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Bed DynSMch14

The problem involves a 170 lb skier moving down a 25 degree slope at 40 ft/s. With a kinetic coefficient of friction of 0.08 between the skis and snow, the governing equations and slip equation are used to find the acceleration, which is then used with kinematics equations to determine how long it will take for the skier's speed to increase to 60 ft/s.

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0% found this document useful (0 votes)

326 views78 pages

Bed DynSMch14

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FMASO

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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 = 1 s.

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
◦
F : (100 lb) sin 20 − f

100 lb
= a
32.2 ft/s2

◦
F : N − (100 lb) cos 20 = 0

No slip: a = 0
Solving, we find that N = 94.0 lb, f = 34.2 lb, a = 0.

To check the no slip assumption, we calculate the maximum friction

force
fmax = µs N = 0.2(94.0 lb) = 18.8 lb.

Since f > fmax , we conclude that our no slip assumption is false. The
governing equations are now known to be

◦ 100 lb
F : (100 lb) sin 20 − f = 2
a
32.2 ft/s

◦
F : N − (100 lb) cos 20 = 0

Slip: f = 0.12N

Solving we have N = 94.0 lb, f = 11.3 lb, a = 7.38 ft/s2 .

1 2 1
To find the distance we have d = at = (7.38 ft/s2 )(1 s)2 = 3.69 ft.
2 2
d = 3.69 ft.

(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

F : 112 kN − 9.3(9.81) kN

= (9,300 kg)a
Solving, we find that
a = 2.23 m/s2 .
Using kinematics we can answer the questions

a = 2.23 m/s2 ,

v = at = (2.23 m/s2 )(3 s) = 6.70 m/s,

1 2 1
h= at = (2.23 m/s2 )(3 s)2 = 10.0 m.
2 2

(a) 6.70 m/s, (b) 10.0 m.

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 + 2t 2 .

(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

2
t
F : 100 kN + 2 kN
s

−9.3(9.81) kN = (9, 300 kg)a

Solving, we find that
a = (0.943 m/s2 ) + (0.215 m/s4 )t 2 .

Using kinematics we can answer the questions

a = (0.943 m/s2 ) + (0.215 m/s4 )t 2

1
v = (0.943 m/s2 )t + (0.215 m/s4 )t 3
3

1 1
h= (0.943 m/s2 )t 2 + (0.215 m/s4 )t 4
2 12
Evaluating these expressions at t = 3 s,

(a) v = 4.76 m/s, (b) d = 5.69 m.

Solution: We use one governing equation and one kinematic

relation
◦ 30 lb
Fx : F cos 20 = a,
32.2 ft/s2

v = (20 ft/s) = a(2 s).

Solving, we find a = 10 ft/s2 , F = 9.91 lb.

Problem 14.5 The coefficient of kinetic friction between F

the 30-lb box and the horizontal surface is µk = 0.1. The 20⬚
box is at rest when the constant force F is applied. Two
seconds later, the box is moving to the right at 20 ft/s.
Determine F .

Solution: We use two governing equations, one slip equation, and F = 10 N 78,48 N
one kinematic relation

◦ 30 lb 20°
Fx : F cos 20 − f = 2
a,
32.2 ft/s

◦
Fy : N − F sin 20 − 30 lb = 0,

f = (0.1)N, Fr

v = (20 ft/s) = a(2 s).

N
Solving, we find

a = 10 ft/s2 , N = 34.7 lb, f = 3.47 lb, F = 13.6 lb.

Problem 14.6 The inclined surface is smooth. The

velocity of the 114-kg box is zero when it is subjected F
to a constant horizontal force F = 20 N. What is the
velocity of the box two seconds later?
20⬚

Solution: From the governing equation we can find the accelera-

tion of the box (assumed to be down the incline).
◦
F : 14(9.81) N sin 20

◦
−(20 N) cos 20 = (14 kg)a

Solving, we have a = 2.01 m/s2 .

Since a > 0, our assumption is correct.
Using kinematics,

v = at = (2.01 m/s2 )(2 s) = 4.03 m/s.

v = 4.03 m/s down the incline.

Solution: From the governing equations and the slip equation, we

can find the acceleration of the box (assumed to be down the incline).
◦
F : 14(9.81) N sin 20 − f

◦
−(20 N) cos 20 = (14 kg)a,

◦
F : N − 14(9.81) N cos 20

◦
−(20 N) sin 20 = 0.

Slip: f = (0.1)N.
Solving, we have

a = 1.04 m/s2 , N = 136 N, f = 13.6 N.

Since a > 0, our assumption is correct.

Using kinematics,

v = at = (1.04 m/s2 )(2 s) = 2.08 m/s.

v = 2.08 m/s down the incline.

170 lb
= a.
32.2 ft/s2

Slip: f = (0.08)N.
Solving yields

a = 11.3 ft/s2 , N = 154 lb, f = 12.3 lb.

Using kinematics, we find

60 ft/s = 40 ft/s + (11.3 ft/s2 )t

Solving yields t = 1.77 s.

Solution: The governing equations and the slip equation are used
to find the acceleration
◦
F : N − (170 lb) cos 25 = 0,

◦
F : (170 lb) sin 25 − f

lb-s2
− 0.015 v2
ft2

170 lb
= a.
32.2 ft/s2

Slip: f = (0.08)N.
Solving yields

N = 154 lb, f = 12.3 lb,

a = (11.3 ft/s2 ) − (0.00284 ft−1 )v 2 .

Using kinematics, we find
dv
a= = (11.3 ft/s2 ) − (0.00284 ft−1 )v 2
dt
60 ft/s t
dv
−1
= dt = t
40 ft/s (11.3 ft/s ) − (0.00284 ft
2
)v 2 0

Performing the integration, we find

t = (5.59 s)[tanh−1 ([0.0159 s/ft][60 ft/s]) − tanh−1 ([0.0159 s/ft][40 ft/s])].

Solving yields t = 6.19 s.

Problem 14.10 The total external force on the 10-kg y

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.)

F
Solution:

F (90i − 60j + 20k) N

a= = = (9i − 6j + 2k) m/s2 . x
m 10 kg

z
v = at + v0 = [(9i − 6j + 2k) m/s2 ](4s) + (−14i + 26j + 32k) m/s.

v = (22i + 2j + 40k) m/s.

Solution:
1
a= [(−20t + 90)i − 60j + (10t + 40)k]N
(10 kg)

a = [(−2t + 9)i − 6j + (t + 4)k] m/s2

Integrate to get the velocity

v = a dt + v0

1 2
v = (−t 2 + 9t − 14)i + (−6t + 26)j + t + 4t + 32 k m/s
2
Integrate again to get the position

r = v dt + r0

1 9
r= − t 3 + t 2 − 14t + 40 i + (−3t 2 + 26t + 30)j
3 2

1 3
+ t + 2t 2 + 32t − 360 k m
6
At the time indicated (t = 4 s) we have

r = [34.7i + 86j − 189.3k] m

Problem 14.12 The position of the 10-kg object shown

in Problem 14.10 is given as a function of time by r =
(20t 3 − 300)i + 60t 2 j + (6t 4 − 40t 2 )k (m). What is the
total external force on the object at t = 2 s?

Solution:
r = [(20t 3 − 300)i + (60t 2 )j + (6t 4 − 40t 2 )k] m

dr
v= = [(60t 2 )i + (120t)j + (24t 3 − 80t)k] m/s
dt

dv
a= = [(120t)i + (120)j + (72t 2 − 80)k] m/s2
dt

F = ma = (10 kg)[(120t)i + (120)j + (72t 2 − 80)k] m/s2

F = [(1200t)i + (1200)j + (720t 2 − 800)k] N

At the time t = 2 s,

F = [2.40i + 1.20j + 2.08k] kN

Solution: Working in components we have

Fx (2000 − 400t 2 ) lb
ax = = = (0.805 − 0.161t 2 ) ft/s2 ,
m 80,000 lb
32.2 ft/s2

Fy (5200 + 440t) lb
ay = = = (2.093 + 0.1771t ) ft/s2 ,
m 80,000 lb
32.2 ft/s2

Fz (800 + 60t 2 ) lb
az = = = (0.322 + 0.0242t 2 ) ft/s2
m 80,000 lb
32.2 ft/s2
4 s
We find the velocity at t = 4 s, by integrating: v = 2 s adt + v0 . In
components this is

1
vx = [0.805][4 − 2] − [0.161][43 − 23 ] + 12 ft/s = 10.6 ft/s,
3

1
vy = [2.093][4 − 2] + [0.177][42 − 22 ] + 220 ft/s = 225 ft/s,
2

1
vz = [0.322][4 − 2] + [0.0242][43 − 23 ] − 30 ft/s = −28.9 ft/s,
3

Thus v = (10.6i + 225j − 28.9k) ft/s.

Problem 14.14 At the instant shown, the horizontal y

component of acceleration of the 26,000-lb airplane due T
to the sum of the external forces acting on it is 14 ft/s2 . If
the pilot suddenly increases the magnitude of the thrust 15°
T by 4000 lb, what is the horizontal component of the
plane’s acceleration immediately afterward? x

Solution: Before

26,000 lb
Fx : F x = 2
(14 ft/s2 ) = 11,304 lb
32.2 ft/s
After

◦ 26,000 lb
Fx : 11,304 lb + (4000 lb) cos 15 = a
32.2 ft/s2

⇒ a = 18.8 ft/s2

Solution: The equation of motion is

F : (2400 kN) − (90,000 kg)(9.81 m/s2 )

−(0.8 kg/m)v2 = (90,000 kg)a

Solving for the acceleration we have
dv
a= = (16.9 m/s2 ) − (8.89×10−6 m−1 )v 2
dt
200 m/s t
dv
= dt = t
100 m/s
2
(16.9 m/s ) − (8.89×10−6 m−1 )v 2 0

Carrying out the integration, we find

t = (81.7 s)(tanh−1 [(0.000726)(200)] − tanh−1 [(0.000726)(100)])

t = 6.01 s

Solution: –F0
t0 2t0
(a) No, the internal dynamics make it impossible to determine the
acceleration of just the cart.
y
(b) Yes, the entire system (cart + water) obeys Newton’s 2nd Law.

5N
F : (5 N) = (10 kg)a ⇒ a = = 0.5 m/s2
10 kg F
(c) The center of mass moves as a “super particle”.

For 0 < t < t0 x

5N
5 N = (10 kg)a ⇒ a = = 0.5 m/s2 (b)
10 kg

v = (0.5 m/s2 )t, s = (0.25 m/s2 )t 2

At t = t0 = 2 s, v = 1.0 m/s, s = 1.0 m

For t0 < t < 2t0 ,

− 5 N = (10 kg)a, a = −0.5 m/s2 , v = −(0.5 m/s2 )(t − t0 ) + 1.0 m/s

s = −(0.25 m/s2 )(t − t0 )2 + (1.0 m/s)(t − t0 ) + 1.0 m

For t ≥ 2t0 , a = v = 0, s = 2.0 m

Solution: Kinematics
1 2 1
a = constant, v = at, s = at , 35 ft = a(2 s)2 ⇒ a = 17.5 ft/s2
2 2

360 lb
Dynamics: Fr = (17.5 ft/s2 ) = 195.6 lb
32.2 ft/s2

195.6 lb
Friction: Fr = (0.8)N ⇒ N = = 245 lb
0.8

Problem 14.18 The mass of the bucket B is 180 kg.

y
From t = 0 to t = 2 s, the x and y coordinates of the
center of mass of the bucket are B

x = −0.2t 3 + 0.05t 2 + 10 m,

y = 0.1t 2 + 0.4t + 6 m.

Determine the x and y components of the force exerted

on the bucket by its supports at t = 1 s.

Solution: x

x = −0.2t 3 + 0.05t 2 + 10, y = 0.1t 2 + 0.4t + 6

vx = −0.6t 2 + 0.1t, vy = 0.2t + 0.4

ax = −1.2t + 0.1, ay = 0.2

Fx = max = (180 kg)(−1.2[1 s] + 0.1) m/s2 = −198 N

Fy = may + mg = (180 kg)(0.2) m/s2 + (180 kg)(9.81 m/s2 )
= 1800 N

Problem 14.19 During a test flight in which a 9000-kg y

helicopter starts from rest at t = 0, the acceleration of its
center of mass from t = 0 to t = 10 s is a = (0.6t)i +
(1.8 − 0.36t)j m/s2 . What is the magnitude of the total
external force on the helicopter (including its weight) at
t = 6 s? L

h
Pat
Solution: From Newton’s second law: F = ma. The sum of W
the external forces is F = F − W = 9000[(0.6t)i + (1.8 − 0.36t)
x
j]t=6 = 32400i − 3240j, from which the magnitude is

√
F = 324002 + 32402 = 32562 (N).

Solution: Integrate the acceleration: v = (0.3t 2 )i + (1.8t − 0.18

t 2 )j, since the helicopter starts from rest. The instantaneous flight
−1
dy dy dx (1.8t − 0.18t 2 )
path angle is tan β = = = . At
dx dt dt (0.3t 2 )

(1.8(6) − 0.18(6) )
2
t = 6 s, βt=6 = tan−1 = 21.8◦ . A unit vector
0.3(6)2
tangent to this path is et = i cos β + j sin β. A unit vector normal to
this path en = −i sin β + j cos β. The weight acts downward:

W = −j(9000)(9.81) = −88.29j (kN).

From Newton’s second law, F − W = ma, from which F = W + ma

= 32400i + 85050j (N). The component tangent to the path is

T = F · et = 32400 cos β + 85050 sin β = 61669.4 (N)

The component normal to the path is

L = F · en = −32400 sin β + 85050 cos β = 66934 (N)

Problem 14.21 At the instant shown, the 11,000-kg y

airplane’s velocity is v = 270 i m/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 L T
parallel to the airplane’s path.) Determine the magnitude x
of the airplane’s acceleration. 15°
Path
15°
Horizontal
D

Solution: Let us sum forces and write the acceleration components

along the x and y axes as shown. After the acceleration components are
known, we can determine its magnitude. The equations of motion, in
the coordinate directions, are Fx = T cos 15◦ − D − W sin 15◦ =
max , and Fy = L + T sin 15◦ − W cos 15◦ = may . Substituting in
the given values for the force magnitudes, we get ax = 4.03 m/s2 and
ay = 16.75 m/s2 . The magnitude of the acceleration is |a| = ax2 + ay2
= 17.23 m/s 2

Horizontal
m = 11000 kg D

g = 9.81 m/s2 mg

|V| = 300 m/s L

T = 120000 N

T
ρ = 4500 m 15° x
◦ ◦
Fx : T cos 15 − D − mg sin 15 = max 15°

L + T sin 15◦ − mg cos 15◦ = may

D
Fy :

ay = V 2 /ρ

Solving, D = 33.0 kN, L = 293 kN mg

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 = y
20t − 1.63t 2 , y = 35t − 0.15t 3 , and z = −20t + 1.38t 2 ,
where t is the time in seconds. The y- axis points upward.
The forces exerted on the plane are its weight, the thrust
vector T exerted by its engine, the lift force vector
L, and the drag force vector D. At t = 4 s, determine
T + L + D.

Solution: There are four forces acting on the airplane. Newton’s

second law, in vector form, given T + L + D + W = (T + L + D) −
mgj = ma. Since we know the weight of the airplane and can evaluate
the total acceleration of the airplane from the information given, we can
evaluate the (T + L + D) (but we cannot evaluate these forces sepa-
rately without more information. Differentiating the position equations
twice and evaluating at t = 4.0 s, we get aX = −3.26 m/s2 , aY =
−3.60 m/s2 , and aZ = 2.76 m/s2 . (Note that the acceleration compo-
nents are constant over this time interval. Substituting into the equation
for acceleration, we get (T + D + L) = ma + mgj. The mass of the
airplane is 360 kg. Thus, (T + D + L) = −1174i + 2236j + 994k (N)

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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 t is 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 = 4 s?
Solution: Since we are working in nonrotating rectangular Carte-
sian coordinates, we can consider the motion in each axis separately.
From Problem 14.23, we have (T + D + L) = ma + mgj. Separating
the information for each axis, we have maX = −1000 + 280t, maY =
4000 − 430t − mg, and maZ = 720 + 200t
Integrating the x equation, we get vx = vx0 + (1/m)(−1000t +
280t 2 /2) and x = vX0 t + (1/m)(−1000t 2 /2 + 280t 3 /6).
Integrating the y equation, we get vY = vY 0 + (1/m)((4000 − mg)t −
430t 2 /2) and y = vY 0 t + (1/m)((4000 − mg)t 2 /2 − 430t 3 /6)
Integrating the z equation, we get vZ = vZ0 + (1/m)(720t + 200t 2 /2)
and z = vZ0 t + (1/m)(720t 2 /2 + 200t 3 /6).
Evaluating at t = 4 s we find the aircraft at (66.1, 137.7, −58.1)m
relative to its initial position at t = 0.

Problem 14.25 The robot manipulator is programmed y

so that x = 40 + 24t 2 mm, y = 4t 3 mm, and z = 0 dur-
ing the interval of time from t = 0 to t = 4 s. The y
axis points upward. What are the x and y components
of the total force exerted by the jaws of the manipulator
on the 2-kg widget A at t = 3 s? A

x
x

Solution:
mg
x = 40 + 24t 2 mm y = 4t 3 mm
Fx
vx = 48t mm/s vy = 12t 2 mm/s

ax = 48 mm/s2 ay = 24t mm/s2

At t = 3 s

ax = 48 × 10−3 m/s2 , ay = 72 × 10−3 m/s2

Fx = max m = 2 kg

Fy − mg = may

Solving,

Fx = 0.096 N

Fy = 19.764 N

Solution: y

dvx
ax = = 400 − 0.8vx
dt
t vx
dvx A
dt =
0 0 (400 − 0.8vx )

vx
1
t= ln(400 − 0.8vx ) y
(−0.8) 0

400 − 0.8vx
(−0.8t) = ln
400
x
x
or 400 − 0.8vx = 400e−0.8t

1
vx = (400)(1 − e−0.8t ) mg
(0.8)

At t = 1 s, vx = 275.3 mm/s
Fx
A similar analysis for vy yields x

vy = 164.8 mm/s at t = 1 s.

At t = 1 s, Fy

ax = 400 − 0.8 vx = 179.7 mm/s2

ay = 200 − 0.4 vy = 134.1 mm/s2

m = 2 Kg

g = 9.81 m/s2

ax = 0.180 m/s2

ay = 0.134 m/s2

Fx : Fx = max

Fy : Fy − mg = may

Solving,

Fx = 0.359 N

Fy = 19.89 N

Solution: The stone travels at an angle relative to the x axis.

1 yd ◦
θ = tan−1 = 1.85
31 yd
The accelerations and distances are related as x
dvx ◦
ax = vx = −(0.01)(32.2 ft/s2 ) cos(1.85 ) = −0.322 ft/s2
dx
0 93.0 ft
vx dvx = (−0.322 ft/s2 )d x,
vx0 0

2
vx0
0− = −(0.322 ft/s2 )(93.0 ft) ⇒ vx0 = 7.74 ft/s.
2

dvy ◦
ay = vy = −(0.01)(32.2 ft/s2 ) sin(1.85 ) = −0.0104 ft/s2
dy

0 3.0 ft
vy dvy = (−0.0104 ft/s2 )d x,
vy0 0

2
vy0
0− = −(0.0104 ft/s2 )(3.0 ft) ⇒ vy0 = 0.250 ft/s.
2

vx0 = 7.74 ft/s, vy0 = 0.250 ft/s.

2 kg
5 kg

Solution: The free-body diagrams are shown. The governing

equations are

Fy left : T − 2(9.81) N = (2 kg)a

Fy right : T − 5(9.81) N = −(5 kg)a

Solving, we find
T = 28.0 N, a = 4.20 m/s2 .

To find the velocity, we integrate the acceleration

v = at = (4.20 m/s)(0.5 s) = 2.10 m/s.

v = 2.10 m/s.

Problem 14.29 The two weights are released from rest.

The horizontal surface is smooth. (a) What is the ten- 5 lb
sion in the cable after the weights are released? (b) How
fast are the weights moving one second after they are
released?

10 lb

Solution: The free-body diagrams are shown. The governing equa-

tions are
5 lb
FxA : T = a
32.2 ft/s2

10 lb
FyB : T − (10 lb) = − a
32.2 ft/s2
Solving, we find

T = 3.33 lb, a = 21.5 ft/s2 .

To find the velocity we integrate the acceleration

v = at = (21.5 ft/s2 )(1 s) = 21.5 ft/s.

(a) T = 3.33 lb, (b) v = 21.5 ft/s.

10 lb

Solution: The free-body diagrams are shown. The governing equa-

tions are
5 lb
FxA : T − f = a,
32.2 ft/s2

FyA : N − 5 lb = 0,

10 lb
FyB : T − (10 lb) = − a,
32.2 ft/s2

f = (0.18)N.
Solving, we find

T = 3.93 lb, a = 19.5 ft/s2 ,

N = 5 lb, f = 0.9 lb.

To find the velocity we integrate the acceleration

v = at = (19.5 ft/s2 )(1 s) = 19.5 ft/s.

(a) T = 3.93 lb (b) v = 19.5 ft/s.

30⬚

Solution: We will first use the kinematic information to find the

acceleration a
a = constant,

v = at,

1 2
d= at
2

1
0.3 m = a(1 s)2
2

a = 0.6 m/s2 .
From the free-body diagrams we have four equations of motion:

FxA : T − µk NA = (14 kg)a,

FyA : NA − (14 kg)(9.18 m/s2 ) = 0

◦
FB : (14 kg)(9.81 m/s2 ) sin 30 − T − µk NB = (14 kg)a,

◦
FB : NB − (14 kg)(9.81 m/s2 ) cos 30 = 0
Solving these equations, we find

T = 36.2 N, NA = 137 N, NB = 119 N, µk = 0.202.

µk = 0.202.

Solution: Assume that no motion occurs anywhere. Then

NB = (45 kg)(9.81 m/s2 ) = 441 N

fB max = µs N = 0.4(441 N) = 177 N.

The blocks will slip as long as F > 177 N. Assume that blocks A and
B move together with the common acceleration a.

FxA : fA = (15 kg)a

FyA : NA − (15 kg)(9.81 m/s2 ) = 0

FxB : F − fA − fB = (30 kg)a

FyB : NB − (30 kg)(9.81 m/s2 ) − NA = 0

Slip at B : fB = (0.35)NB

fA max = (0.4)NA

(a) F = 200 N. Solving we find that a = 1.01 m/s2 , fA = 15.2 N, fA max =

58.9 N. Since fA < fA max , we know that our assumption is cor-
rect (the blocks move together).
(b) F = 400 N. Solving we find that a = 5.46 m/s2 , fA = 81.8 N, fA max =
58.9 N. Since fA > fA max , we know that our assumption is
wrong (the blocks will not move together, but slip will occur
at all surfaces). The equations are now

FxA : (0.35)NA = (15 kg)aA

FyA : NA − (15 kg)(9.81 m/s2 ) = 0

FxB : F − (0.35)(NA + NB ) = (30 kg)aB

FyB : NB − (30 kg)(9.81 m/s2 ) − NA = 0

Solving we find that aA = 3.43 m/s2 , aB = 6.47 m/s2 .

(a) aB = 1.01 m/s2 , (b) aB = 6.47 m/s2 .

(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?
A

5°

Solution: (a) From Newton’s second law, T sin θ = max , and

T sin θ mg
T cos θ − mg = 0. Solve ax = , T = , from which 5° T
m cos θ
a = g tan θ = 9.81(tan 5◦ ) = 0.858 m/s2

800(9.81)
(b) T = = 7878 N
cos 5◦
mg

Problem 14.34 The mass of A is 30 kg and the mass A

of B is 5 kg. The horizontal surface is smooth. The con-
stant force F causes the system to accelerate. The angle F
θ = 20◦ is constant. Determine F .
u

Solution: We have four unknowns (F, T , N, a) and four equations

of motion:
FxA : F − T sin θ = (30 kg)a,

FyA : N − T cos θ − (30 kg)(9.81 m/s2 ) = 0,

FxB : T sin θ = (5 kg)a,

FyB : T cos θ − (5 kg)(9.81 m/s2 ) = 0.

Solving, we find

T = 52.2 N, a = 3.57 m/s2 , N = 343 N,

F = 125 N.

Solution: We have four unknowns (F, T , N, a) and four equations

of motion:
FxA : F − T sin θ − µk N = (30 kg)a,

FyA : N − T cos θ − (30 kg)(9.81 m/s2 ) = 0,

FxB : T sin θ = (5 kg)a,

FyB : T cos θ − (5 kg)(9.81 m/s2 ) = 0.

Solving, we find

T = 52.2 N, a = 3.57 m/s2 , N = 343 N,

F = 194 N.

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 F
how far the crate moves from its initial position in 2 s
if the horizontal force F = 90 lb.
30°

Solution: Denote W = 100 lb, g = 32.17 ft/s2 , F = 90 lb, and

θ = 30◦ . Choose a coordinate system with the positive x axis parallel
to the inclined surface. (See free body diagram next page.) The normal F
force exerted by the surface on the box is N = F sin θ + W cos θ =
30°
131.6 lb. The sum of the non-friction forces acting to move the box
f N
is Fc = F cos θ − W sin θ = 27.9 lb. Slip only occurs if |Fc | ≥ |Nµs | W
which 27.9 > 26.32 (lb), so slip occurs.

The direction of slip is determined from the sign of the sum of

the non friction forces: Fc > 0, implies that the box slips up the
surface, and Fc < 0 implies that the box slips down the surface
(if the condition for slip is met). Since Fc > 0 the box slips up
the surface. After the box slips, the sum of the forces on the box
parallel to the surface is Fx = Fc − sgn(Fc )µkN , where sgn(Fc ) =
Fc W
. From Newton’s second law, Fx = a, from which a =
|Fc | g
g
(Fc − sgn(Fc )µk N) = 2.125 ft/s . The velocity is v(t) = at ft/s,
2
W
a
since v(0) = 0. The displacement is s = t 2 f t, since s(0) = 0. The
2
position after 2 s is s(2) = 4.43 ft up the inclined surface.

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 x axis parallel to

the surface. Denote θ = 30◦ . T

(a) The normal force exerted by the surface on the crate is N =

W cos θ = 120(9.81)(0.866) = 1019.5 N. The force tending to
30°
move the crate is Fc = T − W sin θ, from which the tension
f
required to start slip is T = W (sin θ) + µs N = 1200.3 N. W
N

(b) After slip begins, the force acting to move the crate is F =
T − W sin θ − µk N = 101.95 N. From Newton’s second law,
F 101.95
F = ma, from which a = = = 0.8496 m/s2 . The
m 120
velocity is v(t) = at = 0.8496t m/s, since v(0) = 0. The position
a
is s(t) = t 2 , since s(0) = 0. When the crate has moved
2
2(2)
2 m up the slope, t10 = = 2.17 s and the velocity is
a
v = a(2.17) = 1.84 m/s.

Problem 14.39 The coefficients of friction between the

load A and 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

Solution: The load is on the verge of slipping. There are two

unknowns (a and N ). The equations are

Fx : µs N = ma, Fy : N − mg = 0

Solving we find

a = µs g = (0.4)(9.81 m/s2 ) = 3.92 m/s2 .

a = 3.92 m/s2 .

Solution: The load is on the verge of slipping. There are two

unknowns (a and N ).

Forward: The equations are

Fx : µs N − mg sin θ = ma,

Fy : N − mg cos θ = 0
Solving we find

a = g(µs cos θ − sin θ )

◦ ◦
= (9.81 m/s2 )([0.4] cos 20 − sin 20 )

a = 0.332 m/s2 .

Rearward: The equations are

Fx : −µs N − mg sin θ = −ma,

Fy : N − mg cos θ = 0
Solving we find

a = g(µs cos θ + sin θ )

◦ ◦
= (9.81 m/s2 )([0.4] cos 20 + sin 20 )

a = 7.04 m/s2 .

Solution: Set g = 9.81 m/s2

B
First analyze the motion before it gets to point B.
◦
F : mg sin 30 = ma 30°

◦ ◦ ◦ t2
a = g sin 30 , v = (g sin 30 )t, s = (g sin 30 )
2 mg
When it gets to B we have

◦ t2
s = 2 m = (g sin 30 ) ⇒ t = 0.903 s
2 30°
◦
v = (g sin 30 )(0.903 s) = 4.43 m/s
Now analyze the motion after it hits point B.
N
◦
F : mg sin 30 − 2000 N = ma

dv 2000 N mg
◦
a=v = g sin 30 −
ds m
0 0.1 m 2000 N

◦
v dv = g sin 30 − ds
4.43 m/s 0 m 30°
2000 N
(4.43 m/s)2 ◦ 2000 N
0− = g sin 30 − (0.1 m)
2 m
N
Solving the last equation we find m = 19.4 kg

Problem 14.42 The force exerted on the 10-kg mass s

by the linear spring is F = −ks, where k is the spring
constant and s is the displacement of the mass relative k
to its position when the spring is unstretched. The value
of k is 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 dv ds dv
= = v.
dt ds dt ds

Solution: The equation of motion is −ks = ma

dv k v s k v2 v2 k s2
a=v =− s⇒ vdv = − ds ⇒ − 0 =−
ds m v0 0 m 2 2 m 2
Thus
k 2 40 N/m 2
v = ± v02 − s = ± (4 m/s)2 − s
m 10 kg

√
v = ±2 4 − s 2 m/s

450
s=− ln(0.1) = 25.9 m.
40

s = 25.9 m.

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 0.5v 2 . (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?

Solution: Choose a coordinate system with s positive downward.

For brevity denote Cd = 0.5, W =200 lb, g = 32.17 ft/s . From
2
W dv
Newton’s second law W − D = , where D = 0.5v2 . Use
g dt

dv 0.5v 2 g Cd v 2
the chain rule to write v =− +g =g 1− .
ds W W
(a) At the instant of full opening, the initial velocity has not
decreased, and the magnitude of the acceleration is

Cd 2
|ainit | = g 1 − v = 24 g = 772.08 ft/s.
W

v dv
(b) Separate variables and integrate: = gds, from which D
Cd v 2
1−
W

Cd v 2 2Cd g
ln 1 − =− s + C. Invert and solve:
W W W
2Cd g

W
v2 = 1 − Ce− W s . At s = 0, v (0) = 100 ft/s, from
Cd
Cd 104
which C = 1 − = −24, and
W
2Cd g

W
v2 = 1 + 24e− W s . At s = 20 ft the velocity is
Cd

 
g


− (20)
 = 28.0 ft/s
v (s = 20) = 2W 1 + 24e W

Solution:

1000 m 1h
v0 = 213 km/h = 59.2 m/s
1 km 3600 s

F = −(80,000 + 2.5v2 ) = (18,000 kg)a

dv 80,000 + 2.5v2
a=v =−
ds 18,000
0 s
vdv ds
=−
v0 80,000 + 2.5v2 0 18,000

1 80,000 s
ln =−
5 80,000 + 2.5v02 18,000

x = 374 m.

Solution: Choose a coordinate system with s positive downward.

Divide the fall into two parts: (1) the free fall until the bungee
unstretched length is reached, (2) the fall to the full extension of the
ds
bungee cord. For Part (1): From Newton’s second law = g. Use
dt
dv
the chain rule to write: v = g. Separate variables and integrate:
ds

v 2 (s) = 2 gs, since v(0) = 0. At s = 60 ft, v(s = 60) = 2 gs =
62.1 ft/s. For Part (2):
From Newton’s second law
T
W dv
W −T = , where T = k(s − 60).
g dt

Use the chain rule to write:

dv gk k
v =g− (s − 60) = g 1 − (s − 60) (s ≥ 60 ft).
ds W W
W
Separate variables and integrate:

k
v 2 (s) = 2 gs 1 − (s − 120) + C. At s = 60,
2W

v 2 (s = 60) = [2 gs]s=60 = 120 g,

gk
from which C = − (602 ) = −8106.8. The velocity is
W

gk 2 60 k gk
v 2 (s) = − s + 2g 1 + s− (602 ).
W W W

When the jumper is brought to a stop, v(sstop ) = 0, from which s 2 +

W
2bs + c = 0, where b = − + 60 , and c = 602 . The solution:
k
√
sstop = −b ± b2 − c = 118.08, = 30.49 ft.

(a) The first value represents the maximum distance on the first
excursion, from which

h = 130 − 118.1 = 11.92 ft

is the height above the river at which he comes to a stop. (b) The
maximum force exerted by the bungee cord is

F = k(s − 60) = 14(118.1 − 60) = 813 lb

Solution:
t 4000
dh dh
(a) v= = 66 − 0.01h, ⇒ dt =
dt 0 0 66 − 0.01h

⇒ t = 93.2 s

dv
(b) a=v = (66 − 0.01h)(−0.01) = −0.66 + 0.0001h
dh
At h = 2000 ft we have

20,500 lb
a = −0.46 ft/s2 ⇒ F = 2
(−0.46 ft/s2 ) = −293 lb
32.2 ft/s

Problem 14.48 In a cathode-ray tube, an electron y

(mass = 9.11 × 10−31 kg) is projected at O with Screen
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 x
the charge of the electron e = 1.6 × 10−19 C (coulombs) O
+ + + +
and the electric field strength E = 15 kN/C. External
forces on the electron are negligible when it is not 30
between the plates. Where does the electron strike the mm 100 mm
screen?
Solution: For brevity denote L = 0.03 m, D = 0.1 m. The time y Screen
L 3 × 10−2 m
spent between the charged plates is tp = = =
V 2.2 × 107 m/s – – – –
−9
1.3636 × 10 s. From Newton’s second law, F = me ap . The x
acceleration due to the charged plates is O + + + +
30 100
mm mm
−eE (1.6 × 10−19 )(15 × 103 )
ap = j=− j = j2.6345 × 1015 m/s2 .
me 9.11 × 10−31

ap 2
The velocity is vy = −ap t and the displacement is y = t . At the
2
ap tp2
exit from the plates the displacement is yp = − = −j2.4494 ×
2
10−3 (m). The velocity is vyp = −ap t = −j3.59246 × 106 m/s. The
time spent in traversing the distance between the plates and the
D 10−1 m
screen is tps = = = 4.5455 × 10−9 s. The vertical
V 2.2 × 107 m/s
displacement at the screen is

ys = vyp tps + yp = −j(3.592456 × 106 )(4.5455 × 10−9 )

− j2.4494 × 10−3 = −18.8j (mm)

c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they
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138
Problem 14.49 In Problem 14.48, determine where Solution: Use the solution to Problem 14.48. Assume that the
the electron strikes the screen if the electric field strength electron enters the space between the charged plates at t =
is E = 15 sin(ωt) kN/C, where the frequency ω = 2 × 0, so that at that instant the electric field strength is zero.
109 s−1 . The acceleration due to the charged plates is a = −
eE
j=
me
(1.6 × 10 −19 C)(15000 sin ωt N/C)
− j = −j(2.6345 × 1015 )
9.11 × 10−31 kg
(2.6345 × 1015 )
sin ωt (m/s2 ). The velocity is vy = j cos ωt + C. Since
ω
2.6345 × 10 15
vy (0) = 0 C = − j = −j1.3172 × 106 . The displace-
2 × 109
(2.6345 × 1015 )
ment is y = j sin ωt + Ct, since y(0) = 0. The time
ω2
spent between the charged plates is (see Problem 14.48) tp = 1.3636 ×
10−9 s, from which ωtp = 2.7273 rad. At exit from the plates, the
2.6345 × 1015
vertical velocity is vyp = j cos(ωtp ) + C = −j2.523 ×
2 × 109
106 (m/s).

2.6345 × 1015
The displacement is yp = j sin(ωtp ) + Ctp = −j1.531
4 × 1018
× 10−3 (m). The time spent between the plates and the screen
is tps = 4.5455 × 10−9 s. The vertical deflection at the screen is
ys = yp + vyp tps = −13j(mm)

Problem 14.50 An astronaut wants to travel from a y

space station to a satellites S that needs repair. She
departs the space station at O. A spring-loaded launch- S
ing device gives her maneuvering unit an initial velocity
of 1 m/s (relative to the space station) in the y direction.
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 x direction. The astronaut intercepts
the satellite by applying a constant thrust parallel to the
x axis. The total mass of the astronaut and her maneu- O x
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?

Solution: The path of the satellite relative to the space station is

xs (t) = 2t + 70 m, ys (t) = 50 m. From Newton’s second law, T =
max , 0 = may . Integrate to obtain the path of the astronaut, using
the initial conditions vx = 0, vy = 1 m/s, x = 0, y = 0. ya (t) = t,
T 2
xa (t) = t . (a) When the astronaut intercepts the x path of the
2m
satellite, ya (tint ) = ys (tint ), from which tint = 50 s . (b) The inter-
cept of the y-axis path occurs when xa (tint ) = xs (tint ), from which
T 2
t = 2tint + 70, from which
2m int

2tint + 70 170
T = (2m) 2
= 2(300) = 40.8 N.
tint 2500

(c) velocity of the astronaut relative to the space station is v =

The
T
i tint + j = 6.8i + j. The velocity of the satellite relative to the
m
space station is vs = 2i. The velocity of the astronaut relative to the
satellite is va/s = i(6.8 − 2) + j = 4.8i + j (m/s)

200 N
A

45°

Solution: For brevity, denote θ = 20◦ , α = 45◦ , F = 200 N,

m = 8 kg. The force exerted by the rope on the collar is Frope =
20°
200(i sin θ + j cos θ) = 68.4i + 187.9j (N). The force due to gravity is
Frope
Fg = −gmj = −78.5j N. The unit vector parallel to the bar, positive
upward, is eB = −i cos α + j sin α. The sum of the forces acting to
move the collar is F = Fc = eB · Frope + eB · Fg = |Frope | sin(α −
θ) − gm sin α = 29.03 N. The collar tends to slide up the bar since
Fc > 0. From Newton’s second law, the acceleration is N
Fg
Fc
a= = 3.63 m/s2 .
m

Problem 14.52 In Problem 14.51, determine the accel- Solution: Use the solution to Problem 14.51. Fc = |Frope | sin(α −
eration of the 8-kg collar A relative to the bar if the θ) − gm sin α = 29.03 N. The normal force is perpendicular to
coefficient of kinetic friction between the collar and the the bar, with the unit vector eN = i sin α + j cos α. The normal
bar is µk = 0.1. force is N = eN · Frope + eN · Fg = |Frope | cos(α − θ) − gm cos α =
125.77 N. The collar tends to slide up the bar since Fc > 0. The
friction force opposes the motion, so that the sum of the forces on the
collar is F = Fc − µk N = 16.45 N. From Newton’s second law,
16.45
the acceleration of the collar is a = = 2.06 m/s2 up the bar.
8

Problem 14.53 The force F = 50 lb. What is the mag- y

nitude of the acceleration of the 20-lb collar A along the
(5, 3, 0) ft
smooth bar at the instant shown?

A
(2, 2, 2) ft
Solution: The force in the rope is F

(5 − 2)i + (3 − 2)j + (0 − 2)k

Frope = (50 lb) x
(5 − 2)2 + (3 − 2)2 + (0 − 2)2 z
The force of gravity is Fgrav = −(20 lb)j (2, 0, 4) ft

The unit vector along the bar is

(2 − 2)i + (2 − 0)j + (2 − 4)k
ebar =
(2 − 2)2 + (2 − 0)2 + (2 − 4)2
The component of the total force along the bar is

Fbar = Fžebar = 14.2 lb

20 lb
Thus 14.2 lb = a ⇒ a = 22.9 ft/s2
32.2 ft/s2

Solution: Use the results from Problem 14.53.

The magnitude of the total force exerted on the bar is

F = |Frope + Fgrav | = 48.6 lb

The normal force is N = F 2 − Fbar 2 = 46.5 lb

The total force along the bar is now Fbar − 0.2N = 4.90 lb

20 lb
Thus 4.90 lb = a ⇒ a = 7.89 ft/s2
32.2 ft/s2

Problem 14.55 The 6-kg collar starts from rest at posi- y

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
B
F = −40i + 70j − 40k (N). How long does it take to go
from A to B?
Strategy: There are several ways to work this problem.
One of the most straightforward ways is to note that the A
motion is along the straight line from A to B and that
only the force components parallel to line AB cause x
acceleration. Thus, a good plan would be to find a unit
vector from A toward B and to project all of the forces
acting on the collar onto line AB. The resulting constant F
force (tangent to the path), will cause the acceleration of z
the collar. We then only need to find the distance from
A to B to be able to analyze the motion of the collar.

Solution: The unit vector from A toward B is eAB = et = 0.333i

+ 0.667j − 0.667k and the distance from A to B is 0.3 m. The free W = –mgJ
body diagram of the collar is shown at the right. There are three forces
acting on the collar. These are the applied force F, the weight force
W = −mgj = −58.86j (N), and the force N which acts normal to
the smooth bar. Note that N, by its definition, will have no component N
tangent to the bar. Thus, we need only consider F and W when finding
force components tangent to the bar. Also note that N is the force that
the bar exerts on the collar to keep it in line with the bar. This will be
important in the next problem.
F
The equation of motion for the collar is Fcollar = F + W + N = ma.
In the direction tangent to the bar, the equation is (F + W)·eAB = mat .

The projection of (F + W) onto line AB yields a force along AB

which is |FAB | = 20.76 N. The acceleration of the 6-kg collar caused
by this force is at = 3.46 m/s2 . We now only need to know how long
it takes the collar to move a distance of 0.3 m, starting from rest,
with this acceleration. The kinematic equations are vt = at t, and st =
at t 2 /2. We set st = 0.3 m and solve for the time. The time required
is t = 0.416 s

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 B toward A — opposing the motion).
This will give us a new acceleration, which will result
in a longer time for the collar to go from A to B.

Solution: We use the unit vector eAB from Problem 14.55. The W=–mgJ
free body diagram for the collar is shown at the right. There are four
forces acting on the collar. These are the applied force F, the weight
force W = −mgj = −58.86 j (N), the force N which acts normal to µ kN
the smooth bar, and the friction force f = −µk |N|eAB . The normal N
force must be equal and opposite to the components of the forces F
and W which are perpendicular (not parallel) to AB. The friction force
is parallel to AB. The magnitude of |F + W| is calculate by adding F
these two known forces and then finding the magnitude of the sum.
The result is that |F + W| = 57.66 N. From Problem 14.55, we know
that the component of |F + W| tangent to the bar is |FAB | = 20.76 N.
The force tangent to the bar is FAB = (F + W) · eAB − µk |N|
Hence, knowing the total force and its component tangent to the bar, we
= 10.00 N. The acceleration of the 6-kg collar caused by this force
can find the magnitude of its component normal to the bar. Thus, the
is at = 1.667 m/s2 . We now only need to know how long it takes the
magnitude of the component of |F + W| normal to the bar is 53.79 N.
collar to move a distance of 0.3 m, starting from rest, with this accel-
This is also the magnitude of the normal force N. The equation of
eration. The kinematic equations are vt = at t, and st = at t 2 /2. We set
motion for the collar is Fcollar = F + W + N − µk |N|eAB = ma.
st = 0.3 m and solve for the time. The time required is t = 0.600 s
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 2m
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).
4m

Solution:
L
2 2m
db dL d2b db
(a) Note that b2 + (2)2 = L2 , so b =L and b 2 +
dt dt dt dt

d2L dL 2
=L 2 + . b
dt dt

Setting b = 4 m, dL/dt = −0.2 m/s and d 2 L/dt 2 = 0, we obtain

d 2 b/dt 2 = −0.0025 m/s2 . The crate’s acceleration toward the T
right is a = 0.0025 m/s2 .
α
From the free-body diagram,
mg
T sin α + N − mg = 0, (1)
µk N N
T cos α − µk N = ma, (2)

where α = arctan(2/4) = 26.6◦ . Solving Eqs (1) and (2), we

obtain T = 282.3 N.

(b) Solving Eqs (1) and (2) with a = 0, we obtain T = 282.0 N.

y
300 mm

= R 2 − y 2 . Differ-
Solution: The horizontal displacement is x2 2 2
dx dy dx d x
entiate twice with respect to time: x = −y , +x 2 =
dt dt dt dt
2 2
dy d y d2x 1 y 2
− −y , from which. = − + 1
dt dt 2 dt 2 x x
2
dy y d2y d2x
− . Substitute: = −1.3612 m/s . From New-
2
dt x dt 2 dt 2
ton’s second law, Fh = max = −1.361(0.4) = −0.544 N

Problem 14.59 The 1-kg collar P slides on the vertical y

bar and has a pin that slides in the curved slot. The
vertical bar moves with constant velocity v = 2 m/s. The y = 0.2 sin π x
y axis is vertical. What are the x and y components of
P
the total force exerted on the collar by the vertical bar
and the slotted bar when x = 0.25 m?
x

Solution: 1m

dx
vx = = 2 m/s, constant
dt mg

d2x
ax = 0 =
dt 2 Fx

y = 0.2 sin(π x)

dx
vy = 0.2π cos(π x) Fy
dt
2
dx d2x
ay = −0.2π 2 sin(π x) + 0.2π cos(π x)
dt dt 2

when x = 0.25 m,

ax = 0

π 2
ay = −0.2π 2 sin (2) m/s
4

ay = −5.58 m/s2

Fx : Fx = m(0)

Fy : Fy − mg = may

Solving, Fx = 0, Fy = 4.23 N

Strategy: You know that the tangential component of x

the car’s acceleration is zero. You can use this condi-
tion together with the equation for the profile of the
road to determine the x and y components of the car’s
acceleration.
Solution:
2 2
dx dy
(1) v= + = 100 km/h = 27.78 m/s, const.
dt dt θ
(2) y = 0.0003x 2
dy dx
(3) = 0.0006x
dt dt y
2 2
d y dx d2x
(4) = 0.0006 + 0.0006x 2
dt 2 dt dt

The component of acceleration parallel to the path is zero. ay

dy θ Fx
tan θ = = 0.0006 x x
dx ax
At x = 200 m, θ = 0.1194 rad

θ = 6.84◦ Fy
(5) ax cos θ + ay sin θ = 0

Solving eqns (1) through (5) simultaneously, we get Fx : Fx = max

ax = −0.054 m/s2 , vx = 27.6 m/s Fy : Fy = may

ay = 0.450 m/s2 , vy = 3.31 m/s Solving, Fx = −73.4 N

m = 1360 kg Fy = 612 N

Problem 14.61* The two 100-lb blocks are released Solution: The relative motion of the blocks is constrained by the
from rest. Determine the magnitudes of their accelera- surface separating the blocks. The equation of the line separating
tions if friction at all contacting surfaces is negligible. the blocks is y = x tan 70◦ , where y is positive upward and x is
positive to the right. A positive displacement of block A results in
Strategy: Use the fact the components of the a negative displacement of B (as contact is maintained) from which
d 2 sA d 2 sB
accelerations of the blocks perpendicular to their mutual sA = −sB tan 70◦ , and from which 2
= − 2 tan 70◦ . Thus (1)
interface must be equal. dt dt
aA = −aB tan 70◦ .

From Newton’s second law: for block A, (2) Fy = −W +

P cos 70◦ = maA , for block B, (3) Fx = P sin 70◦ = maB ,
A W aB
from which aA = − + ft/s2 . Use (1) to obtain aA =
m tan 70◦
g aA
− = −28.4 ft/s2 and aB = − = 10.34 ft/s2 , where
B 1 + cot2 70◦ tan 70◦
aA is positive upward and aB is positive to the right.

70°
Q
P

c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they
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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 A to fall 1 ft?
Solution: Use the results of the solution to Problem 14.61. Denote
P
by Q the normal force at the wall, and by P the normal force at the B
contacting surface, and R the normal force exerted by the floor on fQ A f
block B. For aA positive upward and aB positive to the right, (1) W
Q f
aA = −aB tan 70◦ so long as contact is maintained. From Newton’s P
second law for block A, (2) Fx = Q − P sin 70◦ + f cos 70◦ = 0, W
fR
Fy = −W + fQ + f cos 70◦ + P cos 70◦ = maA . For block B:
R
(3)
(4) Fx = P sin 70◦ − f cos 70◦ − fR = maB , (5) Fy = −W +
R − P cos 70◦ − f sin 70◦ = 0. In addition: (6) f = µk P , (7) fR =
µk R, (8) fq = µk Q. Solve these eight equations by iteration: aA =
−24.7 ft/s2 , aB = 9 ft/s2 . Check: (1) The effect of friction should
reduce the downward acceleration of A in Problem 3.61, and (2) for
µk = 0, this should reduce to the solution to Problem 14.61. check.

aA 2 2
The displacement is y = t ft, from which, for y = −1 ft, t = − =
2 aA
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
24 in 30 in
earth-fixed coordinate system is placed with its origin 30 x
in above the ground, at the height of the vehicle’s center 30 in 24 in
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
y
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 y is 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.)

Solution: This analysis follows that of Example 14.3. The

equation for velocity used to determine how far down the vehicle
compresses its springs also applies as the vehicle rebounds. From
Example 14.3, we know that the vehicle comes to rest with vY = 0
and y = 1 ft. Followingthe Example, the velocity on the rebound is
v 0
given by 0 Y vY dvY = 1 32.2(1 − 6y) dy. Evaluation, we get vY =
11.3 ft/s. (+y is down). Note that the vertical velocity component on
rebound is the negative of the vertical velocity of impact.

Solution:

B = ρoil Vg y

W = ρSTEEL Vg
B
4 3 D
V = πr
3
Dy
dvx x
Fx : ms = −dx = −1.6vx Dx v
dt

dvy
Fy : ms = B − W − dy W
dt

dvy
ms = (ρoil − ρSTEEL )Vg − 1.6vy Integrating, we get
dt

Rewriting the equations, we get 1 v

ln(a + bvy )|0y = t
b
dvx 1.6
Fx : =− vx a + bvy
dt ms ln = bt
a
vx t
dvx 1.6
=− dt
vxo vx ms 0 a + bvy = aebt

1.6 t a bt
ln(vx )|vvxx0 = − t| vy = (e − 1)
ms 0 b

Substituting numerical values for a and b, and setting t = 0.1 s

1.6
−
v x = vx0 e m s t vy = −0.461 m/s

Substituting, we have ms = 0.114 kg

vx0 = 2 m/s. At t = 0.1 s,

vx = 0.486 m/s

dvy (ρoil − ρSTEEL )Vg 1.6

Fy : = − vy
dt ms ms

Let a = (ρoil − ρSTEEL )Vg/ms = −8.608

1.6
b=− = −14.147
ms

dvy
= a + bvy
dt
vy t
dy
= dt
0 a + bvy 0

dx
Solution: From the solution to Problem 14.64, = vx =
dt
1.6
−m t
v x0 e s = vx0 ebt where vx0 = 2 m/s and ms = 0.114 kg.

dy a
Also, = vy = (ebt − 1)
dt b

where a = (ρoil − ρSTEEL )Vg/m = −8.608

1.6
b=− = −14.147
ms

Integrating the vx and vy eqns, noting that x = 0, y = 0, at t = 0,

Solving at t = 0.1 s,

we get
x = 0.1070 m = 107.0 mm
vxo
x= (ebt − 1) y = −0.0283 m = −28.3 mm
b

a bt a
y= (e − 1) − t
b2 b

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.

Solution: Since the speed is constant, the tangential acceleration

is zero. We have

Ft = mat = 0,

v2 1200 lb (20 ft/s)2
Fn = man = m = 2
= 373 lb.
R 32.2 ft/s 40 ft

Ft = 0, Fn = 373 lb.

Solution: We first find the tangential acceleration and use that to

find the velocity at B.

Ft = mat ⇒ 60 N = (100 kg) at ⇒ at = 0.6 m/s2 ,

v s
dv v2
at = v ⇒ vd v = at d s ⇒ = at s,
ds 0 0 2

π
vB = 2at sB = 2(0.6 m/s2 ) 200 m + [50 m] = 18.3 m/s.
2
The normal component of the force is
v2 (18.3 m/s)2
Fn = man = m = (100 kg) = 668 N.
R 50 m

vB = 18.3 m/s, Fn = 668 N.

Problem 14.68 In a test of a sun-powered car, the

mass of the car and driver is 100 kg. The car starts
50 m
from rest on the track at A, moving toward the right. B
The tangential force exerted on the car (in newtons) is
given as a function of time by Ft = 20 + 1.2t. Deter- A
mine the magnitude of the car’s velocity and the normal
component of force on the car at t = 40 s. 200 m

Solution: We first find the tangential acceleration and use that to

find the velocity v and distance s as functions of time.

Ft = (20 + 1.2t)N = (100 kg) at

dv
at = = 0.2 + 0.012t
dt

v = 0.2t + 0.006t 2

s = 0.1t 2 + 0.002t 3
At t = 40 s, we have v = 17.6 m/s, s = 288 m.

For this distance the car will be on the curved part of the track. The
normal component of the force is
v2 (17.6 m/s)2
Fn = man = m = (100 kg) = 620 N.
R 50 m

vB = 17.6 m/s, Fn = 620 N.

an = rω2 = r(αt)2 = (10 m) (0.2 rad/s2 )2 t 2 = (0.4 m/s4 )t 2

(a)

At t = 0

Ft = mat = (72 kg)(2 m/s2 ) = 144 N, Fn = man = 0

F = Ft2 + Fn2 = 144 N.

(b)

At t = 10 s

Ft = mat = (72 kg)(2 m/s2 ) = 144 N,

Fn = man = (72 kg)(0.4 m/s4 )(10 s)2 = 2880 N

F = Ft2 + Fn2 = 2880 N.

(a) F = 144 N, (b) F = 2880 N.

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 A is supported horizontally by the smooth
slot and the string attached at B. Determine the tension A
in the string and the magnitude of the horizontal force 2 rad/s2 4 rad/s B
exerted on the slider by the slot.
0.6 m

Solution:

ω = 4 rad/s

α = 2 rad/s2 et
4 rad/s
R = 0.6 m
en 0.6 m
m
m = 0.5 kg T
0.2 rad/s2
Fn : T = mRω2 F

Ft : F = mRα

Solving, T = 4.8 N, F = 0.6 N

0.6 m

Solution:

R = 0.6 m
T

ω = 4 rad/s N
ω = 4 rad/s
en et
m = 0.5 kg

α=0 0.6 m

Fn : N cos 45◦ + T sin 45◦ = mRω2

Ft : −N sin 45◦ + T cos 45◦ = mRα = 0

Solving, N = T = 3.39 N

Problem 14.72 The 32,000-lb airplane is flying in the y

vertical plane at 420 ft/s. At the instant shown the angle
θ = 30◦ and the cartesian components of the plane’s
acceleration are ax = −6 ft/s2 , ay = 30 ft/s2 .
u
(a) What are the tangential and normal components
of the total force acting on the airplane (including x
its weight)?
(b) What is dθ/dt in degrees per second?

Solution:

32,000 lb
F= (−6i + 30j) ft/s2 = (−5963i + 29814j) lb
32.2 ft/s2

Ft = F · (cos 30◦ i + sin 30◦ j) = 9740 lb

(a)
Fn = F · (− sin 30◦ i + cos 30◦ j) = 28,800 lb

(b) 28800 lb
an = = 28.98 ft/s2
(32000 lb/32.2 ft/s2 )

v2 v2 (420 ft/s)2
an = ⇒ ρ= = = 6087 ft
ρ an (28.98 ft/s2 )

420 ft/s 180◦
v = ρ θ̇ ⇒ θ̇ = = 0.0690 rad/s = 3.95◦ /s
6087 ft π rad

Solution: The distance to the outer ring is 100 m.

r
(a) At a distance r the weight would be W= (353 N) =
100 m
(3.53 N/m)r
W = (3.53 N/m)r.

(b) At the center, r = 0

W = 0.

Problem 14.74 Small parts on a conveyor belt moving

with constant velocity v are allowed to drop into a bin. θ
Show that the angle θ at which the parts start sliding
1 v2
on the belt satisfies the equation cos θ − sin θ = ,
µs gR R
where µs is the coefficient of static friction between the
parts and the belt.

Solution: The condition for sliding is Ft = −mg sin θ + f = 0,

where −mg sin θ is the component of weight acting tangentially to f
the belt, and f = µs N is the friction force tangential to the belt.
From Newton’s second law the force perpendicular to the belt is N −
v2
mg cos θ = −m , from which the condition for slip is −mg sin θ +
R N θ
v2 1 v2
µs mg cos θ − µs m = 0. Solve: cos θ − sin θ =
R µs gR W

Strategy: Notice that the vertical acceleration of the

mass is zero. Draw the free-body diagram of the mass
and write Newton’s second law in terms of tangential
and normal components. m

Solution: T

◦
F↑ : T cos 30 − mg = 0
30°

◦ v2 v2
Fn : T sin 30 = m =m
ρ L sin 30◦
Solving we have
mg ◦ ◦
T = , v 2 = g(L sin 30 ) tan 30
cos 30◦

mg
(32.2 ft/s2 )(4 ft) sin2 30◦
v= = 6.10 ft/s
cos 30◦

Problem 14.76 In Problem 14.75, determine the mag-

nitude of the velocity of the mass and the angle θ if the
tension in the string is 50 lb.

Solution: T

F↑ : T cos θ − mg = 0
θ

v2
Fn : T sin θ = m
L sin θ

mg (T 2 − m2 g 2 )L
Solving we find θ = cos−1 , v=
T Tm

Using the problem numbers we have mg

1 slug 32.2 ft/s2
θ = cos−1 = 49.9◦
50 lb

[(50 lb)2 − (1 slug 32.2 ft/s2 )2 ]4 ft
v= = 10.8 ft/s
(50 lb)(1 slug)

55°

Solution: Choose a Cartesian coordinate system in the vertical y

plane that rotates with the mass. The weight of the mass is W =
−jmg = −j98.1 N. The radial acceleration is by definition directed TA
inward: TB
v2 θB
an = −i = −9i m/s2 . The angles from the horizontal are θA =
R θA
90 + 35 = 125◦ , θB = 90◦ + 55◦ = 145◦ . The unit vectors paral-
◦ ◦

lel to the strings, from the pole to the mass, are: eA = +i cos θA + x
j sin θA . eB = +i cos θB + j sin θB . From Newton’s second law for
the mass, T − W = man , from which |TA |eA + |TB |eB − jmg =
2 –j mg
v
−i m . Separate components to obtain the two simultane-
R
ous equations: |TA | cos 125◦ + |TB | cos 145◦ = −90 N|TA | sin 55◦ +
|TB | sin 35◦ = 98.1 N. Solve:

|TA | = 84 N. |TB | = 51 N

Problem 14.78 The 10-kg mass m rotates around the

vertical pole in a horizontal circular path of radius R =
1 m. For what range of values of the velocity v of the
mass will the mass remain in the circular path described?

Solution: The minimum value of v will occur when the string

B is impending zero, and the maximum will occur when string A is
impending zero. From the solution to Problem 14.77,

v2
|TA | cos 125◦ + |TB | cos 145◦ = −m ,
R

|TA | sin 125◦ + |TB | sin 145◦ = mg.

These equations are to be solved for the velocity when one of the
string tensions is set equal to zero. For |TA | = 0, v = 3.743 m/s. For
|TB | = 0, v = 2.621 m/s. The range: 2.62 ≤ v ≤ 3.74 m/s

Solution: The equations of motion are

Fy : T cos θ − mg = 0

v2
Fn : T sin θ = man = m
R
Solving we have
v2 (50 m/s)2
R= = = 700 m
g tan θ (9.81 m/s2 ) tan 20◦

R = 700 m.

Problem 14.80 An airplane of weight W = 200,000 lb 15°

makes a turn at constant altitude and at constant velocity
v = 600 ft/s. The bank angle is 15◦ . (a) Determine the
lift force L. (b) What is the radius of curvature of the L
plane’s path?

Solution: The weightis W = −jW = −j(2 × 105 ) lb. The normal

v2
acceleration is an = i . The lift is L = |L|(i cos 105◦ +
ρ
◦
j sin 105 ) = |L|(−0.2588i + 0.9659j).

(a) From Newton’s second law, F = L + Wn = man , from

which, substituting values and separating the j components:
2 × 105
|L|(0.9659) = 2 × 105 , |L| = = 207055 lb.
0.9659

(b) The radius of curvature

2 is obtained from Newton’s law:
v
|L|(−0.2588) = −m , from which
ρ

W v2
ρ= = 41763.7 ft.
g |L|(0.2588)

(a) What are the tensions in the strings A and B? B

(b) If string A is cut, what is the tension in string B
immediately afterward? m 45°
A

Solution: y
TB
(a) Fx = TB cos 45◦ − TA = 0,

Fy = TB sin 45◦ − mg = 0. TA
45°
x
Solving yields TA = 19.6 N, TB = 27.7 N.

(b) Use Normal and tangential components. mg

Fn = man : TB

v2
TB − mg cos 45◦ = m .
ρ
en
But v = 0 at the instant of release, so

TB = mg cos 45◦ = 13.9 N.

Problem 14.82 The airplane flies with constant veloc-

ity v along a circular path in the vertical plane. The
radius of the airplane’s circular path is 2000 m. The
mass of the pilot is 72 kg.

(a) The pilot will experience “weightlessness” at the

top of the circular path if the airplane exerts no
net force on him at that point. Draw a free-body
diagram of the pilot and use Newton’s second law
to determine the velocity v necessary to achieve
this condition.
(b) Suppose that you don’t want the force exerted
on the pilot by the airplane to exceed four times
his weight. If he performs this maneuver at v =
200 m/s, what is the minimum acceptable radius
of the circular path?

Solution: The FBD assumes that the seat pushes up on the pilot.
If the seat (or shoulder straps) pushes down, we will us a negative
sign for N .
mg

v2
Dynamics: F↑ : N − mg = −m
ρ

√
(a) N =0⇒v= gρ = (9.81 m/s2 )(2000 m) = 140 m/s

(b) The force will push down on the pilot

v2 v2 N
N = −4mg ⇒ −5mg = −m ⇒ρ= ρ = 815 m
ρ 5g

m
ω0
B
Solution:

◦
F↑ : N cos 40 − mg = 0

◦ v2 (R sin 40◦ ω0 )2
F← : N sin 40 = m =m
ρ R sin 40◦
Solving we find mg

mg g
N= , ω0 = 40°
cos 40◦ R cos 40◦

⇒ ω0 = 9.81 m/s2
= 5.06 rad/s
0.5 m cos 40◦ N

Problem 14.84 The force exerted on a charged particle Solution: (a) The force F = q(v × B) is everywhere normal to the
by a magnetic field is F = qv × B, where q and v are velocity and the magnetic field vector on the particle path. Therefore
the charge and velocity of the particle, and B is the the tangential component of the force is zero, hence from Newton’s
dv
magnetic field vector. A particle of mass m and positive second law the tangential component of the acceleration = 0, from
dt
charge q is projected at O with velocity v = v0 i into
which v(t) = C = v0 , and the velocity is a constant. Since there is
a uniform magnetic field B = B0 k. Using normal and
no component of force in the z-direction, and no initial z-component
tangential components, show that (a) the magnitude of of the velocity, the motion remains in the x-y plane. The unit vec-
the particle’s velocity is constant and (b) the particle’s tor k is positive out of the paper in the figure; by application of the
v0
path is a circle of radius m . right hand rule the cross product v × B is directed along a unit vector
qB0 toward the instantaneous center of the path at every instant, from which
F = −|F|en , where en is a unit vector normal to the path. The nor-
mal component of the acceleration is an = −(v02 /ρ)en , where ρ is the
y radius of curvature of the path. From Newton’s second law, F = man ,
from which −|F| = −m(v02 /ρ). The magnitude of the cross product
can be written as |v × B| = v0 B0 sin θ = v0 B0 , since θ = 90◦ is the
v2
angle between v and B. From which: qv0 B0 = m 0 , from which the
ρ
0 mv0
x radius of curvature is ρ = . Since the term on the right is a
O qB0
constant, the radius of curvature is a constant, and the path is a circle
mv0
with radius .
qB0

Strategy: The velocity vector of the mass is perpen- 0

dicular to the string. Express Newton’s second law in

terms of normal and tangential components.
θ

Solution: Make a hypothetical cut in the string and denote the L0

tension in the part connected to the mass by T. The acceleration normal
m
v2 θ
to the path is . The instantaneous radius of the path is ρ = L0 − Rθ,
ρ
v02
from which by Newton’s second law, Fn = T = m , from R v
L0 − Rθ (a) (b)
v02
which T = m
L0 − Rθ
T

Problem 14.86 The mass m is attached to a string

that is wrapped around the fixed post of radius R. At
t = 0, the mass is given a velocity v0 as shown. Neglect
external forces on m other than the force exerted by the
string. Determine the angle θ as a function of time.

Solution: Use the solution to Problem 14.85. The angular

dθ v0
velocity is = . From Problem 14.85, ρ = L0 − Rθ, from
dt ρ
dθ v0
which = . Separate variables: (L0 − Rθ)dθ = v0 dt.
dt (L0 − Rθ)
R 2
Integrate: L0 θ − θ = v0 t, since θ(0) = 0. In canonical form θ 2 +
2
L 2v0 t
2bθ + c = 0, where b = − , and c = . The solution: θ = −b ±
R R
√ L0 L0 2
2v0 t
b2 − c = ± − . The angle increases with time,
R R R

L0 2Rv0 t
so the negative sign applies. Reduce: θ = 1− 1−
R L20
Check: When Rθ = L0 , the string has been fully wrapped around the
2Rv0 t 2Rv0 t
post. Substitute to obtain: 1 − = 0, from which = 1,
L20 L20
which is the value for impending failure as t increases because of the
imaginary square root. Thus the solution behaves as expected. check.

Solution: This problem has several steps. First, we must write z

Newton’s second law and find the acceleration of the aircraft. We
then integrate the components of the acceleration (separately) to find
the velocity components as functions of time. Then we evaluate
the velocity of the aircraft and the force acting on the aircraft at
t = 2s. Next, we find a unit vector along the velocity vector of the
aircraft and project the total force acting on the aircraft onto this
direction. Finally, we find the magnitude of the total force acting on
the aircraft and the force component normal to the direction of flight.
We have aX = (1/m)(−1000 + 280t), aY = (1/m)(480 − 430t), and
aZ = (1/m)(720 + 200t). Integrating and inserting the known initial
velocities, we obtain the relations vX = vX0 + (1/m)(−1000t +
280t 2 /2) (m/s) = 20 + (1/m)(−1000t + 140t 2 ) (m/s), vY = 35 +
(1/m)(480t − 215t 2 ) (m/s), and vZ = −20 + (1/m)(720t + 100t 2 )
(m/s). The velocity at t = 2s is v = 16i + 35.3j − 14.9k (m/s) and the
unit vector parallel to v is ev = 0.386i + 0.850j − 0.359k. The total
force acting on the aircraft at t = 2s is F = −440i − 380j + 1120k N.
The component of F tangent to the direction of flight is Ft =
F ž ev = −894.5 N. The magnitude of the total force acting on the
aircraft is |F| = 1261.9 N. Thecomponent of F normal to the direction
of flight is given by Fn = |F|2 − ( Ft )2 = 890.1 N.

Problem 14.88 In Problem 14.87, what is the instanta-

neous radius of curvature of the plane’s path at t = 2 s?
The vector components of the sum of the forces in the
directions tangenial and normal to the path lie in the
osculating plane. Determine the components of a unit
vector perpendicular to the osculating plane at t = 2 s.

Strategy: From the solution to problem 14.87, we

know the total force vector and acceleration vector acting
on the plane. We also know the direction of the velocity
vector. From the velocity and the magnitude of the
normal acceleration, we can determine the radius of
curvature of the path. The cross product of the velocity
vector and the total force vector will give a vector
perpendicular to the plane containing the velocity vector
and the total force vector. This vector is perpendicular
to the plane of the osculating path. We need then only
find a unit vector in the direction of this vector.
Solution: From Problem 14.87, we know at t = 2 s, that an =
Fn /m = 2.47 m/s2 . We can find the magnitude of the velocity |v| =
41.5 m/s at this time. The radius of curvature of the path can then be
found from ρ = |v|2 /an = 696.5 m.

The cross product yields the desired unit vector, i.e., e = (F × v)/|F ×
v| = −0.916i + 0.308j − 0.256k

(a)

(b)

Solution: mg
ρ = 60 m, g = 9.81 m/s 2

◦ ◦
F↑ : N cos 15 − Fr sin 15 − mg = 0
Fr

◦ ◦ v2 15°
F← : N sin 15 + Fr cos 15 = m
ρ

Fr = 0.4N
N
Solving we have v = 21.0 m/s

Problem 14.90* The freeway off-ramp is circular with

60-m radius (Fig. a). The off-ramp has a slope β (Fig. b).
If the coefficient of static friction between the tires of
a car and the road is µs = 0.4 what minimum slope β
is needed so that the car could (in theory) enter the off-
ramp at any speed without losing traction? (See Example
14.8.)

Solution: mg

F↑ : N cos β − Fr sin β − mg = 0

Fr
v2
F← : N sin β + Fr cos β = m
ρ β

F = µN
µmg
Solving we have Fr = .
cos β − µ sin β N

If we set the denominator equal to zero, then we will always have

enough friction to prevent sliding.

1 1
Thus β = tan−1 = tan−1 = 68.2◦
µ 0.4

We would also need to check the low-speed case, where the car might
slip down the ramp.

v2
Solution: From Newton’s second law; N − W = −m from
W
R
v2 dv
which N = m g − . The acceleration tangent to the path is ,
R dt

dv µk N dv v2 f
from which =− , and = −µk g − = 4.25 m/s2
dt m dt R N

Problem 14.92 A car traveling at 30 m/s is at the bot-

tom of a depression. The coefficient of kinetic friction
between the tires and the road is µk = 0.8. The instan-
taneous radius of curvature of the car’s path is 200 m.
If the driver applies the brakes and the car’s wheel lock,
what is the resulting deceleration of the car in the direc-
tion tangential to its path? Compare your answer to that
of Problem 14.91.
dv µk N
Solution: Use the solution to Problem 14.91: =− . From
dt m f
v2
Newton’s second law, N − W = m , from which
R
W
N
v2
N =m g+ ,
R

dv v2
and = −µk g + = −11.45 m/s2
dt R

Solution:

m = 160 kg P
en 400 m

R = 400 m
et
Along track motion:

at = 2 + 0.2t m/s2

Vt = V = 2t + 0.1t 2 m/s mg
ez
s = t 2 + 0.1t 3 /3 m

Forces at impending slip

en
|F + f| = µk N at impending slip

|F + f| = F 2 + f 2 since f ⊥ F V 2/R

Force eqns.

Ft : F = mat R = 400 m
f

Fn : f = mv 2 /R m = 160 kg N

Fz : N − mg = 0 g = 9.81 m/s2

µs = 0.8 F

F 2 + f 2 = µs N
V 2/R
f
at = 2 + 0.2t
en
et
v = 2t + 0.1t 2

Six eqns, six unknowns (F, f, v, at , N, t)

Solving, we have

t = 14.4 s F = 781 N

v = 49.6 m/s f = 983 N

N = 1570 N

at = 4.88 m/s2

(at t = 14.4 s)

u
x

Solution:

r = 12 − 0.4t 2 , θ = 0.02t 3

ṙ = −0.8t, θ̇ = 0.06t 2

r̈ = −0.8, θ̈ = 0.12t

Set t = 2 s

Fr = m(r̈ − r θ̇ 2 ) = (12 kg)(−0.8 − [10.4][0.24]2 ) m/s2

= −16.8 N
Fθ = m(r θ̈ + 2ṙ θ̇ ) = (12 kg)([10.4][0.24] + 2[−1.6][0.24]) m/s2
= 20.7 N

Problem 14.95 A 100-lb person walks on a large

disk that rotates with constant angular velocity ω0 =
0.3 rad/s. He walks at a constant speed v0 = 5 ft/s along ω0
0
a straight radial line painted on the disk. Determine the
polar components of the horizontal force exerted on him
when he is 6 ft from the center of the disk. (How are
these forces exerted on him?)

Solution:
r = 6 ft, ṙ = 5 ft/s, r̈ = 0, θ̇ = 0.3 rad/s, θ̈ = 0

100 lb
Fr = m(r̈ − r θ̇ 2 ) = (0 − [6 ft][0.3 rad/s]2 )
32.2 ft/s2
= −1.68 lb

100 lb
Fθ = m(r θ̈ + 2ṙ θ̇) = (0 + 2[5 ft/s][0.3 rad/s])
32.2 ft/s2
= 9.32 lb
The forces are exerted as friction between the disk and the
man’s feet.

Determine the radial and transverse components of the

force exerted on A by the robot’s jaws at t = 2 s.

Solution: The radial component of the acceleration is

2
d2r dθ
ar = −r . Fθ
dt 2 dt Fr
The derivatives: θ

dr d
= (1 − 0.5 cos 2π t) = π sin 2π t,
dt dt
mg
d2r d
= (π sin 2π t) = 2π 2 cos 2π t;
dt 2 dt From Newton’s second law, Fr − mg sin θ = mar , and Fθ −
mg cos θ = maθ , from which
dθ d
= (0.5 − 0.2 sin 2π t) = −0.4π cos 2π t.
dt dt
Fr = 0.4ar + 0.4g sin θ = 9.46 N.
d2θ d
= (−0.4π cos 2π t) = 0.8π 2 sin 2π t. The transverse component of the acceleration is
dt 2 dt

From which d2θ dr dθ
aθ = r +2 ,
dt 2 dt dt
[ar ]t=2 = 2π 2 cos 4π − (1 − 0.5 cos 4π )(−0.4π cos 4π )2 ,
from which [aθ ]t=2 = (1 − 0.5 cos 4π )(0.8π 2 sin 4π ) + 2(π sin 4π )
= 2π 2 − 0.08π 2 = 18.95 m/s2 ; (−0.4π sin 4π ) = 0, and

θ(t = 2) = 0.5 rad. Fθ = 3.44 N

Problem 14.97 A 50-lb object P moves along the spi- P

ral path r = (0.1)θ ft, where θ is in radians. Its angular
position is given as a function of time by θ = 2t rad, r
and r = 0 at t = 0. Determine the polar components of
the total force acting on the object at t = 4 s. u

Solution:
θ = 2t, θ̇ = 2, θ̈ = 0, r = 0.1θ = 0.2t, ṙ = 0.2, r̈ = 0

At t = 4 s, θ = 8, θ̇ = 2, θ̈ = 0, r = 0.8, ṙ = 0.2, r̈ = 0

Thus

50 lb
Fr = m(r̈ − r θ̇ 2 ) = (0 − [0.8 ft][2 rad/s]2 )
32.2 ft/s2
= −4.97 lb

50 lb
Fθ = m(r θ̈ + 2ṙ θ̇ ) = (0 + 2[0.2 ft/s][2 rad/s])
32.2 ft/s2
= 1.24 lb

vr2 (10 m/s)2 [2 m]2 [1 m]2
− = (6.28 rad/s)2 −
2 2 2 2

vr = 14.8 m/s.

Problem 14.99 The smooth bar rotates in the horizon- v0

tal plane with constant angular velocity ω0 = 60 rpm.
The spring constant is k = 20 N/m and the unstretched
length of the spring is 3 m. If the radial velocity of the k
1-kg collar A is vr = 10 m/s when its radial position is
r = 1 m, what is its radial velocity when r = 2 m? (See A
Active Example 14.9.)

Solution: Notice that the only radial force comes from the spring. r 3m
Write the term
d2r dvr dvr d r dvr
= = = vr
dt 2 dt dr dt dr
The angular velocity is

2π rad 1 min
ω = 60 rpm = 6.28 rad/s.
rev 60 s
The equation of motion in the radial direction is
k
Frm : −kr = mar ⇒ ar = − r
m
Then

d2r k d2r k
ar = − rω 2
= − r ⇒ = r ω 2
−
dt 2 m dt 2 m

vr 2 m
d2r dvr k k
2
= vr = r ω2 − ⇒ vr dvr = ω2 − rdr
dt dr m 10 m/s 1 m m

vr2 (10 m/s)2 20 N/m [2 m]2 [1 m]2
− = (6.28 rad/s)2 − −
2 2 1 kg 2 2

vr = 12.6 m/s.

Solution: T

L = 0.6 m

m = 2 kg
θ

m
Fr = mar

d2L
−T + mg sin θ = m − Lw 2
dt 2 θ
mg

Fθ = maθ er
eθ
dL
mg cos θ = m 2 w + Lα
dt

dL d2 L
However = 2 =0
dt dt

dw
Lα = L w = g cos θ
dθ
w π/4
g
wdw = cos θdθ
0 L 0

w2 g π/4 g π
= sin θ = sin
2 L 0 L 4

w = 4.81 rad/s

|v| = Lw = 2.89 m/s

− T + mg sin θ = −mLw 2

T = m(g sin θ + Lw 2 )

T = 41.6 N

Solution: For this problem, Newton’s second law in polar coordi-

eθ
nates states er

Fr = mg cos θ − N = m(d 2 r/dt 2 − rω2 ) and θ

Fθ = −mg sin θ = m(rα + 2ω(dr/dt)). θ

W N

In this problem, r is constant. Thus (dr/dt) = (d 2 r/dt 2 ) = 0, and the

equations reduce to N = mrω2 + mg cos θ and rα = −g sin θ. The

first equation gives us a way to evaluate the normal force while the sec-

ond can be integrated to give ω(θ). We rewrite the second equation as

dω dω dθ dω g
α= = =ω =− sin θ
dt dθ dt dθ r
ω60 g 60◦
and then integrate ω0 ωdω = − sin θdθ. Carrying out the
r 0

integration, we get

2
ω60 ω2 g ◦ g
− 0 =− 0 =−
(− cos θ)|60 (1 − cos 60◦ ).
2 2 r r

Noting that ω0 = v0 /R = 3.5 rad/s, we can solve for ω60 = 2.05 rad/s
and v60 = Rω60 = 8.20 ft/s.

Problem 14.102 The 1-lb block is given an initial

velocity v0 = 14 ft/s to the right when it is in the position
θ = 0, causing it to slide up the smooth circular surface.
Determine the normal force exerted on the block by the
surface when θ = 60◦ .
Solution: From the solution to Problem 14.101, we have N60 =
2 + mg cos 60◦ or N = 1.02 lb.
mrω60

45°

Solution: In terms of the angle θ shown, the transverse component

of his weight is mg cos θ. Therefore

Fθ = maθ : θ

2 0
d θ dr dθ
mg cos θ = m r 2 + 2 . (1)
dt dt dt

Note that

d2θ dw dw dθ dw
= = = w,
dt 2 dt dθ dt dθ

So (1) becomes
eθ
er
dw
g cos θ = r w.
dθ mg
Separating variables,

g
wdw = cos θdθ. (2)
r

At A, θ = 45◦ and w = vA /r = 17/6 = 2.83 rad/s. Integrating (2),

wB 90
◦
g
wdw = cos θdθ,
2.83 r 45◦

we obtain wB = 3.00 rad/s. His velocity at B is rwB = 18.0 m/s.

vB = 18.0 m/s.

Solution: From Newton’s second law for the radial component

−mg sin θ ± µs N = −mRω2 , and for the normal component: N −
dω dω θ
mg cos θ = mRα. Solve, and note that α = =ω = 1 = const,
dt dθ
ω = 2θ, since ω(0) = 0, to obtain −g sin θ ± µs (g cos θ + Rα) =
2 f N
−2Rθ. For α = 1, R = 1, this reduces to ±µs (1 + g cos θ) = −2θ + W
g sin θ. Define the quantity FR = 2θ − g sin θ. If FR > 0, the block
will tend to slide away from O, the friction force will oppose the
motion, and the negative sign is to be chosen. If FR < 0, the block
will tend to slide toward O, the friction force will oppose the motion,
and the positive sign is to be chosen. The equilibrium condition
is derived from the equations of motion: sgn(FR )µs (1 + g cos θ) =
2θ − g sin θ
(2θ − g sin θ), from which µs = sgn(FR ) = 0.406 .
1 + g cos θ
Since Fr = −3.86 < 0, the block will slide toward O.

Problem 14.105* The 1/4-lb slider A is pushed along

the circular bar by the slotted bar. The circular bar lies in
A
the horizontal plane. The angular position of the slotted
bar is θ = 10t 2 rad. Determine the polar components of
θ
the total external force exerted on the slider at t = 0.2 s.

2 ft
2 ft

Solution: The interior angle β is between the radius from O to the

slider A and the horizontal, as shown in the figure. The interior angle
formed by the radius from C to the slider A and the line from O to
the slider is β − θ. The angle β is found by applying the law of sines:
2 2
= from which sin θ = sin(β − θ) which is satisfied
sin θ sin(β − θ) A
by β = 2θ. The radial distance of the slider from the hinge point is
r 2 2 sin 2θ θ
also found from the sine law: = , r= ,
sin(180 − β) sin θ sin θ β
from which r = 4 cos θ. The radial component of the acceleration O
2
d2r dθ dθ d
is ar = 2 − r . The derivatives: = (10t 2 ) = 20t. 2 ft
dt dt dt dt
2 ft
d2θ dr dθ d2r
= 20. = −4 sin θ = −(80 sin θ)t. 2 = −80 sin θ −
dt 2 dt dt dt
(1600 cos θ)t 2 . Substitute:

d2θ dr dθ
The transverse acceleration is aθ = r +2 .
[ar ]t=0.2 = [ar = −80 sin(10t 2 ) − (1600 cos(10t 2 ))t 2 dt 2 dt dt
Substitute:
− (4 cos(10t 2 ))(20t)2 ]t=0.2
[aθ ]t=0.2 = [aθ = (4 cos(10t 2 ))(20)
= −149.0 ft/s2 .
+ 2(−80 sin(10t 2 ))(t)(20)(t)]t=0.2 = 23.84 ft/s2 .
From Newton’s second law, the radial component of the external
force is The transverse component of the external force is

W W
Fr = ar = −1.158 lb. Fθ = aθ = 0.185 lb
g g

c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they
currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
168
Problem 14.106* The 1/4-lb slider A is pushed along
the circular bar by the slotted bar. The circular bar lies
in the vertical plane. The angular position of the slotted
bar is θ = 10t 2 rad. Determine the polar components of
the total force exerted on the slider by the circular and
slotted bars at t = 0.25 s.
Solution: Assume that the orientation in the vertical plane is such FromNewton’s second law for the radial component Fr − mg sin θ =

that the θ = 0 line is horizontal. Use the solution to Problem 14.105. W
ar , from which Fr = −1.478 lb The transverse component of
For positive values of θ the radial component of acceleration due to g
gravity acts toward the origin, which by definition is the same direction the acceleration is (from Problem 14.105)
as the radial acceleration. The transverse component of the acceleration
due to gravity acts in the same direction as the transverse acceleration. [aθ ]t=0.25 = [(4 cos θ)(20)
From which the components of the acceleration due to gravity in
the radial and transverse directions are gr = g sin θ and gθ = g cos θ. + 2(−80 sin θ)(t)(20)(t)]t=0.25 = −52.14 ft/s2 .
These are to be added to the radial and transverse components of
acceleration due to the motion. From Problem 14.105, θ = 10t2 rad
Newton’s second law for transverse component Fθ − mg cos θ =
From
W
aθ , from which Fθ = −0.2025 lb
[ar ]t=0.25 = [−80 sin θ − (1600 cos θ)t 2 g

− (4 cos θ)(20t)2 ]t=0.25 = −209 ft/s2 .

Problem 14.107* The slotted bar rotates in the hori-

ω0
zontal plane with constant angular velocity ω0 . The mass
m
m has a pin that fits into the slot of the bar. A spring
holds the pin against the surface of the fixed cam. The Cam
surface of the cam is described by r = r0 (2 − cos θ ).
Determine the polar components of the total external
k
force exerted on the pin as functions of θ .
θ

Solution: The angular velocity is constant, from which θ =

ω0 dt + C = ω0 t + C. Assume that θ(t = 0) = 0, from which C =
2
d2r dθ
0. The radial acceleration is ar = 2 − r . The deriva-
dt dt
dθ d d2θ dr d
tives: = (ω0 t) = ω0 , 2
= 0. = (r0 (2 − cos θ)) =
dt
dt dt dt dt
dθ d2r d
r0 sin θ = ω0 r0 sin θ, = (ω0 r0 sin θ) = ω02 r0 cos θ.
dt dt 2 dt
Substitute: ar = ω02 r0 cos θ − r0 (2 − cos θ)(ω02 ) = 2r0 ω02 (cos θ − 1).
From Newton’s second law the radial component of the external
force is

Fr = mar = 2mr0 ω02 (cos θ − 1).

d2θ
The transverse component of the acceleration is aθ = r 2 +
dt
dr dθ
2 . Substitute: aθ = 2r0 ω0 sin θ. From Newton’s second
2
dt dt
law, the transverse component of the external force is

Fθ = 2mr0 ω02 sin θ

c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they
currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
169
Problem 14.108* In Problem 14.107, suppose that the
unstretched length of the spring is r0 . Determine the
smallest value of the spring constant k for which the pin
will remain on the surface of the cam.
Solution: The spring force holding the pin on the surface of the
cam is Fr = k(r − r0 ) = k(r0 (2 − cos θ) − r0 ) = kr0 (1 − cos θ). This
force acts toward the origin, which by definition is the same direction
as the radial acceleration, from which Newton’s second law for the
pin is F = kr0 (1 − cos θ) = −mar . From the solution to Problem
14.107, kr0 (1 − cos θ) = −2mrω02 (cos θ − 1). Reduce and solve: k =
2mω02 . Since cos θ ≤ 1, kr0 (1 − cos θ) ≥ 0, and 2mr0 ω02 (cos θ − 1) ≤
0. If k > 2mω02 , Define Feq = kr0 (1 − cos θ) + 2mrω02 (cos θ − 1). If
Feq > 0 the spring force dominates over the range of θ, so that the
pin remains on the cam surface. If k < 2mω02 , Feq < 0 and the radial
acceleration dominates over the range of θ, so that the pin will leave
the cam surface at some value of θ. Thus k = 2mω02 is the minimum
value of the spring constant required to keep the pin in contact with
the cam surface.

Problem 14.109 A charged particle P in a magnetic y

field moves along the spiral path described by r = 1 m,
θ = 2z rad, where z is in meters. The particle moves
along the path in the direction shown with a constant x
speed |v| = 1 km/s. The mass of the particle is 1.67 ×
10−27 kg. Determine the sum of the forces on the particle
in terms of cylindrical coordinates.
P
z

1 km/s

Solution: The force components in cylindrical coordinates are

given by

d2r 2
Fr = mar = m rω ,
dt 2

dr
Fθ = maθ = m rα + 2 ω ,
dt

d2z dr d2r
and Fz = maz = m 2
. From the given information, = 2 =
dt dt dt
0. We also have that θ = 2z. Taking derivatives of this, we
dθ dz
see that =ω=2 = 2vz . Taking another derivative, we get
dt dt
α = 2az . There is no radial velocity component so the constant
magnitude of the velocity |v|2 = vθ2 + vz2 = r 2 ω2 + vz2 = (1000 m/s)2 .
Taking the derivative of this expression with respect to time,

dω dvz dvz d2z
we get r 2 2ω + 2vz = 0. Noting that = 2 and
dt dt dt dt
dω dvz
that α = , we can eliminate from the equation. We get
dt
α
dt
ω
2r 2 ωα + 2 , giving (2r 2 + 1/2)ωα = 0. Since ω = 0, α =
2 2
0, and az = 0. Substituting these into the equations of motion,
we get ω = 4(1000)2 5 (rad/s)2 , and
2 Fr = −mrω2 = −1.34 ×
10−21 m/s2 , Fθ = 0 and Fz = 0

Solution: The total force acting on part A in cylindrical y

2 d r
coordinates is given by Fr = mar = m − rω2 , Fθ = eθ
dt 2 Fmanip.
2 er
dr d z
maθ = m rα + 2 ω , and Fz = maz = m 2 . We are A
dt dt
given the values of every term in the right hand side of these equations. j
(Recall the definitions of ω and α. Substituting in the known values, 25°
we get Fr = −5.06 N, Fθ = 8.64 N, and Fz = 10.0 N. These W = –mgj
i x
are the total forces acting on Part A, including the weight.
To find the forces exerted on the part by the manipulator, we
need to draw a free body diagram of the part and resolve the
weight into components along the various axes. We get F=
Fmanip + W = ma where the components of F have already
been determined above. In cylindrical coordinates, the weight is
given as W = −mg sin θer − mg cos θeθ . From the previous equation,
F = Fmanip − mg sin θer − mg cos θeθ . Substituting in terms of
the components, we get ( Fmanip )z = 11.5 (newtons), ( Fmanip )θ =
44.2 (newtons) and ( Fmanip )z = 10.0 (newtons).

θ = 0.15t 2 rad,

dθ
= 0.3t rad/s,
dt

d2θ
= 0.3 rad/s2 .
dt 2

r = 0.5(1 + sin θ) m,

dr dθ
= 0.5 cos θ m/s,
dt dt
2
d2r d2θ dθ
= 0.5 cos θ − 0.5 sin θ m/s2 .
dt 2 dt 2 dt

z = 0.8(1 + θ) m,

dz dθ
= 0.8 m/s,
dt dt

d2z d2θ
= 0.8 2 m/s2 .
dt 2 dt

Evaluating these expressions at t = 2 s, the acceleration is

2 2
d2r dθ d θ dr dθ d2z
a= −r er + r 2 + 2 e θ + 2 ez
dt 2 dt dt dt dt dt

= −0.259er + 0.532eθ + 0.24ez (m/s2 ).

Therefore

F = ma

= −1.04er + 2.13eθ + 0.96ez (N).

and use it to estimate the maximum force during that

F 8
interval of time. m
a 7
g
Solution: Use a numerical solver to work problem 14.111 for a 6
series of values of time during the required interval and plot the mag- –
nitude of the resulting force as a function of time. From the graph, the 5
n
maximum force magnitude is approximately 8.4 N and it occurs at a e 4
time of about 4.4 seconds. w
t 3
o
n 2
s
1
0 .5 1 1.5 2 2.5 3 3.5 4 4.5 5
Time (s)

Problem 14.113 The International Space Station is in

a circular orbit 225 miles above the earth’s surface.
(a) What is the magnitude of the velocity of the space
station?
(b) How long does it take to complete one revolution?

Solution: The radius of the orbit is

r0 = RE + 225 mi (32.2)[(3960)(5280)]2
=
2.21 × 107
= 3960 + 225 mi
= 25200 ft/s(17200 mi/h).
= 2.21 × 107 ft.
(b) Let T be the time required. Then v0 T = 2π r0 ,
(a) From Eq (14.24), the velocity is
2π r0
so T = = 5500 s (1.53 h).
gRE2 v0
v0 =
r0

Problem 14.114 The moon is approximately 383,000

km from the earth. Assume that the moon’s orbit around
the earth is circular with velocity given by Eq. (14.24).

(a) What is the magnitude of the moon’s velocity?

(b) How long does it take to complete one revolution
around the earth?
Solution: EARTH
MOON

RE = 6370 km = 6.37 × 106 m

383000 km

r0 = 383000 km = 3.83 × 108 m

Period = 2π r0 /v0
gRE2
v0 = = 1020 m/s
r0 Period = 2.36 × 106 s = 27.3 days

Solution: We have

r 0 v0 2 1+ε
ε= − 1, rmax = r0 , r0 v0 = rmax vmax radius .
gRE 1−ε
Solving we find
rmax − r0 13,400 km − 6700 km
ε= = = 0.333,
rmax + r0 13,400 km + 6700 km

RE 2 (6370 km)2
v0 = g(1 + ε) = (9.81 m/s2 )(1.333) = 8900 m/s,
r0 6700 km

r0 6700 km
vmax radius = v0 = (8900 m/s) = 4450 m/s
rmax 13400 km

(a) v0 = 8900 m/s, (b)vmax radius = 4450 m/s.

Problem 14.116 A satellite is given an initial velocity

v0 = 6700 m/s at a distance r0 = 2RE from the center
of the earth as shown in Fig. 14.18a. Draw a graph of
the resulting orbit.

Solution: The graph is shown.

2.88RE

5.13RE 2RE

r = 42.2×106 m = 42,200 km.

Problem 14.118* You can send a spacecraft from the Moon’s Elliptic
earth to the moon in the following way. First, launch orbit orbit
the spacecraft into a circular “parking” orbit of radius Parking
r0 around the earth (Fig. a). Then increase its velocity orbit v0
in the direction tangent to the circular orbit to a value
v0 such that it will follow an elliptic orbit whose maxi-
mum radius is equal to the radius rM of the moon’s orbit r0 r0
around the earth (Fig. b). The radius rM = 238,000 mi. rM
Let r0 = 4160 mi. What velocity v0 is necessary to send
a spacecraft to the moon? (This description is simplified
in that it disregards the effect of the moon’s gravity.)
(a) (b)

Solution: Note that

RE = 3960 mi, r0 = 4160 mi, rM = 238,000 mi

First find the eccentricity:

1+ε r M − r0
rmax = rM = r0 ⇒ε=
1−ε rM + r0
Then use eq. 14.23

r 0 v0 2 (1 + ε)g 2grM
ε= − 1 ⇒ v0 = RE = RE
gRE 2 r0 r0 (r0 + rM )

Putting in the numbers we have v0 = 35,500 ft/s

velocity v0 in the direction shown. Show that the polar

equation for the resulting orbit is r0
RE
r (ε + 1) cos2 β
= ,
r0 [(ε + 1) cos2 β − 1] cos θ − (ε + 1) sin β cos β sin θ + 1
2
r0 v 0
where ε = − 1.
gRE2

Solution: We need to modify the solution in Section 14.5 to From conditions (3) and (4) and Eq. (5),

account for this new initial condition. At θ = 0 (see Fig. 14.17) 1 gR 2

B= − 2 2 E2
r0 r0 v0 cos β
dr
vr = = v0 sin β
dt and

and sin β
A=− .
r0 cos β
dθ
vθ = r = v0 cos β.
dt Substituting these expressions for A and B into Eq (2) yields the
desired result.
Therefore Eq (14.15) becomes

dθ
r2 = rvθ = r0 v0 cos β (1)
dt

Following the same steps that led to Eq. (14.21) in terms of u = 1/r

yields

gRE2
u = A sin θ + B cos θ + . (2)
r02 v02 cos2 β

At θ = 0,

1
u= . (3)
r0

Also, note that

dr d 1 1 du 1 du dθ
vr = = =− 2 =− 2
dt dt u u dt u dθ dt

du
= −r0 v0 cos β ,
dθ

where we used (1). Therefore, at θ = 0

du
−r0 v0 cos β = v0 sin β. (4)
dθ

From (2),

du
= A cos θ − B sin θ. (5)
dθ

Solution:

(a) 60 mi/h = 88 ft/s.

dv dv
a= = v.
dt ds

Integrating,

0 112
vdv = ads,
88 0

we obtain a = −34.57 ft/s2 . The magnitude of the friction

force is

3250
f = m|a| = (34.57)
32.2

= 3490 lb.

(b) The Normal force is the car’s weight, so

f 3490
µs = =
N 3250

= 1.07.

Problem 14.121 Using the coefficient of friction

obtained in Problem 14.120, determine the highest speed
at which the NSX could drive on a flat, circular track of
600-ft radius without skidding.

Solution: The free body diagram is at the right. The normal force
is equal to the weight and the friction force has the same magni-
mg
tude as in Problem 14.120 since f = µs N . The equation of motion
in the radial direction (from the center of curvature of the track to
the car) is Fr = mv 2 /R = f = µs N = µs mg. Thus, we have that R
mv /R = µs mg or v 2 = µs Rg. Inserting the numbers, we obtain v =
2

101.8 ft/s = 69.4 mi/h.

N f

40°

Solution: (a) The force in coupling 1 is

F3
◦
F1 = 10,000 g sin 40 = 631 kN. N
F2
mg
The force on coupling 2 is F1 N N
mg mg
F2 = 20,000 g sin(40) = 126.1 kN. N N N
mg mg mg

The force on coupling 3 is

F3 = 30,000 g sin 40◦ = 189.2 kN.

(b) From Newton’s second law, F1a − mg sin 40◦ = ma. Under
constant acceleration up the mountain, the force on coupling 1 is

F1a = 10,000a + 10,000 g sin 40◦ = 75.1 kN.

The force on coupling 2 is F2a = 2F1a = 150.1 kN.

The force on coupling 3 is F2a = 3 F1a = 225.2 kN

Solution: Assume that the mass of the spacecraft is negligible

compared to mass of the asteroid. With constant downward velocity,
mgs
the thrust balances the gravitational force: 0.01 − mgs = 0, where m
is the mass of the space craft. With the change in thrust, this becomes
0.005 − mgs = m(−5 × 10−6 ) N/kg2 . Multiply the first equation by
0.005, the second by 0.01, and subtract: The result:
T

0.01(5 × 10−6 )
gs = = 1 × 10−5 N/kg2
(0.01 − 0.005)

Problem 14.124 A car with a mass of 1470 kg, includ-

ing its driver, is driven at 130 km/h over a slight rise
in the road. At the top of the rise, the driver applies the
brakes. The coefficient of static friction between the tires
and the road is µs = 0.9 and the radius of curvature of
the rise is 160 m. Determine the car’s deceleration at the
instant the brakes are applied, and compare it with the
deceleration on a level road.
Solution: First, note that 130 km/h = 36.11 m/s. We have a sit-
uation in which the car going over the rise reduces the normal force mg
v0
exerted on the car by the road and also reduces the braking force. The CAR
free body diagram of the car braking over the rise is shown at the
N f
right along with the free body diagram of the car braking on a level
surface. For the car going over the rise, the equations of motion are
Ft = −f = mat , where f is the friction force. The normal equation
is Fn = N − mg = mv 2 /R. The relation between friction and nor- 160 m
mal force is given as f = µs N . Solving, we get at = −1.49 m/s2 .

For the car braking on a level surface, the equations are N − mg =

0, f = µs N , and −f = max . Evaluating, we get ax = 8.83 m/s2 .
Note that the accelerations are VERY different. We conclude that at
130 km/h, a rise in the road with a radius of 160 m is not “slight”. mg
v0
The car does not become airborne, but if the radius of curvature were
smaller, the car would leave the road.

N f

Problem 14.126 The aircraft carrier Nimitz weighs

91,000 tons. (A ton is 2000 lb.) Suppose that it is
traveling at its top speed of approximately 30 knots (a
knot is 6076 ft/h) when its engines are shut down. If
the water exerts a drag force of magnitude 20,000v lb,
where v is the carrier’s velocity in feet per second, what
distance does the carrier move before coming to rest?

Solution: The force on the carrier is F = −Kv, where

F gK
K = 20,000. The acceleration is a = =− v. Use the
m W
dv gK
chain rule to write v =− v. Separate variables and inte-
dx W
gK gK
grate: dv = − dx, v(x) = − x + C. The initial veloc-
W W
6076 ft 1h
ity: v(0) = 30 = 50.63 ft/s, from which C =
1h 3600 s
gK
v(0) = 50.63, and v(x) = v(0) − x, from which, at rest,
W

W v(0)
x= = 14,321 ft = 2.71 mi
gK

20°

Solution: Choose a coordinate system with the origin at the wall

and the x axis parallel to the plane surface. Denote θ = 20◦ . Assume T T µ NA
NA
that slip has begun. From Newton’s second law for block A:
µ NA A B
µ NB
(1) Fx = −T + mA g sin θ + µk NA = mA aA ,
NA WA
NB WB
(2) Fy = NA − mA g cos θ = 0. From Newton’s second law for
block B:

(3) Fx = −T − µk NB − µk NA + mB g sin θ = mB aB ,

(4) Fy = NB − NA − mB g cos θ = 0. Since the pulley is one-to-

one, the sum of the displacements is xB + xA = 0. Differentiate
twice:

(5) aB + aA = 0. Solving these five equations in five unknowns, T =

49.63 N, NA = 92.2 N, NB = 460.9 N, aA = −0.593 m/s2 ,

aB = 0.593 m/s2

Problem 14.128 In Problem 14.127, if A weighs

20 lb, B weighs 100 lb, and the coefficient of kinetic
friction between all surfaces is µk = 0.15, what is the
tension in the cord as B slides down the inclined surface?
Solution: From the solution to Problem 14.127,

(1) Fx = −T + mA g sin θ + µk NA = mA aA ,

(2) Fy = NA − mA g cos θ = 0. For block B:

(3) Fx = −T − µk NB − µk NA + mB g sin θ = mB aB ,

(4) Fy = NB − NA − mB g cos θ = 0.

(5) aB + aA = 0 Solve by iteration:

T = 10.46 lb ,

NA = 18.8 lb, NB − 112.8 lb, aA = −1.29 ft/s2 ,

aB = 1.29 ft/s2

where m is the mass of the projectile and A is the cross-

sectional area of the tube.
Solution: The force acting on the projectile is F = pA where p is
the instantaneous pressure and A is the area. From pV γ = K, where
γ
K = p0 V0 is a constant, and the volume V = Ax, it follows that
K
p= , and the force is F = KA1−γ x −γ . The acceleration is
(Ax)γ

F K
a= = A1−γ x −γ .
m m

The equation to be integrated:

dv K dv dv dx dv
v = A1−γ x −γ , where the chain rule = =v has
dx m dt dx dt dx
been used. Separate variables and integrate:

K K x 1−γ
v2 = 2 A1−γ x −γ dx + C = 2 A1−γ + C.
m m 1−γ

When x = x0 , v0 = 0, therefore

K A1−γ 1−γ
v2 = 2 (x 1−γ − x0 ).
m 1−γ

γ γ
Substitute K = p0 V0 = p0 Aγ x0 and reduce:

γ
2p0 Ax0 1−γ
v2 = (x 1−γ − x0 ). Rearranging:
m(1 − γ )

2p0 Ax0γ 1 1
v= − γ −1
m(γ − 1) x γ −1 x
0

Solution: Denote the tension in the cord near the wall by TA . From
Newton’s second law for the two blocks: T T
mA
WA WB
Fx = T A = + aA . mB
g g
wB
WB
For block B: Fy = TA − WB = aB . Since the pulley is one-
g
to-one, as the displacement of B increases downward (negatively) the
displacement of A increases to the right (positively), from which xA =
−xB . Differentiate twice to obtain aA = −aB . Equate the expressions
toobtain:
WA WB WB
a + = WB + a, from which
g g g

WB 20 32.17
a=g =g = = 4.02 ft/s2
WA + 2WB 160 g

Problem 14.131 The 100-Mg space shuttle is in orbit

when its engines are turned on, exerting a thrust force
T = 10i − 20j + 10k (kN) for 2 s. Neglect the resulting
change in mass of the shuttle. At the end of the 2-s burn,
fuel is still sloshing back and forth in the shuttle’s tanks.
What is the change in the velocity of the center of mass
of the shuttle (including the fuel it contains) due to the
2-s burn?
Solution: At the completion of the burn, there are no external
forces on the shuttle (it is in free fall) and the fuel sloshing is caused
by internal forces that cancel, and the center of mass is unaffected.
The change in velocity is

2 T 2(104 ) 2(2 × 104 ) 2(104 )

v= dt = i− j+ k
0 m 105 105 105
= 0.2i − 0.4j + 0.2k (m/s)

Solution: Break the path into two parts: (1) The path from the
base to the top of the ramp, and (2) from the top of the ramp until
impact with the water. Let u be the velocity parallel to the surface of
the ramp, and let z be the distance along the surface of the ramp.

du 8
From the chain rule, u = −g sin θ, where θ = tan−1 =
dz 20
◦
21.8 . Separate variables and integrate:
u2 = −(2g sin θ)z + C. At the base of the ramp

mi 5280 ft 1h
u(0) = 25 = 36.67 ft/s,
h 1 mi 3600 s
√
from√which C = (36.672 ) = 1344.4 and u = C − (2g sin θ)z. At
z = 8 + 20 = 21.54 ft u = 28.8 ft/s. (2) In the second part of the
2 2
dvy
path the skier is in free fall. The equations to be integrated are =
dt
dy
−g, = vy , with v(0) = u sin θ = 28.8(0.3714) = 10.7 ft/s, y(0) =
dt
dvx dx
8 ft. = 0, = vx , with vx (0) = u cos θ = 26.7 ft/s, x(0) =
dt dt
g
0. Integrating: vy (t) = −gt + 10.7 ft/s. y(t) = − t 2 + 10.7t + 8 ft ·
2
vx (t) = 26.7 ft/s, x(t) = 26.7t. When y(timpact ) = 0, the skier has hit
the water. The impact time is timpact + 2btimpact + c = 0 where b =
2

10.7 16 √
− , c = − . The solution timpact = −b ± b2 − c = 1.11 s, =
g g
−0.45 s. The negative values has no meaning here. The horizontal
distance is

x(timpact ) = 26.7timpact = 29.7 ft

Problem 14.133 Suppose you are designing a roller-

coaster track that will take the cars through a vertical
loop of 40-ft radius. If you decide that, for safety, the
downward force exerted on a passenger by his seat at the
top of the loop should be at least one-half the passenger’s
weight, what is the minimum safe velocity of the cars at
the top of the loop?

40 ft

Solution: Denote the normal force exerted on the passenger by the

seat by N . From Newton’s second law, at the top of the loop −N −
2
v N v2 g
mg = −m , from which − = g − = − . From which:
R m R 2

3Rg
v= = 43.93. = 44 ft/s
2

m
A

m
0
40
m
m
0
10
Solution: The angular velocity of the spool relative to the bar is
α = 6 rad/s2 . The acceleration of the collar relative to the bar is
d2r NH
= −Rα = −0.05(6) = −0.3 m/s2 . The take up velocity of the
dt 2
spool is
A

vs = Rα dt = −0.05(6)t = −0.3t m/s. T

The velocity of the collar relative to the bar is

dr
= −0.3t m/s.
dt
The velocity of the collar relative to the bar is dr/dt = −0.3t m/s.
The position of the collar relative to the bar is r = −0.15t 2 + 0.4 m.
d2θ
The angular acceleration of the collar is = 6 rad/s2 . The angu-
dt 2
dθ
lar velocity of the collar is = 6t rad/s. The radial acceleration is
2 dt
2
d r dθ
ar = 2 − r = −0.3 − (−0.15t 2 + 0.4)(6t)2 . At t = 1 s the
dt dt
radial acceleration is ar = −9.3 m/s2 , and the tension in the string is

|T | = |mar | = 9.3 N

Problem 14.135 In Problem 14.134, suppose that the

coefficient of kinetic friction between the collar and the
bar is µk = 0.2. What is the tension in the string at
t = 1 s?
Solution: Use the results of the solution to Problem 14.134 At
t = 1 s, the horizontal normal force is

d2θ dr dθ
NH = |maθ | = m r 2
+2 = 2.1 N,
dt dt dt

from which the total normal force is N = NH2 + (mg)2 From New-

ton’s second law: −T + µk NH2 + (mg 2 ) er + NH eθ = mar er +

maθ eθ , from which −T + µk NH2 + (mg)2 = mar . From the solution

to Problem 14.152, ar = −9.3 m/s2 . Solve: The tension is

T = 11.306 N

Solution: For comfort, the passenger should feel the total effects
of acceleration down toward his feet, that is, apparent side (radial)
accelerations should not be felt. This condition is achieved when the
tilt θ is such that mg sin θ − m(v 2 /ρ) cos θ = 0, from which

v2
tan θ = .
ρg

Problem 14.137 To determine the coefficient of static

friction between two materials, an engineer at the U.S.
National Institute of Standards and Technology places a
small sample of one material on a horizontal disk whose 200 mm
surface is made of the other material and then rotates
the disk from rest with a constant angular acceleration
of 0.4 rad/s2 . If she determines that the small sample
slips on the disk after 9.903 s, what is the coefficient of
friction?

Solution: The angular velocity after t = 9.903 s is ω = 0.4t =

3.9612 rad/s. The radial acceleration is an = 0.2ω2 = 3.138 m/s2 . The
tangential acceleration is at = (0.2)0.4 = 0.08 m/s2 . At the instant
before slip occurs, Newton’s second law for the small sample is F =
µs N = µs mg = m an2 + at2 , from which

an2 + at2
µs = = 0.320
g

t
Solution: The radial position is r = 2 + 1 . The radial veloc-
π
dr 2
ity: = .
dt π
dθ
The radial acceleration is zero. The angular velocity: = 2. The
dt
angular acceleration is zero. At θ = 120 = 2.09 rad. From Newton’s
0

second law, the radial force is Fr = mar , from which

2
dθ
Fr = − r er = −10.67er N
dt

The transverse force is Fθ = maθ , from which

dr dθ
Fθ = 2 eθ = 2.55eθ N
dt dt

Problem 14.139* In Problem 14.138, suppose that the

curved bar lies in the vertical plane. Determine the radial
and transverse components of the total force exerted on
A by the curved and slotted bars at t = 0.5 s.

Solution: Assume that the curved bar is vertical such that

the line θ = 0 is horizontal. The weight has the components:
W = (W sin θ)er + (W cos θ)eθ . From Newton’s second law: Fr −
W sin θ = mar , and Fθ − W cos θ = maθ ., from which Fr −
g sin 2ter = −r(dθ/ dt)2 er , from which

t
Fr = −2 + 1 (22 ) + g sin 2t ,
π

at t = 0.5 s, Fr = −1.02 N . The transverse component Fθ =

2 8
2 (2) + g cos 2t = + g cos 2t . At t = 0.51 s,
π π

Fθ = 7.85 N

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