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*12–192.
SOLUTION
At
,
Ans.
Ans.a
=2
(
-
0.75)2
+
(1.5)2
+
(8)2
=
8.17 ft
>
s2
az=8
au=0+2(1.5)(0.5) =1.5
ar=0-3(0.5)2=-0.75
v=2(1.5)2+(1.5)2+(24)2=24.1 ft>s
vz=24
vu=3(0.5) =1.5
vr=1.5
u
$=0r
$=0z
$=8
u
#=0.5
r
#=1.5 z
#=24
u=1.5
r=3z=36
t=3s
u
$=0r
$=0z
$=8ft>s2
u
#=0.5 rad>sr
#=1.5 ft>sz
#=8tft>s
u=0.5 trad
r=3ft
z=4t2ft
F
or a short time the arm of the robot is extending at a
constant rate such that when
and where tis in seconds.
Determine the magnitudes of the velocity and acceleration
of the grip Awhen t=3s.
u=0.5trad,z=14t22ft,
r=3 ft,r
#=1.5 ft>s
r
z
A
u
SOLUTION
#
0.5 t2fit=1=
12–193.
The double collar C is pin connected together such that one
collar slides over the fixed rod and the other slides over the
rotating rod AB. If the angular velocity of AB is given as
u
#
=
(e0.5 t2) rad
>
s, where t is in seconds, and the path defined by
the fixed rod is r = |(0.4 sin
u
+ 0.2)| m, determine the radial
and transverse components of the collar’s velocity and
acceleration when t = 1 s. When t = 0,
u
= 0. Use Simpson’s
rule with n = 50 to determine
u
at t = 1 s.
A
0.6 m
0.2 m
0.2 m 0.2 m
u
C
B
r
203
12–194.
The double collar C is pin connected together such that one
collar slides over the fixed rod and the other slides over the
rotating rod AB. If the mechanism is to be designed so that
the largest speed given to the collar is 6 m
>
s, determine the
required constant angular velocity
u
#
of rod AB. The path
defined by the fixed rod is r = (0.4 sin
u
+ 0.2) m.
SOLUTION
r=0.4 sin u+0.2
r
#=0.4 cos u u
#
v
r=
r
#=
0.4 cos u u
#
v
u=
r u
#
=
(0.4 sin u
+
0.2) u
#
v
2=
v
r
2+
v
u
3
(6)
2=
[(0.4 cos u)
2+
(0.4 sin u
+
0.2)
2
](u
#
)
2
36
=
[0.2
+
0.16 sin u](u
#
)
2
The greatest speed occurs when
u=90°.
u
#
=
10.0 rad
>
s Ans.
A
0.6 m
0.2 m
0.2 m 0.2 m
u
C
B
r
12–195.
If the end of the cable at Ais pulled down with a speed of
determine the speed at which block Brises.2m>s,
SOLUTION
C
D
2m/s
A
*12–196.
SOLUTION
For B:
Require
Set
The positive root is .Thus,
Ans.
Ans.vAB=5.93 m s:
0.6050i=-5.3281i+vA>Bi
vA=vB+vA>B
vB=5(1.0656) =5.3281 m>s
vA=0.5(1.0656)3=0.6050
t=1.0656 =1.07 s
u=1.1355
u=t20.125u2+2.5u=3
0.125t4+2.5t2=3
sA+sB=d
sB=2.5t2;
vB=5t;
aB=5m>s2;
s
A=0.125 t4:
v
A=0.5t3:
aA=-1.5t2=1.5t2:
2aA=aC=-3t2
2v
A=vC
The motor at Cpulls in the cable with an acceleration
, where tis in seconds.The motor at Ddraws
in its cable at . If both motors start at the same
instant from rest when , determine (a) the time
needed for , and (b) the velocities of blocks A and
when this occurs.
d=0
d=3m
aD=5m>s2
aC=(3t2)m>s2
d=3m
A
D
C
B
B
12–197.
SOLUTION
Ans.yA=24 ft>s
6(4) =yA
6yB-yA=0
6sB-sA=l
5sB+(sB-sA)=l
The pulley arrangement shown is designed for hoisting
materials. If BC remains fixed while the plunger Pis pushed
downward with a speed of determine the speed of the
load at A.
4ft>s,
B
A
C
P
4ft/s
SOLUTION
12–198.
If the end of the cable at A is pulled down with a speed of
5m
>
s, determine the speed at which block B rises.
A
5 m/s
B
12–199.
Determine the displacement of the log if the truck at C
pulls the cable 4 ft to the right.
SOLUTION
Since , then
Ans.
¢
s
B=-
1.33 ft
=
1.33 ft
:
3¢sB=-4
¢sC=-4
3¢sB-¢sC=0
3sB-sC=l
2sB+(sB-sC)=l
C
B
SOLUTION
6
*12–200.
Determine the constant speed at which the cable at A must
be drawn in by the motor in order to hoist the load 6 m
in1.5 s.
C
A
B
D
SOLUTION
6
12–201.
Starting from rest, the cable can be wound onto the drum of
the motor at a rate of vA = (3t2) m
>
s, where t is in seconds.
Determine the time needed to lift the load 7 m.
C
A
B
D
211
SOLUTION
12–202.
If the end A of the cable is moving at vA = 3 m
>
s, determine
the speed of block B.CDA
vA fi 3 m/s
B
Ans:
vB=0.75 m>s
SOLUTION
12–203.
Determine the time needed for the load at B to attain a
speed of 10 m
>
s, starting from rest, if the cable is drawn into
the motor with an acceleration of 3 m
>
s2.
C
A
v
A
B
*12–204.
The cable at A is being drawn toward the motor at vA = 8 m
>
s.
Determine the velocity of the block.
SOLUTION
C
A
vA
B
214
12–205.
If block A of the pulley system is moving downward at 6 ft
>
s
while block C is moving down at 18 ft
>
s, determine the
relative velocity of block B with respect to C.
A
C
SOLUTION
12–206.
Determine the speed of the block at B.
A
B
6 m
/
s
12–207.
Determine the speed of block if the end of the rope is
pulled down with a speed of .4 m>s
A
SOLUTION
Position Coordinates: By referring to Fig.a, the length of the cord written in terms
of the position coordinates and is
Time Derivative: Taking the time derivative of the above equation,
Here,.Thus,
Ans.vA=-133 m>s=1.33 m>s c
4+3vA=0
vB=4 m>s
vB+3vA=0(+T)
sB+3sA=l+2a
sB+sA+2(sA-a)=l
sB
sA
A
4 m/s
B
SOLUTION
*12–208.
The motor draws in the cable at C with a constant velocity
of vC = 4 m
>
s. The motor draws in the cable at D with a
constant acceleration of aD = 8 m
>
s2. If vD = 0 when t = 0,
determine (a) the time needed for blockA to rise 3 m, and
(b) the relative velocity of block A with respect to block B
when this occurs.
C
A
B
D
12–209.
The cord is attached to the pin at Cand passes over the two
pulleys at Aand D.The pulley at Ais attached to the
smooth collar that travels along the vertical rod. Determine
the velocity and acceleration of the end of the cord at Bif at
the instant the collar is moving upwards at
which is decreasing at 2ft>s2.
5ft>s,sA=4ft
s
A
C
B
D
s
B
3ft3ft
12–210.
SOLUTION
Ans.a
A=
s
$
A=-
2.4375
=
2.44 ft
>
s2
c
#
4=-
2(4)(s
#
A)
(42+9)1
2
sA=4ft
22sA
2+32+6=16
At sB=6ft, s
#
B=4ft>s, s
$
B=3ft>s2
s
$
B=- 21s
#
A
2+sAs
$
A2
1sA
2+921
2
+21sAs
#
A22
1sA
2+923
2
s
$
B=-2s
#
A
2(sA
2+9)-1
2-a2sAs
$
Ab(sA
2+9)-1
2-a2sAs
#
Abca-1
2b(sA
2+9)-3
2a2sAs
#
Abd
s
#
B=- 2sAs
#
A
(sA
2+9)1
2
2a1
2b(sA
2+9)-1
2a2sAs
#
Ab+s
#
B=0
22sA
2+32
+sB=l
The 16-ft-long cord is attached to the pin at Cand passes
over the two pulleys at Aand D.The pulley at Ais attached
to the smooth collar that travels along the vertical rod.
When the end of the cord at Bis pulled
downwards with a velocity of and is given an
acceleration of Determine the velocity and
acceleration of the collar at this instant.
3ft>s2.
4ft>s
sB=6ft,
s
A
A
C
B
D
s
B
3ft3ft
SOLUTION
12–211.
The roller at A is moving with a velocity of
v
A
=4 m>s
and
has an acceleration of aA
=
2 m
>
s
2
when
xA=3 m.
Determine the velocity and acceleration of blockB at this
instant.
4 m
A
vA fi 4 m/s
xA
