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Problem 14.1 In Active Example 14.1, suppose that
the coefficient of kinetic friction between the crate and
the inclined surface is µk=0.12. Determine the dis-
tance the crate has moved down the inclined surface at
t=1s.
20⬚
Solution: There are three unknowns in this problem: N,f, and
a. We will first assume that the crate does not slip. The governing
equations are
110
Problem 14.2 The mass of the Sikorsky UH-60A heli-
copter is 9300 kg. It takes off vertically with its rotor
exerting a constant upward thrust of 112 kN.
(a) How fast is the helicopter rising 3 s after it takes
off?
(b) How high has it risen 3 s after it takes off?
Strategy: Be sure to draw the free-body diagram of
the helicopter.
Solution: The equation of motion is
Problem 14.3 The mass of the Sikorsky UH-60A heli-
copter is 9,300 kg. It takes off vertically at t=0. The
pilot advances the throttle so that the upward thrust of its
engine (in kN) is given as a function of time in seconds
by T=100 +2t2.
(a) How fast is the helicopter rising 3 s after it takes
off?
(b) How high has it risen 3 s after it takes off?
Solution: The equation of motion is
t
Problem 14.4 The horizontal surface is smooth. The
30-lb box is at rest when the constant force Fis applied.
Two seconds later, the box is moving to the right at
20 ft/s. Determine F.
F
20⬚
Solution: We use one governing equation and one kinematic
relation
Problem 14.5 The coefficient of kinetic friction between
the 30-lb box and the horizontal surface is µk=0.1. The
box is at rest when the constant force Fis applied. Two
seconds later, the box is moving to the right at 20 ft/s.
Determine F.
F
20⬚
Solution: We use two governing equations, one slip equation, and
Fx:Fcos 20◦−f=30 lb
32.2 ft/s2a,
Fy:N−Fsin 20◦−30 lb =0,
f=(0.1)N,
v=(20 ft/s)=a(2s).
Solving, we find
a=10 ft/s2,N =34.7lb,f =3.47 lb,F=13.6lb.
F = 10 N 78,48 N
N
Fr
Problem 14.6 The inclined surface is smooth. The
velocity of the 114-kg box is zero when it is subjected
to a constant horizontal force F=20 N. What is the
velocity of the box two seconds later?
F
20⬚
112
Problem 14.7 The coefficient of kinetic friction between
the 14-kg box and the inclined surface is µk=0.1. The
velocity of the box is zero when it is subjected to a con-
stant horizontal force F=20 N. What is the velocity of
the box two seconds later?
F
20⬚
Solution: From the governing equations and the slip equation, we
can find the acceleration of the box (assumed to be down the incline).
Problem 14.8 The 170-lb skier is schussing on a 25◦
slope. At the instant shown, he is moving at 40 ft/s. The
kinetic coefficient of friction between his skis and the
snow is µk=0.08. If he makes no attempt to check his
speed, how long does it take for it to increase to 60 ft/s?
Solution: The governing equations and the slip equation are used
to find the acceleration
F ր:N−(170 lb)cos 25◦=0,
32.2 ft/s2a.
a=11.3 ft/s2,N =154 lb,f =12.3lb.
114
Problem 14.9 The 170-lb skier is schussing on a 25◦
slope. At the instant shown, he is moving at 40 ft/s. The
kinetic coefficient of friction between his skis and the
snow is µk=0.08. Aerodynamic drag exerts a resisting
force on him of magnitude 0.015v2, where vis the mag-
nitude of his velocity. If he makes no attempt to check
his speed, how long does it take for it to increase to
60 ft/s?
32.2 ft/s2a.
60 ft/s
dv
dt=t
Problem 14.10 The total external force on the 10-kg
object is constant and equal to F=90i−60j+20k(N).
At time t=0, its velocity is v=−14i+26j+32k(m/s).
What is its velocity at t=4 s? (See Active Example 14.2.)
y
兺 F
Solution:
Problem 14.11 The total external force on the 10-kg
object shown in Problem 14.10 is given as a function of
time by F=(−20t+90)i−60j+(10t+40)k(N).
At time t=0, its position is r=40i+30j−360k(m)
and its velocity is v=−14i+26j+32k(m/s). What is
its position at t=4s?
Problem 14.12 The position of the 10-kg object shown
in Problem 14.10 is given as a function of time by r=
(20t3−300)i+60t2j+(6t4−40t2)k(m). What is the
total external force on the object at t=2s?
116
Problem 14.13 The total force exerted on the 80,000-
lb launch vehicle by the thrust of its engine, its weight,
and aerodynamic forces during the interval of time from
t=2stot=4 s is given as a function of time by F=
(2000 −400t2)i+(5200 +440t)j+(800 +60t2)k(lb).
At t=2 s, its velocity is v=12i+220j−30k(ft/s).
What is its velocity at t=4s?
m=(2000 −400t2)lb
32.2 ft/s2=(0.805 −0.161t2)ft/s2,
3[0.0242][43−23]−30ft/s =−28.9 ft/s,
Problem 14.14 At the instant shown, the horizontal
component of acceleration of the 26,000-lb airplane due
to the sum of the external forces acting on it is 14 ft/s2.If
the pilot suddenly increases the magnitude of the thrust
Tby 4000 lb, what is the horizontal component of the
plane’s acceleration immediately afterward?
15°
x
T
y
Solution: Before
Problem 14.15 At the instant shown, the rocket is trav-
eling straight up at 100 m/s. Its mass is 90,000 kg and
the thrust of its engine is 2400 kN. Aerodynamic drag
exerts a resisting force (in newtons) of magnitude 0.8v2,
where vis the magnitude of the velocity. How long does
it take for the rocket’s velocity to increase to 200 m/s?
200 m/s
dv
dt=t
118
Problem 14.16 A 2-kg cart containing 8 kg of water
is initially stationary (Fig. a). The center of mass of the
“object” consisting of the cart and water is at x=0. The
cart is subjected to the time-dependent force shown in
Fig. b, where F0=5 N and t0=2 s. Assume that no
water spills out of the cart and that the horizontal forces
exerted on the wheels by the floor are negligible.
(a) Do you know the acceleration of the cart during
the period 0 <t<t
0?
(b) Do you know the acceleration of the center of mass
of the “object” consisting of the cart and water
during the period 0 <t<t
0?
(c) What is the x-coordinate of the center of mass of
the “object” when t>2t0?
t
F
F0
y
x
(a)
Problem 14.17 The combined weight of the motorcy-
cle and rider is 360 lb. The coefficient of kinetic friction
between the tires and the road is µk=0.8. The rider
starts from rest, spinning the rear wheel. Neglect the
horizontal force exerted on the front wheel by the road.
In two seconds, the motorcycle moves 35 ft. What was
the normal force between the rear wheel and the road?
32.2 ft/s2(17.5 ft/s2)=195.6 lb
Problem 14.18 The mass of the bucket Bis 180 kg.
From t=0tot=2 s, the xand ycoordinates of the
center of mass of the bucket are
x=−0.2t3+0.05t2+10 m,
y=0.1t2+0.4t+6m.
Determine the xand ycomponents of the force exerted
on the bucket by its supports at t=1s.
x
B
y
Solution:
Problem 14.19 During a test flight in which a 9000-kg
helicopter starts from rest at t=0, the acceleration of its
center of mass from t=0tot=10 s is a=(0.6t)i+
(1.8−0.36t)jm/s2. What is the magnitude of the total
external force on the helicopter (including its weight) at
t=6s?
T
L
Path
y
Solution: From Newton’s second law: F=ma. The sum of
the external forces is F=F−W=9000[(0.6t)i+(1.8−0.36t)
120
Problem 14.20 The engineers conducting the test
described in Problem 14.19 want to express the total
force on the helicopter at t=6 s in terms of three forces:
the weight W, a component Ttangent to the path, and
a component Lnormal to the path. What are the values
of W,T, and L?
Solution: Integrate the acceleration: v=(0.3t2)i+(1.8t−0.18
t2)j, since the helicopter starts from rest. The instantaneous flight
0.3(6)2=21.8◦. A unit vector
tangent to this path is et=icos β+jsin β. A unit vector normal to
this path en=−isin β+jcos β. The weight acts downward:
Problem 14.21 At the instant shown, the 11,000-kg
airplane’s velocity is v=270 im/s. The forces acting
on the plane are its weight, the thrust T=110 kN, the
lift L=260 kN, and the drag D=34 kN. (The x-axis is
parallel to the airplane’s path.) Determine the magnitude
of the airplane’s acceleration.
y
x
LT
Horizontal
Path
15°
15°
D
mg
Problem 14.22 At the instant shown, the 11,000-kg
airplane’s velocity is v=300i(m/s). The rate of change
of the magnitude of the velocity is dv/dt =5 m/s2. The
radius of curvature of the airplane’s path is 4500 m, and
the yaxis points toward the concave side of the path.
The thrust is T=120,000 N. Determine the lift Land
drag D.
Solution:
g=9.81 m/s2
T=120000 N
Fx:Tcos 15◦−D−mg sin 15◦=max
Fy:L+Tsin 15◦−mg cos 15◦=may
ay=V2/ρ
Solving, D=33.0kN,L=293 kN
L
T
y
x
15°
15°
mg
D
T
15°
Problem 14.23 The coordinates in meters of the 360-
kg sport plane’s center of mass relative to an earth-
fixed reference frame during an interval of time are x=
20t−1.63t2,y=35t−0.15t3, and z=−20t+1.38t2,
where tis the time in seconds. The y- axis points upward.
The forces exerted on the plane are its weight, the thrust
vector Texerted by its engine, the lift force vector
L, and the drag force vector D.Att=4 s, determine
T+L+D.
y
x
z
122
Problem 14.24 The force in newtons exerted on the
360-kg sport plane in Problem 14.23 by its engine, the
lift force, and the drag force during an interval of time
is T+L+D=(−1000 +280t)i+(4000 −430t)j+
(720 +200t)k, where tis the time in seconds. If the
coordinates of the plane’s center of mass are (0, 0, 0)
and its velocity is 20i+35j−20k(m/s) at t=0, what
are the coordinates of the center of mass at t=4s?
Problem 14.25 The robot manipulator is programmed
so that x=40 +24t2mm, y=4t3mm, and z=0 dur-
ing the interval of time from t=0tot=4 s. The y
axis points upward. What are the xand ycomponents
of the total force exerted by the jaws of the manipulator
on the 2-kg widget Aat t=3s? A
x
y
y
x
Problem 14.26 The robot manipulator described in
Problem 14.25 is reprogrammed so that it is stationary
at t=0 and the components of its acceleration are ax=
400 −0.8vxmm/s2,ay=200 −0.4vymm/s2from t=
0tot=2 s, where vxand vyare the components of
robot’s velocity in mm/s. The yaxis points upward.
What are the xand ycomponents of the total force
exerted by the jaws of the manipulator on the 2-kg
widget Aat t=1s?
124
Problem 14.27 In the sport of curling, the object is to
slide a “stone” weighting 44 lb into the center of a target
located 31 yards from the point of release. In terms of
the coordinate system shown, the point of release is at
x=0,y =0. Suppose that a shot comes to rest at x=
31.0 yards, y=1 yard. Assume that the coefficient of
kinetic friction is constant and equal to µk=0.01. What
were the xand ycomponents of the stone’s velocity at
release?
y
31 yd
Curling
stone
Solution: The stone travels at an angle relative to the xaxis.
Problem 14.28 The two masses are released from rest.
How fast are they moving at t=0.5 s? (See Example 14.3.)
5 kg
2 kg
Solution: The free-body diagrams are shown. The governing
equations are
Problem 14.29 The two weights are released from rest.
The horizontal surface is smooth. (a) What is the ten-
sion in the cable after the weights are released? (b) How
fast are the weights moving one second after they are
released?
5 lb
Problem 14.30 The two weights are released from rest.
The coefficient of kinetic friction between the horizon-
tal surface and the 5-lb weight is µk=0.18. (a) What is
the tension in the cable after the weights are released?
(b) How fast are the weights moving one second after
they are released?
5 lb
Problem 14.31 The mass of each box is 14 kg. One
second after they are released from rest, they have moved
0.3 m from their initial positions. What is the coefficient
of kinetic friction between the boxes and the surface?
30⬚
Solution: We will first use the kinematic information to find the
From the free-body diagrams we have four equations of motion:
128
Problem 14.32 The masses mA=15 kg and mB=30 kg,
and the coefficients of friction between all of the surfaces
are µs=0.4 and µk=0.35. The blocks are station-
ary when the constant force Fis applied. Determine
the resulting acceleration of block Bif (a) F=200 N;
(b) F=400 N.
F
B
A
Solution: Assume that no motion occurs anywhere. Then
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