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Kinematics: The speed of block A and B can be related using the position
coordinate equation.
(1)
Taking time derivative of Eq. (1), we have
Potential Energy: Datum is set at fixed pulley C.When blocks A and B (pulley D) are
at their initial position, their centers of gravity are located at s
A
and s
B
. Their initial
gravitational potential energies are
,
, and
. When block B (pulley
D
) rises 5 ft, block A decends 10 ft. Thus, the final position of blocks A and B (pulley
D
) are
and
below
datum. Hence, their respective final
gravitational potential energy are
,
, and
.
Thus, the initial and final potential energy are
Kinetic Energy: The mass moment inertia of the pulley about its mass center is
. Since pulley D rolls without slipping,
. Pulley C rotates about the fixed point
hence
. Since the system is at initially rest, the initial kinectic
energy is
. The final kinetic energy is given by
Conservation of Energy: Applying Eq. 18–19, we have
Ans.
y
4
=
21.0 ft
>
s
0
+
(
-
60
s
A
-
25
s
B
)
=
1.0773
y
2
A
+
(
-
60
s
A
-
25
s
B
-
475)
T
1
+
V
1
=
T
2
+
V
2
=
1.0773
y
2
A
+
1
2
(0.01941)(
-
y
A
)
2
+
1
2
(0.01941)(2
y
A
)
2
=
1
2
a
60
32.2
b
y
2
A
+
1
2
a
20
32.2
b
(
-
0.5
y
A
)
2
+
1
2
a
5
32.2
b
(
-
0.5
y
A
)
2
T
2
=
1
2
m
A
y
2
A
+
1
2
m
B
y
B
2
+
1
2
m
D
y
2
B
+
1
2
I
G
v
2
D
+
1
2
I
G
v
2
C
T
1
=
0
v
C
=
y
A
r
C
=
y
A
0.5
=
2
y
A
v
D
=
y
B
r
D
=
y
B
0.5
=
2
y
B
=
2(
-
0.5
y
A
)
= -
y
A
I
G
=
1
2
a
5
32.2
b
A
0.5
2
B
=
0.01941 slug
#
ft
2
V
2
= -
60(
s
A
+
10)
-
20(
s
B
-
5)
-
5(
s
B
-
5)
= -
60
s
A
-
25
s
B
-
475
V
1
= -
60
s
A
-
20
s
B
-
5
s
B
= -
60
s
A
-
25
s
B
-
5(
s
B
-
5)
-
20(
s
B
-
5)
-
60(
s
A
+
10)
(
s
B
-
5) ft
(
s
A
+
10) ft
-
5
s
B
-
20
s
B
-
60
s
A
y
A
+
2
y
B
=
0
y
B
= -
0.5
y
A
¢
s
A
+
2
¢
s
B
=
0
¢
s
A
+
2(5)
=
0
¢
s
A
= -
10 ft
=
10 ft
T
s
A
+
2
s
B
=
l
0.5 ft
A
C
D
0.5 ft
B
Ans:
v
A
=
21.0 ft
>
s
978
© 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
*18–64.
The door is made from one piece, whose ends move along
the horizontal and vertical tracks. If the door is in the open
position,
u
=
0
°
, and then released, determine the speed at
which its end A strikes the stop at C. Assume the door is a
180-lb thin plate having a width of 10 ft.
C
A
B
5 ft
3 ft
u
Solution
T
1
+
V
1
=
T
2
+
V
2
0
+
0
=
1
2
c
1
12
a
180
32.2
b
(8)
2
d
v
2
+
1
2
a
180
32.2
b
(1
v
)
2
-
180(4)
v
=
6.3776 rad
>
s
v
c
=
v
(5)
=
6.3776(5)
=
31.9 m
>
s
Ans.
Ans:
v
c
=
31.9 m
>
s
979
© 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
18–65.
The door is made from one piece, whose ends move along
the horizontal and vertical tracks. If the door is in the open
position,
u
=
0
°
, and then released, determine its angular
velocity at the instant
u
=
30
°
. Assume the door is a 180-lb
thin plate having a width of 10 ft.
C
A
B
5 ft
3 ft
u
Solution
T
1
+
V
1
=
T
2
+
V
2
0
+
0
=
1
2
c
1
12
a
180
32.2
b
(8)
2
d
v
2
+
1
2
a
180
32.2
b
v
G
2
-
180(4 sin 30
°
)
(1)
r
IC
-
G
=
2
(1)
2
+
(4.3301)
2
-
2(1)(4.3301) cos 30
°
r
IC
-
G
=
3.50 m
Thus,
v
G
=
3.50
v
Substitute into Eq. (1) and solving,
v
=
2.71 rad
>
s
Ans.
Ans:
v
=
2.71 rad
>
s
980
© 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
18–66.
SOLUTION
Selecting the smaller root:
Ans.
k
=
100 lb/ft
0
+
0
=
0
+
2
c
1
2
(
k
)(8
-
4.5352)
2
d
-
200(6)
T
1
+
V
1
=
T
2
+
V
2
CD
=
4.5352 ft
CD
2
-
11.591
CD
+
32
=
0
(2)
2
=
(6)
2
+
(
CD
)
2
-
2(6)(
CD
) cos 15°
The end A of the garage door AB travels along the
horizontal track, and the end of member BC is attached to a
spring at C. If the spring is originally unstretched, determine
the stiffness k so that when the door falls downward from
rest in the position shown, it will have zero angular velocity
the moment it closes, i.e., when it and BC become vertical.
Neglect the mass of member BC and assume the door is a
thin plate having a weight of 200 lb and a width and height
of 12 ft. There is a similar connection and spring on the
other side of the door.
15
7 ft
12 ft
A
B
C
D
2 ft
6 ft
1 ft
Ans:
k
=
100 lb
>
ft
981
© 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
18–67.
The system consists of a 30-kg disk, 12-kg slender rod BA,
and a 5-kg smooth collar A. If the disk rolls without slipping,
determine the velocity of the collar at the instant
u
=
0
°
.
The system is released from rest when
u
=
45
°
.
Solution
Kinetic Energy. Since the system is released from rest, T
1
=
0. Referring to the
kinematics diagram of the rod at its final position, Fig. a, we found that IC is located
at B. Thus, (
v
B
)
2
=
0. Also
(
v
A
)
2
=
(
v
r
)
2
r
A
>
IC
; (
v
A
)
2
=
(
v
r
)
2
(2) (
v
r
)
2
=
(
v
A
)
2
2
Then
(
v
Gr
)
2
=
(
v
r
)
2
(r
Gr
>
IC
); (
v
Gr
)
2
=
(
v
A
)
2
2
(1)
=
(
v
A
)
2
2
For the disk, since the velocity of its center (
v
B
)
2
=
0, (
v
d
)
2
=
0. Thus,
T
2
=
1
2
m
r
(
v
Gr
)
2
2
+
1
2
I
Gr
(
v
r
)
2
2
+
1
2
m
c
(
v
A
)
2
2
=
1
2
(12)
c
(
v
A
)
2
2
d
2
+
1
2
c
1
12
(12)
(
2
2
)
d c
(
v
A
)
2
2
d
2
+
1
2
(5)(
v
A
)
2
2
=
4.50(
v
A
)
2
2
Potential Energy. Datum is set as shown in Fig. a. Here,
S
B
=
2
-
2 cos 45
°
=
0.5858 m
Then
(y
d
)
1
=
0.5858 sin 30
°
=
0.2929 m
(y
r
)
1
=
0.5858 sin 30
°
+
1 sin 75
°
=
1.2588 m
(y
r
)
2
=
1 sin 30
°
=
0.5 m
(y
c
)
1
=
0.5858 sin 30
°
+
2 sin 75
°
=
2.2247 m
(y
c
)
2
=
2 sin 30
°
=
1.00 m
Thus, the gravitational potential energies of the disk, rod and collar at the initial and
final positions are
(V
d
)
1
=
m
d
g
(y
d
)
1
=
30(9.81)(0.2929)
=
86.20 J
(V
d
)
2
=
m
d
g
(y
d
)
2
=
0
(V
r
)
1
=
m
r
g
(y
r
)
1
=
12(9.81)(1.2588)
=
148.19 J
(V
r
)
2
=
m
r
g
(y
r
)
2
=
12(9.81)(0.5)
=
58.86 J
(V
c
)
1
=
m
c
g
(y
c
)
1
=
5(9.81)(2.2247)
=
109.12 J
(V
c
)
2
=
m
c
g
(y
c
)
2
=
5(9.81)(1.00)
=
49.05 J
0.5 m
2 m
A
C
B
30
u
982
© 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
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