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Problem 14.33 The crane’s trolley at Amoves to the
right with constant acceleration, and the 800-kg load
moves without swinging.
(a) What is the acceleration of the trolley and load?
(b) What is the sum of the tensions in the parallel
cables supporting the load?
5°
A
m,T=mg
cos θ, from which
a=gtan θ=9.81(tan 5◦)=0.858 m/s2
(b) T=800(9.81)
cos 5◦=7878 N
mg
Problem 14.34 The mass of Ais 30 kg and the mass
of Bis 5 kg. The horizontal surface is smooth. The con-
stant force Fcauses the system to accelerate. The angle
θ=20◦is constant. Determine F.
F
A
B
u
Solution: We have four unknowns (F,T,N,a)and four equations
130
Problem 14.35 The mass of Ais 30 kg and the mass
of Bis 5 kg. The coefficient of kinetic friction between
Aand the horizontal surface is µk=0.2. The constant
force Fcauses the system to accelerate. The angle θ=
20◦is constant. Determine F.
F
A
B
u
Solution: We have four unknowns (F,T,N,a)and four equations
Problem 14.36 The 100-lb crate is initially stationary.
The coefficients of friction between the crate and the
inclined surface are µs=0.2 and µk=0.16. Determine
how far the crate moves from its initial position in 2 s
if the horizontal force F=90 lb.
30°
F
Problem 14.37 In Problem 14.36, determine how far
the crate moves from its initial position in 2 s if the
Solution: Use the definitions of terms given in the solution to
Problem 14.36. For F=30 lb, N=Fsin θ+Wcos θ=101.6 lb, and
Problem 14.38 The crate has a mass of 120 kg, and
the coefficients of friction between it and the sloping
dock are µs=0.6 and µk=0.5.
(a) What tension must the winch exert on the cable to
start the stationary crate sliding up the dock?
(b) If the tension is maintained at the value determined
in part (a), what is the magnitude of the crate’s
velocity when it has moved 2 m up the dock?
30°
Solution: Choose a coordinate system with the xaxis parallel to
the surface. Denote θ=30◦.
move the crate is Fc=T−Wsin θ, from which the tension
(b) After slip begins, the force acting to move the crate is F=
T−Wsin θ−µkN=101.95 N. From Newton’s second law,
F=ma, from which a=F
m=101.95
120 =0.8496 m/s2. The
velocity is v(t) =at =0.8496tm/s, since v(0)=0. The position
is s(t) =a
2t2, since s(0)=0. When the crate has moved
2 m up the slope, t10 =2(2)
a=2.17 s and the velocity is
v=a(2.17)=1.84 m/s.
T
Problem 14.39 The coefficients of friction between the
load Aand the bed of the utility vehicle are µs=0.4
and µk=0.36. If the floor is level (θ =0), what is the
largest acceleration (in m/s2) of the vehicle for which
the load will not slide on the bed?
A
u
132
Problem 14.40 The coefficients of friction between the
load Aand the bed of the utility vehicle are µs=0.4
and µk=0.36. The angle θ=20◦. Determine the largest
forward and rearward acceleration of the vehicle for
which the load will not slide on the bed.
A
u
Solution: The load is on the verge of slipping. There are two
unknowns (aand N).
Forward: The equations are
Rearward: The equations are
Problem 14.41 The package starts from rest and slides
down the smooth ramp. The hydraulic device Bexerts
a constant 2000-N force and brings the package to rest
in a distance of 100 mm from the point where it makes
contact. What is the mass of the package?
2 m
A
Solution: Set g=9.81 m/s2
a=vdv
ds =gsin 30◦−2000 N
m
0
4.43 m/s
vdv =0.1m
0gsin 30◦−2000 N
mds
0−(4.43 m/s)2
2=gsin 30◦−2000 N
m(0.1 m)
Solving the last equation we find m=19.4 kg
30°
N
2000 N
Problem 14.42 The force exerted on the 10-kg mass
by the linear spring is F=−ks, where kis the spring
constant and sis the displacement of the mass relative
to its position when the spring is unstretched. The value
of kis 40 N/m. The mass is in the position s=0 and
is given an initial velocity of 4 m/s toward the right.
Determine the velocity of the mass as a function of s.
Strategy: : Use the chain rule to write the acceleration
as dv
dt=dv
ds
ds
dt=dv
dsv.
k
s
Solution: The equation of motion is −ks =ma
k
0
s2
134
Problem 14.43 The 450-kg boat is moving at 10 m/s
when its engine is shut down. The magnitude of the
hydrodynamic drag force (in newtons) is 40v2, where
vis the magnitude of the velocity in m/s. When the
boat’s velocity has decreased to 1 m/s, what distance
has it moved from its position when the engine was shut
down?
Problem 14.44 A sky diver and his parachute weigh
200 lb. He is falling vertically at 100 ft/s when his
parachute opens. With the parachute open, the magnitude
of the drag force (in pounds) is 0.5v2. (a) What is the
magnitude of the sky diver’s acceleration at the instant
the parachute opens? (b) What is the magnitude of his
velocity when he has descended 20 ft from the point
where his parachute opens?
Problem 14.45 The Panavia Tornado, with a mass of
18,000 kg, lands at a speed of 213 km/h. The decelerat-
ing force (in newtons) exerted on it by its thrust reversers
and aerodynamic drag is 80,000 +2.5v2, where vis the
airplane’s velocity in m/s. What is the length of the air-
plane’s landing roll? (See Example 14.4.)
136
Problem 14.46 A 200-lb bungee jumper jumps from
a bridge 130 ft above a river. The bungee cord has an
unstretched length of 60 ft and has a spring constant
k=14 lb/ft. (a) How far above the river is the jumper
when the cord brings him to a stop? (b) What maximum
force does the cord exert on him?
Problem 14.47 A helicopter weighs 20,500 lb. It takes
off vertically from sea level, and its upward velocity
is given as a function of its altitude hin feet by
v=66 −0.01 hft/s.
(a) How long does it take the helicopter to climb to an
altitude of 4000 ft?
(b) What is the sum of the vertical forces on the heli-
copter when its altitude is 2000 ft?
32.2 ft/s2(−0.46 ft/s2)=−293 lb
Problem 14.48 In a cathode-ray tube, an electron
(mass =9.11 ×10−31 kg) is projected at Owith
velocity v=(2.2×107)i(m/s). While the electron is
between the charged plates, the electric field generated
by the plates subjects it to a force F=−eEj, where
the charge of the electron e=1.6×10−19 C (coulombs)
and the electric field strength E=15 kN/C. External
forces on the electron are negligible when it is not
between the plates. Where does the electron strike the
screen?
O
x
y
+ + + +
– – – –
Screen
30
mm 100 mm
138
Problem 14.49 In Problem 14.48, determine where
the electron strikes the screen if the electric field strength
Solution: Use the solution to Problem 14.48. Assume that the
electron enters the space between the charged plates at t=
9.11 ×10−31 kg j=−j(2.6345 ×1015)
Problem 14.50 An astronaut wants to travel from a
space station to a satellites S that needs repair. She
departs the space station at O. A spring-loaded launch-
ing device gives her maneuvering unit an initial velocity
of 1 m/s (relative to the space station) in the ydirection.
At that instant, the position of the satellite is x=70 m,
y=50 m, z=0, and it is drifting at 2 m/s (relative to
the station) in the xdirection. The astronaut intercepts
the satellite by applying a constant thrust parallel to the
xaxis. The total mass of the astronaut and her maneu-
vering unit is 300 kg. (a) How long does it take the
astronaut to reach the satellite? (b) What is the magni-
tude of the thrust she must apply to make the intercept?
(c) What is the astronaut’s velocity relative to the satel-
lite when she reaches it?
y
x
S
O
Problem 14.51 What is the acceleration of the 8-kg
collar Arelative to the smooth bar?
20°
45°
200 N
A
a=Fc
m=3.63 m/s2.
F
g
8=2.06 m/s2up the bar.
Problem 14.53 The force F=50 lb. What is the mag-
nitude of the acceleration of the 20-lb collar Aalong the
smooth bar at the instant shown?
y
(5, 3, 0) ft
(2, 2, 2) ft
A
32.2 ft/s2a⇒a=22.9 ft/s2
140
c
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Problem 14.54* In Problem 14.53, determine the
magnitude of the acceleration of the 20-lb collar Aalong
the bar at the instant shown if the coefficient of static
friction between the collar and the bar is µk=0.2.
Solution: Use the results from Problem 14.53.
32.2 ft/s2a⇒a=7.89 ft/s2
Problem 14.55 The 6-kg collar starts from rest at posi-
tion A, where the coordinates of its center of mass are
(400, 200, 200) mm, and slides up the smooth bar to
position B, where the coordinates of its center of mass
are (500, 400, 0) mm under the action of a constant force
F=−40i+70j−40k(N). How long does it take to go
from Ato B?
Strategy: There are several ways to work this problem.
One of the most straightforward ways is to note that the
motion is along the straight line from Ato Band that
only the force components parallel to line AB cause
acceleration. Thus, a good plan would be to find a unit
vector from Atoward Band to project all of the forces
acting on the collar onto line AB. The resulting constant
force (tangent to the path), will cause the acceleration of
the collar. We then only need to find the distance from
Ato Bto be able to analyze the motion of the collar.
x
F
y
z
A
B
Solution: The unit vector from Atoward Bis eAB =et=0.333i
Problem 14.56* In Problem 14.55, how long does the
collar take to go from Ato Bif the coefficient of kinetic
friction between the collar and the bar is µk=0.2?
Strategy: This problem is almost the same as prob-
lem 14.55. The major difference is that now we must
calculate the magnitude of the normal force, N, and then
must add a term µk|N|to the forces tangent to the bar (in
the direction from Btoward A— opposing the motion).
This will give us a new acceleration, which will result
in a longer time for the collar to go from Ato B.
In the direction tangent to the bar, the equation is (F+W)·eAB −
µk|N|=mat.
Problem 14.57 The crate is drawn across the floor by a
winch that retracts the cable at a constant rate of 0.2 m/s.
The crate’s mass is 120 kg, and the coefficient of kinetic
friction between the crate and the floor is µk=0.24.
(a) At the instant shown, what is the tension in the cable?
(b) Obtain a “quasi-static” solution for the tension in the
cable by ignoring the crate’s acceleration. Compare this
solution with your result in (a).
4 m
2 m
Solution:
L
2 m
142
Problem 14.58 If y=100 mm, dy
dt =600 mm/s, and
d2y
dt2=−200 mm/s2, what horizontal force is exerted on
the 0.4 kg slider Aby the smooth circular slot?
300 mm
A
y
Problem 14.59 The 1-kg collar Pslides on the vertical
bar and has a pin that slides in the curved slot. The
vertical bar moves with constant velocity v=2 m/s. The
yaxis is vertical. What are the xand ycomponents of
the total force exerted on the collar by the vertical bar
and the slotted bar when x=0.25 m?
y
1 m
P
x
y = 0.2 sin x
π
Solution:
Problem 14.60* The 1360-kg car travels along a
straight road of increasing grade whose vertical profile
is given by the equation shown. The magnitude of the
car’s velocity is a constant 100 km/h. When x=200 m,
what are the xand ycomponents of the total force acting
on the car (including its weight)?
y = 0.0003x2
y
Problem 14.61* The two 100-lb blocks are released
Solution: The relative motion of the blocks is constrained by the
144
Problem 14.62* The two 100-lb blocks are released
from rest. The coefficient of kinetic friction between all
contacting surfaces is µk=0.1. How long does it take
block Ato fall 1 ft?
0.284 s
Problem 14.63 The 3000-lb vehicle has left the ground
after driving over a rise. At the instant shown, it is mov-
ing horizontally at 30 mi/h and the bottoms of its tires
are 24 in above the (approximately) level ground. The
earth-fixed coordinate system is placed with its origin 30
in above the ground, at the height of the vehicle’s center
of mass when the tires first contact the ground. (Assume
that the vehicle remains horizontal.) When that occurs,
the vehicle’s center of mass initially continues moving
downward and then rebounds upward due to the flexure
of the suspension system. While the tires are in con-
tact with the ground, the force exerted on them by the
ground is −2400i−18000yj(lb), where yis the vertical
position of the center of mass in feet. When the vehicle
rebounds, what is the vertical component of the velocity
of the center of mass at the instant the wheels leave the
ground? (The wheels leave the ground when the center
of mass is at y=0.)
y
24 in
30 in 24 in
30 in
x
Problem 14.64* A steel sphere in a tank of oil is given
an initial velocity v=2i(m/s) at the origin of the coor-
dinate system shown. The radius of the sphere is 15 mm.
The density of the steel is 8000 kg/m3and the density
of the oil is 980 kg/m3.IfVis the sphere’s volume,
the (upward) buoyancy force on the sphere is equal to
the weight of a volume Vof oil. The magnitude of the
hydrodynamic drag force Don the sphere as it falls
is |D|=1.6|v|N, where |v|is the magnitude of the
sphere’s velocity in m/s. What are the xand ycom-
ponents of the sphere’s velocity at t=0.1 s?
x
y
146
Problem 14.65* In Problem 14.64, what are the xand
ycoordinates of the sphere at t=0.1s?
x=vxo
y=a
b2(ebt −1)−a
bt
x=0.1070 m =107.0mm
Problem 14.66 The boat in Active Example 14.5
weighs 1200 lb with its passengers. Suppose that the
boat is moving at a constant speed of 20 ft/s in a circu-
lar path with radius R=40 ft. Determine the tangential
and normal components of force acting on the boat.
R
Problem 14.67 In preliminary design studies for a sun-
powered car, it is estimated that the mass of the car
and driver will be 100 kg and the torque produced by
the engine will result in a 60-N tangential force on the
car. Suppose that the car starts from rest on the track at
Aand is subjected to a constant 60-N tangential force.
Determine the magnitude of the car’s velocity and the
normal component of force on the car when it reaches B.
B
A
200 m
50 m
Solution: We first find the tangential acceleration and use that to
Problem 14.68 In a test of a sun-powered car, the
mass of the car and driver is 100 kg. The car starts
from rest on the track at A, moving toward the right.
The tangential force exerted on the car (in newtons) is
given as a function of time by Ft=20 +1.2t. Deter-
mine the magnitude of the car’s velocity and the normal
component of force on the car at t=40 s.
B
A
200 m
50 m
Solution: We first find the tangential acceleration and use that to
find the velocity vand distance sas functions of time.
148
Problem 14.69 An astronaut candidate with a mass of
72 kg is tested in a centrifuge with a radius of 10 m. The
centrifuge rotates in the horizontal plane. It starts from
rest at time t=0 and has a constant angular acceleration
of 0.2 rad/s2. Determine the magnitude of the horizontal
force exerted on him by the centrifuge (a) at t=0; (b) at
t=10 s.
10 m
Solution: The accelerations are
Problem 14.70 The circular disk lies in the horizon-
tal plane. At the instant shown, the disk rotates with
a counterclockwise angular velocity of 4 rad/s and a
counterclockwise angular acceleration of 2 rad/s2. The
0.5-kg slider Ais supported horizontally by the smooth
slot and the string attached at B. Determine the tension
in the string and the magnitude of the horizontal force
exerted on the slider by the slot.
4 rad/s2 rad/s2
0.6 m
B
A
Solution:
