CH 15 - icivil-Hu

PROBLEM
15.1
The motion ofa cam is definedby the relation 0=4t3 - l2t2 +15,
where d is expressedin radiansand / in seconds.Determinethe angular
coordinate,the angularvelocity, and the angularaccelerationof the cam
when(a) t = 0, (b) / = 6 s.
SOLUTION
o=(4t-l2l+15)radians
.=49=hzt2 -2+t\radts
d
t
\
/
o=d'-(24t-24\rad/s1
dt
(a)t=0,
(b) t = 6s,
d = 0+0+15
d = 15.00radians{
rr=0+0
at= 01
a =0-24
a -- 24.0 rad/s21
d = (4)(6)3- (rz)(o)'+1,5= 447
a = (tz)(e)'
(24)(6)= 288
a - (24)(6) 24 = t2o
0 = 44'l rad I
a = 288 rad/s1
a - 120.0rad/s21
PROBLEM
15.12
In Prob. 15.11,determinethe velocity and accelerationof point B at the
instant shown, assumingthat the angular velocity is 3.38 rad/s and
decreases
at the rateof 5.07 rad,/s'
SOLUTION
rro = (toomm)i+ (624mn)j + (z+omm)r - (0.1n)i + (0.624
m)j + (o.z+rn)r<
r r , , = ( r 0 0m m ) i + ( 3 1 2f f n ) j = ( 0 . 1 m ) i + ( 0 . 3 1 2m ) j
1 o -, f i o l t ' * ( 0 6 2 0 ) ' - ( o - r 4=] '0 . 6 7 6 m
Angular velocity.
=
^=
+0.624i+
o.z4k)
t,,o ffito.u
o = (0.5rad/s)i+ (3.12rad/s)j+ (r.z raors)r
Velocity ofpoint B.
VB=6JxrB/o
i
j
kl
I
4 i0 . 1 2-j 0 . 1 5 6 k
0.5 3.12 t.2l- -0.374+
0.10.3120 I
vu - -(o.tttnts)i + (0.1200r/s)j (0.1560r/s)k<
AngularAcceleration.
o - !r*o=
""
Ioo
loZlo.ri
+ 0.624j
+ 0.24k)
'
0.676
- (+.oaraorsr)j- (r.sraalsr)x
a = -(0.7srad/s2)r
15.12CONTINUED
PROBLEM
Acceleration of poinl B.
Z R = c , x f R t O + ( t Jx V B
a,
I t
=
]
j
k l
0.75 -4.68 -1.8
|
I 0.1 o.3r2 o I
i
+
j
k
0.5 3.12t,
l
I
0.37440.12 -0.t56J
- 0.37128j
+ 1.22813k
= 0.5616i- 0.18j+ 0.234k- 0.63072i
"v")r.<
"v.')l* (r.+oz
ov")i- (o.ssr
", = -(o.oolr
PROBLEM
15.17
The belt shownmoves over two pulleys without slipping.At the instant
shownthe pulleys are rotatingclockwiseand the speedof point ,Bon the
belt is 4 nr/s,increasingat the rate of 32 nls2. Determine,it this instant,
(a) the angul velocity and angularaccelerationof eachpulley, (b) the
accelerationofpoint P on pulley C.
SOLUTION
Let vB and a, be the belt speedandacceleration.
Theseare given as ya = 4 m/s and aR - 32 n/s2
Theseare alsothe speedandtangentialaccelerationofperiphery ofeach pulley providedno slippingoccurs.
(a) Angular velocity and angular acceleratian of eachpulley.
Pulley A.
? - 1 6 0m n = 0 . 1 6 0m
v,
Q'') l = " ' L rA
ctt
''. =!-LrA
v- !-ru
4
0.160
- 25rad/s
a,r=25.orad/s)a
au - 32
= 2oorad/s2
ro
0.160
Pulley C.
a,t=2oorad/s2)a
rc = 100mm = 0.100m
v - v ^ 4
--!_
(t).
=--.L= = 40rad/s
' =
rc
r.
0.100
arc = 40.0 rad/s) {
32
= 32orad/s2
a '. = 9 ! : - a u rc
r.
0.100
ac = 32orad/s2) 1
(b) Accelerationofpoint P onpulley C.
Pc = 100mm = 0.100m
(or), = ou = 32 rn/s2[
"u = -F)' =l6omrsi.
lo,l - L Pc
Pc
0.100
o,)l + (o,)' = $t
t^p=#
. 160'= 163.2
ntsz
p=n.31"
ap = 763.2rrls2 7
ll.3l" <
15.28
PROBLEM
The systemshown is held at rest by the brake-and-drumsystemshown.
After the brake is partially releasedat t = 0, it is observedthat the
cylinder moves 5 m in 4.5 s. Assuming uniformly acceleratedmotion,
determine(a) the angular accelerationof the drum, @) the angular
velocity ofthe drum at / = 3.5 s.
,t
SOLUTION
, u -- Lort'
motion.
(a) Assumeuniformlyaccelerated
ou=1=
For the drum,
?P
=0.4e383m/s'
i
er = aA = 0.49383rn/s2, r = 250mm = 0.25m
at-ra,
a,
"-;=
0.49383
O25
= |.97531radls2
,
(b)
a = 1.975radls2
) I
a = ao + at -- 0 +(1.97531)(3.5)
= 6.91rad/s,
ro = o.9l radls) {
PROBLEM
15.39
Collar I movesup with a velocity of 3.6 ft/s. At the instantshownwhen
I = 25o, determine(a) the angularvelocity ofrodlE, (b) the velocity of
collar,B.
SOLUTION
q:l
/
v
qrV
-
o
i/T.
A",t
/
Fl"
AV
+
A
\),cJ
vB=vA+vRtA
Draw velocity vectordiagram.
q = 180" 60'- 65' = 55'
Law of sines.
lBr,t
sin60'
ig,t
-
tB
_
sin65o
v.,sin60"
-
sln p
-
,.1
smp
3.6sin60'
sin55'
= 3.806ft/s = 45.67in.ls
o tD =
(a)
vs s
45.6j
fn,,t
20
2.26 tad/s
a,ra-2.28ndls)a
(r)
vB
v , s i n6 5 o
sin55"
3 . 6 s i n6 5 '
- J.9EIVs
s i n5 5 '
vr - 3.98tusI
30"<
15.50
PROBLEM
The intermediategear B rotates with an angular velocity of 20 rad/s
clockwise.Knowing that the outergear C is stationary,determine(a) the
angularvelocity of the inner geafiA, (b) the angularvelocity of the arm
AB.
SOLUTION
Label the contactpoint betweengearsA andB as 1, the centerof
geu B as2, andthe contactpoint betweengearsB and C as -1.
GearA:
Arrr,AB:
Gear.B:
v, = 3at,I
(l)
v2 - 4.5auul
(2)
I
(3)
vt = v2
t.JoB I
(4)
v,-vr+3oul
GearC:
Data:
From(5),
I
vt = t.)oc I
(5)
atu = 20 rad15,6" = Q
Y:=0'
From (4),
vz= -3a,t= -(l)(zo)J= eoin./sl
From(3),
= -lol = m in.rsf
vr= -60- (l.5x2o)
From(l),
at = vt /3 = -39 rad/s
)
ro; - 30'0 rad/s) {
From (2),
eol=+.sauu! ,^, = -#=
-13.33
rad/s
a,ts = 13.33rad/s) 1
PROBLEM15.65
In the position shown, bar lB has an angular velocity of 4 rad/s
clockwise.Determinethe angularvelocity ofbars BD andDE.
r_r0rr.___f6ir.
Z .l in.
SOLUTION
Bar,48. Rotationaboutl.
ya = a,tn t ru,, = (-4k) , (10i) = -40i
In units ofin./s
.BarED. RotationaboutE.
Yn = @oexro,u = l.norkx (-6i + 2.4i) = -Z.4ronri - 6and
Translationwith B + Rotationabout8.
Bar BD.
Yorn = aao x rr," = a4ok x (-+i) = +atoi
vD=vB+vDtB
-2.4aori - 6rrti
- -40i' kttyj
Components:
!:
-6ato, = -49
i:
-2.4atou = 4auo
.,, -EAt@=
aot = 6.6667radls
-4.oorad/s
aps - 6.67 rad/s) {
asn = 4.00 ra&s) 1
PROBLEM
15.75
A 60-mm-radiusdrum is rigidly attachedto a l0O-mm-radiusdrum as
shown.One of the drumsrolls without sliding on the surfaceshown,and
a cord is wound aroundthe otherdrum. Knowing that endE ofthe cord is
pulled to the left with a velocity of 120mm/s, determine(a) the angular
velocity of the drums,(b) the velocity of the centerof the drums,(c) the
lengthof cord wound or unwoundper second.
SOLUTION
Sincethe drumrolls without slidine.its instantaneous
centerlies at D.
Ye=Ya=l20mm/s.v/ - vi!/Da,
@=
v -R
= _
rBrD
yB - rRtDa
120
100- 60
- Jra(vs
ro = 3.00rad/s){
v, = (roo)(:)= 3oomm/s
v; = 300 mnr/s.*
{
Since v, is greaterthan vu, cord is beingwound.
vA - YB = 300 - 120 = 180mrn/s
(c)
Cord woundper second= 180.0mm {
15.82
PROBLEM
Knowing that at the instant shown the argular velocity of rod DE is
2.4 rad/sclockwise,determine(a) the velocity of collarl' (6) the velocity
of point B.
SOLUTION
aor = 2.4 tad/s)
Rod DE. (Rotat\on about E)
vo = (ro)aou = G)Q.4)= 14.4in.^1
Collar A. (Ptectilinearmotion)
Locate the instantaneouscenter (point O of bar IDB by noting that
velocity directionsat pointsI andD areknown. Draw lC perpendicular
to v, andDC perpendicularto vr.
1 4 . 4i n . i s
r^
= 5 ' o r a.u. s/\l
a 4 D B= Z ; =
4*
(a\
vu = (CA)auo, = (?.5X3.6)= 27 in.ts
v ,r = 27'0 in./s- {
4
talrr'B = --:7.5
(r)
B F = 4 + 3 I a n p = 5 . 6\ n .
CF =3in.
t a n (' D = -
CB=
= -
BF
5.6
cr\' + (nr)'
Q = 26.2-
3 2+ 5 . 6 2
= 6.353in.
vB
= 22.9in.ts
= (cn)ano, = (6.353X3.6)
90' - p = 61.3'
v n = 22.9in./s\61.8" {
15.101
PROBLEM
UsinsthemethodofSec. 15.7,solveProb.15.65.
SOLUTION
v" = (n\a = (tox4)= 4o in./s
Ya = vr I
tanP
' =Z!,
,le
6 '
P =2r'8or"
vo = vo I f
center(point Q, noting that velocity directions
Locatethe instantaneous
known.
Draw BC perpendicularto v s andDC'
points
B
and
D
are
at
perpendicularto vr.
!- =
tO in.,
"3 -- anq
CD =
CB =
in.
- 10.7703
cosP
,ro=&=9=+.uor.),
vo = (co)au,
DE=
u^. =
aao = 4-00 radls ) a
= (10.7703)(4)= 43.081in.is
6
cosB
= 6.4622in.
r-z - 43'081=
6.67rad/s)
DE
6.4622
,','
= 6.67rad/s) {
PROBLEM
15.110
The motion of the 3-in.-radiuscylinder is controlledby the cord shown.
Knowing that end t of the cord has a velocity of 12 in./s and an
acceleration of 19.2 in.is2, both directed upward, determine the
acceleration(a) of pointA, (b) ofpoint 8.
SOLUTION
Velocityanalysis.
Pointl is theinstantaneous
centerofrotationofthe cylinder. vc = vE = 12 in.ls.
v c = 2 r a, = + = d b = 2 r a d l s )
ra' = (Z)(z)' = 12 in./s2
Accelerationanalysis.
a n = f a nI- " . = [ ( , . )l J, . l @ , ) , *] = t * f l . [ ( " . ) -, ]
* J
uc=N,+l(*,),I I * [{,.,,,),
- f =[,u*)+lz,a l1+12,.,
*l
[",1].[(".),
- l * [ o ,t ] + [ z *+ ]
[ r . zf ] + l ( " " ) , * f = l o u
From (l),
19.2= 6a,
Components+l:
ot
a = 3.2 rad/s2)
aG = aA + ("o,u), * (ro,n),
[,"1]=l"u-l+[.a f ] =1,,'-l
t,. I I =[o,* ]* [r.oin.r',t ]* [rzi'.r" - ]
ac = 9.6 in.ls2
From which
and
a,t = 12 in.ls2
t,c = 12.00in.7r: -
(a)
(b)
ac = 9.6 in./s2 1
{
aa = ac + (roo), * (roo),
=[s.o
1]+l,a-l+l,of l] =ln.u
1l+[r.o- ]+[rzf ]
=
[r.o
in.rs'-
]
*
[z.ai".lrt ]]
as = e.e0in./s2
<
Z 14.0.
15.118
PROBLEM
,tn,-1
y-'
r.,
--?tt
.A-t.;l
f
\ a - "- /
f t
i0 'r.,r'J
/ , , e
At
Arm l8 has a constantangularvelocity of 16 rad,/scounterclockwise.
the instant when d = 90., determinethe acceleration(a) of collar D,
(6) of the midpoint G of barBD.
r2rr""
*-
SOLUTION
0 = 90"
Geometry and velocity analysis.
p-r7.4s8"
, i n' p = $ = 0 . : ,
200
centerlies at oo.
vo andv, areparallel,thusthe instantaneous
6o
aao=0
anu - 16 radls,
dn = 0,
Acceleration analysis.
aan = 0
a u - f t l a o ,t ] * [ o o , j uI I - o * [ { o o t t r o, )]'= r s : o o . ' , v,, '
PointD moveson a straightline.
2o=aD-
(o,u\, = l6oa
Bau, I I
"u - ) + lzoocos
("o,r), = luorfi, I I +lzoocos
Ba'uo- ] = o
aD = LB +(o,u),*(oo'u),
Resolveinto components'
. f : 0 - - 1 5 3 6+0 2 l O c o s p a s p".r " (")
l*:
ffi
= S 0 . 5r0a8d / s) 2
= -4331rnrn/s'
ao - 0 - 60auo+0 = -(60)(80.503)
ao = 4.83rn/s2.-
<
* ]+[zoso| ]
= lzoaro-]+[t00cos/ar, | 1= 1z+rs
(uo,,),
.- =o
- Ito;", I f +froo",s
(uo,,),
B,,fio
]
(r)
^o = ^" * (^ r,o),* (u'o ),
- [ r s : o|o] + [ z + r .s- ] ' [ r o s 'o] * o
.- *
=
] fzoto-"v.' J]
[z+ts-ttvr'
'72.5"
ac = 8.05 ntts2;Y
a
PROBLEM
15.128
Knowing that at the instant shown rod lB has a constantangular velocity
of 6 rad/s clockwise,determine(a) the angularaccelerationof member
BDE, (b) the acceleration
ofpoint E.
soLuTtoN
Velocity analysis.
a,ts = 6 rad/s )
vu = (ta)anu =(4.5)(6) = 27 in./s
vR=vR+,
vD=vD+
The instantaneouscenter of bar BDE lies at co.
Then,
=0
an
and
.
uJ.^ = -
vr,
=-
Zj
CD
Acceleration analysis.
vD = vB = 27 in-/s
,,
= J. raors
\
,
a,ca = 0
=
zu= Qa)afou
I ] = rozin.r.,1
[(+.s)(o)'
ao =f(co)a,o* I * llco)d, lf =1r","- I +
[(r)(r),J]
=fsagp
- l+ fsri".r.'J]
aru =l4.5aro-)+flt.zsauo l)
* ln.zsrlo-l+f+sr,fi,l)
= f+.saue
- ]+[rr.zsar,
f]
aD = ^B + aD/B
Resolve
intocomponents.
+J: -81 = -162+ ll.25auo,
aso = 7.20radJs2
) 1
=feauo
*l+fzz.sauo
a,u,u
*l.LnSr, l)
l]+lzz.s"fi,
= [(r)(z.z)
-l+lQz.s)(t.2)
1]+[0- ]. [ol]
=
in.ls,[oe.a
]. [rozin./s,f]
^E= uB+ ^u,u=lrazin.rs,
-].[rezm.rc, f
Jl+[o+.ain.lrt
]
= 64.3 i1.752*
at = 64.811.752
1
PROBLEM15.148
Pin P is attachedto the wheel shownand slidesin a slot cut in bar BD'
The wheel rolls to the right without slipping with a constantangular
velocity of20 rad/s.Knowing that x = 480 mm when B = 0, determine
the angularvelocity of the bar and the relative velocity of pin P with
respectto the rod for the given data.
(a) 0 =0, Q) e =90".
SOLUTION
Coordinates.
xA = (xA)o+r0, yn = r
xs=0,/a=r
xc = x,q,lc =0
xP=xt+esin9
yp--r+ecose
( r r ) o = + S O . - = 0 ' 4 8m
Data:
r = 200 mrn = 0.20m
e = l 4 0 m m = 0 . 1 4m
arc = a,tc
Velocity analysis.
),
an
= aso
),
y p = v A + v p 1 1= l r a t p - * ] + [ e a r r . 5 d ]
vr, =lxrrouof ]+[(ecosa)ato-* I
v,o =lucosB-*l
+ tasinBf l
Use v, = Yp' + l ptF andresolveinto components'
+
(r + ecos0)an6 = (ecos0)at6p + (cos/)a
(l)
*1,
(esin?)ong
-$inB)u
(2)
= xpaso
PROBLEM
15.148CONTINUED
(a) 0 = 0.
r; = 0.48 m,
xr = 0.48 m,
hn7 =!:osg -- o'14.
xp
0.48
a,tc = 20 radls
B - rb.26
Substitutinginto Eqs.(l) and (2),
( o . z o+ o . r + ) ( z o )= 0 . 1 4 a t u+o ( c o s 1 6 . 2 6 . ) z
(l)
0 - 0.48atso- (sin16.26.)r.
Solvingsimultaneously,
u - 6.53rnls,
@) e -90".
e)
asn = 3.8Iradls,
atuo -3.glrad/s
v p1p= 6-53nls Z
)1
16.26"1
. r p - 0 . 4 8+ ( 0 . 2 0' |1L1I + 0 . r + = 0 . 9 3 4 1 m
6
)
l
f=o
Substitutinginto Eqs.(l) and (2),
(0.20)(20)= u
(r)
u=4m/s
= 0.934r6atso
(0.14X20)
oso = 2.9973rad/S,
e)
oap - 3.00 rad/s ) {
vplr = 4.00 rn/s *
<
PROBLEM
15.161
At the instantshownthe lengthofthe booml,B is being increasedat the
constantrate of 0.6 ff:/sandthe boom is being loweredat the constantrate
of 0.08 rad/s.Determine(rr) the velocity of point B, (6) the acceleration
of ooint -B.
SOLUTION
y n, = ra = (18)(0.08) = 1.44ft/s \
Velocityof coincidingpoint B' on boom.
varuoo*- 0.6 ftls -{
Velocityof point B relativeto theboom.
60.
30"
a = vB, + va/boom
\q) Velocityof point B.
5
:
*1 '
(vs)., = 1.44cos60"
+ 0.6cos30'= 1.23962ftJs
( u , ) . , - - 1 . 4 + . i n 6 0 ' + 0 . 6 s i n 3 0 -' - 0 . 9 4 7 0 8f f : / s
,, = Fzts*y * lnntrty = r.560
ft/s
a"p - -?:!9:,
1.23962'
p = ' 3't.4.
vr = r.560
rvs\ 37.4.
{
Acceleration
of coincidingpointB' onboom.
as,= ra2 = (rr)(o.os)' = o.lts2 ftls27
a&boom= 0
Acceleration of B relative to the boom.
Coriotisacceleration.
\b) Acceleration of point B.
30"
2ura= (2X0.0s)(0.6)= 0.osoff/s, \
60.
LB = sR, + ^B,ooon+ zau
+ '
-0.051767
(ar), = -o.ttszcos30'+0.096cos60'=
m/s2
*f,
-0.14074
(""), = -o.ttsz.in30"-0.096sin60'=
m/s2
., = F.osnolt *1uuol
p = 6s.8.
""8' = :::::r.
0.051767
= o.l5oo
fvs2
ar = 0.1500
ff/s2z
6s.s"
<
PROBLEM
15.173
Pin P is attachedto the wheel shownand slidesin a slot cut in bar BD.
The wheel rolls to the right without slipping with a constant angular
velocity of 20 rad,/s.Knowing that -x = 480 mm when d = 0, determrne
(a) the angularaccelerationofthe bar and (b) the relativeaccelerationof
pin P with respectto the bar for the given data.
0 =o.
SOLUTION
Coordinates.
xn = \x,r)o+ r0,
xs=Q,
!a=r
-D
lt = r
x C= x A ,
lc=0
=
xp xA+ esine,
e
yp = r + eCOSe
(rr)o = +SO-- = 0.48m
Data:
r = 200 mm = 0.20m
e=140mm=0.14m
0=0
r
I
rp=480mm=0.48m
Velocity analysis.
ou.^ = 20 radls
),
aao = ano
)
v, = (r + e)ron, -
= (0.20
+ o.l4x2o)
= 6.8rnls*
v p , = f x p a s|o+ l e a *
vr,, =fucosB-
-l
]+ [rrsinpf ]
e -014
t a n B=
xp
0.48
p = t6.260"
Use v" = v p, + v p/F andresolveinto components.
Solving(l) and(2),
+- : 6.8 = O.l4atao+ ucosp
(l)
- usinB
+f : 0= 0.48aro
(2)
oro = 3.8080rad./s,
r = 6.528 mls
15,173CONTINUED
PROBLEM
d'tc = 0'
Acceleration analysis.
oao = ano )
roln= ro)u = (o.r+)(zo)'= SOr/s' I
^,t = 0
sp = aA + ant = 56 mls2I
-)+l,,alo *f *1""',Il
,,' =lr/,uolf +l"au
* I + [10.+s)(:.8080f
- ]
= lo.+t.,uo
!) + lo.t+auo
+[(.r+)(:.aoso)'
] J
* ] + [o.leo+
=lo.+wuo
"v" lf +lo.t+auo
]
+[2.o3ol
lr/s' I ]
^pE= licosB*
I + [';.nB | ]
Cor io I is acceler ation.
= [+r.ztrrvs' \p]
zarou= (z)(z.tos0)(6.528)
Use a, = up,+ s ptF*lzturu
\
/]
andresolveinto components'
+. : 0 = 0.14aBD- 6.9604+ icos p + 49.717sin8
or
o.l4aBD + icos B = -6.9602
(3)
+J: 56 = 0.48auo+ 2.0301+ isin B + 49.7l7cosp
O.4ktBD- isin P = 6.2415
i = -8.43 rnlsz
aao = 8.09 rad/s,
Solving(3) and(4),
or
\a)
(r)
(4)
crro = 8.09rad/s2) {
Nptr = 8.43 mls2 7
16.26' <

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