
H' two or the members ore coll1neor,
a
~
~
0
Chapter 4
II'·J
..'A nalysis
of
J
.
_______.:
~C
.•
11
I!
6
I
the flXCe in the fh1'rd
z.er-o.
mernb~r muGi' be
K
p
'fQs) Determine fhe forces in ecich bc.ir of the truss shown in Fi9
p:. -403 . Hirif: ffr-st ' determi'n e wh·1ch bqrs c.orry no load us'1n9 .
pr 1noiple devebpecl ·11"\ Prob. "402.
P
ce -cF-o
Structures
£fvc0
P"' co S1n6<:>· t OEs1rnso·
G1hce : CO"' DE·
p
CO
+
,.11::
c
QCO s 1n60°
00
o .e,77p =OE
-
C
co =ao "'o.!!!J77P
F
I
l
.
~Fh~O
BF =- BDO?S 60•
J3F
~ 0.577 P
f3f ...
cos 60 •
0.577 p (0.5)
BF" O. !l.99P
- T
·I
\
I
.i
'I······
I';::
.
~
·1
54
55
~Fv " O: 8Cs1n6d+BDs1n-ao• • 1000+ABS!n30•
@pt. 0
40+) Oeterrroine
the f()l"C96
o. seD+o.~ec,. Qooo - · · -
showr> 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'<'er truss shown in .Fig · P-...,...,..,
""',
conipute
~
@
pl·"
.
~Fv•O
10001
b
AB1>1nao •1ooos1n 60
The con ti \ever trvss in fig . P- 406
in
966 (o.B66)
1orce 1n memben:; AB, 1;3e, ~OE.
AB " 1 7S,e lb - .. - T
a
96.6 lb· ..-T
@
pt. s.
0
0C = i00 1b·· · - T
force
co•
CE"" 2020(0.<:>) +1732t 8$. (o.!J) = 3 17.?l lb -· . - C
~,AO• o.B66(A0)
~fy
the
CDs1n6'0°• ec81060• t 1000
COcos60· t A.C t BC ()Ot;(it).
. !<Av c a .5('100)= .so lb .
:li!fh"'O
AC ·CD
"406.)
.
~fh•O
"'0·5AB
@:Cr o:
-c
I!
...c
.,o.e~,C1 o0) "'96.61b-T
~ .s in.+
o. 2!3 eo + o~ ec - 1000
ao - o. +.s~ ., 1.soo
0.7!!J
BD • 2800 lb ~- -- T
A0 = 1o0 lb
RAv
+
CD
cosao·
f':.C = o.966 A0 -
/
. @ pi.C
~fh - 0
100lb
hi~
U61~
<::it D ""--E. Find
Rototio" of A")(•S,
Uv-o
.
SE
"'.
@ pt./\
..a.e
ao
1000 lb
£fv•O
.£P..,•0
ASs1nao"
=100 :
Ae-~lb-· · T
0
OE. '"'" 6'0° • 100Q
PE• 116+1b -· -·
lll!rn · O: AC. AP>COS"°
AC~ 200ocOS.90.
/\C =173Q lb -- .. C ·
I
1cxx>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<M'n
in.
FIB. P-+11
_ ..:i
oi c
·
.
. . If the loods we re opp1·l6U
"".E '~"_tead
o~ 13 \_D, specrfy whioh memberG' would
hove their internol force ohcmged.
100lb
tone-"' :10
+
411.)
::iOOO(!j)
p_,._v ..
£Fi,,•O
~
coMidering the whole' f1qu"e:
R....v(ro)., 2000(19) t -1<XX7(10)
o.
~~
...ae.1 Col'l:'pule the force ir'I eoch member of the Wo"ren tru(;a
'3()00 1b.
pl.
or
4-~50 lb.
3o
e
pt .
@
-e-. 33.69.
izooolb
@
pl . /\ ,
~~e
~
£fv•O
IV.v " ABs1n 6C)
'
z fv "o : Aesinoo· " 12000 ~ ecGines¢'
13C :o4910Stn60° - :2000
'
s1noo·
l\6c 4250
o.966
/\£> "' 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-
<;:1
=400o
~o201b---T
BC c<f.> oo'
2600(0.9)
t
Ct= ..
t
/\C -= CAJ CX:foq:l
Q.+55 "'
t
CE
'20~0 (o.~) t CE
,93.112·
~= ~ · Oe" f3~g·
'.20
-'CJ I
•
~i=-..·o
C0 61nsa.1!2· - BC
::S:fh •O
CJ'O c C O COG .s~. 12
1300 t ~45E> - 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:>s<OO') "' 2.100
AC
=Av
co c so(e/10) ·
-4' )
BO"' 1500 lb (eomp~iori)
.
10- Determine 1he forc.e in eoc;h member ci' the PrcrH
fI
in fig. p..
.
roo rvss eho-Nn
410
= 126 lb(icnGi<>O)
o+ 0,~fl'
lb
.,.
,.,c
ec;
~
~-AC ,. Ae
6
AC • 12S
1900(6) = Av (32)
Av = 2700 lb .
••• CD
AC
CE.
BD • 160 lb
Av=~~
(c;ornpres1>1on)
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 :
=
<L
in rne Of'1g1nol o><es
lt:>. (tens1~)
A0 " 160 lb..( CO("lprcss1on)
co
.
U6iog Hew Axos
Z:Fy • O,
AB-BO ~ Of' "' +soo lb - ··-C
""10
-ll-
1;00
BE s1n60• - ~(sinoo") •O
c
lb - ··- T
Compute . the force in eoch member of the \ru(;S shown in
f ig. P-:-.+10. . If the loads oi 8 ~ D ore 1 .shined vertically down -
@ pt . E,
41f.)
+
word to odd. to the loads ot C ·~ J: 1 will the~ b6 on"/ chonge
in \he reactions? Which membel"s , i,f ony, ""ould undergo o
ohonge in ln\er-nol force?
~r.-
4001b
@ pt.
DF
~fv"O
~rv
_., - T
SQ)lb
I
I
~
-o .
DE = !100 Jb
f
OF
- ~or
=
/·
..Je(7!2o)
OF • 1609.97~1610lb- .. -c.
.j.ta.) Determine the iorce
r
·in each member of
Ghown
P-+13.
the crone
coi;1ne LOW:
o2 = b2 t c2 - .::zbc Cos!\
@' pt.
(l'.)nsidering the whole figure •
:a;:f... ·o
R>.v t ~FV
R.-.v t
'IOOlb
be= 60+ 90_ 2(6)(9)cos120'
P,pv " 1600 lb
+ eoocao) eeo lb
i;ubst the volue o f ~ ..."
b ...
..:-©
·" /
R.-.v (oo) '"100(-40) t ..+C(a:x>) t
@pl. A,
~ad)
!Ct=h=O
~80 "'~ /\B
G
1,.
in
BO " 98% ~ -49~. 5
~ 1600-BSO .. 720 Ii:>
"""'1'9~ 1b-.:1C
(
@pl~~··:.:::..
.....!..-,.. W +~lie
'W
'<5
~ (-t9Q)
R>.v cA8.sin6a.4g·
"6 ~ .,993. 9 1 '<:: 99$ lb- ..
c
{s--{9f3s) - 100 t ec
lb - .
-:;.I
-.C
~fy "'O
J:>C "{e co +czoo
A0 COS68.4S •
996
700 = ...L cot
COGGa.+s·
IC .. 440.ee ~
440lb -· ·-
"9
T/
~b
czoo - coe 1118.0?>
/
~ 11120 lb-..- C
. o.eeo • 0.966SC -"@ .
12000 "o.s66(o.:!,6
~ 13 100lb -·'-T
/'
1n
=A8
@k
/\8 • BOcosao' + BC cos 60•
-<D
C~ - 440+ ~(1100') .. 1141-76 ~ 1440 lb-..-T
62
EC= 6000.!16 ~ 6000 lb-.. -yc
2,
o.s
~Fh•O
t o.~ ec
ac) + o.sec
BD .. .o.&66(6000) "'10.392 lb
AB"' 12000 lb- .. -CI'
1aa:io c o .966 eo
•
EUbsri . 2 1'11 ,
AB" 13100o::>s2:a.a9• ,. 120.23.-49 lb
BO
~fv aQ
eoGinac) • ecs1r-i60•
J..Cs•n23.a9· .. s200
/\C cos 2a.a9•
1il -08,
6"1•(1a.oa)Lf92 - a(13.oS)(9)cos/\
Ar. ~3 .39··
c.ont. ()t'l 8,
£.fv=o
~h-9
@ pt . c,
~fh-0
c
t
In 1 ,
/\C"' 13()99.64
= 400 tac
SC - 701 .o+ -=::: 700
zfv•O
/\C
2
e,
" ..,.00 t QOO t 2()()t 800
~Mr"o
Rrv
6 = b 2 +9 2 -2b(9)cosA-©
ba."' o-.i +ell.- :zoccosa
63
~ 104ootb-!.'.'.c
+i4-.) Determine \he force
~russ $hown ·,n fig . P-""'14.
the force in mernbers fH, Of".~ DG
truss shown in fig . P-41S.
in member /\6, SO, !iic. CD of the
+16.) Solve for
or the
0
p
conGidering the whole,
~Fv "'o
R.iw t R11v "' 300
t
~ t 900
R>.v + Rtiv .. 1500
£.M,., "'0
@)
~v
900 lb
Av(3G)"' 300(2.1)
slope ot A'e • 1
slope at BO .. 1/ 3
slope a t CD " "4./3
at A ,
t
300(1e") t 900(9)
R>Jv(:a6) = aoo(9) i
~
rR:-
R.-.v = 1500- 900 .. 000 lb.
~fv•:
FHV = FH s1n4s•
Av= 600 lb .
fH "' 900/sin-4!1°
FH
,
Y-li A6: GOO
AG
- ·.-r C
@ f,
~FH
~
~
•o .
Ya
FH
.
Ae" 048.~3 lb
=1127l2.79 'I: 1270lb -·· -C
@ pt. i=,
2Fy•O
= · ::i/,.io
~vco.
Y-ta FH -
DF
OF " 946.'6 "" 947 lb-· ·- C
300 t 4/!EI
i:.Fx =o
(@ pl. 6 ,
141
AB
I
BO = 632.40 lb -··-C
*
.;i; Fy
~o
/\B ~
BC ~ 400 lb
- ··-T ·
.J
~·
~rv • o
.
1/"10 BD t BC
co = ec ·
lb -··-C
CD= 125
+foDG t F6 .. 900
../.e 06
)
t 6CO ., 900
1::?0 "'::ioo ( s)
't
06 = 37.!' 1b - ·· ~r
64
Y..rro
OF
J4w(941)
f6r:: .59S . .B6 ~ 600lb
('3)..W
~fy "' O
f6 t
Y-12(1210) • t=G .t
OF• -1Wfr210)
ot C,
3/..fto BO "
:aoo6e) t
Rttv .. 900 tb.
65
900(21
UGing the mS-thocl of .oections, determine the force in rnemberG ao ,CO, ~.CE of the roof fru~S shown in f (g . P--t17.
411'.'>
. 1@.lb
£MJ!·O
1f..; oO
or(ao) - RGV(2o)
o/.Ji BC .. -+00
ac • + 47. 21 ""4+S 1b --·-C
1lz'
Of" BOolb -· ·-C
LMoao
360lb
CE(~o) t 1000(10)-t RGH (!lo}. RGv(3o)
ce~ecd.:ao) - 1000(10)- 600(:20)
/'ll!l,
BO
.
!20
ce ..
i
·.!c
E
to determine
.+1Q.) Use the me thocl orr- secfion6'
.
ao,co,. ~CE of the Worren Trow.
,,
.
1001b -··-T
~t.'lc•O
FV,v (12)
go "'
GOOOlb
=BO ( 9)
com:idering ihe whole f'igur-e ,
120(1&
~M..,•o
9
~o .. 160 10 -· .-c
~i:v .
Rv•(ao) •'1<J00(1o)t~e)t
.-
!2000(1!1)
0
J:l.vE
w
•t-7!50 lb .
p.,...v "' 3/,5 C O
co ..
b/?J (HlO)
I
c e
e-e.66 n.
, I
sin60•(•)
B ~- -··· -""B;:;...0_--1
CO"' !ZOO lb- ··- C
'
...:;Mo ' O
P-Av ( !2.4)
GE
::
er; (9)
= 11ZO(~~
C.
- 9
CE ~ 31ZOlb - · ·-T
41EJ-) The warren . iruss looded os show" in
by
a roller ol C
pu-\e \he
force
CE 10'
'f<lOOlb
~
a ri1n9e a* 6.
in \he
~ember-G
Sy
F•9· P~+1a is
suPPof"'-!ed
the rn&thod of seolions,oom-
BCr OF,
~Cl'- ·
BO "
cons·1derin9 the ~ho\& flgure.,
.:tfh:O : ~0t1
-
%Mc=0
t;,00('10) t 900(10) t
~Me.•0
~Mc ·O
Bo(e.66) =-3ooo(s)t •nso(1o)
6P • .32soo/e.65
600 lb .
100o(~O) .c -406(~+
.37.stZ .9
CO • IZ1~S l b ....:. . · -
-co(e.66) + 31svJ(e-GG} c aooo(s)
co ..
c
.
co " 1Zo20 . ~ lb:::: ~o:zo tb -··-T
RG" - 900 lb .
100Dlb
67
66
1.S000-(32500. 11+)
-~-6'6
27+:2.!5 lb - ·· -T
JtGv (-40)
-'fOOlb
l'
::!!,Mi; aO
=+750(5)
Ci: "' 257.SO/ei.66
Cc
3000(10)~"!000(6)
co· _a12s,9"'1o
lb- ·· -C
~Mo=-0
( 9 -66)CI:
Rvt=(1s)- co(1o)a
using Method of Secfl()r"I • @
CtfECKEO:
.+20:) l)etennm 1he fOCce. in th& mernberG Of' 06. ,.,..1:6 of U\8
ttowe in>S:~ GhoNO in fig. P-.ov.IO·
::E.Ftt'"C
3/.sao=ae ; BO.· ~a(1roo)
!f.l•!ltCJOI>
BO~
£MG•O: Of(9) • 11.100<12)
Pf =!l900 \\:> - · · -C
!2000 lb .
.2:f.... •0 ~ 12£>0 t OIAJ 06 • a100
D6•16001b - · · -C
~f'h·O ~ Df t -4{900•1:6
:zsOo t
~F '" 4fi.J(~) t 2..:.00 t 1200..:. gdoO
-f'/:5 BF t
f:6 "' .+oOOlb - --T
~ 'in Fig. P -<tff, determine t~ f()f'Ge in Br
o(j~ts ~ then chec\( Hi~ result using me-l~ of
section~. Hint: .To' opp\y the rnelhod of secli~, fi~ ob\olO the
<4!M-)
rnettioO
'(()loe .of
ae
by incopection.
cons"ic:Jen°ng the whole figure.
1~ • l:F
bF ... JV+[ata00-1.~001
for 1he truGS
by the
EF .. 3!lOOlb
-=N-0
-4/9(1.500~ .. ~6
BF - al500 lb -· ·- 0
-MS-) In the Fink tru"
&hown in Fig. P- ~, the web members
~ Et= ore pec-pendicola~ 1o the inclined
members ot tne4r midpOin~:
IJIOe the mehld of' 6e0fi0os. to determine the forc.e in mernberG Of
PE~\i._CE .
.
.
'
{tr-
..e'11•0
Rott .. 1!/.00 lb
~F,,.=0
RJ.,v t Rav
=~.it<)0112J:X>
RAv t Rov "3600
~I
- ·©
~o·O
-Wt>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<A "-""
tru1:is shown i'n Fag p-+31
or' 06 I '!,., E? fOI" the Po...i..-r
.
' ""'
of joinls connot determ"ine the
ces in oil bors of the fon fink truss i9 fig. P - -+.126. Then use
the method of 6ections to compute' the forc.e in borS f H,Gti,'11\EK.
If
?£)01b
e panels
R.i.v
~fv -o, conGider the whole f .gure,
~1nce : p.1 • R>.v
·
· 1
.
2R1 • 7(3C>)
R1
9
I
ot l!s' • 2o0'
'
R/\v =1os "
0
ton~·· o/ao
a"" 11.:a:2'
cos -ao' = 3o/c
c - :3-4.64'/
ton60·
EG=
17.a2
c
~~(162) - 2~(ag. 1)
-'-t)
u= 10'
1::6 - 110.7 -:::! 1"10 1<.·a ~ -·-T
..C:MG • O
· RAv(1s):-ac(ao)tao(~)t 125 DF(a.e)
coosidering the wrde r.·gure,
EK
R.-,v • l<ov
I .
2 FV.v
c
1600
i:t-.v = eoolb "'Rov
R.>.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
<:::..92.7
f<jps _ ..
-~
c
H
I!
In Flg._-a
sanao· =DE
~Mi:~o
fl-\~ eoo(w)+ !loo(!2.s)- 1oo(w) - :zoo(1:.z.s) + 2oo(s)
. AE • 30-10 • 20'
·
.£.~·o .
fV.v('2.o)tW0('.1'.1·S-20) - rt-1(10) t 100(20) t~(w-1.!J)
+ Wo(12.o - 1s}
In fig :2,
~
t~ or(~)
OF• 161 , 87 ~ 162 Kips - .. -
zMA=O
'.1()() (7.!J) t 200 ( 16) + '.100 (12.;;..s) • GH ( A.i=/)
~80
to c.ompuf the c
.
e . P-+a2
rorce
-shown .in fig
. rn members
"faa.) Use the method of sec t'torw
1£, ;..o, BC,
of tn t
e
ru GS
GH. '1.00 ( 7 •!!>)\ t ~fr;) t 200 (22.~)
Ij
17.:aa
Gt-I • .519-61'
""~!20 lb
-· · - T
I
71
. II
70
I
,. ,.
~Mc-o
R.-.v(110)t•~M(16)=
"'8 =
¥
o/..JS9
A8(20)
[ 30(~)]
[s(0o)-e(1t0]
·
AB~ 70.7S ""' 70.9 KipG
::!EMB •O
RAv (l&J_ •
f /;.,·,.\
~M ..
or
~·
•o
ii...v(s11)• ~400(:M) +a~16)
_I
zfv•O
o/t<X) t %PP• ~AB
0
ao= %[%97(!S9~-~J
.:i!Mt;.=O
30(a<>)'" 1!Z(2o)t
OF .. ~.+s
BG - ~ AJ!J + ""f~ BO
BC- ~ (s91 o) t +fe (.sooo)
V..,;; ,.o..e(-ia)-o/,m Ae(16)
Aa • 5909.31 c:::,.!591011;>-· ·-C
Be"' 6:4001b - .. -c .
·
~Me"O
R->.v(e) t a,.S A0(9)
L~ .AO(Hl)
/\0 • 3000 It:> - .'·-T
-t.aa.) Cornp:ite the fOl"G6~ ir. bars- AB, N;, OF, "OE
of the sciss~
frur;s .eho.vr. in fig, P-.+99.
~M1·0
K
RAv (60) •
,
.
1~(sD)tn(..o) t 111(30)+1~(.u>)
+ 1!l(10)
JV.v ..
,
~v
J
aoK
~:.
'48'
"
P.Jv
"
12
-.0
LJ
10
~
~4
"
5
~:
A
...e Kips; - · · - i
t!2(10)t~ OF(a·2 -~-f)
+ ~ 0~(10)
~ni-o
Af'Mo•o
RAv(16) t
AC~
/\C = #}4· ( ao)
BD "'5o01.o+ ~ .!5000 1b-· ·- T
~
-+,.f+';
AC - 4€1-02 ;:,
RAv • 36001b
.i....:..... c
.
.5
73
-c
~ ""1l.5 ~ips- ..
i.
1b
.+21.) Use the method
members Of, EF, ~Eo
Prob . +11
for:-e
of ~dions
to compute the
in ~he
.of lhe cantilever '\'russ described in
Z.M,...•o
. Fv(so) • ...00(10) t 200(10) t 000(30)
t 200(30)
'*'
page 92. ·
Fv = 720 lb .
tan-& = '2q/.,0 "
o. 6
-& "'26.57 °
tori20.s1· " a/w
OE" 40/.9
o~
~Mo
0
10 '
0
CE(10) = 720('20)
f,..·-nolb
. CE "
1~
lb -··-T
~Meo c~1...:in9
F)
BO ot
7'20(-1-0) "':z.00(20) t 800(20)+
y.fij" 80(i<>)
BO = 491.93 lb - ··-C
~M p ~o (Re.solV.ng CD ~tC)
200(20)
•
%ME • 0
11..rs co(.ia)
co'" 1118.03 lb ..,.....-c
soo('2o) ,
the force acting
in rnember-s Of, EF. ~ EG of the tioYVe truss descr'1bed in fig .
4Z3.) Use the method of sections to deterr,..iine
to
(R~solving
t
OF ot
o)
3/..f13 Of (40/3) .. 100('20) t '200(10)
P-409
on poge 91.
OF, ,. 360.SG lb -··-T
0
£MA•
2
:£MA• O
Hv (-4-0) = 400(30)t 1000(W)t ~(1~)
Hv "' 900 lb
o / (Resolving EF ot E)
EF (2.0) =- 200 (10) t 2.00( 20)
{5""
0
EF .. 335.41
lb. - · - T
.:lf.Me.·" 0 (Resolving OF o+ H)
zMr •O
Hv(U>) ~ 4t00(10)
E6 (W) • 100(30) t 2.00('1.0) t 200(10)
''=""-
EG "'450 lb - ··- C
%MF
422) Reier •o the descr1~od in Prob. 41'2. on poge 92 , ~ compute the force in f'(lemben; 80, CO. it.-, CE by '\'he method of
seot;on6.
. 74
I
11
h.-=::.L..::.,H
G 10'
-4«> tlv
~o
Hv (10) • E!G (10 tonao·)
OF
c
t
OF(s1n.30')(20)
1500 lb:-·· - C
~MH • 0 (l<eGolving l=F
ol
E)
400(to) - EF s •nao· ( 20)
~F " +oo lb -··-C
E& "' 1G4.S.4.5 lb-··-T
-4-2-1-.) For the truss shown '1n Fig. P- 424, determ·1ne the forGe in
BF . by the rnethod of joints ~ then check· th·1s resul1 using
the me~ hod of .secf1on<0 . Hint : 1o oppI'I th e mefhod o f 6edions
75 .
t
!
•*'
'
-427.) Determine t~ f~
fif"St obto1n the vo\uc of 0E by ins~tion.
-r,
Av(1&) -1~(12) t 2400(9)
1200lb
lb
81"
"'
·~::~
Cv "' 1600 lb.
J_
N;
~fy
1/$>
1e<>0 lb - ·-T
. '
•o,
DE t 1~ • 1~ t £,()()
ot Jdtot 0,
£Me •o (Re501ving coot c)
1/-11o DE (18) " ·aA; (12)(co)
r&(; g
~n. .. o ~
otF,~~f
Cf
ot c,
=0,
OE" 632.<t-6 lb -·· -T
AC .. 3/~ AB •
i200~
~Mf
8'
~so
l\v
of the nooelle
c.,(1&)H200(~)" ~24)t 1200(1e)
"'" .. 4-/~ A0
AB.,. 2500 lb -··-C
.q'
1t'
+
Av "' 2000 lb -··
~y•O
' ,; e ,.-+.
~lb
.i:Mo~o
A,
bor-G 60,CD, '1Ao.. Of
troGS shO"m in Fio. P-.+f.7.
As o~\e.
ot
in
£Fy,.O,
,._C .. Cf "1500 lb-·· - T
4/5
1to0
CO= .5()()1b-'·- T
co = 34ts BO t Y..i;o DE
BO= 360.55 lb-··-T
~F.,. •o, CF • 3/e> BF '· "
428.) U~
the mc:rlhod of GCC{iOl'lS fo deier...,...ine the force in mCJm bers Of. F6, 1t..,61 ofthe triangular t10V'le trvGS shown in Fiq.
P-~ . Hint : fir'Gi. d6icrmine by inGpection . the forces in the web
members of the right s·ide of the tru!:>s .
1~ (5/.a) ~ ef=
BF "' ~50011:>-· ·-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 ><1pG - .. -c
BH . 17S lb
.:e:Fy u
BD(32J
.,ei=,. '"0; BH "AH"' 17.5 lb
O i Av - 100 - 100 "'o
Av "' 200 lb
CH " Bti • 17.5 lb
78
79
Isolating bar CE
...+3.) The frome shOwn in fig. P-.++3. is hinged to rigid sup. i:;orts ot /\ ~ E . find iho Q?fTipo;\ont~ of tho hingos forcos ot
:f;F.,. =o
Cv
Ev -Dv
t
C v = 360 - '2.64
Cv = 96 lb ( downwor-d wiih rospec! to.
/'\ "'- E '6i... tho fon:;os in rnembors . BC !,... 80.
1e:>lb
zt.V.=O: 1'20(4) - EH (4) "'0
e:1-1 ... 1w lb
~
le..
.s'
(~loting
c:
G
i
At-1 '" 120 lb
leO
c.tic -4-0 lb { io wthe
ifh rei:pect
BH ~ 40 lb to tho right
l3C ., 100 1,b - · - T
+++) The frornc ~fuwn in fig . P- +'14 is suppodod by o hingod ot /\ ·&. o rol1er ot E . Compute tho. horizontal ~ vorti. ~I componen\.s o.fi the hingo furc.os o t B ~ C os -they od
upon rnombor AC~~~
The frame shown 1n F19 . P-+<Ts is supported by
god ot E i..., o rollor at D. Computo the
col components Of the h ingo force ot C
Dv
o~
240(9) - 0v (~) - 0
lev
~,
°" =
12' 1'2'\0lb
.
.0
Ev"' 1so lb
lc:oclating bar DB
Dv
lc;olohng bor AB
2..0
%Maco
B!-+.s
. Ev (0) = 240(s)
BO
Isolat ing bar BO
360\b .
""'
"
-+'
~"
l
c
~Ma•O
w/
~pee! +o oo.- N:)
Bv " 120 lb ( upword
µ ..
·rev
D
~Fy"O
240(4)
Bv c 160 lb
~Mo "'O. : 240(3) - Bv(6) "'0
2'
:EM.-. .. o
Bv(6)
DH
(6)Av = 2"1<>(2.)
Av=aolb
=
Cv 16o-go
Cv"' 70 lb
ZMe "' O
CH (3)
c
70(.3) t 90(7)
CH"' 260 lb.
81
80
= 9o lb
:::E:.Mo c.O
E:v >= 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<;ied. together ot 0 in Fig. P- +'l-0 · Compute the. compoi:onts
of the reaction ol /\ ~ then find the fon:;c,s acting 1n ban;
AB'&.._ AC . Hinf: f irst 1·Gol0Je each kuG~ ow o freo. body .
Av(20)-A>.(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
. :<E'MR:i "'o
R2 (10) .. 200(14) t 2000(7) - 300(4-)
lsoloiing loft tr;;ss
~
R~
e
2120 lb .
.zM ~2 ° 0
R:i (10) " 200'.:>(3) + 300( 14) - 1..00( 4-)
R:-l
82
100 l'o/fl
::f::Mcco
@member BC
-t'
m>lb
10'
c
2.40 lb - .. -T
447.)
1
Fl9. P- .+40
10'
.:EF'~ ~o : .320- AC - AB COS
I 10
111
f=;;:::~;:;::=;.::::;=200:;:::1=bA;n;:::::!:=:::;:c:4::~100
=:'%:'.::Jo
'*2o - !\8G1n .36.87• ~o
/
·"
'l-40ib
. . 377.+ 'fb.
A
%Fy=O
10'
(300)2
400lb
@ Joint A,
A
(~)2 t
iG
comf)()!;od of throe GC<Jrnont~ . It is supporlod by four vorti~I roodions ~ joinod. by two frictionlot;i; hinges- Ootcrm1ne tho values of ihc reodionG' .
Av
l'C
8,, 2 t BH 2
c
A boom carrying the . loads shown
446.)
2':N\o: O
Av(40) -36CJC20) t AH(30) cO
At-1(30) =- ¥Q(W) - 4 2.0(40)
I
8
=
AH - 32.0lb
0
)
=0
= 11 ao
lb.
83
c
.:::EMA,.0
449,) The bridge shown in Fig. P-449 rons'1sts of tvvo end
i;ecf1onG, each weighing 200 tons with ccntcr of grovlty ot
G, hingod '\o a 1,.1niforrn center. span y.1cighing 120 tons. COm-
3000(9) +1000(6) t Fi1 (4')=
Fv(12)
.
fv(3)-fH "'0250 -@
puto ~he rcoctionG ot A, B, E, .~ F .
· ~MF
4'
=O
3000(s) - 1000(6}t Ai-I (4)
= Av (12)
4'
1 -- - '' - -
@member
0v
@member CO
Gil
E(ft I
20
I
c
-40
0
Chi
CH
Dv "' 100 tons
~Fx
'/fl
.;
E~·
• .
30
•
0
BH
. I
Ql member CO;
CH
C
~C>v
DH
0
'
Fv (so) =- z.oo(ro)-100(1.0)
Fv
01-1 " CH=1$00 lb
~I
@ membor OF,
fv"' 40 ~onG
I
D
~
.2'.Mi. =O: OH(+)= F~ (4}
.Z:Me "'O
Av (SD)
FH ~ 01-1 "1li00 lb
20o(20) - BO(W)
Av " 48 tons·
"
EH
EH =3000\b
F
Bv (so) ~ · 200(30) t 60(70)
8v = 232 -tons
F.;
1000 lb is .subjecfocl 1o o
per 0 ·as Gho~n ·,n fig . P · 450 . Ncglecf1ng the weights . of the suppor \'Ing members, dotorrnine
the components of the hinge forces o~ /\ &,., F.
300 lb
p rcss;u..-e = 300 l_b/H.
= 300 1b/f1,
l(
::EFx •O:
F~ t
?>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<XJO -
.:i::Mo ='O
..:!!M p"O
D
•
~Me"O
C11(10) .. 3000(5)
C.-(6o) "' 120(30) t 60(W)
Cv = 00 t onG
Ot1
CH
-®
ce.
Dv(oo)"' 1~(30) t 60(4<>)
Ov
Cv
C
..::tMc • O
Av(3) - A H .. 2.Q.SO
·
fv(::i) = 0250 t. 1500
Fv
= 32.SO
lb
Av(3)"' 2QSO t 4500
Av == 2250 lb.
10 'fl
- 3000\b.
i
84
85
J
l'
~
ti'
451.) 1'hc lrome stiown in fig . P- -4-.51 1$ hinged o\ E. !t.._ rollor
svppork0 o\ A. Oetor~1nb horizon\ol ~ vorticol c.omponenti;
of tho hinge fon:;C;s ot B, C, """D .· l'leg\cct tho woightli of tho
mcmbon>.
)
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<'12)
t'
. .°"
T
= 010 lb .
.:i!fx =O: D.i "'8H c.400 lb
6'
~Fy"'-0
Bv
.ll!Fy -=O
s.... .. w
c:
·
.SO t e10 - 300
Bv = 580 lb.
Cv =s<eib.
86
87
block weighing W lb is plo~d upon o plane in on a ngle , -&- with the. horizontol. Discuss wha t
'f"ill hoppen ·it the angle of f6ctlon ¢ IE> (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.<:Juol to fr- lhe ·blool<. w ill sfill not .sh'de, down
b~- having -e- vqlJol to¢ th~ sldstern will 1.Sfill be Incquil;br1.um ~· here the fr1ot ioriol f~ree IG In. 'its minimum .
(c.) If ·¢ Is less thon tr. t hen sli'pping ocot..irs tJeoa;'.se
the fr1'ctioriol force. '1~ not enou.9h to hold t he blooK.
-'l).5.} /\
clined at
,·
__
P_· _
.,
.s1n,sg,o+•
P
=
~OO l b
sin 11io.g6°
60.:>-o l b .
R
sinoo«96·
.
ooolb
s in
~g.o+·
P=300fb .
o rolJ9h horiz.ontol
. '.surface, for .whioh the COCJfficietnt Of friction jG Q,40. Def.
the force. P reql) ire.ct
cause ~otion to ' lmpc-nd '1f opph'od the block. ( oY hor1'z.onto 11 ':1 , or lb) downward o t 00 •
with 'the horizonta l : :·(~) w hot minimum force '1s n::qu;~
.506·) /\ -tOO- lb block Is rest1'ng on
c.)
.to
to start the motiO-,;.
.Zf;oc:::O :
P= 160lb.
400(COSi5') .t f-'(.8.oO)(cos45")=0
F = 70 .71 lb
sotl) The 200 lb block s hown 1'n f'19. P-008 hos 1mpe,nd1
'
n9
motion up the plane caused by the f-ic?ri:wrrfa/ (Orce of
"l-00-lb . Pd · fhe c.oe(ficient . of .sfcd lo frlo~·ion between · the
p
- "IOQ/b
.sin 2-1,e · .- sin. 68,2'
O·
F
•1 '
contoot Gurfacei; .
I
.t::f.1,3 .=0: /'I"" 40o(s1nzio') t
•.•·
i
- - .sin. zie.z• ·
p --- 400lb
Glr'\.u~·
400
2oo(ros30')
/b ..
~F-x. =O: F c 4oo{cos13d) - .zoo(simio')
F . . 246·11 lb :
/'/ "'.373.21
,,«. •.yr,
:
246.41,
'373 · 2.1
0
0.66
'The: blockG s hown in fig. P--50.9 ore conne-otcd b~ f'Je"Xiblc
ine-xfen~1b/e, cords passln9 ove r fr1Ct1onlcss pul/;ys. At A fhe
coofflotenf of' (rict ion ore f. ~ o.oo ~ fk · o. 20 while at /3
-:09,)
400lb
.sin go •
Pmin -= 1-+8.5 lb.
88
, f~ey ore fs • O...,.~ ~ [1<.. • O,i30 · Compute. the rnagnifude- ~ direc f1on of fhe rriot ron force odin.9 Of'\ eoch plOoK.. .
l.!!10 .) What wei9hj W is necc-ssary to stod the sysiem o f'
bloof<G Ghow n In. Fig , P -..510 rnov1'ng to the r ight ? ·The. ooef-
89
)
:-.·.
I
.l
I.
I
fio1'e nt of frich.on
I
1s 0 .10, ~
the
· to
pulle!jS are, oGSumed
•
be fr1o t 1onlesG' .
sol.
or
blf.) find fhe Jeos ~ volue
p required fo c.ause fhe Sl.dS.fem ·of blocl<s .show In . Fig. P -s11 to have. •'mpe,r;d c'nq rrot 1~11.
to the left . The. coefficien t
fr1.chon.
o.i under ~ooh
fBO
.!'og .
TA/\
W
;.#"'
A
.
~~~
TA·
I~·
JO()lb
I
I
~ f.9•0 : N" =.?JOO(cosa6.0•)
HA. 240 lb.
F&>of f3locl<..J : .
· ·. f,... • f"- N11 • (0.2)(140)
f,.. • +8 lb .
~f'H·O: H 11°2oo(c.o.s.s~.1)
c
a6
~
~
'""
1
N
T
~lt co: T tf-Pcos~ cO
100
sin.,1.~1·
67.Mt(M.)(900-Psin<><,)-Pc.osO(•O
His= 1~0 lb
. ·.Fe • f,_ Ne • (o.?.)(110)
rs
·,g
block. .
.
fA
.
or
of Block /\ :
r . 67:3Z lb.
127.32-o.2Ps1nC\ - Pcos<>{ "o
F'(0·~SlflO(tOOSC() • 127.32 Jb.--+
Jb.
dP • P(o.a,OD&0( - '5'tt10() c 0
d&
o.~GQS<>c - s rn"'< cO
p.2 =~ ~ ton<><
·~ . .510·
. i.
.sin 79.1,9 •
p: · 127. .'J~
o.11s1no< tCOS"<
'i
P .. 12.4 . 6 lb .
COS<>(
c
O(' • 11-31 •
A homogeneous J:>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+.<f..s1n&- - R1
;,.
8:H.+sinf> - 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,
W<;!li .• 100IP ·
(,,
.
'f:>lCOS 60
a
f- 'tJ•COS.30 •
)I- ~ cosoo: - 0.677'0
00630
~Ms•O
p1(1) "' r(1)
,. p
p =JA-H ~
·'
I
•
F
~F~ • D: ttsin60' t rs1n30' -P • 100
N (s1n60°t 0.5773s1n30';0.sn'3)""100
H • 100/o.s773 c 17,:,. 2 z lb
.
s1s.) Bloc.I<. ./\
I p
Ir\
P )IAl'l=(0.5773)(173,~2) : 100 1b:
Fe>D 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 blool<-6 ~A we\ghs 400
lb. Find the- rnox . wei,ght ol 13 thot con be storted up the
fnohne qy oppl~1n9 to A a ri,ght word ror1zoniol force P
of .soolb.
j•
cf.,y"o : H = .+OO t C sln.ao·
~~o: soo =Coor;ao· t F
..soo •ccoi;ao· t '(o, ;,o) ( 4Q'.>t csinao·)
.:3eo ... c (1 .016) - cc .:i74 .02 Jb .
We -
.!)74.o.e ( s 1n+l:)·a·)
.sin 76'7'
We = .26.3.7 lb.
94
95
526 J ;4i ladder ~on long we;ghs iO lb ~ '•ts c enter of grova~
'1s 8f1 from. the bottom. The. !odder ~s placed ogo;ns+ o vertic;ol wall GO traf i ~ mo~ on orgle of f:-0°
fhe. 9ro(;,lnd . HO...,
for l)p . the !odder con Cl 160 lb man o l1'mb befOre the !odder
w/
.t.MI\• 0
W(L-1)
tfll(~)-Hi(fl.) c o
o.+yiY1.-1) +o.Q(~)~ - 2 pf .. o
· o.-tL -oA t Q.4-2 co
i6' on -the venae of slipp ng . The ore le of fn;oi ion of o 11
r--
A
o.-+L-12 = O
i.F,y~O ; 'It•
f1 t F1.
£F>i =o ! N1 c t-l!l. ·
, F1 •t-li1 • O.QH1 = o.2N,,
r. =b
o.f
R:z.
(10,11.at)
~ lY"'/1,()'
L~.e>m
@!:I <m'l( -tb - ~·Tern ?s·x
®(~·0.) c-m{x-;ic .)
{._y - 11.n} • - fon1.s· (x-10)
is• H
I ~lb
o;ub,, the volue of.!:! :
bef!
I
c
A(S1!1)
~;lb,
i- I
f1 = f!J.
w-= F1 t F1.
con -
tod Gur foce J{; 15 •
Tan n;'l( -17,32 ~ - Tofl1.S.X t 10Ton 15 •
is·Jt 0.0)
o.2N1 ( 2)
(Ton 1s• +. Ton 1S·)
Ri[lt!
W • 0.4N1
@ ~M,., ·o
+o(& - acos60°)•160{x-s)
:it - 20
x --s ft .
"'fO (1) ,. 160)( - eoO
.s2.5.) /\ uniror rn lodde-r t6ff. long~ weigh;ng W I~ is plo~
: w/ one end on the ground ~ the- o ther ho_nd oga1ns+ a Y~. J o l f Th~ angle of' friofion. o t oil contoot surfaces I~
t ICO
w
.
'
t ,/, th I ct
2d. Find the:. minimum volue. of ~he ongle -e- o w;~ e. 0
dcraY\ be dcSin~a · w/ tho horiz.onial be.fore s), ltpp 1ngocou~.
· ,•
.
.
I
w
iS-
,'
,I'
I'
~I
I
,
!-f~W"-':cr--- ------1
@ (~-.!h)~-m(X-'X~ . .
(~- Li;ine-) 0 - TonW'{x -L<XJ~~)
0
or
!
.
of ~il ibr•um
~IJ)
5 )TonuiL?&tLS.•~
\ Ton70 Tonl!O
t
0) ~=m?.tb
_y =Ton 70 •.X
'
I
.
~(Tan10' t 1onzd) = To nzo"L0%'9- i LGin~
(Ton7o'
Ton 1ZO.L0%9-t l~1ne- - L co~&- •O
y
Ton ~o· t Ton 10•
-4::
f3 - 1'47.6 lb- ft
• o: ti2 cos1s· c N, c.os 1s·
N'L · ~ N1 cosu;•
coio1,·
l'l!l
~f!:j"O:
-
N1&1n75' ·1 H2 s1ri1s' .. .300
~m'" .ltj, ax7~·) sin 11;' ~ 300
'\ cosw·
N1(s1n 1.r't cos1s'ta'1'1s') = aoo
li1
N1 • 289.78 lb
.'.Nz•77. 6slb
:. F2 .. jt1,,
·
~(tonisJ(~76) · -: To"!1s ' (n6s)
11"77.6/b.
Fii=zo.0 1b.
2.3935
2
s ze.) In stead of o covple; d eterm;n'e fhe. minimum horlzonfa:I
force p oppli'e.d tonge.n~io11!:1 to the Jefl of the, top
the
qy li'nder described In. Prob . .s27 to stort the c,1:1llnder rotallng
or
Tofl -&- ., 1.1g11s
-ft- e ,.50 o
counterolock.w'1se,.
Ton w· Lc.oG~ t J..G1ne- • ..!::.coseTon 70~ Ton 20·
~f~
··· f1 =,.M.N.
21on20•)
(f:s9-
@~Ms·Ow('l(- ~-~a-) .o
tTonzo •)
2 L~1n &- 'LG1n&{ron]0° tTan20 -
si'n & •
.
Tan 10• t Ton 20•
•
.J!M" "13
13 " f.:o (1.s) t F, (1.s)
13 .. (20.s)(1.s)t(200.7e)(1.s)
1l Ton 1.0' LcoG& t 2 Ls 1n& • Leo~(>
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-"'°'"<Ton60·-2ism _ L<:oe"<) • 0
p • 20.s t n.6
T0t11G"t 'btl60 '
pc 9,0.+ Jb
sw.) ;\s shown in Fig p - 259, a homogeneol)s o,:il incler 2fi In
diameter~ we-1ghing 120lb is acted l)pon b~ a ver\ieol forc.e
p. Oe,,\errnine ihe mo_gn ;fude.. of P necc.ssory to ~;tart . the.
qyl;nde.r tl)m'1ng . /\G~vme that
o ..30
•·. e- ~16-7'
t 2.tCO!P.Ton60' -12.J{"n"( =)!.c:cis..(ronH•+ra,,¢)
GUI¥.~:.·.
Ton1,• flonoo·
p " 103. 021b
'."
..ss.69 lb
1.06b
In.
plank wff. long is place? .ir;
In f(g · .P-.S.32, two, b loc.N; eoch we19· h;na
· ~ f.50 Jb o· ~connected Qy a l)fl• for'ITI hor1:z.onfo I bar which weighs 1ooJb. If th
. -of'.\ '11> 1.s· undo..- c:ooh block. find p d; ted e- onII_
9 le, o f nnof1
ith
·
o e. <+.s •·1nol1·ne. tho~ will oOl.u;·e ;mpend;ng
rroho"recto the-para
le-f1.
.S.32.)
.
p(H o.B66) ~ 120 (o.U6)
1
J(Tan11~·t fQt"160°) = 2LC06"lTon60°-12~no<.
")( • IZL. ~O(Tan60•- 12l'ino<
Pl 1 n) ; w('!:)
c lose- the lood
1i-npe.nds.
f11.
o)
A
',..
',..td·'X
p' con
be pl oced
• st.
)
,..!- ,.. \ ~9.£t.!:!
,,.,,,,,. '
.,
·
' p
')(
© Isolating, t he. bor
P
horiwntol posi its endg restin9 on two me.lined p lane os ..shown
0
w/
fig. p-530. The, o~le- of fr1d;on i~
t 0n.
1. 461' .. 0 • nu.
:J.
e>{ 0 36.2.
Tancx. .,
c~-~;)--m(x-~ .)
~1 · iZOlb
£M11 ..0
5.30.) /\
.
y tilt.''~: -ton60'(x-2t.eosC<)
J-
p
~
>
~·m';\tb
20· .
"'
Oe-fc-rminc:. ~v;
tttl
\
fso
~~
"-.
,/
p
r?
":';,,, ""
10 ~~',
aof
@ 010CI<,.,,
.-,J (:.ei,'6)
I .u'
I!!
---.. . ffi•:)
to eooh end befOre .slipP1
:.
lf«)
'·
200
~·
C
~~ fonJis·~
y -ton6s"JI (!:J-O) "-TorHo· ( x-10)
Ton25.,X .. - TaneO'x t ~on"(10)
Ton6s·11; -Ton<K>·ll t10Tan..a'
')( (Ton£.s• t Tan.~o')
x c 9.2+
zMs =o:
.:£f,y • o: N· ?,()(}Ginti;' tllOO•rn+s•
(1c1-o) s -1on86'(11-10)
= 10 Tan60°
~=2.&1
p(g.2+ :. (10-11))~0
zMi;cO:
P(2.01 - !:1)~0
2.01-~ · o
._9.2+-10,x-O
,...
)( =
:X(Ton66°tTon-t<f)c 10 Ton"\O'
.Y
o. 76 !.!.+
= ~.81
~r,. ~o: re tc~~ - ~i;· .. p
pc 7..s.g lb
53~.) ".A unifo~ bar /\B, weigh;~g 4.24- lb , is fo?tencd b~ a
fr1~f 1onle66" pin to o bloc-k. we.9 hin3 200 lb 06· shown . Al the
v~hcol wolf' f o.268 wh'de under the- blook, f•o.~o . Oeterrn•n~ th~ for= p neeaaJ
rl mohon lo fh< nght.
I
ff\ total Length 2L 1s
ploc.e.d oG shown in fig . ~-5'1 with ·,~ ends Ir\ co"toct
the,.· inci•"noo po~ . lhc. orig le of f r lcti'on '1i; 1-5°. Defen"nlne.
,;Jf
sa1.) /\ uniform . ploril\ of weighl W
w/
98
2.00 lb
(o. 268) ( 2u t-t) ~ o • p
H " 282.84' lb.
#
ft.
e
p
i?1:J
99
zfi. ·O: R1 cos-4'5' = R:z cos 1s •
"'+
·. ~
%.F!:I •
.
:EF.!1 ·~ : We • R2 sin"s ·. -
• 200 t Av
H = 101.6 lb
P" AH
+-
We• 2.73.Q./b
2IZ'i.,OOS+s' t(O·U&)t\6'eo&tG. -Hs'i..,s1n+6' -o
In Figui;c- ,determine the volu~ of P. just surficieni io
Gt~rt_ the.10 wedQe under the -+oo-.lb plock. The ongle. oF'
n:- ion 1s 20' for all contact .surfaces.
537.)
f
p = 2.86-'t t (o.~)(101.6)
p . +27lb
4ti(-f-OOG-t&') tfs Lea&i-6·-lis siri•s'L •O
fi 1
R,A400 .
1.tg.13 "'0.-'18tte
-~
Ne= 289. 4 lb
AH • He• 289.'ff6 :,wte = (o.ua)(zs~.+)
Fe· n.6 lb .
~<
fl.it
. ·.p~·o:
.
{!,¥.'le)·
/
'
R
.
·
R
.
£f;, • O'
2R&•n (~q•) ·P
.
.2 R[S•fl?JC.OS~ +51noYac.os¢}• P
.2R[o•nt(·Gin"f'L)(i) -+(cos3X~)~~·o
dP
P
d"<
.
.
-.s1n~s1nc<J:z
.
-t
2
.,(
coso/2 co.sl" • o
2
cos~ cos¢ ·= s 1n¢ s,ir1~a
cos ~
_ .sin¢
.s1noV2
cof:¢
Tan ¢ " col °"'~
sao.) · /n Figur:e, determine the minimum weight ol l'locl\ B thoi
w·i11 Keep ·it ot tc.bl o force P Giorfs t:Jlock /I up the, incline
surface, of f3 . The- we ight
/I ;~ 100 lb ~ the onglo of fric~ ion for oll s urfoces in coniooi is 13..
or
p
~1. •. ~1.:9 . 2 lb
-. P • 41g.'Z. &1n~·
Av - +i'f'tfe =-,s01 .6 lb
WE06ES
53s .) t4i wod<ae- is used to 10plit logs . ff ¢ is ihe angle of fhc. tion between th-e wt;4~.~ ~ fhe- log, delt . the mo;>' . ongle1.o< of thewe00e. so that · ii- '1rn re-main embedded on the.- log·
p
R1 s in .+s.0
Ws •(as6.-t)s1n7s'-(141,+a)sin4r.'
o:N
i..fx uO :
.. · o
COS7&"
H
/\v
tOO
.
~
frn+2)cOG+~· ~ Re - 386.4 lb
r,
f600~fi00
p
•
"°
'-f
F1
...
N~'
100
tOO ~R1
~
P.1 ~ 141 . 42 lb.
'
6 in 70•
p~.341 .?
lb
S38) In Fi9 · 537, dofermine, the volue of P act ing to fhe left
thot is r,..,.,uired
to pull the wvuge
-"'
· t firom under the 400- Jb
"""'
ou
Ploc/< .
~-.......----.-.'
p~:
of~ .
FBD
~Ht
,..
~
J
- ~
Fa
~~
.
Ll
. •
P
r.
Fe
11.
R.
-13.1__ • 400
sl'l
.cm 100•
xr
R2 • -"01.7 lb
_ P_ smac•
""
391 · 7
-"'m7o•
.._
P•
203.1
lb.
-
sag.) Jf t he wedge described in lltuG-tration had o weighl
of 4001b' whot value of p would be required (o.) to
start the- wedge under the block , ~ (b} to pull the
wedge. out from under the: bl~.
p
.
101
.
F
I()()
II
~ !7~· -
.
.i.
11.t-
"t ..
fl)
Ii+ R.t
ongie of friohon. .' 1~ •
Rt • 601 .1 lb .
~"°"*'
R3 •
=·
10~.01 lb
£F~-o:
.
P • (109,&1)sinao• 1-l13.z) fNl~
Pc =; 91.sli>
P,.. " Pt.-P1
" g1,5 -73.~
p,.. t:: 18-'' lb
.s+1) De_for~inc:. the for·ce P required lo starf the wedge
~hown 1n f1~ure
io
p.
contact
P·&tl .
reo of B
P
+76 .7 lb
R1 • Rt cos1~,. - sob
R. coi; &•
--+--
~fy ..o: R, cos1s' - R~ sinr& • 2000
, .su~titu+e- R.1 :
"· 111 . fr.1.~1.s·- i;oo) cosu;· - R2sin15·2000
CO' 1i:•
t Rt.
.Sff'\ 19•
:
R~ - -Rq&>~·
~F.!1 •0 :
+<X>
Go111;• •
R,CO'~· - R, .s1ni:' (r,•ni,).-w
.eol:)
..S11"11~ ·
c0&1.s•
R, (cos,•-s1ns'tonH:
0
)
..
c
Rt • 11ss."f lb.
p
QO+s•
s-fO~ As shown i11 Fig P-sw, t wo blooks we1gh1ng ~oolb ~ resflfl9 on a hon2.ontol .surfbce- aret to be- pushed opar1 by a 30•
~t
Now
'1 '" 1s
0
for a ll confoot .surfbcos.
value of P jf; required_. ta Gtort rnovemCl'lf or ihe blook.s ~
wo4fd th'1i; .
onswe-r be chC11"<3ed 'if the woight of
one
bl~ were increa&'ed to 3oolb f
- 11.55.-+
~
p - .Q.of-3 . + lb
.sin oo·
s+a.) Whoi force P mus• be appliOd to the wedgc.G' shown ;I"\
figlJ('e P- .s+.£ to start them .unc:k.r the block ? the angle ol
fr0tion for oil c.ontoof surf~ is 10·.
FDD
of/\
of the
.
4'
100
J3!_ .. ..!E2""1 1&°
.
R1
s
""'~
1a.~
P.1 ·~ ·
~Fa • O : z~ ..sin30• • P
p-
lb
.
1Z.(13.~).s1n110•
P. •
715.~
lb
103
102
.
R2 ( a.~-+6) ~ :it1!166
.+ 11.1 lb ,
-"? p .. 7~.g lli.
wed9e.. The angle- of f,..jot1.on
1
.
Rir (cos16· co1-t6•) - 1509cot j5• - Re &1n1e"c ~
R2 (c.o&1s·cohs"-sfm5•) = ~ooo t soo cotu~·
~
R, (o.97a) "-too
R1
Gurfaces
.e"-,. "0: Rtc.oG1e· -Rt&in1e•., ~
1ooolb
~f).·o : ~.sln1>'1e-tc=is··o
.
The ongle or friction for oil
16 f5~ ·
To adjui;t the 'v6rt 1eol position. of o pos.1t1on of o column
supporting a ')J)O() lb lood • two-!!! ' wedges are usod os shown
.in f igure P-s+a. Detorm1ne the fbrc.o P necessary to 1Siort
the wodge-s if the angle:, of friction a.I oil surfaces is · 20~
Negleot the fr1clion · of lhe· rollers .
J>+3.)
(.son7o' oos"<. - cos1o's•" "')
(~1nW'c.o.So\
- COS1Z0'61'l"'<)
(.sorno' - s rn20· ) COS"<. • (oos10' t ooa20°).s1n"<
.sin o<
SJn70' - s rn 20·
co.s llO •
...
- o.4-66"'
005 70 ' t
ODS"<
Ton o< = o. +663
o<
2S "
><
SSIW'IRE - THR~~ .sCRrew
s-fSJ
~
Rsln6o• =
p
R
20 00
= 23og.4 lb
_ _.:_P_ . _R_
s1n&S
..sin66'
;\
fhrooded jook.screw hos 0 pitch Ot O.S II'\
.single
or 1.7.S.in. The, coer(i'o icni or Gtafrc rriof1on
or Q.JS, ~or f\1net;c , ( r icl1.ori of 0.10 (o) Oeforrnln~
force
P applied of the end of o levor 2 r1 , Jong which will .sfor+
.'.if'fln9 Q we!9ht
2 tons (b) whoi volue- of p wlll Keep fhc
jOOl<sCf"'~-N f urnln9 ?
f1<. • 0 ·10 t on
0 .10
(o/
~
tan-Cc~
(b)
~o.~
2lr
¢ c5.71 0
'&n r
--$-·ton~(~)
P · V-:,,~ ian(l/>t!t)
·
-&- ~-/on-• (o.0'166)
0 meon roel;L>S
the
or
4:
p = 208 il • .3 Jb
-fT- :- :2.6·
h • 0·15
o • .2('j. •£-+in
ton1c0.1 .S
W•
1c8.53'
.
2ton' • 4000/b'
P = -t.2.6
Jbs.
ton
P • .Yi£...
(-e t ¢>)
a
P • (+000)(1 .1s) fa n(:z.6t8·53')
12+
.
p • -57. ...,. Jb.s;.
~.) The- d 1'-tonoc::. be.fwei:-n o~oceni
·1hreo~~ jocl<Gcrew.
R1 (6irr'IO -~) ~
R1 ..
1000 Gin100 •
9134 . s
(.sin10"-«.)
_P
_ _ • ·_R_1__ ; s~bf1 tule- R.1 .•. -. P ·
.-oneo• .
s1n(<1oto<}
P
=
1000
s1n(10--<)
-+ p(s1n(10--<))
son(wto<.)
(S1n1o'c0s -< - r;ono'•nnO() 1000
( s in 20 •cos-1· - oosw·&•n-<)
~
~
1000
1000
oquolu~ P • 1poo ~
151('1
or rri"ot1on
{7-
.271"
70- -<
o .10.
tone- • ~:(z) •
:z1r
c
0. 15.9
f - b.JO -
tan1
9,oa •
x.....ill,_
• o. 1667 rt •
1.21-A.._
f
•0.10
·-5.71'
•·n~o t-<
f "Wr
to" (cp t&) 12000
104
threods on a /r;ple,r-oo;us 1.;: zi11 , the
lood 0011 be raised b~
mean
whof
~e.-t1r19 0 momeot or 2000 · ib - n ~
Pifch e % in ( srn~ ·,~ ·,s lnple lhreodcd muli_iplj t~ pih::l-d~~
· ·. Pitoh : '2 in
~ll
61n(110t"\)
.s1~t<><)
coetT'.01e,f"\t
i6' % ;n. The
QOOOlb-n • W{o.1667rt} 1-an (.s.11°t9 ,0.3')
lb·-n : w(o.o+30-s1aan)
105
,.~ -"'lb
w=
o.o+.3&5733'fl_
.iis .shown ;r. fig p - s+7, o squore threadw 5cre,w 'is
used ;n o vi Ge to O"Jler'f a pres&ure cf' i tons. Jf H-.e. 69~'N l1;
daJblo f/)re,ado:I ~ ho6 pitch of.o.125 In.. ' o mea(\ d oomekr. cf
1.6 in . dotermt'ne fhe- torque tN:iit l'nUb~ be oppl;ed ot E3 to
create. fhi(.' presl;ure . AsslA'l'le- the coefr•o;eni of' rnclion.. cf
fr1d10" to be.
0.1.s .
Pillih. =- 0.2.sin ( 6inoo d ouble. ihreoc:le.d
j-o.1s
rnultiplij b'.:j 2)
fanf : o.1s
to" fr
f> = a.53"
tons•. -teOO Jbs ·
·
T • Wr ton (tp t B-)
-& •
o.s it')
-
- 0.0.531
£n b ·~ 1r\)
r:y fo Sfarf fi(11n9 0 vertiool O'Xial loocl of' 4-0, 000 Jb. what
moment is n~ory ; to 6fari. iowerinq ihe load ;' ,
P. _1_ =-1.... ·· ,"
·
Dm • Oo - .!!... - a - ~ = e.e I"
,
P .. o.'f' in.
I~
.:n(1 ..f)
:ar
. ton.& .. o.o+GS
-fr~ 2-6
T~ - Wr ~Of' ( tp t~) .
" ..-000? lb ( 1-"f' ll
T~
z
a.+ ton& • .Q±_
&-
:t
·
•
.·. ""' • 1..+
,
JI"\ -
f ..0.1
c
T~ ~
tonlfi. 0·1
T
eELT
-t-&)
"f0000(1.+1'.t") ton.f.s.71-~.6·)
253 .6
1 '.4- tvt"'nS
around o stoi'1ono"1 hon:zontal drum ji; UfPOC/ fO '5Upp:ll"'1 0 hGavy wei9hi • If lhe coern' olent of r...:;ct;on ic; o.+ what wC.9ht 00'1 bo -suppar-lod h9
4
106
olhcr
end
or th~
"'
2
.&s~) /\ rope vvrapped
t1.S7 lb·
t ~(o:;, oround a post wi II .support a
wo19ht of 41<700 lb when o force of' .so lb is exerted of the o ther of"'d · CbJ&-rn1T-e H-.e.- coofTi.0tent of' (ricf10" .
"°'b
on9Je or ocntocl : no·
q
~~
Ln
-
f~ -
Ln 80 •
f
l.n
~o · - f (rw·J( Io·)
J (1~·~n)
= ·~
~
f • o .a-1<3
12..s7
ss 2 .) /\ boot c:J1erts o . pull of "'1'000 lb on its haweer w/c Is
wrapped obovt o capGion on the dock. Jr the
o.ao, how
rnon~ Ii.Ams must the hawser mo~ around 1he. oopstari rotrot
1he pull al the other docs rct 01<cecd -50 lb.
f•
Ln ]!._ .. j {J"
fa
/
" Ln 1VT1Z • l.n~ • ln eo • 1+.61 rod-
f
/+.67 x
0.30
1~
0 .30
= 8217.1 •
~~7. ( ~ .:360• = ~.33
-
furne .
553) /\ torque of 2-tO lb -11 ~c~s on ~he. b.ro~rum ahown in·fl9
P-ssa . If !he- brake bond 1s 1r.. con loot ..,..,,H, the. brokedrvm
thro1.19h. zso· ' the ~m-o•.ent ol fnOt iol'\.. 1·,s oao. Oe-t~ine
the (of"G<:/ p Of tne- encJ O( the broke le.Yer.
e~
"'· 2"° 1b · fi
-T1 · c(o.'!>) ('60'• V~)
12
.Ji_· -3.701. .·. T2 ,, _ T _1 _
~MA
FRICTION
~el"'Hng a .so - lb ro~ at k
T1· .J!>O (23·1+)
.·. W • 115 f lb ,
buf T1 • W
Ti
Jb-n-
.sso. /'\ rope- mok.'1'"'9
eJ~ ·(.so)e0.+(~"Teo)
•
tp •.s.71•
1 ~~ ton {,s.71• t 2.6·)
11 =le
1
= 681.82 lb- ff
. T~ • Wr ton (</J
w
tan- • (o. 0531) " a.o'37 °
0
:z.:;;inD
1
2
-tOOOlb
• ~(1.s.,,.Y< ;.P~) ton(a.sa t.J.~3'7')
T = 102.a lb - rf
o+&) A .single- fhre,aded s~are. screw hos a !Z. Y.z ftv-eode /1rch.
ihe. root d1omefc:r Is 2.6 i" \, fhe, 01.1t.sldo d1amo-ter ;~ a'" · 1he.
c.oo£ri'Oie'.nt ·or rrd•ol'\'•:~,"r o.10. De-fc:.rniine the momel"\I ~Ga­
TPI
·
·0.4
..IL· e-f" -
l
..S"t7.)
'1
angle. of conlool ·1~ (a6o t 90) = "hS0°
. f
woolb-~
rope-?
c
3 .70Z
M
(T, -Tz) (8 iA)l.-fP"") •
240 lb- 11
To - T2 = .360 lb
T, - T,
'360 lb
3.102
T1 (o.7~gq) " 360 -
107
T1 ~ +93.23lb
' ·· ~ ·
)j~.
.::!M.11=0
p(16~ ... ~) : 11 (2~11.!f:l)
HZ!f\.
p(p1a3fl);.. +9a.~3 (0.1667)
p"' 61-7 lb ·
.ss+) /n fig p-ss+, Hit:, c0efl:-1(i1e,r)t of f Nolt01"1. tG 0.20 be.tween -\he
rapo ~ the- fhed drum ~ be.-lween. oll surf'occs in. con foe/. Cbtet"
mine the, rninimum weight W to pre-vent dovvnp lalle mohol"I. ol the
,
'.
.flV!p 0
(.::s)T' t T2.(2.) - (2)fa -(a Jr1 cO
ST t , - 2Ta · 311 c CJ
.:, (""/~) - 2 (o.az~)w - .:3(~0()) co
a
0 · 06 1tW
)
w=
1~ l b bod~ ·
vi .
1Fy •O
lt
rl,
'
.
~
T.e -f,
c
wS..na6.e1·
T~ -(o. -i) (WC<X036·t'1) • Wson36.&/
~.,.
1,
w cos 36-67 •
~fx•O :
l\11.
T 2 = 0.16 W
rlc
...!!~
. o .: H2. • 1000 COGa6.e1· +rt 1
rh "' eoo
/ ' '
bi.it T1 • 1.B7h
fi,= 0·76W
H1
t
rl1
.£1',..~o: T, 1 f1t f ._• 1000&i1'36-87 °
,
11 t
JA-111
+JA-rt.. -
600
-w~,j6.r,7• j 1.a7(0·76W) t (o.z) wc,os -;i6.&7 ' t o.~(600.tWC»!.~•lil
N2•.soqttt1 j
·
=600
· '
1.4·!lWto:16w+o·16w=600-16b
1·7'f-W · ·.,."f'Olb
I
w
~
f 5.2. g 112 .
r.-om
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<JOcurn:m• forces. 300 lb ( 3, -'4. 6) ~ '400 lb (-2.-+, -s); 2001b(-'4-. !I
-3).
or= D
Q)M~na;ms
f()RCE
y
-4
.IC
aoc>.lb. ii
B • ..OOlb. -2
c•20011'1 -+
A•
....
(D)
6
7.81
-5
6°7i
-.3
7,07
!!S
r'ORCE.S
olST.
1
llOOMP.
{15.2..f.
-1~.23 ~-4!> - 298.<>6
-11.iMS .1'41. - -EH.R7
-117.H
1DTAl..f~1
I
f t .. ~- .Aie ._6_
y
1l
300/7,91 ""
Av/a
=(-117.H)
.". J\,. " 115.24 lb.
30Q/Mn • l'w/--4- .•• Av
a<:io/7.e1 " A¥6 .·.
n
·z
Y
o
Fon:i
.
"
Systemb
in
Space
~· ..fL 9~·
x
Y
a 117-~Qt;.~
.. o.::ig.s .
6·71
..s;_.
D
l!.
Cos-"y"' 226·2o/.f!96.00
.. Q.762.
eos-ez ..
WO
7,07
1s2.-t0/~96·90
. ·.·Cx ,. -11.:L1S; Cy= 141 -~; Cr."' -&\-.87
= o.&13
:. poi'n\109 bock"Mln::ls,
~ .. ... 115.2+-110-23-113.15 - -117.14 lb .
\..up-Nord to ihe ldl.
.;Ey
n
- 153 . ~ t :Z38.45t
1l
(;o66;. •h/R
••• Bll .. -119.23 ; B,. = 230.16 ; 8%. c -1!190 .QS
Chapter 6
2
t 226.2.f. +(-1~.-tij
R" 296.90 lb.
-1.5::1.65 lb .
l\z." 230..+7 lb.
_fu_• ..fu:_- ~. JL· ~
')14
I
R.2·~11."t:t:I'"+&•
0
L.
~26.2-fo -1.52.~
o•· ll2+y•n:.•
to get OiGfor'lGe(O) use
)(
ltalMP.
Y~P.
-153.65 23().47
141·++ • 226.:Mlb.
~z·
I
- 230.+7 - 298.06 - &t. 87 == - 1.52 - ~ lb .
603-) Determine -the mognitude of the resuHont, its poinh119 , ond
'
ils d 1recf;o" cosines for the follo..,.,,ing system of non -coplonor.
concurrent forces. 100 1b(2.3,4); .300 1b(-3, -4~s); QOOlb(o.0.-1-).
F~CE
COMl'\:)r'li;rtrsoet' o
')(
2
lb· -3
A•1QClb.
s~300
C=-
!ZOO lb.
I y
li!.
DIST.
(0)
3
+
.6·:305
-+
0
0
TOTAL{<1
to got {D) u.sc o 2
5
7.07
+
4
,....'(
D
37.1+ ;
Bx/le .. By/y ~
- 127,-?QI -169 .7.;I
Ay
c
Bz/l:
5s.71 :
'·~ I
i
0
0
~00
- 90.156 -114.()2 "tt36.44
.. , r
1'
:. c)I. -. o ;cy =o; Cz
I
.. 2.00
Ar.= 1-+.:za
" 300/7. 07
•• Bx .. -127.296; Br " -169· 73; Bz - 21 ~2.'t6
110
2H'.16
= x~ t y•o~; 2
..b_" ~ ~ -6.z._ - _jQQ_
x
y
z
5.385
:.
FoRC.E.S
')( o::>MP. YCOMP•. !:.c.oMP.
74 :28
37-1"1- .55.71
111
I
I
I
\
, ;.
.-
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
6().5.) Three concun-ent
£><
0
37.14 -127· 298 .. 0 - -oo.1se lb.
a
•
£.7 "' -1H-.02 lq;
1+. ~0 1" 212.16 " '2/Jo ... 4a6.441b.
it passes ih rough the p0in~ (s , 2~3) .
l)ete.m11ne ·1he mogn'1 \ude of i~ third force f ti...., the angles it
makes witn the reference oxes.
:z.21 -
..,olue cf:~ is ol.so 20lb ~
p.2 =~?<)2t(~y)ll•(~~)2
~ (-90._1 se) 12 t (-114.02) ~ t (-..e6.+4)
2
· R "'-507,7 lb .
po;nbnq forworcl \...clO'Nfl \o· the lefl .
Cos~ · .. ~:1</f'\ = ,go.1ss/.so7.7 ,. o.17s
Cor;.tty "'~,../~ = 1H- .02/.!001.1 • o.~2s .
Q)s-&'i!.
£:.z/R ~ 4e6.-M-/.507,7 .. ().958
•
or
- 3,):
COMPOHENT!O OF D
y
z:.
X·
DIST.
(0}
FORCES
llCOMP. YCOMP. .Z.OOMP .
.s
+
-.3
7.07
11.!3·15
-5
0.77
- .3
...,,39
-:l13.66 182 .44 -228.05
'21/2,63 -H1.a2 - 166·98
A .~~lb.
4
B • 4«> lb. -6
C= 300 lb.
TOTAL(.;)
+
-2
62.12
141.+4
,l(
7-0~'l·
i!
y
212 .56 -479'.9
f
y
i!
•
4
4.se
e.73
Q.2 20 10 .
.s
.3
6.16
16.23
4,37
6·"'1-9
2.6
?
2.5
_fr_ .. _fr_ ".EL~ E£..
y
)(
4,se
i!
-,
17.47
g .74
?'
,3 • .!'>4
F z • f x2 t fy 2. t fz 2.
· - (:-22.-%)2 t(- B.3'0)
:. Pit =8.73; Py " 4-.37; P2 " 17°<t7
-&.. •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/<J3.G8
7:S. 6
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<irnent of the force oboo.1+ each
6()7.)
.•. 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" - <t7.9.9 lb.
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
obout the X ""z aJ(es .
zMy • -.594."'f lb - fl .
- 1+.3 (1) t
:. r)l .• 7.....,, 1b.;
'f:
-F~(,.) - Fy (2)
~le" -1.306·2 lb-ft .
~M>1 " Fi!!.(1)~1~.6 lb -f\.
d ".s.3BS fl .
s
=-16a.3(4)-3:26.S(2)
: • .f,." 163.3 lb . ; Fy • 326.slb; F.w;•163.3 IP .
' eoch of the coond1no1e axes.
d'J.· :z"+3•t+2
lb-ft.
%My • Fx(+) t Fz~2)
rt .
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 .
10
I>'
''
'
'''
lb ..
1-..(.M>.0)
I
·'.· /\D • 12236.1 lb.;
'•
D>< • 2£XX> lb ; 01:. • O
d111a. 6.708 fl .
.__ J____
"&,1
_L_ ,._J,~:
.!1-',-'0 3'
AB = 2494-. .3 lb.
D ,,/~
/
I! J'
d"'c ·vk~-0)2 t (6-o)'' • (o-(-3))
1
d>.c ~ 7,35 ft .
d.a.o .. vf.3.-o)2 -+(f,-o)1 + (s-o) 1 ·
ct,.., .. e.361 n.
,. (O.o,&)
~-
116
1000(3)-0.,(e>
117
..
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)
I ,
..tlllo •a
(~c00-1ooo)(s)-C,,(8)
£Moii.c" 0
By(') - 2otxh) ·o
'
.
AO ... .Qx_
M
6'
8.1'9., ...
Dy• 37S lb .
...L&..· ~
7.3$
·~
•
1
d.-.c • /(s-o)*-.(+<>) t (s-o) • 1a'
1
d>-o • Jte-o)2 -+ (~) 2 t(o-(-.t)) • 9.16!1'
O
Oy(S) -(WcXHooo)(3) a O
Cy:: 62$
. By • 1000 lb.
___J,k_
~
6.70&'
~Mc ·
•o
1
e.367
AC • 76!5,6 lb .
6
r.o .. .522.9 lb.
/\B • 1116 lb -·- C
· The point(; B, C, II., D of the con Ii lever frorneworl'- Ghown in
fig . P"-617 ore ottoched too vedicol wall . The 4<Xrlb lood ii; porollel to the z. pxis , "&., the 1200 lb lpoCl is vertical·. Compute the fon:::e
in eoch member.
Isolate l op View,
617.)
!Y
d~ "v't-10--0)2. t(o-o)~
I
0 (c),,,q)
,,,I
dN:- .. /(10-or
7
f
Isolafo
(o~(-4-))2 .; 10.n '
t
co-o)" + C-+- o) 2
c;~
• 10:11'
C"(10)-1<l00(")-800(3) =o
Ct-= g-+o lb .
.tMc.o • "
1100(0) -ex (10) - 8y(~) eo
C ·":._--;-
;_,../(o,o ,'1-)
s ~ (10) t Br,(a)
1~1b
1.sola+e Top View,
~Mc-o
.001b
a,.(s)- ~(.+) -100(10). o
s~
AB
£.Mc.o,e •
o
2'Me • O
1200(10) - 0.(6)• 0
o~
- 2ooolb.
...t:Q_ . _Qz_ <
11.66
10
~
10
l\D • 2332 lb - ·-T
•
"1eoo lb .
1~00
-10
10-77
. c~
.AS " 1615.S lb-·- C
.2000(<1-) - -'f-00(10)-C~(il) •
o
c." .soo lb
118
e
:. ey
~ 6/s s,.
a,. (10)
f
-
AC ·1260 lb-·-,.
+
6
·ZM c •O
1000(-4-)- ~(3)-o,. (10) •6
sub,, lo 1
°" . 160 lb
6/.s 8JC (S) • 1S6ai
13,. • 1ooolb. :. By " 1:zoo lb.
,'. ;\e, .: 17.S.S lb-·-C
B.i ·· BOO lb-
"
1\0 • 160
~----S-1\0 ... 193,3 lb -· - T
Golve Prob. 618 ·,f the 11oolb lood in fa'g.610 acts horiz.on\olly Ou~ward from/\ in the direotion from E to-Nord /\.
61g.)
o"'e
10
AC • .538 . .!5 lb
s.77~
8-ia
----s-
from Prob. "71.. 610
_&__ · ~
10· 77
/\C •
----:j"2
£13600 - ··-(j}
~~...fu_ · ~·k._
.I.
lsolot .e Front V iew,
~
.=::Mo=O
-·-c
=
e.17s
n.,
d.-v
o"'o = 9.168 0.
119
= 12
n.
;
lsolote
lsolote front Viow,
Isolate Front View,
Top View,
~t.1c•O, -Oy(&.66) t By<'.-+), •o.
90t1Dlb
"
By = 6 .66/+ Oy
-· -©
Z.Ma•O, ~(...)-Cy(...)-()y(12.66) •o
Cy(4)t Oy(12.'6) - 8000
••
' I
I
·,
- ~-
.
. •'
I
: a'
lsolote Right Side View,
..... :
!
Cir
.iMc:,o • 0
0,. (10h By(3) - 1100(.+)
c
~"10
Qr -C..(10) t
"'0
s
P,.71!5
~
.+
;,/~ Bit - sutw. to©
6/s a,. (a) • 6000
/'C
J3.G 811 "6800
e,.- .500 lb.
:. 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.
:I
d.-..e • J.o-o)• t 60-0)~ • (o-(-8))2
·1
jf
d-'6 = 12· 806,
d/\C - J..&-o)at (10·0)2. ~ (6-o)•
8
•
. d>.& ~ (o++))'l tb0-0)11 t (o--O)e
10
/\C=009.07 1b -C ;
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
eooo
= &.66/,... ( ...53) ,;. 980.74.5 lb.
AB •10.56.a lb- ·-C
-0~(10) -B..(+) •Bz(.3)+ 1100(0) co
. /\0 • 1200 .
g.165
0
~~
&
Cy =.!5 66.25 lb
in@, Oy = -+/!!> (s".2.5) • ""'.!53 lb.
in©, By
1r- -6.-
:. By "
109't t
©
Cy<'.<4-) t +/.,Cy(12.G6) • 6000
B~(l.)tB.!(3)-1100(t)= O
00
•
+Is Cy - · -@ t;u~. to
in©,
c,.- 80 lb.
/\B .. __fu_-~ · ~
()y •
-o
c,..(lo)·~)t~(:l) - 3'400 .
Bit (10) i-By(:l) " 6000 -·-©
£Ms:O, Dy(s)-Cy(.ot) •O
llOOO
&z
....:..- ®
(0,0;6)
d.-G"' t+-11'2 ,
10.7703 ft.
<JN;• " (o- o)• +(10-o)e ~ (o-(-+))'
d1'C • 10:1103
o,....s ..
dAO
("""·o,o)
fl.
(S.&6-0)1 +(10-o)• t(o-(-s))
c
1+,1.....
2
!
fi.
./>z
'/
,/'
(9,0,6)
120
121
~
:[.
l&alote P.ight Side View,
Isolate Front View,
lsolote front View,
eQOlb
I
Vt
Of
Ito
Dy
zMc •o .J>
-01 (,-zo) t OOO(f>) - Sy(1")
--+00(6 -~o)t
:. By· 19g,g lb.
LMo•O
• Oy
gy
.£Mo •O !)
6/
C1 (-.) t 1200(2)- By(10) •
By(S) t C,...(13)-1200(9); 0
•O
.!I
.subs. <Y lo@.
- 2400. -4-0C>Zo . -2001 .4
/\C .. _k.~
1+.1+.t
10
1•M'ta
{?QQO ~Cy(13)(1o)-Cy(-+) • 2400
Zo • 0.997 '
: . C1 , 100.1 lb
12a:x? - Cy(JZc) - Cy(.+) = 2400
~Fy•O
Oy
t 199.'~.
-Cy(ao)' .. -9600
Dz • [-4<X:>(o.gg1)]/10
'· Dz = 39.66 tb
200.1-000·0
Dy .. -4{)0 lb
%Me.•0
Cy~
:. Br
.
/\0 • J..t60) 2
t
~t. t(39.06)f.
1<0 • [200.1(el]/4<:lqf°
/\0
a
432.7 lb-·-C
0y (){o)- Cy(S) "0
~o • +1
0
14-1.+2
a
s
I
12.00<0
/\C
1+-i+!Z
10
/\0 - 471 .3\b -C
/\C
_6!L..
120.1 lb.
csu.) In fig .P-62.1, ·,r Ps1irolb. ~ tne coorcl'tnates ci Dare 'f.o.. S'
~ Zo - 2 fl , compute lht:i force in each leg of the tripod .
I
Dy; .!!512.lb .
32()
----:;o--
6
!
c~ 160.~ lb. .
~
~"' ~
tCz ·
320lb.
6000 -(a20)(13) =-· 360 lb .
.5
Cr +DY t 0y -12<l? .. 0
320t Dy •368-1200-0 -
•
_!1;._.~- ~
1·+.1~
o
0y(w)-Cy(4) • 2100 - · -©
St • 6000 -Cy(13) - ·- ©
-+soo- 2190.e .. o
...
_!'_4__ ~:--~ .. __
o a s'
i;- - -- -
p
d~ ~ /co-o) t(10-o)'2
d.....a
t(o -l-a))2
= 12 .006
n.
11.358
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"'<l? lb ifl any member .
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 " /<o-(;2)) t(G-0)1 i(O-(-e)) 1
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
1l
1331,3(.5) -
. .1YL
+
'
Ax (g) = O
/\-;, ""' 7.39.61 lb. '
:. /\x • &+1.4·.5 lb ; /\y· 1QM.36 lb ;.A.z •1202.91 lb.
IGOlote
lsolote· R1gh l' Side View,
checl<.
it OA 7 24QO lb
PA
7.39.61
~
'7 •.o>tf>8
Front VieN,
CV- , 2 767. 2.5 lb.
: . -this conno! be acaipl beoause
the hrri1f
.~
.,
""'" ,..• •'
: • .so1e
c c.
,,.'
LMr ·O
(By•!w)t
l\"f(-4-) - l3y(.e) - O
~. I
192+.~6(+) •By(!>)
By•
I
l
f
l
Gt ·.
6
doe&
noi
ei<~
..,
~.()61
-c1 (+) •o
.
j g .5195.8
lb .
2
6
B
W •(17ai.9H)(:9) • 5191;.g lb.
7.~11
.!I
~
6
'°°'
1 ,6'
4
C>
11000
~
1
zM...,e • o, - 100!>(..}
·,r DC· t"KJO •
~-~·2t_·.Q:_
Bz
P,(gh.·G"1de \/t'eW I
Fr-on! View .
"I
w(lf) - c.,c;) =o
.
t
.az_-fu:'.....·
1'131.92+ lb.
6'
a.,.,sgs.30 lb; Sy=110b.1" ;ez·1+08-~ lb .
- C(4")1 100o(6) - J3(g)
B
.ZF;r eO,
A
C1 • 1006.9 lb .
ot the r~h l side view,
c~ ~
.tMc · O, /\y(-t) - By(/;) •O
/\y ·• ..9/-t(17061G)
diecl<. ·,f DA "'7 2-400
: .. C1 • 1331 .:3 lb
~ '223:Z.1 lb
o
L
2
9
B ~ 400 1b T
1000 - 400 - -400
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<lO lbT
A tB t C - 1000 • O
7.~
W
+
Z:M,....,, - o
•·,r oa • 2..+<lO ,
~-
·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<e
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).
0
,,r~-
:•'
I
•
c ~~:'"""'° '. e·
1: ------
From@,
AC. [4-M(1eU/.ot
AD ~ 91.66 lb
IC • 12GO lb.
c. · ~·2s(eo).
Retell"•r'lg lo Front Yiew ,
in CY ,
S()()(-t) - 13;!. (3)
Z:Mo 0
6
Bi. (6) t 62:(3) - C..(10) ·, 0
s.. (6) t Bz (a) · '!-200 -- - @
0
0x(+)- 8i'.(.!!J)
r
0CO
c
800
Sl!.c 400 lb.
.£Mc.• 0 ';J
B,. (+) - Bz (3) - 0,.(10) ·CJ
in ' ©,
-!100(10) t Sy (a) : 9350
0y = 14-SO lb .
- ·-@
1'00.@~®
. B,. (+) - ~ ~ 000
B,. (4 ) ~uG) ~ 4'200
.630.) The Boom BE
I
,,
of lhe .st.IT· leg
derr1c 1<.. shown in Fig. P - 630
IS rotal<!ld fOrwol"tl :.! Jo' measured In o horrzon\ol pkme- . The mast
h"TO'
f
@ Cr · G / 2 :
- B
9 .1(.5
@ 0,. • D~h.
t\B ·,, vedicol "'- '1s suppor\ed in o .socl<et ot /\ . The points I\,
C. """- 0 ore In the .some horizonta l plane . Oe.!emi;ne ~he forces
in ihe legG
~ 130 \..._ th9 components of the bsarinq reocrion d A
Ci _-c,.
.6IL.
f ront View,
.!2!_ •
6
EL •..Tu+
ec
2
: Oz " o.:/+
subso.@t....© lo ©
(o.,.f~· t C../2)::. -(c,..+O•}IO + +2.!50
o. +C•
.£1Ac,o • 0,
O
I
• .soo' -·-@
deo ·dee
·o
e~ (10)-1 By~) · g::iso - ·-©
.t.Me •O,
• c 1)(3) - (c. t0.)(10) ~ 1700(2.5) co
(0
1
(o.,. • c 1)(3) - (c. i o.)(~ o) • 42.&0 ~ o- · :.@
126
·1
ij
c
a.5 o,.-+ e.-' c.· = +2~0
g. (1o) +By l~) -11oo(e-M)
-+-~o 1b
ax'· .so0 lb.
1I I 1·
•; ' "'~
I
eolb
;\O. [eo(0.1~6 )1/s
81< (10) = .sooo
.,' , ,
/ -~.
c.(3) ' c,. (+) . 0
subs. @ to@
1200(5) ~ 1000 (f,) • 800(') - .13l9) •O
C ,:;, 600 lb - T
o,.(G) -
c,. ... ·.s.2.!!J o,. -' · - @
~M,..·O
:='<M.-..v"'O
.c,. (•) ·o
0u+: 0.a • 0•/+ ~ C• ~ c,.
11
127'
= ...[i1s·
11
l~olofe
lsolote Front
' t - •-
I'
~·
dei: s ~lf • (:20· !5)• , (10-0)• = ~4:Zs
R19ht ~ide View,
deo
Ill
2 t(20-o).,. t(o -(;-16)) 2
= ,Ao-(-10))
lsolote front View I
Koo
c
l sol ot.~ Righi s ide. View ;
•
I
0
·I.
0
1;•;-
c;;. I?< c..O
I
10
0,
c.,
~o·O, C,(1lJ)- 1~(s) =O
±"'c;.o~o,
Cy
/\y(!i)-~(!!H &.~) cO
/\y
.i<f.M,..~O
= 1~ lb .
{fi5
9
G83 lb .
s
. ~
:. K
/\y(-i.)t Ex
l
Oy • 1~ lb
Oy ~ 866 1-·-©
£Me·(),
.
:. AO
-AJ.(1s)-t (c,.-. o~)(+s) t .soo(s.~)-(CytD,Xs)-o
"(c,. • 0,.)(1$) -o -· -@
. {77-5
subc. i.
o"
~
""M ~ (:z17.7t.~;~
~ 183
~
®1
~ 103
-:;a-
• • o~
= 200.1 ,1b .
:Z27.7
1.sE,.
:a;,.+ 4C/30Er
10<;tJ(2)- (1 .~ !:,.)(~) - {)y(t)- E,x '0
Bu•: Ey = 1.s E,.
Bu~:
<4E•
e,,;;~
t
• e,.f0
Dy/:20; D•/10
Refe1'1"'ing -\o ~~igh~ Side View,
.£Me •O: /\z6s)+(DL tCi!.) (1s) +Cy(,;)- Oy(s) - .soo ~s) • O
Az ; -(221.1 t61)(1s) -: 683(.s) t 103(1!1) t 2500
•••
20,. · ·2()0() - ·
:. r.,. "6000- 0r)/a.
lb .
>.
Ey-=
-@
21::"
.s1.1bs. fo©
0)1 • Otl:z -.s1.1bs . ~o@
(160C>:z-!?.l..) ~
€y
A~(+) I 1ocv.l - Oy ~ 8000
lb .
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
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)
~<1000
10'
/'<fl.
---~r~y-----
~Mo · 0;1oao<_ro)-A.,..60)- Ex(.s)•o
.·. ec " 1~ 10.
15
Cy(s)d)y(s) -soo(s.~)·o
'
•
.h.:.
... "
o
=o
The boom of the c}erricl'- shown IP Pig P-f>32.
rotated bock.
~· from the Y.Y plone . Oe.terrri1ne the forces in BC ~ BD ~ the
component~ of 'lh6 beoring reoc\iOf'\ a\ \he ~\- /\ of the~ AB
c;32.)
&(•o,.,.o)
·w;
.
ir.(~,ll,·!>) ~z • smao"(to) • ,!;f} .
I
0
1- " 10 COS 'a()
fltl(X)Jb
i~.+s· -1tJ/;;:D
"'n'°"• -p()/t.o
I
:
..
,,,
1
__ ••Jfli
-~
'.I.. - o·--71,. "
~·
0 ·-
- - ----
(10,0.0)
(0. 0,11• .,..)
ton fiO"
8.66.fi .
.·. 00"17.32
n.
i5A • 10 n.
10/CA.. :. l5A,. :zo ft .
..17.3t/6A :.
C.CSE()' ,.
·
•
:. N5 .. 200.
dBA ...)jo-w)l+ (26-o)•+ (o-o)' " 2Cfi. Ell • <5A t x
dso • 16o-c)" t (w-o'f +67.32·0"'f : 28-280. b • 10~ s.~<;
dee
·,/60-o't+Gi:>'o)'it(o-(;17.!Jt)) 2 •
26.~& {1 .
~ 1s.~ fi .
-oc .. 1otoo oo· . 17;~2 n.
0C = S77.3~6
160iote fron\ View,
t.&.'26
:.
2a
ec "816.3 lb.
Chop tcr
7
&, CentorG
or
'" Eq.©
Cy tOy•1132
~ • 1732- 577.316 " 11.s+-'&+ lb
UCO
I
:
I
co9
I
s·
°' -__1~~-- ~--- :
Cy
Ay
,·
f&,26
Refern~ to F~n1 View: ~f,.,, 0
A. - c.. -o~ "'0
• 0.
-A1 (10)•2¢00(11uc}~ o
"268,i:i?i8 lb.
Cit.• .&77.31G (17.~2)
1·
• sn. 3-'v.l lb.
I
Oz· 11~.66+(17.:32)
Q .. ,.99.96 lb
- ·-@
Oy • 1132
o~
'LO
- 2000(a.1;~) • o
I
0, • 11s+.~(10)
w
A,..-~ lb
;t"tv1.-. ~ 0 (£,
Gt t
c. • sn.3~(10/i~)
Ax .. ue."5a +.s11."M1Z
Ay • 3732 lb
(Ct• Dy) 1 0
'}.()
.
8.'40'
Or
~Mc.o.o
G rovity
••• 00~1632 .7 lb .
BO • 1154.60-+
.~ -
Centroid<>
I
w
loolo+e Righ~ Sida View,
Dz
,, ~-:
=99.9 _qr; lb .
Rel. \o P.'ogh t ~ide View : £:Fe ·O
"
Dz - A~ .-cz ' 0
Ail.
~
0
De
Oy
__.. _- ---------- c,.
"•
i?-32 .
A
l\y
G1
a
11
999-96 - 4:99 .96
Al-
a
.SOO
lb
C.
17.31,
~tvlo s9; ·3732.(17.32)+Cy(~.~) 1icn?(na2)=0
Cy(3"!-.c.-+) • 1999S·Hll'.J
130
131
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<oc.oso()(ocos0<d0()
- 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.
r'ig. P-n-4.
value shou ld ·the
1+4,!593)(
1f ,
~.666s ,,,~
or the f\onge
width
t 2. 5
b "
40 t 0·8 b
T;t_l.)(
•(¥~£)(~~ - (~<2 )(%)
.IC.~
_:rv«~
8
!l
~!J
B
..
T~ 3°
" .9~
1
r% - V-'#
c
6.57S
r~
>t$..6
~ •
relative to the Y
OJCiS ·,~
of the '6Clme fbrm
OG ~
e
20
y~
• "'
f-2 <1~M _'f11(t>)~2~j "f%(111)(~)(3J~-1'1~-r~(cx11x%-1~J
,
12-,. -
o
111'
. )<
12
·
-~ -~
12T
-!57.6
x • +.at-1.
~ =[%(12)(c;)cafa ·"~ : [!111 (4>)(H1)(Y~ .,5)
1-ig •
r,% - rX2 .
e(v~)( -t~-,) - (irr,Ya )(rlz)
vr,% ij
01<
t<-x 11 with respeot to the X 011 is .
by cutting o semicicle of d iorneler- r ft'Om o quo.-+er c irc le of'
r.
B
Jr • s .oa in .
the curve y2 •
b-1oin.
_
:g • 7.;Ja,76!l
nts.) Locai'e t he centro'1cl of the shoded oreo enolosed b~ t he curve
y 11 •ox'!+... the stroigh~ l•ne .shown in Fig . P-726'. Hint : Observe thof
72~.) Locate fhe centroid o f the shoded oroo in Fig. P - 72'1 crcate::I
[ J~C
't/11-)
~(12 - 1jc;>)
1-t.+.593
be chonged so tho+ the
2b '"20
,.
1118.531
iri .
1+4-.s9ag; 12(1e)(c;)- '111(c;)(c)(o/s) - v(+)"(12 -
centroid of the ON:!O Is 1.s in . obove the bage ?
[ 1(9) t b(1i2.~) .. 1(0)(•1-11) • b(1)(0.~)
..'10
,
e
j( ~ 7 . 7+
".
T- section of Prob.n+ shown in Fig . P - 7•4. To ...mot
6 in .
T~)~ i< • 111(1e~)-~~c;)
(2/~(G') t H)- T~•t(<f-) - T~t (1s-~)
..
b "6.53 in .
121.) Refer to the
in.
- - 12{E1) - Yt.(c)(c;)- lT/.f (a)«(va) -
--tr-"'-.-TTT77'77i'-+-
in • .. 2 in b 2
ba "
2.69
~
n-+.) find the coordinores of the centroid or the shoded or-ea
shown'"
,
[ ~ (-.)(eiJ['lf -aJ c[MbX6U[1/!tb]
95 . 333
-y : 6'-9.599
y•
in.
·
g ~ +.s(s)(a) t ('h)(1.s)(c;)(2/5.c;) - Tr(,)""(r; - ~)
24.4314-2
49,397 if " 261,659
no.) The cen t roid
~ + .s(G")- T(~)~V:., Y11(t.s)(<;){Yt(1.s)t+.s) ·
+(+.s)(c;)(;.~s)-: v~L)(W)
x • 3.o+ in.
in .
g - .s.+1
the shoded area in Fig. P-723.
)( 24."431.+2. 714.15
<fS.387 j' ... Q.5'4.322
Y: "'
or
y
. ·1
I
3c;
g : .::ifl.
" o.637 r
121.) Loco~e
•he cen~roid of" the culvert ..shown in Fi'g .
tegrot1on '1s unnecesso r~
·,r !he oreo
P-727 .
H1'nf : In-
·,s subd'1v;ded 'i nto the e lement
to be- found In Table v 11-1
- l -- - 1 ,
--t
[1(Y11 )(,X&)rn2(')- o/gf...x•~1lr ,r(reWt~)('J9) . 111(6)(~)
- (~Is) (-+ )(1a) (g/a ·1>)]
-j
6'
'59~
c
204
.H ·a.sf'!.
6.57~C
(ff
136
·
II
137
I
·1
[1(,) 1 12 (1 ) 1 ~12)(t)]g • t(,)~s) t12(t)(7) 112(1)(M)
:ao~-111
~
- .s.1 in .
.ZM:ico • 1 (<.)(J,s) t ( 1)(7.!!)(-'·6.s)
· L.M~o •
731 .) Two 10 in.-15.3
fig. P - 731.
1119.) /\
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 .
. .. we - ~ - b
~)
2
2nt1
2
- bh%.
.:hruLg& -~bh'-00
~&
:
h(n•1¥nt1)
• 6.,Xo.r.)(.a.2a)•(~"f..o.7)
27-7GS
~(ent1)
(n+1)
[111t3.7.5 tll;(o.e)1(~)] ~ .: l!l("to.211 - 1.,5)I3:1!l(o.-,s~o.211)
ant-i
~e • 2~,;
g = !ZG!l,<t-03
E .. 9:~
in .
-,.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\
~ -
<t-b
.!!l;
• 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;.
-
b)
g._
v • 'Vs Tr•
A • 2T. ~ . Le.n9fh of oul'Ve
to one-holf'
Yor """"bol.,;d , 20. ~ . "
.c'eT ..::!J2. • .a..bh
'Q.2
Y=~f.a .Tr~
v ~ 21f·r~a
'5.
" .. ~ 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<ra
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 -
t1on .
y • .J.r!._
[T/j! (~•- 1.s•~ '1 = ~ (•)' fW?J - %{t·!S)2f4,.(~e)J
6
A = l~
7.of<!!.) Cornp~e.
Totol Gurfoce Area •
A.1c.a.- T3 .T r 2
¥
t
lrit
..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.'<J' ~ - 15(/. 66
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
¥
_
:__
- _
rirn .
[ 4(4) -2 (1:'&21(!1)]
g.
4(4)(-2-) - 2(1·~"~)(+-%)
11.5 'j = 1 s.s
g- = 1.61 in.
v. 2T .(1. 61 •10) ,11.~
10•
V = 939.9 in~
V e o.+ess rl
1
!2f. S.+1 .
cp~plete
or the GI-laded oreo of' Prob. 72!5.
779.92 in.?>
m
] for length
r3
i
.
lit.!~ t f•s,H~ ~ ~ ~6.s)t +.S(•)>6(') • ~1.st.,.,,,(J1.i:1&' )(~)
.
19.997
t
T(%
_g =3·9+ in.
i
'
·1
238 lb .
142
(6-~)
g ~ 78 :a2g
w • o.4BS.S ( 490)
w"
(1H~tJ4•t1fi ti'~)
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+<f.77 in.t
Compute. the. .surfoce. oreo ~ volumS generated by o
.
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
fhe- cone. . Loco•e fhe centroid of the
result of Prob .
~II: [413f(4~11 ~}~] (-.9<) lb/ff 5)
Wss. ~ 76 . 02 lb.
.
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<is; t hrough ihe opex porollel
to the base · U~ the trons;fer formula ~ ihe- r&:ult- Ot illuG .
Prob · 602. .
I,.= bh°Y.96
I"'!,. t Ads • bhf36 t bh/2 (2/3h) 2
• bh3/36 t a.bh}'g
-~·
. ~~
--
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
f/ &1n'~de­
• r4/.+ J; Yt(1-cc.s.i~)de-
I~ • r~
J~
z
Iy •
Iy • (
f
0"
• r'Ys [-e- - s11·1 2&/tl:
• r•/11 [(1-o) - (4>in2lr-s1no)]
'2
Tr"';/e
f ~ dA
1
f'cas~ fd<Jdf • {,,.for p ' of 00$2-6-dfr • _!," f~ (cos~de)];
• fo" r~ (cos'-~dtr) • r'/+ foT (1t(;•nlle-2 d&
T
. ri"er~ - ~J
it
0
~ r'"/e [cr-o) - (casalf -coso)]
Iy ::
146
lfr./e
147
.
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% )]: • [ <y,(ti -0):1 ;- [O/b"<~)J{»-<>)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<es.
= 30in?
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-
923.) /\n oquilo\ero\ triong\e hos
:Wl"l~ol ,._ vo<i1eal oxe'" ore equal .
h - Jb1 - (~)1 - o.8"b
i,, " b(o.9"bf = 1.80+)1.1()-tb+
:I1
b
12.5(6)3
t
~(HZ.!i)(<;)(<O t y3.,)2
r)t
~
il<A
t
Ille ..
t ..c
o.Wb(ti/~t
.
1iZ
t
of 1,, 1-,. 11 .
cf side o .
= 1.9o+it10' b+
ton ao
Y-n
•WLYh
=t>Mlh
b
c
c
I•
5
0[2(0~/t)]~ -t ( 1h(a)(o.JY,4) ]4
1~
AT
tonqle shown ip fig . P-82S
1
.Oclined ol on ' onqle-fJ- "' sin - ¥~ . Hinf : ReE>ONe the fiqure
<it. C .
.38Y •
Y
b•7•.5
in.
d • 1!1 -7.5 = 7.5 in.
.n.
h = c.osa6.87(7.s) =
c.oG .s3.1a· =
a
fx.; -
12.!! (•) 3
t
'12
2
a/10
=~in.
(12..5)(<> )('15 .,) = 225 in~
~
152
in.
~
2
• [1Cs)3 t 1 (s)(o.-t7) ] 2
n
lir ~ 37G.~ io ~
u,;og Tobie Vlll- 2 Proper-fie, of Gtrvcturol Section.2
9•,.'"•,,i •: j,. - 38.8 1n~ ; y -1.G."? in . ; A~ 1~ 1n.
!,,
~ 36.97• = h/d
12°(1)(G) t[8(1)(11.s)'t .S(-1)(9.5)] 2
= 9.g7
I,, .. ~(f,, •Ad•)
= 1{12~ H2.ft)(t.97f +[~ (1)(2.53Y] 2
u
ton 53.13 =10,tb
~[e(1) t !S(1)]2i-12{1) " :39 in~
A,y · ~ ~Ay
the 1oin . by~ in. reoobout the X ax1~ to which if is
K. • 1ojs1n53:t1\ " 12.s
tf
-two 8 x ~ x 1 \n. angles riveted to o 12><1 in. web plate . Determine
the moment of inerlia w'1\h re~pect to the cenh'Oidol )( Olli~.
toe moment <:A inedia of
ifl\o parls A. B,
>
A,. .+.s(:a/.+) +[3.is(3.4)]2 ~ e .62!> ·.n~
The built - up S«:fion shown in Fig. P-627 is composed of
,.•'
825.) Cornpvte
4
i 1 =iis."M in:4
1/2 (o)
827.)
12
iri
= +.s(<V+f t [¥t(a&f t ¥+(3.s)(1.37?S)2 ] z
b/o
h "' a-1!/2
o
a
.~~
8'0/ei -= 60•
:3GOO
i .... +2.11 in~
ly = ~(fy t Ad 2 )
12
-& •
0
a
.. 3/+(+.r.)3 t (3.5(3/4l t 3,5(3/•)(2.'925)2] 2
12
12
2
3
o .M§b(l:Vt)
12
momenl
Compute lhe
paei;ing through
900
i,. ~ zO~ • /\d~) .
cf 1iierlia with reGJ)CCt toan ax~G
two oppo&ifo opeJ!eG cf o mgu\or hB"<l90""\
ez+.)
24'75 t 22.!i ,i-
0
36
;
2-.1-' in~
3li
826.) The. cross GBCtion shown in . Fig. P-026 ·,s t hot of o sfructurol member "'1\oWn OS 0 Z SeC~ion. Determine the values
~
.
f,.,.. •
= 1 (12} 3 t12(1 )(2.<n) 2 • [3S.6 t 13 ( 1.se) 2]
z
12
4&
in.
j,.
= 37<..97
itl :+
s20.) T""o 12 in. 20.7 lb chonnelc; ore Iott iced toge+her to form
the Geet.Ori Ghown in Fig. P- 626. Oeten-nine how for oport the
c honnels sha..ild be placed so oso to moke I. equol to f y for
the Gechoo. (tieqlecf the loHice bars which ore 1i-idlcotecl by
the doGhed linec; .)
163
________ i IYo
__________ _
from Table Ytl l -2
Prop. of Shvchm:il A!°'Olys'1s
t. ly x
Olcn"lel Size /\reo
12'20.7
Iir
~
l,.
-(1:2a.1) 2
ly ~
__ J_
. -
(,.03 128.1 3,9 ().1 0
!y • 3019-63 1n.+
r.
2
12e.1~ -~3-9 +6.os(d/2 to.1) ]
d = 1.~a in
929.) Oefermine the d 1s4once d ot which the two 3 in. by s in. rectangles shown in f i9. P- 829 should be spaced so ihal f,. • ly .
•
h =
...;··
.
Using Tobie v111 -2 Properfiec; of strvcturol secl•ons
a·)(+·><1" ; Areo • 111"~ ; 1, .. 11:, 1n•;Jy • 69.6 in"°j i c.a.os. in ~ 'j'• 1.0.s in
1. ·[1(1<4YJ,A2 ~ [1t.6t11(6. 2)~1 + + [(1e(2. 2sl~~ -t 1S(MS)(B.37.6)'l.] 2
r~
f1
<
.
7601:99 ·.n.+
2
[ 14(•) 3],A2 I [G9.G t 11(3.!!S) ]4
!y
c
3021 -1 ·, n~
f.-+
Iy =1.
3
j,.
; [ 8(3) + 6(3)(Glh t 1•!>)2] 2 •
Ad?.
;.[3(S)3 +o;~~[e(a)3
12
+S(3Xcl/2
12
tu)~
plotes riveted to two 12 in 'J.0.7 lb channels.
s---
cl"' 1:20
Strvdural Geetionc;
11 =3. 9 in.4 ;
~
!,.
2C&
in.
too web plote 2.3~ in by S/16'1n . to fQNT'l the
~ ·in fig. P- 030. Compute tho volue of i •.
-
U~ing
in~
t
.
1~
Tobie v111-2 Pl"Qpertie'? of Giruci.,rol sections
•h : Arco - 4.7.s in 2
= 6.a ·in 4 ~ I 1 •11.+ in 4 ;
6>\4>'
!,.
I. " [ 6.3 t 4,75(11.01)~] 4
t
y· o.99·..,_ ~
ii... or:igle column
·1c; composed
1.99 in
Prob. 826, ore riveted
1o o
12 bt1 in. plate to form the seoi ion
12
12
.
[ +.i;(a;..,.)9 t +.!>(::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<OG. 3.S in:"'
031.) /\ plo te
2
833-) Four Z bo~. eoch hovin9 the Gize ~ properties determined ·,n
2
'~~·~t t
6(0.5)(11.1s) ] ....
[o.s(a.!:>)'5 t o.6(:a.s)(g ,;g)~]+
r .... :2667.5.5
11 e(3,9t~.oa(+:i) ]i +[1C~~)3] 2
Iy " gs6.87
y • 12 in.
3
f!JA61''1.3.s) .. [
1
J. • 1610. 9 ° 111~
12 •. !O.Tlb
ore connected
plate~ ongle girder-
I.. ..
/\reo •G.os 1 n~~lx = 120.1in
y • o it._ il • o.1·1n .
" 128.1(2) +[%~) 3 + 1 6(1)(6.~)'2] 2
12"- 20.1 lb :
or four 6 by +by 1h ir'l. ongles
930.) The ~hod legs
From Tobie Vlll-2 Propert ie!O of'
--<1>" lb
I
I
+f[ 2..2S(10)"412} (2)
832.) Determi~ the controidol momen•s of inertio of /he built_:
up calumn section shown in Fig. P-832.. It is camposed of two 1o"x1·
""
L •
• 3(1)(.ot.7?1)i) +
1
Jy • 1i1)3 t (1.26(18)3 ]2. +(1(~)3+1(0)(+.5)2] "t
1
12
1
t [~fi)'& + 3(1)(•)~] 4
Jy "[:i.9 I ~03(d/iz +0·7)t] 2
=iy
r
]+ t [-i(i)3
I. - 7~&+.65 in~
------- ---~---------
+Ad ~
12
12
f,6(1)-a t 8(1)(,,1s)1
~
-
L
'
i.c t A~"
h-
j,. • 1(1+)3 +(1e(2.2st + 10(2.25)(0.375)~]2 •
'
'i
1,. "1291.31s in:t
11 ~ ~ t [ :a.s (3/±l3 t 3.s(3/+)(o.01st' J +
12
1~
t[-"1+\+.s)3
1
t
s/-t(+.s)(s.s)e.]+ +[3.!5(3/-+t
12
155
t 3/4(a.s)(6.m;)
2 ]+
iy
I= MK'
t Md2.
= Yt Mr'" t M(r)1.
.591.6 in .~
&
I\ 10 in. 1!1.3-lb chonnel is welded to the fop of' o 1+ WF 34
oio sho-Hn 1n fiQ. P-03+. The wide non9e beam has an overoll height of 1+ in. on oreq of 10.00 in~. 'lt-. 1
339. 2 in ~ C:Ompu-le
9 "". the moment of inertia obovt lhe centroidal ;>( all.i6'.
From Tobie v111-2 ProperheG of 5trvchJral ~ ·
I = s/2 Mr'Z
834.)
beort'l
)l
or
10·- 1s.s lb
Area • +..ot1 ·,n :l.
Area• 10 '1n~
1,. =~.01n~ !I... ly 2_.310:4'
1"=339.2 in~ S.,. t•o.2<t in.
x ·0.6411"1,.., y · 0
9
A1y £Z.Ay
14.'4-7y '-10(7) t 4.47(13.6)
i.3
t
4 .47("t.!l6)Z t
f 1' = 476.01
I .. MK 2 t Md 2
2
2
~
= % Mr t Mr
I • 7/s Mr11.
866.) 9y using the transfer formula 'lili., the result
Prob. 003, determine the moment of' inedio of' o rod w ·dh re~pect to on oxii;;
through the center
gravity . perperx:llculor to the rod.
~
Y"' g.o+in.
I,. "
(Tr\
or
'-14.47in 2
/\T • 10t4.'47
eoo.) By using the transfer forrl'lulci ~the re!OuH of Prob·. B62,
determine the momen~ of inertia of o homogenea><0 sphere of
rno~ M ~ rad'u; s r w1\h respect ~o a tangent .
3~.2 t
10(2.04)
_L:~
---1
2
in~
83S.) Two 10in fS.3lb chonMIS Ol"C weldeO toqether as sho-Nn in FK).
of Le
P-835 : Compute t\;:le,volues
AT y
Q
Ya ML2
I"
t
t
'I
Md 2
Ya ML : 1 +
I .. 1 ML2
tor orrongemen+s (o) ~(b). Eoch
M(1-12)
2
~
From Tobie Ylll-2 10 '-15.3 lb
Areo
I"
2
chonnel web '1io .'0 24 in. thick..
o.)
or
Ty"' 2.31n:+
i( .. o.C04 in. ""' y e 0
4..+1 111 2 ;
.. zAy
&...l• =66.9 in~
- 4.41(2) y "'4.47(!5)t + .-t1(9.6)
Y"
using the tronsfer formula~ the result of' Proh.861,determine the moment of inedio of the rectangular porollelepiped
shown in Fiq. 8 - 29. with re~ct to a medlon 1i·n e orthe z face.
Tak.e the median line porollel to the )( axis.
860.) By
1.3.1n
'
I~ =".9t4.47(2.'3)z ~ 2.3 t 4.47(2.3)2
lJ< = 116.5 '1n.-+
b.)
" MK.2 2t_Md2
.
= M(b +c 2)(1.42) + M(C/:z)2
" 1A2 M(b 2 t c 2 ) t 114 Mc
I : 1,42 M(b 2+.+c2)
b
2
2(4.47)y :. .ot.47(5) t 4,47(10.6+)
y • 1.82 ·1 n.
I, ~ COG.9
t 't.47(2.8~) 2 t
2.3
t
4.-n(2.e2)
2
Ill " 140.3 in:"
8'04.) By using the tronsf'er formula ~ the result of Prob. 060,de-
termine the moment of inertia of'o homogeneous r ight circu lor cylinder Obovt on O~iG through on element on
surface.
The cylinder hoc; o rnass M 'b,, o rodiuG r .
as
869.).
0
Determine the morneot
or inerlio o\ the rectongulo r porellel-
ep1ped !Ohown in Fi9 .P-8GG or f ig 0-29 v./1th re.s·pect to o n ox•G" t hrough
one edge porollel \o the Y o.xis .
I c MK 2 +Md 2 =1.l.;2 M( o 2 +cz)+M(o12.)2+(c/1f
.. 1/t~ M(o~+c 2 ) + M [./(Ola)• 1(c;,d•
~ 1A2 M (o~1c 2 ) + M(o}'+ t c)l..f.)
I r % M(a 2 +c 2 )
157
156
J
f
·l
l
I
Y• 4% [1011- 8 3]
012.) Determine the moment of inedio of o hollow steel cylinder
w'1th respect to its geomeiric o)(i6. Tho cylinder iG 1 fl long ~ hos
Of" ov\sicle diorneter of af\ ~on inside diorneter of 2 fl. Steel
wei9hs
4QJ
lb per n~
.
--3£'~·
2
Ill.
V•1f(1.9•-1 2)(1)" 3:Q27 n~
w= 4SO_lbj,r((M83fi'} ""S32.3Slb
-
W" ~Olb/f\1 (v)
v21 mt for fue hole
Determine the momeni of inertia of \he c~t -iron flywheel
in F'.Q· :-078 with res!)C'ct to the o,.ic; of rotoliOl"I. The flywr.i,,I
hot;; 61>' ell1pt1col spok~. 3x+ in. 1n cross section which may be a:ms1dered os slender roc1~ . O:lGt iron wO.ghs 400 lb per-cu.fl.
+I
Ws "(.+.90)] (1.5)(2)(23)
1728
.. ~ (1138.03/32.2) [::10
or 20 in
outside diameter
~ 16 in· inside diorneter, compute thcS mo.men-\ of inert ia~ -the
lo
o d a·ometer.
w
=- 450 lb/ft 3
= .50.45 lb .
Ir= Y2 M(R1 tr~)
I= 1.l(;z ML:i
:. 1.42(~2.~)(6') +{4-0/32.2)(1)'
1 = +.97 fl -lb-.seci
158
Ws = globL i for one spol\es
Wr •(+SO)lf ("30 1 -28 2 ) 12 ,. 1138.83 lb.
1729
874·) A solender rod 6 f\ \of\9 rotates about on o)tiS perpendicul
io it ot 0 p01nt Qf) . from one end. The rod we19hG . 40\b. Compute
the moment of ·,nertia obout the 011'1s of roto\ion .
wei9h~ 450 lb per cu. fl .
Given·. Oo # 2c/':.. R=-1oin .
6m
Wh = 'lST(R.•- r 1 ) L i hub .
1
'"1/t M(R2 +r 1 )
rodiu<0 of' gyration w '1t h res;-pect
w, • i•(R' - r')L ;
·~lfR h
\12•1fr'h-rTl1.""Tr'h
1
V1"lR h -m1
:""*
87.S.) for' 0 hollow cos\-iron sphere
7.35 in.
shew:
r Is. ~ NIR2
M~gv
I': Y2 [ m1 R2 - mt r 1]
2
M ·'f,irh(~~-:· 2 )
.::. 1/~ [ ~TR1 h(~)-llrr h(r~)1
4
4
lSlfh [ R -r ]
1
"' 1h glfh (R'- r')(R +r•)
1
i
=-
878.)
der of' rodiuc; R from which is drilled o .concentric hole of n:r
·diuc; r . Denote the mass of ihe resulting hollow cylinder by M,
"'-, the moSG" per unH volume by 't .
v-t\. m, for the cyhnder
..532 -35
K
I · 97.11 ft-lb - sec'
973,) By using the rnethocl d1scus&ed in Prob . s10.determ.1ne tha
moment of ir-edia.w1th respect to the9eomefr1c axico.of'o cylin -
R•
.....
1,...~_,,.2 (1_+4_)....,(.-3-2.......
2)
l : 6.i fl-lb-.SCJC(
.,.
=~O lb/;:r' (3.927~
32.:2
h
~ -
K"
.
w = 192.+.23 lb
1
2
I . Y2 M(R' tr') = 4ft(19i-+.2"'J [(1.5) +(1) ]
1fl.
"' 1.183 ff.3
(12)•
Cod ·,ron
Ir "'
2~. s
1
+2si)A+t1
fl -lb-sec(
Ih • 11.l M(R« t r 1 )
- r~(1:::i7.++/32.i)[(s~+2 2.Vi<KJ
In "o.+290 fHb-.sec2
Is ... n(i t Md 2 )
1
c n (1A2 ML'+ Md )
=6 [~2(.sG..+s_h2.2)(23A~)1. +(.!56.1-o/32.2)(16.5A2) 2 ]
Is=
2.3 .11
t1-lb-.sec 2
·
I"' Ir
e.
t Ih •Is
206.e + 0°4298 t 23.11
1 ... 230.34
ft - lb-sec~
159
.~
0.i a ccr~oin .sfrdch of track, tr01ns run oi 60mph. Ho.,.
for boc\<. of' o stopped lroin should o warning torpodo bo plocod
to i;ignol on on coming troin? .A.ssumo thdt tho brokes oro ap2
plied at once ~ retard ihc lroin· ol the uniform role; of 2n por soc -
1002.)
.
(}ivon:
Vo~
60mph
0
-2fl/so=-"
Q
•·
..
I.
Vo .. 60 x s 2so ~ eeft/~ec
f-vo' • 20s
3600
-SI/ = -2(.2)S
.s .. 1936
n or
-
o.36G7 mi .
1003·) /\ sfono· is thrown xcdicolfy Upwor-d ~ returns f0 Oorth
1osoc. Whof was i~s initial volocity 'I.._ how high did
Given
t=1osoc .
I(\
if go?
y(vo col
-Vo
c
-9.Q1 (10)
= 99.1 mis
or
-vo" -;32.11+(10) "' 321.7+ fl/s
Choptcr 10
S
Roctilineor lronsloiion
s
U
Viz gt~= ~(.9:g1)(10)'1
¥129+~
D
c
~ (3.2,174)( 10) 2
4-90 .S
C
I
m.
or
1609-7 ft,
n
dropped From the top of o tower oo
high of the
some instant thot o second boll ·,s thrown upward frorn tho ground
with on inihol vo\oci~ of -40 fl/soc . Whon t+.._ Whore cb H~y pass,
'ii<... wi\h what ro lotivo veloc1~y ?
.
100-+.)
.~
co
/\ boll is
I
!
l
1
!I
!
~ ' vo.l t ~gt2
•
"!Ot - ~(M.:i)t.
2
-
..
-©
ii
80-h ~ y~ (3.2.2)l ~
h ~ eo-16.1t' - ·· -@
substitute 2 to 1
80-~ .. 4-0t - Y12~
t:
12SOC .
Qo - 16 .1(2) 2 ~ . 15. 6 from
hs
tho
botlom
s' = 80-15.6 - G-+.-4- from the top
Vf1 -Yot
Yf1
2
-.0-32. 2(2)
Yf, • 312. ~ OZ)
11
I
: ot
e
'=
- 24-4fl/GOG
64.+
ft/soc
~II
!
160
161
"
j\.
r·p·
ar
what ic; tho <k;plh
a
~
i~ fOoe.) ~
1llfOI "" .5SOC iofor Ibo 60~
1r lhG «>loc.ily of sound ~ mo n ~GO(; ,
1CtJ5,) /\ sfono Ir; droppOd
~ splat;tl -. . heOl"d-
O
the~?
v~
h"40.3t -
tl
..5600- ttmt, ~,,._1 tl
'IHI0(-5-l,..) = 16-1
. Y'· .
fhc souod
d .. fl2I0(41-~)"---@
&ubsfilule 1 lo~~- .
fq..t_=
0"' '12~'
d"'
·1'-1(3.793)
=
~(3£2)t,.r --- @
•
-b :!:J_b_*~-40--C
231.'3
fl_
..9 =3~ Sl/scc
'IOCO-n = h
tr
1
t - tt.Ar.c ~ lhG 1!s[ ~
l~-L~
I
fer t1hG. f'._.j~
.
I •
I
t
£
I'\
h
fl .
~ 386. 4
.second ball
'1-He(l-2)- '1~ (::n)l(l - 2)~
h ~ 2-tel:-~ -116!.~ tEftt - 64 --·-
2nd boll
Vo " 1!09. 69
n /soc .
t2
~10.)
t • :2.S36 -sco
storic 'rs thrown vc;rlicolly up frol'l'l fho ground wifh 0 velo city of 300 fl/soc . How long mus~ one wo'it boforc dry;>pplng 0 Gee
/\
or
1-sl ·s tono
©
.S '" Vot
• I
15400 =312 l
t".S SCC- :.h: f()0() - 14i(li)~ : 600fl.
162
386'. 4 • Vo (9.~+ - 4) - 16.1 ( 9."64-4) 2
s1onc f rorn •he bp of 0 600 n low~r ·.( fhc two ~fones o ro
co ch othor ~co fl fri:Jm the fop
the fow er .'7
(~)l L
tooo-~~- 21-St -~ -.u-V t *I.: - G+
11!JGQJ~·
fl frorn tho ground f'
ft I9GC
Bt ~uodroHc E~ t ~ 9.41-G+soc s,.....
1ooo.-16 l~ - · - - @
SJ1il6f'itiuille 1 fo 2
-~bi
n
386: 4
Vo, • 193. :2
396.+· 193.:2 ~ - 16.1
A Sto.nci. droppc;di from o cvpflvG balloon of on elevotioo of
100QI a ~ ~ ~ llailcr of'Olho:-.,sbio CG ~oafad vcri;colDy up'f<Ofd fioro> tho 9'0 nd w ith ...,doolly of He fl per .«c. If 9 ·l'i 3'2
R per scx;Z. ~ 11>:-~ '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 - .. -<!)
0 per sec-_ After :- sec, arothc,r ball is shat vort1cally ·,nto tho air.
~hot ind1al vcloody muGt t ho 2nd ball havo il'I ordor to moot tho
for- tho .stone
IOr
P, ~i°G
t • 1. -+ sec .
h" +9·3 (1.+) - 16.1 (1.4):2
h =36.064 n .
h" 96.(; t -96'.6 - 16.1 t. ~ t 32.:2t h .. 120.et - 1.:;.H ll - 112.1 _ .. - @
"fSX:-
16-1 ~ ~
112. 7 "' 80.!! t
h~ oo.6(t.-1)- 16.1 ( t - 1):l
fq- you gi;t t, = +..a45 liOCd= t(S..1 (+.QH5)S - 3.53 -'31 ff0)6) Rapoot Prob - tooS if tho sound of tho splosh i& ~ ofter
+t80- tt20fp ~ 16... t.
subst. 1 to !l
4S:st -~." 129.~ l ·- ~ -112.7
for lhc 2nd stone
By~
c
• 96.6 ft/soc
t " t ime for tho 1st -stono
t -1 ~ timo ta- lhc 2nd Gtono
for tho 1"t .sfono
h ~ 46.3 t - Yll (.3::<>.2) t'
subsfifulc; t lo ~
1120(+-t..) '
with
Vo, • +B ..3 n/soc
d • ~(M_t')t.,S --·@
' l.
th~ ground
o velocity o.f +e .::1 fl por sec. Ono soc.end lotor orothcr .stono ·,9
thrown 'V'ortic.olly upword wif.h o volocity of 96.6 fl per soc_ How
for obovo from tho ground w i ll .stono bo ot tho .some lovol ?
f....csmc
'115 c t'100/soc
t. "" time b- hslorio lo drCJfPOd
f.... -'1 • i'ilne IOr-i1hc;5GQOd iobG hGotvJ:;(Or-60Urld " d c"sl
tor ihc.stono d-= -'i~
ct• 1t20(.s-t,..)-6)
·I
stone is thrown vorlicolly upwon:l from
•o pos~
2nd Gtorio
t Y~gl:z
200 .. )[z@a.<2) (11.100 -
t')"
-400 ~ 300t - 1i;.11,?
:200 • 1 G.1 [(t1.1ee)'J - ~(17.1 BB. l') _ t'
By~uodrotic · t ~ 1 7. 1ss soc.
200"' 'l-7.S6.30 - 5.S3.4's t ' -+16.1i,'z
i..__
'JJ
0-= .;sSG, 38- SS3.45t'11 6 .11.•!l
t __
< 1.44 &cc
~Y_ '""
*":__u_a_d ro11c
• · you 90 \ •
_ _ __ _ _
__ _
163
t :
13,1;~ i;cc.
8y ~uodrotic fonnulo you 90\ vf "' 2-40 ,H/i;
~40 " Yf ( t1 th)
~
t 1 ~i'.1.tts:-+1
1011.) / , ship boing lounohcd GlidoG down the ways v./1lh o consiont
41 - tg
occ.olorotion . She takos B soc to .slido tho first foot . How IMq .,...·,11
~al<o to s lide down the woyio 'if thC.r length ·1s 62.5 f\ .'
Q'40 •
,Sc G!l~f'\
~61-0 = vf(41-:!j-)
t1 d2
sho
s • 1 ft when t a soc .
..s • V12~t:l • Y2a~:l
1 c Y2 a (e)Q
E
vf= ee H/s
tJso or 90+ 00 fU,
Bfl fi {s x a~e. 'x
vf (-41 - ta)
1~
n
mile
mph .
60
-!'>2&0R
$
Cl - o.031~.5
i
Qoo.sec or .3min , 2osoc.
c
2G-40
B
n/sec,•
?J'v\013.) /\n automobi le starting from rest s peeds up to
-40
fl persa:.
w1\h o constant occolcratlon of -+ft per soc-2, runs o t th'1s Gpcod for •
2
time, k. finally c.omcc: to rest w'1\h a dccelorotian or ~n por sc<? ·
If the total disfonco f rovclc.d is 1000 fl , find the ~o\ol time r0qd ·
0
£0\'n :
6 ivon: Vo=O
.
a•-tf'Vi;'.I.
y s '\ONs.
o•- sfU
..
~
Yf = -+efl/G
~
0=4-r+/s"
a -~n/s•
VJ..' 1016.) An oufomobi le. mo¥1ng at a conG\ant '<'elocity of -+5 ft por
s;ec .paGGos o gaool1ne -station . Two GOCCnciS IC\ter, onoH·lO". outorn0bile leoves the 9o"°line stotion 1*.._ oooolcratcs at the c:Onstanl
retie of 6fl porsoc 2 • How .soon wi ll the sc.c.ond o utomobile overto~c the first' ?
Given : V1 • +.s fi/s
o
~eq'd :
40 "'4 t1
.G. •
Req'd : total Time
-
.,
Y.i (-+X 10)
2
•
I
~ooH .
t ' = 1s. 79 - e
when t • 2600.
e10CI'.)
ta
~
top Gpcocl ?
•
o • &fl/," Soln: vr -v/,·ot.
6iven: d· 'le m 'i le
t1. 41 soc
Req<:l : tnO.)C. ¥pcocl in mph
,.,__ disti'nce trove. I ot
thl6 top .speed
Yf • !;;11/i11
- vr - st,
1~0) /\ porhclo mo-.-es In
= ta -
40t whero
Q
cc.loro Hon ?
61von : s - t
12 1') ts2
+Vft1 -
24(1l)
-
tip- t
3-
+
=
98 n;s~
6froighl- line occordln9 to the low
ihc s Ii; in f1 g,.._ t in GC.COnd6 . (o) Whon
+ot
ago in comes to
dur:n9
re~;t, what
l = '51SOCS,
the 4th
is 'its oc-.
Rcq'd : a 1 ve-1. b ~ avo. vel. !!o...
c~ occclc.-ot ion .
~
165
164
2
(X)rflpu~c fhc veloc'.ity .(b) find rhc ave. veloo.•ty
G"OCOnds . (s;) 'NrY:ri the pci0t iole
-
~ + Yfh
'o
Ql~
Q -
G
S1 tS.itSa • o,e mi le.
26+o •
ei"' - t.}ti; t
3
'
Roqa : vlr..,a
-'If= - 6t&
G3 .; Yf ta -
sea .
t• fGOC.
.3-+soc .
th'1G
=1s.79
.
$.von: .s .
=-+ot2 -"" t2: 16 Gee .
10~ 16
~ ~uodrotic
1he motion of o pariicle- is g ivon by the equation s • ~t+ - t.
Qt~ whores Is iri foot 1t..., t inGCXX>nd6 . Compute tho values of v
~tSo.tS3
t
'Jl>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 (<O) (tl1-+tt+)
time
t!I - is :: 8-sec
.s~ =-tole) - 'Y2 (J!,)(s)'.I. = 1GoH.
b'"IO
trovol
d - Y2 .o (t - 2) 2
19t "! 12
- -1-0 ~ - s
·,t
6H/s
4 i; t .. d ~©
-
t1 • 106CC
·' V" S~ 2 - S2 • "!Ct"
Vf-Vo • -ota
doc:>G
~
Q>l'n : . Y=d/t
2
Vf·Vo • at1 ·
d = ~ooon .
/
= 41 Vf - n_•
,,I
' I
>~~
' .p
solh: .a§
dl
"'v
..,.. t • c
2
=
= fq / .sa;e- •
ton-sf
l -~
- Bl\ I~., ~ ~
.
.+o
;; (s) 2 - +o .. ss ft/s
" 3l
-
c~
- b)
>
et; "" J(v"-'9)
t =3. 6.S1 sec
Wr• ~e on C:><prcssion rototinq -x ~ y, ~ by· -succossivo
d ifforontiotion -Ghow tha,t Ye= -XY/\/JTi<ttr->J '!,..__ ae. = J(0.11Ah11tl<~ t
hllY"/~. Compute Ve 'II,.,, o a from these relations.
£,'/J>I
Tan
z
e- = -Ye_/...;" s ince Ve is
Ya = -v.... tan&
cli tfcronf1a~c
O"'
'
o~f°" Oe
~=VA
+l<OJ\
~ :.pz+n;z
--.!X•th 2
-j.
-x'1.VA'2.
~~"'Ya
h
-Y~xYA..(x 2 +'ri~)-;f(~y.)
~t.
1
1
2
-tYA (x th )-11 YA
( -..G"'th•}s
2
Req~
f:.ol 'n'-
: s-t, v- t, ~o-t
St9=v2- -
__Q§__""' dt
..fGt"
~
a
""'~ "Q£.
i;ec, 2.&
,;e-
~ 9+9ton~
s= aton'l7"' :::ise0"-e- ded<> ~ aGeG1-e- de3 sec &-
dt
diGf'm;r;cji b y -jhG
~ - ·u
&Von :
t-1
'l-.2
Reqa : v-t, .i;-t, v-~
3S'='£(%)3 t11
Solh : d'l'/dl: 4t
35 "" ;r,(.:(v)3 't 1!
~...rr
'/-=~ i-C =.2i~t-C
.
2
3s
c=o;;f i.-1
""°
:ztL - t -=-.fo/.w
dt&/dt = d~
3
-S = ~t
-t C
1
a-
= 2 (1)"
=4..F'f
"'"
t 1· .
(3s- 1)« "'~-k
tC
3
3"' !l
~ ilho~
i
!I
o~"'i-t
+ ac
c" 1/a
167
166
ft/80C~
'<l"'-.t • .s1...t.,., ~6 s~ 1
2
;vc3
Tho moi1on of O llJllllfiiCb j 5
o is in ft per GCG/z ..._ t in
8
iroHaliiorii a •-4.t
Is
.s=1fl
""' v= ~{per soc -11:ion. t - 1 -soc- De\onrilfl(; ·llOO rc.DofiorlS ~
102.5.)
*here
VA2
The roctilinoar motion of o particle> 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}(]9<-1.,e 4 tr1rlll(10~:µ1~9•t12 2)
.; (;,<:,7 f'f/s
Gi...on" .s-= v g_-9
"~ £2'f' - '4{1l)2- .. 1>( 2)
d'YdJt "'"3~~-8Jt +6
~(x 2 th~}'!>
the s -t,v -t, 'ii<.,
S:>Bn.7
is ~
a: Vdo/<JJI'.
3
ft per sec· find
)t•- -f-)l2 i- 4'X where v
ctt
2
= ')(CV,
Ux'-111"-)3 '1~'"'+h2
2
feiof por sc.ca;d "-. ;ic; l.s ill
Cocnpuio fho vol.Jic of tho occobrotm ~ ~=· dt _
by Y =
'%? ( xo. +h 2 ) -'12 (2x dx) - _Qy
dt
cJt;
Tho volooii'Y of o porhGlc; ~ olonq tho -X ollits is do-
Retfd :
l
Qe"" XOA .
3
f.ne.d
fGOt-
1
. VB = XVA(x 1 + h e)~YQ
.
.
dVa = Oe =d;c VA(xo.th 2tl'2 + xcJVA (x-zth 2fYll
cH
Cit
I
at
Qe=V.AQ.
·• ---
t 9
dt
~=2
xll +hci-= z.2 i L ~~ + h-y @
=(v'--g)2
v 2- 9
IOU.)
6iwon: ..., ... -r.3 -~...,.s;x
downwo!"d
=
di;
'fhod:
=Va j
1
3
1022.) Checl-. tho answers tc Tilus. Prob. 1018 by t he followlng mo-
1
c
Y£•S-t-9
..SL
·- 4-0=:3t.'2.
;~A
11-
..tsa-n t ~J
.t = In
. 1-a-3 J
Y =a.l 2 - -to
o = atll-40
c .)
t
]l
i4" governed by the equation o = - ~,
-tihcro o ·;s in f eyt per .soc~ 1J.., s le In feet · 'Nhon t "1.scc., s • + fi "-._
10£6.) The mOtion of a particle
v=2ft perecc. Pet ermino tho relation bot woon v..._t,.s1to..,t , v,s.
Given :
0/5 2
Q • -
~
V •
4dt
.solh :
.+t
ods "VdV
YdV -%~ ds·
- 1
+c
1
V Q = 1' S- tC
"if
S ~ "'f 'ro..V•2 .; C =O
Yi c - 1ct+ 60
S2" - 1ot/t t 60t
ds
~Sa/IZ tC
12t
s = 16,NQ
£
=
(t%e) 3 /z
aH6"
G 1Ve'1 '.
S .. Ya (')(10) = 600.
126 - 4
2
t•O, 9='1 , V·•O
Roqo : v-t , s- t, v-s '
sotn: d%1; - Gv"e
. d'0;.(v ~ dt
I
d%~ ~ 9t.~
0 • GvVe.
,,.
' ,'
dV(v)·Y: • Gdt. I
'.2.Vve = 6t +c '
s = 93 1;
.,.
6
= 3t
+6
5 ".:s("o/3)a +~
"'~ v ~<Z t b
-SI (9)
s
i f t . 0 '-. v - 0 ••• c .. o .
012
'l V
• Gt
v"e ".3t
v = 9te
t:
·.stj1a
i
governed by the r elation a "4!2,
where a ~,G
~ t '1s in GOC · When t 11> :zero, v • 2 ft/s· """s • +foot .
fsnd the valuoio of v li.._S whon t = 2 sec .
G1von : o -+t £
{l. ~ ~ +c ~ c • £ .•. v " +ta t 2
3
3
3
t•O, V• ~,5".f
V"' 4-}.$ (a) H' = 1~.67 f"-1/B
Reqa : v ~s when t .. ~ dS/dt "+t.,% t 2
••. s =4t.~1Z t .et t4
Solh : d%t • +t 2
s = +t).(1Z +et t c £ =+(2)/'.12 t 2(tZ) ++
v"' +t~ +c -,r t·o~s = + ; c=4
s =1:a.:a3 rL
'1G
0
168
-.0
:
:
I
I
I
I
I
I
t.
.S1 •..st•/t
10a+.) Thc ' rnotion of o porticlc .Gtorfinq from rest' jc; 9ovcrnod
by the o-t curve shown in fig. P-1034. S t<ofch the v-t &:. s-t
CUfVcd. Dofcrmlno the cl1splacemont ot t • g soc.
o (ft/s•)
Os·B • ·+/3(t -o)
C12 ~ -60
12
/.O'l · - 4,(,t. t to
8
Y2 = -2./a t 2 t wt - 60
II
• - o/..3(0) 2 t 2o(g) -60
I
'
o parf 1olc
~~
51•2.&t.
C1
-+-----'.,__~:..;;.
tel;)
in· 0/s 2
.sc...• ::'.!.2.!.~t-120
Vi •!It
-iv"~
1028.) The motion of
- -
S,•01;...l•O:.c, · O
"if"s .. c;'k..,.t"" o ••• C " 6
3
#I)
ll
"
'TI
·~
8
---a- t C
G" 9t
3
•-40ff
al 6.seG
-4
yr
'10
Ca " - 120
2
S1 = -5t t60t.-i.u>
S• 'h(+)(t O)
givon by a " G v " where a le; Inn
per- &'OV;z ""v ic; in fl per "'OC · When t Is zero, ~"6 t..., v •O. find thb
relot s"Qn.s botwccn v ~ t , s 'I.., t, v ~ s .
1027.) The rnot ;on of .o roriiclc
t C2
S1 •.G.,• 4IO t·+
40 " -.s(4) 1 +60(+) +c~
- 3 --
c
to ..scolc.
. Y· -'o/a(t-4)
'l(t1'')
~ .rs
e
112' t
Vil."" 1~/s
.curves approximately
dG
dt
+ ,. 2/3 ( +) 3/11. tC
c = -+/~
+t ~ 2/3 .s ;lfiz t (- 4./3)
12t = ~GWct _ +
3
v/f. "' -as-1
..
$"
t=1, G·+,-v:ct
R~q'd: v-t, s-t, v-!;,
"1033.) From the v-t curvo in Fig. P-1033, determine the distanco
tr-ovelcd in 4 Gee~ olso in 6scc. /\lso sketch the o-t ~ s.-i
v, .
6
.9
V2
·"'°His
V(fl/')
" --
2/li't t t Vt
if V1 •O Jo...t•o :. C, • O
v, • t :r. = ' 2 • 36 nIs
3' -
- :--it~ -:-'(«~
- «-- - -
" .t
\ "\
....0
~ ~
I
I
I
Ot = -4./3t t 2.0
11
Vt • -+/, t t !lot tC.t.
Vt· t"
H' v-ll. -v, =9
Si;
""- t • 6
36 • -+/,(<.)tt20(6)tC2
t.'/a tC,;lf s, •o Jrr. t·o:. c,·o
ttls =(6°th - 7Z. n.
&1::
169
104-4.) f.lfl elevofor weighing 3220 lb st~ls from resl t..,. ocquircs
an upward velo01ty of 6000 per min in o distonco of 2of1. lf 1ho
oc.cclerot1on ·, s conslont . Whol ii;; \ho \cns1on in lhe olcvotor ooble?
Yt .. -f/,gf1 t ~ot - '°
.!;1. =
2
-1./g t3 t z({'d
-
'7Gt t C1.
I!.-. t ~ 6
2
~. :Z/g (') 3 ~ 10 (6) - 60(.~) t C1.
.G1. • G1 _- 72.
72. •
c~_;
w= 3220
120
s;,. - 220
n
1040.) /\n objoct attains o veloc"ity of 16 fl por ~ by mo"lnq
in a siroi9h\ N'ie wi\h oh occc!Gro\ion 'Hhich vo..;oi; un-1fonnly
from :zero to 6 fl por GCG2 In 65CC- Cornputo its ·101\ial "°'ocity"'
tho c;hat'lgo In dlGplocomcnt durinq the 6 sex; imCN"ol . Solve
t>y uGlng rnot1on curve<: "' ohcck. by calcµlw;.
q
,45 "' v(tsd1) .. AQrOO.H • t,.
"' -'1>("·-o) t(&l')/~]('~·')
8
0
1::.$
=D
2
6
tC.
V* 16 \.., t =b
·:·
v = Skzt.. - s
A mon weighi ng 1b1 \b is in on olevafot"' 'rnoving upward w/
on occolcrotio n of 8 pcrsc.C~- (0) Whof prossuro docs he O)(ori
on tho fl oor
the ol<5volol"' ?(b) What will tho pros.suro bo ·,f lho
c lovolol"' ·," dcsoonding w ith the some accclorol ion ?
104.s.)
or
n
Soln :
Gi-.en:
lb
a~ 'T- W .. W/g (a)
o • 8f!/s 2
T - 161 • 16%2.~ (a)
Rcqo : a -) prossuro ho 6i<Ori
T c 201 lb .
b~ pros&urc ·,f' elcvofor
b.) T -w "' (o)
iG doscc.ndin9 · v'<f lho somo
T - 161 "-(1'Ya2.~(e)
Wm= 161
6iven:
Vo· - efl/i;
11>'t1· Tho oc.c- of on
ob'p:;f dcO'GOGCS uniformly lll)Trl iHl W scc;.7.
to 7.0r'O In 6 ~ ot w/v h"rnc a, . .~ty i{; 10 fl pei-~- Bnd thv initiol vclx1~y k._ tho chongo it'l d:splocoi:ncnl during tho 6 !>CGin\orvol . fulvo by ui;'1nq rno\ion cuNCh ~ ~ by ca\culvs -
. o~ -ei6lt.·6)
8
·
v = ~ Wl'".tet - 14
Goin:
if v ·'4
"- t • o
A~
'
0. - 41.3t t0
=v(t1.- l 1) t
=
(,
-)4(" - o)
/:;.f, " 12fl -
i fv~10S...tc6
Aroa A.-t • h
t 6/i. (• )(1/.5 • '
)
P.eq<:J: Coeff1c'1eol of l'-ine-tiC
O.Ll\lh
"°
2
=
a-= 8
frloi1on
p - F =mo .
11 ·
Qo(100)
ff/s~
60-161 (f-) "1'~2 .2 (e)
/"- :0,12422
Delcrm1no . tho force P thOi w i ll gi-lc iho body in fig- P - 1647 on
occoler-of1011 of' 6fi pcr ooc,<. The coofflciont of Nriohc friclion 1s o2.
0 ;ven : o "' c0n/6e. f"M .sol'n = F• (:a22-Py)o.2
RcqCI : force P
"(3Q2 -% P) oQ
+~ P - (:3Q~ - % P)o.t. = a~2.~ (6)
4/~P - 64.+ t O· ~/.s P ~ 60
p .: 135,22. lb -
171
I
q
I
7''1'~
ye. • ~as -
thorc fore ,C· -1~
170
~
~
~ -100 1! .
'\047.)
Yo • ;+fl/f'
t
2/a t :2 t et t
52!2%.v2 ( 2.-5)
"
T = .34-70 lb
'1f \I cYo \.., t =O
" =-
T - 3220
1oofl,.siorlinq from t"'CGi- Compufc tho oooffiOient of 1<-1noiic fricA 1on
between the blook. ~ the ground .
-8
0
wh (a)
T-W ~
acoolcl"Ofion
T "' 121 lb.
10<1e) The blovk In f ig - p - 1 0~ roaohos o vclocdy of 40 f{ per sco In
16 ., eM.(r,) 2 t C
c"
w
= 2CJ(00)
a • 2.5 ft/s<1
w19
t
01
v ·"'11d
I
'/
Y -= 600 ft/min
5· 200 .
RoqCl:
~.2. = -o/0 t • t 1ot~ - "ot t 120
~ = -1/g l9) 3 t 10(9) 2 - l>O(~) t120
( ~60)2
1
Given:
I
I
I.
l
II
t
)
o+s.) A rnognclic porllclc .•.•~;ighing a., gr<irns is pullod through a
.OOloroid with an accc\crof1on
6 moto~ per GCGe · Compufo the fO~
in ~c acting on tho particlo . Nole : 1 lb ·"'\&t9rom' ""-1.n .=2,f>tOTl.
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<s; touoh . ,
Gr~en : -fl- ~ 30•
0 • (Q,OQ
>
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 t<.inoh·c frlcf1on '1s o.:so under A If...,, 0.10 undor- 8 . De tcrm·mc the fof'CO In the &pr1ng .
G1¥on :
"-fa= o.+
P•2001b
Roq'd: Timo to elapGcxl
5,... o.ao
unt'tl
.solution : 8Ginao• -
tho blool<-6 \ouch
OA-8<:.06.90.
0
~.2
J8. 0.10
as
Qa =4.% fi/s ~
j\Ginao· -
a,..
o-Q /\ c,$'30. 3
10..sa
')/g~~ a,.,
O/s 2
.S()
t /\ .,
- 200 t % (..oo)- 4h(4<)())(o.a) ts t 4'?l£Ma
'° 1 2~ a --Q)
ot B, S/a(600) - s • +ls(60::))(0.1) .. 600,.32.20
·. cs -~aal aXz ("l-9.S)ill
/ · ~ts ,= Y2aA tll .. ~ (10.~ate)
I
0
- .s6 l .5
2
f 2,"f'7,5t~ • ,5.2'5t£
from
t · = +. 2.3 .sec.
os2.)
Detcr
m1nc
the
occcbrotron
of the bodies In Fig. p-1o.s2
1
1
tho focccl drum ·, ~:/.smooth """ A ls hoovicr thon 13.
w,_- T.. w... ( 0 /9) -©
1- We ·We ( o/g) -@
l " We t Ws(~/~)
W>. - We
G -= 12. 420 i'56
1
·,r
( WA-We)g ~a
~
-s
1
,subi;f tiutc
-W13 ( 0 /9) :w'.o. (Gl/g)
~ 18,b.3 0
31Q
to~
/ 312 - 12420 - ,sc; : 18·6a CA
Q.56 • .g1.asa
.S ~ 1!2.42(8-24) t.$6 ° 1S8.4 lb.
or
1os1.) The cooffici onl
!<Jnclro frictior\ under blcx;k. A in Fi<J _P-1os7
'1s o.-ao ~under blod<-. B ii: o."20 · f ind ~ho ocx:.olcratiorr of 1hc Gystorn
.._ ~he fcnG'1on if\ covh c.horci .
o\ ..c, :300 -12
10:53.) Rorcrring to f\g . f> - 1os2, aGSumc A woighs QOO lb ~ 13 ~el9hs
t\t
. +he acx::clorotion of the bx:lies ·,r tho cocffidtont of 1<-i'notic
frsCtion ;, 0.10 bot the cablo i.... tho fl~od dn.im .
w,., ~ 200 lb ) • 0.10
T~fie c e ro- 0 •1 (r)
Te - 10 0 • 1ooh.e.~
-=-
ai A,
~~.t.a
= ~1.11a
11 "1.s.ga +".11a
- ·
to@, 11'!-&-s ·99~~ .1101) -1.31-.<::+ • c.!.11 a
1~ -210.6~ c 9,o/i'OI - ·
172
a
©
T, -1a::>G1f'l<J0·- 100Q:>&.'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<J&.30° (o.!l)
100 \b. poi
we" 100 lb
..so1n :
i
a " s.24 f'-l/.s:z
.sub!;t. 2 lo 1
f1
+
from 2 , Jfl " 100 t 1ooh 2 .2 a ·
GUb6f1tuto. 2 lo +
7.~9sx10-"'(19.69)
f " o. Oo+85 lb
.
1051 ,) Two blockb A '-8 arc roloosocl frow rost on a '30° 1nohno w
fhoy orol?!Ofl afXlri. Tho oocffioic.n ~ ot friclion urdor the uppor blodlt\
Is o,Q ~that under tho lower blool<.. 8 '6 O·f. Cornpuf o tho olapGc.d
~ = O.'J..
200-1;a7lj .. 200/3fM! 0 - ..
173
·· ©
·®
~
T~
" 210.G2 t g,32 a
Sob!.i1luto to 1
300- 210.1;2 t ,9 . s~a
ag.3e
:
.soo~,a
a
= 1a64a
n4e
a,. ·4.a
11 = 1s.ge t ::1.11(.of..a)
d
go.01 lb.
1£ ,. 210.bf t 9.3~ (.;.,a)"
2"5. :3G ft) .
>
. 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.<3
xe
7(
/ er/ - 257.~
·o
e
19:1.2 S1n3o't -
'/ll la2.2) ! 2
- 1<0.1 l ~
~ t ..
'!-=Yo cos&t • 193 ~
11os) Rcpoot Prob . 11o+
\
fl. or
743. 97
to the
n.
o modor w/ a muzzle veloci~y of .sooft/.soc;
ot ro· w·1th the hof' i.u:>nt al. PcAcrrri1no tho positl0t1
of tho .sholl "" i+G rosulion\ volooty
wi ll
'lO .soc offor flr'1ng .
·
v,,: soo fVs
t . 20.SOC ·
= 2:220. 2~
1
.
;
V-= j_V_x_«_t_Vl-,,- •
J
'2!>0• t
2 10.99~
·• :327.13
:zg
n/s .
1'1L
....
- .. . . ' '
-e-·so·
_g CoGf>
''
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<; on9.1na1 level . Also dcicrrnine the mox'1mum height oHoin .o d by the projc.ct ilc .
_(ya!~)
h .. Vos1n9-t - Y1Z9tt
Y. • Vo O'.).Se-t,
.-'
h . •.. \ at (x,o) i h =o
1
/I. pr'Qioctile ·,s f ired with on indiol voloc'ity of 19a.2 O/soc ·
upward ot on ongic or 30' to lhc hor-i:zont.o l from 0 po·1nt 257.6 n.
above o level ·plain . Whot hor1zon1al dic;tonco wi ll ·.t e<:Nor boforo
)(
t
fl.. (32.~
.so\n:
ay ~uoorotlo fof'mulo
)( "' 334. 63
(0,0)
ft .
1104.)
~cqa :
__
,.- .- l .· ..
fl;$ .
h= V~ s 1n 1fJ- • sooes1n2-ro• "" 2911 • .5
.
n.
- Yx -·soo~oo· a Q - Vx ~ 1260f1/s.
Yy - .soo.s1n<00' • - 32.2 (,.20)
Vy = 210, 99
Vo· 193.~ H/s
'J.57.' - 96· ' I:. t 1'1·1 ~ ll
x
)( ~vocose-l "' 1 9:1. 2cos-30·(~)
x .. sooo ft .
y =Vo sint:tt - 1n9i 2
2
'/ =.soos1nfi0' (20)-Yq(a2.~)(20)
-fr = 60°
0ivon :
/l57,,
How high
s o\n : y=voGiriet 1 Vi29i ll
2.su, = 1Q'3.2s•nsot t Y2 (,9:v:z)t 2
.so\'n : x =Yo c.os~ t
x "' soo ~ f,(/('20)
6iven :
projectile ·1s f.r"Od dowl"lwon::i at 30•
ao•
.._
·at ·,.:1so?
ir the
Ssec.
c.osao•(s) = 1338.5:3. fl .
hori zen ta I .
1103.) /\ shell lcovcs
d ;rcc,tcd upward
2
• 1338.S3 fi .
- 2 5 7,, ~ 9G.~ t
y · -743. 97
~)
x .
ton-e-)
'/I. "' Vocose-(2vo s1rie-)
01(x/a,h) ; t= t ./1Z
9
'l:zgt -- t,
• v0 l2s1n2-&
9
~ 2Vos·1~
9
.
.
h= Vo.sin& [2.Y~~ne- -.V£g(v.o;•nerzJ
h "'
Voe.s,n 2-G- - Vo 2 s1ri!eg
~9
2
h =_Voit sin e-
-1.9
1"
176
177
lj
shOwn 111 f:g. P-1107 is ju'°* to cleor tho wotorfilled gop· find tho to\<.e -off vclociiy Vo.
1107.) The
c;:Jf"
/"°
solh:
~ ·Y/x
;gy '•J'. - .. ©
-tf,f c.Vo .s1n:adt - YL(~fl·,2H'-··©
.sub6i ituto l' fo 1,
-u.!l - y{, srn30 ( 2o/~)
- 16·1 ( 20;V0 )
-y
t .
-'t.t Vo • 6440
~'-'""
Vo
c
14114
x .. 77. 8' (.s.51) :. 429. ft.
.
r·.,fl.
1< .. 6oft -tr ~6Cl
~"
= " •• 7i0t_- 1~.1t 2 - ·
~.SIZ.~
n.
_n.
Retfd:
Vo
'Aro %s+:s -
..._t'
ft.
cos-I== x/5
3/w =
x4st..s =VoS111-&t
Y.
j
= €41 ...+4.f1 .
l
-Y4!!g.f.2 -· · -(j)
2
~. '
~ 4 • 4 = Yo £9Glt'I~
--VoG1 oe-"'G4·4~
.
<,gJ
SUb6fitufo .3 s.,_ 1 /
'°·'
1
·I
I
-eo.s =64.4t -
16.H «
t • .S 6oC.
'· t
60 ,. Yo COG~/>
241-++ = Vo case. (s)
t • 1£0/vo
h • Vo 6111-6-t - Y~ 9t q
..
10-.s - Vo s1nGo(12D/y0 ) - 1,.1 (1t0;(,0 )
Yo
~
cose- • <tS .~ee
6 111 e- : Tone- "
vqVo~
~ 2.318'40
Yo 6 1r'l .!13.137
4e.4ff/soc.
the distoncc s a1 whion o ba ll thrown w/ o
1
velocity Yo of 100 fl. par soc, of on angle ft-' ton- 3/4 will .strike
1109.) Deter mine
ihc lricilnc shown in
S •
241."M = Vo ca;&t . - .. - @
: . Yo.
. I
solh: x cVoros&t / ..
Yo=
-s=J(41Z9l +(14t.<JB)!l.
1110.~ In Fi'g. P-1109, o ball thrown dOV'ln tt'lo 1nolino .stn1<os 'it at
a _drsfaflce s • f.s-t.s
If the ball rlsoc. io a moximurn heiqht
h 6+.4fi · 000¥0 the point of release, CO'Tlputc '1t.s ·1niiia l velocity
Yo 1f., inolination -fr.
Given: h ·<0+.+f1.
Sol h : sin~= H/s
s = t.s+:.s rt.
=
H - go.s
!'
~oqct
ga.9 v/
fl.
·©
'/. =77.86t -··@
1"':S, 3f. • 77.8H ; y •fS.%t - ·8)
J · __%.
/'
· ·
~ L.s.g.s (.s.61)" 14i'.·9S
y.
-80.s
Given" h =1ofl.
= 64-·4
0ivon :
-e"
~e:=-·
Vo
p
6
100
/ .
-
:...--fr
c
.sa 1::17'
~
'
ao,s N/s
couso the proj cetilo to
!I
.Vo "
.
t,·, t
I!
l[
I
ft/s
tan - • '3/+
-fJ : :38.67°
RcqCI : distonco .s
)(
•
178
'4...,. /.,.a.teB
1111.) Rof~ to flg.P-1111 ""'- find 0( to
p61nt B in v1-0Ctly 4 sec. What i s the distance x (' .
f.q. P-1109.
,...-1---\
- I
.
I
ft/Gee .
1100.) /\ boll is thrOWfl so thot ·, t .just clearG a 1oft forico "oft
owoy. If il loft tho hon.d -'fl otx>Vo the ground 'Iii..... of o on9le of
60' to. the hor.1%on"tol, whot wo.s the iriii io l veloe1ty of the boll?
.
t "" .s.s1 Goe.
)( "'~00 C.O!i.38.&7 t
- · ·-@
~·
~6t.7i0;(-16.1 t~
-ts,g.s,.t'
-y"' 1oo~l11 3e. &7t -1,.1 t 2
Vo oosao·t
17.~£ =VoGOSO!O·t
t • 20/vo
t~-t,
Y. =-Vo GOS O-t
h "Vo61n&t - Y~glt
~-
-y =Vo,s111&t - 1/Lgt~
179
5.) /\ pof'f1cl~ hm; such a cur-vi linear mo t 1.on t h0 t I·+f; )'. COQr . f ' d r
3
1no c 1~
c11nod bv
, inches~ t i'n so d
,
r x =..s t - 1ost w h ere x ·'" in
condi;; . Whon t ~ QG'cc, the total accclcrohon is 75 ,·n per. soc~ .
If tho Y component of accclcroh on '1s constan t t._ the parti cle
storf~ from rest at the orig in whon t ~o, dcf crrri1no the t ofol
velooty when t •.4scc.
·
·
111
o coni>tor'lt volocity of w fl pcrsoc,stor·
h'""3 from the posrt ion shown in Flq. P- 1111. foiel -& in order for
t.ho projectile to ha · ~he boat .s .sec. ofter .starting. under tho
q:indition qivcn . How h'1gh ii; the h'ill obovc t be wotcr ?
111ll.) Boot /\ moves w ith
d~ v t - 100 : Q.O(s)
,,rt/sec
Given: )( , -sf3- 1ost
.Soln : x
t=2soc, a - 7-S in/&:
t
0
0.
8' t
t •2
=4soo.
-sP- 1os t
= 1si 2
0)(
;
. Rcci'd : V
e
No. g ·1rol'1 irdined at .::12 • .
.
I
Given: V0 60 fVs
.Solh :
2
)( ~ 100
.
h/
RcqCI: rio. of Jron
.
'f.
= Yo~ s 1n2e-
= .31 · 7~ •
iron.
1114.) /\ stono hos on initie1l velooity of 100 ft per.sec up to the
right at .::io· w·rth the horizontal . The components of o~lcrof
oro canstont Qt Ox = -4 fl per sccfl ~ O~ = - QO f1 Per'"'-seoll. C.Om --6-
Use No. 9
put o the ho.-:1:z.onto l 9h;toncc covered unti l the ..stone reaches
o point
60
fl
below 'its or'1ginol clovotion .
Given: Vo• 100 ft/i;;
-e--·~ 30 • , ·a ", - -+ ft/s 12
oy "-~o
fl/s
2
,
2
Rcqa: horimnto l di.stance
Sbln : y = Vos1n&t - '/12 gt 2
..'oo = 100.sin30t - 'h. (QO) i
-60 ~.so t
- 10t
By ~uod(l)tic
valu•c
of t
~
-;.="\-47.GQfl .
Vy e Oyt
2
V = Vy. t t Yy2
.
..
.
:2.25
th/sec .
Reqa : .Spcc;p i.n rpm.
&:irn:
011
=~
r ·
-
8<XX>"~
s
Y =2ooff/s
V = lfO H
t1 "6-+rps (6~)
c
.382 rprn
1119.)
. .At fhe boHorri of a Ioop th c .sp eod of 01 olrplane Is 4-00 mph
Thi~ COUSC-G 0 norma l occclcrotion
g .9 fl pcr .gx2. Detorm1no the
ro~1us of the loop.
or
V c ~rnph
Ro9'd : RodluG
You get the
135 in/G
1110.). The. normal acce lorotion of o po"tic lc on tho rim of'o ulley
10 ft 111 d 1amctcr '1Gconstant ot 8<XX> f l
2
·
p
..,,..,,.,., · f th
.
pcr<SCC · Octcrm1nc ihc-s1"""-"-' 0
c pul1cy 1n rpm .
011 =ggftj~>!
IZ
=
= "l-5 (4) = 180 in/G
~ /.?!S e t180::z
V =
Given:
2
= ~ in/so:;"
200 = 1 (10) t1
x =Vocos&t t Y2ot
.
y. = 1oo cosao(6) t V~(-4)(f) IZ
y = 60fl .
= f;O"to:r"
0'1':'en : d" 10f1
On = 8000 fl/sz
g
a!Z.~(100)= GOiz s 1n!Z&
I
2
=
~75"
V-x = 1St IZ- 10S
Vx = 1S(4) - 10S
V
= 60
= Ja.,.-:r.toy~
Oy
t e-t€cc.;
111s.) H Is desired to pitch o golf ball across a trop to o grocn
1ooft owoy. What is the;; best c\l..lb to use H' the initial velocity
of tho ball i" 60ft per soc? .Assume that tho boll stops deod ofter .striking the green, which is·on tho some lc;;bel as the point
· from which tho boll IG struck . ,.Assumo t he clubs hove slopes
groduafod ot ir'ltcl"vals of 6. Q) thot a No.1 ·iron hos a foce "1n chnc.d ot BO· to ]hd ground, the tio. 2 iron ot 74 ·, c1c ., clown too
- .1os
30l
O;ic = 3o(q,)
a
li'
'l
=
Yx
Sa in :
Or1
= v"/r
.9('32.e)r- [4oox.s~ox 1;8600Je.
r - 1187. 64 ft .
I
l
l
:1
II
I
.1
6 sec .
181
l80
I
:]
I
11
1120.)
A padiclc
mo'«iG on o cif'COlor path of 20 fl rodius;;
coo thot
'of ..s ..
"its arc dl<donco frorn o fi,lcd po'1nt on the path i10 gi..,en
-1ot whcro .G ·,~ in tho foot s,..._ t . In sccondG. O?mputo -tho total
•f'
occ.elcrot ion at tho end of 2 sc.c.
61~on: rc2ofl
Seth: d6 •12tll - 10 • V
s•+t~-1ot
-at""'
t• ~GeG ·
Req8 :
1!Z (£)
41
path. .
-10 •3e fl/GOC
.sfi._=. 24t rOt
dt
a
2
Ot. " 2+(«-) .. 48 n/s
•
ao .. vo/r • (;ge))/w
2
1121.) /\ porlidc .'1s lllOVinq
a
H • \l&Ls1n «4 •
fl.Oy
tan- 1 o."1$
c
,'I'"
~6.~7
0
2
(WO)
Q:::
~a./· ta/ . ~" c t
c
(4/e)
•
a,.
=11. " 2 rt;,,.•
10"
~.l.-.
~-
l!}' ·
"4rAid
-1''-"-w-tO ft.
-
Vy
"T
, '~~)<;
-
. \ ',,
~
\.
- --
.
~(20)
2C0(1t/s) t - Yt.(izo)t z
'40 "1rot - 1ol 2 - · -1ot
f60t i 640 ~ o
tL - 16t t64 -0 - (t-&)(t- 6) " 0
olonq a curvod po\h. /\t a ooriain
*-
•
.•. ta8G(IC.
V~ -
'1,()()(:t./.s) • -HZ(i)
v. -~4ff/,
Ori " Oy COS-&- - a,. .sine• 20<PGO -1a~1nO
Vy " wo(+M - ~(e)
Vy • O ; Y·24ff/!;
0'1 .. QO f t/.s e
an vy,.. - r " 247to
r· 2&. 8 fl
a
0 41 •Ota t On11.
(11 .c;~2)c. Ot2 t (+.+)(
41
,!
CH " 10· 8 H/s
1
1122 .) A s tone is throWri w"ith on ·1nit'ial vctoo'1ty of 100 rt pcr60C .
upwardo.t w· to the rori:wntal. Compvtc the rodius of cur'w'Oturc
of i\s pclh at the point where ' it is .soft horiwntally frOfl"I it' initial po-
Ms.) /\ podiclc moves on o circlo in occorclonco w'1th tho cquo tion G ~ t•-~t whcro .sis tho diGploccrnent in foct mOOGurod
along the ciroulor path ~ t it0 in sooond.s . Two .seconds of·
for start1nq frorn rest the total ocoelorotion
HlO partiolo is
48-IT ·ft por 6!Xi!7 . Compute tho radius of the; oirclo.
..si \ion.
Given : Yo; 1oofl/s
a•48~ft/sa
Rcx:iCI : rod iuc;; of tho
Roq'd: Ct \.,,On
. ,r
I'
\
·
H ·Vo.sin ~l - Y~yi 1 " b«> •
86.7 fl/Gt:e;t.
inGtont whon -the .slope of the poth Is o.1s. 0)1 c Gft per S!Ctjz ~ ay
- 10 fl per soo41 • Compute -the voluos of ab t,.., On at this · instant~
st<Otoh how the path curves .
61.ven: Ox = 6 fi/s<
Soln: On " Oy Cos& - a,. G1""°
On ,. 10 cosa~.67 ° - ~ so'l 36.07°
2
01 . 10(1/5
Qn ,. ,....+ fl/c; 2
slopo - 0 .75
/· r
~-
.
61'vcn : Yo· ux:>ft/s
o.. • -112. ff/.st1, Or· -120 fl/st1
m• -+/a
RcqCI : rodi'ui;i of curvc..duro
Soln :
On ~ .72·2 fl/s
2
a« • at« t Ont. .. (46) 2 t. (72.2)
0
1tu.) A stono I 1os on init iol v61ocify of ~oofl per soc. uP to·
1ho ri9ht ot o slopo of + to .3. Tho c.onpononl-s of occelerotion
ore constont a\ . Oic. • -1~ f1 J?OI"' -scc/z.11.., Oy " - :zo ft pci:-.soc ~ C.ompt.1le the rodius of curvatur-e ot tho stort 81<, ot ihc top of fho
p-60•. x ··. sofl
Req'.d : ~ 1u<0 of
Curvofurc
Vy - 100Slfl60 • - 32. 2(1)
Vy "..s+.4 ft/s
y ... ~ 50 ct(..s-t.4)l ..
ton &
".Vy/y,.
"
73 - 89
ft/s .
S<t-.4/so • 1.088
b'olh: Oy "9 c 32.:2 ft/s'
x "Vo cos~t
.so :. 100 cos roo• t
t
~
On· v'lr
1.soc.
Vx -100 GOS60• • 0
v~ •$0
On ~ Oy cose - a., sin-eOn • :92. ~ Q:>s +7."11 • .. 21.79
ft/s
r -c-n~.6~)~1.79
r - 2so.6(; fl -
or
Given : .s- t+- et ; l • !ZGCG . .
C1i-Glo
&>tn: .s -t+-at
3
Vt = -"t(t)a-8 .. 4(a) - 6 ~ 2+
Q,, •
>./'/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<13GCC·
horizontal piano obout
;
the 120-lb ball
0.901 m/s
Is tcrccd to rotate
o vorii°'I
fO"Go on tho &holl bccomo zoro?
<3ivon ; w- 20 lb
tcn/-tSGGi 9 =:a2fi/stl
Rccid : toriGion ~force
t.......spoed when F8 •0
Gl:>ln: r .. Lsine= -i sin30· = 12 n
H - s0.2s lb
T =IZOto.s(.s!M!S) n • 1 re..Y,;-t ~ "'1-/t rps
a round
+he
v~
for
160/s
rl ::.o /
Tcos:a0- T1sinso• - W "' 0-©
~f11
o.sT
~. 1
)(= 1.Sft.
r " 1.s tlZ • .:a.s ft .
To"& =v~/9 ,.
tM :30. ..
40ubGtitutc to
"' 8·07 (60)
t
~·,,_ 2ov/12(a2)
=o.::i125V~
v - 6.079 f-1 /s
v
= ~trn
6.019 • [-.?ir (2)
n ..
from 1, T • (20 t o.s li)/o.g6t;
"/'32.11. (~.s)
o.s,, T - ~: 20
o.s(2s.1)
·0
Tslnso t rleo&:ac>'- 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_.
pa<X1 of ro\ofion n .1i" rpm to 1noline eao h corcl at :30• w1'th the
vortiool .
Given : 1.. • :af\.
v ~r
J~.2 ton30'
1130.) In Fl9· P-113",
lCOG.S7.67•
T = 187 lb
r = L.s1"& = 1.s (.sin &7. ,1•)
-f :30°w/
r =o.471ft .
g!.
1 • O _. By~uo d rotic.
0
ton.:3o' " v y 32.1(0:+71)
zirj ~tan~o·
1 • 2 11'
,9 L
-
ton-&:
Rcqo : vcioci ty
f>olh :
.G•n~~ " v c.ose-/..9 L
1-cos 2 e- - v "cos& .:: 0
0 cor"d form 0 copor1'cx:I ic; 1 sc;c . Dotcrmmc tho
if the cord rotates Inclined
:.o•
-& •
v/g Ls •~fr
2
C0611& t s.re~co&&
,3!I.'l (.1.~)
the
6ivori ·. t , 1 sec .
=
-to"~ "' v2/ qr
concontratod al the end of
n 1col pendulum for whicn
n],/Go
29 rprY'l.
2
o.s(20to.6tt) .. 0.86(; rl
0·8"
=
80
'I
I
n = 22 rprn.
i
I
184
185
Ii
:1
1t31.) /\ body of weigh\ W ros\s on the. smooth inclinod surfoco of the fromo shown in fl9. P-1131. A poq attached to
the frorno forces t ho body to rotate w/ ·.~ about thv VCl"tlcol axis . Ootcrrri1nc the speed in rpm ot which tho tension in the
cord
·,i; CGluol
to the
wc1ghl of
the body.
~~
r =GOS3o·(s)
~r""o;
r .. w
•
,.o·;
Ts;t'l:;o· t Hs in6o•-=- W
Tcas:30'- ti cos6o·
.
•
tt:ss.) Wh~t counter:--c'1ght W will mointoin tho Corla's$ en<Ji
governor 1n the poGihon show in Fig. P-mis at o ro•ional spood
n • 120 rpm·. Eooh plyboll wo·1ghs 16·1 lb. Neglect iho weight
the other links.
·
.SOl'n : s1nzio· • dAo - d "s"
r•.st1 ·~· ·-o.s;·
'J·2irrn•2v (o.1'~)
wv' .1,.1 ,.2&)~
v•<a.28 fl/s
or
V°
:S2.2 O•.S)
.3g.+e lb
.81o •O
"''!:!.£:.
©
gr
wsin30 t rt s 1n 60 cw ·
N = w- o.sw ~ o.577+ w
(6)AB.S1n1s•
A•LJAti
CEJ
v • s.91:z ff/s
V~~lfrn = 8.~rn! cH(-!5co.sao·)ri/(;()
n .. 19.79 rym ~ ~o rpm.
1132) The hammer of on impaot · tes;tinq machine weighs "..-.+lb.
/\s shown ,;, flq P-1132, 'it is attached to the oriel of o li~ht rod
,,.. A br-q whioh Is p~vo\od too honzontol oxis ot A . {o) What is tho
bcor-(ng roootion <», :Hio pivo~ on instat'lf aftor bolnq · rolcased fn::rn
the 91~d' p001\ion 1 (.~)What Is the beorinq rooc\ion just before impoot ot e ii thG vdooi~ of the haromcr is thon -s.g ft por6CG?
. ·'
6iv6n :
I
.
~ '
T cos~o
1. =4f1 .
T
Reqa : (o)Roocfon
,
le ..'
, ._.,,.
also
at (b}
.c
c ""'"""
radius of o roi lroad curve, tho oITocl of o ~o
lb wo19ht Is observed to be 20.7 on o .sprin9 svole .su.sporidGd from
the roof of on expenmenta l c.ar rounding the ~NC at· +omph .
'Htlot i.s the rod1us of the ourve i'
Given : v • 40 mpl'i - 66.6 ft/s
cos &- =2ofao. 7
1 = 20.1 lb : w ~ !20 lb
-t:r - 1-t.943°
RpqCI : rod;us
-'l'!f,. =- o i TG1n& "WtT'~r M.1s1n14>q"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 weighir1<3 :a2ro lb trovels a:, long tho rood at o oonstont t;pcod of 30 0 per soc. What iG the prc6SUrc on tho wheels of the cor- when it is at tho crest of the hill wh4 fl ? .Ai the what 6peod will the road eressul"O be xero?
lf'.nl : tho rodius of cur-voturo by t1' c(d'Y/cix 11y[ 1 t(d#J)t) 11]~
e>-c 'I
T = 81.8\b . == R .
r .. 400 . 5
co-s+s• (2) • W
W • G3.781 lb .
1137~ The segment of rood poss1n9 ovor the crest ·of o hi II iG definod
b~ T > w ~wv71r
- , ........ +[,41-.-t(.s.9)~]/322(4)
Tcos& :W
"1-S.1
~~ : M
7+.:%10 • R.
0
l\8GOS4'0• i/\BC0$415• •W
100 lb· lhe cranl<S AD \....BC aro of longht r.,. 18 in. 11.... rotoic at :aoo
.
Soln: a) Tcasaa •W
W~6•1-.+lb
1136.) Tho side rod
30.~s.smcoo"(10) - 16,1 .sinae~~
M • +s.1 lb.
·~
o.866
subsotitll\e to•.
'f/cosa-0· - 0,5774-.,f cos6o• = .)l(vY3~.~ (s) cosao
0
a
Given : r+rt. ; v • ao O/.s
.
We •..3220lb
X =20
" '10
±.. - 100
2x ~
Reqa: pref.Suro R
'l
IJ.._ Gpeod whon r-ood
y" • - 2/100
possurc ;,
ZOl"O
Goin: .of.
£--L
10 100
;l(«-+o)(
~ :
12Aoo
4Ao - 1.(1l1J)/,.A
;.>:tc.v
.....
I'. son.
R+ wv1l/,,.r :::.w
/':.I
R c aa20 -[aQ2.0(ao)~/a2'.2 (.so)
t+oo-o .·.(x-2o)(x-U>)·O
R., 14~0 lb .
·I
186
187
'if R• o
y;
W'_v 2
e
,gr
Given: Wp • 100 lb
-.. vtJ. gr ; 32.Q (so) "'1610
'J4
s
Y
0
Soth :
tar"& .. -v~r
v· 460 mph
40.1~ ff/G
A boy ronning o foot race rounds o f1at curvo of .soft.rod.
· fho ronG ol the ro.ic. of1s rnph, ot -NhDt onglc v4 the v~riicoJ :'ii~
1141.)
tone-
ReqGI : prossuro
fon<t
V •1Smph
Rcgl:i:
-e-
1142.) /\ darcaevil drives
=
o. 3006
o rnotOl"'Cyolc around
o 01rcular vorti-
What r~ the offoot of frovcli1"1<3 a1 o greafor spcccl?
Given -. c:l • 1ooft.
Solh:
f .. o.6
17- " tan-10.0 .. ::11·
z
2
Rc;qCJ: Ym'in
ton'& - q r/v
• 32.~(so) .. o.f.V
V"" 51 . 0 fi/s • 51.& [::1~00/s.280] c::is.a mph.
,-a-
ot eo rnph?
@ivcn : r:: "2000
n.
-tft a.s i'r) - .sG.s ,;,
Req'd : the .supcrolcvotion
e. for o t rocl<- ·
or
tontr
0.1102.
=
-e-,.
~.s,
0
". 60 mph
r -o.
9r
90
3872 lb
32.2(2.00)
to.nf • o~ g
RoqCI : friction Force
9..
R
~jtj3220) 2
4-Q .
+(3872) 2
-
..s03s.gs 1b .
.Sino/" F/iz. ~.S1n42°"' f/s~.95
f =3368.37 lb.
1f'+6.)
find fhc, on9lc of bonk:in9 fOr' a hf9hwoy curve ot :300 fl. ·
rodiu5 ~csignod to aa:omodotc cars troveling of 100 mph, if' H'O
rooff101cnt of friction betwoon the t ires ~ the r0ocl IG o.t;o.
Whot Is the rated Gpccd of tho curvo (
6 1vcri:
Gain :
r=~ft . µ_=0.60
ton¢=o.G
V = 100 mph.
¢ .. .30,9c;
Rcqli: roto9~pced
ton(1+6-)=v'l9r
fon (ao.96 Hr)
cfioo )( .s2eof3600J
2
3!Ml (300)
11+'1') An oirplano rnakc o iurn in o horiwntol plonc w'1thout
.Sides.l ip ot 4ao rnph . At what onqlo must the plane be bon~ed irthc
rad;uG of the turn is .1 rn't\c,? Jr~ho pilot weighs 1.00\b,whcrl presOl<Ort
:z
~ .. 3Q~O [ 60 x 88A;o] •
r = ~oon ;-e- • 30·
fan-& : eh
e/s6-'S
e = 6.79 in
he
Solh:
<01vcn: W 3220 lb
3<"B6 t ir "' ~S.82 ..
-& ... 34.86°
0. 1202 -
clOGG
= 462 ..97 lb
0
v=60mph.
sure
1soll + 438 2
when the car ilO froyo\lfng ot 60 mph. ihc c.oofll"oiont of frivt10n
botweon tno tire' ' t he rood iG o.<~o.
or
a roilrood trock is the number cl'
inoi"i<x ihot t ho atft~tdC roi I (1> ro1·Ged to provCf\t side ihroGt On tho
wheel flongCG of con; ro1.>nd1'n9 the curve ot rote speod · Dctcrrri1no
iho Gt.Jpor elo'<af10'n c, for a traok hov'1ng a gauge o{' 4 ft 8 Y~· ··ri ·Of
2000 fl. radius '*1 a rate speed of 60 mph . Whof is tho flan9c proGGUre p on the ~hoolG of o 100,000 lb. cor that rounds t~e curvo
~r
= 71. 07 °
A car ..,..,c1(Jhinq .3220 lb rounds o curvo of 200 f1 rQdlu.s bQni<od at on on9lc
ao·. find tho fr1'clion foroe ootinq on the t/r-es
114.s)
col woll 100 {\in diamoio"" Tho f bot. fire~~ wall ls o.<'10· Who~ i<>
the m'1n'1mum .£Opccd that will prcvonf plG Gliding down the wall?
.At whot onqlc 'tl·.it the motor cyole bo inollnod to the hori:i:ontoll
11""'3-) The GupOl"OICvofion
N ~~
_:1:l.'L = !l.q2(1so) .. 43" lb.
= ~.92
"fr
=~5l<.S280j3600]faM(so)
2
32.~ (s~ao)
r .. 1 mile
1
ho inotino his bod'( ?
Given: r~.sofl
·.sotn : tan& -v%r
·&aox,9280@6oo]
on his
-scat f'
188
I
.1
tone= Yr~r
fun.34-·U'c Vr~!l.2(300J
Yr Ii! ,,,. .;,2,2(300) +on 34-86.
Yr= gQ,024-H/s
Vr = a~.024' x 6%~ =; .s6 mph .
189
I
:1
11-47.) The roted spoed of ·o highwoy curvo of '200 f1 ro~ius is
30mph. If the coofflciont of fr;cf ion bct~con the tire~ ~ tho
rood :,G O,(,(), whot is tno maximum spocd at wh'ich a car co n
round tho curve wi ~hout .i;k.1.dc::ling?
@iven :
Solh :
r~ 200 n. /}- a0,6
torrfr . vo/gr
·-(:] c
1'). 7.3 0
ton 1 o.,
c
ao.g<;
11~) The coc(ffoicnt or friction bOI. thy rood ~ tho f'1 ""s. of the
CXJf' shown in f l9 P-11~ is Q.<;O . Tho oor w.01(jis 3'2{'0 lb . H Is rou ndlr'9 the CUr'VC of ~-ft rod.uJS at maximum speed- What is tho
voluo of tho fr1chofi foreo ooHng und~ coch whoo! ? how high
o~o tho rood must/ tho contet\ of qrovity be to limit +his; ma 'ltlinum ~~ by th~ tOndcncy to overfuf"n ?
+57Q9.48 '•fl30(~·'H7) t 5799,+8
(t){;.3C(a) -322ot»in30•(a)
row
Sotn:
- 'i3970.7 lb .
~~-o. . t11 (+.9333) =.~22oco~20(2.411) t a0w.1.s1n20(2."f11)
't
a010,1 a:is20(3) -a220~1n:z.0(3)
H1 .3024,G1 lb
F1 ~ O.G (38H,61) • 2294.B lb at outor wheels
r
~Me·o
rtr(4.&33.3) ~322oc.as~(~.411) tag10 .1s1n20 (1Z,4"1) t 3nos1ri2o(s)
- 3970,7 COG ~·(3)
H'l. c S59.87 lb
f~ ".S5~.e1(0.t;) - .335.02 lb 1nnc,.., whoo l s
~8 ~ o, rit.(-+.8333) =3200 ~30°(2.+11)
t~799 · 1'8 G1030(2A17) t 3'2'20 s 1ri ao· (5) ·
ion 4- o.G
l/J "'.30.96°
- 5799.+s cosao·cs)
ion(~ 1~) • v2./9r
H~ • n".4" lb
ton (ao.96" 1 ao') vo/32,2 (soc)
v - 110,129 fl/6
W'i" • 3'220 (110.,eq) <l. =.s 799, <113 lb
»
gr
f'+/, ..
t11 - 4963. lb
F1 •JAN1 • o.G(49G3) '" ~977.8 lb
at tho ouier -Nhoel~
p. "0 · 6
lk.qo : f rnohon "-hi~h
.
3e,2 (s~)
V • 84.11 ( 60/86) = ..S7. 3S mph .
w-3220 lb
• o.6
~" ~0.96
yf. c 19853--49
c
'
ton~
WV/gr • a2ro(HJ&58,19)
fan (ao.q~·+ 1,,73•) " vV32.2 (1200)
v 04.11 fl/s
r esooft
~h
~In'.
v%2 2 esOO)
0
ton ( 4t&) -= vo/gr
Given :
r=.sooft
tan(~-ttJ) ~v 2/9r
tao (30.9' t 20) =
#
~
orao
G1v0r1 : J_A "0·'
RcqCJ: FfnGf.;o.,
ton-&
RcqCI : rnoximum GpOXJ
Is bon~d ot !lo· 1nfoad
¥(. 3220
•[3ox.S~80/3600]J3!2,:2(200J
ton fr = 0.301
Vr· aomph
·,r t ho rood
11-49.) Ropoat Prob.114&
OG 'hown In Fig. P- 1M·&.
32,Q (soo)
l
Ii
'I
f:z. "fH:z. • ~' (12<i, 4,;) •435.96 lb
at the
outOf' .....-tioo IG
!i
2.41'''Vh
tM .gc.g6"
h .. 4,0278
ft .
I
.I
~l\"0
N, (4.6333) "afl20C"5acl(~.+11)
li
190
191
II
~I
flywheel 6fl. in diomefor occelorotes from rosl oi i he
constant rote or 4 rprn por sec. Compute the normal ~ tonqcntlol c.ompononts of ihe occelcrotion of o particle on ihc r im of
1hc flywhee l. oftor 10 sec.
61vefl : d " 6 fi .
Rcql:J : the norma l ~ tonqcnhol
1202.) /\.
0( • 4
rpm/sec .
tho oc.cc.lcrotion
components of
t ~ 1oscc .
Goin :
o< ~ 4 ~ :L
/l}l'ln~
o(
4
o(,,
x .1.!.CQQ. x
JP'f
1.P'lfn
60sec
o. ~ rodAcc~
urA
cur
o.4~(10)
uT "'4·2 rod/s
'·
Qn cref2
~
"' 3 (4.Q)
C h optcr 12
P.ototion
I
~­
I
'
i
c
.52.92
.
n/s
11
Ot =- ~ = :a(o.42)
Ot
= 1.26 fl/s 2 ·
1~03 .) Tho rim of c .so-it1. wheel on o broke.shoe testing machine
hos o .gpc.ccl of EIJ mph when the broke Is dropped . Tt comes fo
rest ofter the r im hos trovolcd a rincor distonco of 600 n. v.kol
f
ore the c.o:'\Gfont ongulor occ.elcratlon ~
tho whc.ol makes in comlnq to rest?
61vcn: d "..SO in
Roq~ : o< is...._'17
v =60mph
I
I
.s "600 fl.
.SOlh:
r = 25/i!l
2 .oe.s ft .
o<. ~
.S"'r'&
600 .. (2,083) (Ji}- = 26B rod
fr = :?S0 rod '1- 1 rc~n rod
-$- =- 4.5. 64 rev .
V =r ur.
(~ xf>?.>/Go)
w
2.oe5 lli.
'll( ~ 42.21' rod/s
192
number
~"lJJ/t2~&
-(42•24)t
=
the
193
e
~o( (~88)
- 3 .10
rod/sc.ct
or revolut i
\ '
12as.) Whon the angular vclooHy of o -+fl diomolor pulley
12 ·," 3 rod/s.
tho totol oc.colorofion of a p:i111i on it!O rim '1s 30 fl pcr-soc • Ootorrnine the O~lar ao::.olorat 1on
\he pulley OI th\G °instant '.
0iven ~ d = 4f\.
RcqCi : o<
or
ur=:arod/s
a=sonN1
Soln
1209·) The .stop pu lloys shown
cro&G~ tx:H . If t~ onqulor
who+
is
for A to trove! 1BO flfrorn 1"'<%f?
who~ chstancc will D rn<NC while t\ Is mavinq 240 fl~
6"1von ·. o<c • Qrad/s 2
Solh :
;
4
ti~c
·~c 180{l.
s. : ? "."hen s-.._
=
On • ru/ 2
-
c
ot'1+"1e
0<
SA ~
~ 12 rocJ/sz..
Octcrm1no tho hor1ionk1I ~ vertical components of' tho oca:ilorot ion of paint B on tho r im of' the f)ywhoel ~hown in F~- P-im>
Ai t ho givon po51tion,ur•+radpcr..soo. ~0<. =1!l rod pcr-.scc1:, both
o\oc1<.wiooG1:.0cn: uf· 4rod/G
o<." 1'2 rad/s 2
~rou
h
'3
~
L"-.3ft
Rcqa : hori-z:onto l ~
vcn•ool compononh:;· of
the accdcroHon
Ot"ro<.. "' 3 (12) ..
a"· rur
0
=
2
•
36
f*/s
2
3 ("1-)2 " 48 fl /G
~ at" t On 2
a= 6on/s
or "- 2 roo/s
. a ·1o fl/G 2
cliomcfer
2
Ot
•
J~!L
t 46
2
10
2
=
Cit 2
~3
Ori ~
t
"(.:ir) 2 t (4r} ~
100 = ~r 2 t1Gl
r
d
=
4(1~
=
-tn .
2
194
Q
2
2n
r
'/£ (z.) (7. 1.s) 2
-
60
rod .
t ~.
o
1n.
b.) f)A • S.../r " 24<>/(aoM)
-{)A~ 96 rad
't1A " 1'€ O(" t !J
- 2
t . 0.48 6CC
<><c
B-o • Y2~ct~ .. Y2(2)(0.48/
o<.. _ .. z. 67 rod/s,,_ .
SA ~
ro<.. • r (3)
a
a2
rodk:z.
1, SO<"' "2(2)
2
soc.
~ • 1/1(2.61)t1.
H1
12ce.) A pulley ha-s a consbnt onqular occc\ercrllOI'\ of ..3 rod pcrs~:.2.
When tho onqulor vcloci~y iG 2 rod por 6C.C, tho+ total ocoolcral1on
2
of o point on the rim of \he pullcy is 10-!1 pcrGC.0 - Com~~c +119
cl16rriotvl" of the pulloy .
Sol'n: On = r~ = r (1.)2
Given: o< • .3rod/s
~ 18 1n.
a~ ..11l0<,..
2
V1<.(4)J:'1
~ .. r&o ~ .3 (60) ~ 100 f'I .
so1n :
~
= 7,75
Y2 <;::l(c. t 2
120 =
if' the rodii of pulley a ore chonqcd
RcgCl: tA .~st> whcl"l S.,:2<10'
~
I
Y20( t 2 ,._
-ffo ..
~epc.of Prob. 1209
~-a
Qp =120rod
t
--&v :
t,. • 6 ' 71 &CC .
1210 ~
0
.,.180 • 2&A
s..._ =1oon .
~
I
't:t
go · Y2 (-1-)t...2
6ivcn :
.solh:
Y
re-.._
itA • gorad
1200.)
:
0
r.so::;~
n.
b) -t}A -
2
Ol·rcx.
R~(;I
!.140
o<,...' 2o!..c
°<A. 2(2) . 4roc1/5 2
2
Ot .. 24fi/s
'24 • a<><. -
4
4'~
by
!l (3)e..
.an"18fVs 2
o~ • 0t 2 t On 2
302
~u1rccl
Rcq'd : l,.. = :'
)
1n fl9. P- 120g ore c.onncctcd
occ.olcrotion of c i's Q rod
re-..._
=P- 160 &
~ =
2
-&..._ = Y.e <>< t,.,
7'2
6-o ~ 71.<Jl rod.
So • rfJ0
So • .3(n.91)
39'-12 -e,.,
rad ,
.
7rJ. ~ Y2 ( 2 .i;7)
1
t,..
t.... • 7. 34 sec.
.
of o pu lley is dcfl.f'lcd by Lh
1 ·
r c· r·o 1011ori
& " 2t +_
6' whc;ro-tf fs rYIOOGurod In radius 'l.t. t ·
...1_
"') •n sccon(.A.!>. COmputc
ctvoluhcs ot f onqulor· voloc:ify ' ongular accclerot 1on at the inston w en ~ 4GCC.
121.3.) The rotai 1on
.30 ~
-th i
t
0 '1 ven ·.
& ~
t
zl 4
Re<:ja : ongulor vcloofy
- 30t'116
Ol"Jgulor acc.oloroti'on
~ 4-GCC-
195
~
sol'n : d&/dl :
ur = 8(4)3 -60(4)
rod/.s
:212
uf " Sl3 -6ot
c1ur/ot : o<. "24l.~ - 60
o< w 24(4) 2 - i;;o "324 rod/s
.
121"!'.) Tho ,r ota tion
a \lrN°h col is governed by the cquo i ion iJf•
.[[ ;w is fn rod ;ans per Gccond ~ t i..l in seconds . -e ~ 2 rad whcr
4
t • 1 sec. G:imputc ihe values of -6' ~ 0( at the inslant when t ~ a sec.
or
6i...cn :
we -+-lr
Solh :
~ = 2 rod
Rcqo : -e~cx
di%t
,/ r
·'
o<.
= Bfa(~,. - o.6,7
.:3,
c ;,
i t "'/?t
c
;f or. + rod/-s
(0)
ur . 2
'
d&
-&
-, r -e ~ o
=
=
I)) •
fl. t
4
~
t · o :. c
oz
,, _ _
" - .
a S-t.
J2t
.!---'t
'
of
ur~ ~ · ;
o Ovwhee l decreases un i -
4'2
£
(6 )'~
-6" • HIO rad
c
4-('
t ·6
S(<;)- 6~ tC
2
)
-
t
1 ~(')
I
·l
ca 1~
23/3
f} :
=
t 4t
if i
1 -t(SJ
10. ,1 rod .
"2.GCC .
uf·uf. t ·D
uJ • et - t &/1
ul -
Hl
11
t 12
roo/s
!I
-o
196
rev .
dB'/dt "' Bt -{ 2/:2 t H!
2
-fi- = 4{ -t'/6 t 12t 1 C
it fT •O 1o.._ t • o .·. c • 0
8- · f j t·6
c
iT
23.~
A
i7
or • st - t'lR t c
·e- ~ t%
t 1 1+ C)'t
t%1 t .+t t c
2Tr~
or
duf'/c::lt "8-t
vo
t t (U;cc.
= e ract/.soc.
l t.H
-&e. -146.<n rad
-f7 • 1%.~ rad )(~
1n-v
•
Glope. ~ e-a - 1
•
<><-& c - 1(t -a)
a
f
c
60·'l-4<S
2
--f- - " --
t
t
~7
fo~ly, from 8 rod per scc in 6scc of wh ic h fonc ifs on9ulor vc
lov1ty 1.s 42 rod per.sec.' Compute the ini.fio l angular velocity ~
ih~o( number
revoluf1onb mode during lhe 6 sec inforvo l .
,('"'«
c ·1'
cl&/cft -
't
-9/!e i" - "f6. 3.9 l
-&t =6B/6(1) 2-efie(1f - -+s.SE(7) t60.443
The angular accclorotion
1211.)
t c.
I
2
'& .. f'
-1'.S. ~8
0
when
2
W ;:: t jl t "!'
\
=66/6 t
ufi • (;6/3t -·8/6i 1 - -t.s.33
= 1.164 7 rod / st.
4- -
"frt
G9/..s ("r) - 8/,(+)'-
4 )".:
2 ( 3 ) _ •It
ur
... c -o
C=60,443
6~ t - 8/, t 2. t c
11'2 · W11!.... t·f
\S/fl.
Q,(;{;1
t31t - o.GC.7
c
t •o
d fhjd t = 69fo t - 8/6 {2- 'tS.3-3
.3
•J-
(ont u>hcn t - 2 Goe.
Ga°ln : o<.- gt
6 1ven: 0<, - 2t
.dvi
/ dt - ~ ,. 2t
4rod/.r.
duT = 1t cit .
or" t7
::..ill.§.§.
W,· 3,.1zt2.
w;• f '~ % t
/\ body rot<('k;s occorclir:i9 to tho rolotion <><. -~2t , whoro 0(
is In rodians pc;' ,socOnd ~ t '1 ~ in .scconclG'. w= ~ rod per .soc ~
tJ is zero when {1G :zero . Q::mpvto tho val ~us of or' ~ & o\ the ins-
Rcqa :
;r-e, •o ,
- ett'8
o( •
5
-3/<;(1) 3 • 68/6 (4t - 8f!g (+) - "'5.33(4) tC
1216)
or-
~ O( ..
B;idl-' - +.s.33t i c
if -B-1 =t:T,. "" t ·+
-&- "' 1a.2 rod
• o<. "' 'h. (4) t - •It
o( .
-&, ·-~i
.3"( • 3 ' - St HI£
\
..
C• -
0/S
=
=:at:
1
-~ • 69/6 f;" -
2 " 4(1) 31t('l./8) t C
-e-
ir
•
0&/dt " W"4.[t
06' • + l w dt
31
TJ · +t. 9.(#3) tc
if fr• !l. roo u>hcn t • 1 scc.
uf" 4Jt
tho number of revoluhons throuqh w/c 0 pu l lc~ will rototo from rcsi
ifs angular occclcrof1 on is incrcoscd
urn formly ffl?rn :zero to 12 rad por sa/· durinq 4 sec ~ i hcn uniformly decreased lo <f rod per" sec/' dlJr inq Jhe ne:i1. i -3 sec .
f ~ •1{
slope,· + - 12 • -iv3
t °'. at
7· '1'
l·
e;if~..--; ---o< - 12 • - 6/s(t-1)
de/
/{0 t
- 2
1216.) 0olerminc
ur • el~ -6ot
I
197
ii
..•
I.
I"
Who! force P will g i-10 the sys\om of bodios .shown 1n fj'9 P0 vcloci'\y of 30 fl per soc a fl er rnovin9 20
rorn rosi f'
1406.)
t
nr
1406
t
~
'4fJ'
100
,,,,,,
>
y•,, 205
· a 301/2(20)
4
=2:z.sfl/s~
~ f~1
p = .agg,34 -
ll/az.2
Zf)l / T1 -100 (0.2) • 600(22.~
T1
•
so(o.2) • .so(22.sl
a2,2
69.SS lb.
Jc
...
1
£f.1S/"O
Ti ..
Choptcr 14
Worl'\ .~
a~u,6 t
0
2ocicas-.s (0.t.) t eoor.in.+.s
-[~(2t.oil/at.d
Tc - 399.3+ lb.
Ho1.) fuid fhc veloci ty of body A in ffq P- 1.+01 ofter ·,t hos moved
10 fl from r-est . N:.Gumc the pu lley~ to be weightless ~ H--ictionless.
Energy
1 i....2
2 Tt • 300 t 9,32 o,... - · - @
..... a,_3
2(~-12.412a.._) .. ~.g.320,...
-4a>- 24. MC:V- .. .3CO t g,32 o,...
l .
100 " ..:l-4 ·165CIA .
°" - s,..
2.<J3
H/G2
v,..~ !lQ
v,.. ~ ~---2(~.93)(10) .
,.
v,.. " 7.65 ft/s
:I
I
'. I
T2 • 200 - 6.21 Qa
T~ • W0 - 6 , L1 (20,..)
i
l.
Ti - .eoo - u ..;20,.. -· ·-@
!
198
199
11
:I
"
1408·) Through who! distance w·,11 body A
. chonginq its velocdy frorn 6
n par.soc
in fig.
or
1-+10.) In what d1'.. .fonco w·dt block. A
P-1400 m&1e in
lo 12 n per'6eG·
.,;.
City of 12
Ft'g. P - 104<3 attolfl
~t ?
fl por sec, storflnq from
1 "'-.ll
.!I ""-- 2
1ZT1
300 t
=
+·6' QA
_ ..
1ZT~ ·OD+ t 1fl.. 4~ CIA - · · - $
-@
4 &._3
4 ~1
£(200 - ~.210A) d 300t4.66QA
L { W4 - 18 .G<l-O,Y • -ao+ t 11'. 44' QA
400 - 12 .~IZQA = 300 t 4.G6 DA
400- 37.2.8 0A
a.-. .;, s.es ft/s 2
7
- YA1 - 2as,...
VA2
12
'.i
1 -
£,...
c
304- t12.42
°"
0..., •
IZ. Og
fl/s ~
·Yt...i " 2as...
12
z(s.es)s...,
= g.IZ.3
G
1()4 • 4g,7QA
100 - 17.• 00 OA
%
velo-
0
IL c
fl.
2 (fl,<XJ)
GA
£A" 3"\..41fi.
•;I
I
.:j1
Gubsfdule
4
s,... 3
Zo=~l5Lt6L
£(£o+-'0·6+o ... ) ~ so4 +12.+~ a"
4-00 - 37. 20('.M =::1D4t11Z.4{?0A
104
=~.7
°"'
o"' • .e.09 n/.svA' • !ZOG
y,., ·~ .~1Z(.t.oq)(1~)
VA
2
7.08
f'i/.s
Ve
to 1,
1400 , a .11 (0) 1 t
Ille • 304 t 12.~ O" - ·· : ~ +
:,'
;_•
4,6,
v,./'·
v,., . . 11.33 n/s.
II
"17
i
zr~Jost6" -8'
i!.~ ~ ei+xL
2~dt/ot = llxdx/dt
. J'
11
~Vt? • XVA
. Rw - ~ uJ~g (v 2-v,, L)
100 (11. -8) = 2oo;t+.4 v8 "'
l
t
aa:1/6 4, 4 v,.,Z
-·- -(!)
14<X> ~ 3: ~1Va~ t . 4..66VAa
.
1-41~.) Fr'nd th.c veloci ty o( body A, in ft'q . P-1411 oner I~ hos
m~v~, Gtarting from rest ai tho g iver) paGitlon, for 9 flalonq fhe
fr1cf 1 onl es~ surface shown
xz10~ ~ z:~
2/\d:11/at • u.ot/ot
2ot- 9,32 as
m £04-- g,~£ (La..,)
•
TL.
T~
-
·
~ ~~
Zf
=
)<VA• ZVs
2
15 10
2
~ '~ t 0 ~ -
-
·17
10
I
f.
i
I
I
i
J
I
11
200
201
l
';
800 = 1,5.SVA
2
t t.33Vs
2
- -@
subsiduto 1 fo f,
eioo • 1.sB (rA-Ve''/ a,) t £.33 Ve£
t1/s
1417.) I'.. W:cigh ~ iGdropped f ~mo p<:>&ition Jus;f obovo, but ~al- to!.1chlng, a $prlng . Show thc rno:><•mum dcf0rrnot1on produood wi II be
· twiec that if the samo wo"19ht IG gnxluo lly low~ up:m the ~prlrig,
Ve .. 1Lo.S<f
~1<.b1 • w'1
b, • 2W/K
1
W • l<.bt.
6~ - W/1<..
~lb•
•(tYJ/ty64'J • R
&
IZ
SOb" 1• ~v"' «.t 1.si> Ys t 0;11' Vis
= 1Zf>e
1+16.) A block wc"ighing Q6,6 lb is droPPCct from o height of 4ft .
upon o spring whose modulus is 100 lb per in . Whot voloci~wlll
tho block hove ot tho instant the spring is dcformod 4 In.?
61ven:
so1n :
W= 96.6 lb
f.
.soo • 1,ssVA" t ~. a3Ve~ - · ·-©
~ituto 1to1l
soo: u;s(a6YA)"~) +£.3:3Ve £
h ·-tft ..
Ve. c 1£.S fi/~-
K• 100 lb/iri
•
x1 t y<!
6a+in
I
•10~
,,
6t+Yt '.' 1o<r
RcqCJ ! Velocity
I
xVe • -7VA
VA ( ~ (<;Vs/~) t
-· .
-(j}
.z. u.1,49 (v'-v., ")
IZ
100 (10 -9) t .so(6) • 100/<A-·tvA• HaJ/h'!'.+Vsc: t'5o/'~AVc
Ye •Ve
1414.) Repeat Proh tt1a when i< c
t'·
10
1
f
~
. v • 1.s ..32 fl/s.
2yd"1at =o
6Ve = -SVA
RW c
"wv/:zq
.
7s an
~xclx/at t
w(htb)-12 K.b"
gM; (4t4A~) - Y2(100)(12)(4A2) • q6.6/6+.+V
1419.) A 600 lb bk~;k s li'dcs down on inc.line having o .slope of .3
vertic.al to<r horizonto/. It .s.1ort.s frorn rest~ oflcr moving 4fecL
~lril<o<.: a spring whose modulus i&100 lb po-n . If fhe cocf'fic1cnt .
of 1<.;neiic frlclion j1; o,w , find the rno;icimum vcloc"1fy of the blocK.
Given :
Gol'n:
~f\
~
... •
_,....,.-~"b
~
W=60olb· ·
en·
- ef. - y = 6H
Glope; .3/4
K • 100 lb/fl
/).l<., Q.Q
d " 4fl ·
Rcqo :
mo)( . velocity
)(2ty1.<101.
2i<cl:it/dt t tydy /dt =O
xVa : -yv"
8Ye = - 6VA
202
......_
Y2 K6.r = wvj'2q
. 600(4 tb) [3/_s - 4/s(o.t.J] - 100/.2 bL~(600/64.f}v f
[w(.a/s) - W(4/s)(o.ti] (dtli) 2G+o
t
1os6
-soob..
203
£
g.32v ..
d iffcrentiolo vokx:;•tY. with ~pc.ct.
~6+-1oob · -· g,3~
Neglc.cl fricl ion of hie 40-lb collar ogo·1ns l ii!> vertica l gu·1c1c
tho velocity of the collar offer 'ii ho" fallen j fl s tarting frol't'l rest in the position ~hown in Fig P- 1423 . 1hc unsfrotched
lcn9lh of l he i;pr;nq is 3f't .
L " ~ .3 2 t 3~ , -f,243 ft .
1-423.)
tob,
·'°" compute
6 .. 2.~+fl ((or mo><. velociiy)
G00(4u.6+) [ a/s- 4/5(9;2)] - so(e.6+)' " g,32v'
Y • 12-26
what
fl/s.
&•
or
muGt the 600 lb blocli:.
Prob. 1+1Q
s lido from re"t boforc louchin9 the spnnq ·,f Its -velocity i" 10 fVG
at the in~o nt the .sprinq i" doforYr'IOd 3 fl f Assume the spring c.ons-
I
tont jc; changed to .30 lb/in .
f·
1420.) lhrough
cliGfoncc
Given :
Solh:
w • 6001b
[600(3/s)
1ofl/,
6 • 3 fl
.Y·
Rc:qCJ :
d
=6.67 fl.
distonco
suspc;ndcd fron1 o verticol s pring (Fig P- ·
1-t!Z1.) whose modulus ic; K.. lb por fl . The weight iG pv llcd dD'Nn s f1.
frorn ifo c:quilit:riu"1 P,OG°it1011 ~ thon rc lcoGOd . Determine '1h; velo1421.) /\weight of W lb
'1G
I
gh'!
·,c 'rtG
of
"'
vt =¥
v -sJ "-o/w
ft .
332 .21 ft 'ls 2
142-4.)
fl/s .
Repeat Prob. '\423 if ihe unstrefched lon9ht. of \ho' sprin9 is 2ft.
0 1vcn :
Soln:
W·"IOlb
L~~3c1!3"-= 4 . 241-3ft.
Rc:qCI : Velocity
b, •
4 . 24.::i - 2 • 2.2....3
ft.
"'I0(:3) t Y2(.sx12)(2. ~3) 2 =(40/6'f-.4)y 2 ·
v • ~0. 88s fl/s
3 fl
-to(+)- 1/2(sx12)(3)'J. =("f<);i+.+)(v/ 02
=~HZ •
Yt'J ~ 259.1
Y1.. = 16. 1
·
20
.aest)
.
fi/s
142.s.) Tho ~r i~ Fig. P - 1+1s i& movlnq toward 1hc bumper .spr'1ng
1'4212.) Tho rigid hor"1x.on tal .shown in fig- P1+~22 is supp::irtcd .by 2
Gpring" . ,A. 300 lb weigh t iG the plocod upon tho bar ot 8 /\ .wclder1
blow prQ.fccts the weight toward/\ with an initia l velocity of M\jt;.
Whot
• .!!
·.Qn.
Y2 = 18. 23
I
,·
.s-.:9
Vz 2
'*K8" "'wvyrg
"'-
=(-10fa+.4)Y 2
"f()(4)- !/2(.sx12)(2)2 .. (40fi;+A)(v,ii- 1 ".~s")
city when if re\or,iis to the oquilibr1urn position .
w
6t ..
a'
(600(10)''_1/6..,..4
I\.· 30 lb/i'n
n.
v • 16 . 36.5 fl/s.
L • ~ 3• t"1'1
-ooo(~~)(o.z~ (d t3) {30(1~)(~)~/1
1~+ (d t3) • ~s.s1 .7
4.243 -3 ,. 1.243
40(3) t V2(sx12)(1.24~) 2
veloerty when it rcochcs t...? tleglccl fr iction Yt._, tho wei-
Inc bar .
"
V1 - 19.9
Yt - -t .~1 fi/s
204
~ h~ o 1<1nctic e nergy of 1~cm in-lb. The rn:::iin bumper shield
(?a) i_s c.onnccfod to the l'flClin ~pr1hg, which hos a modulus of1000
lb per 1n. lhc two oux1'liary bumper Ghlelds (b) orc 12 in . tehind 00 ~
ore atl'och~ to secondary springs, coch of which has o modu lus of
soon por 1n . \IVl'\o1 the CQr iG' b'r ought to rest, whal will hove been
'
fuc: grcat&ci mO'<'cmont aa ? What pcrconiago of the erorgy hos ~n .
obGOr:-bcd by the rnoin Gprlng ? .
·
Gr~on:
&>In :
KE = 100,DOO in- lb
l<a ~ 1000 lb/.n
1<-1> ~ .!'()() 1
b/ 1n
P.cqCt : .s I!.., /.
205
J[
I•
I
I
/2(1000)(12t6) 2
1
t
'/2(2)(~)6
.500 ( 12 £ t 2.46 t 6
2
1
e .. output /1·n put
=100, 000
0.9
t EOObt • 1001 000
)
hp(~) - 3.5. 7S
1ooob2 t 120006 - ~aooo ,, o
b'- t12b - 28 ·o
6
-12 t J12• - .ot(1)(:-2&)
~ (1)
0.9+ hp/3Sas
hp " 33.G1 hp .
~jn .
•
Powor Output = 33.61(0.1+61<..w) ~ 25.1 KW.
e~fcr5 o hydraulic react.on turb'1no w/ 0 vcloeify
1431.) Wofor
nper GCC·
G
1· • K~/Kt::
= gaooo;1oaooo .. ga Y,-
·
1429.) /\ troin wc.1gh1.nq 100 tons iG being pulled up to 2t. 9rado . lhc
troin res~fonCC '1G constant ot 10 lb per ton . 1hc Spead of fh e fra in
iG . inorroGod frorri ro fl p<]r .sec to 40 ft pc;r i;cc. In a distan co of 1axfl.
find the rnox1inurn ~rGC fOV'<Cr dc-<c lopcd ·
6rvon :
w-1oo ~onG
1a>o
ful'ri :
ZJ.. s
·
R• 1010Aon$
c
hp.
A
a
S ~ 6 i 12 = 2 +12 = 14 in .
2
KE~ Y21<.(12tb)
2
/<.Es " Y2 (1000)(12-t2) = 90000
s
hp /3C,J,73
=
-100(10)-100(2000)(.2;100~ x1000=100<'20002(402-.eJ}
[F
n.
6+.+
RCCJO : Po-Hen in +.Ip- ,, r f a, 726 . 71 lb .
. Hp .., fV {EVZ6.n ( 40l}/sso
Hp ~ ~::i4-<o7
' I
.
.
.
1+.30.) Wcrlor nowG . through 0 nozz le 1 if). in d1omotcr under 0 heocl
of' 4CO fl ~o dr'1"1'.: 0 ,/urb'1ne . The turbiM '1 G o/J/, . offi(:;'ienf ~ ji; conna:;tcd fo Cl generator w/c ·, ~ 941. emcien~. w1·ia1 '1s the power ovlput
G
in 1<-i lowoHs ?
G i'ven : d • 1 il"I,
.Rc'1Cl : Power G\Jtput
·
vf/zg •
J/.f ][<too]
4-00
y = [2(:32,l1)(400)
p
160.S
~
136 .136
Vt a tf!h;
e • 001.
h .. 3ff.
Req~
: Hp Oufput
Se in:
V..j49 " V1ftg
(12)~(su) • 4l
i
bt
l'f
µ (3it,:2) t 3 + H
ti ~ ~·9876 fl.
Hp1U-~& .. u)H •
tip :
e
=
3
100x10
3
498.7G 1<'10
(f.907'1) = +98.76 x103 n-lb
fNQ. )( 1 hp
sso f.+..-)£
=
725.S
E
906. B4 hp.
.
I
l
l
Pcutpul /Pinput
hp
I
I
lb.
·1
11
rJ /s
II
hp• fV • [l3 M3G(160.sTI/sso
hp = 3~;L 73
206
Q
V1 • fZ fl/5
Poutpu/
PONc" "' ~Qh • fv .
f'J/ ~ lS AYh
2
f ; "12.-tgv(1M)
lower w'd h
O. 8 -= Poutput / 906.64 hp
~oln:
of 12
or n
volocity
4
pci-~cc . If
100,000 lb of woicr flow through fhc turbine e,och .sc:;coricl, compute
ihc. h:>rscpowc,... output. A'-Sumc the ~urbino jc; 80 f. e.ffic1ent .
6 1vcn :
::irt
%ter()) ... 1ooe.:1 fb/s
wfa~ (v -v0 2 )
2
=
·,t
~ loavc.G
207
!.
1s'o.3.) /\ 300 lb block is 1·n confo ..... t w ith o level plane who
so coefficient of 1<.1nef 1C. friction is 0.10 . If the block is octcd
upon by o hori zontal . force of ..so lb, what time will c lopso before the block re.aches
veloc°ity of 40.a f1 . per sec, stor ting from roGt ? If the ..SO-lb force·,~ thon removed, how much
o
\onger wil l tho block conilnuo
0 ivon :
>
.Goin :
to move?
·
W • ~lb, ' f &.90 1t:>
;(.C • 0,1, V"' 48.3 fl/G
~Fi< =o
[so - 300(0.1~ t • 300/32.~ (48.3)
Rcqo: time
t • ~.s sec .
.300(0.1) t
300/a2.~ (40.3)
c
t : 1s sec .
A hor1".i:.ontol forco of 300 lb pu.shes o ~oO-lb block up
on incl ino whose s lopo is· 3 ver!i'cal to -hor·1z.ontol. If f ... "' o.2.,
1504.)
Chapter
lmpulGc
~
15
Momentum
determine the I imc required to incrcoso f ho velocity of t he
block frorn 10 to so
por sec . .
Given:
f>ol'n:
f
n
f".300lb~W-= ~OJ lb.
rn=%. i )A... " o'.2
v1 ~1ofl/G,Y~~soft/.s
Rc.q(:j: t -1mo
,
~
o/.s(~(4ls)
·
[300-1!20-160(02)] t ,, 2D'.lp~2 (so-10)
148t
t
c
240,4$
= 1° 68
soc.
1543.) 0 1
·root central ·1mpocf occurs bet. a 100 - lb body moving
or
to tho right of .s n per 6cx:i ~ 0 body
weight w moving to
tho loP ot .3 ft por .GOC· The cooff1·c·1cnt of rest ituilon e='o,5.
Aflcr 1inpact. the 100 lb body rebounds to t'1o loft ai 2 ft/soc.
Petcrrri1no the weight w of the ofhc.r. body .
G ivcn :
W,.= 100 1b ,
Sol'n :
e c o.s
V"= ~ n/s , v,; ~ 2(1/.s
Ws =W, Ya c.=ift/G
Rcq'd : Wc(ght
w,..v" tWeVa
c
WAv"' 1
wave
1oo(s) - .3W • 100(:-2) t wve'
e
700 "sW+Wve' -.·-GJ
= va' -v..:/-tA - V~
o..5" Va'H/.s-C-.a)
Ye= Z(I/~
208
209
·!
·i
''~l .
t
subsfi-luto Ve
'o
154-6.) The S!:-JS\ern shown in Fig. P - 1s4.s is used to defer' -
equation 1 ,,
mine tho c.o::ffic1onl of resii lut ion . If ball A iG refeo.s~ from
700 = 3W • W(IJ..)
rest ~ball B GWings through -&= 53.1• oflor ooing struck, determino & .
Soln: eos·6o .. h/10
C.0S Sl!l,1• = h/e
h -8ft.
ti'="'t.eff
b=10-.5 ..
C" 9-4.e = 3.2{1 .
v" =.[29h · ~1-2-(3-2.2.......)(,,....s
Ve' " J2(~:Z:2)(:3.Z) = 14.35 ff/s
= 17-9+ M
Wl'..V/\ =w"v,: t WaVfJ' .
·W"HOlb
f\ golf boll iG droppod . f rorn () height Of 2~. fl ,upoh
Q hordonod. .stcol plate; . 1ne cooffi·c ienf of rOGti\ufion IS 0.894;
find the ho1gh~ tC> which the boll rebounds on the ffrs~, ind,
v...._ third bounces.
>
1544.)
h • .20 ft , c= o.s94
Re.q'c:l: h, , h2 ~ha
e =~ h1fii
•
.5n.
(o.eg4l(h1) "h2
h~ - 1~.70 ft .
Given;
.soln !
0
(o.eg4)2(hz)
=
h2
30(17,g4) • 36V,...' t 20(14.35)
h:3 = 10.21 fl .
f %, •Jh3/h2
1
Vl'.,
0.094 = f1/~o _,.. h1 = 1.s.gs.fl .
.
.
The balls A'*' B in f (g. P-1545 ore oHochod to sh ff rods
of ncgligiblo - weight. Ball /\ 'i.G roloosccl from rci;t ~ a llo:"'od
1545.)
to Giril<-0 a. If tho cocffic1ant of rc'1.1tution le o.6 , dci~rrn1ne
the; onglo tT 1hrou9h w hich ba II B wi II .9W1ng· If tho
lo.Gts for 0,01 sex;,.: dlGO find the aveirago impact force
.
/
r~1' .
6 1ven :
! ·
'3<./ n ·
1mpoct
"
8.37
n/s
e = f/a'-yl'..' )/(:./A -va) • (14,35 - S.37)/(17.94-0)
e - o.333
1547.) A . ball is t hrown at on ongle e- w ith the normal ~o 0 sm?'t ~ wall, os shown in f ig- P- H.57 . It rebounds ot on onglo
orrcsfi tul'ion is
& wdh the normal. Show thot !he coclTi'c1'eni
oxpres.sc.d by o • tan 6-/tan-e-'.
·
·
Q,
IH
i
' ,i7 ! -e',,../
"(jY
,,,,,.,,,,,,,»;;,,,,,»,,,,
I.
.soln :
,D
c
C().&60 = h/10
vv..v,...
c = 4,61
h·0-4.61 ; 3,'.S.9 ft .
COG tr -=. 3.3.9/a • O· 42375
2ova' -' ·-(!)
o.G .(v& -y,..,'. )/@1. 0'f- O)
to,17 = Ye'-Vp..' - - ·
-@
-tr =6.S"
from 2, Y,...'• Ye/ -10.77
su ~titutc fo 1,
.s39,2. ~ e,o(ve'- 10.77) -t QOVe"
'> •.30,Q "
ao ve' - 3 .' Z3.112ova'
86P~
=..sova' -- vs'"'
11. 1226
e =(-)t'(s111-e-)(-cos-e-')
vs' · f29Z ; h· a-c
+ MVs'
©
cose-
from 1, Y/ ~ - V1.S111G-/s 1ne-'
I
17,2Q6 • ~~a2.2)c
~ w,..,vA' tWsYa'
.s:;e.2 • 3ov,..· f
,'
~ -v/~e-/y1
sin&'
h =.Sf\ · j b • 10-S = .S fl
YA - ~ = j2(a2.2)(s .. 11.g4f\/s
30(17,94) " 3oV,..'
'~ ...
1
mY1 cos& • mVi''cose-'
V1 sin &'· \I.I .Srn9- - · ·-
e
I
~ (vi'' -v,')/(v -w.)
?o'·'
=
tone cot -e'
V{QJ!.e
v " ton-&/tan&'
a
or .so
1548.) /\
is thrown w·dh
velocity
fl per .sec d1·r cctcd oi 60 w ith lhc horizontal agoinst o .smooth vcrhcal woll . ·
1hc ball i~ role.a.God from o -~·1 tion . 40A the woll ond 6 (lobovo
the level ground 'i frovcls in o verhcol plono. The coefficien t
rcstiluhon between ball ~ woll iGo.6. How for rrom /he; wolf
doeG the ball Glr'rkc fhc ground ?
Given:
Re<:jo: d1's~oncc frorn the wolf
Yo " so ft /s
when /he ba II .sf~i kc the ground .
R"' 40ff
or
fl/.s
;'
210
2ll
......
"
.ti t
..
'
1~.) As .shown in f(q. P-1549, o 4<Hb ball moving horizont a lly
to iho right with a velocity of e fl parsec, collldos obtlquoly w/ .
a 30-lb ball rnovinq up to the left at 30• to tho horiwntol ·ai 1l>
sol'n:
fl pcrsc.c-. If tho cooffo:::iont of rcGi°ifution i~ o.6 , oo~erm'1nc .the
a""unt \.,. d1'rochon <£. the Yeloc'1ty of each bp ll d1'rodly aflcr .
t1
-L
--.. o·---
· lmpoot.
34,07 = 11.1s1nfb·"
Fl~Yocose- l
16.1 !2 ~ 0 .~t
"!<> = sac.osoo t
t: 1,6SGG·
ycYof;1n&l - Y2Ql
1
y-
~.so
H~
t•-
1
2'.
07
-o
"
)
o< ~ ton-1 e,11./2s
,'
• 10.16
}:
ti
I
c
{
=
from rz, ~ -1otv1; sub.st. to©
t
.30·(~o•v1;)
'°'l • '40V1le t ~ t
-2'39,8" 70V1~
"1.,: ::-~..+a
ac>V1 ~
n;,
-fJ ~
Y4<' " 3,.+3 fi/~
to tho loft
I
'
•
I I
I
fan· 1·s,4,57
0
371 27.
~G)e~
16 .. t g.Q:z.
ft/s
~.. " Vit MN;' t '/1 M1Vi 11
e • Ve'-v.,'
v1 -v'J.
t.1,\4 t-Ma'ff • M.V.1' +MtV/
l<E1
~ .ee. 66
•
1
',iM,V1't'/tM2V1
1
1
.~
~
V30 •a.~~ fl/& up to right ot
37,27• w/ the ~zonfal
a. .2 nJs
tor1·1 s.2:1.s
f
Vty -v,; •s rt/~
"'..;id
& • fan-1 V•)' /,..,;
0
n/s .
y 62 ~ 11.1+
1
• nljV1y t msV2 y
~
26, 3 1sin18.16' .. Ve2.Y
Vs~
-©
w.'
· e ~ ve2 "/. /vs1~
•
..
&
L,,,,. . . . ~
y82 y
-
from a. , v.; 10 t (-3....3) - 6 •.s7 ft/.s
v,' = J~'-+v{yi.
1
Y1' •Y30 ·~'·51 +51 • 8, f6 nJs
.
Vf!ZY.- " 15
m,'°"1 t!ni.V2y
Y/ •
/<~:].,,·
O." = Y&!l'j./~6.31~10.1"
f
-&,; - v,x')/8 t 10GO!i30•
:.a6 (10&rn3(/)
,l
H/s
•
2 :r. . Ile 26.31
,
60.t • 40V1'Jt
c
V1a1o('ffr;
10 ::v1~ -v1 /
x. 10.3 n.
~, -(sosin6o).2 c.·~(32.'1)(2.a.01)
1
Vy e..e ft/s'
.
f v•
_
in.V1,. t ma '4 )I .. n'1j Y1x t IT1% Vix
'4<'(&) - 30(10 ~"311J • -4<1 Vix +3(JV;.,_
~0, 2 --wv,~ t 3tJV(,. · - - · -©
0.6
y,2 _ y0y 1. •.20yS '.,
"Y81· -~ \JQS
I £
~-~·
e ..(vt.; -v,.,.'Vv1,. -v•~
ScG ·
x -Vocos~t ·
• 1 ~, 1 Q)S £8,66 (1.222)
V:,..' - Yox ii • 2 a~s
It
v./ -(socos6o) .. o
Yx • 250/c
.,
J(8.£.); - +(161'1)(-34-.07)
t • 1.222
fl
6 t28.01 = 3+.010-
8.2_+
--a:.
.t 116·1 t z
2(16.1)
,-A(1 •6) - 1/.2
.sin.,.,.
, , (32-2)(1.6)
Q0.07
w,·10lt> ....-ao11>
.l-- -" 1 1
7 • Vosin~t t hgt
212
213
Chapter 2:
bor CABO supported os shown ·.n Fig . B is octed
1.) f\ r '1gid
·
J
?l=, . '
upon by two equal hor·1zon-\ol fore.es P applied a t c t.._D. G::ilCl)lo\e the reoctionG that will be induced ot the points of.support. /\ssume l · -+ft , a .. afl, b'" 2H
h><. Rb= -Ro • MSP
~·
I
PROBLEMS .
.·
)
.,
:.:-: r
t
~
2.)
\
Owing to the weight W of the locomotive
GhO"<n
in Fig· e:. t~
reootionG at ihe two points of support A 'i;.,.B w'dl eoch be eqool to W/2 . When the locomotive ·,s pulling o iro'1n !t.,, t~e
drowbor pl)ll P i<0 juGt equal · to the total friction ot the pt.s.
A St.. B, determine the magn.dudes of
vertical
·reootions Ro 1b.. Rb.
Ans. Ra " w - Pb J Rb= W t Pb
'T
~
T
2Gi""
+he
or contact
214
215
/\ boci\ i6 mm1ed uniformly oloog o conol by \V't'O horses pulling with forces P• 200 lb ~ ~-= 24<'.>lb octing under
on angle e1. • roo (Fig . A). Deierrnine th<9 rnogri1tude of the
resl)l\ont pul\' on \he boo\ ~-\he ongle f.> 'b...,"'< oGlelhown in
the figure .
/\no . P.-= .:3S2 lb ~ fd 33• ; -Y =· 2( .
.a.)
0
f'.
vedicol lood P is c;upportecl by o t r ior'\qulor tirock.et
os shown in Fig.G · Flncl the forcdS +rorismitled to the bolts
/\ Zt._B. f\ssume thot the bolt B f'Hs loosely in o vertical slot
in the plote .
/\ns. Ro"' 1.2s.P : Rb = o ...,s p
5 .)
<r
..q
For the particular position shown ·, n Fig .c the connecting rod BA of on engine e~ds o force P = !500 lb on the
cronk.p1n ot /\. R~t09lve this force iflto two rectonQvlor cornponen\s Ph """Pv17 ~9tiog horizontolly ~ ver\icol\y, o\ /\ .
/\ns . Ph
= "'tOS ID
P'V' "' 11T lb .
.
\
Chapier 3;
1.) /\ weightle66 bor /\B ·,c; 6Vpported ·,n o verticot plone bf
a h'1nge ot /\ ~ o tie bor DC, os shown. Oeterrri1ne, graphi-
colly . the 0"'1al forc.e S induced in the tie bar by the action
a vediool lood P applied ot 0 .
Ans. ~· " QP (tene1on)
9
217
216
of
T
of roclil)<i r ; 12 ·1n . ~weight ~ s ~oo lb ·,G to be pul, \eo C/'ler o cvrb of he"19ht h • 6 in. by o hor .1zontal force P op plied to the ~ncl of o strinq wound oround the circumference
of the roller . f 1hd the mognitucle of P ~1..i"1"'cl to Giort the-
i
2) /\ roller
roller
O'ler
the cvrb .
l\C &...
.
p\ted
the a><iol
r.
r
1 ~ .:::11
!Ir.... S« 1"nduc--' . th b
....... \(\ e OrG
in he F19vre oue to the acti~ of the hori :i:.on.tol o
ec .
t
l~oo at C . Th6 baro ore hinged together at C o. to ti...._:.
fovndotion at A "'-.. B .
""
•c
Ans. S1" 182 lb.(tentoion); S!t . 6'40lb,(comp.)
Ans. Pa 2-00 1b .
10001:>
~
,.
1,
,t,
.i!J.) !In electric -. light f1")(t1..1re of weight '1':1.
~ 40 lb .l'il SUpporied OS
&hewn· DeteMT11ne the tensile forces' c t.' c_ · ii...._ ·
. r- h .
. .
~ ""-"'-':! rn nic: wires BA ~
8C Ir terr ongles 0f1nclinofion Ore OG&hown .
eorry
1r
<if~il.i br-1 un'I .
Al60 f ind the
~ssure
/\ns. S1 " 2g.;s lb ; S2
R bdween
the bol\ 1*.. the plar,le.
,r
'I
'
3 .) A bOll of wc19ht w reGt upon 0 .smanh hor-izonfol plane oncl
hos aHochecl to ·
con tor two Gfring~ AB ~ AC .,...\iich poss oYer ft"iationleGG pulley~ ot B ~C !!:...
toads P &i-..\S!, t"e1>pecti--ely, oG Ghown .
the strong AB ·,c; hon%.Ol"\tol , f"ir'd ~heanqle o<.
thoi the string /\C,.~ rt;ia~ w ith tho. hor-i :t:ont o l when the ball i s
·,n a poGdlon of
4) Determine
Alis. cos« - P/6( ; R .. w -/~Lp'J.
218
'\".
219
" 2C>.7 lb .
(
Chopter
.+: .
1.) Determine the reoction o\ /\
I'
·
i
bo OEcl
!ii... the 1orce
S 1n 1he
r
ue
. to the oction of the loods P ft.,~ opp\ied to the crone. os shown.
Neglec\ the weight
the crone !..._ ossume ideal hinges ot A. D,
tt.,_ E . Assume tho\ P'>.!lOOlb, ~ • 300lb, o .. eft
.
. ..
I
I Ans. Ro = 1140 lb: s "1~5 lb comprCGSlOn
1-o ..,-f--0
--1
..
.
c
or
"
+.) A plone figure - four frorne ABO~ i~ ~upporfed on on inc li ned
plane ~ looded os shown . Colcu la~e the oxi~t force induced in
the member BO.
Ans . Sbd .. 106 .7 lb, tenG"ion
e
the oxiol force• ·,no bon; 1.2.3,4, ~5 of the pl~ne
trus~ s;;upporied ~ looded os GhCJwn.
Ans. S, • -P ; S2 "+P ; .
2.) Oelermine
0
Sa• -o.,P ; 5.t"'
o.....+2 P ; Se • - 0.33.3 P.
a) Two beoms AB 'l.t..,BC. joined toqeiher by o hinqe B, aro supf)O("ted by four bol"G, hinged at their ends . Determ.me the force produced ·,n eoch of the1:1e bore clue to the ochan of' t he load p,. 500 lb. The
dimension of' the structure are as shown. Ans. So= 186 lb, Comp : ;
Sd " Se : 2Gs lb. Compress'1on ~ Sc '° G2 lb • tens.ion .
p
3 .) Oeiermine the for-ce S ·1n the bor /\S of \he plane truGG looded 1..,, supported 06 shown .
An{; · S .. o.-+3 P.
Chapter S',
Ki the necessary coefficie nt of friction between ~i res &.,,
roadway to enable the four wheel drive ouiomob'1le to clirnb 0 30 'l.
grode?
A ni;. p ~ o.3
w
1.) Whot
220
221
s~
. ht"' W, t.. Wt. ore connected by
2 ) Two block. hav .ing weig
fi .o 1Gtnn9
·
r ~
. on hori:r.on\ol planes OS s how n . If lhe .ongle
rest
. of r1cr1on
lhe I ior-t
ch bloc!<. ·1~ f F1.nd the mog0-Yiude g..._ d1~tron o . . eas
ea
tied 'to the upper bloc!<- that will induce s liding .
force p opp
Ans. Pmin • (w1.tW2) s in'('.
r
'
".
I
, I
A solid right clrcula · cone of altitude h = 12 fr1 . &... radius d'
base 1 =3in . has i ts cen~erof grovity Con ·its geometric oidG at :lhe
di 6tance h/+ c 3 in . o~ t~ base . This cone rests on on incl ined ·
plane AB, which makes ·on 01191e of 30° with the hori zon ta I &_for
. which_th~ c.eefflc1en~ of' fr-idion is µ.=o.f,. A hori~ntol force P
is applied to the vertex 0
the cone &..obts in the 'vertical plane
of the f 1gu"' os shQl.vn . find the mo.ximurn 8tc:._ O') inimurn values of '
P consistent with equ'dibriurT) of the cone
the weight w =
10 lb .
Ans. Pma.>< ,,. 4.61 lb ; Prn1n" 0 ..590 lb.
or
·,r
3
l'
t
I
. Ior
) A .smooth cir-cu
CY
r nder of weight 1Q
&. rod.1us r is suppor•
the .some
werq. .
.r en
tre
1
rodiu~ ~
·
· ....1- ooch of
ted
by two oomlcirculor cylinUCl
ht ""'!2 OS Ghown . If the coomoent of' stoi1c fr-d1~ betwe
•· of the sem1c
. 1rculor
.
plane
cyl1.nders ~the hor1:wntol
.
t h on
fl at '\'faces
.
r
bet
cen
the
eyl1nderG
em
fi
which \hey rest is»-= 1/1 &.-. r-19 ion . w . .-.· t
b . between
•--t-...1
GelveG .1<:> neglOU
=- / chtermine the rnox1mum · <.111s ance b l ../ \
th
nter.s B ~c for which equilibrium w ·111 be p0(;SI e w, ou
e
ce
.
· t I Plone .
the middle cylinder,. touching
the honi:on
·
/
'
/\nG. bmox
2.03 I
°
Chapter 6 !
1.) In the case of' lhe tripod shown. there is no f'r ict ion between
the ends o f !he leqs &...., the floor on whic h they rest. To prevent
or
~
I'
s lipping
the legs their ends one connected by strings olon9
the lines AS, BC, '&._AC . Determine then the tensi le force S in ·
eoch ol these strings if each leg makes 30• with the
Pis o vertical load .
Ans . S = 1/g P
vertical~
p
I
jl
0
1/
Referring Lo the figure the coefT1.cients or fr1'dion ore OS follow: 0.25 a l. the floor. Q.3 ot the woll, ti\.,, 0 · 2 between b locks. f ind
u.._
. .
·~lue of'a horirontol force P opplied to the lower
rne m1n1mum v....
.
.. .
A
P. . ,. e .2 1b
. b lock. thot will hold the {;)'Sfem in equ1hbnurn .
ns. m1n· 1
.+.)
I
II
:l
2.) De ler rnine the forces pr oduced in the bors 1 to 6 , inclusive ,
ol lhe s poce truss shown , 0¥<";'19 to the action of four vertical
load c; P oppli'ed o.s shc:><vn . Ac eo ~ A'C' 0 ·o ' ore two squares
wi fh paralle l sides ol len gths a ~ 20 , res pec f ive l y, lY.. the
222
223
\,•·•
'
\
,·
~
distonoe between hori zontol plones ACBO
A'B 'c'o' is 2a.
f\ns. S -== S =- 0.2.5 P; S3 =s,.. • -1.06P; Ss=Sio "0.
1
p
2
p
111£---..10
.5.) ': pulley A of rodius o is Gupporf~d frorn \he foce of o
-.ert1col wall by two braces f\8&.._AC ioge-l~r w'tth a tie bar AO
a s shown. A fl e"ible cord
·1G fostef1(jld to the wolf at E,pas:
~AF
a'
~ver -the pulley, &.... can"'1es ot its end F a load ~· find the ·
tensile for.ve S produood 1·n ihe h e bor AO ·,r ..o • 1oolb a = 6 ·
b • .+ fl
= 'I: l'1
~
'
in.
, C · 7 :Z ' r ·
Ans. S = .!53 lb .
995
3) Find the tension 1 i.n each of \he guy wires BO
~BE of the
______ --=-::~··----· · ·-·-
/\n<o. T .== ..+.3etons , cach-
crof\e loaded o s shown .
e
p
f
..
Chapter 7:
1~
/\
ght
+)
A s trut AB o\\oched to the foce of o verhcol wollo\ A by
y
sph<'ricol hinge £\ands perpendicular \o the woll &... ·,s sup ported by two guy w ires of> shoV'l'll . At B . in o plane porollel to
\he wail , two fof>'".es P
os shown. -E< be'1ng hor1"z.on1ol bi,_P; ver-
1iool . Usin9
duced in the mernoors ·,f p " soo lb ~ \Q
the
oxiol forces pro-
""' . . 2.0 1
A~G. ".n ; Ye= Zc = 0:67 in.
ter of gravity .
o
~~
the rn~ihod of (l1()!Ylent~, flixl
homogeneous slender wire 12 in . long is bent in lwo "r ion9I~ as shown . Determine the coordinates of '1\G cen-
A
+"
0
:i
+·
...
e·
"
7
4-00 lb .
Ans . S1 = 103 lb ; S2 = 162.5 lb; G3"' - 1 , 600 lb.
225
224
2 .) A steel shofl of or-culor crosG section hoG a circulor steel
hub presGed onto it OS shown . for tho dimension6 .;;hown in tho
flguf'e; de\er mine the diG\anc.e 1-c f rorn \he \er\ end
\he short
~o the center or ·g rovity c of the composite body.
Ans. k;,r 6°2.8 in.
or
,.
! "·::
.5.) Deter~ine i~· coordinate Ye of the center of grovi ~y of
a steel r1vef having the dimensions shown in Fig. H . Assume
the head of the rive~ to be hemispherico l.
ti
16
/\ homogeneous body consists of o c'1rculor- ey\indricol porhon
rodius r oHoched too . hemispherical podion of rodius
r OS shQ'Nn . Oeterrnine \he height h of \he cy\inclricol pod ion ·,f'
HIO center of gravity ol \he composite body lies o t the center
3 .)
·I
Ans. h
A
i
·I
Ans. Ye• 1.1s in.
"'/
Ye
or
c of ~the
cir-culor ,yiore fore or the hemi sphero.
.
f
I
r/{2
T
41·'
I
Chapter 8:
j
1.) f 1'r.d
. J
4 .) A homogeneous body cos'u;ts
the polor moment of inertio of on isosceles t r iongle
having bose b ~ oltitude h w ith respei:.t to its ope)( A .
An~ . Ja
of o righ ~ c irculor coni col por-
=bh3/4 t hbj'46
or
. hon oHoched too hem·1spher'1ca l porhon
rodiu ~ r os shown . Determ ine the qltihx:le h
\he cone
the .~nter of
9rovdy of the, composite bocly coincides w ith the centerC ot
the arculor base
the cone :
Ans. h ,. % r.
or
·,r
or
.,
226
227
i
I
2) find the p0\or moment cf ined io of
&~ ~olculate. the moment d(' inertia oft~ 0 ~0 of the angle
section hoVlng the dimensions shown in fig . /\ with respect
1o o c;entroidal oici'°. porollel to the y. o.xiG
Y 11-/
t\nG. I.,. "' s.~ in~
ihe shaded oreo
sho-.vn ·,n figure v-i1\h respeot to poin~ 0 .
Ans . J 0 = 0.27.+ ,. ....
0
3 .) Ft'nd the polar moment of inertia of the oreo
or
a circ.ulor
sector of rodiuG r 'iJ.... centrol angle ex w'dh respect to ds
center .
Ans. Jo "o< r""/4 ·
.,.
.i
Chapter 10:
1.) A body starts . to move verticol ~upword under the influence of gravity with on Initial velocity Xo = 2ofps. Find
(o~ lhe m()'J(imum height to which ·.t w'ill riGC. ?.,_ (b) the time
requirro for ·,t to return to 'tis initial po,Mion . 1ok.e the slot'\.
flng pOint os the origin so thot Xo • o ~ neglect ~ir resist ·
Ans . (o~ /Cmo,. • 6.2 f1: (b) t • 1.24 sec.
.
;
or
-...) Calculate the moment
inert ia of the shaded /\rca in
fig. 0 w·1th r espect to the -x - a'X 1S.
Ans. I,. c :26.83 in.""'
.
228
229
' ,
.
. i:!ll .
}~,
>·
~n:»1 n is moving dow n o slope of Q.CX)B with o veloc'dy of
30 rnph . /\t o certoin ins1ont the engineer oppli"e s the brokes
'lit.._ proch..iccs o to to~ resistance to rnohon equal to one - tent h
2.) A
o( the weigM of the troin . Whoi distonce x w'1ll the troin
· trove I before stopping?
/\ns. 'J. = 327 ft . .
The troin tr-ovel i1> from stat ion A. to station B which ·,s
1 t<.m apart in o min'1mum time of one minule. If lhe t roin
stods frorn red at sta tion A , &... occclerotcs o t 2 ..5 m/sec~
continues ot consfont speed ~ decele r"Otes ot Q.5 m/s 2 unti I ·, t sfop a t sto1 ion 5, f1ncl the r:no)( ;mum · speed ·,n Km/hr-.
How long did ·, t trove l at . this top .speed .
4 .)
Ans . s ~ 60. 76 1<..m/hr. ;
t
=
7.6.39 S .
.I
.,
1
I
3.) The depth of the' crater of the tool Volcano was colclJloted
in the
following ~·o hner : From a helicopter flying vertically up-
won:J ol 6 m/s, ·a ,smoll bomb
'NOS
rcleosed ot the inston~ the
9r
.5.) ."The motion of o portio le \s 9iven by the eqlJofion s= 4t
-'3t 2 +s t + 6 where s ·,G in fl . k. t iG in sec . C:Ompute ihe
volues ot V ~ o when t • 1 sec.
helicopter was zorn. obove the crater surfoce. The oound
eitplosion wos ~ord 9 ~econds later. If thc•·speed of sound
·,s 335 m/s' what ·, ~ the depth
!he croter ?
I
or
Ans. v=11 O/s ;
Ans . depth" 240.2 m.
o =16 fl/s 2
3
l
!
.\'
I•
!
230
231
l!
11
i
· Chapter 11 :
1.) /\ motorcycle ?H.. rider of totol weight W = 50() lb travel
·,I"! 0 vedicol plane with cor.stont speed v = ~mph olo~q
the cif"'Culor curve /\B of rQdius r "' 100011 , os shown. f ind
the reaction R eicerted ori the motorcycle by the trocl' os
·,1 posses the crest C of the ourve.
/\ns. R "' 432 lb.
~) In figure ~low, o hammer of we'19ht W - 2 lb s 1orls from
rest ot /\ &... shdes down 0
for which the
r-r:· . 1
of fr . t . c.oen ICICl'lr
nc ion is µ. = 0.2. F.ind -the distonce x to fh
. t D
.t
c poin.
w here ·,t ht .s the ground .
Ans. :>< = 14-.4 ft.
roor
4 .) In whot proportion will the maximum range of o pf'Qjec tile be increased ·, f the initial velocity ·,G inoreosed by
10 per cent ?
The coefTlci.e0t of frlcfion ootwcc.n wet Olipholt po..,. ......-.
'it.., 1he tires of on outo~bile iG found to hove the .volue µ..
• o.w. At wl-Jt constant speed v con the automobile fro vel around o curve of' rodius r- eoofl . w/out skidding ·1 f
ihe rood ·,5 level ?
Ans. Ymo, = 49 mph .
Ans. 21 per cent .
. 2 .)
ihe pilot of on oirplone f1y1ng hon':zontol ~
ly with constant speed V"' ~ rnph ot on elevation h =20000
obove a level ploin wishes to bo;nb a forget B on the grotJnd . .At who~ ongle-e>- below the hori:c.ontol should he .see
.s) In. the figure,
the torget ot the instant of rel.easing
to score
1
1
,..
o
hit? Neglect oir resistonce .
v
9-=(~--- -
' -'"' .
;',,,_
',
B
/flT7777777ij/)7/777777/,IT:ll/; /)7
232
233
the
bomb in order
Ans.
-e- .. 22• 12 •
;I
'.!
3.) f\ slender but rigid semicircular wire Ot roi.Jius r iG
Gupported in its own vertical piano by o hinge ot 0 z.... o
smooth peg /\ as shown. It tho p09 Gtorts from O ~ rnoves
with c.onstont s peed 'lo· olonq the hor1.z.ontol '/.. 0')('1s·, find
the onglJlor vcloc"ity .& of tho w~rc ot tho '1nston ~ when fJ,, 60·.
/\ns. -6 = Yo/r .
Chapter 12:
lhc · ormoture o f on ·eloct r ic motor hos on ongulor
speed n"' 1800 rprn ol tho ·1nstont when the power is cut
ofT. (o~ If ·,I cornes to rest in 6 sec, colculote the ongulor
dcceler o tion o< assuming -\hot ·,t ·,s consiont . (n) How rnon'y complete rcvohAions docs the ormoture rnoke dur·1ng
2
tn1s ponod?
/\ns .(o) ex= 101T scc- ; ( b) go rev.
1.)
For tho figure s hown w,...,. 8 rocl/G ; o<,... "' -1.s rod/st ; t ,.
2socs .. find the veloe1\y ~ occclcrotion of blocK C .
4:)
Ans. Ve= 12s rnm/s ; Oc: 37.5 mm/s2
¢" 100mm
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2.) Considering the system in the f 1gur-c , dct~rm·1nc tho
voluo of fJ- for which the negative onqulor occoloro\ iol\
-& of the bor OA hos 'its rnox imurn voluc.
A ns. fT., 30•
1-...e
~I
-P--t
0
.!!.) For
~he
234
I
the figurn shown , o<. =
60
rod/s2.
~
Ans.
blocK C is rising .
235
S'" 6 m · How fosl
Ve
c
14.7 m/s.
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3.) When o boll
wo'1 ght W roGt On 0 spring of constant K.
producos o static doncction of 1 in. How mvch will the
some boll cornpross the .Gprin9 ·,f ·,+ i~ droppc.d fromo height
·,t
h,. 1 ft? Hcg\oct tho mac;;G of tho spring
Ans.
<S:: 6 in.
ll
Chapter 1.+ : 'i
1.) I\ bloc!<- of 'f"'Cight W ·,s in given on inif1ol vel<;x:ity Vo
010119 o rovgh ~rizontol p!ono ~ is brought to rost by friction in o distonco x . Octor mine the coefficient of f rictiori,
oGcouminq thal -.t ·,s independent of vcloe1\y.
2 /.
Ans. p. •Yo/ 29x
11
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+.)I\ smoll b lock. of weight w=10 1b is given on 'i n itial veloc ity Yo•10 fps down tho inclin~ plonc shown in figure . If
the cocrfici"cnt of frichon botwccn the pl one ~ .tho blocK·
·,s µ. == o.a, f ind tho velocity v of -the block. al B oflcr ·,+ hos
\reveled o distoncod ~=.soft .
Aru;. v=29.Sfps .
2.) Determine
the dynamica l deflection
o that ·will
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bo pro-
duce<::l ot the center of o simply supportod boom by allowing o '4;000 lb woi9ht to drop onto ·,t from o hoight of'
4 in . Whon gradually opplied, the same load produces o
static deflection of 0·1 in . NC.(Jlect the mosoG
tho boorri .
Ans. c5 • 1.00 in .
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or
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237
236
.~r
If tho syst~rn in figurv is re1eoscd from roc;t '1n tho eonflgurotion shown · by solid lineG, fi'nd tho rno1<imum d iGtonce
!I.)
h -\hot tho weight P will foll . Neglect frichon "*..assume
·thot the pulleys A '*-. B aro very smo ll .
Ans. h • 4P~t/(4~~-p2) .
3) A mon wc.1ghing 1so lb runs~ jumpi; from o p'1cr into
boo~ wi~h
o
o horiz.ontol velocit y Y1-= 10 fps. A~uming .tha t the
impoct i'7 en t ire ly ploshc, fi'nd ~he velocit y w i th whic h. the
man ~ boot will move owoy from the pier ·, f the boot weigh
200 lb.
Ans . v. "4.3 fps .
,.:
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Chapter 1.5:
' ,
A locomotive weighing 60 tons hos o volocity of' 10 mph~
txlcks into o froight o:J.r weighing 10 tons thot•1s ot rest on o
level trocK · Aftif' c.ouphng is rnodc, with wbot voloe1ty v
will the enfiro .s~,stem c.ontinue to move? Neglect oll friction.
Ans· .v " 8.57 mph.
. 1.)
4) A golf boll dropped from rest onlo o corncn t Giclewolk
rebour:ids eight-tonths of the hciqht through which 'it foll .
Heglcding o'1r ros istoncc, determine the cocfT1c icnt of .
Ans. e "'o.9.
rc:Gtit ution .
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2.) A ""'ood block wc'1ghing. .5 1b ros;ts on o ..smooth horizon -
tol surfocc . /\ revolver bullet weighing 1h oz i~ .Shot hor·, :z:.on\olly into tho s ide of the block . If the b.lock. oltoins a
velooity of 10 f p~ whot WOG the muzzle vcloc'1ty V
the
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or
Ir
11
bullot?
Ans . v
s1010
fps.
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239
238
ll
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s.) In +he figure shown, a small oar of we igh~ w starts
from ~t 6t A "'·rollG w ithout frict ion ' olorig on inclined
plonc to B where ·,t ic; ..str'i k.os.· o b lock. ol~o of' wo'1ght W
~ in itial ly a t reGf. /\ssvming o p lost ic irnpoor ~. B , tho
·car '/,t,... block w ill movc. frorn B to C o,G one podiclc. If the .
. cooff1cicnt
fric tion between th~ .b locK. ~piano :1c; ,µ.. ...
1/tZ. , colcl) lo te the cliGfo nce 'X to point£ where tho bodicG
c:Omo to rost .
· An.s. x"" 14.2 ft .
or
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240
1.