690
*16–60.
The slider block C moves at 8 m
>
s down the inclined groove.
Determine the angular velocities of links AB and BC, at the
instant shown.
SOLUTION
2 m
2 m
A
B
45
691
16–61.
Determine the angular velocity of links AB and BC at the
instant
u=30°
. Also, sketch the position of link BC when
u=55°
, 45°, and 30° to show its general plane motion.
Ans:
v
BC =2.31
rad>s
d
v
AB =3.46
rad>s
d
SOLUTION
Rotation About Fixed Axis. For link AB, refer to Fig. a.
vB=VAB *rAB
vB=(
v
ABk)*j=–
v
AB i
General Plane Motion. For link BC, refer to Fig. b. Applying the relative velocity
equation,
v
B
=v
C
+V
BC
*r
B
>
C
–
v
AB i=6j+(
v
BCk)*(–3 cos 30°i+3 sin 30°j)
–
v
ABi=–1.5
v
BCi+(6 –2.5981
v
BC)j
Equating i and j components,
0=6–2.5981
v
BC
; v
BC =2.3094 rad>s=2.31 rad>s
d Ans.
–
v
AB =–1.5(2.3094)
; v
AB =3.4641 rad>s=3.46 rad>s
d Ans.
u=30°
A
B
C
1 ft
3 ft
vC 6 ft/s
u
16–62.
SOLUTION
. At the instant shown. vBis directed to the left. Also, at the same
instant, point Eis moving to the right with a speed of
.
Velocity Equation:.tfel eht ot detcerid si hcihw,ereH
Applying Eq. 16–15, we have
Ans.vA=32.0 rad s
–120 =40 –5vA
A
:
+
B
c120
;d=c40
:d+c5vA
;d
vB=vE+vB>E
vB>E=vArB>E=5vA
2 (20) =40 in. >s
vE=vBrCE =
8(15) =120 in. >s
The planetary gear Ais pinned at B. Link BC rotates
clockwise with an angular velocity of 8 rad/s, while the outer
gear rack rotates counterclockwise with an angular velocity
of 2 rad/s. Determine the angular velocity of gear A.
15 in.
C
BC =8rad/s
ω
20 in.
D
A
B
=2rad/s
ω
16–63.
If the angular velocity of link AB is
determine the velocity of the block at Cand the angular
velocity of the connecting link CB at the instant
and f=30°.
u=45°
vAB =3 rad
>
s,
SOLUTION
dAns.
Ans.
Also,
dAns.
Ans.vC=2.20 ft s;
vCB =2.45 rad s
(+c)0=-5.196 +2.12vCB
a:
+b–vC=3–2.12vCB
–vCi=(6 sin 30°i–6 cos 30°j)+(vCB k)*(3 cos 45°i+3 sin 45°j)
vC=vB+v*rC>B
vC=2.20 ft>s;
vCB =2.45 rad>s
(+c)0=-6 cos 30° +vCB (3) sin 45°
(:
+)–vC=6 sin 30° –vCB (3) cos 45°
B
vC
;
R
=C6
30°cS+DvCB (3)
45°bT
vC=vB+vC>B
3ft
2ft
AB =3rad/s
ω
=45°
θ
=30°
φ
C
A
*16–64.
SOLUTION
Ans.
Also:
Ans.vC=2.40 ft>s
–vCi=0+(4k)*(0.6j)
vC=vB+v*rC>B
vC=2.40 ft>s
(;
+)vC=0+4(0.6)
vC=vB+vC>B
C
B
v0.3 ft
A
The pinion gear A rolls on the fixed gear rack B with an
angular velocity v = 4 rad>s. Determine the velocity of
the gear rack C.
16–65.
SOLUTION
Ans.
Ans.
Also:
Ans.
Ans.vA=2ft>s:
vAi=8i+20k*(0.3j)
vA=vB+v*rA>B
v=20 rad>s
–4=8–0.6v
–4i=8i+(vk)*(0.6j)
vC=vB+v*rC>B
vA=2ft>s:
(:
+)vA=8–20(0.3)
vA=vB+vA>B
v=20 rad>s
(:
+)–4=8–0.6(v)
vC=vB+vC>B
The pinion gear rolls on the gear racks. If Bis moving to the
right
at and Cis moving to the left at , determine
the
angular velocity of the pinion gear and the velocity of its
center
A.
4ft>s8ft>s
C
B
V0.3 ft
A
16–66.
SOLUTION
Equating the icomponents yields
(1)
Ans. (2)
For points Oand C,
Thus,
Ans.vO=0.667 ft>s:
=[0.6667i]ft>s
=-4i+
A
–3.111k
B
*
A
1.5j
B
vO=vC+v*rO>C
v=3.111 rad>s
3=2.25v–4
3i=
A
2.25v–4
B
i
3i=-4i+
A
–vk
B
*
A
2.25j
B
Determine the angular velocity of the gear and the velocity
of its center Oat the instant shown.
3ft/s
4ft/s
A
O0.75 ft
1.50 ft
45
697
16–67.
Determine the velocity of point on the rim of the gear at
the instant shown.
A
SOLUTION
Equating the icomponents yields
(1)
(2)
For points Aand C,
Equating the iand jcomponents yields
Thus, the magnitude of vAis
Ans.
and its direction is
Ans.u=tan–1C
A
vA
B
y
A
vA
B
xS=tan–1
¢
3.2998
3.9665
≤
=39.8°
vA=2
A
vA
B
x2+
A
vA
B
y2=23.96652+3.29982=5.16 ft>s
A
vA
B
x=3.9665 ft>s
A
vA
B
y=3.2998 ft>s
A
vA
B
xi+
A
vA
B
yj=3.9665i+3.2998j
A
vA
B
xi+
A
vA
B
yj=-4i+
A
–3.111k
B
*
A
–1.061i+2.561j
B
vA=vC+v*rA>C
v=3.111 rad>s
3=2.25v–4
3i=
A
2.25v–4
B
i
3i=-4i+
A
–vk
B
*
A
2.25j
B
3ft/s
4ft/s
A
O0.75 ft
1.50 ft
45
698
*16–68.
Knowing that angular velocity of link AB is
v
AB
=
4 rad
>
s, determine the velocity of the collar at C
and the angular velocity of link CB at the instant shown.
Link CB is horizontal at this instant.
45
CB
350 mm
SOLUTION
vB={–2 cos 30°i+2 sin 30°j } m/s
vC=–vC cos 45°i–vC sin 45°j
v
=
v
BCk
r
C
>
B
={–0.35i} m
v
C
=v
B
+v*r
C
>
B
–vC cos 45°i–vC sin 45°j=(–2 cos 30°i+2 sin 30°j )+(
v
BCk)*(–0.35i )
–vC cos 45°i–vC sin 45°j=–2 cos 30°i+(2 sin 30°–0.35
v
BC)j
Equating the i and j components yields:
–vC cos 45°=–2 cos 30°
vC=2.45 m>s
Ans.
–2.45 sin 45°=2 sin 30°–0.35
v
BC
v
BC =7.81 rad>s
Ans.
Ans:
vC=2.45 m>s
v
BC =7.81 rad>s
699
Ans:
v
C=
24.6 m
>
s
T
700
16–70.
The angular velocity of link AB is
v
AB
=
5 rad
>
s.
Determine the velocity of block C and the angular velocity
of link BC at the instant
u
=
45° and f
=
30°. Also, sketch
the position of link CB when
u
=
45°, 60°, and 75° to show
its general plane motion.
SOLUTION
vB=VAB *rAB
=(5k)*(–3 cos 45°i – 3 sin 45°j )
=
•
15
2
2
2 i – 15
2
2
2 j
¶
m
>
s
General Plane Motion. For link BC, refer to Fig. b. Applying the relative velocity
equation,
v
C
= v
B
+ V
BC
*r
C
>
B
vC i=
°
15
2
2
2i–15
2
2
2j
¢
+(vBC k)*(2 sin 30° i–2 cos 30°j)
vC i=
°
15
2
2
2 +
2
3 vBC
¢
i +
°
vBC –15
2
2
2
¢
j
Equating j components,
O = vBC –15
2
2
2
; vBC = 15
2
2
2
rad
>
s = 10.6 rad
>
s d Ans.
Then, equating i components,
vC = 15
2
2
2 +
2
3
°
15
2
2
2
¢
=28.98 m
>
s=29.0 m
>
sS Ans.
f
A
B
3 m
vAB
5 rad/
s
u
701
16–70. Continued
Ans:
v
BC =10.6 rad>s
d
vC=29.0 m>sS
16–71.
The similar links AB and CD rotate about the fixedpins
at A and C. If AB has an angular velocity
v
AB
=
8 rad
>
s, determine the angular velocity of BDP and
the velocity of point P.
300 mm
B
A
D
C
700 mm
300 mm
300 mm300 mm
6060
vAB 8 rad/s
SOLUTION
v
D
=v
B
+v*r
D
>
B
–vD cos 30°i–vD sin 30°j=–2.4 cos 30°i+2.4 sin 30°j+(
v
k)*(0.6i)
–vD cos 30°=–2.4 cos 30°
–vD sin 30°=2.4 sin 30°+0.6
v
vD
=2.4 m>s
v
=– 4 rad>s
Ans.
v
P
= v
B
+v*r
P
>
B
vP=–2.4 cos 30°i+2.4 sin 30°j+(–4k)*(0.3i–0.7j)
(
vP
)
x
=–
4.88 m
>
s
(
vP
)
y=0
vP
=4.88 m>s d
Ans.
703
*16–72.
If the slider block A is moving downward at
vA
=4 m>s,
determine the velocities of blocks B and C at
the instant shown.
SOLUTION
v
B
=v
A
+v
B
>
A
S
v
B
=
4
T +
v
AB
(0.55)
(
S
+
)
vB=0+vAB(0.55)
a3
5
b
(
+
c
)
0 =–4+vAB(0.55)
a4
5
b
Solving,
v
AB =9.091 rad>s
vB=3.00 m>s
Ans.
v
D
=v
A
+v
D
>
A
vD=4+[(0.3)(9.091) =2.727]
T
4
Q
3
5
v
C
=v
D
+v
C
>
D
vC=4+2.727 +
v
CE(0.4)
S T
Q
5
43 h 30
°
(
S
+
)
vC=0+2.727
a
3
5b
–vCE (0.4)(sin 30°)
(
+
c
)
0 =–4+2.727
a4
5
b
+vCE(0.4)(cos 30°)
vCE
=5.249 rad>s
vC
=0.587 m>s
Ans.
Also:
v
B
=v
A
+v
AB
*r
B
>
A
vBi =–4j+(–vABk)*
e–4
5
(0.55)i+
3
5
(0.55)j
f
4
5
3
250 mm
400 mm
300 mm
300 mm
E
B
C
D
A
30
vA 4 m/s
704
*16–72. Continued
vB=
v
AB(0.33)
0=–4+0.44
v
AB
v
AB =9.091 rad>s
vB
=3.00 m>s
Ans.
v
D
=v
A
+v
AB
*r
B
>
A
vD=–4j+(–9.091k)*
e–4
5
(0.3)i+
3
5
(0.3)j
f
vD=51.636i–1.818j6 m>s
v
C
=v
D
+v
CE
*r
C
>
D
vCi=(1.636i–1.818j)+(–
v
CEk)*(–0.4 cos 30°i–0.4 sin 30°j)
vC=1.636 –0.2
v
CE
0=–1.818 –0.346
v
CE
v
CE =5.25 rad>s
vC
=0.587 m>s
Ans.
Ans:
vB
=3.00 m>s
vC
=0.587 m>s
vB
=3.00 m>s
vC
=0.587 m>s
705
16–73.
If the slider block A is moving downward at vA
=
4 m
>
s,
determine the velocity of point E at the instant shown.
SOLUTION
2.424 b
4
5
3
250 mm
400 mm
300 mm
300 mm
E
B
C
D
A
30
vA 4 m/s
Ans:
vE=4.00 m>s
u=52.7°
c
706
16–74.
The epicyclic gear train consists of the sun gear Awhich is
in mesh with the planet gear B.This gear has an inner hub C
which is fixed to Band in mesh with the fixed ring gear R.If
the connecting link DE pinned to Band Cis rotating at
about the pin at determine the angular
velocities of the planet and sun gears.
E,vDE =18 rad>s
The velocity of the contact point Pwith the ring is zero.
bAns.
Let be the contact point between Aand B.
dAns.vA=vP¿
rA
=36
0.2 =180 rad>s
vP¿=36 m>sc
vP¿j=0+(–90k)*(–0.4i)
vP¿=vP+v*rP¿>P
P¿
vB=90 rad>s
9j=0+(–vBk)*(–0.1i)
vD=vP+v*rD>P
BA
100 mm
R
C
E
600 mm
D
707
16–75.
If link AB is rotating at
v
AB
=
3 rad
>
s, determine the
angular velocity of link CD at the instant shown.
SOLUTION
v
B
=v
AB
*r
B
>
A
v
C
=v
CD
*r
C
>
D
v
C
=v
B
+v
BC
*r
C
>
B
(
v
CD
k)*(–4 cos 45°i+4 sin 45°j)=(–3k)*(6i)+(
v
BCk)*(–8 sin 30°i–8 cos 30°j)
–2.828
v
CD =0+6.928
v
BC
–2.828
v
CD =–18 –4
v
BC
Solving,
vBC
=–1.65 rad>s
vCD
=4.03 rad>s
Ans.
A
B
8 in.
30
45
4 in.
D
C
vCD
6 in.
vAB
Ans:
v
CD =4.03 rad>s
*16–76.
If link CD is rotating at
v
CD
=
5 rad
>
s, determine the
angular velocity of link AB at the instant shown.
SOLUTION
v
B
=v
AB
*r
B
>
A
v
C
=v
CD
*r
C
>
D
v
B
=v
C
+v
BC
*r
B
>
C
(–
v
AB
k)*(6i)=(5k)*(–4 cos 45°i+4 sin 45°j)+(
v
BCk)*(8 sin 30°i+8 cos 30°j)
0=–14.142 –6.9282
v
BC
–6
v
AB =–14.142 +4
v
BC
Solving,
v
AB =3.72 rad>s
Ans.
v
BC =–2.04 rad>s
A
B
8 in.
30
45
4 in.
D
C
vCD
6 in.
vAB
709
16–77.
The planetary gear system is used in an automatic
transmission for an automobile. By locking or releasing
certain gears, it has the advantage of operating the car at
different speeds. Consider the case where the ring gear Ris
held fixed, , and the sun gear Sis rotating at
. Determine the angular velocity of each of the
planet gears Pand shaft A.
vS=5 rad>s
vR=0
SOLUTION
Ans.
Ans.vA=200
120 =1.67 rad>s
vC=0+(–5k)*(–40j)=-200i
vC=vB+v*rC>B
vP=-5 rad>s=5 rad>s
0=-400i–80vpi
R
S
P
A
vS
vR
40 mm
Ans:
v
P=5 rad>s
v
A=1.67 rad>s