PROBLEM 11.37
A small package is released from rest at A and
moves along the skate wheel conveyor ABCD.
The package has a uniform acceleration of
4.8 m/s 2 as it moves down sections AB and CD,
and its velocity is constant between B and C. If
the velocity of the package at D is 7.2 m/s,
determine (a) the distance d between C and D,
(b) the time required for the package to reach D.
SOLUTION
(a)
For A
B and C
D we have
v 2 = v02 + 2a( x − x0 )
2
vBC
= 0 + 2(4.8 m/s 2 )(3 − 0) m
Then, at B
= 28.8 m 2 /s 2
(vBC = 5.3666 m/s)
2
vD2 = vBC
+ 2aCD ( xD − xC )
and at D
d = xD − xC
(7.2 m/s)2 = (28.8 m 2 /s 2 ) + 2(4.8 m/s 2 )d
or
d = 2.40 m 
or
(b)
For A
B and C
D we have
v = v0 + at
Then A
B
5.3666 m/s = 0 + (4.8 m/s 2 )t AB
t AB = 1.11804 s
or
and C
7.2 m/s = 5.3666 m/s + (4.8 m/s 2 )tCD
D
tCD = 0.38196 s
or
Now, for B
C, we have
xC = xB + vBC t BC
or
3 m = (5.3666 m/s)t BC
or
t BC = 0.55901 s
Finally,
t D = t AB + t BC + tCD = (1.11804 + 0.55901 + 0.38196) s
t D = 2.06 s 
or
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47
PROBLEM 11.39
As relay runner A enters the 20-m-long exchange zone with a
speed of 12.9 m/s, he begins to slow down. He hands the baton to
runner B 1.82 s later as they leave the exchange zone with the
same velocity. Determine (a) the uniform acceleration of each of
the runners, (b) when runner B should begin to run.
SOLUTION
(a)
For runner A:
At t = 1.82 s:
x A = 0 + (v A )0 t +
1
a At 2
2
20 m = (12.9 m/s)(1.82 s) +
1
a A (1.82 s) 2
2
a A = −2.10 m/s 2 
or
Also
At t = 1.82 s:
v A = (v A ) 0 + a A t
(v A )1.82 = (12.9 m/s) + ( −2.10 m/s 2 )(1.82 s)
= 9.078 m/s
For runner B:
vB2 = 0 + 2aB [ xB − 0]
When
xB = 20 m, vB = v A : (9.078 m/s) 2 = 2aB (20 m)
or
aB = 2.0603 m/s 2
aB = 2.06 m/s 2 
(b)
For runner B:
vB = 0 + aB (t − t B )
where t B is the time at which he begins to run.
At t = 1.82 s:
or
9.078 m/s = (2.0603 m/s 2 )(1.82 − t B ) s
t B = −2.59 s
Runner B should start to run 2.59 s before A reaches the exchange zone. 
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49
PROBLEM 11.52
At the instant shown, slider block B is moving
with a constant acceleration, and its speed is
150 mm/s. Knowing that after slider block A
has moved 240 mm to the right its velocity is
60 mm/s, determine (a) the accelerations of A
and B, (b) the acceleration of portion D of the
cable, (c) the velocity and change in position of
slider block B after 4 s.
SOLUTION
xB + ( xB − x A ) − 2 x A = constant
From the diagram
Then
2vB − 3v A = 0
(1)
and
2aB − 3a A = 0
(2)
(a)
First observe that if block A moves to the right, v A → and Eq. (1)  v B → . Then, using
Eq. (1) at t = 0
2(150 mm/s) − 3(v A )0 = 0
(v A )0 = 100 mm/s
or
Also, Eq. (2) and aB = constant  a A = constant
v A2 = (v A )02 + 2a A [ x A − ( x A )0 ]
Then
When x A − ( x A )0 = 240 mm:
(60 mm/s) 2 = (100 mm/s) 2 + 2a A (240 mm)
or
aA = −
40
mm/s 2
3
a A = 13.33 mm/s 2
or

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66
PROBLEM 11.52 (Continued)
Then, substituting into Eq. (2)
 40

2aB − 3  −
mm/s 2  = 0
 3

aB = −20 mm/s 2
or
(b)
a B = 20.0 mm/s 2

From the diagram, − xD − x A = constant
vD + v A = 0
Then
Substituting
aD + a A = 0
 40

aD +  −
mm/s 2  = 0
3


or
(c)
We have
vB = ( vB ) 0 + a B t
At t = 4 s:
vB = 150 mm/s + ( −20.0 mm/s 2 )(4 s)
or
Also
At t = 4 s:
x B = ( xB ) 0 + ( vB ) 0 t +
a D = 13.33 mm/s 2

v B = 70.0 mm/s

1
aB t 2
2
xB − ( xB )0 = (150 mm/s)(4 s)
+
1
(−20.0 mm/s 2 )(4 s) 2
2
x B − (x B )0 = 440 mm
or

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67
PROBLEM 11.55
Block C starts from rest at t = 0 and moves downward with a constant
acceleration of 4 in./s2. Knowing that block B has a constant velocity of 3 in./s
upward, determine (a) the time when the velocity of block A is zero, (b) the
time when the velocity of block A is equal to the velocity of block D, (c) the
change in position of block A after 5 s.
SOLUTION
From the diagram:
Cord 1:
2 y A + 2 y B + yC = constant
Then
2v A + 2vB + vC = 0
and
2a A + 2aB + aC = 0
Cord 2:
(1)
( y D − y A ) + ( y D − y B ) = constant
Then
2vD − v A − v B = 0
and
2aD − a A − aB = 0
(2)
Use units of inches and seconds.
Motion of block C:
vC = vC 0 + aC t
where aC = −4 in./s 2
= 0 + 4t
Motion of block B:
vB = −3 in./s;
Motion of block A:
From (1) and (2),
aB = 0
1
1
v A = −vB − vC = 3 − (4t ) = 3 − 2t in./s
2
2
1
1
a A = −aB − aC = 0 − (4) = −2 in./s 2
2
2
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71
PROBLEM 11.55 (Continued)
(a)
Time when vB is zero.
3 − 2t = 0
Motion of block D:
From (3),
vD =
(b)
t = 1.500 s 
1
1
1
1
v A + vB = (3 − 2t ) − (3) = −1t
2
2
2
2
Time when vA is equal to v0.
3 − 2t = −t
(c)
t = 3.00 s 
Change in position of block A (t = 5 s).
1
a At 2
2
1
= (3)(5) + (−2)(5)2 = −10 in.
2
Δy A = ( v A ) 0 t +
Change in position = 10.00 in. 
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72
PROBLEM 12.27
A spring AB of constant k is attached to a support at A and to a
collar of mass m. The unstretched length of the spring is .
Knowing that the collar is released from rest at x = x0 and
neglecting friction between the collar and the horizontal rod,
determine the magnitude of the velocity of the collar as it
passes through Point C.
SOLUTION
Choose the origin at Point C and let x be positive to the right. Then x is a position coordinate of the slider B
and x0 is its initial value. Let L be the stretched length of the spring. Then, from the right triangle
L = 2 + x 2
The elongation of the spring is e = L − , and the magnitude of the force exerted by the spring is
Fs = ke = k (  2 + x 2 − )
x
cos θ =
By geometry,
 + x2
2
ΣFx = max : − Fs cos θ = ma
− k (  2 + x 2 − )
a=−

v
0
v dv =
k
x
x−
2
m 
 + x2
x
 + x2
2
= ma




0
 0 a dx
x
v
1 2
k
v =−
m
2 0


x
x−
2
x0 
 + x2

0
0

k 1 2
2
2 
=
−
−
+
dx
x
x





m  2
x

0
k
1 2
1

v = −  0 −  2 − x02 +   2 + x02 
m
2
2

k
v2 =
2 2 + x02 − 2  2 + x02
m
k
=   2 + x02 − 2  2 + x02 +  2 

m 
)
(
(
)
answer: v =
k
m
(
)
2 + x02 −  
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339
PROBLEM 13.54
The elevator E has a weight of 6600 lbs when fully loaded and is
connected as shown to a counterweight W of weight of 2200 lb.
Determine the power in hp delivered by the motor (a) when the elevator
is moving down at a constant speed of 1 ft/s, (b) when it has an upward
velocity of 1 ft/s and a deceleration of 0.18 ft/s 2 .
SOLUTION
(a) Acceleration = 0
Elevator
Counterweight
Motor
ΣFy = 0: TW − WW = 0
ΣF = 0: 2TC + TW − 6600 = 0
TW = 2200 lb
Kinematics:
TC = 2200 lb
2 xE = xC , 2 x E = xC , vC = 2vE = 2 ft/s
P = TC ⋅ vC = (2200 lb)(2 ft/s) = 4400 lb ⋅ ft/s = 8.00 hp
P = 8.00 hp 
aE = 0.18 ft/s 2 , vE = 1 ft/s
(b)
Counterweight
Elevator
Counterweight:
ΣF = Ma : TW − W =
W
(aW )
g
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576
PROBLEM 13.54 (Continued)
TW = (2200 lb) +
(2200 lb)(0.18 ft/s 2 )
(32.2 ft/s 2 )
TW = 2212 lb
Elevator
ΣF = ma
2TC + TW − WE =
2TC = (−2212 lb) + (6600 lb) −
−WE
(a E )
g
(6600 lb)(0.18 ft/s 2 )
(32.2 ft/s 2 )
2TC = 4351 lb
TC = 2175.6 lb
vC = 2 ft/s (see part(a))
P = TC ⋅ vC = (2175.6 lb)(2 ft/s) = 4351.2 lb ⋅ ft/s
= 7.911 hp



P = 7.91 hp 
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577
PROBLEM 13.72
A 1-lb collar is attached to a spring and slides without friction along
a circular rod in a vertical plane. The spring has an undeformed
length of 5 in. and a constant k = 10 lb/ft. Knowing that the collar is
released from being held at A determine the speed of the collar and
the normal force between the collar and the rod as the collar passes
through B.
SOLUTION
W
1
=
= 0.031056 lb ⋅ s 2 /ft
g 32.2
For the collar,
m=
For the spring,
k = 10 lb/ft l0 = 5 in.
 A = 7 + 5 + 5 = 17 in.
At A:
 Δ −  0 = 12 in. = 1 ft
 B = (7 + 5) 2 + 52 = 13 in.
At B:
 B −  0 = 1.8 in. =
2
ft
3
Velocity of the collar at B.
Use the principle of conservation of energy.
TA + VA = TB + VB
Where
TA =
1 2
mv A = 0
2
1
k ( A −  0 ) 2 + W (0)
2
1
= (10)(1) 2 + 0 = 5 ft ⋅ lb
2
1 2 1
TB = mvB = (0.031056)vB2 = 0.015528vB2
2
2
1
VB = k ( B −  0 ) 2 + Wh
2
VA =
2
1
2
 5
(10)   + (1)  − 
2
3
 12 
= 1.80556 ft ⋅ lb
=
0 + 5 = 0.015528vB2 = 1.80556
vB = 14.34 ft/s 
vB2 = 205.72 ft 2 /s 2
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612
PROBLEM 13.72 (Continued)
Forces at B.
2
Fs = k ( B −  0 ) = (10)   = 6.6667 lb.
3
5
sinα =
13
5
ρ = 5 in. = ft
12
mvB2
man =
ρ
(0.031056)(205.72)
5/12
= 15.3332 lb
=
ΣFy = ma y : Fs sin α − W + N = man
N = man + W − Fs sin α
 5
= 15.3332 + 1 − (6.6667)  
 13 
N = 13.769 lb
N = 13.77 lb 
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613
PROBLEM 14.42
In a game of pool, ball A is moving with a velocity v 0 of magnitude
v0 = 15 ft/s when it strikes balls B and C, which are at rest and aligned
as shown. Knowing that after the collision the three balls move in the
directions indicated and assuming frictionless surfaces and perfectly
elastic impact (i.e., conservation of energy), determine the magnitudes
of the velocities vA, vB, and vC.
SOLUTION
Velocity vectors:
v 0 = v0 (cos 30° i + sin 30° j)
v0 = 15 ft/s
vA = vA j
vB = vB (sin 45° i − cos 45° j)
vC = vC (cos 45° i + sin 45° j)
Conservation of momentum:
mv 0 = mvA + mvB + mvC
Divide by m and resolve into components.
i: v0 cos 30° = vB sin 45° + vC cos 45°
j: v0 sin 30° = vA − vB cos 45° + vC sin 45°
Solving for vB and vC ,
vB = 0.25882v0 + 0.70711v A
vC = 0.96593v0 − 0.70711v A
Conservation of energy:
1 2 1 2 1 2 1 2
mv0 = mv A + mvB + mvC
2
2
2
2
Divide by m and substitute for vB and vC .
v02 = v 2A + (0.25882v0 + 0.70711vA ) 2
+ (0.96593v0 − 0.70711vA ) 2
= v02 − v0 vA + 2v A2
vA = 0.5v0 = 7.500 ft/s
vA = 7.50 ft/s 
vB = 0.61237v0 = 9.1856 ft/s
vB = 9.19 ft/s 
vC = 0.61237v0 = 9.1856 ft/s
vC = 9.19 ft/s 
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899
PROBLEM 14.108
In a game of pool, ball A is moving with a velocity v0 when it strikes
balls B and C which are at rest and aligned as shown. Knowing that
after the collision the three balls move in the directions indicated and
that v0 = 12 ft/s and vC = 6.29 ft/s, determine the magnitude of the
velocity of (a) ball A, (b) ball B.
SOLUTION
Conservation of linear momentum. In x direction:
m(12 ft/s) cos 30° = mv A sin 7.4° + mvB sin 49.3°
+ m(6.29) cos 45°
0.12880vA + 0.75813vB = 5.9446
(1)
In y direction:

m(12 ft/s) sin 30° = mvA cos 7.4° − mvB cos 49.3°
+ m(6.29)sin 45°
0.99167vA − 0.65210vB = 1.5523
(a)
Multiply (1) by 0.65210, (2) by 0.75813, and add:
0.83581 vA = 5.0533
(b)

(2)
vA = 6.05 ft/s 
Multiply (1) by 0.99167, (2) by –0.12880, and add:
0.83581 vB = 5.6951 
vB = 6.81 ft/s 
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991
PROBLEM 14.51
In a game of billiards, ball A is given an initial velocity v 0 along
the longitudinal axis of the table. It hits ball B and then ball C,
which are both at rest. Balls A and C are observed to hit the sides of
the table squarely at A′ and C′ , respectively, and ball B is
observed to hit the side obliquely at B′. Knowing that v0 = 4 m/s,
v A = 1.92 m/s, and a = 1.65 m, determine (a) the velocities v B
and vC of balls B and C, (b) the Point C ′ where ball C hits the
side of the table. Assume frictionless surfaces and perfectly elastic
impacts (that is, conservation of energy).
SOLUTION
Velocities in m/s. Lengths in meters. Assume masses are 1.0 for each ball.
( v A )0 = v0i = 4i,
Before impacts:
v A = −1.92 j,
After impacts:
(v B )0 = ( vC )0 = 0
v B = (vB ) x i + (vB ) y j,
vC = vC i
v 0 = v A + v B + vC
Conservation of linear momentum:
i: 4 = 0 + (vB ) x + vC
j: 0 = −1.92 + (vB ) y + 0
(vB ) x = 4 − vC
(vB ) y = 1.92
1 2 1 2 1 2 1 2
v0 = v A + vB + vC
2
2
2
2
Conservation of energy:
1 2 1
1
1
1
(4) = (1.92) 2 + (1.92)2 + (4 − vC )2 + vC2
2
2
2
2
2
vC2 − 4vC + 3.6864 = 0
vC =
4 ± (4) 2 − (4)(3.6864)
= 2 ± 0.56 = 2.56
2
or
1.44
Conservation of angular momentum about B′:
0.75v0 = (1.8 − a)v A + cvC
cvC = (0.75)(4) − (1.8 − 1.65)(1.92) = 2.712
c=
2.712
vC
If vC = 1.44,
c = 1.8833
If vC = 2.56,
c = 1.059
off the table. Reject.
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914
PROBLEM 14.51 (Continued)
(vB ) x = 4 − 2.56 = 1.44,
Then,
v B = 1.44i + 1.92 j
Summary.
v B = 2.40 m/s
(a)

53.1° 
vC = 2.56 m/s


c = 1.059 m 
(b)
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915