Solution manual engineering mechanics dynamics 12th edition r 1

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R1–1. The ball is thrown horizontally with a speed of
8 m>s. Find the equation of the path, y = f(x), and then
determine the ball’s velocity and the normal and tangential
components of acceleration when t = 0.25 s.

y

vA ϭ 8 m/s
A

Horizontal Motion: The horizontal component of velocity is yx = 8 m>s and the
initial horizontal position is (s0)x = 0.
+ B
A:

sx = (s0)x + (y 0)x t
x = 0 + 8t

[1]

Vertical Motion: The vertical component of initial velocity (y 0)y = 0 and the initial
vertical position are (s 0)y = 0.
(+ c )

sy = (s0)y + (y0)y t +

y = 0 + 0 +

1
(a ) t2
2 cy

1
( -9.81) t2
2

[2]

Eliminate t from Eqs. [1] and [2] yields
y = -0.0766x2

Ans.

The vertical component of velocity when t = 0.25 s is given by
yy = (y 0)y + (ac)y t

(+ c )

yy = 0 + ( -9.81)(0.25) = -2.4525 m>s = 2.4525 m>s T
The magnitude and direction angle when t = 0.25 s are

y = 2y2x + y2y = 282 + 2.45252 = 8.37 m>s
u = tan - 1

yy
yx

= tan - 1

Ans.

2.4525
= 17.04° = 17.0° c
8

Ans.

Since the velocity is always directed along the tangent of the path and the
acceleration a = 9.81 m>s2 is directed downward, then tangential and normal
components of acceleration are
at = 9.81 sin 17.04° = 2.88 m>s2

Ans.

an = 9.81 cos 17.04° = 9.38 m>s2

Ans.

479

x

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R1–2. Cartons having a mass of 5 kg are required to move
along the assembly line with a constant speed of 8 m>s.
Determine the smallest radius of curvature, r, for the
conveyor so the cartons do not slip. The coefficients of static
and kinetic friction between a carton and the conveyor are
ms = 0.7 and mk = 0.5, respectively.

+ c ©Fb = m ab ;

8 m/s

r

N - W = 0
N = W
Fs = 0.7W

+ ©F = m a ;
:
n
n

0.7W =

W 82
a b
9.81 r

r = 9.32 m

Ans.

R1–3. A small metal particle travels downward through a
fluid medium while being subjected to the attraction of a
magnetic field such that its position is s = (15t3 - 3t) mm,
where t is in seconds. Determine (a) the particle’s
displacement from t = 2 s to t = 4 s, and (b) the velocity
and acceleration of the particle when t = 5 s.

a)

s = 15t3 - 3t
At t = 2 s, s1 = 114 mm
At t = 4 s, s3 = 948 mm
¢s = 948 - 114 = 834 mm

b)

Ans.

v =

ds
= 45t2 - 3 2
= 1122 mm>s = 1.12 m>s
dt
t=5

Ans.

a =

dv
= 90t 2
= 450 mm>s2 = 0.450 m>s2
dt
t=5

Ans.

480


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*R1–4. The flight path of a jet aircraft as it takes off is
defined by the parametric equations x = 1.25t2 and
y = 0.03t3, where t is the time after take-off, measured in
seconds, and x and y are given in meters. If the plane starts
to level off at t = 40 s, determine at this instant (a) the
horizontal distance it is from the airport, (b) its altitude,
(c) its speed, and (d) the magnitude of its acceleration.

y

x

Position: When t = 40 s, its horizontal position is given by
x = 1.25 A 402 B = 2000 m = 2.00 km

Ans.

and its altitude is given by
y = 0.03 A 403 B = 1920 m = 1.92 km

Ans.

Velocity: When t = 40 s, the horizontal component of velocity is given by

#
yx = x = 2.50tΗt = 40 s = 100 m>s
The vertical component of velocity is
#
yy = y = 0.09t2Η t = 40 s = 144 m>s
Thus, the plane’s speed at t = 40 s is
yy = 2y2x + y2y = 21002 + 144 2 = 175 m>s

Ans.

Acceleration: The horizontal component of acceleration is
$
ax = x = 2.50 m>s2
and the vertical component of acceleration is
$
ay = y = 0.18tΗt = 40 s = 7.20 m>s2
Thus, the magnitude of the plane’s acceleration at t = 40 s is
a = 2a2x + a2y = 22.502 + 7.202 = 7.62 m>s2

Ans.

481


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R1–5. The boy jumps off the flat car at A with a velocity of
v¿ = 4 ft>s relative to the car as shown. If he lands on the
second flat car B, determine the final speed of both cars
after the motion. Each car has a weight of 80 lb. The boy’s
weight is 60 lb. Both cars are originally at rest. Neglect the
mass of the car’s wheels.

v ¿ ϭ 4 ft/s
5

B

Relative Velocity: The horizontal component of the relative velocity of the boy with
12
respect to the car A is (y b>A)x = 4a b = 3.692 ft>s. Thus, the horizontal
13
component of the velocity of the boy is
(y b)x = yA + (y b>A)x
+ B
A;

[1]

(y b)x = - yA + 3.692

Conservation of Linear Momentum: If we consider the boy and the car as a system,
then the impulsive force caused by traction of the shoes is internal to the system.
Therefore, they will cancel out. As the result, the linear momentum is conserved
along x axis. For car A
0 = m b (y b)x + m A yA
+ B
A;

0 = a

60
80
b(y b)x + a
b(-y A)
32.2
32.2

[2]

Solving Eqs. [1] and [2] yields
Ans.

yA = 1.58 ft>s
(y b)x = 2.110 ft>s
For car B
m b (y b)x = (m b + m B) yB
+ B
A;

a

60
60 + 80
b (2.110) = a
b yB
32.2
32.2
Ans.

yB = 0.904 ft>s

482

13
12

A


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R1-6. The man A has a weight of 175 lb and jumps from
rest at a height h = 8 ft onto a platform P that has a weight
of 60 lb. The platform is mounted on a spring, which has a
stiffness k = 200 lb>ft. Determine (a) the velocities of A
and P just after impact and (b) the maximum compression
imparted to the spring by the impact. Assume the coefficient
of restitution between the man and the platform is e = 0.6,
and the man holds himself rigid during the motion.

A
h

Conservation of Energy: The datum is set at the initial position of platform P. When
the man falls from a height of 8 ft above the datum, his initial gravitational potential
energy is 175(8) = 1400 ft # lb. Applying Eq. 14–21, we have
T1 + V1 = T2 + V2
0 + 1400 =

1 175
a
b(y M) 21 + 0
2 32.2

(y H)1 = 22.70 ft>s
Conservation of Momentum:
mM (y M)1 + mP (y P)1 = mM(y M)2 + mp (y p)2
(+ T)

a

175
175
60
b(22.70) + 0 = a
b (y M) 2 + a
b(y p) 2
32.2
32.2
32.2

[1]

Coefficient of Restitution:

( + T)

e =

(yP)2 - (yM)2
(yM)1 - (yp)1

0.6 =

(yP)2 - (yP)2
22.70 - 0

[2]

Solving Eqs. [1] and [2] yields
(y p)2 = 27.04 ft>s T = 27.0 ft>s T

(y M)2 = 13.4 ft>s T

Ans.

Conservation of Energy: The datum is set at the spring’s compressed position.
60
Initially, the spring has been compressed
= 0.3 ft and the elastic potential
200
1
energy is (200) A 0.32 B = 9.00 ft # lb. When platform P is at a height of s above the
2
datum, its initial gravitational potential energy is 60s. When platform P stops
momentary, the spring has been compressed to its maximum and the elastic
1
potential energy at this instant is (200)(s + 0.3)2 = 100s 2 + 60s + 9. Applying
2
Eq. 14–21, we have
T1 + V1 = T2 + V2
1 60
a
b A 27.04 2 B + 60s + 9.00 = 100s2 + 60s + 9
2 32.2
s = 2.61 ft

Ans.

483

P


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R1–7. The man A has a weight of 100 lb and jumps from
rest onto the platform P that has a weight of 60 lb. The
platform is mounted on a spring, which has a stiffness
k = 200 lb>ft. If the coefficient of restitution between the
man and the platform is e = 0.6, and the man holds himself
rigid during the motion, determine the required height h of
the jump if the maximum compression of the spring is 2 ft.

A
h

Conservation of Energy: The datum is set at the initial position of platform P. When
the man falls from a height of h above the datum, his initial gravitational potential
energy is 100h. Applying Eq. 14–21, we have
T1 + V1 = T2 + V2
0 + 100h =

1 100
a
b(yM)21 + 0
2 32.2

(yH)1 = 264.4h
Conservation of Momentum:
mM (yM)1 + mP (yP)1 = mM (yM)2 + mP (yP)2
(+ T)

a

100
100
60
b (264.4h) + 0 = a
b(yM)2 + a
b(yP)2
32.2
32.2
32.2

[1]

Coefficient of Restitution:
e =

(+ T)

0.6 =

(yp)2 - (yM)2
(yM)1 - (yp)1
(yp)2 - (yM)2

[2]

264.4h - 0

Solving Eqs. [1] and [2] yields
(yp)2 = 264.4h T

(yM)2 = 0.4264.4h T

Conservation of Energy: The datum is set at the spring’s compressed position.
60
Initially, the spring has been compressed
= 0.3 ft and the elastic potential
200
1
energy is (200) A 0.32 B = 9.00 ft # lb. Here, the compression of the spring caused by
2
impact is (2 - 0.3) ft = 1.7 ft. When platform P is at a height of 1.7 ft above the
datum, its initial gravitational potential energy is 60(1.7) = 102 ft # lb. When
platform P stops momentary, the spring has been compressed to its maximum and
1
the elastic potential energy at this instant is (200) A 2 2 B = 400 ft # lb. Applying
2
Eq. 14–21, we have

T1 + V1 = T2 + V2
1 60
a
b A 264.4h B 2 + 102 + 9.00 = 400
2 32.2
h = 4.82 ft

Ans.

484

P


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*R1–8. The baggage truck A has a mass of 800 kg and is
used to pull each of the 300-kg cars. Determine the tension
in the couplings at B and C if the tractive force F on the
truck is F = 480 N. What is the speed of the truck when
t = 2 s, starting from rest? The car wheels are free to roll.

Neglect the mass of the wheels.
+ ©F = ma ;
:
x
x

A
C

B
F

480 = [800 + 2(300)]a
a = 0.3429 m>s2

+ )
(:

v = v0 + ac t
Ans.

v = 0 + 0.3429(2) = 0.686 m>s
+ ©F = ma ;
:
x
x

TB = 2(300)(0.3429)
TB = 205.71 = 206 N

+ ©F = ma ;
:
x
x

Ans.

TC = (300)(0.3429)
TC = 102.86 = 103 N

Ans.

R1–9. The baggage truck A has a mass of 800 kg and is
used to pull each of the 300-kg cars. If the tractive force F
on the truck is F = 480 N, determine the acceleration of
the truck. What is the acceleration of the truck if the
coupling at C suddenly fails? The car wheels are free to roll.
Neglect the mass of the wheels.
+ ©F = ma ;
:
x
x

A
C

F

480 = [800 + 2(300)]a
a = 0.3429 = 0.343 m>s2

+ ©F = ma ;
:
x
x

B

Ans.

480 = (800 + 300)a
a = 0.436 m>s2

Ans.

485


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R1–10. A car travels at 80 ft>s when the brakes are

suddenly applied, causing a constant deceleration of
10 ft>s2. Determine the time required to stop the car and
the distance traveled before stopping.

+ b
a:

v = v0 + ac t
0 = 80 + (-10)t
t = 8s

+ b
a:

Ans.

v 2 = v20 + 2ac (s - s0)
0 = (80)2 + 2(-10)(s - 0)
s = 320 ft

Ans.

R1–11. Determine the speed of block B if the end of the
cable at C is pulled downward with a speed of 10 ft>s. What
is the relative velocity of the block with respect to C?
C
10 ft/s

3sB + sC = l
B

3vB = -vC
3vB = -(10)
Ans.

vB = -3.33 ft>s = 3.33 ft>s c

A+TB

vB = vC + vB>C
-3.33 = 10 + vB>C
vB>C = -13.3 ft>s = 13.3 ft>s c

Ans.

486


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*R1–12. The skier starts fom rest at A and travels down

the ramp. If friction and air resistance can be neglected,
determine his speed vB when he reaches B. Also, compute
the distance s to where he strikes the ground at C, if he
makes the jump traveling horizontally at B. Neglect the
skier’s size. He has a mass of 70 kg.

A

50 m

vB
B

4m

s

Potential Energy: The datum is set at the lowest point B. When the skier is at point
A, he is (50 - 4) = 46 m above the datum. His gravitational potential energy at this
position is 70(9.81) (46) = 31588.2 J.
Conservation of Energy: Applying Eq. 14–21, we have
TA + VA = TB + VB
0 + 31588.2 =

1
(70) y2B
2

yB = 30.04 m>s = 30.0 m>s

Ans.

Kinematics: By considering the vertical motion of the skier, we have
(+ T)

sy = (s0)y + (y0)y t +

1
(a ) t2
2 cy

4 + s sin 30° = 0 + 0 +

1
(9.81) t2
2

[1]

By considering the horizontal motion of the skier, we have
+ B
A;

sx = (s0)x + yx t
s cos 30° = 0 + 30.04 t

[2]

Solving Eqs. [1] and [2] yields
s = 130 m

Ans.

t = 3.753 s

487

C

30Њ


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R1–13. The position of a particle is defined by
r = 551cos 2t2i + 41sin 2t2j6 m, where t is in seconds and
the arguments for the sine and cosine are given in radians.
Determine the magnitudes of the velocity and acceleration
of the particle when t = 1 s. Also, prove that the path of the
particle is elliptical.

Velocity: The velocity expressed in Cartesian vector form can be obtained by
applying Eq. 12–7.
v =

dr
= {-10 sin 2ri + 8 cos 2rj} m>s
dt

When t = 1 s, v = -10 sin 2(1)i + 8 cos 2(1) j = (-9.093i - 3.329j} m>s. Thus, the
magnitude of the velocity is
y = 2y2x + y2y = 2( -9.093)2 + (-3.329)2 = 9.68 m>s

Ans.

Acceleration: The acceleration express in Cartesian vector form can be obtained by
applying Eq. 12–9.
a =

dv
= {-20 cos 2ri - 16 sin 2rj} m>s2
dt

When t = 1 s, a = -20 cos 2(1) i - 16 sin 2(1) j = {8.323i - 14.549j} m>s2. Thus,
the magnitude of the acceleration is
a = 2a2x + a2y = 28.3232 + ( -14.549)2 = 16.8 m>s2

Ans.

Travelling Path: Here, x = 5 cos 2t and y = 4 sin 2t. Then,
x2

= cos2 2t
25

[1]

y2
= sin2 2t
16

[2]

Adding Eqs [1] and [2] yields
y2
x2
+
= cos2 2r + sin2 2t
25
16
However, cos2 2 r + sin2 2t = 1. Thus,
y2
x2
+
= 1 (Equation of an Ellipse)
25
16

(Q.E.D.)

488

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R1–14. The 5-lb cylinder falls past A with a speed
vA = 10 ft>s onto the platform. Determine the maximum
displacement of the platform, caused by the collision. The
spring has an unstretched length of 1.75 ft and is originally
kept in compression by the 1-ft-long cables attached to the
platform. Neglect the mass of the platform and spring and
any energy lost during the collision.

vA ϭ 10 ft/s

A

3 ft

Potential Energy: Datum is set at the final position of the platform. When the
cylinder is at point A, its position is (3 + s) above the datum where s is the maximum
displacement of the platform when the cylinder stops momentary. Thus, its
gravitational potential energy at this position is 5(3 + s) = (15 + 5s) ft # lb. The

1
initial and final elastic potential energy are (400) (1.75 - 1)2 = 112.5 ft # lb and
2
1
(400) (s + 0.75)2 = 200s2 + 300s + 112.5, respectively.
2

k ϭ 400 lb/ft

1 ft

Conservation of Energy: Applying Eq. 14–22, we have
©TA + ©VA = ©TB + ©VB
1
5
a
b A 102 B + (15 + 5s) + 112.5 = 0 + 200s2 + 300s + 112.5
2 32.2
s = 0.0735 ft

Ans.

R1–15. The block has a mass of 50 kg and rests on the
surface of the cart having a mass of 75 kg. If the spring
which is attached to the cart and not the block is
compressed 0.2 m and the system is released from rest,
determine the speed of the block after the spring becomes
undeformed. Neglect the mass of the cart’s wheels and the
spring in the calculation. Also neglect friction. Take
k = 300 N>m.

k
B
C

T1 + V1 = T2 + V2
[0 + 0] +

1
1
1
(300)(0.2)2 = (50) v2b + (75) v2e
2
2
2

12 = 50 v2b + 75 v2e
+ )
(:

©mv1 = ©mv2
0 + 0 = 50 vb - 75 ve
vb = 1.5ve
vc = 0.253 m>s ;
vb = 0.379 m>s :

Ans.

489

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*R1–16. The block has a mass of 50 kg and rests on the
surface of the cart having a mass of 75 kg. If the spring
which is attached to the cart and not the block is
compressed 0.2 m and the system is released from rest,
determine the speed of the block with respect to the cart
after the spring becomes undeformed. Neglect the mass of
the cart’s wheels and the spring in the calculation. Also
neglect friction. Take k = 300 N>m.

k
B
C

T1 + V1 = T2 + V2
[0 + 0] +

1
1

1
(300)(0.2)2 = (50)v2b + (75) v2e
2
2
2

12 = 50 v2b + 75 v2e
+ )
(:

©mv1 = ©mv2
0 + 0 = 50 vb - 75 ve
vb = 1.5 ve
vc = 0.253 m>s ;
vb = 0.379 m>s :
vb = vc + vb>c

+ )
(:

0.379 = -0.253 + vb>c
vb>c = 0.632 m>s :

Ans.

490


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R1–17. A ball is launched from point A at an angle of 30°.
Determine the maximum and minimum speed vA it can
have so that it lands in the container.

vA
A

30Њ
1m

B

2.5 m
4m

Min. speed:
+ b
a:

s = s0 + v0 t
2.5 = 0 + vA cos 30°t

A+cB

s = s0 + v0 t +

1
a t2
2 c

0.25 = 1 + vA sin 30°t -

1
(9.81)t2
2

Solving
t = 0.669 s
vA = A vA B min = 4.32 m>s

Ans.

Max. speed:
+ b
a:

s = s0 + v0 t
4 = 0 + vA cos 30°t

A+cB

s = s0 + v0 t +

1
a t2
2 c

0.25 = 1 + vA sin 30° t -

1
(9.81) t2
2

Solving:
t = 0.790 s
vA = (vA)max = 5.85 m>s

Ans.

491

C
0.25 m


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R1–18. At the instant shown, cars A and B travel at speeds
of 55 mi>h and 40 mi>h, respectively. If B is increasing its
speed by 1200 mi>h2, while A maintains its constant speed,
determine the velocity and acceleration of B with respect to
A. Car B moves along a curve having a radius of curvature
of 0.5 mi.

vB ϭ 40 mi/h

B

A
vA ϭ 55 mi/h

vB = -40 cos 30°i + 40 sin 30°j = {-34.64i + 20j} mi>h
vA = {-55i} mi>h
vB>A = vB - vA
= (-34.64i + 20j) - (-55i) = {20.36i + 20j} mi>h
yB>A = 220.362 + 202 = 28.5 mi>h

Ans.

u = tan-1 a

Ans.

(aB)n =

20
b = 44.5° a
20.36

y2B
402
=
= 3200 mi>h2
r
0.5

(aB)t = 1200 mi>h2

aB = (3200 sin 30° - 1200 cos 30°)i + (3200 cos 30° + 1200 sin 30°)j
= {560.77i + 3371.28j} mi>h2
aA = 0
aB = aA + aB>A
560.77i + 3371.28j = 0 + aB>A
aB>A = {560.77i + 3371.28j} mi>h2
aB>A = 2(560.77)2 + (3371.28)2 = 3418 mi>h2 = 3.42 A 103 B mi>h2
u = tan-1 a

3371.28
b = 80.6° a
560.77

Ans.

Ans.

492

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R1–19. At the instant shown, cars A and B travel at speeds
of 55 mi>h and 40 mi>h, respectively. If B is decreasing its
speed at 1500 mi>h2 while A is increasing its speed at
800 mi>h2, determine the acceleration of B with respect to
A. Car B moves along a curve having a radius of curvature
of 0.75 mi.

vB ϭ 40 mi/h

B

A

vA ϭ 55 mi/h

(aB)n =

(40)2
v2B
=
= 2133.33 mi>h2
r
0.75

aB = aA + aB>A
2133.33 sin 30°i + 2133.33 cos 30°j + 1500 cos 30°i - 1500 sin 30°j
= -800i + (aB>A)x i + (aB>A)y j
+ )
(:

2133.33 sin 30° + 1500 cos 30° = -800 + (aB>A)x
(aB>A)x = 3165.705 :

(+ c )

2133.33 cos 30° - 1500 sin 30° = (aB>A)y
(aB>A)y = 1097.521 c
(aB>A) = 2(1097.521)2 + (3165.705)2
aB>A = 3351 mi>h2 = 3.35 A 103 B mi>h2

Ans.

u = tan-1 a

Ans.

1097.521
b = 19.1° a
3165.705

493

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*R1–20. Four inelastic cables C are attached to a plate P
and hold the 1-ft-long spring 0.25 ft in compression when
no weight is on the plate. There is also an undeformed
spring nested within this compressed spring. If the block,
having a weight of 10 lb, is moving downward at v = 4 ft>s,
when it is 2 ft above the plate, determine the maximum
compression in each spring after it strikes the plate.

Neglect the mass of the plate and springs and any energy
lost in the collision.

v

2 ft
k ϭ 30 lb/in.
k¿ ϭ 50 lb/in.
P

k = 30(12) = 360 lb>ft
k¿ = 50(12) = 600 lb>ft
0.75 ft

Assume both springs compress;
T1 + V1 = T2 + V2
1
1
1
1 10
(
)(4)2 + 0 + (360)(0.25)2 = 0 + (360)(s + 0.25)2 + (600)(s - 0.25)2 - 10(s + 2)
2 32.2
2
2
2
13.73 = 180(s + 0.25)2 + 300(s - 0.25)2 - 10s - 20

(1)

33.73 = 180(s + 0.25)2 + 300(s - 0.25)2 - 10s
480s2 - 70s - 3.73 = 0
Choose the positive root;
s = 0.1874 ft 6 0.25 ft

NG!

The nested spring does not deform.
Thus Eq. (1) becomes
13.73 = 180(s + 0.25)2 - 10s - 20
180 s2 + 80 s - 22.48 = 0
s = 0.195 ft

Ans.

494

C
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R1–21. Four inelastic cables C are attached to plate P and
hold the 1-ft-long spring 0.25 ft in compression when no
weight is on the plate. There is also a 0.5-ft-long undeformed
spring nested within this compressed spring. Determine the
speed v of the 10-lb block when it is 2 ft above the plate, so
that after it strikes the plate, it compresses the nested
spring, having a stiffness of 50 lb>in., an amount of 0.20 ft.
Neglect the mass of the plate and springs and any energy
lost in the collision.

v

2 ft
k ϭ 30 lb/in.
k¿ ϭ 50 lb/in.
P

k = 30(12) = 360 lb>ft
k¿ = 50(12) = 600 lb>ft

C

0.75 ft

0.5 ft

T1 + V1 = T2 + V2
1 10

1
1
1
a
bv2 + (360)(0.25)2 = (360)(0.25 + 0.25 + 0.20)2 + (600)(0.20)2 - 10(2 + 0.25 + 0.20)
2 32.2
2
2
2
v = 20.4 ft>s

Ans.

z

R1–22. The 2-kg spool S fits loosely on the rotating
inclined rod for which the coefficient of static friction is
ms = 0.2. If the spool is located 0.25 m from A, determine
the minimum constant speed the spool can have so that it
does not slip down the rod.

5

S
0.25 m

A

4
r = 0.25 a b = 0.2 m

5
+ ©F = ma ;
;
n
n

+ c ©Fb = m ab ;

3
4

3
4
v2
b
Ns a b - 0.2Ns a b = 2a
5
5
0.2
4
3
Ns a b + 0.2Ns a b - 2(9.81) = 0
5
5
Ns = 21.3 N
v = 0.969 m>s

Ans.

495

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z

R1–23. The 2-kg spool S fits loosely on the rotating inclined
rod for which the coefficient of static friction is ms = 0.2. If
the spool is located 0.25 m from A, determine the maximum
constant speed the spool can have so that it does not slip up
the rod.

5

S
0.25 m

A

4
r = 0.25 a b = 0.2 m

5
+ ©F = ma ;
;
n
n

+ c ©Fb = m ab ;

v2
3
4
b
Ns a b + 0.2Ns a b = 2a
5
5
0.2
4
3
Ns a b - 0.2Ns a b - 2(9.81) = 0
5
5
Ns = 28.85 N
v = 1.48 m>s

Ans.

*R1–24. The winding drum D draws in the cable at an
accelerated rate of 5 m>s2. Determine the cable tension if
the suspended crate has a mass of 800 kg.

D

sA + 2 sB = l
aA = - 2 aB
5 = - 2 aB
aB = -2.5 m>s2 = 2.5 m>s2 c
+ c ©Fy = may ;

3
4

2T - 800(9.81) = 800(2.5)
T = 4924 N = 4.92 kN

Ans.

496


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R1–25. The bottle rests at a distance of 3 ft from the center
of the horizontal platform. If the coefficient of static friction
between the bottle and the platform is ms = 0.3, determine
the maximum speed that the bottle can attain before
slipping. Assume the angular motion of the platform is
slowly increasing.

©Fb = 0;

N - W = 0

3 ft

N = W

Since the bottle is on the verge of slipping, then Ff = ms N = 0.3W.
©Fn = man ;

0.3W = a

W
y2
ba b
32.2
3

y = 5.38 ft>s

Ans.

R1–26. Work Prob. R1–25 assuming that the platform
starts rotating from rest so that the speed of the bottle is
increased at 2 ft>s2.

3 ft

Applying Eq. 13–8, we have
©Fb = 0;

N - W = 0

N = W

Since the bottle is on the verge of slipping, then Ff = ms N = 0.3W.
©Ft = mat ;

©Fn = man ;

0.3W sin u = a
0.3W cos u = a

W
b(2)
32.2

[1]

W
y2
ba b

32.2
3

[2]

Solving Eqs. [1] and [2] yields
y = 5.32 ft>s

Ans.

u = 11.95°

497


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R1–27. The 150-lb man lies against the cushion for which
the coefficient of static friction is ms = 0.5. Determine the
resultant normal and frictional forces the cushion exerts on
him if, due to rotation about the z axis, he has a constant

speed v = 20 ft>s. Neglect the size of the man. Take
u = 60°.

z

8 ft
G

u

+a

©Fy = m(an)y ;

N - 150 cos 60° =

150 202
a
b sin 60°
32.2 8

N = 277 lb
+b©Fx = m(an)x ;

-F + 150 sin 60° =

Ans.
150 202
a
b cos 60°

32.2 8

F = 13.4 lb

Ans.

Note: No slipping occurs
Since ms N = 138.4 lb 7 13.4 lb

*R1–28. The 150-lb man lies against the cushion for which
the coefficient of static friction is ms = 0.5. If he rotates
about the z axis with a constant speed v = 30 ft>s, determine
the smallest angle u of the cushion at which he will begin to
slip up the cushion.

z

8 ft
G

u

+ ©F = ma ;
;
n
n
+ c ©Fb = 0;

0.5N cos u + N sin u =

150 (30)2
a
b
32.2
8

-150 + N cos u - 0.5 N sin u = 0
N =

150
cos u - 0.5 sin u

(0.5 cos u + sin u)150
150 (30)2
=
a
b
(cos u - 0.5 sin u)
32.2
8
0.5 cos u + sin u = 3.493 79 cos u - 1.746 89 sin u
u = 47.5°

Ans.

498


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R1–29. The motor pulls on the cable at A with a force
F = (30 + t2) lb, where t is in seconds. If the 34-lb crate is
originally at rest on the ground when t = 0, determine its
speed when t = 4 s. Neglect the mass of the cable and
pulleys. Hint: First find the time needed to begin lifting
the crate.

A

30 + t2 = 34
t = 2 s for crate to start moving
(+ c )

mv1 + ©

L

Fdt = mv2

4

0 +

(30 + t2)dt - 34(4 - 2) =

L2

[30 t +

34
v
32.2 2

1 34
34
t ] - 68 =
v
3 2
32.2 2

v2 = 10.1 ft>s

Ans.

R1–30. The motor pulls on the cable at A with a force
F = (e2t) lb, where t is in seconds. If the 34-lb crate is
originally at rest on the ground when t = 0, determine the
crate’s velocity when t = 2 s. Neglect the mass of the cable
and pulleys. Hint: First find the time needed to begin lifting
the crate.

A

F = e 2t = 34
t = 1.7632 s for crate to start moving
(+ c )

mv1 + ©

L

F dt = mv2

2

0 +

L1.7632

e 2 t dt - 34(2 - 1.7632) =

34
v
32.2 2

1 2t 2 2
e
- 8.0519 = 1.0559 v2
2
1.7632
v2 = 2.13 ft>s

Ans.

499


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R1–31. The collar has a mass of 2 kg and travels along the
smooth horizontal rod defined by the equiangular spiral
r = (eu) m, where u is in radians. Determine the tangential
force F and the normal force N acting on the collar
when u = 45°, if force F maintains a constant angular motion
#
u = 2 rad>s.

F

r

u

r = eu
#
#
r = euu
#
#
$
r = e u(u)2 + e u u
At u = 45°
#
u = 2 rad>s
u = 0
r = 2.1933 m
r = 4.38656 m>s
$
r = 8.7731 m>s2
#
$
ar = r - r (u)2 = 8.7731 - 2.1933(2)2 = 0
$
# #
au = r u + 2 r u = 0 + 2(4.38656)(2) = 17.5462 m>s2
tan c =

r

A B
dr
du

= e u>e u = 1

c = u = 45°
©Fr = m a r ;

-Nr cos 45° + F cos 45° = 2(0)

©Fu = m a u ;

F sin 45° + Nu sin 45° = 2(17.5462)

N = 24.8 N

Ans.

F = 24.8 N

Ans.

500

r ϭ eu


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*R1–32. The collar has a mass of 2 kg and travels along the
smooth horizontal rod defined by the equiangular spiral
r = (eu) m, where u is in radians. Determine the tangential
force F and the normal force N acting on the collar when
u# = 90°, if force F maintains a constant angular motion
u = 2 rad>s.

F

r

u

r = eu
#
#
r = eu u
#
$
$
r = e u (u)2 + e u u
At u = 90°
#
u = 2 rad>s

#
u = 0
r = 4.8105 m
#
r = 9.6210 m>s
$
r = 19.242 m>s2
#
$
ar = r - r(u)2 = 19.242 - 4.8105(2)2 = 0
$
# #
au = r u + 2 r u = 0 + 2(9.6210)(2) = 38.4838 m>s2
tan c =

r

A B
dr
du

= e u>e u = 1

c = u = 45°
+ c ©Ft = m a t ;

-N cos 45° + F cos 45° = 2(0)

+ ©F = m a ;
;

u
u

F sin 45° + N sin 45° = 2(38.4838)
Nt = 54.4 N

Ans.

F = 54.4 N

Ans.

501

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R1–33. The acceleration of a particle along a straight line is
defined by a = (2t - 9) m>s2, where t is in seconds. When

t = 0, s = 1 m and v = 10 m>s. When t = 9 s, determine
(a) the particle’s position, (b) the total distance traveled, and
(c) the velocity. Assume the positive direction is to the right.

a = (2t - 9)
dv = a dt
t

v

dv =

L10

L0

(2t - 9) dt

v - 10 = t2 - 9t
v = t2 - 9t + 10
ds = v dt
s

L1

t

ds =

s - 1 =

s =

L0

A t2 - 9t + 10 B dt

1 3
t - 4.5t2 + 10t
3

1 3
t - 4.5t2 + 10t + 1
3

Note v = 0 at t2 - 9t + 10 = 0
t = 1.298 s and t = 7.702 s
At t = 1.298 s, s = 7.127 m
At t = 7.702 s, s = -36.627 m
At t = 9 s, s = -30.50 m
a) s = -30.5 m

Ans.

b) stot = (7.127 - 1) + 7.127 + 36.627 + (36.627 - 30.50) = 56.0 m

Ans.

c) v|t = 9 = (9)2 - 9(9) + 10 = 10 m>s

Ans.

502


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R1–34. The 400-kg mine car is hoisted up the incline using
the cable and motor M. For a short time, the force in the
cable is F = (3200t2) N, where t is in seconds. If the car has
an initial velocity v1 = 2 m>s when t = 0, determine its
velocity when t = 2 s.

M

v 1 ϭ 2 m/s

17

8

15

3200t2 - 400(9.81)a

+Q©Fx¿ = max¿ ;

8
b = 400a
17

a = 8t2 - 4.616

dv = adt
2

y

L2

dv =

L0

A 8t2 - 4.616 B dt
Ans.

v = 14.1 m>s
Also,
mv1 + ©

L

F dt = mv2
2

400(2) +

+Q

L0

3200 t2 dt - 400(9.81)(2 - 0) a

8
b = 400v2
17

800 + 8533.33 - 3693.18 = 400v2
v2 = 14.1 m>s

R1–35. The 400-kg mine car is hoisted up the incline using
the cable and motor M. For a short time, the force in the
cable is F = (3200t2) N, where t is in seconds. If the car has
an initial velocity v1 = 2 m>s at s = 0 and t = 0, determine
the distance it moves up the plane when t = 2 s.

M

v 1 ϭ 2 m/s

17

8

15

©Fx¿ = max¿ ;

3200t2 - 400(9.81)a

8
b = 400a
17

a = 8t2 - 4.616

dv = adt
t

y

L2
v =

dv =

L0

ds
= 2.667t3 - 4.616t + 2

dt
s

L2

A 8t2 - 4.616 B dt

2

ds =

L0

A 2.667t3 - 4.616t + 2 B dt

s = 5.43 m

Ans.

503