lOMoARcPSD|12889239
공업역학 동역학 연습문제 cp15 솔루션
공업동역학 (Silla University)
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lOMoARcPSD|12889239
15–1. The 12-Mg “jump jet” is capable of taking off
vertically from the deck of a ship. If its jets exert a constant
vertical force of 150 kN on the plane, determine its velocity
and how high it goes in T 6 s, starting from rest. Neglect
the loss of fuel during the lift.
150 kN
(+ c )
i
M(vY)1
฀0
'
&Y DT M(vy)2
150 103 (6)
12 103 (9.81)(6) 12 103 v
฀ v 16.14 ms 16.1 ms
(+ c )
v v0
Ans.
AC T
฀ 16.14 0
A(6)
฀ A 2.690 ms2
(+ c)
S S0
฀S 0
v0 T
0
฀ S 48.4 m
1
A T2
2 C
1
(2.690)(6)2
2
Ans.
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lOMoARcPSD|12889239
15–2.
A 0.5-kg particle is acted upon by the force
F 2U2i — (3U
3)j
(10 U2)k N, where t is in
seconds. If the particle has an initial velocity of
v0 5i
10j
20k ms, determine the magnitude
of the velocity of the particle when U 3 s.
Principle of Impulse and Momentum:
Nv1
j
(U1
U2
FEU Nv2
0.5(5i
10j
20k)
v2 41i 35j
(0
82.2 ms
5 2U2i 3U
62k ms
The magnitude of v2 is given by
v2 2 v2 Y 2
3s
v2 Z 2
3j
v2 [ 2 2 41 2
2
3 10 U 4 k 6 0.5v2
35 2
62 2
Ans.
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lOMoARcPSD|12889239
15–3. A 2.5-kg block is given an initial velocity of 3 ms
up a 45° smooth slope. Determine the time for it to travel
up the slope before it stops.
1 2.5(3)
M(2X€)1
i
'T1
2.5(9.81) N
T2
&X DT M(2X€)2
( 2.5(9.81) sin 45°)T 0
T 0.432 s
Ans.
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lOMoARcPSD|12889239
15–4.
The baseball has a horizontal speed v1 when it is struck by the bat B. If it then travels away
at an angle T from the horizontal and reaches a maximum height h, measured from the
height of the bat, determine the magnitude of the net impulse of the bat on the ball.The ball
has a mass M. Neglect the weight of the ball during the time the bat strikes the ball.
Given:
M
0.4 kg
v1
35
h
50 m
T
60 deg
g
9.81
m
s
m
2
s
Solution:
Guesses
v2
20
m
s
1 N˜ s
Impx
Impy
10 N˜ s
Given
1
2
M v2 sin T
2
§ v2 ·
¨
¸
¨ Impx ¸
¨ Imp ¸
© y¹
Mgh
M v1 Impx
Find v2 Impx Impy
v2
M v2 cos T
36.2
m
s
0 Impy
M v2 sin T
§ Impx ·
¨
¸
© Impy ¹
§ 21.2 ·
¨
¸ N˜ s
© 12.5 ¹
§ Imp.x ·
¨
¸
© Imp.y ¹
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24.7 N˜ s Ans.
lOMoARcPSD|12889239
15–5.
A man hits the 50-g golf ball such that it leaves the tee at an
angle of 40° with the horizontal and strikes the ground at
the same elevation a distance of 20 m away. Determine the
impulse of the club C on the ball. Neglect the impulse
caused by the ball’s weight while the club is striking the ball.
v2
C
SOLUTION
+ )
(:
sx = (s0)x + (v0)x t
20 = 0 + v cos 40°(t)
(+ c)
s = s0 + v0t +
1 2
at
2 c
0 = 0 + v sin 40°(t) -
1
(9.81)t2
2
t = 1.85 s
v = 14.115 m>s
( +Q)
mv1 + ©
0 +
L
L
L
F dt = mv2
F dt = 10.052114.1152
F dt = 0.706 N # s au 40°
Ans.
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40⬚
lOMoARcPSD|12889239
15–6.
A train consists of a 50-Mg engine and three cars, each
having a mass of 30 Mg. If it takes 80 s for the train to
increase its speed uniformly to 40 km>h, starting from rest,
determine the force T developed at the coupling between
the engine E and the first car A. The wheels of the engine
provide a resultant frictional tractive force F which gives
the train forward motion, whereas the car wheels roll freely.
Also, determine F acting on the engine wheels.
v
A
F
SOLUTION
(yx)2 = 40 km>h = 11.11 m>s
Entire train:
+ )
(:
m(vx)1 + ©
L
Fx dt = m(vx)2
0 + F(80) = [50 + 3(30)] A 103 B (11.11)
F = 19.4 kN
Ans.
Three cars:
+ )
(:
m(vx)1 + ©
L
Fx dt = m(vx)2
0 + T(80) = 3(30) A 103 B (11.11)
T = 12.5 kN
E
Ans.
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lOMoARcPSD|12889239
15–7.
A man kicks the ball of mass M such that it leaves the ground at angle T with the horizontal
and strikes the ground at the same elevation a distance d away. Determine the impulse of his
foot F on the ball. Neglect the impulse caused by the ball ’s weight while its being kicked.
Given:
M
200 gm
T
30 deg
d
15 m
g
9.81
m
2
s
First find the velocity vA
Solution:
vA
Guesses
1
m
s
t
1s
§ g · t2 v sin T t
¨ ¸
A
©2¹
Given
0
§t ·
¨ ¸
© vA ¹
Find t vA
vA cos T t
d
t
1.33 s
vA
I
M vA
I
13.03
m
s
Impulse - Momentum
0I
M vA
2.61 N˜ s
Ans.
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lOMoARcPSD|12889239
15–8.
If the jets exert a vertical thrust of T = (500t3>2)N, where t
is in seconds, determine the man’s speed when t = 3 s. The
total mass of the man and the jet suit is 100 kg. Neglect the
loss of mass due to the fuel consumed during the lift which
begins from rest on the ground.
T
SOLUTION
Free-Body Diagram: The thrust T must overcome the weight of the man and jet before
they move. Considering the equilibrium of the free-body diagram of the man and jet
shown in Fig. a,
+ c ©Fy = 0;
500t3>2 - 100(9.81) = 0
t = 1.567 s
Principle of Impulse and Momentum: Only the impulse generated by thrust T after
t = 1.567 s contributes to the motion. Referring to Fig. a,
(+ c)
m(v1)y + ©
Lt1
t2
Fy dt = m(v2)y
3s
100(0) +
L1.567 s
a 200t5>2 b 2
3s
500t3>2dt - 100(9.81)(3 - 1.567) = 100v
- 1405.55 = 100v
1.567 s
v = 11.0 m>s
Ans.
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lOMoARcPSD|12889239
15–9.
Under a constant thrust of T = 40 kN, the 1.5-Mg dragster
reaches its maximum speed of 125 m>s in 8 s starting from
rest. Determine the average drag resistance FD during this
period of time.
FD
SOLUTION
Principle of Impulse and Momentum: The final speed of the dragster is v2 = 125 m>s.
Referring to the free-body diagram of the dragster shown in Fig. a,
+ )
(;
t2
Fx dt = m(v2)x
Lt1
1500(0) + 40(103)(8) - (FD)avg (8) = 1500(125)
m(v1)x + ©
(FD)avg = 16 562.5 N = 16.6 kN
Ans.
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T ⫽ 40 kN
lOMoARcPSD|12889239
15–10.
P
The 50-kg crate is pulled by the constant force P. If the crate
starts from rest and achieves a speed of 10 m>s in 5 s, determine the magnitude of P. The coefficient of kinetic friction
between the crate and the ground is mk = 0.2.
30⬚
SOLUTION
Impulse and Momentum Diagram: The frictional force acting on the crate is
Ff = mkN = 0.2N.
Principle of Impulse and Momentum:
(+ c)
t2
Fy dt = m(v2)y
Lt1
0 + N(5) + P(5) sin 30° - 50(9.81)(5) = 0
m(v1)y + ©
N = 490.5 - 0.5P
+ )
(:
(1)
t2
Fx dt = m(v2)x
Lt1
50(0) + P(5) cos 30° - 0.2N(5) = 50(10)
m(v1)x + ©
4.3301P - N = 500
(2)
Solving Eqs. (1) and (2), yields
N = 387.97 N
P = 205 N
Ans.
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lOMoARcPSD|12889239
15–11.
When the 5-kg block is 6 m from the wall, it is sliding at
v1 = 14 m>s. If the coefficient of kinetic friction between
the block and the horizontal plane is mk = 0.3, determine
the impulse of the wall on the block necessary to stop the
block. Neglect the friction impulse acting on the block
during the collision.
v1
SOLUTION
Equation of Motion: The acceleration of the block must be obtained first before
one can determine the velocity of the block before it strikes the wall.
+ c ©Fy = may ;
N - 5(9.81) = 5(0)
N = 49.05 N
+
: ©Fx = max ;
-0.3(49.05) = - 5a
a = 2.943 m>s2
Kinematics: Applying the equation v2 = v20 + 2ac (s - s0) yields
+ )
(:
v2 = 142 + 2(- 2.943)(6 - 0)
v = 12.68 m>s
Principle of Linear Impulse and Momentum: Applying Eq. 15–4, we have
t2
m(vx)1 + ©
+ )
(:
Lt1
Fxdt = m(vx)2
5(12.68) - I = 5(0)
I = 63.4 N # s
Ans.
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14 m/s
6m
lOMoARcPSD|12889239
15–12.
For a short period of time, the frictional driving force acting
on the wheels of the 2.5-Mg van is FD = (600t2) N, where t
is in seconds. If the van has a speed of 20 km>h when t = 0,
determine its speed when t = 5 s.
FD
SOLUTION
Principle of Impulse and Momentum: The initial speed of the van is v1 = c20(103)
c
m
d
h
1h
d = 5.556 m>s. Referring to the free-body diagram of the van shown in Fig. a,
3600 s
+ )
(:
m(v1)x + ©
Lt1
2500(5.556) +
v2 = 15.6 m>s
t2
Fx dt = m(v2)x
L0
5s
2
600 t dt = 2500 v2
Ans.
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lOMoARcPSD|12889239
15–13.
The 2.5-Mg van is traveling with a speed of 100 km>h when
the brakes are applied and all four wheels lock. If the speed
decreases to 40 km>h in 5 s, determine the coefficient of
kinetic friction between the tires and the road.
SOLUTION
Free-Body Diagram: The free-body diagram of the van is shown in Fig. a. The frictional
force is Ff = mkN since all the wheels of the van are locked and will cause the van
to slide.
Principle of Impulse and Momentum: The initial and final speeds of the van are
m
1h
1h
m
v1 = c100(103) d c
d = 27.78 m>s and v2 = c40(103) d c
d = 11.11 m>s.
h 3600 s
h 3600 s
Referring to Fig. a,
(+ c)
t2
Fy dt = m(v2)y
Lt1
2500(0) + N(5) - 2500(9.81)(5) = 2500(0)
m(v1)y + ©
N = 24 525 N
+ )
(;
t2
Fx dt = m(v2)x
Lt1
2500(27.78) + [- mk(24525)(5)] = 2500(11.1)
m(v1)x + ©
mk = 0.340
Ans.
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lOMoARcPSD|12889239
15–14.
The force acting on a projectile having a mass m as it passes
horizontally through the barrel of the cannon is
F = C sin (pt>t¿). Determine the projectile’s velocity when
t = t¿ . If the projectile reaches the end of the barrel at this
instant, determine the length s.
s
SOLUTION
+ )
(:
m(vx)1 + ©
0 +
- Ca
L0
L
Fx dt = m(vx)2
t
C sin a
pt
b = mv
t¿
pt t
t¿
b cos a b 2 = mv
p
t¿ 0
v =
Ct¿
pt
a 1 - cos a b b
pm
t¿
When t = t¿ ,
v2 =
2C t¿
pm
Ans.
ds = v dt
L0
s
ds =
s = a
s =
L0
t
a
pt
C t¿
b a 1 - cos a b b dt
pm
t¿
t¿
pt t¿
C t¿
b ct sin a b d
pm
p
t¿
0
Ct¿ 2
pm
Ans.
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lOMoARcPSD|12889239
15–15.
The package of mass M is released from rest at A. It slides down the smooth plane which is
inclined at angle T onto the rough surface having a coefficient of kinetic friction of Pk. Determine
the total time of travel before the package stops sliding. Neglect the size of the package.
Given:
M
5 kg
T
30 deg
Pk
h
3m
g
9.81
m
2
s
0.2
Solution:
On the slope
On the flat
v1
2g h
v1
M v1 P k M g t2
0
t1 t2
t
t
7.67
m
s
t1
t2
5.47 s
v1
g sin T
v1
Pk g
Ans.
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t1
1.56 s
t2
3.91 s
lOMoARcPSD|12889239
15–16.
The twitch in a muscle of the arm develops a force which
can be measured as a function of time as shown in the
graph. If the effective contraction of the muscle lasts for a
time t0 , determine the impulse developed by the muscle.
F
F ⫽ F0 ( t )e⫺t/T
T
SOLUTION
t0
t0
t
F0 a b e t>Tdt
T
L0
I =
L
I =
F0 t0 - 1t>T2
te
dt
T L0
F dt =
I = -F0 c Te - t>T a
t0
t
+ 1b d
T
0
I = - F0 c Te - t0>T a
t0
+ 1b - T d
T
I = TF0 c 1 - e - t0>T a1 +
t0
bd
T
Ans.
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t
lOMoARcPSD|12889239
15–17.
From experiments, the time variation of the vertical force on a runner’s foot as he strikes
and pushes off the ground is shown in the graph.These results are reported for a 1-N
static load, i.e., in terms of unit weight. If a runner has weight W, determine the
approximate vertical impulse he exerts on the ground if the impulse occurs in time t5.
Units Used:
3
ms
10
s
N
Given:
W
800 N
t1
25 ms
t
t2
50 ms
t3
125 ms
t4
200 ms
t5
210 ms
F2
3.0 N
210 ms
F1
1.5 N
Solution:
Area
Imp
1
1
1
t1 F 1 F 1 t2 t1 F 1 t4 t2 t5 t4 F 1 F 2 F 1 t4 t2
2
2
2
Area
W
N
Imp
321 N˜ s
Ans.
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lOMoARcPSD|12889239
15–18.
The 40-kg slider block is moving to the right with a speed of
1.5 m>s when it is acted upon by the forces F1 and F2. If
these loadings vary in the manner shown on the graph,
determine the speed of the block at t = 6 s. Neglect friction
and the mass of the pulleys and cords.
F2
F1
F (N)
SOLUTION
40
The impulses acting on the block are equal to the areas under the graph.
30
+ )
(:
m(vx)1 + ©
L
F1
20
Fx dt = m(vx)2
10
40(1.5) + 4[(30)4 + 10(6 - 4)] - [10(2) + 20(4 - 2)
+ 40(6 - 4)] = 40v2
v 2 = 12.0 m>s ( : )
F2
0
Ans.
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2
4
6
t (s)
lOMoARcPSD|12889239
15–19.
The tennis ball has a horizontal speed v1 when it is struck by the racket. If it then travels away
at angle T from the horizontal and reaches maximum altitude h, measured from the height of the
racket, determine the magnitude of the net impulse of the racket on the ball. The ball has mass
M. Neglect the weight of the ball during the time the racket strikes the ball.
Given:
m
s
v1
15
T
25 deg
h
10 m
M
g
180 gm
9.81
m
2
s
Solution:
Free flight
v2 sin T
2g h
v2
2g h
v2
sin T
33.14
Impulse - momentum
M v1 Ix
0 Iy
I
M v2 cos T
M v2 sin T
2
2
Ix Iy
I
Ix
M v2 cos T v1
Ix
8.11 N˜ s
Iy
M v2 sin T
Iy
2.52 N s
8.49 N s
Ans.
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m
s
lOMoARcPSD|12889239
15–20.
A jet plane having a mass M takes off from an aircraft carrier such that the engine thrust
varies as shown by the graph. If the carrier is traveling forward with a speed v, determine
the plane’s airspeed after time t.
Units Used:
3
Mg
10 kg
kN
10 N
3
Given:
M
v
7 Mg
t1
2s
km
hr
t2
5s
40
F1
5 kN
F2
15 kN
t
5s
Solution:
The impulse exerted on the plane is equal to the area under the graph.
Mv v1
1
1
F 1 t1 F 1 F 2 t2 t1
2
2
v
M v1
1
ªF1 t1 F1 F2 t2 t1 º¼
2M ¬
v1
16.11
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m
s
Ans.
lOMoARcPSD|12889239
15–21.
If it takes 35 s for the 50-Mg tugboat to increase its speed
uniformly to 25 km>h, starting from rest, determine the
force of the rope on the tugboat.The propeller provides the
propulsion force F which gives the tugboat forward motion,
whereas the barge moves freely. Also, determine F acting on
the tugboat. The barge has a mass of 75 Mg.
F
SOLUTION
1000
25a
b = 6.944 m/s
3600
System:
+ )
(:
my1 + ©
L
F dt = my2
[0 + 0] + F(35) = (50 + 75)(103)(6.944)
F = 24.8 kN
Ans.
Barge:
+ )
(:
mv1 + ©
L
F dt = mv2
0 + T(35) = (75)(103)(6.944)
T = 14.881 = 14.9 kN
Ans.
Also, using this result for T,
Tugboat:
+ )
(:
mv1 + ©
L
F dt = mv2
0 + F1352 - 114.88121352 = 15021103216.9442
F = 24.8 kN
Ans.
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lOMoARcPSD|12889239
15–22.
A crate of mass M rests against a stop block s, which prevents the crate from moving
down the plane. If the coefficients of static and kinetic friction between the plane and the
crate are Ps and Pk respectively, determine the time needed for the force F to give the crate
a speed v up the plane.The force always acts parallel to the plane and has a magnitude of F
= at. Hint: First determine the time needed to overcome static friction and start the crate
moving.
Given:
m
M
50 kg
T 30 deg g 9.81
2
s
m
P s 0.3
v 2
s
a
300
N
s
Pk
0.2
Solution:
t1
Guesses
Given
1s
NC M g cos T
NC
t2
1N
1s
0
a t1 P s NC M g sin T
0
t2
´
µ
¶t
§ t1 ·
¨ ¸
¨ t2 ¸
¨N ¸
© C¹
a t M g sin T P k NC dt
Mv
1
Find t1 t2 NC
t1
1.242 s
t2
1.929 s
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Ans.
lOMoARcPSD|12889239
15–23.
The 5-kg block is moving downward at v1 = 2 m>s when it
is 8 m from the sandy surface. Determine the impulse of the
sand on the block necessary to stop its motion. Neglect the
distance the block dents into the sand and assume the block
does not rebound. Neglect the weight of the block during
the impact with the sand.
v1
8m
SOLUTION
Just before impact
T1 + ©U1 - 2 = T2
1
1
(5)(2)2 + 8(5)(9.81) = (5)(v2)
2
2
v = 12.687 m>s
1 + T2
mv1 + ©
L
F dt = mv2
5112.6872 I =
L
L
F dt = 0
F dt = 63.4 N # s
Ans.
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2 m/s
lOMoARcPSD|12889239
15–24.
The 5-kg block is falling downward at v1 = 2 m>s when it is
8 m from the sandy surface. Determine the average
impulsive force acting on the block by the sand if the
motion of the block is stopped in 0.9 s once the block strikes
the sand. Neglect the distance the block dents into the sand
and assume the block does not rebound. Neglect the weight
of the block during the impact with the sand.
v1
8m
SOLUTION
Just before impact
T1 + ©U1 - 2 = T2
1
1
(5)(2)2 + 8(5)(9.81) = (5)(v2)
2
2
v = 12.69 m>s
1 + T2
mv1 + ©
L
F dt = mv2
5(12.69) - Fang (0.9) = 0
Fang = 70.5 N
Ans.
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2 m/s
lOMoARcPSD|12889239
15–25.
The jet plane has a mass M and a horizontal velocity v0 when t = 0. If both engines
provide a horizontal thrust which varies as shown in the graph, determine the plane’s
velocity at time t1. Neglect air resistance and the loss of fuel during the motion.
Units Used:
3
Mg
10 kg
kN
10 N
3
Given:
M
250 Mg
v0
100
t1
15 s
a
200 kN
b
2
m
s
kN
2
s
Solution:
t1
´
2
M v0 µ a b t dt
¶0
v1
1
v0 M
M v1
t1
´
2
µ a b t dt
¶0
v1
121
m
s
Ans.
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lOMoARcPSD|12889239
15–26.
As indicated by the derivation, the principle of impulse and
momentum is valid for observers in any inertial reference
frame. Show that this is so, by considering the 10-kg block
which rests on the smooth surface and is subjected to a
horizontal force of 6 N. If observer A is in a fixed frame x,
determine the final speed of the block in 4 s if it has an initial
speed of 5 m>s measured from the fixed frame. Compare the
result with that obtained by an observer B, attached to the x¿
axis that moves at a constant velocity of 2 m>s relative to A.
L
6N
F dt = m v2
10(5) + 6(4) = 10v
v = 7.40 m>s
Ans.
Observer B:
+ )
(:
m v1 + a
L
F dt = m v2
10(3) + 6(4) = 10v
v = 5.40 m>s
B
x¿
5 m/s
Observer A:
m v1 + a
x
2 m/s
SOLUTION
+ )
(:
A
Ans.
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lOMoARcPSD|12889239
15–27.
The winch delivers a horizontal towing force F to its cable
at A which varies as shown in the graph. Determine the
speed of the 70-kg bucket when t = 18 s. Originally the
bucket is moving upward at v1 = 3 m>s.
A
F
F (N)
600
360
v
B
12
SOLUTION
Principle of Linear Impulse and Momentum: For the time period 12 s … t 6 18 s,
F - 360
600 - 360
, F = (20t + 120) N. Applying Eq. 15–4 to bucket B, we have
=
t - 12
24 - 12
m A vy B 1 + ©
Lt1
Fy dt = m A vy B 2
t2
18 s
(+ c)
70(3) + 2c 360(12) +
L12 s
(20t + 120)dt d - 70(9.81)(18) = 70v2
v2 = 21.8 m>s
Ans.
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24
t (s)
lOMoARcPSD|12889239
15–28.
The winch delivers a horizontal towing force F to its cable
at A which varies as shown in the graph. Determine the
speed of the 80-kg bucket when t = 24 s. Originally the
bucket is released from rest.
A
F
F (N)
600
360
v
B
12
SOLUTION
Principle of Linear Impulse and Momentum: The total impluse exerted on bucket B
can be obtained by evaluating the area under the F–t graph. Thus,
t2
1
I = ©
Fy dt = 2 c 360(12) + (360 + 600)(24 - 12) d = 20160 N # s. Applying
2
Lt1
Eq. 15–4 to the bucket B, we have
m A vy B 1 + ©
(+ c)
Lt1
Fy dt = m A vy B 2
t2
80(0) + 20160 - 80(9.81)(24) = 80v2
v2 = 16.6m>s
Ans.
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24
t (s)
lOMoARcPSD|12889239
15–29.
The train consists of a 30-Mg engine E, and cars A, B, and C,
which have a mass of 15 Mg, 10 Mg, and 8 Mg, respectively.
If the tracks provide a traction force of F = 30 kN on the
engine wheels, determine the speed of the train when
t = 30 s, starting from rest. Also, find the horizontal
coupling force at D between the engine E and car A.
Neglect rolling resistance.
C
Principle of Impulse and Momentum: By referring to the free-body diagram of the
entire train shown in Fig. a, we can write
m A v1 B x + ©
Lt1
Fxdt = m A v2 B x
t2
63 000(0) + 30(103)(30) = 63 000v
v = 14.29 m>s
Ans.
Using this result and referring to the free-body diagram of the train’s car shown in
Fig. b,
+ )
(:
m A v1 B x + ©
Lt1
Fx dt = m A v2 B x
t2
33000(0) + FD(30) = 33 000 A 14.29 B
FD = 15 714.29 N = 15.7 kN
A
E
D
SOLUTION
+ )
(:
B
Ans.
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F
30 kN
lOMoARcPSD|12889239
15–30.
The crate B and cylinder A have a mass of 200 kg and 75 kg,
respectively. If the system is released from rest, determine
the speed of the crate and cylinder when t = 3 s. Neglect
the mass of the pulleys.
B
SOLUTION
Free-Body Diagram: The free-body diagrams of cylinder A and crate B are shown in
Figs. b and c. vA and vB must be assumed to be directed downward so that they are
consistent with the positive sense of sA and sB shown in Fig. a.
Principle of Impulse and Momentum: Referring to Fig. b,
(+ T)
m(v1)y + ©
Lt1
t2
Fy dt = m(v2)y
75(0) + 75(9.81)(3) - T(3) = 75vA
vA = 29.43 - 0.04T
(1)
From Fig. b,
(+ T)
m(v1)y + ©
Lt1
t2
Fy dt = m(v2)y
200(0) + 2500(9.81)(3) - 4T(3) = 200vB
vB = 29.43 - 0.06T
(2)
Kinematics: Expressing the length of the cable in terms of sA and sB and referring
to Fig. a,
sA + 4sB = l
(3)
Taking the time derivative,
vA + 4vB = 0
(4)
Solving Eqs. (1), (2), and (4) yields
vB = - 2.102 m>s = 2.10 m>s c
vA = 8.409 m>s = 8.41 m>s T Ans.
T = 525.54 N
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A
lOMoARcPSD|12889239
15–31.
The block of mass M is moving downward at speed v1 when it is a distance h from the sandy
surface. Determine the impulse of the sand on the block necessary to stop its motion. Neglect
the distance the block dents into the sand and assume the block does not rebound. Neglect the
weight of the block during the impact with the sand.
Given:
M
6 kg
v1
3
h
8m
g
9.81
m
s
m
2
s
Solution:
Just before impact
Collision
M v2 I
v2
0
I
2
v1 2g h
M v2
m
s
v2
12.88
I
77.3 N˜ s
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Ans.
lOMoARcPSD|12889239
15–32.
The block of mass M is falling downward at speed v1 when it is a distance h from the sandy
surface. Determine the average impulsive force acting on the block by the sand if the motion of
the block is stopped in time ' t once the block strikes the sand. Neglect the distance the block
dents into the sand and assume the block does not rebound. Neglect the weight of the block
during the impact with the sand.
Given:
M
6 kg
v1
3
't
0.9 s
m
s
h
8m
g
9.81
m
2
s
Solution:
Just before impact
Collision
M v2 F 't
v2
0
F
2
v1 2g h
M v2
't
v2
12.88
F
64.4 N
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m
s
Ans.
lOMoARcPSD|12889239
15–33.
The log has a mass of 500 kg and rests on the ground for
which the coefficients of static and kinetic friction are
ms = 0.5 and mk = 0.4, respectively. The winch delivers a
horizontal towing force T to its cable at A which varies as
shown in the graph. Determine the speed of the log when
t = 5 s. Originally the tension in the cable is zero. Hint: First
determine the force needed to begin moving the log.
T (N)
1800
T
200 t2
3
t (s)
A T
SOLUTION
+
: ©Fx = 0;
F - 0.5(500)(9.81) = 0
F = 2452.5 N
Thus,
2T = F
2(200t 2) = 2452.5
t = 2.476 s to start log moving
+ )
(:
m v1 + ©
L
F dt = m v2
3
0 + 2
L2.476
200t 2 dt + 211800215 - 32 - 0.41500219.81215 - 2.4762 = 500v2
t3 3
400( ) `
+ 2247.91 = 500v2
3 2.476
v2 = 7.65 m>s
Ans.
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lOMoARcPSD|12889239
15–34.
The 50-kg block is hoisted up the incline using the cable and
motor arrangement shown. The coefficient of kinetic
friction between the block and the surface is mk = 0.4. If the
block is initially moving up the plane at v0 = 2 m>s, and at
this instant (t = 0) the motor develops a tension in the cord
of T = (300 + 120 2 t) N, where t is in seconds, determine
the velocity of the block when t = 2 s.
v0
30
SOLUTION
+a©Fx = 0;
( +Q)
NB - 50(9.81)cos 30° = 0
m(vx)1 + ©
L
NB = 424.79 N
Fx dt = m(vx)2
A 300 + 120 2t B dt - 0.4(424.79)(2)
L0
- 50(9.81)sin 30°(2) = 50v2
2
50(2) +
v2 = 1.92 m>s
2 m/s
Ans.
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lOMoARcPSD|12889239
15–35. A railroad car having a mass of 15 Mg is coasting at
1.5 ms on a horizontal track. At the same time another car
having a mass of 12 Mg is coasting at 0.75 ms in the
opposite direction. If the cars meet and couple together,
determine the speed of both cars just after the coupling.
Find the difference between the total kinetic energy before
and after coupling has occurred, and explain qualitatively
what happened to this energy.
+ )
(:
iMv1 iMv2
15 000(1.5)
12 000(0.75) 27 000(v2)
v2 0.5 ms
Ans.
1
(12 000)(0.75)2 20.25 kJ
2
41 1
(15 000)(1.5)2
2
42 1
(27 000)(0.5)2 3.375 kJ
2
b4 41
20.25
42
3.375 16.9 kJ
This energy is dissipated as noise, shock, and heat during the coupling.
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Ans.
lOMoARcPSD|12889239
15–36.
The 50-kg boy jumps on the 5-kg skateboard with a horizontal velocity of 5 m>s. Determine the distance s the boy
reaches up the inclined plane before momentarily coming
to rest. Neglect the skateboard’s rolling resistance.
s
30⬚
SOLUTION
Free-Body Diagram: The free-body diagram of the boy and skateboard system is
shown in Fig. a. Here, Wb,Wsb, and N are nonimpulsive forces. The pair of impulsive
forces F resulting from the impact during landing cancel each other out since they are
internal to the system.
Conservation of Linear Momentum: Since the resultant of the impulsive force along
the x axis is zero, the linear momentum of the system is conserved along the x axis.
+ )
(;
mb(vb)1 + msb(vsb)1 = (mb + msb)v
50(5) + 5(0) = (50 + 5)v
v = 4.545 m>s
Conservation of Energy: With reference to the datum set in Fig. b, the
gravitational potential energy of the boy and skateboard at positions A and B are
A Vg B A = (mb + msb)ghA = 0 and A Vg B B = (mb + msb)ghB = (50 + 5)(9.81)(s sin 30°)
= 269.775s.
TA + VA = TB + VB
1
1
(mb + msb)vA 2 + A Vg B A = (mb + msb)vB 2 + A Vg B B
2
2
1
(50 + 5) A 4.5452 B + 0 = 0 + 269.775s
2
s = 2.11 m
Ans.
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lOMoARcPSD|12889239
15–37.
The 2.5-Mg pickup truck is towing the 1.5-Mg car using a
cable as shown. If the car is initially at rest and the truck is
coasting with a velocity of 30 km>h when the cable is slack,
determine the common velocity of the truck and the car just
after the cable becomes taut. Also, find the loss of energy.
30 km/h
SOLUTION
Free-Body Diagram: The free-body diagram of the truck and car system is shown in
Fig. a. Here, Wt,WC, Nt, and NC are nonimpulsive forces. The pair of impulsive forces
F generated at the instant the cable becomes taut are internal to the system and thus
cancel each other out.
Conservation of Linear Momentum: Since the resultant of the impulsive force is
zero, the linear momentum of the system is conserved along the x axis. The initial
m
1h
speed of the truck is A vt B 1 = c 30(103) d c
d = 8.333 m>s.
h 3600 s
+ )
(;
mt A vt B 1 + mC A vC B 1 = A mt + mC B v2
2500(8.333) + 0 = (2500 + 1500)v2
v2 = 5.208 m>s = 5.21 m>s ;
Ans.
Kinetic Energy: The initial and final kinetic energy of the system is
T1 =
=
1
1
m (v ) 2 + mC(vC)12
2 t t1
2
1
(2500)(8.3332) + 0
2
= 86 805.56 J
and
T2 = (mt + mC)v22
=
1
(2500 + 1500)(5.2082)
2
= 54 253.47
Thus, the loss of energy during the impact is
¢E = T1 - T2 = 86 805.56 - 54 253.47 = 32.55(103) J = 32.6 kJ
Ans.
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lOMoARcPSD|12889239
15–38.
A railroad car having a mass of 15 Mg is coasting at 1.5 m>s
on a horizontal track. At the same time another car having a
mass of 12 Mg is coasting at 0.75 m>s in the same
direction. If the cars meet and couple together, determine
the speed of both cars just after the coupling. Find the
difference between the total kinetic energy before and after
coupling has occurred, and explain qualitatively what
happened to this energy.
SOLUTION
+ )
(:
©mv1 = ©mv2
15 000(1.5) + 12 000(0.75) = 27 000(v2)
v2 = 1.167 m>s
Ans.
T1 =
1
1
(15 000)(1.5)2 + (12 000)(0.75)2 = 20.25 kJ
2
2
T2 =
1
(27 000)(1.167)2 = 18.39 kJ
2
¢T = T1 - T2
= 20.25 - 18.39 = 1.86 kJ
Ans.
This energy is dissipated as noise, shock, and heat during the coupling.
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lOMoARcPSD|12889239
15–39.
The winch on the back of the jeep A is turned on and pulls in the tow rope at speed vrel.
If both the car B of mass MB and the jeep A of mass MA are free to roll, determine their
velocities at the instant they meet. If the rope is of length L, how long will this take?
Units Used:
Mg
3
10 kg
Given:
MA
2.5 Mg
MB
1.25 Mg
vrel
2
L
m
s
5m
Solution:
m
s
vA
Given
0
MA vA MB vB
vA vB
t
L
vrel
t
1
vB
m
s
Guess
1
2.5 s
vrel
§ v.A ·
¨ ¸ Find v.A v.B
© v.B ¹
§ vA ·
¨ ¸
© vB ¹
Downloaded by ?? ? (rlaalswo2559@gmail.com)
§ 0.667 · m
¨
¸
© 1.333 ¹ s
Ans.
lOMoARcPSD|12889239
15–40.
The 200-g projectile is fired with a velocity of 900 m>s
towards the center of the 15-kg wooden block, which rests
on a rough surface. If the projectile penetrates and
emerges from the block with a velocity of 300 m>s, determine the velocity of the block just after the projectile
emerges. How long does the block slide on the rough surface, after the projectile emerges, before it comes to rest
again? The coefficient of kinetic friction between the surface and the block is mk = 0.2.
900 m/s
Before
300 m/s
SOLUTION
Free-Body Diagram: The free-body diagram of the projectile and block system is
shown in Fig. a. Here, WB, WP, N, and Ff are nonimpulsive forces. The pair of impulsive forces F resulting from the impact cancel each other out since they are internal to
the system.
Conservation of Linear Momentum: Since the resultant of the impulsive force along
the x axis is zero, the linear momentum of the system is conserved along the x axis.
+ )
(:
mP A vP B 1 + mB A vB B 1 = mP A vP B 2 + mB A vB B 2
0.2(900) + 15(0) = 0.2(300) + 15(vB)2
(vB)2 = 8 m>s :
Ans.
Principle of Impulse and Momentum: Using the result of A vB B 2, and referring to the
free-body diagram of the block shown in Fig. b,
+ )
(:
m A v1 B y + ©
Lt1
t2
Fy dt = m(v2)y
15(0) + N(t) - 15(9.81)(t) = 15(0)
N = 147.15 N
+ )
(:
m A v1 B x + ©
Lt1
t2
Fxdt = m(v2)x
15(8) + [- 0.2(147.15)(t)] = 15(0)
t = 4.077 s = 4.08 s
Ans.
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After
lOMoARcPSD|12889239
15–41.
The block has a mass of 50 kg and rests on the surface of the
cart having a mass of 75 kg. If the spring which is attached to
the cart and not the block is compressed 0.2 m and the system
is released from rest, determine the speed of the block
relative to the ground after the spring becomes undeformed.
Neglect the mass of the cart’s wheels and the spring in the
calculation. Also neglect friction. Take k = 300 N>m.
B
C
SOLUTION
T1 + V1 = T2 + V2
(0 + 0) +
1
1
1
(300)(0.2)2 = (50)(vb)2 + (75)(vc)2
2
2
2
12 = 50 v2b + 75 v2c
+ )
(:
©mv1 = ©mv2
0 + 0 = 50 vb - 75 vc
vb = 1.5vc
vc = 0.253 m>s ;
vb = 0.379 m s :
Ans.
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lOMoARcPSD|12889239
15–42.
The block has a mass of 50 kg and rests on the surface of
the cart having a mass of 75 kg. If the spring which is
attached to the cart and not the block is compressed 0.2 m
and the system is released from rest, determine the speed
of the block with respect to the cart after the spring becomes
undeformed. Neglect the mass of the wheels and the spring
in the calculation. Also neglect friction. Take k = 300 N>m.
B
C
SOLUTION
T1 + V1 = T2 + V2
(0 + 0) +
1
1
1
(300)(0.2)2 = (50)(vb)2 + (75)(vc)2
2
2
2
12 = = 50 v2b + 75 v2c
+ )
(:
©mv1 = ©mv2
0 + 0 = 50 vb - 75 vc
vb = 1.5vc
vc = 0.253 m>s ;
vb = 0.379 m>s :
vb = vc + vb>c
+ )
(:
0.379 = - 0.253 + vb>c
vb c = 0.632 m s :
Ans.
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lOMoARcPSD|12889239
15–43.
The three freight cars A, B, and C have masses of 10 Mg, 5 Mg,
and 20 Mg, respectively. They are traveling along the track
with the velocities shown. Car A collides with car B first, followed by car C. If the three cars couple together after collision, determine the common velocity of the cars after the two
collisions have taken place.
20 km/h
A
SOLUTION
Free-Body Diagram: The free-body diagram of the system of cars A and B when they
collide is shown in Fig. a. The pair of impulsive forces F1 generated during the collision cancel each other since they are internal to the system. The free-body diagram
of the coupled system composed of cars A and B and car C when they collide is
shown in Fig. b. Again, the internal pair of impulsive forces F2 generated during the
collision cancel each other.
Conservation of Linear Momentum: When A collides with B, and then the coupled
cars A and B collide with car C, the resultant impulsive force along the x axis is zero.
Thus, the linear momentum of the system is conserved along the x axis. The initial
speed of the cars A, B, and C are
A vA B 1 = c 20(103)
A vB B 1 = c 5(103)
m
1h
da
b = 5.556 m>s
h 3600 s
1h
m
da
b = 1.389 m>s,
h 3600 s
and A vC B 1 = c 25(103)
m
1h
da
b = 6.944 m>s
h 3600 s
For the first case,
+ )
(:
mA(vA)1 + mB(vB)1 = (mA + mB)v2
10000(5.556) + 5000(1.389) = (10000 + 5000)vAB
vAB = 4.167 m>s :
Using the result of vAB and considering the second case,
+ )
(:
(mA + mB)vAB + mC(vC)1 = (mA + mB + mC)vABC
(10000 + 5000)(4.167) + [- 20000(6.944)] = (10000 + 5000 + 20000)vABC
vABC = - 2.183 m>s = 2.18 m>s ;
Ans.
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5 km/h
B
25 km/h
C
lOMoARcPSD|12889239
15–44.
The cart has mass M and rolls freely down the slope. When it reaches the bottom, a
spring loaded gun fires a ball of mass M1 out the back with a horizontal velocity vbc
measured relative to the cart. Determine the final velocity of the cart.
Given:
M
3 kg
M1
0.5 kg
vbc
0.6
h
1.25 m
g
9.81
m
s
m
2
s
Solution:
v1
2g h
M M1 v1
vc
v1 M vc M1 vc vbc
§ M1 ·
¨
¸ vbc
© M M1 ¹
vc
5.04
m
s
Ans.
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lOMoARcPSD|12889239
15–45.
The block of mass m is traveling at v1 in the direction u1
shown at the top of the smooth slope. Determine its speed
v2 and its direction u2 when it reaches the bottom.
z
v1
x
y
u1
h
SOLUTION
There are no impulses in the v direction:
v2
mv1 sin u1 = mv2 sin u2
u2
T1 + V1 = T2 + V2
1
1
mv12 + mgh = mv22 + 0
2
2
v2 = 3v21 + 2gh
sin u2 =
Ans.
3v21 + 2gh
v1 sin u1
u2 = sin - 1 £
3v21 + 2gh
v1 sin u1
≥
Ans.
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lOMoARcPSD|12889239
15–46.
The boy B jumps off the canoe at A with a velocity vBA relative to the canoe as shown. If
he lands in the second canoe C, determine the final speed of both canoes after the motion.
Each canoe has a mass Mc. The boy’s mass is MB, and the girl D has a mass MD. Both
canoes are originally at rest.
Given:
Mc
40 kg
MB
30 kg
MD
25 kg
vBA
5
T
m
s
30 deg
Solution:
vA
Guesses
Given
0
1
m
s
vC
m
s
Mc vA MB vA vBA cos T
MB vA vBA cos T
§ vA ·
¨ ¸
© vC ¹
1
Find vA vC
Mc MB MD vC
§ vA ·
¨ ¸
© vC ¹
§ 1.856 · m
¨
¸
© 0.781 ¹ s
Ans.
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lOMoARcPSD|12889239
15–47.
20 km/h
The 30-Mg freight car A and 15-Mg freight car B are moving
towards each other with the velocities shown. Determine the
maximum compression of the spring mounted on car A.
Neglect rolling resistance.
A
SOLUTION
Conservation of Linear Momentum: Referring to the free-body diagram of the freight
cars A and B shown in Fig. a, notice that the linear momentum of the system is conserved along the x axis. The initial speed of freight cars A and B are
m
1h
1h
m
(vA)1 = c 20(103) d a
b = 5.556 m>s and (vB)1 = c10(103) d a
b
h 3600 s
h 3600 s
= 2.778 m>s. At this instant, the spring is compressed to its maximum, and no relative
motion occurs between freight cars A and B and they move with a common speed.
+ )
(:
mA(vA)1 + mB(vB)1 = (mA + mB)v2
30(103)(5.556) + c -15(103)(2.778) d = c30(103) + 15(103) dv2
v2 = 2.778 m>s :
Conservation of Energy: The initial and final elastic potential energy of the spring
1
1
1
is (Ve)1 = ks12 = 0 and (Ve)2 = ks22 = (3)(106)smax2 = 1.5(106)smax2.
2
2
2
©T1 + ©V1 = ©T2 + ©V2
1
1
1
c mA(vA)12 + mB(vB)12 d + (Ve)1 = (mA + mB)v22 + (Ve)2
2
2
2
1
1
(30) A 103 B (5.5562) + (15) A 103 B A 2.7782 B + 0
2
2
=
1
c 30 A 103 B + 15 A 103 B d A 2.7782 B + 1.5 A 106 B smax2
2
smax = 0.4811 m = 481 mm
Ans.
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10 km/h
k ⫽ 3 MN/m
B
lOMoARcPSD|12889239
15–48.
Blocks A and B have masses mA and mB respectively. They are placed on a smooth surface and
the spring connected between them is stretched a distance d. If they are released from rest,
determine the speeds of both blocks the instant the spring becomes unstretched.
Given:
mA
40 kg
mB
60 kg
d
2m
k
180
Solution:
N
m
Guesses
vA
1
m
s
vB
1
m
s
Given
mA vA mB vB
momentum
0
energy
1 2
kd
2
§ vA ·
¨ ¸
© vB ¹
Find vA vB
1
2 1
2
mA vA mB vB
2
2
§ vA ·
¨ ¸
© vB ¹
§ 3.29 · m
¨
¸
© 2.19 ¹ s
Ans.
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lOMoARcPSD|12889239
15–49. A 4-kg projectile travels with a horizontal
velocity of 600 ms before it explodes and breaks into two
fragments A and B of mass 1.5 kg and 2.5 kg, respectively. If
the fragments travel along the parabolic trajectories shown,
determine the magnitude of velocity of each fragment just
after the explosion and the horizontal distance E" where
segment A strikes the ground at C.
vA
600 m/s
45⬚
A
C
Conservation of Linear Momentum: By referring to the free-body diagram of the
projectile just after the explosion shown in Fig. a, we notice that the pair of
impulsive forces F generated during the explosion cancel each other since they are
internal to the system. Here, WA and WB are non-impulsive forces. Since the
resultant impulsive force along the x and y axes is zero, the linear momentum of the
system is conserved along these two axes.
NvY N" vA Y
N# vB x
4(600) 1.5v" cos 45°
2.5v# cos 30°
2.165v# 1.061v" 2400
(+ c)
NvZ N" vA Z
(1)
N# vB y
0 1.5v" sin 45° 2.5v# sin 30°
v# 0.8485vA
(2)
Solving Eqs. (1) and (2) yields
v" 3090.96 ms 3.09(103) ms
Ans.
v# 2622.77 ms 2.62(103) ms
Ans.
By considering the x and y motion of segment A,
(+ c)
TZ T0 Z
v0 Z U
60 0
1
BZ U2
2
1
9.81 U"$ 2
2
3090.96 sin 45° U"$
4.905U"$ 2 2185.64U"$ 60 0
Solving for the positive root of this equation,
U"$ 445.62 s
and
+ )
(;
TY T0 Y
E" 0
v0 Y U
3090.96 cos 45° 445.62 973.96 3103 4 m 974 km
30⬚
vB
60 m
D
dA
+ )
(:
B
Ans.
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dB
lOMoARcPSD|12889239
15–50. A 4-kg projectile travels with a horizontal
velocity of 600 ms before it explodes and breaks into two
fragments A and B of mass 1.5 kg and 2.5 kg, respectively. If
the fragments travel along the parabolic trajectories shown,
determine the magnitude of velocity of each fragment just
after the explosion and the horizontal distance E# where
segment B strikes the ground at D.
vA
600 m/s
45⬚
A
60 m
Conservation of Linear Momentum: By referring to the free-body diagram of the
projectile just after the explosion shown in Fig. a, we notice that the pair of
impulsive forces F generated during the explosion cancel each other since they are
internal to the system. Here, WA and WB are non-impulsive forces. Since the
resultant impulsive force along the x and y axes is zero, the linear momentum of the
system is conserved along these two axes.
NvY N" vA Y
N# v# Y
4(600) 1.5v" cos 45°
2.5v# cos 30°
2.165v# 1.061v" 2400
(+ c )
NvZ N" vA Z
(1)
N# v# Z
0 1.5v" sin 45° 2.5v# sin 30°
v# 0.8485vA
(2)
Solving Eqs. (1) and (2) yields
v" 3090.96 ms 3.09(103) ms
Ans.
v# 2622.77 ms 2.62(103) ms
Ans.
By considering the x and y motion of segment B,
(+ c )
TZ T0 Z
v0 Z U
1
BZ U2
2
60 0 2622.77 sin 30° U#%
4.905U#% 2
1
9.81 U#% 2
2
1311.38U#% 60 0
Solving for the positive root of the above equation,
U#% 0.04574 s
and
+ )
(:
TY T0 Y
E# 0
v0 Y U
2622.77 cos 30° 0.04574 103.91 m 104 m
30⬚
vB
C
D
dA
+ )
(:
B
Ans.
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dB
lOMoARcPSD|12889239
15–51.
1.5 m/s
The 20-kg package has a speed of 1.5 m>s when it is delivered to the smooth ramp. After sliding down the ramp it
lands onto a 10-kg cart as shown. Determine the speed of the
cart and package after the package stops sliding on the cart.
2m
SOLUTION
Conservation of Energy: With reference to the datum set in Fig. a, the gravitational
potential energy of the package at positions (1) and (2) are
A Vg B 1 = mgh1 = 20(9.81)(2) = 392.4 J and A Vg B 2 = mgh2 = 0.
T1 + V1 = T2 + V2
1
1
m(vP)12 + (Vg)1 = m(vP)22 + (Vg)2
2
2
1
1
(20)(1.52) + 392.4 = (20)(vP)22 + 0
2
2
(vP)2 2 = 6.441 m>s ;
Conservation of Linear Momentum: Referring to the free-body diagram of the
package and cart system shown in Fig. b, the linear momentum is conserved along
the x axis since no impulsive force acts along it. The package stops sliding on the cart
when they move with a common speed. At this instant,
+ )
(;
mP(vP)2 + mC(vC)1 = (mP + mC)v
20(6.441) + 0 = (20 + 10)v
v = 4.29 m>s ;
Ans.
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lOMoARcPSD|12889239
15–52.
The free-rolling ramp has a mass of 40 kg. A 10-kg crate is
released from rest at A and slides down 3.5 m to point B. If
the surface of the ramp is smooth, determine the ramp’s
speed when the crate reaches B. Also, what is the velocity of
the crate?
3.5 m
A
30
SOLUTION
Conservation of Energy: The datum is set at lowest point B. When the crate is at
point A, it is 3.5 sin 30° = 1.75 m above the datum. Its gravitational potential energy
is 1019.81211.752 = 171.675 N # m. Applying Eq. 14–21, we have
T1 + V1 = T2 + V2
0 + 171.675 =
1
1
1102v2C + 1402v2R
2
2
171.675 = 5 v2C + 20 v2R
(1)
Relative Velocity: The velocity of the crate is given by
vC = vR +vC>R
= -vRi + 1vC>R cos 30°i - vC>R sin 30°j2
= 10.8660 vC>R - vR2i - 0.5 vC>Rj
(2)
The magnitude of vC is
vC = 2(0.8660 vC>R - vR22 + 1- 0.5 vC>R22
= 2v2C>R + v2R - 1.732 vR vC>R
(3)
Conservation of Linear Momentum: If we consider the crate and the ramp as a
system, from the FBD, one realizes that the normal reaction NC (impulsive force) is
internal to the system and will cancel each other. As the result, the linear momentum
is conserved along the x axis.
0 = mC 1vC2x + mR vR
+ )
(:
0 = 1010.8660 vC>R - vR2 + 401 -vR2
0 = 8.660 vC>R - 50 vR
(4)
Solving Eqs. (1), (3), and (4) yields
vR = 1.101 m>s = 1.10 m>s vC = 5.43 m>s
Ans.
vC>R = 6.356 m>s
From Eq. (2)
vC = 30.866016.3562 - 1.1014i - 0.516.3562j = 54.403i - 3.178j6 m>s
Thus, the directional angle f of vC is
f = tan - 1
3.178
= 35.8°
4.403
cf
Ans.
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B
lOMoARcPSD|12889239
15–53. Block A has a mass of 2 kg and slides into an open
ended box B with a velocity of 2 ms. If the box B has a
mass of 3 kg and rests on top of a plate P that has a mass of
3 kg, determine the distance the plate moves after it stops
sliding on the floor. Also, how long is it after impact before
all motion ceases? The coefficient of kinetic friction
between the box and the plate is mL 0.2, and between the
plate and the floor mL 0.4. Also, the coefficient of static
friction between the plate and the floor is mT 0.5.
2 m/s
A
Equations of Equilibrium: From FBD(a).
D j'Z 0;
/# (3
2)(9.81) 0
/# 49.05 N
When box B slides on top of plate P, ('G)# mL/# 0.2(49.05) 9.81 N. From
FBD(b).
D j'Z 0;
/1 49.05 3(9.81) 0
j'Y 0;
9.81 ('G)1 0
/1 78.48 N
('G)1 9.81 N
Since ('G)1 ('G)1 max mT /1 0.5(78.48) 39.24 N, plate P does not move.
Thus
T1 0
Ans.
Conservation of Linear Momentum: If we consider the block and the box as a system,
then the impulsive force caused by the impact is internal to the system. Therefore, it
will cancel out. As the result, linear momentum is conserved along the x axis.
N" (y")1
+ )
(:
N3 (y3)1 (N"
2(2)
0 (2
N3) y2
3) y2
y2 0.800 ms Principle of Linear Impulse and Momentum: Applying Eq. 15–4, we have
N(yY)1
+ )
(:
5(0.8)
j
(U1
U2
'Y EU N(yY)2
9.81(U) 5(0)
U 0.408 s
Ans.
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B
P
lOMoARcPSD|12889239
Block A has a mass of 2 kg and slides into an open
15–54.
ended box B with a velocity of 2 ms. If the box B has a
mass of 3 kg and rests on top of a plate P that has a mass of
3 kg, determine the distance the plate moves after it stops
sliding on the floor. Also, how long is it after impact before
all motion ceases? The coefficient of kinetic friction
between the box and the plate is &K 0.2, and between the
plate and the floor &€K 0.1. Also, the coefficient of static
friction between the plate and the floor is &€S 0.12.
2 ms
A
Equations of Equilibrium: From FBD(a),
C i&X 0;
."
2)(9.81) 0
(3
." 49.05 N
When box B slides on top of plate P. (&F)" &K ." 0.2(49.05) 9.81 N. From
FBD(b).
C i&Y 0;
.0
i&X 0;
49.05
3(9.81) 0
(&F)0 0
9.81
.0 78.48 N
(&F)0 9.81 N
Since (&F)0 (&F)0 max &S € .0 0.12(78.48) 9.418N, plate P slides. Thus,
(&F)0 &K € .0 0.1(78.48) 7.848 N.
Conservation of Linear Momentum: If we consider the block and the box as a
system, then the impulsive force caused by the impact is internal to the system.
Therefore, it will cancel out. As the result, linear momentum is conserved along
x axis.
M2 (22)1 (M!
M! (2!)1
+ )
(:
0 (2
2(2)
M2) 22
3) 22
22 0.800 ms Principle of Linear Impulse and Momentum: In order for box B to stop sliding on
plate P, both box B and plate P must have same speed 23. Applying Eq. 15–4 to box
B (FBD(c)], we have
i
M(2X)1
+ )
(:
'T1
T2
&X DT M(2X)2
9.81(T1) 523
5(0.8)
[1]
Applying Eq. 15–4 to plate P[FBD(d)], we have
i
M(2X)1
+ )
(:
3(0)
'T1
9.81(T1)
T2
&X DT M(2X)2
7.848(T1) 323
[2]
Solving Eqs. [1] and [2] yields
T1 0.3058 s
23 0.200 ms
Equation of Motion: From FBD(d), the acceleration of plate P when box B still
slides on top of it is given by
i&X MAX ;
9.81
7.848 3(A0)1
(A0)1 0.654 ms2
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B
P
lOMoARcPSD|12889239
15–54.
Continued
When box B stop slid ling on top of box B, (&F)" 0. From this instant onward
plate P and box B act as a unit and slide together. From FBD(d), the acceleration of
plate P and box B is given by
i&X MAX ;
7.848 8(A0)2
(A0)2 0.981 ms2
Kinematics: Plate P travels a distance s1 before box B stop sliding.
+ )
(:
฀ S1 (20)0 T1
0
1
(A ) T21
2 01
1
(0.654) 0.30582 0.03058 m
2
The time t2 for plate P to stop after box B stop slidding is given by
+ )
(:
24 23
0 0.200
(A0)2 T2
T2 0.2039 s
( 0.981)T2
The distance s2 traveled by plate P after box B stop sliding is given by
+ )
(:
224 223
0 0.2002
2(A0)2 S2
2( 0.981)S2
S2 0.02039 m
The total distance travel by plate P is
S0 S1
S2 0.03058
0.02039 0.05097 m 51.0 mm
Ans.
The total time taken to cease all the motion is
TTot T1
T2 0.3058
0.2039 0.510 s
Ans.
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lOMoARcPSD|12889239
15–55. The 75-kg boy leaps off cart A with a horizontal
velocity of W 3 ms measured relative to the cart.
Determine the velocity of cart A just after the jump. If he
then lands on cart B with the same velocity that he left cart
A, determine the velocity of cart B just after he lands on it.
Carts A and B have the same mass of 50 kg and are
originally at rest.
v¿
B
Free-Body Diagram: The free-body diagram of the man and cart system when the
man leaps off and lands on the cart are shown in Figs. a and b, respectively. The pair
of impulsive forces F1 and F2 generated during the leap and landing are internal to
the system and thus cancel each other.
Kinematics: Applying the relative velocity equation, the relation between the
velocity of the man and cart A just after leaping can be determined.
vN v"
+ )
(;
vN"
vN 2 v" 2
3
(1)
Conservation of Linear Momentum: Since the resultant of the impulse forces along
the x axis is zero, the linear momentum of the system is conserved along the x axis
for both cases. When the man leaps off cart A,
+ )
(;
NN vN 1
0
N" vA 1 NN vm 2
0 75 vm 2
50 vA 2
N" vA 2
vm 2 0.6667 vA 2
Solving Eqs. (1) and (2) yields
vA 2 1.80 ms 1.80 ms vN 2 1.20 ms Using the result of vm 2 and considering the man’s landing on cart B,
+ )
(;
NN vm 2
75(1.20)
N# vB 1 NN
0 75
v 0.720 ms 50 v
N# v
Ans.
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A
lOMoARcPSD|12889239
15–56.
Two boxes A and B, each having a mass of 80 kg, sit on
the 250-kg conveyor which is free to roll on the ground. If
the belt starts from rest and begins to run with a speed of
1 m>s, determine the final speed of the conveyor if (a) the
boxes are not stacked and A falls off then B falls off, and (b)
A is stacked on top of B and both fall off together.
B
SOLUTION
a) Let vb be the velocity of A and B.
+ )
(:
©mv1 = ©mv2
0 = (160) (vb ) - (250) (vc)
+ )
(:
vb = vc + vb>c
vb = -vc + 1
Thus, vb = 0.610 m>s :
vc = 0.390 m>s ;
When a box falls off, it exerts no impulse on the conveyor, and so does not alter the
momentum of the conveyor. Thus,
a) vc = 0.390 m>s ;
Ans.
b) vc = 0.390 m>s ;
Ans.
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A
lOMoARcPSD|12889239
15–57.
The 10-kg block is held at rest on the smooth inclined plane
by the stop block at A. If the 10-g bullet is traveling at
300 m>s when it becomes embedded in the 10-kg block,
determine the distance the block will slide up along the
plane before momentarily stopping.
300 m/s
A
30
SOLUTION
Conservation of Linear Momentum: If we consider the block and the bullet as a
system, then from the FBD, the impulsive force F caused by the impact is internal
to the system. Therefore, it will cancel out. Also, the weight of the bullet and the
block are nonimpulsive forces. As the result, linear momentum is conserved along
the x œ axis.
mb(vb)x¿ = (mb + mB) vx œ
0.01(300 cos 30°) = (0.01 + 10) v
v = 0.2595 m>s
Conservation of Energy: The datum is set at the blocks initial position. When the
block and the embedded bullet is at their highest point they are h above the datum.
Their gravitational potential energy is (10 + 0.01)(9.81)h = 98.1981h. Applying
Eq. 14–21, we have
T1 + V1 = T2 + V2
0 +
1
(10 + 0.01) A 0.25952 B = 0 + 98.1981h
2
h = 0.003433 m = 3.43 mm
d = 3.43 > sin 30° = 6.87 mm
Ans.
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lOMoARcPSD|12889239
15–58.
A ball having a mass of 200 g is released from rest at a
height of 400 mm above a very large fixed metal surface. If
the ball rebounds to a height of 325 mm above the surface,
determine the coefficient of restitution between the ball
and the surface.
SOLUTION
Before impact
T1 + V1 = T2 + V2
0 + 0.2(9.81)(0.4) =
1
(0.2)v21 + 0
2
v1 = 2.801 m>s
After the impact
1
(0.2)v22 = 0 + 0.2(9.81)(0.325)
2
v2 = 2.525 m>s
Coefficient of restitution:
(+ T)
(vA)2 - (vB)2
(vB)1 - (vA)1
0 - ( -2.525)
=
2.801 - 0
e =
= 0.901
Ans.
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lOMoARcPSD|12889239
15–59.
The 5-Mg truck and 2-Mg car are traveling with the freerolling velocities shown just before they collide. After the
collision, the car moves with a velocity of 15 km>h to the
right relative to the truck. Determine the coefficient of
restitution between the truck and car and the loss of energy
due to the collision.
30 km/h
10 km/h
SOLUTION
Conservation of Linear Momentum: The linear momentum of the system is
conserved along the x axis (line of impact).
The initial speeds of the truck and car are (vt)1 = c 30 A 103 B
and (vc)1 = c10 A 103 B
1h
m
da
b = 8.333 m>s
h 3600 s
m
1h
da
b = 2.778 m>s.
h 3600 s
By referring to Fig. a,
+ )
(:
mt A vt B 1 + mc A vc B 1 = mt A vt B 2 + mc A vc B 2
5000(8.333) + 2000(2.778) = 5000 A vt B 2 + 2000 A vc B 2
5 A vt B 2 + 2 A vc B 2 = 47.22
(1)
Coefficient of Restitution: Here, (vc>t) = c15 A 103 B
1h
m
da
b = 4.167 m>s : .
h 3600 s
Applying the relative velocity equation,
(vc)2 = (vt)2 + (vc>t)2
+ )
(:
(vc)2 = (vt)2 + 4.167
(vc)2 - (vt)2 = 4.167
(2)
Applying the coefficient of restitution equation,
+ )
(:
e =
(vc)2 - (vt)2
(vt)1 - (vc)1
e =
(vc)2 - (vt)2
8.333 - 2.778
(3)
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lOMoARcPSD|12889239
15–59. continued
Substituting Eq. (2) into Eq. (3),
e =
4.167
= 0.75
8.333 - 2.778
Ans.
Solving Eqs. (1) and (2) yields
(vt)2 = 5.556 m>s
(vc)2 = 9.722 m>s
Kinetic Energy: The kinetic energy of the system just before and just after the
collision are
T1 =
=
1
1
mt(vt)1 2 + mc(vc)1 2
2
2
1
1
(5000)(8.3332) + (2000)(2.7782)
2
2
= 181.33 A 103 B J
T2 =
=
1
1
m (v ) 2 + mc(vc)2 2
2 t t2
2
1
1
(5000)(5.5562) + (2000)(9.7222)
2
2
= 171.68 A 103 B J
Thus,
¢E = T1 - T2 = 181.33 A 103 B - 171.68 A 103 B
= 9.645 A 103 B J
= 9.65 kJ
Ans.
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lOMoARcPSD|12889239
15–60.
Disk A has a mass of 2 kg and is sliding forward on the
smooth surface with a velocity 1vA21 = 5 m/s when it strikes
the 4-kg disk B, which is sliding towards A at 1vB21 = 2 m/s,
with direct central impact. If the coefficient of restitution
between the disks is e = 0.4, compute the velocities of A
and B just after collision.
SOLUTION
Conservation of Momentum :
mA (vA)1 + mB (vB)1 = mA (vA)2 + mB (vB)2
+ )
(:
2(5) + 4( - 2) = 2(vA)2 + 4(vB)2
(1)
Coefficient of Restitution :
+ )
(:
e =
(vB)2 - (vA)2
(vA)1 - (vB)1
0.4 =
(vB)2 - (vA)2
5 - ( -2)
(2)
Solving Eqs. (1) and (2) yields
(vA)2 = - 1.53 m s = 1.53 m s ;
(vB)2 = 1.27m s :
Ans.
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(vA)1 = 5 m/s
(vB)1 = 2 m/s
A
B
lOMoARcPSD|12889239
15–61.
Block A has a mass of 3 kg and is sliding on a rough horizontal
surface with a velocity 1vA21 = 2 m>s when it makes a direct
collision with block B, which has a mass of 2 kg and is
originally at rest. If the collision is perfectly elastic 1e = 12,
determine the velocity of each block just after collision and
the distance between the blocks when they stop sliding. The
coefficient of kinetic friction between the blocks and the
plane is mk = 0.3.
(vA)1
A
SOLUTION
+ )
(:
a mv1 = a mv2
3122 + 0 = 31vA22 + 21vB22
+ )
(:
e =
1vB22 - 1vA222
1 =
1vB22 - 1vA22
1vA21 - 1vB21
2 - 0
Solving
1vA22 = 0.400 m>s :
1vB22 = 2.40 m>s :
Block A:
Ans.
Ans.
T1 + a U1 - 2 = T2
1
13210.40022 - 319.81210.32dA = 0
2
dA = 0.0272 m
Block B:
T1 + a U1 - 2 = T2
1
12212.4022 - 219.81210.32dB = 0
2
dB = 0.9786 m
d = dB - dA = 0.951 m
Ans.
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B
lOMoARcPSD|12889239
15–62.
If two disks A and B have the same mass and are subjected
to direct central impact such that the collision is perfectly
elastic 1e = 12, prove that the kinetic energy before collision
equals the kinetic energy after collision. The surface upon
which they slide is smooth.
SOLUTION
+ )
(:
©m v1 = ©m v2
mA (vA)1 + mB (vB)1 = mA (vA)2 + mB (vB)2
mA C (vA)1 - (vA)2 D = mB C (vB)2 - (vB)1 D
+ )
(:
e =
(1)
(vB)2 - (vA)2
= 1
(vA)1 - (vB)1
(vB)2 - (vA)2 = (vA)1 - (vB)1
(2)
Combining Eqs. (1) and (2):
mA C (vA)1 - (vA)2 D C (vA)1 + (vA)2 D = mB C (vB)2 - (vB)1 D C (vB)2 + (vB)1 D
1
Expand and multiply by :
2
1
1
1
1
m (v )21 + mB (vB)21 = mA(vA)22 + mB (vB)22
2 A A
2
2
2
Q.E.D.
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lOMoARcPSD|12889239
15–63.
Each ball has a mass m and the coefficient of restitution
between the balls is e. If they are moving towards one another
with a velocity v, determine their speeds after collision. Also,
determine their common velocity when they reach the state of
maximum deformation. Neglect the size of each ball.
SOLUTION
+ )
(:
©m v1 = ©m v2
mv - mv = mvA + mvB
vA = - vB
+ )
(:
e =
(vB)2 - (vA)2
vB - vA
=
(vA)1 - (vB)1
v - (- v)
2ve = 2vB
vB = ve :
Ans.
vA = - ve = ve ;
Ans.
At maximum deformation vA = vB = v¿ .
+ )
(:
©m v1 = ©m v2
mv - mv = (2m) v¿
v¿ = 0
Ans.
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v
v
A
B
lOMoARcPSD|12889239
15–64.
The three balls each have a mass m. If A has a speed v just
before a direct collision with B, determine the speed of C
after collision. The coefficient of restitution between each
pair of balls e. Neglect the size of each ball.
v
B
A
SOLUTION
Conservation of Momentum: When ball A strikes ball B, we have
mA (vA)1 + mB (vB)1 = mA (vA)2 + mB (vB)2
+ )
(:
my + 0 = m(vA)2 + m(vB)2
(1)
Coefficient of Restitution:
+ )
(:
e =
(vB)2 - (vA)2
(vA)1 - (vB)1
e =
(vB)2 - (vA)2
v - 0
(2)
Solving Eqs. (1) and (2) yields
(yA)2 =
y(1 - e)
2
(yB)2 =
y(1 + e)
2
Conservation of Momentum:When ball B strikes ball C, we have
mB (vB)2 + mC (vC)1 = mB (vB)3 + mC (vC)2
+ )
(:
mc
v(1 + e)
d + 0 = m(vB)3 + m(vC)2
2
(3)
Coefficient of Restitution:
e =
+ )
(:
e =
(vC)2 - (vB)3
(vB)2 - (vC)1
(vC)2 - (vB)3
v(1 + e)
- 0
2
(4)
Solving Eqs. (3) and (4) yields
(vC)2 =
v(1 + e)2
4
(vB)3 =
v(1 - e2)
4
Ans.
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C
lOMoARcPSD|12889239
15–65.
1-kg
is traveling
horizontally
at
*15–56. The
A 1-lb
ballball
A is Atraveling
horizontally
at 20 fts
20
ms itwhen
it strikes
a 10-kg
B that
rest.IfIf the
when
strikes
a 10-lb
blockblock
B that
is is
atatrest.
coefficient of restitution between A and B is
is Ee 0.6,
0.6, and
the coefficient of kinetic friction between the plane and the
block is &K 0.4, determine the distance block B slides on
the plane before it stops sliding.
+ )
(:
iM1 v1 iM2 v2
0 (1)(v!)2
(1)(20)
10(v")2 20
(v!)2
+ )
(:
E (v")2
(v!)1
0.6 7 10(9.81) N
(10)(v")2
(v")2
20
(v!)2
(v")1
&F 0.4 N 0.4(10)(9.81)
39.24 N
(v!)2
0
. 10(9.81) N
(v!)2 12
(v")2
Thus,
(v")2 2.909 ms (v!)2 9.091 ms 9.091 ms Block B:
41
i51
2
42
1
(10)(2.909)2
2
D 1.078 m
39.24D 0
Ans.
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lOMoARcPSD|12889239
15–66.
Block A has a mass MA and is sliding on a rough horizontal surface with a velocity vA1 when it
makes a direct collision with block B, which has a mass MB and is originally at rest. If the
collision is perfectly elastic, determine the velocity of each block just after collision and the
distance between the blocks when they stop sliding. The coefficient of kinetic friction
between the blocks and the plane is Pk.
Given:
MA
5 kg
g
9.81
m
2
s
MB
4 kg
e
vA1
m
3
s
Pk
1
0.35
Solution:
Guesess
vA2
3
m
s
vB2
5
m
s
d2
1m
Given
2
MA vA1
§ vA2 ·
¨ ¸
¨ vB2 ¸
¨d ¸
© 2¹
MA vA2 MB vB2
Find vA2 vB2 d2
e vA1
§ vA2 ·
¨ ¸
© vB2 ¹
vB2 vA2
§ 0.33 · m
¨
¸
© 3.33 ¹ s
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d2
d.2
2
vB2 vA2
2gP k
1.602 m
Ans.
lOMoARcPSD|12889239
15– 67.
Disks A and B have masses MA and MB respectively. If they have the velocities shown,
determine their velocities just after direct central impact.
Given:
MA
2 kg
MB
4 kg
e
0.4
vA1
2
m
s
vB1
5
m
s
Solution:
Given
MA vA1 MB vB1
e vA1 vB1
§ vA2 ·
¨ ¸
© vB2 ¹
vA2
Guesses
Find vA2 vB2
1
m
s
vB2
1
m
s
MA vA2 MB vB2
vB2 vA2
§ vA2 ·
¨ ¸
© vB2 ¹
§ 4.533 · m
¨
¸
© 1.733 ¹ s
Ans.
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lOMoARcPSD|12889239
15–68.
The two billiard balls A and B are originally in contact with one another when a third ball C
strikes each of them at the same time as shown. If ball C remains at rest after the collision,
determine the coefficient of restitution. All the balls have the same mass. Neglect the size of
each ball.
Solution:
Conservation of “x” momentum:
mv
v
2m v' cos ( 30 deg)
2v' cos ( 30 deg)
( 1)
Coefficient of restitution:
v'
v cos ( 30 deg)
e
( 2)
Substituiting Eq. (1) into Eq. (2) yields:
v'
e
2v' cos ( 30 deg)
e
2
3
2
Ans.
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lOMoARcPSD|12889239
15–69.
A 300-g ball is kicked with a velocity of vA = 25 m>s at point A
as shown. If the coefficient of restitution between the ball and
the field is e = 0.4, determine the magnitude and direction u
of the velocity of the rebounding ball at B.
vA
v¿B
30
u
B
SOLUTION
Kinematics: The parabolic trajectory of the football is shown in Fig. a. Due to the
symmetrical properties of the trajectory, vB = vA = 25 m>s and f = 30°.
Conservation of Linear Momentum: Since no impulsive force acts on the football
along the x axis, the linear momentum of the football is conserved along the x axis.
+ )
(;
m A vB B x = m A v Bœ B x
0.3 A 25 cos 30° B = 0.3 Av Bœ B x
A v Bœ B x = 21.65 m>s ;
Coefficient of Restitution: Since the ground does not move during the impact, the
coefficient of restitution can be written as
(+ c)
e =
0 - A v Bœ B y
A vB B y - 0
- A v Bœ B y
0.4 =
- 25 sin 30°
A v Bœ B y = 5 m>s c
Thus, the magnitude of vBœ is
v Bœ = 2 A v Bœ B x + A vBœ B y = 221.652 + 52 = 22.2 m>s
and the angle of vBœ is
u = tan - 1 C
A v Bœ B y
A
B
v Bœ x
S = tan - 1 ¢
5
≤ = 13.0°
21.65
25 m>s
Ans.
Ans.
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A
lOMoARcPSD|12889239
15–70.
Two smooth spheres A and B each have a mass m. If A is
given a velocity of v0, while sphere B is at rest, determine
the velocity of B just after it strikes the wall. The coefficient
of restitution for any collision is e.
v0
A
SOLUTION
Impact: The first impact occurs when sphere A strikes sphere B. When this occurs, the
linear momentum of the system is conserved along the x axis (line of impact).
Referring to Fig. a,
+ )
(:
mAvA + mBvB = mA(vA)1 + mB(vB)1
mv0 + 0 = m(vA)1 + m(vB)1
(vA)1 + (vB)1 = v0
+ )
(:
e =
(vB)1 - (vA)1
vA - vB
e =
(vB)1 - (vA)1
v0 - 0
(1)
(vB)1 - (vA)1 = ev0
(2)
Solving Eqs. (1) and (2) yields
(vB)1 = a
1 + e
b v0 :
2
(vA)1 = a
1 - e
b v0 :
2
The second impact occurs when sphere B strikes the wall, Fig. b. Since the wall does
not move during the impact, the coefficient of restitution can be written as
+ )
(:
e =
e =
0 - C -(vB)2 D
(vB)1 - 0
0 + (vB)2
c
1 + e
d v0 - 0
2
(vB)2 =
e(1 + e)
v0
2
Ans.
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B
lOMoARcPSD|12889239
15–71. The 2-kg ball is thrown at the suspended 20-kg
block with a velocity of 4 m/s. If the coefficient of restitution
between the ball and the block is F 0.8, determine the
maximum height h to which the block will swing before it
momentarily stops.
System:
+ )
(:
4 m/s
jN1 v1 jN2 v2
0 (2)(v")2
(2)(4)
(20)(v#)2
10(v#)2 4
(v")2
+ )
(:
B
A
F (v#)2 (v")2
(v")1 (v#)1
0.8 (v#)2 (v")2
4 0
(v#)2 (v")2 3.2
Solving:
(v")2 2.545 ms
(v#)2 0.6545 ms
Block:
Datum at lowest point.
51
71 52
1
(20)(0.6545)2
2
72
0 0
I 0.0218 m 21.8 mm
20(9.81)I
Ans.
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h
lOMoARcPSD|12889239
The 2-kg ball is thrown at the suspended 20-kg
15–72.
block with a velocity of 4 m/s. If the time of impact between
the ball and the block is 0.005 s, determine the average normal
force exerted on the block during this time. Take F 0.8.
System:
+ )
(:
(2)(4)
0 (2)(v")2
B
A
(20)(v#)2
10(v#)2 4
(v")2
+ )
(:
4 m/s
jN1 v1 jN2 v2
F (v#)2 (v")2
(v")1 (v#)1
0.8 (v#)2 (v")2
4 0
(v#)2 (v")2 3.2
Solving:
(v")2 2.545 ms
(v#)2 0.6545 ms
Block:
+ )
(:
N v1
0
j
(
' EU N v2
'(0.005) 20(0.6545)
' 2618 N 2.62 kN
Ans.
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h
lOMoARcPSD|12889239
15–73. The pile P has a mass of 800 kg and is being driven
into loose sand using the 300-kg hammer C which is
dropped a distance of 0.5 m from the top of the pile.
Determine the initial speed of the pile just after it is struck
by the hammer. The coefficient of restitution between the
hammer and the pile is F 0.1. Neglect the impulses due to
the weights of the pile and hammer and the impulse due to
the sand during the impact.
C
0.5 m
P
The force of the sand on the pile can be considered nonimpulsive, along with the
weights of each colliding body. Hence,
Counter weight: Datum at lowest point.
71 52
51
0
72
300(9.81)(0.5) 1
(300)(v)2
2
0
v 3.1321 ms
System:
5
jN v1 jN v2
300(3.1321)
(v$)2
5
0 300(v$)2
800(v1)2
2.667(v1)2 3.1321
F (v1)2 (v$)2
(v$)1 (v1)1
0.1 (v1)2 (v$)2
3.1321 0
(v1)2 (v$)2 0.31321
Solving:
(v1)2 0.940 ms
Ans.
(v$)2 0.626 ms
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lOMoARcPSD|12889239
The pile P has a mass of 800 kg and is being driven
15–74.
into loose sand using the 300-kg hammer C which is dropped
a distance of 0.5 m from the top of the pile. Determine the
distance the pile is driven into the sand after one blow if the
sand offers a frictional resistance against the pile of 18 kN.
The coefficient of restitution between the hammer and the
pile is F 0.1. Neglect the impulses due to the weights of
the pile and hammer and the impulse due to the sand during
the impact.
C
0.5 m
P
The force of the sand on the pile can be considered nonimpulsive, along with the
weights of each colliding body. Hence,
Counter weight: Datum at lowest point,
71 52
51
72
300(9.81)(0.5) 0
1
(300)(v)2
2
0
v 3.1321 ms
System:
5
jN v1 jN v2
300(3.1321)
(v$)2
5
0 300(v$)2
800(v1)2
2.667(v1)2 3.1321
F (v1)2 (v$)2
(v$)1 (v1)1
0.1 (v1)2 (v$)2
3.1321 0
(v1)2 (v$)2 0.31321
Solving:
(v1)2 0.9396 ms
(v$)2 0.6264 ms
Pile:
52
j 623 53
1
(800)(0.9396)2
2
800(9.81)E 18 000E 0
E 0.0348 m 34.8 mm
Ans.
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lOMoARcPSD|12889239
15–75.
The 1-kg ball is dropped from rest at point A, 2 m above the
smooth plane. If the coefficient of restitution between the
ball and the plane is e = 0.6, determine the distance d
where the ball again strikes the plane.
A
2m
B
SOLUTION
30⬚
Conservation of Energy: By considering the ball’s fall from position (1) to position (2)
as shown in Fig. a,
TA + VA = TB + VB
1
1
2
2
m v + (Vg)A = mBvB + (Vg)B
2 A A
2
1
2
0 + 1(9.81)(2) = (1)vB + 0
2
vB = 6.264 m>s T
Conservation of Linear Momentum: Since no impulsive force acts on the ball along
the inclined plane (x¿ axis) during the impact, linear momentum of the ball is conserved along the x¿ axis. Referring to Fig. b,
mB(vB)x¿ = mB(v¿B)x¿
1(6.264) sin 30° = 1(v¿B) cos u
v¿B cos u = 3.1321
(1)
Coefficient of Restitution: Since the inclined plane does not move during the
impact,
e =
0 - (v¿B)y¿
(v¿B)y¿ - 0
0.6 =
0 - v¿B sin u
-6.264 cos 30° - 0
v¿B sin u = 3.2550
(2)
Solving Eqs. (1) and (2) yields
u = 46.10°
v¿B = 4.517 m>s
Kinematics: By considering the x and y motion of the ball after the impact, Fig. c,
+ )
(:
sx = (s0)x + (v¿B)x t
d cos 30° = 0 + 4.517 cos 16.10°t
t = 0.1995d
(+ c)
sy = (s0)y + (v¿B)y t +
(3)
1 2
at
2 y
-d sin 30° = 0 + 4.517 sin 16.10°t +
4.905t2 - 1.2528t - 0.5d = 0
1
(- 9.81)t2
2
(4)
Solving Eqs. (3) and (4) yields
d = 3.84 m
Ans.
t = 0.7663 s
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d
C
lOMoARcPSD|12889239
15–76.
A ball of mass m is dropped vertically from a height h0
above the ground. If it rebounds to a height of h1, determine
the coefficient of restitution between the ball and the
ground.
h0
h1
SOLUTION
Conservation of Energy: First, consider the ball’s fall from position A to position B.
Referring to Fig. a,
TA + VA = TB + VB
1
1
2
2
mvA + (Vg)A = mvB + (Vg)B
2
2
1
2
0 + mg(h0) = m(vB)1 + 0
2
Subsequently, the ball’s return from position B to position C will be considered.
TB + VB = TC + VC
1
1
2
2
mvB + (Vg)B = mvC + (Vg)C
2
2
1
2
m(vB)2 + 0 = 0 + mgh1
2
(vB)2 = 22gh1 c
Coefficient of Restitution: Since the ground does not move,
(+ c)
e = -
e = -
(vB)2
(vB)1
22gh1
- 22gh0
=
h1
A h0
Ans.
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lOMoARcPSD|12889239
15–77.
The cue ball A is given an initial velocity (vA)1 = 5 m>s. If
it makes a direct collision with ball B (e = 0.8), determine
the velocity of B and the angle u just after it rebounds from
the cushion at C (e¿ = 0.6). Each ball has a mass of 0.4 kg.
Neglect the size of each ball.
(vA)1 ⫽ 5 m/s
A
B
30⬚
C
SOLUTION
u
Conservation of Momentum: When ball A strikes ball B, we have
mA(vA)1 + mB(vB)1 = mA(vA)2 + mB(vB)2
0.4(5) + 0 = 0.4(vA)2 + 0.4(vB)2
(1)
Coefficient of Restitution:
e =
+)
(;
(vB)2 - (vA)2
(vA)1 - (vB)1
0.8 =
(vB)2 - (vA)2
5 - 0
(2)
Solving Eqs. (1) and (2) yields
(vA)2 = 0.500 m>s
(vB)2 = 4.50 m>s
Conservation of “y” Momentum: When ball B strikes the cushion at C, we have
mB(vBy)2 = mB(vBy)3
(+ T)
0.4(4.50 sin 30°) = 0.4(vB)3 sin u
(vB)3 sin u = 2.25
(3)
Coefficient of Restitution (x):
e =
+ )
(;
(vC)2 - (vBx)3
(vBx)2 - (vC)1
0.6 =
0 - [ -(vB)3 cos u]
4.50 cos 30° - 0
(4)
Solving Eqs. (1) and (2) yields
(vB)3 = 3.24 m>s
u = 43.9°
Ans.
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lOMoARcPSD|12889239
15 –78.
The three balls each have the same mass m. If A is released from rest at T, determine the
angle I to which C rises after collision. The coefficient of restitution between each ball is
e.
Solution:
Energy
0 l 1 cos T m g
vA
1
2
m vA
2
2 1 cos T g l
Collision of ball A with B:
m vA 0
m v'A m v'B
e vA
v'B v'A
v'B
1
( 1 e) v'B
2
Collision of ball B with C:
m v'B 0
m v''B m v''C
e v'B
v''C v''B
v''.C =
Energy
1
2
m v''c 0
2
0 l 1 cos I m g
1§ 1 ·
4
¨ ¸ ( 1 e) ( 2) 1 cos T
2 © 16 ¹
1 cos I
4
§ 1 e · 1 cos T
¨
¸
© 2 ¹
I
ª
acos «1 ¬
1 cos I
4
§ 1 e · 1 cos T »º
¨
¸
© 2 ¹
¼
Ans.
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1
2
( 1 e) v.A
4
lOMoARcPSD|12889239
15–79.
The sphere of mass m falls and strikes the triangular block
with a vertical velocity v. If the block rests on a smooth
surface and has a mass 3 m, determine its velocity just after
the collision. The coefficient of restitution is e.
SOLUTION
Conservation of “x¿ ” Momentum:
m (v)1 = m (v)2
( R +)
45°
m( v sin 45°) = m(vsx¿)2
(vsx¿)2 =
22
v
2
Coefficient of Restitution (y¿ ):
(+ b )
e =
(vb)2 - A vs y¿ B 2
e =
vb cos 45° - [ - A vsy¿ B 2]
A vsy¿ B 1 - (vb)1
v cos 45° - 0
A vsy¿ B 2 =
22
(ev - vb)
2
Conservation of “ x ” Momentum:
0 = ms (vs)x + mb vb
+ )
(;
0 + 0 = 3mvb - m A vsy¿ B 2 cos 45° - m(vsx¿)2 cos 45°
3vb -
22
22
22 22
(ev - vb)
v
= 0
2
2
2
2
vb =
1 + e
v
7
Ans.
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lOMoARcPSD|12889239
15–80.
Block A, having a mass m, is released from rest, falls a
distance h and strikes the plate B having a mass 2m. If the
coefficient of restitution between A and B is e, determine
the velocity of the plate just after collision. The spring has
a stiffness k.
A
h
B
k
SOLUTION
Just before impact, the velocity of A is
T1 + V1 = T2 + V2
0 + 0 =
1
mv 2A - mgh
2
(+ T )
e =
vA = 22gh
22gh
(vB)2 - (vA)2
e22gh = (vB)2 - (vA)2
(+ T )
(1)
©mv1 = ©mv 2
m(vA) + 0 = m(vA)2 + 2m(vB)2
(2)
Solving Eqs. (1) and (2) for (vB)2 yields;
(vB)2 =
1
3
2gh(1 + e)
Ans.
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lOMoARcPSD|12889239
15–81.
The girl throws the 0.5-kg ball toward the wall with an
initial velocity vA = 10 m>s. Determine (a) the velocity at
which it strikes the wall at B, (b) the velocity at which it
rebounds from the wall if the coefficient of restitution
e = 0.5, and (c) the distance s from the wall to where it
strikes the ground at C.
B
vA
A
10 m/s
30
1.5 m
C
SOLUTION
Kinematics: By considering the horizontal motion of the ball before the impact,
we have
+ )
(:
sx = (s0)x + vx t
3 = 0 + 10 cos 30°t
t = 0.3464 s
By considering the vertical motion of the ball before the impact, we have
(+ c)
vy = (v0)y + (ac)y t
= 10 sin 30° + ( -9.81)(0.3464)
= 1.602 m>s
The vertical position of point B above the ground is given by
(+ c )
sy = (s0)y + (v0)y t +
(sB)y = 1.5 + 10 sin 30°(0.3464) +
1
(a ) t2
2 cy
1
( - 9.81) A 0.34642 B = 2.643 m
2
Thus, the magnitude of the velocity and its directional angle are
(vb)1 = 2(10 cos 30°)2 + 1.6022 = 8.807 m>s = 8.81 m>s
u = tan - 1
1.602
= 10.48° = 10.5°
10 cos 30°
Ans.
Ans.
Conservation of “y” Momentum: When the ball strikes the wall with a speed of
(vb)1 = 8.807 m>s, it rebounds with a speed of (vb)2.
+ )
(;
mb A vby B 1 = mb A vby B 2
mb (1.602) = mb C (vb)2 sin f D
(vb)2 sin f = 1.602
(1)
Coefficient of Restitution (x):
e =
+ )
(:
0.5 =
(vw)2 - A vbx B 2
A vbx B 1 - (vw)1
0 - C -(vb)2 cos f D
10 cos 30° - 0
(2)
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3m
s
lOMoARcPSD|12889239
15–81. continued
Solving Eqs. (1) and (2) yields
f = 20.30° = 20.3°
(vb)2 = 4.617 m>s = 4.62 m>s
Ans.
Kinematics: By considering the vertical motion of the ball after the impact, we have
(+ c)
sy = (s0)y + (v0)y t +
1
(a ) t2
2 cy
- 2.643 = 0 + 4.617 sin 20.30°t1 +
1
( - 9.81)t21
2
t1 = 0.9153 s
By considering the horizontal motion of the ball after the impact, we have
+ )
(;
sx = (s0)x + vx t
s = 0 + 4.617 cos 20.30°(0.9153) = 3.96 m
Ans.
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lOMoARcPSD|12889239
15– 82
The ball of mass mb is thrown at the suspended block of mass mB with a velocity of vb. If the time
of impact between the ball and the block is 't, determine the average normal force exerted on the
block during this time.
Given:
mb
mB
kN
3 kg
25 kg
3
10 N
vb
e
m
s
4
g
9.81
m
2
s
't
0.85
0.005 s
Solution:
vA
Guesses
1
m
s
vB
1
m
s
F
1N
Given
mb vA mB vB
momentum
mb vb
restitution
e vb
momentum B
0 F't
§ vA ·
¨ ¸
¨ vB ¸
¨F¸
© ¹
Find vA vB F
vB vA
mB vB
§ vA ·
¨ ¸
© vB ¹
§ 2.61· m
¨
¸
© 0.79 ¹ s
F
3.96 kN
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Ans.
lOMoARcPSD|12889239
15–83.
The ball of mass mb is thrown at the suspended block of mass mB with velocity vb. If the
coefficient of restitution between the ball and the block is e, determine the maximum height h to
which the block will swing before it momentarily stops.
Given:
mb
vb
3 kg
mB
e
25 kg
4
m
s
g
9.81
m
2
s
0.85
Solution:
vA
Guesses
1
m
s
vB
1
m
s
h
1m
Given
mb vA mB vB
momentum
mb vb
restitution
e vb
energy
1
2
mB vB
2
§ vA ·
¨ ¸
¨ vB ¸
¨h¸
© ¹
Find vA vB h
vB vA
mB g h
§ vA ·
¨ ¸
© vB ¹
§ 2.61 · m
¨
¸
© 0.79 ¹ s
h
32.04 mm
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Ans.
lOMoARcPSD|12889239
15–84.
A ball is thrown onto a rough floor at an angle u. If it rebounds
at an angle f and the coefficient of kinetic friction
is m, determine the coefficient of restitution e. Neglect the size
of the ball. Hint: Show that during impact, the average
impulses in the x and y directions are related by Ix = mIy .
Since the time of impact is the same, Fx ¢t = mFy ¢t or
Fx = mFy .
y
u
x
SOLUTION
0 - 3 - v2 sin f4
( + T)
e =
+ )
(:
m1vx21 +
v1 sin u - 0
Lt1
e =
v2 sin f
v1 sin u
(1)
t2
Fx dx = m1vx22
mv1 cos u - Fx ¢t = mv2 cos f
Fx =
(+ T)
mv1 cos u - mv2 cos f
¢t
m1vy21 +
Lt1
(2)
t2
Fy dx = m1vy22
mv1 sin u - Fy ¢t = - mv2 sin f
Fy =
mv1 sin u + mv2 sin f
¢t
(3)
Since Fx = mFy, from Eqs. (2) and (3)
m1mv1 sin u + mv2 sin f)
mv1 cos u - mv2 cos f
=
¢t
¢t
cos u - m sin u
v2
=
v1
m sin f + cos f
(4)
Substituting Eq. (4) into (1) yields:
e =
sin f cos u - m sin u
a
b
sin u m sin f + cos f
f
Ans.
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lOMoARcPSD|12889239
15–85.
A ball is thrown onto a rough floor at an angle of u = 45°. If
it rebounds at the same angle f = 45°, determine the
coefficient of kinetic friction between the floor and the ball.
The coefficient of restitution is e = 0.6. Hint: Show that
during impact, the average impulses in the x and y
directions are related by Ix = mIy . Since the time of impact
is the same, Fx ¢t = mFy ¢t or Fx = mFy .
y
u
x
SOLUTION
(+ T )
+ )
(:
e =
0 - [-v2 sin f]
v1 sin u - 0
m(vx)1 +
Lt1
e =
v2 sin f
v1 sin u
(1)
t2
Fx dx = m(vx)2
mv1 cos u - Fx ¢t = mv2 cos f
Fx =
(+ c )
mv1 cos u - mv2 cos f
¢t
m A vy B 1 +
Lt1
(2)
Fydx = m A yy B 2
t2
mv1 sin u - Fy ¢t = -mv2 sin f
Fy =
mv1 sin u + mv2 sin f
¢t
(3)
Since Fx = mFy, from Eqs. (2) and (3)
m(mv1 sin u + mv2 sin f)
mv1 cos u - mv2 cos f
=
¢t
¢t
cos u - m sin u
v2
=
v1
m sin f + cos f
(4)
Substituting Eq. (4) into (1) yields:
e =
sin f cos u - m sin u
a
b
sin u m sin f + cos f
0.6 =
sin 45° cos 45° - m sin 45°
a
b
sin 45° m sin 45° + cos 45°
0.6 =
1-m
1+m
m = 0.25
f
Ans.
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lOMoARcPSD|12889239
15–86.
A pitching machine throws the ball of weight M towards the wall with an initial velocity
vA as shown. Determine (a) the velocity at which it strikes the wall at B, (b) the velocity
at which it rebounds from the wall and (c) the distance d from the wall to where it strikes
the ground at C.
Given:
M
0.5 kg
a
3m
vA
10
m
s
b
1.5 m
e
0.5
T
30 deg
g
9.81
m
2
s
Solution:
Guesses
vBx1
h
1
m
s
vBy1
t1
1m
1
m
s
1s
vBx2
t2
1
m
s
vBy2
d
1s
1
m
s
1m
Given
vA cos T t1
vBy2
d
vBy1
vBx2 t2
§ vBx1 ·
¨
¸
¨ vBy1 ¸
¨ vBx2 ¸
¨
¸
¨ vBy2 ¸
¨ h ¸
¨
¸
¨ t1 ¸
¨ t2 ¸
¨
¸
© d ¹
a
b vA sin T t1 1 2
g t1
2
vA sin T g t1
vBy1
h vBy2 t2 1 2
g t2
2
vA cos T
h
e vBx1
vBx1
vBx2
0
Find vBx1 vBy1 vBx2 vBy2 h t1 t2 d
§ vBx1 ·
¨
¸
© vBy1 ¹
8.81
m
s
Ans.
§ v.Bx2 ·
¨
¸
© v.By2 ¹
4.62
m
s
Ans.
d
3.96 m
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Ans.
lOMoARcPSD|12889239
15–87.
Two smooth disks A and B each have a mass of 0.5 kg. If
both disks are moving with the velocities shown when they
collide, determine their final velocities just after collision.
The coefficient of restitution is e = 0.75.
y
(vA)1
x
B
SOLUTION
+ )
(:
(vB)1
©mv1 = ©mv2
3
0.5(4)( ) - 0.5(6) = 0.5(yB)2x + 0.5(yA)2x
5
+ )
(:
e =
(vA)2 - (vB)2
(vB)1 - (vA)1
0.75 =
(vA)2x - (vB)2x
4 A 35 B - ( - 6)
(vA)2x = 1.35 m>s :
(vB)2x = 4.95 m>s ;
(+ c)
mv1 = mv2
4
0.5( )(4) = 0.5(vB)2y
5
(vB)2y = 3.20 m>s c
vA = 1.35 m>s :
vB = 2(4.59)2 + (3.20)2 = 5.89 m>s
u = tan-1
3.20
= 32.9˚ ub
4.95
6 m/s
Ans.
Ans.
Ans.
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5 4
3
4 m/s
A
lOMoARcPSD|12889239
15–88.
Two smooth disks A and B each have a mass of 0.5 kg. If
both disks are moving with the velocities shown when they
collide, determine the coefficient of restitution between the
disks if after collision B travels along a line, 30°
counterclockwise from the y axis.
y
(vA)1
x
B
SOLUTION
(vB)1
©mv1 = ©mv2
+ )
(:
3
0.5(4)( ) - 0.5(6) = -0.5(vB)2x + 0.5(vA)2x
5
- 3.60 = -(vB)2x + (vA)2x
(+ c)
4
0.5(4)( ) = 0.5(vB)2y
5
(vB)2y = 3.20 m>s c
(vB)2x = 3.20 tan 30° = 1.8475 m>s ;
(vA)2x = - 1.752 m>s = 1.752 m>s ;
+ )
(:
e =
e =
(vA)2 - (vB)2
(vB)1 - (vA)1
- 1.752 -( -1.8475)
4(35) -( -6)
= 0.0113
6 m/s
Ans.
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5 4
3
4 m/s
A
lOMoARcPSD|12889239
15–89.
Two smooth disks A and B have the initial velocities shown
just before they collide at O. If they have masses mA = 8 kg
and mB = 6 kg, determine their speeds just after impact.
The coefficient of restitution is e = 0.5.
y
13 12
5
A
vA
O
SOLUTION
vB
+b©mv1 = ©mv2
-6(3 cos 67.38°) + 8(7 cos 67.38°) = 6(vB)x¿ + 8(vA)xœ
e =
(+b)
0.5 =
(vB)2 - (vA)2
(vA)1 - (vB )1
(vB)xœ - (vA)xœ
7 cos 67.38° + 3 cos 67.38°
Solving,
(vB)xœ = 2.14 m>s
(vA)xœ = 0.220 m>s
(vB)yœ = 3 sin 67.38° = 2.769 m>s
(vA)y¿ = -7 sin 67.38° = -6.462 m>s
vB = 2(2.14)2 + (2.769)2 = 3.50 m>s
vA = 2(0.220)2 + (6.462)2 = 6.47 m>s
Ans.
Ans.
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3 m/s
7 m/s
x
B
lOMoARcPSD|12889239
15–90.
If disk A is sliding along the tangent to disk B and strikes
B with a velocity v, determine the velocity of B after the
collision and compute the loss of kinetic energy during
the collision. Neglect friction. Disk B is originally at rest.
The coefficient of restitution is e, and each disk has the
same size and mass m.
B
v
A
SOLUTION
Impact: This problem involves oblique impact where the line of impact lies along x¿
axis (line jointing the mass center of the two impact bodies). From the geometry
r
u = sin - 1 a b = 30°. The x¿ and y¿ components of velocity for disk A just before
2r
impact are
A vAx¿ B 1 = - v cos 30° = - 0.8660v
A vAy¿ B 1 = - v sin 30° = - 0.5 v
Conservation of “x¿ ” Momentum:
mA A vAx¿ B 1 + mB A vBx¿ B 1 = mA A vAx¿ B 2 + mB A vBx¿ B 2
(R +)
m(- 0.8660v) + 0 = m A vAx B 2 + m A vBx¿ B 2
(1)
Coefficient of Restitution (x¿ ):
e =
e =
A vBx¿ B 2 - A vAx¿ B 2
A vAx¿ B 1 - A vBx¿ B 1
A vBx¿ B 2 - A vAx¿ B 2
(2)
- 0.8660v - 0
Solving Eqs. (1) and (2) yields
A vBx¿ B 2 = -
23
(1 + e) v
4
A vAx¿ B 2 =
23
(e - 1) v
4
Conservation of “y¿ ” Momentum: The momentum is conserved along y¿ axis for
both disks A and B.
( +Q)
( +Q)
mB A vBy¿ B 1 = mB A vBy¿ B 2;
mA A vAy¿ B 1 = mA A vAy¿ B 2;
A vBy¿ B 2 = 0
A vAy¿ B 2 = - 0.5 v
Thus, the magnitude the velocity for disk B just after the impact is
(vB)2 = 2(vBx¿)22 + (vBy¿)22
=
D
a-
2
23
23
(1 + e) v b + 0 =
(1 + e) v
4
4
and directed toward negative x¿ axis.
Ans.
Ans.
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lOMoARcPSD|12889239
15–90. continued
The magnitude of the velocity for disk A just after the impact is
(vA)2 = 2( vAx¿)22 + (vAy¿)22
=
=
2
23
(e - 1) v d + ( -0.5 v)2
D 4
c
v2 (3e2 - 6e + 7)
A 16
Loss of Kinetic Energy: Kinetic energy of the system before the impact is
Uk =
1
mv2
2
Kinetic energy of the system after the impact is
Uk œ =
=
2
2
2
1
23
1
m B v (3e2 - 6e + 7) R + m B
(1 + e) v R
2
A 16
2
4
mv2
A 6e2 + 10 B
32
Thus, the kinetic energy loss is
¢Uk = Uk - Uk œ =
=
m v2
1
mv2 (6e2 + 10)
2
32
3mv2
(1 - e2)
16
Ans.
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lOMoARcPSD|12889239
15–91.
The billiard ball of mass M is moving with a speed v when it strikes the side of the pool
table at A. If the coefficient of restitution between the ball and the side of the table is e,
determine the speed of the ball just after striking the table twice, i.e., at A, then at B.
Neglect the size of the ball.
Given:
M
200 gm
m
s
v
2.5
T
45 deg
e
0.6
Solution:
Guesses
v2
Given
1
m
s
T2
e v sin T
e v2 cos T 2
v
¨§ 2 ·¸
¨ v3 ¸
¨ ¸
¨ T2 ¸
¨ T3 ¸
© ¹
1 deg
v3
1
v2 sin T 2
T3
v cos T
v3 sin T 3
Find v2 v3 T 2 T 3
m
s
1 deg
v2 cos T 2
v2 sin T 2
§ v2 ·
¨ ¸
© v3 ¹
v3
v3 cos T 3
§ 2.06 · m
¨
¸
© 1.50 ¹ s
1.500
m
s
§ T .2 ·
¨ ¸
© T .3 ¹
Ans.
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§ 31.0 ·
¨
¸ deg
© 45.0 ¹
lOMoARcPSD|12889239
15–92.
Two smooth billiard balls A and B
each have mass M. If A strikes B
with a velocity vA as shown,
determine their final velocities just
after collision. Ball B is originally at
rest and the coefficient of
restitution is e. Neglect the size of
each ball.
Given:
M
0.2 kg
T
40 deg
vA
1.5
e
0.85
m
s
Solution:
Guesses
Given
M vA cos T
e vA cos T
vA sin T
§ vA2 ·
¨ ¸
¨ vB2 ¸
¨T ¸
© 2¹
vA2
1
m
s
vB2
1
m
s
T2
20 deg
M vB2 M vA2 cos T 2
vA2 cos T 2 vB2
vA2 sin T 2
Find vA2 vB2 T 2
T2
95.1 deg
§ v.A2 ·
¨
¸
© v.B2 ¹
§ 0.968 · m
¨
¸
© 1.063 ¹ s
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Ans.
lOMoARcPSD|12889239
15–93.
Disks A and B have a mass of 15 kg and 10 kg, respectively.
If they are sliding on a smooth horizontal plane with the
velocities shown, determine their speeds just after impact.
The coefficient of restitution between them is e = 0.8.
y
5
4
Line of
impact
3
A
10 m/s
x
SOLUTION
8 m/s
Conservation of Linear Momentum: By referring to the impulse and momentum of
the system of disks shown in Fig. a, notice that the linear momentum of the system is
conserved along the n axis (line of impact). Thus,
¿
+Q mA A vA B n + mB A vB B n = mA A vA
B n + mB A vB¿ B n
3
3
¿
15(10)a b - 10(8) a b = 15vA
cos fA + 10vB¿ cos fB
5
5
¿
15vA
cos fA + 10vB¿ cos fB = 42
(1)
Also, we notice that the linear momentum of disks A and B are conserved along the
t axis (tangent to? plane of impact). Thus,
¿
+a mA A vA B t = mA A vA
Bt
4
¿
15(10)a b = 15vA
sin fA
5
¿
sin fA = 8
vA
(2)
and
+a mB A vB B t = mB A vB¿ B t
4
10(8)a b = 10 vB¿ sin fB
5
vB¿ sin fB = 6.4
(3)
Coefficient of Restitution: The coefficient of restitution equation written along the n
axis (line of impact) gives
+Q e =
¿
(vB¿ )n - (vA
)n
(vA)n - (vB)n
0.8 =
¿
cos fA
vB¿ cos fB - vA
3
3
10 a b - c -8a b d
5
5
¿
vB¿ cos fB - vA
cos fA = 8.64
(4)
Solving Eqs. (1), (2), (3), and (4), yeilds
¿
= 8.19 m>s
vA
Ans.
fA = 102.52°
vB¿ = 9.38 m>s
Ans.
fB = 42.99°
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B
lOMoARcPSD|12889239
15–94.
Determine the angular momentum H O of the particle about
point O.
z
1.5 kg
A
6 m/s
SOLUTION
2i - 4j - 4k
2(22 + ( - 4)2 + ( -4)2
≤ = {2i - 4j - 4k} m>s
4m
HO = rOB * mvA
HO = 3
i
0
1.5(2)
2m
B
vA = 6 ¢
rOB = {- 7j} m
4m
j
-7
1.5( -4)
3m
x
k
3 = {42i + 21k} kg # m2>s
0
1.5(- 4)
Ans.
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O
y
lOMoARcPSD|12889239
15–95.
Determine the angular momentum HO of each of the particles about point O.
T
Given:
30 deg
I
60 deg
mA
6 kg
c
2m
mB
4 kg
d
5m
mC
2 kg
e
2m
f
1.5 m
g
6m
h
2m
vA
4
m
s
vB
6
m
s
vC
2.6
m
s
a
8m
l
5
b
12 m
n
12
Solution:
kg˜ m
s
HAO
a mA vA sin I b mA vA cos I
HAO
22.3
HBO
f mB vB cos T e mB vB sin T
HBO
7.18
HCO
h mC¨
H.CO
21.60
§
l
·v g m §
·v
C
C¨
C
¸
2
2¸
2
2
© l n ¹
© l n ¹
n
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2
kg˜ m
s
Ans.
2
kg˜ m
s
Ans.
2
Ans.
lOMoARcPSD|12889239
15–96.
Determine the angular momentum HP of each of the particles about point P.
Given:
T
30 deg
I
60 deg
mA
6 kg
c
2m
mB
4 kg
d
5m
mC
2 kg
e
2m
f
1.5 m
g
6m
h
2m
vA
4
m
s
vB
6
m
s
vC
2.6
m
s
a
8m
l
5
b
12 m
n
12
Solution:
kg˜ m
s
HAP
mA vA sin I ( a d) mA vA cos I ( b c)
H.AP
57.6
HBP
mB vB cos T ( c f) mB vB sin T ( d e)
HBP
94.4
HCP
l
§ n · v ( c h) m §
· v ( d g)
mC¨
C
C¨
C
¸
¸
2
2
2
2
n
n
l
l
©
¹
©
¹
HCP
kg˜ m
41.2
s
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kg˜ m
s
2
Ans.
2
Ans.
2
Ans.
lOMoARcPSD|12889239
15–97.
z
Determine the total angular momentum H O for the system
of three particles about point O. All the particles are
moving in the x–y plane.
3 kg
6 m/s
C
O
900 mm
SOLUTION
4 m/s
A 1.5 kg
HO = ©r * mv
i
= 3 0.9
0
j
0
- 1.5(4)
= {12.5k} kg # m2>s
x
i
k
0 3 + 3 0.6
-2.5(2)
0
j
0.7
0
i
k
0 3 + 3 -0.8
0
0
j
-0.2
3(- 6)
k
03
0
Ans.
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700 mm
200 mm
800 mm
y
B
2 m/s 600 mm
2.5 kg
lOMoARcPSD|12889239
15–98.
Determine the angular momentum H O of each of the two
particles about point O. Use a scalar solution.
y
5m
P
15 m/s
5
A
3
4m
4
2 kg
1.5 m
O
x
2m
SOLUTION
4m
4
3
a + (HA)O = - 2(15) a b(1.5) - 2(15)a b (2)
5
5
30⬚
1.5 kg
10 m/s
= - 72.0 kg # m2>s = 72.0 kg # m2>s b
Ans.
a +(HB)O = - 1.5(10)(cos 30°)(4) - 1.5(10)(sin 30°)(1)
= -59.5 kg # m2>s = 59.5 kg # m2>s b
B
Ans.
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1m
lOMoARcPSD|12889239
15–99.
Determine the angular momentum H P of each of the two
particles about point P. Use a scalar solution.
y
5m
P
15 m/s
5
A
3
4m
4
2 kg
1.5 m
O
x
2m
SOLUTION
4m
3
4
a +(HA)P = 2(15) a b (2.5) - 2(15) a b (7)
5
5
30⬚
1.5 kg
10 m/s
= -66.0 kg # m2>s = 66.0 kg # m2>s b
Ans.
a + (HB)P = -1.5(10)(cos 30°)(8) + 1.5(10)(sin 30°)(4)
= -73.9 kg # m2>s b
B
Ans.
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1m
lOMoARcPSD|12889239
15–100.
The small cylinder C has a mass of 10 kg and is attached to
the end of a rod whose mass may be neglected. If the frame
is subjected to a couple M = 18t2 + 52 N # m, where t is in
seconds, and the cylinder is subjected to a force of 60 N,
which is always directed as shown, determine the speed of
the cylinder when t = 2 s. The cylinder has a speed
v0 = 2 m>s when t = 0.
z
60 N
5 4
0.75 m
v
x
C
SOLUTION
M
(Hz)1 + ©
L
Mz dt = (Hz)2
2
3
(10)(2)(0.75) + 60(2)( )(0.75) +
(8t 2 + 5)dt = 10v(0.75)
5
L0
8
69 + [ t3 + 5t]20 = 7.5v
3
v = 13.4 m>s
Ans.
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3
(8t2
y
5) N m
lOMoARcPSD|12889239
15–101.
At the instant S 1.5 m, the 5-kg disk is given a
speed of W 5 ms, perpendicular to the elastic cord.
Determine the speed of the disk and the rate of shortening
of the elastic cord at the instant S 1.2 m. The disk slides
on the smooth horizontal plane. Neglect its size. The cord
has an unstretched length of 0.5 m.
A
r
k ⫽ 200 N/m
B
v ⫽ 5 m/s
Conservation of Energy: The intial and final stretch of the elastic cord is
T1 1.5 0.5 1 m and T2 1.2 0.5 0.7 m. Thus,
51
71 52
72
1
Nv1 2
2
1
1
1
LT1 2 Nv2 2 LT2 2
2
2
2
1
5(52)
2
1
1
(200)(12) (5)v2 2
2
2
1
(200)(0.72)
2
v2 6.738 ms
Ans.
Conservation of Angular Momentum: Since no angular impulse acts on the disk
about an axis perpendicular to the page passing through point O, its angular
momentum of the system is conserved about this z axis. Thus,
)0 1 )0 2
S1Nv1 S2N v2 u
v2 u 1.5(5)
S1v1
6.25 ms
S2
1.2
Since v2 2 v2 u 2
v2 S 2, then
v 2 S 2v2 2 v2 u 2 26.7382 6.252 2.52 ms
Ans.
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lOMoARcPSD|12889239
15–102.
Determine the total angular momentum HO for the system of three particles about point O. All the
particles are moving in the x-y plane.
Given:
mA
2 kg
a
900 mm
vA
4
m
s
b
700 mm
mB
3 kg
c
600 mm
vB
2
m
s
d
800 mm
mC
4 kg
e
200 mm
vC
6
m
s
Solution:
HO
§ a · ª« ¨§ 0 ¸·»º § c · ª« ¨§ vB ¸·»º § d · «ª ¨§ 0 ·¸º»
¨ 0 ¸ u m v
¨ ¸
¨ ¸
b u m
e u m v
¨ ¸ « A¨ A ¸» ¨ ¸ « B¨ 0 ¸» ¨ ¸ « C¨ C ¸»
© 0 ¹ «¬ ¨© 0 ¸¹»¼ © 0 ¹ «¬ ¨© 0 ¸¹»¼ © 0 ¹ «¬ ¨© 0 ¸¹»¼
HO
§ 0.00 ·
2
¨ 0.00 ¸ kg˜ m
¨
¸ s
© 16.20 ¹
Ans.
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lOMoARcPSD|12889239
15–103. The two spheres each have a mass of 3 kg and are
attached to the rod of negligible mass. If a torque
. (6F0.2U) N m, where t is in seconds, is applied to the
rod as shown, determine the speed of each of the spheres in
2 s, starting from rest.
0.4 m
0.4 m
M
Principle of Angular Impluse and Momentum: Applying Eq. 15–22, we have
j
()[)1
2[0.4 (3) (0)]
(0
2s
(U1
U2
.[ EU ()[)2
6F 0.2 U EU 2 [0.4 (3) y]
y 6.15 ms
Ans.
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lOMoARcPSD|12889239
15–104. The two spheres each have a mass of 3 kg and are
attached to the rod of negligible mass. Determine the time
the torque - (8T) N m, where t is in seconds, must be
applied to the rod so that each sphere attains a speed of
3 ms starting from rest.
0.4 m
0.4 m
Principle of Angular Impluse and Momentum: Applying Eq. 15–22, we have
((Z)1
2[0.4 (3) (0)]
i
'0
'T1
M
T2
-Z DT ((Z)2
T
(8 T) DT 2[0.4 (3) (3)]
T 1.34 s
Ans.
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lOMoARcPSD|12889239
15–105.
The ball B has mass M and is attached to the end of a rod whose mass may be neglected. If the
rod is subjected to a torque M = at2 + bt + c, determine the speed of the ball when t = t1. The ball
has a speed v = v0 when t = 0.
Given:
M
a
b
10 kg
3
N˜ m
2
s
N˜ m
5
s
c
2 N˜ m
t1
2s
v0
2
L
1.5 m
Solution:
m
s
Principle of angular impulse momentum
t1
´
2
M v0 L µ a t b t c dt
¶0
v1
1
v0 ML
M v1 L
t1
´
2
µ a t b t c dt
¶0
v1
3.47
m
s
Ans.
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lOMoARcPSD|12889239
15–106.
A small particle having a mass m is placed inside the
semicircular tube. The particle is placed at the position
shown and released. Apply the principle
of angular
#
momentum about point O 1©MO = HO2, and show that
the motion
$ of the particle is governed by the differential
equation u + 1g>R2 sin u = 0.
O
u
SOLUTION
c + ©MO =
dHO
;
dt
-Rmg sin u =
d
(mvR)
dt
g sin u = -
d 2s
dv
= - 2
dt
dt
But, s = R u
$
Thus, g sin u = -Ru
$
g
or, u + a b sin u = 0
R
Q.E.D.
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R
lOMoARcPSD|12889239
15–107.
The small cylinder C has mass mC and is attached to the end of a rod whose mass may
be neglected. If the frame is subjected to a couple M = at2 + b, and the cylinder is
subjected to force F, which is always directed as shown, determine the speed of the
cylinder when t = t1. The cylinder has a speed v0 when t = 0.
Given:
mC
a
20 kg
t1
2.5 s
m
v0
3
8N
2
m
s
s
5 N˜ m
b
F
60 N
d
0.75 m
e
4
f
3
Solution:
t1
´
2
mC v0 d µ a t b dt ¶0
ª
1 «´
2
v0 µ a t b dt mC d «¶0
t1
v1
f
§
·
¨ 2 2 ¸ F d t1
© e f ¹
¬
mC v1 d
º
f
§
·F d t »
¨ 2 2 ¸ 1»
© e f ¹
¼
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v1
11.11
m
s
Ans.
lOMoARcPSD|12889239
15–108.
A child having a mass of 50 kg holds her legs up as shown as
she swings downward from rest at u1 = 30°. Her center of
mass is located at point G1 . When she is at the bottom
position u = 0°, she suddenly lets her legs come down,
shifting her center of mass to position G2 . Determine her
speed in the upswing due to this sudden movement and the
angle u2 to which she swings before momentarily coming to
rest. Treat the child’s body as a particle.
3m
G2
First before u = 30°;
T1 + V1 = T2 + V2
1
(50)(v1)2 + 0
2
v1 = 2.713 m>s
H1 = H2
50(2.713)(2.80) = 50(v2)(3)
v2 = 2.532 = 2.53 m>s
Ans.
Just after u = 0°;
T2 + V2 = T3 + V3
1
(50)(2.532)2 + 0 = 0 + 50(9.81)(3)(1 - cos u2)
2
0.1089 = 1 - cos u2
Ans.
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30°
G1
G2
0 + 2.80(1 - cos 30°)(50)(9.81) =
2.80 m
u1
SOLUTION
u2 = 27.0°
u2
lOMoARcPSD|12889239
15–109.
A small particle having a mass 2 m is placed inside the semicircular tube. The particle is
placed at the position shown and released. Apply the principle of angular momentum about
point O (6M0 = H0), and show that the motion of the particle is governed by the differential
equation T'' + (g / R) sin T = 0.
Solution:
6M0
d
H0
dt
R 2 mg sin T
d
v
dt
g sin T
But,
s
2
d
dt
2
s
RT
Thus, g sin T
or,
d
(2 m v R)
dt
R T''
g
T'' §¨ ·¸ sin T
© R¹
0
Ans.
2 mg
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lOMoARcPSD|12889239
15–110.
A toboggan and rider, having a
total mass M, enter horizontally
tangent to a circular curve (T1)
with a velocity vA. If the track
is flat and banked at angle T2,
determine the speed vB and the
angle T of “descent”, measured
from the horizontal in a vertical
x–z plane, at which the
toboggan exists at B. Neglect
friction in the calculation.
Given:
km
hr
M
150 kg
T1
90 deg
vA
70
rA
60 m
rB
57 m
r
55 m
T2
60 deg
Solution:
h
rA rB tan T 2
m
s
T
Guesses
vB
Given
1
2
M vA M g h
2
§ vB ·
¨ ¸
©T ¹
10
Find vB T
vB
1 deg
1
2
M vB
2
21.9
m
s
M vA rA
T
M vB cos T rB
20.9 deg
Ans.
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lOMoARcPSD|12889239
15–111.
The two blocks A and B each have a mass M0. The blocks are fixed to the horizontal rods,
and their initial velocity is v' in the direction shown. If a couple moment of M is applied
about shaft CD of the frame, determine the speed of the blocks at time t. The mass of the
frame is negligible, and it is free to rotate about CD. Neglect the size of the blocks.
Given:
M0
0.4 kg
a
0.3 m
v'
2
M
0.6 N˜ m
t
m
s
3s
Solution:
2a M0 v' M t
v
v' Mt
2a M0
2a M0 v
v
9.50
m
s
Ans.
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lOMoARcPSD|12889239
15–112.
A small block having a mass of 0.1 kg is given a horizontal
velocity of v1 = 0.4 m>s when r1 = 500 mm. It slides along
the smooth conical surface. Determine the distance h it must
descend for it to reach a speed of v2 = 2 m>s. Also, what is
the angle of descent u, that is, the angle measured from the
horizontal to the tangent of the path?
r1 = 500 mm
υ 1 = 0.4 m>s
h
r2
θ
v2
SOLUTION
T1 + V1 = T2 + V2
30°
1
1
(0.1)(0.4)2 + 0.1(9.81)(h) = (0.1)(2)2
2
2
h = 0.1957 m = 196 mm
Ans.
From similar triangles
r2 =
(0.8660 - 0.1957)
(0.5) = 0.3870 m
0.8660
(H0)1 = (H0)2
0.5(0.1)(0.4) = 0.3870(0.1)(v 2 ¿)
v 2 ¿ = 0.5168 m>s
v 2 = cos u = v 2 ¿
2 cos u = 0.5168
u = 75.0°
Ans.
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lOMoARcPSD|12889239
15–113.
An earth satellite of mass 700 kg is launched into a freeflight trajectory about the earth with an initial speed of
vA = 10 km>s when the distance from the center of the
earth is rA = 15 Mm. If the launch angle at this position is
fA = 70°, determine the speed vB of the satellite and its
closest distance rB from the center of the earth. The earth
has a mass Me = 5.976110242 kg. Hint: Under these
conditions, the satellite is subjected only to the earth’s
gravitational force, F = GMems>r2, Eq. 13–1. For part of
the solution, use the conservation of energy.
vB
rA
vA
SOLUTION
(HO)1 = (HO)2
ms (vA sin fA)rA = ms (vB)rB
700[10(103) sin 70°](15)(106) = 700(vB)(rB)
(1)
TA + VA = TB + VB
GMe ms
GMems
1
1
m (v )2 = ms (vB)2 rA
rB
2 s A
2
66.73(10-12)(5.976)(1024)(700)
1
1
= (700)(vB)2
(700)[10(103)]2 6
2
2
[15(10 )]
-
66.73(10-12)(5.976)(1024)(700)
rB
rB
fA
(2)
Solving,
vB = 10.2 km>s
Ans.
rB = 13.8 Mm
Ans.
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lOMoARcPSD|12889239
15–114.
The fire boat discharges two streams of seawater, each at a
flow of 0.25 m3>s and with a nozzle velocity of 50 m> s.
Determine the tension developed in the anchor chain,
needed to secure the boat. The density of seawater is
rsw = 1020 kg>m3.
30⬚
45⬚
60⬚
SOLUTION
Steady Flow Equation: Here, the mass flow rate of the sea water at nozzles A and
dmA
dmB
B are
=
= rsw Q = 1020(0.25) = 225 kg>s. Since the sea water is coldt
dt
lected from the larger reservoir (the sea), the velocity of the sea water entering the
control volume can be considered zero. By referring to the free-body diagram of the
control volume (the boat),
+ ©F = dmA A v B + dmB A v B ;
;
x
A x
B x
dt
dt
T cos 60° = 225(50 cos 30°) + 225(50 cos 45°)
T = 40 114.87 N = 40.1 kN
Ans.
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lOMoARcPSD|12889239
15–115.
The rocket car has a mass MC (empty) and carries fuel of mass MF. If the fuel is
consumed at a constant rate c and ejected from the car with a relative velocity vDR,
determine the maximum speed attained by the car starting from rest. The drag resistance
due to the atmosphere is F D = kv2 and the speed is measured in m/s.
Units Used:
Mg
3
10 kg
Given:
MC
3 Mg
vDR
250
MF
m
s
150 kg
2
c
kg
4
s
k
s
60 N˜
m
2
Solution:
m0
Set F
At time t the mass of the car is m0 c t
MC MF
2
2
k v
k v , then
m0 c t
d
v vDR c
dt
Maximum speed occurs at the instant the fuel runs out.
Thus, Initial Guess:
v
Given
v
4
MF
t
c
37.5 s
m
s
´
1
µ
dv
µ
2
c vDR k v
µ
¶
0
t
t
´
1
µ
dt
µ m0 c t
¶0
v
Find ( v)
v
4.062
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m
s
Ans.
lOMoARcPSD|12889239
15–116.
The 200-kg boat is powered by the fan which develops a
slipstream having a diameter of 0.75 m. If the fan ejects air
with a speed of 14 m/s, measured relative to the boat,
determine the initial acceleration of the boat if it is initially at
rest.Assume that air has a constant density of ra = 1.22 kg/m3
and that the entering air is essentially at rest. Neglect the drag
resistance of the water.
0.75 m
SOLUTION
Equations of Steady Flow: Initially, the boat is at rest hence vB = va>b
p
dm
= 14 m>s. Then, Q = vBA = 14 c A 0.752 B d = 6.185 m3>s and
= raQ
4
dt
= 1.22(6.185) = 7.546 kg>s. Applying Eq. 15–26, we have
©Fx =
dm
(v - vAx);
dt Bx
- F = 7.546(- 14 - 0)
F = 105.64 N
Equation of Motion :
+
: ©Fx = max;
105.64 = 200a
a = 0.528 m>s2
Ans.
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lOMoARcPSD|12889239
15–117.
The chute is used to divert the flow of water, Q = 0.6 m3>s.
If the water has a cross-sectional area of 0.05 m2, determine
the force components at the pin D and roller C necessary
for equilibrium. Neglect the weight of the chute and weight
of the water on the chute, rw = 1 Mg>m3 .
A
0.12 m
C
SOLUTION
2m
Equations of Steady Flow: Here, the flow rate Q = 0.6 m2>s. Then,
Q
0.6
dm
=
= 12.0 m>s. Also,
= rw Q = 1000 (0.6) = 600 kg>s. Applying
y =
A
0.05
dt
Eqs. 15–26 and 15–28, we have
dm
(dDB yB - dDA yA);
dt
- Cx (2) = 600 [0 - 1.38(12.0)]
1.5 m
a + ©MA =
dm
+
: ©Fx = dt A yBx - yAx);
Dx + 4968 = 600 (12.0 - 0)
dm
A youty - yiny B ;
dt
Dy = 600[0 - ( -12.0)]
D
Cx = 4968 N = 4.97 kN
Dx = 2232N = 2.23 kN
Ans.
Ans.
+ c ©Fy = ©
Dy = 7200 N = 7.20 kN
Ans.
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B
lOMoARcPSD|12889239
15–118. The toy sprinkler for children consists of a 0.2-kg
cap and a hose that has a mass per length of 30 gm.
Determine the required rate of flow of water through the
5-mm-diameter tube so that the sprinkler will lift 1.5 m
from the ground and hover from this position. Neglect the
weight of the water in the tube. +w 1 Mgm3.
Equations
of
Steady
Flow:
Here,
2 1
)
4
DM
+w 1 10001. Applying Eq. 15–25, we have
DT
i&Y DM
2"Y
DT
2!Y ;
[0.2
(0.0052)
1.5 m
1.5 (0.03)] (9.81) 100012
1 0.217 10 3 m3s
1
6.25(10 6))
1
6.25 (10 6)฀ )
and
03
Ans.
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[0.2
1.5(0.03)] (9.81)
lOMoARcPSD|12889239
15–119.
The boat of mass M is powered by a fan F which develops a slipstream having a diameter
d. If the fan ejects air with a speed v, measured relative to the boat, determine the initial
acceleration of the boat if it is initially at rest. Assume that air has a constant density Ua and
that the entering air is essentially at rest. Neglect the drag resistance of the water.
Given:
M
200 kg
h
0.375 m
d
0.75 m
v
14
Ua
m
s
1.22
kg
m
3
Solution:
Q
Av
Q
d
m
dt
m'
6Fx
m' vBx vAx
F
6Fx
a
m'
3
S 2
d v
Q
6.1850
m
s
Ua Q
m'
7.5457
kg
s
4
Ua Q v
F
105.64 N
M ax
F
Ma
a
0.528
F
M
m
2
Ans.
s
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lOMoARcPSD|12889239
15–120. The hemispherical bowl of mass m is held in
equilibrium by the vertical jet of water discharged through
a nozzle of diameter d. If the discharge of the water through
the nozzle is Q, determine the height h at which the bowl is
suspended. The water density is rw. Neglect the weight of
the water jet.
h
Conservation of Energy: The speed at which the water particle leaves the nozzle is
2
2
42
v1 . The speed of particle v" when it comes in contact with the
p 2
"
pE2
E
4
bowl can be determined using conservation of energy. With reference to the datum
set in Fig. a,
51
71 52
1
Nv1 2
2
7H 1 42 2
1
N3 2 4
2
pE
v" 720
0 C p2 E4
1622
1
Nv2 2
2
1
Nv" 2
2
7H 2
NHI
2HI
Steady Flow Equation: The mass flow rate of the water jet that enters the control
EN"
volume at A is
rw 2, and exits from the control volume at B is
EU
1622
EN#
EN"
2HI. Here, the vertical force
rw 2. Thus, v# v" EU
EU
C p2 E4
acting on the control volume is equal to the weight of the bowl. By referring to the
free - body diagram of the control volume, Fig. b,
D j'Z 2
EN#
EN"
v# v";
EU
EU
NH (rw2) B
Cp E
1622
2
4
2HIJ rw2 B
NH 2rw2 B
C p2E4
1622
N2H2 4rw 222 B
I 822
p2E4H
1622
p2E4
N2H
8rw 222
C p2E4
1622
2HIJ
2HIJ
2HIJ
Ans.
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lOMoARcPSD|12889239
15–121.
The rope has a mass m' per unit length. If the end length y = h is draped off the edge of the
table, and released, determine the velocity of its end A for any position y, as the rope
uncoils and begins to fall.
Solution:
d
d
m v vDi mi
dt
dt
Fs
At a time t, m
Here, vDi
m' y and
d
v,
v
d
dt
mi
m'
d
dt
y
m' v.
g.
dt
d
m' y v v( m' v)
dt
m' g y
gy
d
2
y vv
dt
gy
vy
Since v
d
dt
dy
y, then dt
v
d
2
vv
dy
Multiply both sides by 2ydy
2
2g y dy
2
´
2
µ
µ 2g y d y
¶
´
2 2
µ 1 dv y
¶
2 3
gy C
3
v
0
at
2
2v y dv 2y v dy
2 2
v y
y
2 3 2 3
g y gh
3
3
2 §¨ y h
g
2
3 ¨
© y
3
v
2 3
gh C
3
h
0
C
2 3
gh
3
2 2
v y
3·
¸
¸
¹
Ans.
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lOMoARcPSD|12889239
When operating, the air-jet fan discharges air
15–122.
with a speed of W# 20 ms into a slipstream having a
diameter of 0.5 m. If air has a density of 1.22 kgm3,
determine the horizontal and vertical components of
reaction at C and the vertical reaction at each of the two
wheels, D, when the fan is in operation. The fan and motor
have a mass of 20 kg and a center of mass at G. Neglect the
weight of the frame. Due to symmetry, both of the wheels
support an equal load. Assume the air entering the fan at A
is essentially at rest.
0.5 m
A
G
vB
1.5 m
EN
rv" 1.22(20)(p)(0.25)2 4.791 kgs
EU
D
C
EN
j'Y EU (v#Y v"Y )
0.8 m
0.25 m
$Y 4.791(20 0)
$Y 95.8 N
Ans.
D j'Z 0;
a j.$ $Z
0.25 m
2%Z 20(9.81) 0
EN
(E$( v# E$( v")
EU
2%Z (0.8) 20(9.81)(1.05) 4.791( 1.5(20) 0)
Solving:
%Z 38.9 N
Ans.
$Z 118 N
Ans.
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lOMoARcPSD|12889239
15–123.
A scoop in front of the tractor collects snow at a
rate of 200 kgs. Determine the resultant traction force T
that must be developed on all the wheels as it moves
forward on level ground at a constant speed of 5 kmh. The
tractor has a mass of 5 Mg.
m
1h
65
6
h 3600 s
1.389 ms. Thus, v%T v 1.389 ms since the snow on the ground is at rest.
ENT
The rate at which the tractor gains mass is
200 kgs. Since the tractor is
EU
Ev
moving with a constant speeds
0. Referring to Fig. a,
EU
Here, the tractor moves with the constant speed of v 5 5 103 Ev
j'T N
EU
v%T
ENT
;
EU
5 0
1.389(200)
5 278 N
Ans.
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lOMoARcPSD|12889239
15–124.
The boat has a mass of 180 kg and is traveling forward on a
river with a constant velocity of 70 km>h, measured relative
to the river. The river is flowing in the opposite direction at
5 km>h. If a tube is placed in the water, as shown, and it
collects 40 kg of water in the boat in 80 s, determine the
horizontal thrust T on the tube that is required to overcome
the resistance due to the water collection and yet maintain
the constant speed of the boat. rw = 1 Mg>m3.
T
SOLUTION
40
dm
=
= 0.5 kg>s
dt
80
vD>t = (70)a
©Fs = m
1000
b = 19.444 m>s
3600
dm i
dv
+ vD>i
dt
dt
T = 0 + 19.444(0.5) = 9.72 N
Ans.
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vR
5 km/h
lOMoARcPSD|12889239
15–125.
Water is discharged from a nozzle with a velocity of 12 m>s
and strikes the blade mounted on the 20-kg cart. Determine
the tension developed in the cord, needed to hold the cart
stationary, and the normal reaction of the wheels on the
cart. The nozzle has a diameter of 50 mm and the density of
water is rw = 1000 kg>mg3.
45⬚
A
SOLUTION
Steady Flow Equation: Here, the mass flow rate at sections A and B of the control
p
dm
= rWQ = rWAv = 1000 c (0.052) d (12) = 7.5p kg>s
volume is
dt
4
Referring to the free-body diagram of the control volume shown in Fig. a,
dm
+
: ©Fx = dt [(vB)x - (vA)x];
- Fx = 7.5p(12 cos 45° - 12)
Fx = 82.81 N
dm
+ c ©Fy =
[(vB)y - (vA)y];
dt
Fy = 7.5p(12 sin 45° - 0)
Fy = 199.93 N
Equilibrium: Using the results of Fx and Fy and referring to the free-body diagram
of the cart shown in Fig. b,
+
: ©Fx = 0;
82.81 - T = 0
T = 82.8 N
Ans.
+ c ©Fy = 0;
N - 20(9.81) - 199.93 = 0
N = 396 N
Ans.
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B
lOMoARcPSD|12889239
15–126.
Water is flowing from the 150-mm-diameter fire hydrant with
a velocity vB = 15 m>s. Determine the horizontal and vertical
components of force and the moment developed at the base
joint A, if the static (gauge) pressure at A is 50 kPa. The
diameter of the fire hydrant at A is 200 mm. rw = 1 Mg>m3.
vB
15 m/s
B
500 mm
SOLUTION
dm
= rvAA B = 1000(15)(p)(0.075)2
dt
A
dm
= 265.07 kg>s
dt
vA = (
dm 1
265.07
)
=
dt rA A
1000(p)(0.1)2
vA = 8.4375 m>s
dm
+
; ©Fx = dt (vBx - vAx)
A x = 265.07(15 -0) = 3.98 kN
+ c ©Fy =
Ans.
dm
(vBy - vAy)
dt
- A y + 50(103)(p)(0.1)2 = 265.07(0- 8.4375)
Ay = 3.81 kN
a + ©MA =
Ans.
dm
(dAB vB - dAA vA)
dt
M = 265.07(0.5(15) - 0)
M = 1.99 kN # m
Ans.
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lOMoARcPSD|12889239
15–127.
A coil of heavy open chain is used to reduce the stopping
distance of a sled that has a mass M and travels at a speed of
v0. Determine the required mass per unit length of the
chain needed to slow down the sled to (1>2)v0 within a
distance x = s if the sled is hooked to the chain at x = 0.
Neglect friction between the chain and the ground.
SOLUTION
Observing the free-body diagram of the system shown in Fig. a, notice that the pair
of forces F, which develop due to the change in momentum of the chain, cancel each
other since they are internal to the system. Here, vD>s = v since the chain on the
dms
ground is at rest. The rate at which the system gains mass is
= m¿v and the
dt
mass of the system is m = m¿x + M. Referring to Fig. a,
0 = A m¿x + M B
+ ) ©F = m dv + v dms ;
(:
D>s
s
dt
dt
0 = A m¿x + M)
Since
dv
+ v A m¿v B
dt
dv
+ m¿v2
dt
(1)
dx
dx
= v or dt =
,
dt
v
A m¿x + M B v
dv
+ m¿v2 = 0
dx
dv
m¿
= -¢
≤ dx
v
m¿x + M
(2)
Integrating using the limit v = v0 at x = 0 and v =
1
2
Lv0
v0
1n v
1
v at x = s,
2 0
s
m¿
dv
= bdx
a
v
L0 m¿x + M
1
2 v0
v0
= -1n A m¿x + M B
s
0
1
M
=
2
m¿s + M
m¿ =
M
s
Ans.
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lOMoARcPSD|12889239
15–128.
The car is used to scoop up water that is lying in a trough at
the tracks. Determine the force needed to pull the car
forward at constant velocity v for each of the three cases.
The scoop has a cross-sectional area A and the density of
water is rw .
v
v
F1
F2
(a)
(b)
v
SOLUTION
F3
The system consists of the car and the scoop. In all cases
©Ft = m
dme
dv
- vD>e
dt
dt
(c)
F = 0 - v(r)(A) v
F = v2 r A
Ans.
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lOMoARcPSD|12889239
15–129.
250 mm
The water flow enters below the hydrant at C at the rate of
0.75 m3>s. It is then divided equally between the two outlets at A and B. If the gauge pressure at C is 300 kPa, determine the horizontal and vertical force reactions and the
moment reaction on the fixed support at C. The diameter
of the two outlets at A and B is 75 mm, and the diameter of
the inlet pipe at C is 150 mm. The density of water is
rw = 1000 kg>m3. Neglect the mass of the contained water
and the hydrant.
30⬚
A
B
650 mm
600 mm
SOLUTION
Free-Body Diagram: The free-body diagram of the control volume is shown in
Fig. a. The force exerted on section A due to the water pressure is FC = pCAC =
p
300(103)c A 0.152 B d = 5301.44 N. The mass flow rate at sections A, B, and C, are
4
Q
dmA
dmB
dmC
0.75
=
= rW a b = 1000 a
b = 375 kg>s
= rWQ =
and
dt
dt
2
2
dt
1000(0.75) = 750 kg>s.
The speed of the water at sections A, B, and C are
vA = vB =
Q>2
0.75>2
= 84.88 m>s
=
p
AA
(0.0752)
4
vC =
Q
0.75
= 42.44 m>s.
=
p
AC
(0.152)
4
Steady Flow Equation: Writing the force steady flow equations along the x and
y axes,
dmA
dmB
dm C
+
: ©Fx = dt (vA)x + dt (vB)x - dt (vC)x;
Cx = - 375(84.88 cos 30°) + 375(84.88) - 0
Cx = 4264.54 N = 4.26 kN
dmB
dmC
dmA
(vA)y +
(vB)y (vC)y;
+ c ©Fy =
dt
dt
dt
Ans.
- Cy + 5301.44 = 375(84.88 sin 30°) + 0 - 750(42.44)
Cy = 21 216.93 N = 2.12 kN
Ans.
Writing the steady flow equation about point C,
dmA
dmB
dmC
dvA +
dvB dvC;
+ ©MC =
dt
dt
dt
-MC = 375(0.65)(84.88 cos 30°) - 375(0.25)(84.88 sin 30°)
+ [ -375(0.6)(84.88)] - 0
MC = 5159.28 N # m = 5.16 kN # m
Ans.
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C
lOMoARcPSD|12889239
15–130.
The mini hovercraft is designed so that air is drawn in at a
constant rate of 20 m3>s by the-fan blade and channeled to
provide a vertical thrust F, just sufficient to lift the hovercraft
off the water, while the remaining air is channeled to produce
a horizontal thrust T on the hovercraft. If the air is discharged horizontally at 200 m>s and vertically at 800 m>s,
determine the thrust T produced. The hovercraft and its passenger have a total mass of 1.5 Mg, and the density of the air
is ra = 1.20 kg>m3.
T
SOLUTION
Steady Flow Equation: The free-body diagram of the control volume A is shown in
dmA
Fig. a. The mass flow rate through the control volume is
= raQA = 1.20QA.
dt
Since the air intakes from a large reservoir (atmosphere), the velocity of the air
entering the control volume can be considered zero; i.e., (vA)in = 0. Also, the force
acting on the control volume is required to equal the weight of the hovercraft. Thus,
F = 1500(9.81) N.
(+ T )©Fy =
dmA
c (vA)out - (vA)in d ; 1500(9.81) = 1.20QA (800 - 0)
dt
QA = 15.328 m3>s
The flow rate through the control volume B, Fig. b, is QB = 20 - QA =
20 - 15.328 = 4.672 m3>s. Thus, the mass flow rate of this control volume is
dmB
= raQB = 1.20(4.672) = 5.606 kg>s. Again, the intake velocity of the control
dt
volume B is equal to zero; i.e., (vB)in = 0. Referring to the free-body diagram of this
control volume, Fig. b,
+ )©Fx = dmB c (vB)out - (vB)in d ; T = 5.606(200 - 0) = 1121.25 N = 1.12 kN
(;
dt
Ans.
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F
lOMoARcPSD|12889239
15–131.
B
Sand is discharged from the silo at A at a rate of 50 kg>s with
a vertical velocity of 10 m>s onto the conveyor belt, which is
moving with a constant velocity of 1.5 m>s. If the conveyor
system and the sand on it have a total mass of 750 kg and
center of mass at point G, determine the horizontal and vertical components of reaction at the pin support B roller support A. Neglect the thickness of the conveyor.
1.5 m/s
G
A
SOLUTION
10 m/s
Steady Flow Equation: The moment steady flow equation will be written about
point B to eliminate Bx and By. Referring to the free-body diagram of the control
volume shown in Fig. a,
+ ©MB =
dm
(dvB - dvA);
dt
750(9.81)(4) - Ay(8) = 50[0 - 8(5)]
Ay = 4178.5 N = 4.18 kN
Ans.
Writing the force steady flow equation along the x and y axes,
dm
+
: ©Fx = dt [(vB)x - (vA)x];
- Bx = 50(1.5 cos 30° - 0)
Bx = | -64.95 N| = 65.0 N :
+ c ©Fy =
dm
[(vB)y - (vA)y];
dt
30⬚
Ans.
By + 4178.5 - 750(9.81)
= 50[1.5 sin 30° - (- 10)]
By = 3716.25 N = 3.72 kN c
Ans.
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4m
4m
lOMoARcPSD|12889239
15–132. The
blows air
airatat6000
162 ft
m33min
min.. IfIfthe
thefan
fanhas
hasa
•15–117.
The fan blows
aweight
weight
N (,
15 kg) of
and
a center
gravity at the
G,
ofof
30150
lb and
a center
gravity
at G,ofdetermine
determine
the smallest
d of
itsitbase
will
smallest diameter
d of diameter
its base so
that
will so
notthat
tip itover.
not
over. The
specific
air is
Thetip
specific
weight
of airweight
is of
0.076
lbft3.11.82 Nm3.
162 m3
3
min
0.45 m
1 min
3 100
ft33s
Then,
2.7 m
s,. Then,
60 s
1
2.7
11.82
DM
2 )
16.98 ms. Also,
+a 1 (2.7) 3.253 kgs.
2
!
9.81
DT
4 (0.45 )
Equations of Steady Flow: Here 1 2
2
G
0.15 m
1.2 m
Applying Eq. 15–26 we have
a
i-/ DM
(D/" 2"
DT
D/! 2! ;฀ 150 2 0.15
D 0.583 m
D
3 3.253 [1.2(16.98)
2
0]
Ans.
150 N
d
2B 16.98 ms
1.2 m
0.15
d/2
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lOMoARcPSD|12889239
15-133.
The bend is connected to the pipe at flanges A
•15–121. The bend is connected to the pipe at flanges A and
and B as shown. If the diameter of the pipe is 0.3 m and
B as shown. If the diameter of the
pipe is 1 ft and it carries a
it carries a discharge
of 1.35 m3s, determine the horizontal
discharge of 50 ft3s, determine the horizontal and vertical
and vertical components of force reaction and the moment
components of force reaction and the moment reaction
reaction exerted at the fixed base D of the support. The
exerted at the fixed base D of the support. The total weight of
total weight of the bend and the water within it is
the bend and the water within it is 500 lb, with a mass center at
2500 N, with a mass center at point G. The gauge pressure
point G. The gauge pressure of the water at the flanges at A
of the water at the flanges at A and B are 120 kNm2
and B are 15 psi and 12 psi, respectively.Assume that no force
and 96 kNm2, respectively. Assume that no force is
is transferred to the flanges at A and B. The specific weight of
transferred to the flanges
at A and B. The specific weight
water is
62.4 lbft3.
of water isw w 10 kNm3.
G
45°
1.2 m
D
Free-Body Diagram: The free-body of the control volume is shown in Fig. a. The
force exerted on sections A and B due to the water pressure is
)
)
&! 0! !! 120 4 (0.32)5 8.4823 kN and &" 0" !" 96 4 (0.32)5
4
4
6.7858 kN. The speed of the water at, sections A and B are v! 1
1.35
19.10 ms. Also, the mass flow rate at these two sections
v" )
!
(0.32)
4
DM !
DM "
10(103)
are
+7 1 (1.35) 1376.1 kgs.
DT
DT
9.81
Steady Flow Equation: The moment steady flow equation will be written about
point D to eliminate Dx and Dy.
a
i-$ DM !
Dv! ;
DT
6.7858(103) cos 45°(1.2)
-$
DM "
Dv"
DT
1376.1(1.2)(19.10 cos 45°)
2500(0.45 cos 45°)
8.4823(103)(1.2)
[ 1376.1(1.2)(19.10)]
-$ 14 454.23 N m 14.45 kN m
Ans.
Writing the force steady flow equation along the x and y axes,
( + c ) i&Y $Y
DM
4 v" Y
DT
2500
v! Y 5 ;
6.7858(103) sin 45° 1376.1(19.10 sin 45°
0)
$Y 25 883.53 N 25.88 kN
+ ) i& DM v (:
" X
X
DT
8.4823(103)
v! X ;
Ans.
6.7858(103) cos 45°
$X 11382.28 N 11.38 kN
B
0.45 m
A
$X 1376.1[19.1 cos 45°
19.1]
Ans.
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lOMoARcPSD|12889239
v" 19.10 ms
F" 6.7858 kN
2500 N
F! 8.4823 kN
v! 19.10 ms
0.45 m
1.2 m
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lOMoARcPSD|12889239
15–134.
Each of the two stages A and B of the rocket has a mass of
2 Mg when their fuel tanks are empty. They each carry
500 kg of fuel and are capable of consuming it at a rate of
50 kg>sand eject it with a constant velocity of 2500 m> s,
measured with respect to the rocket. The rocket is launched
vertically from rest by first igniting stage B. Then stage A is
ignited immediately after all the fuel in B is consumed and
A has separated from B. Determine the maximum velocity
of stage A. Neglect drag resistance and the variation of the
rocket’s weight with altitude.
A
B
SOLUTION
The mass of the rocket at any instant t is m = (M + m0) - qt. Thus, its weight at the
same instant is W = mg = [(M + m0) - qt]g.
+ c ©Fs = m
dme
dv
dv
; -[(M + m0) - qt]g = [(M + m0) - qt]
- vD>e
- vD>eq
dt
dt
dt
vD>eq
dv
=
- g
dt
(M + m0) - qt
During the first stage,
vD>e = 2500 m>s. Thus,
M = 4000 kg,
m0 = 1000 kg ,
q = 50 kg>s ,
and
2500(50)
dv
=
- 9.81
dt
(4000 + 1000) - 50t
2500
dv
= a
- 9.81b m>s2
dt
100 - t
The time that it take to complete the first stage is equal to the time for all the fuel
500
in the rocket to be consumed, i.e., t =
= 10 s. Integrating,
50
L0
v1
dv =
L0
10 s
a
2500
- 9.81b dt
100 - t
v1 = [ -2500 ln(100 - t) - 9.81t] 2
10 s
0
= 165.30 m>s
During the second stage of launching, M = 2000 kg, m0 = 500 kg, q = 50 kg>s, and
vD>e = 2500 m>s. Thus, Eq. (1) becomes
2500(50)
dv
=
- 9.81
dt
(2000 + 500) - 50t
2500
dv
= a
- 9.81 b m>s2
dt
50 - t
The maximum velocity of rocket A occurs when it has consumed all the fuel. Thus,
500
the time taken is given by t =
= 10 s. Integrating with the initial condition
50
v = v1 = 165.30 m>s when t = 0 s,
vmax
L165.30 m>s
dv =
L0
10 s
a
2500
- 9.81b dt
50 - t
vmax - 165.30 = [- 2500 ln(50 - t) - 9.81t] 2
vmax = 625 m>s
10 s
0
Ans.
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lOMoARcPSD|12889239
15–135.
A power lawn mower hovers very close over the ground.
This is done by drawing air in at a speed of 6 m>s through
an intake unit A, which has a cross-sectional area of
A A = 0.25 m2, and then discharging it at the ground, B,
where the cross-sectional area is A B = 0.35 m2. If air at A
is subjected only to atmospheric pressure, determine the
air pressure which the lawn mower exerts on the ground
when the weight of the mower is freely supported and no
load is placed on the handle. The mower has a mass of
15 kg with center of mass at G. Assume that air has a
constant density of ra = 1.22 kg>m3.
vA
A
G
B
SOLUTION
dm
= r A A vA = 1.22(0.25)(6) = 1.83 kg>s
dt
+ c ©Fy =
dm
((vB)y - (vA)y)
dt
P = (0.35) - 15(9.81) = 1.83(0 - ( -6))
P = 452 Pa
Ans.
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lOMoARcPSD|12889239
15–136.
The rocket has mass M including the fuel. Determine the constant rate at which the fuel must
be burned so that its thrust gives the rocket a speed v in time t starting from rest. The fuel is
expelled from the rocket at relative speed vr. Neglect the effects of air resistance and assume
that g is constant.
Given:
M
v
65000 kg
60
t
vr
900
g
9.81
m
s
10 s
m
s
m
2
s
Solution:
A System That Losses Mass: Here,
·
d
§
¨ m0 me t¸ g
dt
©
¹
W
Applying Eq. 15-29, we have
n6
d
d
m v vDE me
dt
dt
Fz
§
·
¸ gt
d
¨ m0 me t ¸
dt
©
¹
mo
vDE ln ¨
v
m0
with
v
integrating we find
M
§
vr
·
¸ g( t)
d
¨ m0 me t ¸
dt
©
¹
vr ln ¨
d
me
dt
vDE
A
m0
§ m0
·1
¨
m0¸
¨ v g t
¸t
v
¨e r
¸
©
¹
A
§ m0
·1
¨
m0¸
¨ v g t
¸t
v
¨e r
¸
©
¹
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A
1047.2
kg
Ans.
s
lOMoARcPSD|12889239
15–137.
P
A chain of mass m0 per unit length is loosely coiled on the
floor. If one end is subjected to a constant force P when
y = 0, determine the velocity of the chain as a function of y.
y
SOLUTION
From the free-body diagram of the system shown in Fig. a, F cancels since it is
internal to the system. Here, vD>s = v since the chain on the horizontal surface is at
dms
= m0v, and the mass of the
rest. The rate at which the chain gains mass is
dt
system is m = m0y. Referring to Fig. a,
+ c ©Fs = m
dms
dv
;
+ vD>s
dt
dt
P - m0gy = m0y
dv
+ v(m0v)
dt
P - m0gy = m0y
dv
+ m0v2
dt
y
Since
dv
P
- gy
+ v2 =
m0
dt
dy
dy
,
= v or dt =
v
dt
dv
P
- gy
+ v2 =
m0
dy
vy
Multiplying by 2y dy,
a2vy2
Since
d A v2y2 B
dy
dv
2P
+ 2v2y b dy = a y - 2gy2 b dy
m0
dy
= 2vy2
(1)
dv
+ 2yv2, then Eq. (1) can be written as
dy
d A v2y2 B = a
2P
y - 2gy2 b dy
m0
Integrating,
L
d A v2y2 B =
v2y2 =
2P
a y - 2gy2 bdy
L m0
2
P 2
y - gy3 + C
m0
3
Substituting v = 0 at y = 0,
0 = 0 - 0 + C
C = 0
Thus,
v2y2 =
v =
P 2
2
y - gy3
m0
3
2
P
- gy
3
B m0
Ans.
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lOMoARcPSD|12889239
15–138. A coil of heavy open chain is used to reduce the
stopping distance of a sled that has a mass M and travels at
a speed of W0. Determine the required mass per unit length
of the chain needed to slow down the sled to ( ¼ )W0 within
a distance Y T if the sled is hooked to the chain at Y 0.
Neglect friction between the chain and the ground.
Observing the free-body diagram of the system shown in Fig. a, notice that the pair
of forces F, which develop due to the change in momentum of the chain, cancel each
other since they are internal to the system. Here, v%T v since the chain on the
ENT
ground is at rest. The rate at which the system gains mass is
Nv and the
EU
mass of the system is N NY .. Referring to Fig. a,
+ ) j'T N Ev
(:
EU
v%T
ENT
;
EU
0 NY
Since
0 NY
.)
Ev
EU
.
Nv2
Ev
EU
v Nv (1)
EY
EY
,
v or EU EU
v
NY
.v
Ev
EY
Nv2 0
Ev
N
A
I EY
v
NY .
(2)
Integrating using the limit v v0 at Y 0 and v 1
4
(v0
v0
1n v
T
Ev
N
3
4EY
v
(0 NY .
1
4 v0
v0
1n NY
.
1
4
NT .
N 1
v0 at Y T,
4
3.
T
.
T
0
Ans.
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lOMoARcPSD|12889239
15–139.
A commercial jet aircraft has a mass of 150 Mg
and is cruising at a constant speed of 850 kmh in level flight
(u 0°). If each of the two engines draws in air at a rate of
1000 kgs and ejects it with a velocity of 900 ms, relative to
the aircraft, determine the maximum angle of inclination u
at which the aircraft can fly with a constant speed of
750 kmh. Assume that air resistance (drag) is proportional
to the square of the speed, that is, '% DW2, where c is a
constant to be determined. The engines are operating with
the same power in both cases. Neglect the amount of fuel
consumed.
u
Steady Flow Equation: Since the air is collected from a large source (the
atmosphere), its entrance speed into the engine is negligible. The exit speed of the
air from the engine is given by
vF vQ
vFQ
When the airplane is in level flight, it has a constant speed of
1h
m
4 236.11ms. Thus,
vQ 5 850(103) 6 3
h 3600 s
+ )
(:
vF 236.11
900 663.89 ms By referring to the free-body diagram of the airplane shown in Fig. a,
+ ) j'Y EN v# Y v" Y ;
(:
$(236.112) 2(1000)(663.89 0)
EU
$ 23.817 kg sm
When the airplane is in the inclined position, it has a constant speed of
m
1h
4 208.33 ms. Thus,
vQ 5 750 103 6 3
h 3600 s
vF 208.33
900 691.67 ms
By referring to the free-body diagram of the airplane shown in Fig. b and using the
result of C, we can write
B j'Y EN
v# Y v" Y ;
EU
23.817 208.332 150 103 (9.81)sin u
2(1000)(691.67 0)
u 13.7°
Ans.
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lOMoARcPSD|12889239
15–140.
The 10-Mg helicopter carries a bucket containing 500 kg of
water, which is used to fight fires. If it hovers over the land in
a fixed position and then releases 50 kg>s of water at 10 m>s,
measured relative to the helicopter, determine the initial
upward acceleration the helicopter experiences as the water
is being released.
a
SOLUTION
+ c ©Ft = m
dm e
dy
- yD>e
dt
dt
Initially, the bucket is full of water, hence m = 10(103) + 0.5(103) = 10.5(103) kg
0 = 10.5(103)a - (10)(50)
a = 0.0476 m>s2
Ans.
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lOMoARcPSD|12889239
15–141.
The earthmover initially carries 10 m3 of sand having a
density of 1520 kg>m3. The sand is unloaded horizontally
through a 2.5 m 2 dumping port P at a rate of 900 kg>s
measured relative to the port. Determine the resultant
tractive force F at its front wheels if the acceleration of the
earthmover is 0.1 m>s2 when half the sand is dumped. When
empty, the earthmover has a mass of 30 Mg. Neglect any
resistance to forward motion and the mass of the wheels. The
rear wheels are free to roll.
P
F
SOLUTION
When half the sand remains,
m = 30 000 +
1
(10)(1520) = 37 600 kg
2
dm
= 900 kg>s = r vD>e A
dt
900 = 1520(vD>e)(2.5)
vD>e = 0.237 m>s
a =
dv
= 0.1
dt
+ ©F = m dv - dm v
;
s
dt
dt
F = 37 600(0.1) - 900(0.237)
F = 3.55 kN
Ans.
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lOMoARcPSD|12889239
15–142.
The 12-Mg jet airplane has a constant speed of 950 km>h
when it is flying along a horizontal straight line. Air enters
the intake scoops S at the rate of 50 m3>s. If the engine burns
fuel at the rate of 0.4 kg>s and the gas (air and fuel) is
exhausted relative to the plane with a speed of 450 m>s,
determine the resultant drag force exerted on the plane by
air resistance. Assume that air has a constant density of
1.22 kg>m3. Hint: Since mass both enters and exits the plane,
Eqs. 15–28 and 15–29 must be combined to yield
dme
dmi
dv
©Fs = m
- vD>e
+ vD>i
.
dt
dt
dt
v
950 km/h
S
SOLUTION
©Fs = m
dm e
dm i
dv
(vD>E) +
(v )
dt
dt
dt D>i
v = 950 km>h = 0.2639 km>s,
(1)
dv
= 0
dt
vD>E = 0.45 km>s
vD>t = 0.2639 km>s
dm t
= 50(1.22) = 61.0 kg>s
dt
dm e
= 0.4 + 61.0 = 61.4 kg>s
dt
Forces T and R are incorporated into Eq. (1) as the last two terms in the equation.
+ ) - F = 0 - (0.45)(61.4) + (0.2639)(61)
(;
D
FD = 11.5 kN
Ans.
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lOMoARcPSD|12889239
15–143.
The rocket has an initial mass m0, including the fuel. For practical reasons desired for the crew,
it is required that it maintain a constant upward acceleration a0. If the fuel is expelled from the
rocket at a relative speed ver, determine the rate at which the fuel should be consumed to
maintain the motion. Neglect air resistance, and assume that the gravitational acceleration is
constant.
Solution:
d
v
dt
a0
n 6Fs =
d
d
m v ver me
dt
dt
mao ver
mg
ver
dm
m
d
m
dt
a0 g dt
Since ver is constant, integrating, with t = 0 when m = m0 yields
§m·
¸
© m0 ¹
ver ln ¨
a0 g t
m
m0
§ a 0 g ·
¨ v ¸t
er ¹
e©
The time rate fuel consumption is determined from Eq.[1]
d
m
dt
m
a0 g
ver
d
m
dt
§ a 0 g ·
¨
¸t
§ a0 g · © ver ¹
m0¨
¸e
© ver ¹
Note : ver must be considered a negative quantity.
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Ans.
lOMoARcPSD|12889239
15–144.
A four-engine commercial jumbo jet is cruising at a
constant speed of 800 km>h in level flight when all four
engines are in operation. Each of the engines is capable of
discharging combustion gases with a velocity of 775 m>s
relative to the plane. If during a test two of the engines, one
on each side of the plane, are shut off, determine the new
cruising speed of the jet. Assume that air resistance (drag) is
proportional to the square of the speed, that is, FD = cv2,
where c is a constant to be determined. Neglect the loss of
mass due to fuel consumption.
SOLUTION
Steady Flow Equation: Since the air is collected from a large source (the
atmosphere), its entrance speed into the engine is negligible. The exit speed of the
air from the engine is
+ )
(:
ve + vp + ve>p
When the four engines are in operation, the airplane has a constant speed of
m
1h
b = 222.22 m>s. Thus,
vp = c 800(103) d a
h 3600 s
+ )
(:
ve = - 222.22 + 775 = 552.78 m>s :
Referring to the free-body diagram of the airplane shown in Fig. a,
+ ©F = dm C A v B - A v B D ;
:
B x
A x
x
dt
C(222.222) = 4
dm
(552.78 - 0)
dt
C = 0.044775
dm
dt
When only two engines are in operation, the exit speed of the air is
+ )
(:
ve = - vp + 775
Using the result for C,
dm
+
: ©Fx = dt
C A vB B x - A vA B x D ; ¢ 0.044775
dm
dm
≤ A vp 2 B = 2 C - vp + 775 B - 0 D
dt
dt
0.044775vp 2 + 2vp - 1550 = 0
Solving for the positive root,
vp = 165.06 m s = 594 km h
Ans.
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lOMoARcPSD|12889239
15–145.
v
The car has a mass m0 and is used to tow the smooth chain
having a total length l and a mass per unit of length m¿. If
the chain is originally piled up, determine the tractive force
F that must be supplied by the rear wheels of the car,
necessary to maintain a constant speed v while the chain is
being drawn out.
F
SOLUTION
dmi
dv
+
: ©Fs = m dt + vD>i dt
At a time t, m = m0 + ct, where c =
Here, vD>i = v,
dmi
m¿dx
=
= m¿v.
dt
dt
dv
= 0.
dt
F = (m0 - m¿v )(0) + v(m¿v) = m¿v2
Ans.
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lOMoARcPSD|12889239
15–146.
Water is discharged at speed v against the fixed cone diffuser. If the opening diameter of
the nozzle is d, determine the horizontal force exerted by the water on the diffuser.
Units Used:
3
Mg
10 kg
Given:
m
s
v
16
d
40 mm
T
30 deg
Uw
1
Mg
m
3
Solution:
Q
S 2
4
d v
m'
Uw Q
Fx
m' ¨ v cos ¨
§
©
§ T · v·
¸ ¸
©2¹ ¹
Fx
11.0 N
Ans.
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lOMoARcPSD|12889239
15–147.
Determine the magnitude of force F as a function of time,
which must be applied to the end of the cord at A to raise
the hook H with a constant speed v = 0.4 m>s. Initially the
chain is at rest on the ground. Neglect the mass of the cord
and the hook. The chain has a mass of 2 kg>m.
A
H
v
0.4 m/s
SOLUTION
dv
= 0,
dt
y = vt
mi = my = mvt
dmi
= mv
dt
+ c ©Fs = m
dm i
dv
+ vD>i (
)
dt
dt
F - mgvt = 0 + v(mv)
F = m(gvt + v2)
= 2[9.81(0.4)t + (0.4)2]
F = (7.85t + 0.320) N
Ans.
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lOMoARcPSD|12889239
15–148.
The truck has a mass of 50 Mg when empty.When it is unloading 5 m3 of sand at a constant rate of 0.8 m3>s, the sand flows
out the back at a speed of 7 m> s, measured relative to the
truck, in the direction shown. If the truck is free to roll, determine its initial acceleration just as the load begins to empty.
Neglect the mass of the wheels and any frictional resistance to
motion. The density of sand is rs = 1520 kg>m3.
a
45⬚
7 m/s
SOLUTION
A System That Loses Mass: Initially, the total mass of the truck is
dme
and
m = 50(103) + 5(1520) = 57.6(103) kg
= 0.8(1520) = 1216 kg>s .
dt
Applying Eq. 15–29, we have
dme
dv
+
: ©Fs = m dt - vD>e dt ;
0 = 57.6(103)a - (0.8 cos 45°)(1216)
a = 0.104 m>s2
Ans.
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lOMoARcPSD|12889239
15–149.
The chain has a total length L 6 d and a mass per unit
length of m¿ . If a portion h of the chain is suspended over
the table and released, determine the velocity of its end A
as a function of its position y. Neglect friction.
h
y
d
A
SOLUTION
©Fs = m
dm e
dv
+ vD>e
dt
dt
m¿gy = m¿y
dv
+ v(m¿v)
dt
m¿gy = m¿ ay
Since dt =
gy = vy
dv
+ v2 b
dt
dy
, we have
v
dv
+ v2
dy
Multiply by 2y and integrate:
L
2gy2 dy =
L
a 2vy2
dv
+ 2yv2 b dy
dy
2 3 3
g y + C = v2y2
3
2
when v = 0, y = h, so that C = - gh3
3
Thus, v2 =
v =
2 y3 - h3
ga
b
3
y2
y3 - h3
2
b
ga
B3
y2
Ans.
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