91962_05_R1_p0479-0512

6/5/09

3:53 PM

Page 479

© 2010 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.

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

91962_05_R1_p0479-0512

6/5/09

3:53 PM

Page 480

© 2010 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.

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

91962_05_R1_p0479-0512

6/5/09

3:53 PM

Page 481

© 2010 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.

*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

91962_05_R1_p0479-0512

6/5/09

3:53 PM

Page 482

© 2010 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.

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

91962_05_R1_p0479-0512

6/5/09

3:53 PM

Page 483

© 2010 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.

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

91962_05_R1_p0479-0512

6/5/09

3:53 PM

Page 484

© 2010 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.

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

91962_05_R1_p0479-0512

6/5/09

3:53 PM

Page 485

© 2010 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.

*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

91962_05_R1_p0479-0512

6/5/09

3:53 PM

Page 486

© 2010 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.

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

91962_05_R1_p0479-0512

6/5/09

3:53 PM

Page 487

© 2010 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.

*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

91962_05_R1_p0479-0512

6/5/09

3:53 PM

Page 488

© 2010 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.

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

91962_05_R1_p0479-0512

6/5/09

3:53 PM

Page 489

© 2010 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.

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

91962_05_R1_p0479-0512

6/5/09

3:53 PM

Page 490

© 2010 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.

*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

91962_05_R1_p0479-0512

6/5/09

3:53 PM

Page 491

© 2010 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.

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

91962_05_R1_p0479-0512

6/5/09

3:53 PM

Page 492

© 2010 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.

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

30Њ

91962_05_R1_p0479-0512

6/5/09

3:53 PM

Page 493

© 2010 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.

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

30Њ

91962_05_R1_p0479-0512

6/5/09

3:54 PM

Page 494

© 2010 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.

*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

0.5 ft

91962_05_R1_p0479-0512

6/5/09

3:54 PM

Page 495

© 2010 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.

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

91962_05_R1_p0479-0512

6/5/09

3:54 PM

Page 496

© 2010 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.

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

91962_05_R1_p0479-0512

6/5/09

3:54 PM

Page 497

© 2010 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.

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

91962_05_R1_p0479-0512

6/5/09

3:54 PM

Page 498

© 2010 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.

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

91962_05_R1_p0479-0512

6/5/09

3:54 PM

Page 499

© 2010 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.

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

91962_05_R1_p0479-0512

6/5/09

3:54 PM

Page 500

© 2010 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.

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

91962_05_R1_p0479-0512

6/5/09

3:54 PM

Page 501

© 2010 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.

*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

r ϭ eu

91962_05_R1_p0479-0512

6/5/09

3:54 PM

Page 502

© 2010 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.

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

91962_05_R1_p0479-0512

6/5/09

3:55 PM

Page 503

© 2010 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.

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

6/5/09

3:53 PM

Page 479

© 2010 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.

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

91962_05_R1_p0479-0512

6/5/09

3:53 PM

Page 480

© 2010 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.

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

91962_05_R1_p0479-0512

6/5/09

3:53 PM

Page 481

© 2010 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.

*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

91962_05_R1_p0479-0512

6/5/09

3:53 PM

Page 482

© 2010 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.

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

91962_05_R1_p0479-0512

6/5/09

3:53 PM

Page 483

© 2010 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.

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

91962_05_R1_p0479-0512

6/5/09

3:53 PM

Page 484

© 2010 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.

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

91962_05_R1_p0479-0512

6/5/09

3:53 PM

Page 485

© 2010 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.

*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

91962_05_R1_p0479-0512

6/5/09

3:53 PM

Page 486

© 2010 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.

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

91962_05_R1_p0479-0512

6/5/09

3:53 PM

Page 487

© 2010 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.

*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

91962_05_R1_p0479-0512

6/5/09

3:53 PM

Page 488

© 2010 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.

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

91962_05_R1_p0479-0512

6/5/09

3:53 PM

Page 489

© 2010 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.

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

91962_05_R1_p0479-0512

6/5/09

3:53 PM

Page 490

© 2010 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.

*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

91962_05_R1_p0479-0512

6/5/09

3:53 PM

Page 491

© 2010 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.

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

91962_05_R1_p0479-0512

6/5/09

3:53 PM

Page 492

© 2010 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.

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

30Њ

91962_05_R1_p0479-0512

6/5/09

3:53 PM

Page 493

© 2010 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.

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

30Њ

91962_05_R1_p0479-0512

6/5/09

3:54 PM

Page 494

© 2010 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.

*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

0.5 ft

91962_05_R1_p0479-0512

6/5/09

3:54 PM

Page 495

© 2010 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.

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

91962_05_R1_p0479-0512

6/5/09

3:54 PM

Page 496

© 2010 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.

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

91962_05_R1_p0479-0512

6/5/09

3:54 PM

Page 497

© 2010 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.

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

91962_05_R1_p0479-0512

6/5/09

3:54 PM

Page 498

© 2010 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.

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

91962_05_R1_p0479-0512

6/5/09

3:54 PM

Page 499

© 2010 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.

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

91962_05_R1_p0479-0512

6/5/09

3:54 PM

Page 500

© 2010 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.

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

91962_05_R1_p0479-0512

6/5/09

3:54 PM

Page 501

© 2010 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.

*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

r ϭ eu

91962_05_R1_p0479-0512

6/5/09

3:54 PM

Page 502

© 2010 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.

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

91962_05_R1_p0479-0512

6/5/09

3:55 PM

Page 503

© 2010 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.

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

## Solution manual accounting information systems 12th edition by romney and steinbart CH01

## Solution manual accounting information systems 12th edition by romney and steinbart CH02

## Solution manual accounting information systems 12th edition by romney and steinbart CH03

## Solution manual accounting information systems 12th edition by romney and steinbart CH04

## Solution manual accounting information systems 12th edition by romney and steinbart CH05

## Solution manual accounting information systems 12th edition by romney and steinbart CH06

## Solution manual accounting information systems 12th edition by romney and steinbart CH07

## Solution manual accounting information systems 12th edition by romney and steinbart CH08

## Solution manual accounting information systems 12th edition by romney and steinbart CH09

## Solution manual accounting information systems 12th edition by romney and steinbart CH10

Tài liệu liên quan