Solution manual for engineering mechanics dynamics 13th edition by hibbeler by finchjem266

Solution Manual for Engineering Mechanics Dynamics 13th Edition by Hibbeler F

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13–1. The 6-lb particle is subjected to the action of its weight and forces F1 = 52i + 6j - 2tk6 lb, F2 = 2 5t i - 4tj - 1k6 lb, and F3 = 5 - 2ti6 lb, where tis in seconds. Determine the distance the ball is from the origin 2 s after being released from rest.

SOLUTION

6 32 .2 ≤(axi + ay j + azk)

©F = ma;

(2i + 6j - 2tk) + (t i - 4tj - 1k) - 2ti - 6k = ¢

Equating components: 6

¢ 32.2 ≤ax = t - 2t + 2 ¢ 32.2 ≤ay = - 4t + 6 ¢ 32.2 ≤az = - 2t - 7 Since dv = a dt, integrating from n = 0, t = 0, yields

2 ¢ 32.2 ≤vx = 3 - t + 2t ¢ ≤ 32.2

s = x

¢ 32.2 ≤vy = - 2t

¢ ≤ 32.2 y

-3+t

¢ 32.2 ≤vz = - t

+ 6t

2 - 7t

¢ + 3t

3 - Dissemination

Wide

instr u ctors

permitted

States sx = 14.31 ft,

Web)

When t = 2 s then,

teaching

2. -

32.2 ≤s z =

Since ds = v dt, integrating from s = 0, t = 0 yields 6

laws

= sz 89.44of

sy = 35.78 ft

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© 2013 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

13–2. The 10-lb block has an initial velocity of 10 ft>s on the smooth plane. If a force F = 12.5t2 lb, where t is in seconds, acts on the block for 3 s, determine the final velocity of the block and the distance the block travels during this time.

v = 10 ft/s

F = (2.5t) lb

SOLUTION +

10 : ©Fx = max; 2.5t = ¢ 32.2 ≤a a = 8.05t dv = a dt n

dv = L0 8.05t dt

L10

v = 4.025t

+ 10

When t = 3 s, laws

v = 46.2 ft>s

teaching .Dissemination copyright Wide A s.

ds = v dt s

or Web) .

ds = L 0 (4.025t s = 1.3417t

+ 10) dt

States

+ 10t

United

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of learning the is not on

use

When t = 3 s, for

World permitted

and

the student (including work

s = 66.2 ft protected is

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Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

13–3. P

If the coefficient of kinetic friction between the 50-kg crate and the ground is mk = 0.3, determine the distance the crate travels and its velocity when t = 3 s. The crate starts from rest, and P = 200 N.

308

SOLUTION Free-Body Diagram: The kinetic friction Ff = mkN is directed to the left to oppose the motion of the crate which is to the right, Fig. a. Equations of Motion: Here, ay = 0. Thus, + c ©Fy = 0;

N - 50(9.81) + 200 sin 30° = 0

N = 390.5 N + : ©Fx = max;200 cos 30° - 0.3(390.5) = 50a

a = 1.121 m>s Kinematics: Since the acceleration a of the crate is constant, + :B

Web) .

instructors

1 s = s0 + v0t +

teaching

copyrightAns. Wide Dissemination . States World permitted

and

v = v0 + act v = 0 + 1.121(3) = 3.36 m>s

laws

not

the learning

act 1

Uniteduse

and

0 + 0 + 2 (1.121) A3 B = 5.04 m

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*13–4. P

If the 50-kg crate starts from rest and achieves a velocity of v = 4 m>s when it travels a distance of 5 m to the right, determine the magnitude of force P acting on the crate. The coefficient of 308

kinetic friction between the crate and the ground is mk = 0.3.

SOLUTION Kinematics: The acceleration a of the crate will be determined first since its motion is known. (: )

2a (s

4 = 0 + 2a(5 - 0) a =

1.60 m>s : Free-Body Diagram: Here, the kinetic friction Ff = mkN = 0.3N is required to be directed to the left to oppose the motion of the crate which is to the right, Fig. a. Equations of Motion: + c ©Fy = may;

N + P sin 30° - 50(9.81) = 50(0)

or laws teaching Web) . copyright Wide Dissemination

N = 490.5 - 0.5P Using the results of N and a,

+ : ©F x

= max;

P cos 30° - 0.3(490.5 - 0.5P)

States

= 50(1.60) United

P = 224 N

u se

World permitted not of learning the is on and Ans. instructors

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13–5. The water-park ride consists of an 800-lb sled which slides from rest down the incline and then into the pool. If the frictional resistance on the incline is Fr = 30 lb, and in the pool for a short distance Fr = 80 lb, determine how fast the sled is traveling when s = 5 ft. 100 ft

SOLUTION + b aFx = max;

800 32.2 a

800 sin 45° - 30 = a = 21.561 ft>s

100 ft

v 2 = v 2 + 2a (s - s ) 1

v 2 = 0 + 2(21.561)(100 2 2 - 0)) 1

v1 = = 78.093 ft>s 800

+ ; aFx = max;

-80 = 32.2a a = -3.22 ft>s 2 v2

laws

= (78.093)

v 2 = v1

teaching Web) Dissemination or

+ 2( -3.22)(5 - 0)

Wide

permitted

+ 2ac (s2 - s1 )

instructors States United of

World

learning

not

on the

v 2 = 7 7 . 9 ft >s

Aisns.

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photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

13–6. If P = 400 N and the coefficient of kinetic friction between the

50-kg crate and the inclined plane is mk = 0.25, determine the velocity of the crate after it travels 6 m up the plane. The crate starts from rest.

30°

SOLUTION

30°

Free-Body Diagram: Here, the kinetic friction Ff = mkN = 0.25N is required to be directed down the plane to oppose the motion of the crate which is assumed to be directed up the plane. The acceleration a of the crate is also assumed to be directed up the plane, Fig. a. Equations of Motion: Here, ay¿ = 0. Thus, ©Fy¿ = may¿;

N + 400 sin 30° - 50(9.81) cos 30° = 50(0) N = 224.79 N laws teaching

Web) or

Using the result of N,

. Dissemination copyright Wide

a = 0.8993 m>s

400 cos 30° - 50(9.81) sin 30° - 0.25(224.79) = 50a

= ©Fx¿

may¿;

instructors World permitted of learning the is not use on States

United

Kinematics: Since the acceleration a of the crate is constant,

v = v0 + 2ac(s - s0)

and student

by the v

= 0 + 2(0.8993)(6 - 0)

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for

(including

of the

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v = 3.29 m>s

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13–7. If the 50-kg crate starts from rest and travels a distance of 6 m up the plane in 4 s, determine the magnitude of force P acting on the crate. The coefficient of kinetic friction between the

30°

crate and the ground is mk = 0.25.

SOLUTION

30°

Kinematics: Here, the acceleration a of the crate will be determined first since its motion is known. 1 at 2

s=s +vt+ 0

6=0+0+

1 a(4 ) 2

a = 0.75 m>s

Free-Body Diagram: Here, the kinetic friction Ff

= mkN = 0.25N is required to be laws

Web)

teaching

directed down the plane to oppose the motion of the crate which is directed up the

Dissemination

Equations of Motion: Here, a y¿ = 0. Thus,

Wide .

plane, Fig. a.

©Fy¿ = may¿;

instructors permitted . World learning on

N + P sin 30° - 50(9.81) cos 30° =

50(0) States

N = 424.79 - 0.5P

the is not

United and

use

Using the results of N and a, ©Fx¿ = max¿;

for student by the

P cos 30° - 0.25(424.79 - 0.5protectedP) - 50(9.81) sinof30° = 50(0.75) assessing

P = 392 N

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*13–8. The speed of the 3500-lb sports car is plotted over the 30-s time period. Plot the variation of the traction force F needed to cause the motion.

v(ft/s) F

80 60

SOLUTION Kinematics: For 0 … t we have dv a = dt = 6 ft>s

60 dv 10 t = {6t} ft>s. Applying equation a = dt ,

6 10 s. v =

2 t (s) 10

For 10 6 t … 30 s, dv

v - 60

= 80 - 60 , v = {t + 50} ft>s. Applying 30 - 10

t - 10

equation v

a = dt , we have

dv a=

1 ft>s

laws

teaching Web) Dissemination copyright Wide .

Equation of Motion: For 0 … t 6 10 s

+ ;

aFx = ma x ; F =

instructors

3500 32.2 ≤(6) =

of United

For 10 6 t … 30 s

+ ;

aFx = max ; F =

World

the is not on and

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3500 ¢ 32.2 ≤(1) =

permitted

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652 lb

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13–9. p

The crate has a mass of 80 kg and is being towed by a chain which is always directed at 20° from the horizontal as shown. If the magnitude of P is increased until the crate begins to slide, determine the crate’s initial acceleration if the coefficient of static friction is ms = 0.5 and the coefficient of kinetic friction is mk = 0.3.

SOLUTION Equations of Equilibrium: If the crate is on the verge of slipping, Ff = ms N = 0.5N. From FBD(a), + c ©Fy

= 0;

N + P sin 20° - 80(9.81) = 0

(1)

+ : ©Fx = 0; P cos 20° - 0.5N = 0 Solving Eqs.(1) and (2) yields

(2)

P = 353.29 N

N = 663.97 N

Equations of Motion: The friction

force developed

between

the crate and its laws

contacting surface is Ff = mkN = 0.3N since the crate is moving. From FBD(b),

teaching Web) Dissemination + c ©Fy

= may ;

: ©Fx = max ; +

N - 80(9.81) + 353.29 sin 20° = 80(0)

353.29 cos 20° - 0.3(663.97) = 80a N = 663.97 N a = 1.66 m>s

instructors

States

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13–10. p

The crate has a mass of 80 kg and is being towed by a chain which is always directed at 20° from the horizontal as shown. Determine the crate’s acceleration in t = 2 s if the coefficient of static friction is ms = 0.4, the coefficient of kinetic friction is 2

m k = 0.3, and the towing force is P = (90t ) N, where t is in seconds.

SOLUTION 2

Equations of Equilibrium: At t = 2 s, P = 90 A 2 B = 360 N. From FBD(a) + c ©Fy = 0; +

N + 360 sin 20° - 80(9.81) = 0

: ©Fx = 0;

N = 661.67 N

360 cos 20° - Ff = 0Ff = 338.29 N

Since Ff 7 (Ff)max = ms N = 0.4(661.67) = 264.67 N, the crate accelerates. Equations of Motion: The friction force developed between the crate and its contacting surface is Ff = mkN = 0.3N since the crate is moving. From FBD(b), + c ©Fy = may ; :

+ N - 80(9.81) + 360 sin 20° = 80(0) N = 661.67 N

laws

Web)

.Dissemination

teaching

©Fx = max ;

360 cos 20° - 0.3(661.67) = 80a

a = 1.75 m>s

Wide

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13–11. The safe S has a weight of 200 lb and is supported by the rope and pulley arrangement shown. If the end of the rope is given to a boy B of weight 90 lb, determine his acceleration if in the confusion he doesn’t let go of the rope. Neglect the mass of the pulleys and rope. S

SOLUTION B

Equation of Motion: The tension T developed in the cord is the same throughout the entire cord since the cord passes over the smooth pulleys. From FBD(a), 90 + c ©Fy = may ;

32.2 b aB

T-90 = - a

(1)

From FBD(b), + c ©Fy = may ;

200 b a

2T - 200 = - a

(2)

32.2 laws

teaching Web) Dissemination copyright Wide .

Kinematic: Es tablis h the po sition-coordinate equation , we have

permitted

Taking time derivative twice yields

2sS +sB

instructors

States

World

on the

not

and

is(3)

learning

United

1 + T2

2aS + aB = 0

use 2

aB = -2.30 ft>s

for by

student

= 2.30 ft>sprotectedsolely

of the

Solving Eqs.(1),(2), and (3) yields aS = 1.15 ft>s T

the T = 96.43 lb

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Ans.

*13–12. The boy having a weight of 80 lb hangs uniformly from the bar. Determine the force in each of his arms in t = 2 s if the bar is moving upward with (a) a constant velocity of 3 ft>s, and (b) a

speed of v = 14t 2 ft>s, where t is in seconds.

SOLUTION (a) T = 40 lb (b) v = 4t

Ans.

a = 8t + c aFy = may ;

2T - 80 =

80 32.2 18t2

At t = 2 s. T = 59.9 lb

Ans. laws

teaching Web) Dissemination copyright Wide .

States United

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and courses

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photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

13–13. The bullet of mass m is given a velocity due to gas pressure caused by the burning of powder within the chamber of the gun. Assuming this pressure creates a force of F = F0 sin 1pt>t02 F

on the bullet, determine the velocity of the bullet at any instant it is in the barrel. What is the bullet’s maximum velocity? Also, determine the position of the bullet in the barrel as a function of time.

F 0

SOLUTION

: ©Fx

= max ;

= ma

t b 0

F0 sin a

F 0 dv pt a = dt = a m b sin a t b 0

v L dv =

pt m b sin a t b dt

Ft pt t v = - a pm b cos a t b d

F0t0

v = a pm b a 1 - cos a t0 b b

Ans.

vmax occurs when cos a t b = -1, or t = t0.

laws

Web)

teaching

.Dissemination copyright

v = pm

max

A s. Wide instruct ors

World

perm itted

2F0t0

F0t0

- sin

pm b c t

United

b b dt

learning

use

for

bd0

the is not and

the student (including

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sin

of the work

i ntegr ity

bat -

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F0t0

b a 1 - cos a

s= a

F0t0

ds =

work

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Ans.

the

and courses

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of any sale will

photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

13–14. The 2-Mg truck is traveling at 15 m> s when the brakes on all its wheels are applied, causing it to skid for a distance of 10 m C

before coming to rest. Determine the constant horizontal force developed in the coupling C, and the frictional force developed between the tires of the truck and the road during this time. The total mass of the boat and trailer is 1 Mg.

SOLUTION Kinematics: Since the motion of the truck and trailer is known, their common acceleration a will be determined first.

+ v = v0 + 2ac(s - s0)

a: b

0 = 15 + 2a(10 - 0) a = -11.25 m>s

= 11.25 m>s

;

Free-Body Diagram:The free-body diagram of the truck and trailer are shown in Figs. (a) and (b), respectively. Here, F representes the frictional force developed when the truck skids, while the force developed in coupling C is represented by T. laws

teaching

.Dissemination

: ©Fx = max ;

Web)

Wide

-T = 1000( -11.25) instructors permitted States World

T = 11 250 N =

11.25 kN

Using the results of a and T and referring to Fig. (b), + c ©Fx = max ;

A s.

United use

11 250 - F = 2000( -11.25) for

of learning the is on

not

and student work (including

protected by theof the assessing

F = 33 750 N =

33.75 kN

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13–15. A freight elevator, including its load, has a mass of 500 kg. It is prevented from rotating by the track and wheels mounted along its sides. When t = 2 s, the motor M draws in the cable with a speed of 6 m>s, measured relative to the elevator. If it starts from rest, determine the constant acceleration of the elevator and the tension in the cable. Neglect the mass of the pulleys, motor, and cables.

SOLUTION 3sE + sP = l 3vE = -vP

A + T B vP = vE + vP>E -3vE = vE + 6 vE = - 6 4

= -1.5 m>s = 1.5 m>s c

A + c B v = v0 + a c t aE = 0.7 5 m>s

1.5= 0 +

+ c ©Fy = may ;

2 c

Web)

Ateachings.

laws .Dissemination

aE (2)

4T - 500(9.81) = 500(0.75)

Wide

AWor T = 1320 N = 1.32 kN

States

United

instructors ldns. permitted

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*13–16. The man pushes on the 60-lb crate with a force F. The force is always directed down at 30° from the horizontal as shown, and its magnitude is increased until the crate begins to slide. Determine the crate’s initial acceleration if the

308

coefficient of static friction is ms = 0.6 and the coefficient of kinetic friction is mk = 0.3.

SOLUTION Force to produce motion: + : ©Fx = 0;Fcos 30° - 0.6N = 0 + c ©Fy = 0;N - 60 - F sin 30° = 0 N = 91.80 lb

F = 63.60 lb

Since N = 91.80 lb, +

: ©Fx = max ;

63.60 cos 30° - 0.3(91.80) = a a = 14.8 ft>s

32.2 ba

2 laws

Web)

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An . Dissemination

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13–17. The double inclined plane supports two blocks A and B, each having a weight of 10 lb. If the coefficient of kinetic friction between the blocks and the plane is mk = 0.1, determine the acceleration of each block. A

SOLUTION Equation of Motion: Since blocks A and B are sliding along the plane, the friction forces developed between the blocks and the plane are (Ff)A = m k NA = 0.1 NA and (Ff)B = mk NB = 0.1NB. Here, aA = aB = a. Applying Eq. 13–7 to FBD(a), we have

a+ F a

y¿

10 = ma y¿; NA - 10 cos 60° = a

N A = 5.00 lb

32.2 b (0)

Q+ F a x¿

10 = ma x¿;

T + 0.1(5.00) - 10 sin 60° = - a

32.2 b a

(1)

From FBD(b), a

y¿

= ma y¿; NB - 10 cos 30° = a

laws

b (0)

= 8.660 lb

States

32.2

a+ F a

x¿

= max¿;

T - 0.1(8.660) - 10 sin 30° =

the student (including work

protected

(2)

2 for by

permitted World

and

use

a = 3.69 ft>s

United

Solving Eqs. (1) and (2) yields

teaching Web) Dissemination copyright Wide .

instructors

a 32.2 b a

Ans.

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T = 7.013 lb

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13–18. A 40-lb suitcase slides from rest 20 ft down the smooth ramp. Determine the point where it strikes the ground at C. How long does it take to go from A to C?

20 ft

30 B 4 ft

SOLUTION

40 + R ©Fx = m ax ;40 sin 30° = a = 16.1 ft>s 2

32.2a

2 2

( + R)v = v0

+ 2 ac(s - s0);

vB = 25.38 ft>s ( + R) v = v0 + ac t ; 25.38 = 0 + 16.1 tAB

laws

Web)

teaching

Dissemination

tAB = 1.576 s R = 0 + 25.38 cos 30°(tBC) +

Wide

instructors

not permitted

States

World

( : )sx = (sx )0 + (vx)0 t

1 ( + T) sy = (sy)0 + (vy)0 t +

United

2 act

4 = 0 + 25.38 sin 30° tBC +

= 1.82 s

1 2(32.2)(tBC)

by the and for student work (including

protected is

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integrity this

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R = 5.30 ft

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Total time = tAB + tB

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sale will

photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

13–19. Solve Prob. 13–18 if the suitcase has an initial velocity down

the ramp of vA = 10 ft>s and the coefficient of kinetic friction along AB is mk = 0.2.

20 ft

308 B 4 ft

SOLUTION + R©Fx = max;

40 sin 30° - 6.928 = a = 10.52 ft>s

40 32.2 a

( + R) v = v 0 + 2 ac (s - s0);

v B = (10) + 2(10.52)(20) vB

= 22.82 ft>s ( + R) v = v0 + ac t; 22.82 = 10 + 10.52 tAB laws

Web)

teaching

tAB = 1.219 s ( : ) s x = (s x )0

Dissemination

+ (v x ) 0 t

( + T ) sy = (sy)0 + (vy)0 t +

ac t 2

instructors World permitted

R = 0 + 22.82 cos 30° (t BC)

learning

2 1

Wide

of United

. the

not

use

4 = 0 + 22.82 sin 30° tBC + 2(32.2)(tBC) t BC

by the student

= 0.2572 s

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assess in g

R = 5.08 ft

Total time = tAB + tBC = 1.48 s

and

(including

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photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

*13–20.

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 =

13200t 2 N, where t is in seconds. If the car has an initial

2 m/s

velocity v1 = 2 m>s when t = 0, determine its velocity when t = 2 s.

SOLUTION

17 8 15

Q + ©Fx¿ = max¿ ;

3200t - 40019.812 a 17 b = 400a

a = 8t - 4.616

dv = adt v

L2 dv = L0 18t

-4.6162 dt

v = 14.1 m>s

Ans.

laws

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States United

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of any sale

will

photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

13–21. 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 =

13200t 2 N, where t is in seconds. If the car has an initial

2 m/s

velocity v1 = 2 m>s at s = 0 and t = 0, determine the distance it moves up the plane when t = 2 s.

SOLUTION

17 8 15

Q + ©Fx¿ = ma x¿;3200t

a = 8t - 4.616

- 40019.812 a 17 b = 400a

dv = adt v

L2 dv = L0 18t - 4.6162 dt

ds v= s

dt = 2.667t 2

- 4.616t + 2

L0 ds = L0 (2.667 t - 4.616t + 2)

dt laws

s = 5.43 m

teaching

or Web)

.Dissemination copyright A s. Wide

States

United

instructors

World permitted

of learning the is not on

and

use

for

the student (including work

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solely

work provided and of

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integrity this

the

and courses

of sale

work

part

any

will

photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

13–22. Determine the required mass of block A so that when it is released from rest it moves the 5-kg block B a distance of 0.75 m up along the smooth inclined plane in t = 2 s. Neglect the mass of the pulleys and cords.

SOLUTION

1 Kinematic: Applying equation s = s0 + v0 t +

2 ac t , we have

0.75 = 0 + 0 + 2 aB A B aB = 0.375 m>s (a +) Establishing the position - coordinate equation, we have 2sA + (sA - sB) = l

3s A - s B = l

Taking time derivative twice yields

From Eq.(1), 3aA - aB = 0

laws

teaching

Web)

(1) Dissemination

aA = 0.125 m>s

3aA - 0.375 = 0

2 copyright

Equation of Motion: The tension T developed in the cord is the sameinstructorsthroughout States

World

learning

United

T = 44.35 N a+ ©Fy¿

= may¿ ;

protected is

may ;

use

for

student

on the is not and work

T - 5(9.81) sin 60° = 5(0.375)the

From FBD(a),

+ c ©Fy =

solely

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work

workprovided and of thisintegrity This is part the

3(44.35) -9.81mA their

= mA

sale will

( -0.125)

destroy

Ans.

Wide

permitted

13–23. The winding drum D is drawing in the cable at an accelerated

rate of 5 m>s . Determine the cable tension if the suspended crate has a mass of 800 kg.

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

= 2.5 m>s c

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

Ans. laws

teaching Web) Dissemination copyright Wide .

States United

instructors permitted . World

of learning the is not on

and

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for

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work

integrity this

the

and courses part

of any sale

will

photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

*13–24.

3>2

If the motor draws in the cable at a rate of v = (0.05s ) m>s, where s is in meters, determine the tension developed in the cable when s = 10 m. The crate has a mass of 20 kg, and the coefficient of kinetic friction between the crate and the ground

v M

is mk = 0.2.

SOLUTION Kinematics: Since the motion of the create is known, its acceleration a will be determined first.

dv a=v

ds =

A0.05s

3>2

Bc(0.05) a 2 bs

1>2

d = 0.00375s m>s

When s = 10 m,

a = 0.00375(10 ) = 0.375 m>s

Free-Body Diagram: The kinetic friction Ff = mkN = 0.2N must act to the left to oppose the motion of the crate which is to the right, Fig. a. Equations of Motion: Here, ay = 0. Thus, + c ©Fy = may;

laws

teaching

Web) .

N - 20(9.81) = 20(0) copyright Wide Dissemination

N = 196.2 N

States

World permitted

instructors

Using the results of N and a, + : ©Fx

of learning the is not United

use

= max;

T - 0.2(196.2) = 20(0.375) T = 46.7 N

by the student work for (including protected of the is solely work assessing this

and

Ans.

workprovided and of integrity This is the and courses part of their destroy any

sale will

13–25.

If the motor draws in the cable at a rate of v = (0.05t ) m>s, where t is in seconds, determine the tension developed in the cable when t = 5 s. The crate has a mass of 20 kg and the coefficient of kinetic friction between the crate and the ground

v M

is mk = 0.2.

SOLUTION Kinematics: Since the motion of the crate is known, its acceleration a will be determined first.

dv a=

dt = 0.05(2t) = (0.1t) m>s

When t = 5 s,

a = 0.1(5) = 0.5 m>s : Free-Body Diagram: The kinetic friction Ff = mkN = 0.2N must act to the left to oppose the motion of the crate which is to the right, Fig. a. Equations of Motion: Here, ay = 0. Thus, + c ©Fy = may;

laws

teaching

Web) .

N - 20(9.81) = 0 copyright Wide Dissemination

N = 196.2 N

States

World permitted

instructors

Using the results of N and a, + : ©Fx

United

of learning the is not on

use

= max;

T - 0.2(196.2) = 20(0.5) T = 49.2 N

by the student work for (including protected of the is solely work assessing this

and

Ans.

workprovided and of integrity This is the and courses part of their destroy any

sale will

13–26. The 2-kg shaft CA passes through a smooth journal bearing at B. Initially, the springs, which are coiled loosely around the shaft, are unstretched when no force is applied to the shaft. In this position s = s¿ = 250 mm and the shaft is at rest. If a horizontal force of F = 5 kN is applied, determine the speed of the shaft at the instant s = 50 mm, s¿ = 450 mm. The ends of the springs are attached to the bearing at B and the caps at C and A.

s¿

s F 5 kN

kCB

3 kN/m

2 kN/m

SOLUTION FCB = kCBx = 3000x FAB = kABx = 2000x + ; ©Fx = max ; 5000 - 3000x - 2000x = 2a 2500 - 2500x = a a dx - v dy

L0 0.2 (2500 - 2500x) dx = L0 2500(0.2) - ¢

v 2

v dv

v 2500(0.2) 2 2 ≤= 2

laws teaching

v = 30 m>s

.Dissemination copyright Wide

or Web) .

instructors

States United

World

of learning the is not on

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the student (including work

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work provided and of

This is

work

integrity this

the

and courses

part

any

sale will

photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

13–27. The 30-lb crate is being hoisted upward with a constant

acceleration of 6 ft>s . If the uniform beam AB has a weight of 200 lb, determine the components of reaction at the fixed support A. Neglect the size and mass of the pulley at B. Hint: First find the tension in the cable, then analyze the forces in the beam using statics.

5 ft B

SOLUTION 2

Crate:

6 ft/s

30 + c ©Fy = may ;T - 30 = a

32.2 b(6)

T = 35.59 lb

Beam: + : ©Fx = 0;

- A x + 35.59 = 0

a+ c ©Fy = 0;

Ay - 200 - 35.59 = 0

+ ©MA = 0;

Ax = 35.6 lb

Ans.

Ay = 236 lb

M A - 200(2.5) - (35.59)(5) = 0

Ans.

MA = 678 lb # ft

Ans. laws

teaching Web) Dissemination copyright Wide .

instructors permitted . World States of learning the is not

United

use

and

the student for (including work by

protected is

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assess in g

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work provided and of integrity this This is and

the courses part

of any sale will

photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

*13–28. The driver attempts to tow the crate using a rope that has a tensile strength of 200 lb. If the crate is originally at rest and has a weight of 500 lb, determine the greatest acceleration it can have if the coefficient of static friction between the crate and the road is ms = 0.4, and the coefficient of kinetic friction

is mk = 0.3.

SOLUTION Equilibrium: In order to slide the crate, the towing force must overcome static friction. + : ©Fx

= 0;

-T cos 30° +0.4N = 0

(1)

+ : ©F = 0; N + T sin 30° -500 = 0 Solving Eqs.(1) and (2) yields: T = 187.6 lb

(2)

N = 406.2 lb

Since T 6 200 lb, the cord will not break at the moment the crate slides. After the crate begins to slide, the kinetic friction is used for the calculation. + c ©Fy = may;

N + 200 sin 30° - 500 = 0

+ : ©Fx

laws

400 lb

500 = ma x ;

200 cos 30° - 0.3(400) =

teaching Web) . copyright Wide Dissemination

States

32.2

permitted

World instructors

of learning the is not

a = 3.43 ft>s

United

use

Ans. and

the student for

(including work protected of the is solely work assessing this workprovided and of integrity This is the and courses part of their destroy any

sale will

13–29. The force exerted by the motor on the cable is shown in the graph. Determine the velocity of the 200-lb crate when t = 2.5 s.

F (lb) 250 lb

M t (s) 2.5

SOLUTION Free-Body Diagram: The free-body diagram of the crate is shown in Fig. a. Equilibrium: For the crate to move, force F must overcome the weight of the crate. Thus, the time required to move the crate is given by + c ©Fy = 0;

100t - 200 = 0

t=2s 250

Equation of Motion: For 2 s 6 t 6

2.5 s, F = 2.5 t = (100t) lb. By referring to Fig. a, 200

+ c ©Fy = may;

100t - 200 =

32.2 a

a = (16.1t - 32.2) ft>s

2 laws

Kinematics: The velocity of the crate can be obtained by integrating the kinematic

equation, dv = adt. For 2 s … t 6 integration limit. Thus,

2.5 s, v = 0 at t =

(+c)

adt

teaching Web) 2 s will be used as the lower

dv = L

States instructors World permitted not

L v

L0 dv = L2 s(16.1t - 32.2)dt

for the student

work (including

v = A8.05t

- 32.2tB 2 sis - 32.2t +work2 2 32.2providedB = A8.05t ftand>s

When t = 2.5 s,

work assessing

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Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

13–30. The force of the motor M on the cable is shown in the graph. Determine the velocity of the 400-kg crate A when t = 2 s.

F (N) 2500

F 5 625 t M

t (s) 2

SOLUTION Free-Body Diagram: The free-body diagram of the crate is shown in Fig. a. Equilibrium: For the crate to move, force 2F must overcome its weight. Thus, the time required to move the crate is given by

+ c ©Fy = 0;

2(625t ) - 400(9.81) = 0

t = 1.772 s

Equations of Motion: F = A625t B N. By referring to Fig. a, + c ©F

= ma ; y

2 A625t B

a = (3.125t

- 400(9.81) = 400a - 9.81) m>s

Kinematics: The velocity of the crate can be obtained by integrating the kinema ic equation, dv = adt. For 1.772 s … t 6

laws

teaching

Web) .

2 s, v = 0 at t = 1.772 s will be used ainthe

lower integration limit. Thus, (+c)

dv = L

States

adt

World permitted instructors

L0 dv = 1.772 s A3.125t

United

use

- 9.81 Bdt by

the and forstudent work (including

learning

not

the is

v = A1.0417t - 9.81tB 1.772 protectedsolely is

work of the

assessing

m sthis

+2provide = A1.0417t - 9.81t d11.587andB of> integrity work

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When t = 2 s,

the

and courses part any

v = 1.0417(2 ) -

9.81(2)t of+ 11.587destroy= 0.301

m>s

heir sale

Ans.

will

Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

13–31. The tractor is used to lift the 150-kg load B with the 24-m-long rope, boom, and pulley system. If the tractor travels to the right at a constant speed of 4 m>s, determine the tension in the rope when sA = 5 m. When sA = 0, sB = 0. 12 m

SOLUTION 12 - s

2s 2 + (12)

= 24 #

- sB + As A + 144 B

as As Ab = 0

- As

-s

asA s

Ab 2 + A s 2

2 #

$ sB

+ 144 B

sAs

= -

AsA 2 + 144 B

2 2

+ 144

- 1 2

a A $ Ab

AsA 2 + 144 B S 2

1.0487 m s 2 2

(4)

B= -C

((5)2 + 144)

+ c ©F y = ma y ;

+ s

+ sAs A

(5) (4) a

AA 2

B 2 1 a #A b

laws or

150(9.81) = 150(1.0487) = > 2 2

. Dissemination Web) copyright

Wide

Tinstructors World permitted of learningthe is

States

1.63 kN

Anots.

United use

for

on and

the student (including work

protected

of the assess in g

work

solely

work provided and of

integrity this

This is

the

and courses part any of sale will

photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

*13–32. The tractor is used to lift the 150-kg load B with the 24-mlong rope, boom, and pulley system. If the tractor travels to 2

the right with an acceleration of 3 m>s and has a velocity of

4 m>s at the instant sA = 5 m, determine the tension in the rope at this instant. When sA = 0, sB = 0.

12 m

SOLUTION

+ 2s 2

12 = s B

+ (12)

= 24

- s B+

A 2

2s s Ab = 0

As + 144 B 2 a

- s B - AsA + 144

B3 2

# 2

+ AsA + 144

asAs Ab

B-

# 2

A b

AsA

#2 sA s A 2

s B= -

A sA2 + 144 B 2

((5) + 144)

+ sAs A 1

(5) (4)

AsA 2 + 144 B S

2 2

aB = - C

#2 -

(4) 2 + (5)(3)

laws teaching

+ 144)

((5)

or .

2 . Dissemination Web) copyright Wide

+ c ©F y = ma y ;

- 1

144 B $

$ asAs Ab = 0

2 = 2.2025 m>s

TT=

States

1.80 kN United

instructors World permitted learning Ans.not on the is

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work

integrity this

the

and courses

any

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part

and

13–33. Each of the three plates has a mass of 10 kg. If the coefficients

18 N

of static and kinetic friction at each surface of contact are ms =

0.3 and mk = 0.2, respectively, determine the acceleration of each plate when the three horizontal forces are applied.

15 N

B A

SOLUTION Plates B, C and D

+ :

©Fx = 0;100 - 15 - 18 - F = 0 F = 67 N Fmax = 0.3(294.3) = 88.3 N 7 67 N

Plate B will not slip. aB = 0

Ans.

Plates D and C +

laws

Web)

teaching

Dissemination copyright Wide

100 - 18 - F = 0

: ©Fx = 0;

F = 82 N

instructors States .

82N

Fmax = 0.3(196.2) = 58.86 N 6

learning

of United

Slipping between B and C. +

for student protected by the of

not

work

assessing

ax = 2.138 m>s :

: ©Fx = max;

the on and is

use

Assume no slipping between D and C,

permitted World

= 20 ax

100 - 39.24 - 18

is solely work provided and of

the work integrity this

This coursesany part the and their of destroy

+ Check slipping between D and C.

: ©Fx

= m ax;

F - 18 = 10(2.138) F = 39.38 N

sale will

Fmax = 0.3(98.1) = 29.43 N 6 39.38 N Slipping between D and C. Plate C: + : ©Fx

= m a x;

100 - 39.24 - 19.62 = 10 ac ac = 4.11 m>s

Ans.

Plate D: + : ©Fx

= m ax;

19.62 - 18 = 10 aD

aD = 0.162m>s :

Ans.

100 N

13–34. Each of the two blocks has a mass m. The coefficient of kinetic friction at all surfaces of contact is m. If a horizontal force P moves the bottom block, determine the acceleration of the bottom block in each case.

B A

(a) B

SOLUTION

Block A: (b)

+ (a) ; ©Fx = max ;

P - 3mmg = m aA P aA = m - 3mg

Ans.

(b) sB + sA = l aA = - aB

(1)

Block A: laws

+ ; ©F x = ma x ;

teachingWeb) (2) or

P - T - 3mmg = m a A

Dissemination

+ Block B:

; ©Fx = max ;

instructors permitted States World .

mmg - T = maB

Subtract Eq.(3) from Eq.(2):

United use

learning the is on not

P - 4mmg = m (aA - aB)

and for

student work (including

by the

protected Use Eq.(1);

Wide

A 2m

of assess in g

- 2mg

work

work provided and of integrity this This is and

the

courses part

of any sale will

the

Ans.

13–35. The conveyor belt is moving at 4 m>s. If the coefficient of static friction between the conveyor and the 10-kg package B is ms = 0.2, determine the shortest time the belt can stop so that the package does not slide on the belt.

SOLUTION + : ©Fx = m ax;

0.2198.12 = 10 a a = 1.962 m>s

+ 1 : 2v = v0 + ac t 4 = 0 + 1.962 t t = 2.04 s

Ans.

laws

teaching Web) Dissemination copyright Wide .

States United

instructors permitted . World

of learning the is not on

use

and

the student for (including work by

protected

of the assess in g

solely

work provided and of

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work

integrity this

the

and courses

part

of any sale will

photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

*13–36. The 2-lb collar C fits loosely on the smooth shaft. If the spring is unstretched when s = 0 and the collar is given a velocity of 15 ft>s, determine the velocity of the collar when s = 1 ft.

15 ft/s s C

1 ft k4 lb/ft

SOLUTION F = kx; s

F s = 4A 2 1 + s

- 1B

- 4A 21 + s - 1B ¢ 2 1 + s ≤ = a 32.2 b a v ds b 4s ds 2 v

¢4s ds -

21 + s

: ©F x = ma x ;

-L 0

≤ = L15 a 32.2 b v dv 1

1 2 2 - C 2s - 43 1 + s D 0 = 32.2 A v - 15 B v = 14.6 ft>s

Ans. laws

teaching Web) Dissemination copyright Wide .

States United

instructors permitted . World

of learning the is not on

and

use

for

the student (including work

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solely

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work provided and of integrity this This is

the

and courses part of any sale will

photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

13–37. Cylinder B has a mass m and is hoisted using the cord and pulley system shown. Determine the magnitude of force F as a function of the cylinder’s vertical position y so that when F is applied the cylinder rises with a constant acceleration Neglect the mass of the cord, pulleys, hook and chain.

d/2

aB.

SOLUTION y + c ©Fy = may ;

2F cos u - mg = maB

where cos u =

d 2 2B

y d 2 2 2F£ 2 y + A 2 B ≥ - mg= maB m(aB + g)2 4y F= 4y

2 Ans.

laws

teaching Web) Dissemination copyright Wide .

instructors permitted States . World

United

of learning the is not on

use

for

and

the student (including work

protected

of the assess in g

solely

work provided and of

This is

work

integrity this

the

and courses

part

of any sale will

photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

13–38. vA

The conveyor belt delivers each 12-kg crate to the ramp at A

2.5 m/s

such that the crate’s speed is vA = 2.5 m>s, directed down along the ramp. If the coefficient of kinetic friction between

each crate and the ramp is mk = 0.3, determine the speed at which each crate slides off the ramp at B. Assume that no tipping occurs. Take u = 30°.

SOLUTION Q +©Fy = may;

NC - 1219.812 cos 30° = 0 NC = 101.95 N

+ R©Fx = max ;

(+ R)

v2=v B

1219.812 sin 30° - 0.31101.952 = 12 aC

aC = 2.356 m>s + 2a 1s - s

vB 2 = 12.52

+ 212.356213 - 02 Ans.

vB = 4.5152 = 4.52 m>s

laws

teaching Web) Dissemination copyright Wide .

States United

instructors permitted . World

of learning the is not on

and

use

the student for (including work by

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work provided and of

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work

integrity this

the

and courses

part

of any sale

will

photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

13–39. An electron of mass m is discharged with an initial

horizontal velocity of v0. If it is subjected to two fields of force for which F x = F0 and F y = 0.3F0, where F0 is

constant, determine the equation of the path, and the speed of the electron at any time t.

++++++++++++++

SOLUTION

+ : ©Fx

= max;

F0 = max

+ c ©Fy = may; Thus,

t vx

dv =

Lv0

0.3 F0 = may

L0 m dt

F0 vx = m t + v0 vy

0.3F

dv =

vy = m t

laws

t+v Cam

+ 1

0.3F0

t am

2 2 + 2F0tmv0 + m v0

F m L0

dx =

F0 t

+ v0 t

L0 0.3F0

0.3F t y = 2m 0 2

t = aB

0.3F0

2m 2m

x= y

0.3F0

+v by

0aB

0.3F0

+v x=

0.3

0aB

0.3F0

*13–40. v

The engine of the van produces a constant driving traction force F at the wheels as it ascends the slope at a constant velocity v. Determine the acceleration of the van when it passes point A and begins to travel on a level road, provided that it maintains the same traction force.

A F u

SOLUTION Free-Body Diagram: The free-body diagrams of the van up the slope and on the level road are shown in Figs. a and b, respectively. Equations of Motion: Since the van is travelling up the slope with a constant velocity, its acceleration is a = 0. By referring to Fig. a, ©Fx¿ = max¿;

F - mg sin u = m(0)

F = mg sin u Since the van maintains the same tractive force F when it is on level road, from Fig. b, + : ©Fx = max;

mg sin u = ma

laws

teaching Web) Dissemination

a = g sin u

Ans. copyright

States United

Wide

instructors permitted . World

of learning the is not on

use

and

the student for (including work by

protected

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solely

work provided and of

This is

work

integrity this

the

and courses

part

of any sale

will

13–41. The 2-kg collar C is free to slide along the smooth shaft AB. Determine the acceleration of collar C if (a) the shaft is fixed from moving, (b) collar A, which is fixed to shaft AB, moves downward at constant velocity along the vertical rod, and (c)

collar A is subjected to a downward acceleration of 2 m>s . In all cases, the collar moves in the plane.

SOLUTION (a) + b©Fx¿ = max¿ ;

219.812 sin 45° = 2aC

(b) From part (a) aC>A = 6.94 m>s

aC = 6.94 m>s

aC = aA + aC>A Where aA = 0 = 6.94 m>s

(c)

C = A + C>A

=2+a C>A

√

laws

(1)

teaching

.Dissemination copyright

From Eq.(1)

+ b©Fx¿

aC>A = 5.5225 m> 2 b

= max¿ ; 219.812 sin 45° = 212 cos 45°+ aC>A2

aC = 2 + 5.5225 = 3.905 + 5.905 T b ; T aC = 23.905

States

United use

+ 5.905 = 7.08 m>s

Web) or

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instructors World permitted learning is not the

on andAns.

5.905 u = tan-1

= 56.5° ud

for by

student work

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of assess in g

sol el y

work

work provided and of

this This is

the

and courses

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of any sale will

Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

13–42. The 2-kg collar C is free to slide along the smooth shaft AB. Determine the acceleration of collar C if collar A is subjected to

an upward acceleration of 4 m>s . 458

SOLUTION + ;

©Fx = max ;N sin 45° = 2aC>AB sin 45°

N = 2 aC>AB + c ©Fy = may ;N cos 45° - 19.62 = 2142 - 2aC>AB cos 45° aC>AB = 9.76514

a=a C

+a AB

C>AB

1aC)x = 0 + 9.76514 sin 45°

= 6.905 ;

(aC)y = 4 - 9.76514 cos 45° = 2.905T aC = 2 16.9052 2 + 12.9052 -1

= 7.49 m>s

Ans.la

or teaching Web) Dissemination

2.905

instructors permitted States . World United

of learning the is not on

and

use

the student for (including work by

protected

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assess in g

work

solely

work provided and of

This is

integrity this the

and courses

part

of any sale will

photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

13–43. The coefficient of static friction between the 200-kg crate and the flat bed of the truck is m = 0.3. Determine the shortest time

for the truck to reach a speed of 60 km>h, starting from rest with constant acceleration, so that the crate does not slip.

SOLUTION Free-Body Diagram: When the crate accelerates with the truck, the frictional force F f develops. Since the crate is required to be on the verge of slipping, F f = msN = 0.3N. Equations of Motion: Here, ay = 0. By referring to Fig. a, + c ©Fy = may;

N - 200(9.81) = 200(0) N = 1962 N

+ = max;

: ©F x

- 0.3(1962) = 200( - a) a = 2.943 m>s

; km

Kinematics: The final velocity of the truck is v = a60

1000 m ba

laws

1 ba

teaching

Web)

.Dissemination

1 km

3600

. in

+ 16.67 m>s. Since the acceleration of the truck is constant, ( ;)

instructors States

v = v0 + ac t

16.67 = 0 + 2.943t

United

protected by

solely

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of learning the is not

use

for

permitted Wide

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student the of the

work

work provided and of integrity

this This is

the

and courses part

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Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage i n a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

*13–44. When the blocks are released, determine their acceleration and the tension of the cable. Neglect the mass of the pulley.

SOLUTION Free-Body Diagram: The free-body diagram of blocks A and B are shown in Figs. b and

c, respectively. Here, aA and aB are assumed to be directed downwards so that they are

10 kg

consistent with the positive sense of position coordinates sA and sB of blocks A and B,

Fig. a. Since the cable passes over the smooth pulleys, the tension in the cable remains constant throughout.

B 30 kg

Equations of Motion: By referring to Figs. b and c, + c ©Fy = may;

2T - 10(9.81) = - 10aA

(1)

T - 30(9.81) = - 30aB

(2)

and + c ©Fy = may;

laws

teaching

2sA + sB = l

Dissemination Web) copyright Wide

Kinematics: We can express the length of the cable in terms of sA and sB by referrin to Fig. a.

instructors permitted States . World

The second derivative of the above equation gives

United use

of learning the is on

2aA + aB = 0

not

and(3) for

Solving Eqs. (1), (2), and (3) yields

student

work

by the

A a

= -

> = 3.773 m s

> 3.77 m s c

protected

B 2

7.546 mis

T = 67.92 N = 67.9 N

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Ans.

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13–45. If the force exerted on cable AB by the motor is

3>2

F = (100t ) N, where t is in seconds, determine the 50-kg crate’s velocity when t = 5 s. The coefficients of static and kinetic friction between the crate and the ground are ms = 0.4 and mk = 0.3, respectively. Initially the crate is at rest.

SOLUTION Free-Body Diagram: The frictional force Ff is required to act to the left to oppose the motion of the crate which is to the right. Equations of Motion: Here, ay + c ©Fy = ma y;

= 0. Thus,

N - 50(9.81) = 50(0) N = 490.5 N

Realizing that Ff

= mkN = 0.3(490.5) = 147.15 N,

+ c ©F

100t

3>2

= ma ; x

- 147.15 = 50a

3>2

a = A2t

- 2.943 B m>s

or laws

Equilibrium: For the crate to move, force F must overcome the static f ictionteachingof

Web) . copyright Wide Dissemination

Ff = msN = 0.4(490.5) = 196.2 N. Thus, the time required to cause the crate to be

on the verge of moving can be obtained from.

+ : ©Fx

States instructors World permitted

3>2

= 0;

100t

- 196.2

Uniteduse

= 1.567 s

learning the is not on

and by the student eq u atio n wo r kd v = a d t

Kinematics: Using the result of a and integrating the kinem ati cfor

(including of

with the initial condition v = 0 at t = 1.567 as the lower inte +

protected is solely

dv=adt

( :)

this work t

This is

v = 0.8(5)

part courses any

- 2.943tB 1.567 stheir

5>2

v = A0.8t When t = 5 s,

- 2.943t +

integrity

the

- 2.943 Bdt and

5>2

and of

provided

2t 3>2

dv = L1.567 s

v = A0.8t

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destroy

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2.152 B m>s

5>2 - 2.943(5) + 2.152 = 32.16 ft>s = 32.2 ft>s

Ans.

13–46. Blocks A and B each have a mass m. Determine the largest horizontal force P which can be applied to B so that A will not move relative to B. All surfaces are smooth.

A P

B u C

SOLUTION Require aA = aB = a Block A: + c ©Fy = 0;N cos u - mg = 0 + ; ©Fx

= max ;

N sin u = ma a = g tan u

Block B:

+ ; ©Fx = max;

P = 2mg tan u

laws

teaching

Web)

.AnDissemination.

P - N sin u = ma

P - mg tan u = mg tan u

Wide

instructors

States United

World

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and

use

for

permitted

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work

integrity this

the

and courses

part

of any sale will

photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

13–47. Blocks A and B each have a mass m. Determine the largest horizontal force P which can be applied to B so that A will not slip on B. The coefficient of static friction between A and

A u

B is ms. Neglect any friction between B and C.

SOLUTION Require aA = aB = a Block A: + c ©Fy = 0;

+ ;

©F x = max ;

N cos u - msN sin u - mg = 0 N sin u + m sN cos u = ma mg N = cos u - ms sin u a = ga cos u - ms sin u b

laws

sin u + ms cos u

teaching Web) Dissemination copyright Wide .

Block B:

+ ;

©F x = max;

P - ms N cos u - N sin u = ma

instructors permitted States . World

sin u + m s cos u

sin u + m

P - mga

= mga

coslearningu

Uniteduse

b andthe is not on

sin u + m s cos u

for student bycos

cos u - ms sin u

work

- ms s the

P = 2mga

cos u - m s sin u

protected b

assessing

is solely work work provided and of integrity this This is

the

and courses

any

sale will

part

of the

Ans.

*13–48. A parachutist having a mass m opens his parachute from an atrest position at a very high altitude. If the atmospheric drag

resistance is FD = kv , where k is a constant, determine his velocity when he has fallen for a time t. What is his velocity when he lands on the ground? This velocity is referred to as the terminal velocity, which is found by letting the time of fall t : q .

SOLUTION mg - kv = m dt

+ T ©Fz = m a z ; t m dv v m

L0 1mg - kv 2 = L 0 dt v

dv k

-v

mg m

mg ln D

= ln

mg A

Ak mg e

2t2

assessing

= mg

-ve

e2t2 k

mg mg

Ak When t : q

e2t2

13–49. The smooth block B of negligible size has a mass m and rests on the horizontal plane. If the board AC pushes on the

block at an angle u with a constant acceleration a0,

determine the velocity of the block along the board and the distance s the block moves along the board as a function of time t. The block starts from rest when s = 0, t = 0.

C A

SOLUTION Q + ©Fx = m ax;

0 = m aB sin f

a=a B

+a AC

B>AC

a =a+a B

Q+

B>AC

aB sin f = - a0 sin u + aB>AC

Thus, 0 = m(- a0 sin u + aB>AC)

sin

aB>AC = a0

B>AC

laws

teaching

Dissemination Web) copyright Wide permitted States instructors .

v sin u dt

learning

of United =s=

a0sin u t dt

use

the is not on and

for by

student work (including the

a0 sin u t

protected of assessing the is solely work work provided and of integrity this This is

the

and courses

of any sale will

part

Ans.

13–50. A projectile of mass m is fired into a liquid at an angle u0 with

an initial velocity v0 as shown. If the liquid develops a frictional or drag resistance on the projectile which is proportional to its velocity, i.e., F = kv, where k is a constant, determine the x and y components of its position at any instant.

Also, what is the maximum distance xmax that it travels?

θ0

SOLUTION + : ©Fx = max ;

- kv cos u = m ax

+ c ©Fy = m ay ;

- m g - k v sin u = m ay

d x

- k dt = m dt

d y

- m g - k dt = m dt

laws

# -k Integrating yields mg In (y +

Dissemination copyright Wide permitted

C 1

mt+

In x =

teaching

instructors

t + C2

# k

States

the

m United

World not

use

student

x = v0 cos u0 e

When t = 0, x = v0 cos u0,

y = v0 sin u0 #

-(k>m)t

protected

work

and

the

of the

for

- (k> m)tassessing

solely

work

provided k

integrity and of this

work

y= -

+ (v0 sin u0 +

)

This is

- m v0

part the

and courses any

cos u0 e-(theirk>m)t + C3

k y=

sale

will

mg t - (v sin u mg m )e-(k>m)t )( 0 0 + k k k

When t = 0, x = y = 0, thus - k>m t

m v0

y= -

mgt k

(

k cos u0(1 - e

x= +

)

Ans.

)

m (v sin u mg )(1 - e-(k>m)t) 0 0+ k k

Ans.

As t : q

m v cos x

max

0 u 0k

Ans.

or .

Web)

13–51. The block A has a mass m A and rests on the pan B, which has a

mass mB. Both are supported by a spring having a stiffness k

that is attached to the bottom of the pan and to the ground. y d

Determine the distance d the pan should be pushed down from the equilibrium position and then released from rest so that separation of the block will take place from the surface of the pan at the instant the spring becomes unstretched.

SOLUTION For Equilibrium + c ©Fy = may;

Fs = (m A + m B)g

Fs eq

= k

(mA + m B)g =

Block: + c ©Fy = may ;

- mA g + N = m A a

Block and pan + c ©Fy = may;

- (mA + m B)g + k(y eq + y) =(mA + m B)a

laws teaching Web) .

Thus, copyright

mA + mB

- (m A + mB)g + kc a

b g + yd = (m

+ m )a B

- mAg +

States mA

Wide

. Dissemination World permitted

instructors

of learning the is not Uniteduse on

Require y = d, N = 0 kd = - (mA + mB)g

Since d is downward,

(m A + m Bwork)g k

by the and for student work (including protected of the is solely work assessing this and of provided

This is

and

integrity

Ans.

the courses part of destroy any

their sale

will

Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

*13–52. A girl, having a mass of 15 kg, sits motionless relative to the surface of a horizontal platform at a distance of r = 5 m from the platform’s center. If the angular motion of the platform is slowly increased so that the girl’s tangential component of acceleration can be neglected, determine the maximum speed which the girl will have before she begins to slip off the platform. The coefficient of static friction between the girl and the platform is m = 0.2.

SOLUTION Equation of Motion: Since the girl is on the verge of slipping, Ff = msN = 0.2N. Applying Eq. 13–8, we have ©Fb = 0;

N - 15(9.81) = 0

N = 147.15 N

v ©Fn = man ;

0.2(147.15) = 15a 5 b v = 3.13 m>s

Ans.

laws

teaching Web) Dissemination copyright Wide .

States United

instructors permitted . World

of learning the is not on

and

use

the student for (including work by

protected

of the assess in g

solely

work provided and of

This is

work

integrity this

the

and courses

part

of any sale will

photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

13–53. The 2-kg block B and 15-kg cylinder A are connected to a light cord that passes through a hole in the center of the smooth table. If the block is given a speed of v = 10m>s, determine the radius r of the circular path along which it travels.

B v

SOLUTION A

Free-Body Diagram: The free-body diagram of block B is shown in Fig. (a). The tension in the cord is equal to the weight of cylinder A, i.e., T = 15(9.81) N = 147.15 N. Here, an must be directed towards the center of the circular path (positive n axis).

Equations of Motion: Realizing that an =

10 ©Fn = man;

= r and referring to Fig. (a),

147.15 = 2a r b r = 1.36 m

Ans. laws

teaching Web) Dissemination copyright Wide .

States

United

instructors permitted . World

of learning the is not on

use

and

the student for (including work by

protected

of the assess in g

solely

work provided and of

This is

work

integrity this

the

and courses part

of any sale

will

photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

13–54. The 2-kg block B and 15-kg cylinder A are connected to a light cord that passes through a hole in the center of the smooth table. If the block travels along a circular path of radius r = 1.5m, determine the speed of the block.

B v

SOLUTION A

v2 Equations of Motion: Realizing that an =

v2 =1.5 and referring to Fig. (a),

v ©Fn = man;

147.15 = 2a 1.5 b v = 10.5 m>s

Ans. laws

teaching Web) Dissemination copyright Wide .

States

United

instructors permitted . World

of learning the is not on

use

and

the student for (including work by

protected

of the assess in g

solely

work provided and of

This is

work

integrity this

the

and courses

part

of any sale will

photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

13–55. P

The 5-kg collar A is sliding around a smooth vertical guide rod. At the instant shown, the speed of the collar is v = 4 m>s,

v 5 4 m/s

308

which is increasing at 3 m>s . Determine the normal reaction of the guide rod on the collar, and force P at this instant.

A 0.5 m

SOLUTION + : ©Ft = mat;

P cos 30° = 5(3) P = 17.32 N = 17.3 N

Ans.

42 + T ©Fn = man;

N + 5 A9.81 B - 17.32 sin 30° = 5 a 0.5 b N = 119.61 N = 120 N T

Ans.

laws

teaching Web) Dissemination copyright Wide .

States United

instructors permitted . World

of learning the is not on

and

use

the student for (including work by

protected

of the assess in g

solely

work provided and of

This is

work

integrity this

the

and courses

part

of any sale

will

photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

*13–56. Cartons having a mass of 5 kg are required to move along the assembly line at 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

8 m/s

between a carton and the conveyor are m s = 0.7 and mk = 0.5, respectively.

SOLUTION + c ©Fb = m ab;

N-W=0 N=W Fx = 0.7W

W 8 ; ©Fn = m an;

0.7W = 9.81 ( r ) Ans.

r = 9.32 m

laws

teaching Web) Dissemination copyright Wide .

States United

instructors permitted . World

of learning the is not on

and

use

the student for (including work by

protected

of the assess in g

solely

work provided and of

This is

work

integrity this

the

and courses

part

of any sale

will

photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

13–57. The block B, having a mass of 0.2 kg, is attached to the vertex A of the right circular cone using a light cord. If the block has a speed of 0.5 m>s around the cone, determine the tension in the cord and the reaction which the cone exerts on the block and the effect of friction.

z A 200 mm

400 mm

SOLUTION r

300

200

=500 ;

300 mm

r = 120 mm = 0.120 m (0.5)

4 + Q©F y = ma y;

T - 0.2(9.81)a 5 b = B 0.2¢ 0.120 ≤ R a 5 b T = 1.82 N (0.5)

+ a©F x = ma x;

Ans.

2 4

NB - 0.2(9.81)a 5 b = - B0.2¢ 0.120 ≤ Ra 5 b NB = 0.844 N

Ans.

Also,

+ : ©Fn = man;

laws

3 Ta 5 b - NB a

teaching Web) Dissemination

4 5b

(0.5)

instructors

Wide permitted

0.120

+ c ©Fb = 0;

4 3 Ta 5 b + N B a 5 b -

States

0.2(9.81) =

. World

of learning the is not

T = 1.82 N

United

on and

use

NB = 0.844 N

for student work by the (including protected

Ans.

of the assess in g

solely

work provided and of

This is

work

integrity this

the

and courses part

of any sale will

photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

13–58. The 2-kg spool S fits loosely on the 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 3 4

S 0.25 m

SOLUTION 4 r = 0.25a

+ ; ©Fn

+ c ©Fb

= 0.2 m 4

= m an;

Na b 5 - 0.2Ns a 5 b = 2a 0.2 b s

= m ab;

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

Ans. laws

teaching

Web) .

States

World permitted

instructors

United by

of learning the is not on

use

and

the student for

(including work protected of the is solely work assessing this workprovided and of This is the

integrity

and courses part of their destroy any

sale will

13–59. The 2-kg spool S fits loosely on the 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 3 4

S 0.25 m

SOLUTION 4 r = 0.25( 5 ) = 0.2 m +

; ©Fn = m an ; Ns( 5 ) + 0.2Ns( 5 ) = 2( 0.2 ) 4 3 + c ©Fb = m a b ;

N s( 5) - 0.2N s( 5 ) - 2(9.81) = 0 Ns = 28.85 N Ans.

v = 1.48 m s

laws

teaching Web) Dissemination copyright Wide .

States United

instructors permitted . World

of learning the is not on

and

use

the student for (including work by

protected

of the assess in g

solely

work provided and of

This is

work

integrity this

the

and courses

part

of any sale

will

photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

*13–60. At the instant u = 60°, the boy’s center of mass G has a u

downward speed vG = 15 ft>s. Determine the rate of increase in his speed and the tension in each of the two supporting cords of the swing at this instant. The boy has a weight of 60 lb. Neglect his size and the mass of the seat and cords.

10 ft

SOLUTION

60 + R©Ft = mat ; 60 cos 60° = 32.2 at

at = 16.1 ft>s 60 15

Q + ©Fn = man ; 2T - 60 sin 60° =

Ans.

32.2 a 10 b

Ans.

T = 46.9 lb

laws

teaching Web) Dissemination copyright Wide .

States

United

instructors permitted . World

of learning the is not on

and

use

the student for (including work by

protected

of the assess in g

solely

work provided and of

This is

work

integrity this

the

and courses

part

of any sale

will

photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

13–61. At the instant u = 60°, the boy’s center of mass G is momentarily at rest. Determine his speed and the tension in each of the two supporting cords of the swing when u = 90°. The boy has a weight of 60 lb. Neglect his size and the mass of the seat and cords.

10 ft

SOLUTION

60 60 cos u = 32.2 ta

+ R© = ma ; t

Q + ©Fn = man;

2T - 60 sin u =

v dn = a ds v

= 32.2 cos u t

60 v

32.2 a 10 b

(1)

however ds = 10du

90°

v dn = L0

60° 322 cos u du

Ans.

v = 9.289 ft>s From Eq. (1) 60 2T - 60 sin 90° =

32.2a

9.289

laws or teaching Web) Dissemination copyright Ans. Wide .

2 T = 38.0 lb

10 b

States United

instructors permitted . World

of learning the is not on

and

use

for

the student (including work

protected

of the assess in g

solely

work provided and of

This is

work

integrity this

the

and courses

part

of any sale will

photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

13–62. The 10-lb suitcase slides down the curved ramp for which the

1 y=

–– x

coefficient of kinetic friction is mk = 0.2. If at the instant it reaches point A it has a speed of 5 ft>s, determine the normal force on the suitcase and the rate of increase of its speed.

6 ft

SOLUTION 1

n=8x

dx = tan u = 4 x 2 x = - 6 = - 1.5

u = - 56.31°

d y dx

1 =4 dy

B1 + a dx b r=

2 dx 2 + Q©Fn = man ;

2 C 1 + (- 1.5) D 23

= 23.436 ft

laws teaching

N - 10 cos 56.31° = a 32

.2 b ¢23 .436 ≤

Dissemination copyright Wide

or .

Web)

Ans.

N = 5.8783 = 5.88 lb

permitted

instructors

States

World

learning

+ R©Ft = mat;

- 0.2(5.8783) + 10 sin 56.31° =

32.2 atUniteduse of

on the is not

the student

a t

protected

(including

for

of the

= 23.0 ft s

work

Ans. is solely work work provided and of integrity this This is

the

and courses

part

of any sale will

13–63. 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°.

8 ft G

SOLUTION

150 20

+ aaFy = m1an2y ;

N - 150 cos 60° =

32.2 a 8 b sin 60°

N = 277 lb

Ans.

+ b aF = m1a 2 x

;

150 a 20 b cos 60°

- F + 150 sin 60° =

n x

32.2

F = 13.4 lb

Ans.

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

teaching Web) Dissemination copyright Wide .

States United

instructors permitted . World

of learning the is not on

use

and

the student for (including work by

protected

of the assess in g

solely

work provided and of

This is

work

integrity this

the

and courses part

of any sale

will

photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

*13–64. 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 off. 8 ft G

SOLUTION 150 1302

+ ; ©Fn = man ; +

0.5N cos u + N sin u =32.2a

c ©Fb = 0;- 150 + N cos u - 0.5 N sin u = 0 150 N = cos u - 0.5 sin u

150 1302 10.5 cos u + sin u2150 1cos u - 0.5 sin u2 = 32.2 a 8 b 0.5 cos u + sin u = 3.49378 cos u - 1.74689 sin u laws

Web)

teaching

u = 47.5°

Dissemination Ans.

States United

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instructors permitted . World

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13–65. Determine the constant speed of the passengers on the amusement-park ride if it is observed that the supporting cables are directed at u = 30° from the vertical. Each chair including its passenger has a mass of 80 kg. Also, what are the components of force in the n, t, and b directions which the chair exerts on a 50-kg passenger during the motion?

u t

SOLUTION

+ ; ©Fn = m an ;

T sin 30° = 80( 4 + 6 sin 30° )

+ c ©Fb = 0;T cos 30° - 8019.812 = 0 T = 906.2 N v = 6.30 m>s ©Fn = m an ; ©Ft = m at; ©Fb = m ab ;

Ans.

16.302

) = 283 N

= 50(

Ans.

7 Ft

Ans.

b - 490.5 = 0

laws

Ans.teach

Fb = 490 N

ing Web) Dissemination copyright Wide .

instructors permitted States . World

United

of learning the is not on

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part

any

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13–66. The man has a mass of 80 kg and sits 3 m from the center of the rotating platform. Due# to the rotation his speed is increased from rest

by v = 0.4 m>s . If the coefficient of static friction between his clothes and the platform is ms = 0.3, determine the

time required to cause him to slip.

10 m

SOLUTION ©Ft = m at ;

Ft = 80(0.4) Ft = 32 N

v2 ©Fn = m an ;

Fn = (80) 3

F = m s Nm = 2(Ft)

+ (Fn)

v 2

0.3(80)(9.81) = A(32)

+ ((80)

)

55 432 = 1024 + (6400)( 9 )

laws teaching

v = 2.9575 m>s

or Web)

Dissemination copyright Wide

dv at = dt

= 0.4

L0 dv =

instructors permitted States . World

0.4 dt

United

of learning the is not on

and

use

v = 0.4 t

the student for (including work

2.9575 = 0.4 t

t = 7.39 s

protected

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solely

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and of

integrity

Ans.

this

work

This is

the

and courses part

of any sale will

13–67. The vehicle is designed to combine the feel of a motorcycle with the comfort and safety of an automobile. If the vehicle

u curved road of radius 100 m, determine the tilt angle u of

is traveling at a constant speed of 80 km>h along a circular

the vehicle so that only a normal force from the seat acts on the driver. Neglect the size of the driver.

SOLUTION Free-Body Diagram: The free-body diagram of the passenger is shown in Fig. (a). Here, an must be directed towards the center of the circular path (positive n axis). km

1000 m

Equations of Motion: The speed of the passenger is v = a 80 h b a 1 km b a 3600 s b = 22.22 m>s. Thus, the normal component of the passenger’s acceleration is given by an =

100

= 4.938 m>s . By referring to Fig. (a),

laws

22.22

9.81m

Dissemination

Wide

nstructors i 9.81m

not permitted

= 0;

World

States

+ c ©Fb

Web)

teaching

cos

N = learning

N cos u - m(9.81) = 0

cos u

United

and

use

u = 26.7°

for by

the is on

student

Ans.

the

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and courses

part

of any sale will

*13–68. The 0.8-Mg car travels over the hill having the shape of a parabola. If the driver maintains a constant speed of 9 m> s, determine both the resultant normal force and the resultant frictional force that all the wheels of the car exert on the road at the instant it reaches point A. Neglect the size of the car.

x2 y 20 (1

6400 ) A

SOLUTION

Geometry: Here, dy = - 0.00625x and d y = - 0.00625. The slope angle u at point dx dx A is given by 2

80 m

dy tan u = dx x = 80 m = - 0.00625(80) and the radius of curvature at point A is

u = - 26.57°

2 3>2

[1 + (dy>dx) ] r=

|d y>dx |

[1 + (- 0.00625x) ] =

|- 0.00625|

x = 80 m = 223.61 m

Equations of Motion: Here, at = 0. Applying Eq. 13–8

andla

with u = 26.57°

Web)

teaching

Disseminationor copyright

800(9.81) sin 26.57 ° - Ff = 800(0)

Wide

r = 223.61 m, we have

instructors

Ff = 3509.73 N = 3.51 kN

States

permitted

Ans.World

. learning

©Fn = man;

800(9.81) cos 26.57° - N = 800Unitedause

thenot b

on and

for student by the

N = 6729.67 N = 6.73 kN

work

Ans.

223.61

protected

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This is

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and courses

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13–69. The 0.8-Mg car travels over the hill having the shape of a parabola. When the car is at point A, it is traveling at 9 m>s

and increasing its speed at 3 m>s . Determine both the resultant normal force and the resultant frictional force that all the wheels of the car exert on the road at this instant. Neglect the size of the car.

x2 y 20 (1

6400 ) A

SOLUTION

Geometry: Here,

dy = - 0.00625x and d y = - 0.00625. The slope angle u at point dx dx

80 m

A is given by dy tan u = dx

x = 80 m = - 0.00625(80)

u = - 26.57°

and the radius of curvature at point A is

2 3>2 C 1 + (dy>dx) D 2 2 |d y>dx |

2 C 1 + (- 0.00625x) D 3>2 =

|- 0.00625|

x = 80 m = 223.61 m

Equation of Motion: Applying Eq. 13–8 with u = 26.57° and r = 223.61 m, we havelaws

Web)

teaching

. DisseminationAns. or

Ff = 1109.73 N = 1.11 kN

©F t = mat;

800(9.81) sin 26.57° - Ff =

800(3)

Wide

9 2instructors ©F n = ma n;

States

800(9.81) cos 26.57° - N = 800a 223.61learning≤on

not permitted World

the is

United and

use

forstudent protected by the N = 6729.67 N =

6.73 kN

of the Ans.

is solely work work provided and of integrity this This is and

the courses part any of

sale will

photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

13–70. The package has a weight of 5 lb and slides down the chute. When it reaches the curved portion AB, it is traveling at 8 ft>s 1u = 0°2. If the chute is smooth, determine the speed of the package when it reaches the intermediate point C 1u = 30°2 and when it reaches the horizontal plane 1u = 45°2. Also, find the normal force on the package at C.

45 8 ft/s θ = 30 20 ft A

SOLUTION 5 5 cos f = 32.2 at

+ b ©Ft = mat ;

= 32.2 cos f

+ a©Fn = man ;

N - 5 sin f =

32.2 (20 )

v dv = a t ds v

Lg v dv = L45°32.2 cos f (20 df) 1 2

laws

Web)

teachi ng

Dissemination v

At f = 45° + 30° =

(8)

= 644(sin f -

sin 45°)

Wide

vC = 19.933 ft>s = 19.9 ft>s 75°,

World permitted .

NC = 7.91 lb

United

of learning the is not on

use

At f = 45° + 45° = 90°

Ans.

and

for student work by the (including

vB = 21.0 ft s

Ans.

protectedsolely assessing is

work

the

work provided and of integrity this This is

45

the

and courses part

of any sale will

13–71. If the ball has a mass of 30 kg and a speed v = 4 m>s at the instant it is at its lowest point, u = 0°, determine the tension in the cord at this instant. Also, determine the angle u to which the ball swings and momentarily stops. Neglect the size of the ball.

u 4m

SOLUTION (4) + c ©Fn = man;

T - 30(9.81) =

30a 4 b

T = 414 N

Ans.

+ Q©Ft = mat;- 30(9.81) sin u = 30at at = - 9.81 sin u at ds = v dv Since ds = 4 du, then u

- 9.81 L 0 s in u(4 d u) =

L4 v dv

laws

teaching Web) Dissemination copyright Wide .

C 9.81(4)cos uD 0 = - 2 (4) 39.24(cos u - 1) = - 8 u = 37.2°

States

instructors not permitted . World

of learning the is United use for

Ans.

and

the student (including work

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*13–72. The ball has a mass of 30 kg and a speed v = 4 m>s at the instant it is at its lowest point, u = 0°. Determine the tension in the cord and the rate at which the ball’s speed is decreasing at the instant u = 20°. Neglect the size of the ball.

u 4m

SOLUTION v2 + a©Fn = man;

T - 30(9.81) cos u = 30a

+ Q©Ft = mat;- 30(9.81) sin u = 30at at = - 9.81 sin u at ds = v dv Since ds = 4 du, then u

- 9.81

L0 sin u (4 du) = L4 u

9.81(4) cos u 2 0 = 2 (v) 39.24(cos u - 1) + 8 =

v dv 1

- 2 (4) 1 2 2 v

laws

teaching Web) Dissemination copyright Wide .

instructors permitted States . World

At u = 20° v = 3.357 m>s

United

learning the is not on

use

at = - 3.36 m>s = 3.36 m>s b T

and

the student by for

= 361 N

(including

protected of assessing the is solely work work provided and of integrity this This is

work

Ans. Ans.

the

and courses

part

of any sale will

photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

13–73. Determine the maximum speed at which the car with mass m can pass over the top point A of the vertical curved road and still maintain contact with the road. If the car maintains this speed, what is the normal reaction the road exerts on the car when it passes the lowest point B on the road?

B r

SOLUTION

Free-Body Diagram: The free-body diagram of the car at the top and bottom of the vertical curved road are shown in Figs. (a) and (b), respectively. Here, an must be directed towards the center of curvature of the vertical curved road (positive n axis). Equations of Motion: When the car is on top of the vertical curved road, it is required that its tires are about to lose contact with the road surface. Thus, N = 0. v

Realizing that an =

r = r

and referring to Fig. (a),

v + T ©F n = man;

mg = m¢ r ≤

+ c © Fn = man; v gr

N - mg = mg

Using

the

an =

result

v = 2 gr

Ans. teaching Web)

component

the normal

. Dissemination or Wide

car accelerationlaws

= g when it is at the lowest point on the road. By referring to Fig. (b),

instructors Stat e s

N = 2mg

of United

not permitted is

World

Ans.

on and

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the student (including work

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© 2013 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, stor age in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to:

Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

13–74. If the crest of the hill has a radius of curvature r = 200 ft,

determine the maximum constant speed at which the car can travel over it without leaving the surface of the road. Neglect the size of the car in the calculation. The car has a weight of 3500 lb.

r 200 ft

SOLUTION 3500 T ©Fn = man;

3500 = 32.2 a 200 b v = 80.2 ft>s

Ans.

laws

teaching Web) Dissemination copyright Wide .

States

United

instructors permitted . World

of learning the is not on

use

for

and

the student (including work

protected

of the assess in g

solely

work provided and of

This is

work

integrity this

the

and courses

part

of any sale

will

photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

13–75. Bobs A and B of mass mA and mB (mA 7 mB) are connected to an inextensible light string of length l that passes through the smooth ring at C. If bob B moves as a conical pendulum such that A is suspended a distance of h from C, determine the angle u and the speed of bob B. Neglect the size of both bobs.

C u

SOLUTION

Free-Body Diagram: The free-body diagram of bob B is shown in Fig. a. The tension

developed in the string is equal to the weight of bob A, i.e., T = mAg. Here, an must be directed towards the center of the horizontal circular path (positive n axis). Equations of Motion: The radius of the horizontal circular path is r = (l - h) sin u.

v Thus, a n = r

= (l - h) sin u . By referring to Fig. a,

+ c ©Fb = 0;

mA g cos u - mB g = 0

u = cos 1 a

+ ; ©F

= ma ;

m g sin u

Ans. vB 2

or laws

(l - h) sin u

teaching Web) m A g (l - h)

vB =

B 2

(1)

sin u

States

2mA - mB

From Fig. b, sin u =

World permitted not

. Substituting this value into Eq. (1),instructors

m g(l - h)

-m 2

- mB ) m m

the student

and work

(including the

protected

£ =B

for

g(l - h)(mA

use

vB =

the is learning

United

Wide copyright Dissemination

≥ solely assessing work of this

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Ans.

the

and courses part of their destroy any

sale will

*13–76. Prove that if the block is released from rest at point B of a smooth path of arbitrary shape, the speed it attains when it reaches point A is equal to the speed it attains when it falls freely through a distance h; i.e., v = 22gh.

h A

SOLUTION + R©Ft = mat;

mg sin u = mat

v dv = at ds = g sin u ds v

at = g sin u

dy = ds sin u

However

L0 v dv =

L0g dy

v 2 = gh v = 2 2gh

Q.E.D.

laws

teaching Web) Dissemination copyright Wide .

States United

instructors permitted . World

of learning the is not on

use

and

the student for (including work by

protected

of the assess in g

solely

work provided and of

This

work

integrity this

the

and courses

part

of any sale will

photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

13–77. The skier starts from rest at A(10 m, 0) and descends the smooth slope, which may be approximated by a parabola. If she has a mass of 52 kg, determine the normal force the ground exerts on the skier at the instant she arrives at point B. Neglect the size of the skier. Hint: Use the result of Prob. 13–76.

y 10 m

1 y

–– x

2 5

x 5m B

SOLUTION 2

dy Geometry: Here, dx

d y

= 10x and dx dy

1 = 10

. The slope angle u at point B is given by

=0 tan u = dx

u = 0°

x=0m

and the radius of curvature at point B is

c1+

C 1 + (dy>dx) D

3>2 2

3>2

|d y>dx |

teaching

Equations of Motion:

permitted

52(9.81) sin u = - 52at

+ b©Ft = mat ; + a©Fn = man;

use

must

be in

the

direction

positive

Kinematics: The speed of the skier can be determined usingby

1 is

102

Here, tan u = 10x. Then, sin u =

This

L10 m

Substituting v

2 2

= 98.1 m >s , u = 0°, and r = 10.0 m into Eq.(1) yields N - 52(9.81) cos 0° = 52a N = 1020.24 N = 1.02 kN

1 +

100 xof

= 9.81 m sale> 2

13–78. A spring, having an unstretched length of 2 ft, has one end attached to the 10-lb ball. Determine the angle u of the spring if the ball has a speed of 6 ft>s tangent to the horizontal circular path.

6 in.

SOLUTION Free-Body Diagram: The free-body diagram of the bob is shown in Fig. (a). If we denote the stretched length of the spring as l, then using the springforce formula, Fsp = ks = 20(l - 2) lb. Here, an must be directed towards the center of the horizontal circular path (positive n axis). Equations of Motion: The radius of the horizontal circular path is r = 0.5 + l sin u.

Since an = r = 0.5 + l sin u , by referring to Fig. (a), + c ©Fb = 0;

20(l - 2) cos u - 10 = 0

+ ; ©Fn = man;

(1)

laws

20(l - 2) sin u =

teaching

Solving Eqs. (1) and (2) yields 32.2

(2) Ans.Dissemi

0.5 + l sin u

Web) or

nation copyright

u = 31.26° = 31.3°

Wide

l = 2.585 ft instructors permitted States . World of learning the is

Ans.no t United

on and

use

the student for (including by

protected

of the assess in g

solely

work provided and of

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work

integrity this

the

and courses

of any sale will

part

work

k 20 lb>ft

13–79. The airplane, traveling at a constant speed of 50 m>s, is executing a horizontal turn. If the plane is banked at u = 15°, when the pilot experiences only a normal force on the seat of the plane, determine the radius of curvature r of the turn. Also, what is the normal force of the seat on the pilot if he has a mass of 70 kg.

SOLUTION +c aFb = mab;

NP sin 15° - 7019.812 = 0 NP = 2.65 kN

Ans.

+ F

;a n

= man;

NP cos 15° = 70a r b r = 68.3 m

Ans.

laws

teaching Web) Dissemination copyright Wide .

States United

instructors permitted . World

of learning the is not on

and

use

the student for (including work by

protected

of the assess in g

solely

work provided and of

This is

work

integrity this

the

and courses

part

of any sale

will

photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

*13–80. A 5-Mg airplane is flying at a constant speed of 350 km>h along a horizontal circular path of radius r = 3000m. Determine the uplift force L acting on the airplane and the banking angle u. Neglect the size of the airplane.

L u

SOLUTION Free-Body Diagram: The free-body diagram of the airplane is shown in Fig. (a). Here, an must be directed towards the center of curvature (positive n axis). km 1000 m Equations of Motion: The speed of the airplane is v = ¢350 = 97.22 m>s. Realizing that an = v

h ≤ ¢ 1 km

≤ ¢ 3600 s ≤

= 97.22 = 3.151 m>s and referring to Fig. (a), r 3000

+ c ©Fb = 0; +

T cos u - 5000(9.81) = 0

(1)

; ©Fn = man;

T sin u = 5000(3.151)

(2) laws teaching

Solving Eqs. (1) and (2) yields

or Web)

Dissemination copyright

u = 17.8°

Wide

T = 51517.75 = 51.5 kN

Ans.

instructors permitted States . World

United

of learning the is not on

and

use

the student for (including work by

protected

of the assess in g

solely

work provided and of

This is

work

integrity this

the

and courses part

of any sale will

13–81. A 5-Mg airplane is flying at a constant speed of 350 km>h along a horizontal circular path. If the banking angle u = 15°, determine the uplift force L acting on the airplane and the radius r of the circular path. Neglect the size of the airplane.

L u

SOLUTION Free-Body Diagram: The free-body diagram of the airplane is shown in Fig. (a). Here, an must be directed towards the center of curvature (positive n axis). km Equations of Motion: The speed of the airplane is v = a 350 2

= 97.22 m>s. Realizing that an = r + c ©Fb = 0;

h b a 1 km b a 3600 s b

97.22

= r

and referring to Fig. (a),

L cos 15° - 5000(9.81) = 0 L = 50780.30 N = 50.8 kN

+ ; ©Fn = man;

1000 m

Ans.

97.22 50780.30 sin 15° = 5000¢

≤

laws

teaching Web) Dissemination copyright Wide . r = 3595.92 m = 3.60 km

Ans.

instructors permitted States . World

United

of learning the is not on

use

and

the student for (including work by

protected

of the assess in g

solely

work

work provided and of

integrity this

This is and

the courses

part

of any sale will

photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

13–82. The 800-kg motorbike travels with a constant speed of 80 km> h up the hill. Determine the normal force the surface exerts on its wheels when it reaches point A. Neglect its size.

2 y

100 m

SOLUTION 1>2

Geometry: Here, y = 22x

dy . Thus,

the hill slope at A makes with the horizontal is

22 1>2

d y

and dx2

3>2

. The angle that

dy u = tan

-1

2 = tan

b 2 x = 100 m

-1

2x1>2

The radius of curvature of the hill at A is given by dy

B1 +

rA =

dxb

3>2

x = 100 m

Free-Body Diagram: The free-body diagram of the motorcycle is shown copyright

Here, an must be directed towards the center of curvature (positive n axis).

Equations of Motion: The speed of the motorcycle is km v=a 80

1000 m

1 km

Thus, a n=

22.22

3600 s

R+ ©Fn

2849.67 = man;

800(9.81)cos 4.045° N = 7689.82 N =

This and

sale

≤

13–83. The ball has a mass m and is attached to the cord of length l. The cord is tied at the top to a swivel and the ball is given a velocity v0. Show that the angle u which the cord makes with the vertical as the ball travels around the circular path must

satisfy the equation tan u sin u = v 0>gl. Neglect air resistance and the size of the ball.

u l

SOLUTION v2

+ : ©Fn

= man;

+ c ©Fb = 0; Since r = l sin u

T sin u = ma r b T cos u - mg = 0 T= mv

2 0

a l

mv0 2 l sin u

cos u

b a sin u b = mg

tan u sin u = v

Q.E.D.

laws

teaching Web) Dissemination copyright Wide .

States United

instructors permitted . World

of learning the is not on

and

use

for

the student (including work

protected

of the assess in g

solely

work provided and of

This is

work

integrity this

the

and courses

part

of any sale will

photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

*13–84.

The 5-lb collar slides on the smooth rod, so that when it is at A it has a speed of 10 ft>s. If the spring to which it is attached has an unstretched length of 3 ft and a stiffness of k = 10 lb>ft, determine the normal force on the collar and the acceleration of the collar at this instant.

y 10 ft/s A 1

y 8 –– x

SOLUTION 2

y=8-

1x 2

dy - dx

= tan u= x 2 x = 2 = 2

u = 63.435° 2 ft

d y

dx 2 = - 1 dy 2

B1 + a dx b R 2 d y 2 dx

(1 + (- 2) )

3 2

= 11.18 ft

|- 1| laws

teaching Web) Dissemination co p yr i g h t

(2) + (6)

OA = 2

= 6.3246

instructors

States

of United use

tan f =

; f = 71.565° f or

assessing 32.2

N = 24.4 lb

This is

+ R©Ft = mat;

any

the

11.18

(10) ba

Ans. 5 b at

32.2

their

at = 180.2 ft>s

= the

and courses part

permitted World

the is not on and

work5 is solely work provided and of integrity

5 cos 63.435° - N + 33.246 cos 45.0°

the student work (including

by protect ed

+ b©Fn = man;

Wide

2 2

sale will

an = v2 r

= (10)

a=2

(180.2) + (8.9443)

= 8.9443 ft>s

11.18

a = 180 ft>s

Ans.

photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

13–85. The spring-held follower AB has a weight of 0.75 lb and moves back and forth as its end rolls on the contoured surface of the cam, where r = 0.2 ft and z = 10.1 sin u2 ft. If the cam is rotating at a constant rate of 6 rad>s, determine the force at the end A of the follower when u = 90°. In this position the spring is compressed 0.4 ft. Neglect friction at the bearing C.

SOLUTION z = 0.1 sin 2u # z = 0.2 cos 2uu $ #2 #

z = - 0.4 sin 2uu #

$ + 0.2 cos 2uu

u = 6 rad>s ##

u =0 $ z = - 14.4 sin 2u $

aF = ma ; z

0.75 FA - 12(0.1 sin 2u + 0.3) = For u = 45°,

teaching

Web)

Wide

permitted

32.2(- 14.4 sin 2u) States

United

instructors World .

learningthe is not on

0.75

FA - 12(0.4) =

laws Dissemination

FA - 12(z + 0.3) = mz

32.2 (- 14.4)

FA = 4.46 lb

use

for

student

protected

the

and work

Ans.

of the

assess in g

solely

work provided and of this This is

work

integrity

the

and courses part of any sale will

Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

13–86. Determine the magnitude of the resultant force acting on a 5-kg particle at the instant t = 2 s, if the particle is moving along a horizontal path defined by the equations r = (2t + 10) m 2

and u = (1.5t - 6t) rad, where t is in

SOLUTION r = 2t + 10|t= 2 s = 14

# r =2

$ r =0 u = 1.5t #

- 6t

u = 3t - 6 t= 2 s = 0

$ u=3 #2

=0-0 =0

- ru

ar = r

laws

au = ru + 2ru = 14(3) Hence,

+ 0 = 42

©Fr = mar;

Fr = 5(0) = 0

©Fu = ma u;

Fu = 5(42) = 210 N

instructors

teaching Web) Dissemination copyright Wide .

States

F = 2(Fr) + (F u) = 210 N

United

of learning the is not on andAns.

use

for

permitted

. World

the student (including work

protected is

solely

work provided and of

This is

of the assess in g

work

integrity this

the

and courses

part

of any sale will

photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

13–87. The path of motion of a 5-lb particle in the horizontal plane is described in terms of polar coordinates as r = (2t + 1) ft and u

= (0.5t - t) rad, where t is in seconds. Determine the magnitude of the resultant force acting on the particle when t = 2 s.

SOLUTION # r = 2t + 1|t= 2 s = 5 ft

r = 2 ft>s

r =0

au = ru + 2ru

= 5(1) + 2(2)(1) = 9 ft>s 5

©Fr = mar;

Fr = 32.2 (- 5) = - 0.7764 lb 5

©F u = ma u; 2

Fu = 32.2

u = 1 rad>s

u = 0.5t - t|t= 2 s = 0 rad u = t - 1|t= 2 s = 1 rad>s $ #2 2 2 = ar = r - ru 0 - 5(1)= - 5 ft>s $ ## 2

(9) = 1.398 lb

laws teaching

.Dissemination

F = 2F r

+ Fu

= 2(- 0.7764)

+ (1.398) = 1.60 lb

Web) or

An . Wide

States

United

instructors World permitted

of learning the is not on

and

use

the student for (including work by

protected

of the assess in g

solely

work provided and of

This is

work

integrity this

the

and courses

part

of any sale

will

photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

*13–88. A particle, having a mass of 1.5 kg, moves along a path defined 2 3

and z = 16 - t 2 m, where t is in seconds. Determine the r, u, and z components of force which the path exerts on the particle

SOLUTION

u=t

u = 2t| t = 2 s = 4 rad>s

u = 2 rad>s

= 3 m>s

z=6-t

$ r =0 $

r = 4 + 3t| t = 2 s = 10 m

z = - 3t

z = - 6t| t = 2 s = - 12 m>s

ar = r - r u = 0- 10(4) = - 160 m>s $ # 2 au = ru + 2ru = 10(2) + 2(3)(4) = 44 m>s

az = z = - 12 m>s ©F = ma ; Fr = 1.5(- 160) = - 240 N r

©F

Ans. Anslaw s.

= ma ; u

= 1.5(44) = 66 N

Web)

teaching

.Dissemination

©Fz = maz;

Wide .

Fz - 1.5(9.81) = 1.5(- 12)

Fz = - 3.28 N

A .

States

United

instructors

World permitted

of learning the is not on

and

use

the student for (including work by

protected is

of the

assess in g

work

solely

work provided and of integrity this This is

the

and courses

part

of any sale will

photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

13–89.

Rod OA rotates counterclockwise with a constant angular velocity of u = 5 rad>s. The double collar B is pin-connected together such that one collar slides over the rotating rod and the other slides over the horizontal curved rod, of which the shape is described by the equation r = 1.512 - cos u2 ft. If both collars weigh 0.75 lb, determine the normal force which the curved rod exerts on one collar at the instant u = 120°. Neglect friction.

A B r

SOLUTION O

# $ Kinematic: Here, u = 5 rad>s and u = 0. Taking the required time derivatives at u = 120°, we have

= 5 rad/s

r = 1.5(2 - cos u)|u = 120° = 3.75 ft # r# = 1.5 sin uu|u = 120° = 6.495 ft>s # 2 )| $ $ 2 u = 120° = - 18.75 ft>s r = 1.5(sin uu + cos uu Applying Eqs. 12–29, we have

r = 1.5 (2 – cos ) ft.

- ru = - 18.75 - 3.75(5 ) = - 112.5 ft>s = ru + 2ru = 3.75(0) + 2(6.495)(5) = 64.952 ft>s

ar = r au

laws

1.5(2 - cos u)

Web) teaching Dissemination

tan c =

dr>du

1.5 sin u

= 2.8867

c = 70.89°

States

World

Equation of Motion: The angle c must be obtained first. instructors

u = 120°

United

learning the is not on

and F a r

= mar ;

- N cos 19.11° =

the student by use

(- 112.5)

Applying Eq. 13–9, we have

work

for

0.75

protected

assessing 32.2

is solely work work provide d and of this integrity

N = 2.773 lb = 2.77 lb

F a

= mau ;

FOA + 2.773 sin 19.11°This

(64.952) = 32.2 and

part the

courses

is 0.75 of destroy any

FOA = 0.605 lb

their sale

will

the

Ans.

Wide . permitted

13–90. z

The boy of mass 40 kg is sliding down the spiral slide at a constant speed such that his position, measured from the top of the chute, has components r = 1.5 m, u = 10.7t2 rad, and z = 10.5t2 m, where t is in seconds. Determine the components of

force Fr, Fu , and Fz which the slide exerts on him at the instant t = 2 s. Neglect the size of the boy.

SOLUTION z

r = 1.5 #

u = 0.7t $

r =r $ u=0

z = - 0.5t

$ ar = r - r(u) $

u = 0.7

= - 0.5

= 0 - 1.5(0.7) 2 = - 0.735

au = ru + 2ru = 0 $ az = z =0 ©Fr = mar;

Fr = 40(- 0.735) = - 29.4 N

©Fu = mau;

Fu = 0

Ans. Ans.la ws

Web)

teaching

Dissemination

©Fz = maz;

Fz -

40(9.81) = 0

Wide

Fz = 392 N instructors permitted States World

Ans. .

United

of learning the is not on

and

use

for

the student (including work

protected is

solely

work provided and of

This is

of the assess in g

work

integrity this

the

and courses part

of any sale will

1.5 m

13–91. The 0.5-lb particle is guided along the circular path using the slotted# arm guide. If the arm has an$ angular velocity u = 4 rad>s

and an angular acceleration u = 8 rad>s at the instant u

= 30°, determine the force of the guide on the particle. Motion occurs in the horizontal plane.

0.5 ft

SOLUTION

0.5 ft

r = 2(0.5 cos u) = 1 cos u

# r

= - sin uu # 2

r = - cos uu #

$ - sin uu $

At u = 30°, u = 4 rad>s and u = 8 rad>s r = 1 cos 30° = 0.8660 ft

# r = - sin 30° (4) = - 2 ft>s .. 2 2 r = - cos 30° (4) - sin 30° (8) = - 17.856 ft>s $ #2 2 2 ar = r - ru = - 17.856 - 0.8660(4) = - 31.713 ft>s $ ## au = ru + 2ru Q + ©Fr

= mar;

= 0.8660(8) + 2(- 2)(4) = - 9.072 ft>s 0.5 (- 31.713) - N cos 30° = 32.2

laws teaching

N = 0.5686 lb States

United

0.5

or Web) .

World

instructors

of learning the is not on

use

+ a©Fu = mau;

F - 0.5686 sin 30° = F = 0.143 lb

32.2 (- 9.072) by the

and student for (including work protected of the Ans. is solely work assessing this

workprovided and of integrity This is the and courses part of their destroy any

sale will

*13–92. Using a forked rod, a smooth cylinder C having a mass of 0.5 kg is forced to move along the vertical slotted path r = 10.5u2 m, where u is in radians. If the angular position 2 of the arm is u = 10.5t 2 rad, where t is in seconds, determine the force of the rod on the cylinder and the normal force of the slot on the cylinder at the instant t = 2 s. The cylinder is in contact with only one edge of the rod and slot at any instant.

u r

SOLUTION # r = 0.5u

r = 0.5u u = 0.5t

r = 0.5u

u=1

At t = 2 s, u = 2 rad = 114.59°

u =2 rad>2

u = 1 rad>s

r=1m

r = 1 m>s

r = 0.5 m>s

tan c =

r 0.5(2) dr>du = 0.5

c = 63.43°

ar = r - ru

= 0.5 - 1(2)

= - 3.5

laws

# au = ru + 2r u =

Web)

teaching

1(1) + 2(1)(2) = 5

Dissemination or

Wide

instructors

permitted

States

World

of learning Ans.not

NC = 3.030 = 3.03 N

United

on the is

use

for student work by the (including

+ b© Fu = ma u;

F - 3.030 sin 26.57° + 4.905 sin 24.59° F = 1.81 N

= 0.5(5)

protected of the assessing is solely work work provided and of integrity this This is

and

Ans.

the

and courses

part

any

sale will

0.5 u

13–93.

If arm OA rotates with a constant clockwise angular velocity of u = 1.5 rad>s. determine the force arm OA exerts on the smooth 4-lb cylinder B when u = 45°. A B r

SOLUTION

u u

Kinematics: Since the motion of cylinder B is known, ar and au will be determined 4

first. Here, r = cos u or r = 4 sec u ft. The value of r and its time derivatives at the instant u = 45° are

4 ft

r = 4 sec u |u= 45° = 4 sec 45° = 5.657 ft #= # r 4 sec u(tan u)u| = 4 sec 45° tan 45°(1.5) = 8.485 ft>s # $ u= 45°

4 Csec u(tan u)u + u Asec u sec uu + tan u sec u tan uu B D $ 2 #

= 4 Csec u(tan u)u + sec uu

+ sec u tan uu

D u= 45°

laws in teaching

Using the above time derivatives,

.Disseminationpermitted

= 38.18 ft>s

3 2 + sec 45°(1.5) +

=4 sec 45° tan 45°(0)

instructors

ar = r - ru $

= 38.18 - 5.657 1.5 #

Unit e d

not

States World learningon of the is

= 25.46 ft> s

Wide Web)

sec 45° tan 45°(1.5)

use 2

and

student

work (including

for

= 25.46the ft>

au = ru - 2ru = 5.657(0) + 2(8.485)(1.5) Fig. a,

assessing solely work

Equatio ns of Motion: By referring to the free-body diagram of the cy linderthe

N cos 45° - 4 cos 45°

provide

2.2d 4

integrity this and of

( 25. 4 6)

work =

This is

©Fu = ma u; N = 8.472 lb

part the

and courses any their of

sale will

FOA = 12.0 lb

shown in

(25.46) 4

32.2 Ans.

13–94. The collar has a mass of 2 kg and travels along the smooth

horizontal rod defined by the equiangular spiral r = 1e 2 m, where u is in radians. Determine the tangential force F and the

normal force N acting on the collar when u = 90°, if the force F maintains a constant angular motion u = 2 rad>s.

SOLUTION u

r=e

r =eu $

r = e (u)

+eu

At u = 90° #

u = 2 rad>s $

u =0 r = 4.8105 laws

teaching

r = 9.6210

= 19.242 - 4.8105(2)

()

r = 19.242

World #

States

# 2

au = ru + 2ru = 0 + 2(9.6210)(2) =

tan c =

instructors

not

38.4838 m>s

the is on and

use

u u

= e >e = 1

by the student for (including work protected of

AduB

c = 45°

assessing

is solely work provided and of

+ c a Fr

= mar;

- NC cos 45° +

F cos 45°

2(0)

This is +

and

;a u

Wide permitted

learning

United

Web) or

Dissemination

a= r-ru

F = 54.4 N

the

this the

course s

part

NC = 54.4 N

an y

their

work integrity

destroy

Ans.

sale will

Ans.

13–95. The ball has a mass of 2 kg and a negligible size. It is originally traveling around the horizontal circular path of radius #0 = 0.5 msuch that the angular rate of rotation is u0 = 1 rad>s. If the attached cord ABC is drawn down through the hole at a constant speed of 0.2 m>s, determine the

A r Bu r0

tension the cord exerts on the ball at the instant r = 0.25 m. Also, compute the angular velocity of the ball at this instant. Neglect the effects of friction between the ball and horizontal plane. Hint: First show that the equation of motion in the u $

0.2 m/s

ru + 2ru = 11>r21d1r u2>dt2 = 0. # When integrated, r u = c, where the constant c is determined from the problem data. direction yields

2au =

SOLUTION $

F= a u Thus,

1d 2 # = mc r dt (r u ) d = 0

mau;

# 0 = m[ru + 2r u]

d(r u) = 0 2

ru=C

laws

(0.5) (1) = C = (0.25) u

teaching Web) Dissemination copyright

# u = 4.00 rad>s Since r = - 0.2 m>s, ar = r

Ans. Wide

instructors not permitted States . World

$ #2

2 - r (u) = 0 - 0.25(4.00) =

Uniteduse of learning the is the student (including work for

- 4 m>s a Fr = mar; - T = 2(- 4)

on a nd

T=8N

by protected of assessing the is solely work work provided and of integrity this This is

Ans.

the

and courses

part

of any sale will

*13–96. A

The particle has a mass of 0.5 kg and is confined to move along the smooth horizontal slot due to the rotation of the arm OA. Determine the force of the rod on the particle and the normal

force of the slot on the particle when u = 30°. The rod is rotating with a constant angular velocity u = 2 rad>s. Assume the particle contacts only one side of the slot at any instant.

u 2 rad/s

0.5 m u

SOLUTION 0.5 r = cos u = 0.5 sec u # # r = 0.5 sec u tan uu $ # $ 2 + sec u(sec uuD u + sec u tan uuF r = 0.5E C (sec u tan uu) tan u # # $ 2 3 2 = 0.5C sec u tan uu + sec uu + sec u tan uuD # $ When u = 30°, u = 2 rad>s and u = 0

r = 0.5 sec 30° = 0.5774 m $

r = 0.5 sec 30° tan 30°(2)

= 0.6667 m>s

r = 0.5C sec 30° tan 30°(2)

$ ar = r - ru

+ sec

laws

= 3.849 -

sec 30° t an 30°(0)D

au =

Q + ©Fr = mar;

. learning

2.667 m>s

ru + 2ru = 0.5774(0) + 2(0.6667)(2) =

Wide

States World permitted instructors not

0.5774(2) = 1.540 m>s 2 = 3.849 m>s

Web)

Dissemination or

30°(2) +

United use

the is on and

N cos 30° - 0.5(9.81) cos 30° = 0.5(1.for540) by the

N = 5.79 N

student work (including

Ans.

protectedsolely assessing is

work

the

and of integrity p r o v id ed

+ R©Fu = ma u;

F + 0.5(9.81) sin 30° -

work5.79 s n 30°

F = 1.78ThisN courses part the and

=this0.5(2.667)

Ans.

destroy

of any sale will

13–97. Solve Problem 13–96 if the arm has an angular acceleration $ # 2

of u = 3 rad/s and u = 2 rad /s at this instant. Assume the particle contacts only one side of the slot at any instant.

· u 2 rad/s

0.5 m u

SOLUTION 0.5 r = cos u = 0.5 sec u # # r = 0.5 sec u tan uu

uu) D u

+ sec u tan uuF

= 0.5E C (sec u tan uu) tan u +

sec u(sec

3 #2 $ 2 #2 = 0.5C sec u tan uu + sec uu + sec u tan uuD # $

When u = 30°, u = 2 rad>s and u = 3 rad>s

r = 0.5 sec 30° = 0.5774 m laws teaching

r = 0.5 sec 30° tan 30°(2) = 0.6667 m>s

$ r

= 4.849 m>s

= 0.5C sec 30° tan

a r =

Dissemination

30°(2)

+ sec 30°(2)

= 4.849 - 0.5774(2)

r - ru

= ru + 2r

= 2.5396 m>s

States

World permitted instructors

. learning

u = 0.5774(3) + 2(0.6667)(2) = 4.3987 m> s

Q + ©Fr = mar;

+ sec 30° tan 30°(3) D

# u

Web)

Wide

the is not

Uniteduse

on and

N cos 30° - 0.5(9.81) cos 30° = 0.5(2.5396)student for

work (including

by the

N = 6.3712 = 6.37 N + R©Fu = mau;

protected of assessing the is solely work work provided and ofintegrity

F + 0.5(9.81) sin 30° - 6.3712 sin 30°this= F= 2.93ThisNis part the

Ans.

0.5(4.3987)

Ans.

and courses of any sale will

13–98. The collar has a mass of 2 kg and travels along the smooth u horizontal rod defined by the equiangular spiral r = 1e 2 m, where u is in radians. Determine the tangential force F and the normal force N acting on the collar when u =# 45°, if the force

F maintains a constant angular motion u = 2 rad>s.

SOLUTION u

r=e #

θ #

r =e u

$ 2 u u r = e (u) + e u #

At u = 45° #

u = 2 rad>s $ u=0 r = 2.1933

r# = 4.38656

laws Web) teaching Dissemination or

$ r = 8.7731 $ # r

Wide

instructors

not permitted

= r - r(u)

States

= 8.7731 - 2.1933(2) = 0

World

. learning

au = r u + 2 r u = 0 + 2(4.38656)(2) = 17.5462 m>s

of United

tan c =

u u

and

use

= e >e = 1

dr a

the is on

the student work (including

r c = u = 45°

assessing

solelywork

the

workprovided and of thisintegrity

Q+ F

2(0)

a r = mar ;

- N C cos 45° + F Thiscosis45° = part

the

and courses any

sale will

+ a aF u =

mau ;

F sin 45° + NC sin 45°

= 2(17.5462)

N = 24.8 N

Ans.

F = 24.8 N

Ans.

r=e

13–99. For a short time, the 250-kg roller coaster car is traveling along the spiral track such that its position measured from the top of the track has components r = 8 m, u = 10.1t + 0.52 rad, and z = 1 - 0.2t2 m, where t is in

r =8m

seconds. Determine the magnitudes of the components of force which the track exerts on the car in the r, u, and z directions at the instant t = 2 s. Neglect the size of the car.

SOLUTION # Kinematic: Here, r = 8 m, r t = 2s, we have

$ = r = 0. Taking

the required time derivatives at $ u

u = 0.1t + 0.5|t = 2s = 0.700 rad

u = 0.100 rad>s

z = - 0.2t|t = 2 s = - 0.400 m z = - 0.200 m>s Applying Eqs. 12–29, we have $2 $ 2 2 ar = r - ru = 0 - 8(0.100 ) = -0.0800 m>s $ # au = r u + 2 ru = 8(0) + 2(0)(0.200) = 0 $

z =0

az = z = 0 Equation of Motion:

laws teaching

or Web)

= 250(- 0.0800) = - 20.0 N

Ans.

Dissemination

Wide

©Fr = mar ; ©Fu = mau ;

Stat e s

©Fz = maz ;

instructors World permitted learning the is

Ans. Fu = 250(0) = 0

United

student

not

use

- 250(9.81) = 250(0)

by the

protected

work

(including of the

= 2.45forkN

and Ans.

assessing is

solely

work provided and of

This is

work integrity this

the

and courses part

of any sale will

*13–100. The 0.5-lb r = 2rc cos# v$elocity

ball is guided along the vertical circular path

u using the arm OA. If the arm has an angular u = 0.4 rad>s and an angular acceleration

u = 0.8 rad>s at the instant u = 30°, determine the force of the arm on the ball. Neglect friction and the size of the ball. u

Set rc = 0.4 ft.

SOLUTION r = 2(0.4) cos u = 0.8 cos u #

r =

- 0.8 sin uu $ #2 r = - 0.8 cos uu #

$ - 0.8 sin uu $

At u = 30°, u = 0.4 rad>s, and u = 0.8 rad>s r = 0.8 cos 30° = 0.6928 ft

# r = - 0.8 sin 30°(0.4) = - 0.16 ft>s $ 2 r

ar = r - ru + Q©Fr

= mar ; au

= - 0.8 cos 30°(0.4) - 0.8 sin 30°(0.8) = - 0.4309 ft>s $ #2 2 2 $ ## = - 0.4309 - 0.6928(0.4)

( - 0.5417)

2ru = 0.6928(0.8) + 2( - 0.16)(0.4)

or .

Dissemination Web) copyright Wide N = 0.2790 lb permitted

= - 0.5417 ft>s

N cos 30° - 0.5 sin 30° =

= ru +

laws teaching

instructors

= 0.4263 ft>s

0.5

States

. World

0.5Uni of learningthe is not

ted

use

FOA + 0.2790 sin 30° - 0.5 cos 30° = by

(0.4263) for

and

student (including work

32.2t he protected

assessing FOA = 0.300 lb

is solely work provided and of

This is

work integrity this

the

Ans.

the

and courses part of any sale will

13–101.

The ball of mass m is guided along the vertical circular path

r = 2rc cos u usin#g the arm OA. If the arm has a constant angular velocity u0, determine the angle u … 45° at which

the ball starts to leave the surface of the semicylinder. Neglect friction and the size of the ball.

r c

u O

SOLUTION r = 2rc cos u # r

= - 2rc sin uu

$ #2 $ r = - 2rc cos uu - 2rc sin uu # $ Since u is constant, u = 0. $

#2 ru =

ar = r -

+ Q©Fr = mar;

2 - 2rc cos uu0 - 2rc cos uu0 =

# - mg sin u = m( - 4r cos uu ) 0 # c 4r c u

tan u =

u = tan

-1

2 - 4rc cos uu0

# 4r u 2 c

≤

¢ g

laws or teaching Web) Dissemination copyright Wide .

States

instructors permitted . World

of learning the is not

United

use

and

the student for (including work by

protected is

work provided and of

This is

of the assess in g

solely

work

integrity this

the

and courses part

of any sale will

photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

13–102. Using a forked rod, a smooth cylinder P, having a mass of 0.4 kg, is forced to move along the vertical slotted path r = 10.6u2 m, where u is in radians. If the cylinder has a constant speed of P

vC = 2 m>s, determine the force of the rod and the normal force of the slot on the cylinder at the instant u = p rad. Assume the cylinder is in contact with only one edge of the rod and slot at any instant. Hint: To obtain the time derivatives necessary to compute the cylinder’s

acceleration components ar and au, take the first and second time derivatives of r = 0.6u. Then,

# for further information,

use Eq. 12–26 to determine u. Also,# take the time derivat$ive of Eq. 12–26, noting that vC = 0, to determine u.

SOLUTION #

r = 0.6 u

vr = r 2

r = 0.6 u

# # vu = ru = 0.6uu

# 0.6u 2

v = r+ a rub # 2

2 = a 0.6ub

+ a 0.6uub

laws teaching

0.621 + u 2

Dissemination copyright

# u=-

0 = 0.72u u + 0.36a 2uu + 2u u ub

At u = p rad,

Web) or

Wide

= 1.011 rad>s

instructors +u

States

permitted World

learning

0.62 1 + p $

(p)(1.011)

u= r = 0.6(p) = 0.6 p m

United

= - 0.2954 rad>s

1+p

r# = 0.6(1.011) = 0.6066

#2 $ r = 0.6(- 0.2954)

ar = r - ru $

work

solely

0.6u tan c = dr>du = 0.6 = u = p

and

the student work (including of the

assess in g

and of

and coursesany their

by for

= - 0.1772 m>s = - 0.1772 - 0.6 p(1.011)This= -is2.104 mpart>s

use

protected

the is not on

work

integrity this the 2

destroy

sale will

c = 72.34°

- N cos 17.66° = 0.4(- 2.104) N = 0.883 N - F + 0.4(9.81) + 0.883 sin 17.66° = 0.4(0.6698) F = 3.92 N

Ans. Ans.

0.6u

13–103. A ride in an amusement park consists of a cart which is supported by small wheels. Initially the cart is traveling in a circular path# of radius r0 = 16 ft such that the angular rate of r

rotation is u0 = 0.2 rad>s. If the# attached cable OC is drawn

u O

the effects of friction. Hint: First show that the equation of

= .. + . . = motion in# the u direction yields #au ru 2ru

11>r2 d1r u2>dt = 0. When integrated, r u = c, where the constant c is determined from the problem data.

-T=¢

# 2

32.2 ≤ ¢r - ru ≤ 400 $ #

(1)

0 = ¢ 32.2 ≤ ¢ru + 2r u ≤ d # #

From Eq. (2), ¢ r ≤ dt ¢r u≤

(2)

r u=c laws

teaching

Web)

Dissemination copyright

Since u0 = 0.2 rad>s when r 0 = 16 ft, c = 51.2.

Statesinstructors

4 ft,

World permitted not

. of

Wide

≤

u = ¢ (4)2 Hence, when r =

learning the is

51.2 $

Uniteduse on

Since r = - 0.5 ft>s, r = 0, Eq. (1) becomes

and

for

student work (including the

-T=

32.2 a 0 - (4)(3.2) b

400 T = 509 lb

protected of assessing the solely is work work provided and of integrity this This is

the

and courses part

of any sale will

· u

inward at a constant speed of r = - 0.5 ft >s, determine the tension it exerts on the cart at the instant r = 4 ft. The cart and its passengers have a total weight of 400 lb. Neglect

Ans.

*13–104. #

$ arm is rotating at a rate of u = 5 rad>s when u = 2 rad>s

The

and u = 90°. Determine the normal force it must exert on the 0.5-kg particle if the particle is confined to move along the slotted path defined by the horizontal hyperbolic spiral ru = 0.2 m.

0.2 —

u r ·

·· u 5 rad/s, u 90 u

SOLUTION p u = 2 = 90° # u = 5 rad>s $

u = 2 rad>s

r = 0.2>u = 0.12732 m

- 2

r = - 0.2 u $ r = - 0.2[- 2u ar = r - r(u) au = r u + 2 r

u = - 0.40528 m>s -3

-2

(u) + u u] = 2.41801 = 2.41801 - 0.12732(5) = - 0.7651 m>s

laws

$ tan c =

Dissemination Web) copyright Wide dr

(

)

- 0.2u

instructors World permitted .

-2

0.2>u p

learning

of United

-1

c = tan

teaching

u = 0.12732(2) + 2(- 0.40528)(5) = - 3.7982 m>s

= - 57.5184°

(- 2 )

Np cos 32.4816° = 0.5(- 0.7651)

use

for

protected

the is not on

student the

and work

of the

assess in g

;

= ©F u

mau

;

= - 0.453 N

F + Np sin 32.4816° =

provided

work

integrity

0.5( - 3.7982) work

This is

F = - 1.66 N

solely

and courses any of

and

this

part the

Ans.

sale will

2 rad/s

13–105.

The forked rod is used to move the smooth 2-lb particle around

the horizontal path in th e shape of a limaçon, r = (2 + cos u) ft. If at all times u = 0.5 rad>s, determine the force which the rod exerts on the particle at the instant u = 90°. The fork and path contact the particle on only one side.

2 ft

r u

· u

3 ft

SOLUTION r = 2 + cos u

= - sin uu

$ #2 $ r = - cos uu - sin uu #

At u = 90°, u = 0.5 rad>s, and u = 0 r = 2 + cos 90° = 2 ft

# r

= - sin 90°(0.5) = - 0.5 ft>s

= - cos 90°(0.5) - sin 90°(0) = 0 $ #2 2

laws teaching

tan c =

= 2 + cos u

dr>du

- sin u

= -2

- 63.43°

u= 90°

United

2 32.2

Statesinstructors not of learning the is on

use

for

and

student (including work

by + c ©Fr

= mar;

- N cos 26.57° =

( - 0.5)

N = 0.0the3472 lb

protectedsolely assessing

; ©Fu = mau;

F - 0.03472 sin 26.57° =is

the

and of thisintegrity

F = - 0.0155 lbprovided 32.2 This is and

( - 0.5) work work

Ans. the

courses part

of any sale will

Dissemination Web) copyright Wide . World permitted

= r - ru = 0 - 2(0.5) = - 0.5 ft>s

13–106. The forked rod is used to move the smooth 2-lb particle around

the horizontal path in the shape of a limaçon, r = (2 + cos u) ft. If at all times u = 0.5 rad>s, determine the force which the rod exerts on the particle at the instant u = 60°. The fork and 2 ft

path contact the particle on only one side.

u· u 3 ft

SOLUTION r=2

+ cos u # = - sin uu

r # 2 $ r = - cos uu #

$ sin uu $

At u = 60°, u = 0.5 rad>s, and u = 0 r = 2 + cos 60° = 2.5 ft

# r = - sin 60°(0.5) = $

r = - cos 60°(0.5)

r = r - ru

- 0.4330 ft>s - sin 60°(0) = - 0.125 ft>s

laws

= - 0.125 - 2.5(0.5)

teaching

au = ru + 2ru = 2.5(0) + 2( - 0.4330)(0.5) = - 0.4330 ft>s r 2 + cos u tan c = = = - 2.887 c = - 70.89° dr>du - sin u u= 60°

- N cos 19.11°

( - 0.75)

F - 0.04930 sin 19.11° =

is ( 32.2

F = - 0.0108 lbprovided

This

and lb

for

student (including work of the

0.4330)sole ly

permitted

not

N =0.04930the

2 protected

World

of learning the is

use

32.2

+a©Fu = mau;

States instructors United

Web)

+ Q©Fr = mar;

- 0.75 ft>s

work

assessing

work

and of this integrity

the

Ans.

and

courses part of their destroy any

sale

will

photocopying, recording, or likewise. For information regarding permission(s), write to:

Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

13–107.

The forked rod is used to move the smooth 2-lb particle around the horizontal path in the shape of a limaçon, r = (2 + cos u) ft.

If u = (0.5t ) rad, where t is in seconds, determine the force which the rod exerts on the particle at the instant t = 1 s. The fork and path contact the particle on only one side.

2 ft

r u

· u

3 ft

SOLUTION u = 0.5t

r = 2 + cos u

# r = - sin uu

u =t

$ 2 r = - cos uu #

$ - sin uu

$ 2 u = 1 rad>s $

At t = 1 s, u = 0.5 rad, u = 1 rad>s, and u = 1 rad>s r = 2 + cos 0.5 = 2.8776 ft

r# = - sin 0.5(1) = - 0.4974 ft>s $

r = - cos 0.5(1) - sin 0.5(1) #2 $ $ ar

= - 1.357 ft>s

= r - ru

tan c =

laws teaching

dr>du

= 2 + cos u

- sin u

= - 6.002

c = - 80.54°

Statesinstructors

u= 0.5 rad

use

for

not

of learning the is

United

2 32.2

and

student (including work

- N cos 9.46° =

( - 4.2346)

=the0.2666 lb

protected

assessing

+ a©Fu = mau;

F - 0.2666 sin 9.46° =

(1.9187)solely

32.2 This is

the

and courses part of any sale

will

the

work

work and of thisintegrity provided

F =0.163 lb

Dissemination Web) copyright Wide . World permitted

= - 1.375 - 2.8776(1) = - 4.2346 ft>s r

Ans.

*13–108. The collar, which has a weight of 3 lb, slides along the smooth rod lying in the horizontal plane and having the shape of a P

parabola r = 4>11 - cos u2, where u is in radians and r is #in feet. If the collar’s angular rate is constant and equals u = 4 rad>s, determine the tangential retarding force P needed to cause the motion and the normal force that the collar exerts on the rod at the instant u = 90°.

r u

SOLUTION 4 r=

1 - cos u #

= - 4 sin u u 2 (1 - cos u)

= (1 - cos u)

- 4 sin u u

- 4 cos u (u)

8 sin u u

+ (1 - cos u) # $ u = 4, u=0

At u = 90°,

+ (1 - cos u)

r=4 r = - 16

laws teaching

$ r = r - r(u)

= 128 - 4(4)

= 64 2

States

World

learning

au = ru + 2 ru = 0 + 2(- 16)(4) = - 128

United

use

r = 1 - cos u dr

Wide

instructors

# 2

. Dissemination Web)

r = 128

$ $

on the

not

and

the student for (including work by

- 4 sin u r

is protected

1 - cos u)

provided integrity work and of this

=Thistan c =

dr (

) du

the

- 4 sin u 2 2

u = 90

= - 4 1is and

courses part any

(1 - cos u)

their sale will

P sin 45° - N cos 45° =

32.2 (64)

- P cos 45° - N sin 45°

= 32.2 (- 128)

Ans.

N = 4.22 lb

Ans.

13–109. The smooth particle has a mass of 80 g. It is attached to an elastic cord extending from O to P and due to the slotted arm guide moves along the horizontal circular path r = 10.8 sin u2 m. If the cord has a stiffness k = 30 N>m and an unstretched length of 0.25 m, determine the force of the guide on the

parti cle when u = 60°. The guide has a constant angular velocity u = 5 rad>s.

SOLUTION

0.4 m

r = 0.8 sin u

# = 0.8 cos u u

r = - 0.8 sin u (u)

# u = 5,

+ 0.8 cos uu

$ u=0

At u = 60°,

r = 0.6928

# r

$ laws teaching

r = - 17.321

or Web)

Dissemination

$ ar = r -

# 2

r(u) = - 17.321 -

0.6928(5) = - 34.641

Fs = ks; Fs = 30(0.6928 - 0.25) = 13.284 N au = ru + 2 ru = 0 + 2(2)(5) = 20

instructors States . World

United

- 13.284 + NP cos 30° =

and

34.641) by for

= mau;

of learning the is not

use

0.08(-

a + ©Fu

5 rad/s

protected solely

F - NP sin 30° = 0.08(20) F = 7.67 N

the work

and of

integrity

provi ded

work

this

This is

the

and courses part

of any sale

will

of the

assess in g

NP = 12.1 N

student work (including

Ans.

permitted

13–110. The smooth particle has a mass of 80 g. It is attached to an elastic cord extending from O to P and due to the slotted arm guide moves along the horizontal circular path r = (0.8 sin u) m. If the cord has a stiffness k = 30 N>m and an unstretched length of 0.25

the force of the guide on the $ m, determine 2 #

particle when u = 2 rad>s , u = 5 rad>s, and u = 60°.

SOLUTION

0.4 m

5 rad/s

r = 0.8 sin u u

# r = 0.8 cos u u $ #2 # u = 5,

r = - 0.8 sin u (u)

+ 0.8 cos uu

$ u =2

At u = 60°,

r = 0.6928 r# = 2

$ r = - 16.521 $

laws

Web) Dissemination or teaching

au= r u + 2 r u =

0.6925(2) + 2(2)(5) = 21.386

- r(u) = - 16.521 - 0.6928(5)

ar = r

Wide

= - 33.841

F s = ks;

Fs = 30(0.6928 - 0.25) = 13.284 N States

- 13.284 + NP cos 30° = 0.08(- 33.841) Uniteduse

+ a©Fu = mau;

F - NP sin 30° = 0.08(21.386)

F = 7.82 N

for

student

learning the is not on and work

by the

protected of assessing the is solely work work provided and of integrity

NP = 12.2 N

instructors permitted . World

Ans.

this This is

the

and courses part

of any sale will

photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

13–111. A 0.2-kg spool slides down along a smooth rod. If the rod has a constant angular rate of rotation # = 2 rad>s in the vertical plane, show that the equations of u $ motion for the spool - 4r - 9.81 sin u = 0 and are

0.8r + Ns - 1.962 cos u = 0, where Ns is the magnitude of the normal force of the rod on the spool. Using the methods of differential equations, it can be shown that the solution of the first of these equations is r = C 1 e-2t + C e2t - 19.81>82 sin 2t. If r, r , and u are

2 rad/s r

evaluate the constants C1 zero when t = 0, determine r at the instant u = p>4 rad.

and C2 to

SOLUTION # Kinematic: Here, u. = 2 rad>s and u = 0. Applying Eqs. 12–29, we have $ # $ $ 2 ar = r - ru2 = r - r(2 ) = r - 4r #

$ ##

au = ru + 2ru = r(0) + 2r(2) = 4r Equation of Motion: Applying Eq. 13–9, we have $

©Fr = mar ;

1.962 sin u =

0.2(r - 4r)

laws

teaching

Web)

Disseminationor

r - 4r -

9.81 sin u = 0

(Q.E.D.)

(1)

©Fu = mau;

Wi de

1.962 cos u -

u. Since

0.8r + Ns

= 2 rad>s, then

equation (Eq.(1)) is given by - 2t

r = C1

+ C2

Thus, #

- 2t

r = - 2 C1 e

+ 2C2

At t = 0, r = 0. From Eq.(3) 0 =

# = 0. From Eq.(4) 0 = - 2 C1 (1) + 2C2 (1) -

At t = 0, r Solving Eqs. (5) and (6) yields

Thus, 9.81 16e

r==

9.81

- 2t

-e

9.81 +

16 e

sin 2t

- sin 2tb 8

9.81 8

- sin 2t)

(sin h 2t p

At u = 2t

9.81 ,

a sin h

*13–112. The pilot of an airplane executes a vertical loop which in part follows the path of a cardioid, r = 600(1 + cos u) ft. If his speed at A (u = 0°) is a

constant vP = 80 ft>s, determine the vertical force the seat belt

must exert on him to hold him to his seat when the plane is upside down at A. He weighs 150 lb.

r 600 (1 + cos u) ft u

SOLUTION r = 600(1 + cos u)|u = 0° = 1200 ft

r = - 600 sin uu u = 0° = 0 $ $ #2 r

= - 600 sin uu

- 600 cos uu

(80)

= -600u

+ a rub #

u = 0°

# 2

= 0 + a 1200ub $

u = 0.06667 $

2vpvp = 2rr + 2a rub a ru + rub 2 $ $ 0=0+ 0 + 2r uu u = 0 # $ 2 2 2 2 ar =r - ru = - 600(0.06667) - 1200(0.06667) = - 8 ft>s $ #

or laws teaching Web) .

States

au = ru + 2ru = 0 + 0 =

150

+ © c

World permitted

instructors

; N

150

32.2

not

of b(

the is learningAns.

113 lb

United use by

on and

the student (including work protected of the is solely work assessing this for

This

workprovided and of integrity is the

and courses part of their destroy any

sale will

13–113. The earth has an orbit with eccentricity e = 0.0821 around thed sun. Knowing that the earth’s minimum distance from the sun is 6

151.3(10 ) km, find the speed at which the earth travels when it is at this distance. Determine the equation in polar coordinates

SOLUTION Ch

GMS

1 r

e = GM

- r v 2 ≤ and h = r

where C =0 ¢1

GMS

e = GMS r0 ¢1 - r0 v0

r0 v 0

≤(r0v0)

r0v0

e = ¢ GMS - 1≤

= e+1

GMS

GMS (e + 1)

= B66.73(10

- 12

)(1.99)(10

)(0.0821 + 1) = 30818 m>s = 30.8

Ans.

km>s 151.3(10 ) laws

Web) or

1 =

GMS rv

GMS

. Dissemination

0 0

66.73(10

- 12

)(1.99)(10

30 )

not

instructors

66.73(10

States 9

= 0.502(10

) cos u + 6.11(10 - 12

b cos u +

151.3(10 )D

)

(30818)

learning

United

- 12

use for

permitted

World 92

Ans.

on the is and

the student (including work

protected

of the assess in g

solely

work provided and of

This is

work

integrity this

the

and courses part

of any sale will

photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

13–114. A communications satellite is in a circular orbit above the earth such that it always remains directly over a point on the earth’s surface. As a result, the period of the satellite must equal the rotation of the earth, which is approximately 24 hours. Determine the satellite’s altitude h above the earth’s surface and its orbital speed.

SOLUTION The

period

of the satellite

around

the

circular orbit

radius

r 0 = h + r e = C h + 6.378(10 )D m is given by

2pr0 v s

24(3600) =

2p C h + 6.378(10 )D vs

2pC h + 6.378(10 ) vs =

(1)

86.4(10 3 )

r 0 = h + r e = C h + 6.378(10 )D m is given by

laws

Web) or Dissemination copyright Wide . States World permitted instructors teaching

The velocity

of the satellite

66.73(10

-12

)(5.976)(10

orbiting around

the circular orbit

)

GMe

yS =

of radius

C r0

of learning

the is

(2)no

yS =

t United

on and

use

for student protected by the of

Solving Eqs.(1) and (2),

work

assessing

= yS = 3072.32 m> 3.07thekm>s

h = 35.87(10 ) m = 35.9 Mm

is solely work work provided and of integrity this This is

the

and courses

any

sale will

part

Ans.

13–115. The speed of a satellite launched into a circular orbit about the earth is given by Eq. 13–25. Determine the speed of a satellite launched parallel to the surface of the earth so that it travels in a circular orbit 800 km from the earth’s surface.

SOLUTION For a 800-km orbit

- 12 v0 =

66.73(10

)(5.976)(10

)

B (800 + 6378)(10 )

= 7453.6 m>s = 7.45 km>s

Ans.

laws

teaching Web) Dissemination copyright Wide .

States United

instructors permitted . World

of learning the is not on

and

use

for

the student (including work

protected

of the assess in g

solely

work provided and of

This is

integrity this

the

and courses

of sale

work

part

any

will

photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

*13–116. A rocket is in circular orbit about the earth at an altitude of h = 4 Mm. Determine the minimum increment in speed it must have in order to escape the earth’s gravitational field.

h = 4 Mm

SOLUTION Circular Orbit: 66.73(10)5.976(10 )

GM e

0 vC = A = B 4000(10 ) + 6378(10 ) = 6198.8 m>s Parabolic Orbit:

2GM e ve = A

2(66.73)(10)5.976(10 )

4000(10 ) + 6378(10 )

= 8766.4 m>s

¢v = ve - vC = 8766.4 - 6198.8 = 2567.6 Ans.

m>s ¢v = 2.57 km>s

laws

teaching Web) Dissemination copyright Wide .

States United

instructors permitted . World

of learning the is not on

and

use

the student for (including work by

protected

of the assess in g

solely

work provided and of

This is

work

integrity this

the

and courses

part

of any sale

will

photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

13–117. Prove Kepler’s third law of motion. Hint: Use Eqs. 13–19, 13– 28, 13–29, and 13–31.

SOLUTION From Eq. 13–19, GMs

1 r = C cos u +

For u = 0° and u = 180°, GMs

1 rp = C +

2 h

GMs

ra = - C +

teaching

Web)

Dissemination

Eliminating C, from Eqs. 13–28 and 13–29, From Eq. 13–31,

laws copyright

2GM s

2a =

instructors

States

Uniteduse

by the protected

T2h2

permitted . World

of learningthe is not on and

forstudent T = h (2a)(b)

Wide

assessing

Thus, 4p a

work

provided

the

integrity

This is part the and courses their

destroy

of any

2 sale will

T2 = a GMs ba

Q.E.D.

photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

13–118. The satellite is moving in an elliptical orbit with an eccentricity e = 0.25. Determine its speed when it is at its maximum distance A and minimum distance B from the earth. 2 Mm

SOLUTION 2

Ch e=

GMe

where C = r0 ¢1 - r0v0

≤ and h = r0 v0. 1

GMe

¢1 - r0 v0 ≤ (r0 v0) r v2

e = GMe r0

e = ¢ GMe

r0 v0

- 1≤

GMe

laws teachingWeb)

GMe (e + 1)

=e+1

. Dissemination

vB = v0 = C 66.73(10 6

rp= 2A 10 B

where r0 =

= 7713 m>s = 7.States71 )(5.976)(10 )(0.25 + 1) km>

Ans.World permitted

8.378(10 )

+ 6378 A 10 B

= 8.378A 10 B

6 8.378(10 )

United

of learning the is on

use

ra =

and

13by.96A 106 B

= m student

-1

2(66.73)(10

2GMe vA =

rp ra

)(5.976)(10 )

for

protected

wor k

the

(7713) = 4628 m>s = 4assessing.63 km>s

8.378(10 )

solely

work integrity

work provided and of

13.96(10 )

this This is

the

and courses part

of any sale will

Wide

of the Ans.

13–119.

The elliptical orbit of a satellite orbiting the earth has an eccentricity of e = 0.45. If the satellite has an altitude of 6 Mm at perigee, determine the velocity of the satellite at apogee and the period.

P A

SOLUTION 6 Mm 6

6(10 ) + 6378(10 ) = 12.378(10 ) m.

Here, rO = rP =

h = 12.378(10 )vP

(1)

and GMe

1 C= r

a1 - r v 2 b P P

66.73(10

C = 80.788(10

)(5.976)(10

)

C = 12.378(10 ) c1 -

-9

- 12

12.378(10 )vP 2

laws(

2.6027

teaching

Web) Dissemination

v P

Wide permitted instructorsnot

GMe

Using Eqs. (1) and (2), States

. World

of learningthe is Unitedon use

c80.788(10

-9

and

d c12.378(10 6)vforP d

by the

P - 12

student work (including

protected 24

2.6027

assessing is

66.73(10

0.45 =

vP = 6834.78 m>s

work and ofintegrity provided this This is

part

and courses any

2GM

the

work

)(5.976)(10 )solely

their

the

des troy

Using the result of v P ,

r v 2 -1 P

sale will

12.378(10 )

2(66.73)(10

- 12

)(5.976)(10

) -1

12.378(10 )(6834.78 ) 6 = 32.633(10 ) m

Since h = rPvP = 12.378(10 )(6834.78) = 84.601(10 ) m >s is constant, rava = h

32.633(10 )va = 84.601(10 ) va = 2592.50 m>s = 2.59 km>s Using the result of h, T=

p (r + r ) 2 P a

rPra

Ans.

Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission

any

form

or by

any

means, electronic, mechanical,

Ans.

photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

*13–120. Determine the constant speed of satellite S so that it circles the earth with an orbit of radius r = 15 Mm. Hint: Use Eq. 13–1.

r 5 15 Mm

SOLUTION y

mm e F= G

y2 0

ms a r b = G

AlsoF = ms me

G r =

ms a r b

Hence

5.976(10

y= A

B66.73(10

- 12

)

) a 15(10 )

b = 5156 m>s = 5.16 km>s

Ans.

laws

teaching Web) Dissemination copyright Wide .

States

United

instructors permitted . World

of learning the is not on

use

and

the student for (including work by

protected

of the assess in g

solely

work provided and of

This is

work

integrity this

the

and courses

part

of any sale

will

photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

13–121. The rocket is in free flight along an elliptical trajectory A¿A. The planet has no atmosphere, and its mass is 0.70 times that of the earth. If the rocket has an apoapsis and periapsis as shown in the figure, determine the speed of the rocket when it is at point A.

r A

6 Mm

SOLUTION

0 0

B - 1 , with r0 = r p = 6A 10 B m and

M = 0.70Me, we have 6

9A 10 B =

2(66.73) (10

-12

6(10)

) (0.7) [5.976(10

6(10

)]

laws

≤-1

)yP

teaching

Dissemination

yA = 7471.89 m>s = 7.47 km>s

instructors States . World

United

permitted

of learning the is not on

and

use

for

Web)

Ans. Wide

the student (including work

protected

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solely

work provided and of

This is

A¿

9 Mm

Central-Force Motion: Use ra = (2 GM>r y

3 Mm

work

integrity this

the

and courses part

of any sale will

13–122. A satellite S travels in a circular orbit around the earth. A rocket is located at the apogee of its elliptical orbit for which e = 0.58. Determine the sudden change in speed that must occur at A so that the rocket can enter the satellite’s orbit while in free flight along the blue elliptical trajectory. When it arrives at B, determine the sudden adjustment in speed that must be given to the rocket in order to maintain the circular orbit.

120 Mm A

10 Mm

SOLUTION C=

Central-ForceMotion: Here,

1 r

11 -

GMe

2 [Eq. 13–21] and

h = r0 v0

[Eq. 13–20]. Substitute these values into Eq. 13–17 gives 1

2 2

GM e

r A1 0

e = GM

- r 2v 2 B A r v B rv 0 GM = GM - 1 0

0 0

(1)

Rearrange Eq. (1) gives GMe

r0 v0

Rearrange Eq. (2), we have

Substitute Eq. (2) into Eq. 13–27,

v0 =

A 2 GMe >r0 v0 B

r0 2A

1+e

-1

ra =

or the first elliptical orbit e = 0.58, from Eq. (4)

= r0

b C 120110

=a 1

+ 0.58

- 0.58

This

Substitute r0 = 1rp21

= 31.89911062 m intoandEqcourses. ( 11 + 0.582166.732110

1vp21 = D

215.976s

31.899110 2 Applying Eq. 13–20, we have rp

31.89

1va21 = a

b 1vp21 =

120110 2

When the rocket travels along the second elliptical orbit, from Eq. (4), we have 1- e

10110 2 Substitute r0 = 1rp22

1+ e

b C 120110

11 + 0.84622166.732110

215.9672110

v p

2 =

10110

)

13–122. continued And in Eq. 13–20, we have 6

1rp22

10110 2

(va22 = c 1ra22 d 1vp22

= c 120110 2 d 18580.252 = 715.02 m>s

For the rocket to enter into orbit two from orbit one at A, its speed must be decreased by ¢v = 1va21 - 1va22 = 1184.41 - 715.02 = 466 m>s

Ans.

If the rocket travels in a circular free-flight trajectory, its speed is given by Eq. 13–25.

v= cD

GMe

66.73110

- 12

215.9762110 2

10110 2 = 6314.89 m>s The speed for which the rocket must be decreased in order to have a circular orbit is 0=

¢v = 1vp22 - ve

= 8580.25 - 6314.89 = 2265.36 m>s = 2.27 km>s

Ans.

laws

teaching Web) Dissemination copyright Wide .

instructors permitted States . World

United

of learning the is not on

use

for

and

the student (including work

protected is

of the

assess in g

solely

work

work provided and of

integrity this

This is

the

and courses

of sale

part

any

will

photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

13–123. An asteroid is in an elliptical orbit about the sun such that its 9

periheliondistance is 9.30110 2 km. If the eccentricity of the

SOLUTION 9

rp = r0 = 9.30110 2 km

ch e = GM

1 =r a 1 s

r v 0

GMs rv 0

r0v0 b a GM b s

e = a GMs - 1b

r v 0

GMs = e + 1 1

(1)

GMs rv 0

laws

b r e +1

2 0

-1

copyr ig ht

States

r 0 1e + 12

2GMs

Web)

teaching

. Dissemination or instructors World permitted Wide

(2)

2 e+1

of learning the is

11 - e2

0.927

United use

a = 10.8110 2 km

for

not on andAns.

the student (including work

protected

of the a s se ss in g

solely

work

work provided and of

integrity this

This is and

the

courses part

of any sale will

photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

*13–124. An elliptical path of a satellite has an eccentricity e = 0.130. If it has a speed of 15 Mm>h when it is at perigee, P, determine its speed when it arrives at apogee, A. Also, how far is it from the earth’s surface when it is at A?

SOLUTION e = 0.130 vp = v0 = 15 Mm>h = 4.167 km>s

GMe

r0 v0

e = GMe = r0 ¢1 - r0v0 ≤a GMe b rv 0

2 0

e = ¢ GMe - 1≤ 2

r0 v0

GMe = e+ 1

laws

Web)

teaching

Wide .

= v02

permitted

instructors

States

learning =

- 12

1.130166.73 2110

32 D 2

= 25.96 Mm

thenot

United

GMe 0 0 2

= e+1

-1

rA 2GM

-1

e +

0 0

work

rA =

work

of ass essing

r0 1e + 12

solely

the

work

and of thisintegrity

This is part the and courses

25.96110 211.1302 1-

and

the student (including for by A

use

protected 2

24 215. 976 211 0 2

C 4.167110

r v

W orl d

destroy

their sale

0.870

of any

will

= 33.71110 2 m = 33.7 Mm vrr vA =

0 0 A

2 = 15125.962110 6 33.71110 2

= 11.5 Mm>h

Ans.

d = 33.71110 2 - 6.378110 2 = 27.3 Mm

Ans.

13–125.

A satellite is launched with an initial velocity v0 = 2500 mi>h parallel to the surface of the earth. Determine the required altitude (or range of altitudes) above the earth’s surface for launching if the free-flight trajectory is to be (a) circular, (b) parabolic, (c) elliptical, and (d) hyperbolic.

-9

Take G = 34.4110 21lb

ft 2>slug , Me = 409110

SOLUTION v

0 = 2500 mi>h = 3.67(10

C h GMe = 0

(a) e = 1

2 slug,

) ft>s

or C = 0

= GMe

r0 v 0 GMe = 34.4(10

-9

= 14.07(10 r0 =

)

)(409)(10

13 2

= 1.046(10 ) ft

laws

teaching

Web)

14.07(10 ) 9

- 3960 = 194(10 ) mi

1.047(10 )

[3.67(10

copyrightAns.

Dissemination Wide

)]

instructors permitted States . World

5280

learning

C h

(b)

e = GM = 1

Uniteduse

of the

not

on and for student work by the (including

GM e (r 0 v 0)a r

2GM 1

2 2

v0 2

b ¢1 - r v 2 ≤ = 1

protected assessing

15 2(14.07)(10 ) 0

GM e [3.67(10 )]

is solely work provided ofintegrity this 9

r0 =

of the

This is

r = 396(103) - 3960 = 392(103) mandi

part

courses

the

Ans.

of any sale will

(c)

e61

194(10 ) mi 6 r 6 392(10 ) mi (d)

Ans.

e71

r 7 392(10 ) mi

Ans.

13–126.

A probe has a circular orbit around a planet of radius R and mass M. If the radius of the orbit is nR and the explorer is B

traveling with a constant speed v0, determine the angle u at which it lands on the surface of the planet B when its speed is reduced to kv0, where k 6 1 at point A.

u A

SOLUTION

When the probe is orbiting the planet in a circular orbit of radius r O = nR, its speed is given by GM r

v = O

= B nR The probe will enter the elliptical trajectory with its apoapsis at point A if its speed is GM decreased to va = kvO =

at this point. When it lands on the surface of the

planet, r = rB = R.

a1 -

= r

v P P

1 R

1 a

r P

GM 2

= nRakA nR b = k

Since h = rava GM

rPvP = h

k2nGMR

learning

vP =

rP Also, 2

2GM

ra = rP rv P P

2GM

2 -1

nR =

is of

P P

2nGMR

rp(rp + nR)

Solving Eqs.(2) and (3),

k n

rp = 2 - k R Substituting the result of rp and vp into Eq. (1),

2-k

1 R

= u = cos

-1

k nR

k na

1-k

Here u was measured from periapsis.When measured from apoapsis, as in the figure

then k n - 1 u = p - cos

-1

photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

13–127. Upon completion of the moon exploration mission, the command module, which was originally in a circular orbit as shown, is given a boost so that it escapes from the moon’s gravitational field. Determine the necessary increase in velocity so that the command module follows a parabolic trajectory. The mass of the moon is 0.01230 Me. 3 Mm

SOLUTION When the command module

is moving around

the circular orbit of radius

r0 = 3(10 ) m, its velocity is GM

66.73(10 )(0.0123)(5.976)(10

)

= B

3(10 )

= 1278.67 m>s The escape velocity of the command module entering into the parabolic trajectory is

2GM

2(66.73)(10

)(0.0123)(5.976)(10

)

0 =B B = 1808.31 m>s

3(10 ) laws

teaching Web) Dissemination copyright Wide .

Thus, the required increase in the command module is

instructors

States

United

. World

of learningthe is not on

and

use

for

the student (including work

protected

of the

assess in g

solely

work

work provided and of integrity this This is

the

and courses part

of any sale will

photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

*13–128. The rocket is traveling in a free-flight elliptical orbit about the earth such that e = 0.76 and its perigee is 9 Mm as shown. Determine its speed when it is at point B. Also determine the sudden decrease in speed the rocket must experience at A in order to travel in a circular orbit about the earth. A

SOLUTION

9 Mm

GMe 1 Central-Force Motion: Here C = r 0 a 1 - r 0 v 2 b [Eq. 13–21] and [Eq. 13–20] Substitute these values into Eq. 13–17 gives

h = r0 v 0

2 2

1 ch

A 1 - GM e B 1 r0 v0 2

r0v0

0 0

GMe

e = GMe = Rearrange Eq.(1) gives

= GMe - 1

(1)

GMe 1 2 1+e =r v 0

Rearrange Eq.(2), we have

laws(

teaching

Dissemination

. Web)

11 +

v0 =

instructors States .

e2 GMe

permitted World (3)

learning

r0 2 Substitute Eq.(2) into Eq. 13–27, r a

Wide

12GMe >r0 v0 2 - 1

United use

student

is not

the work

and

the 1

-1

protected

1+e

of the assess in g

solely

work provided and of

Rearrange Eq.(4), we have

work

integrity

this

1+e 1-e

1 + 0.76 This is 6 part 1 - 0.76 and coursesany their

ra = a

destroy

b C 9110

b r0 = a

the

66.0110

m into Eq.(3)saleyieldswill

Substitute r 0 = r p = 9A 10 B (1 + 0.76)(66.73)(10 p

)(5.976) (10

) = 8830.82 m>s

9(10 6)

Applying Eq. 13–20, we have

9110 2

v a = a ra b np = B 66.0110 2 R18830.822 = 1204.2 m>s = 1.20 km>s

Ans.

If the rocket travels in a circular free-flight trajectory, its speed is given by Eq. 13–25. GM

66.73110

- 12

215.9762110 2

v e = D r0 D 9110 2 = 6656.48 m>s The speed for which the rocket must be decreased in order to have a circular orbit is ¢v = v p - v c = 8830.82 - 6656.48 = 2174.34 m>s = 2.17 km>s

Ans.

13–129.

A rocket is in circular orbit about the earth at an altitude above the earth’s surface of h = 4 Mm. Determine the minimum increment in speed it must have in order to escape the earth’s gravitational field.

4 Mm

SOLUTION Circular orbit: GM

66.73(10)5.976(10 )

vC = B 0 = B 4000(10 ) + 6378(10 ) = 6198.8 m>s Parabolic orbit:

2GM

2(66.73)(10

)5.976(10 )

ve = B

4000(10 ) + 6378(10 ) = 8766.4 m>s

¢v = ve - vC = 8766.4 - 6198.8 = 2567.6

laws Ans.

m>s ¢v = 2.57 km>s

teaching Web) Dissemination copyright Wide .

instructors permitted States . World

United

of learning the is not on

and

use

for

the student (including work

protected

of the assess in g

solely

work provided and of

This is

work

integrity this

the

and courses

part

of any sale

will

photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

13–130.

The satellite is in an elliptical orbit having an eccentricity of e 15 Mm/h

= 0.15. If its velocity at perigee is vP = 15 Mm>h, determine its velocity at apogee A and the period of the satellite.

SOLUTION 6m

1h h

Here, vP = c 15(10 )

d a

3600

s b = 4166.67 m>s.

h = rPvP h = r P (4166.67) = 4166.67rp

(1)

and GMe

C = rP ¢1 - rP vP

66.73(10

≤

-12

)(5.976)(10

)

rp(4166.67 )

C = r p B1 -

R laws(

C = rP B1 -

22.97(10 )

2) teaching Web) DisseminationWide copyright .

GMe

2 Ch

22.97(10 )

instructors

r P

States

R(4166.67 r P) 2

B1 -

0.15 =

66.73(10

)(5.976)(10

permitted

. World

United of learningthe is not by use and on

)

6 rP = 26.415(10 ) m

for protected

student

work

theof the assess in g

work

rA =

work

and of thisintegrity

provided

This is

2 2GM e

solely

26.415(10 ) 2(66.73)(10

the

any

rP vP

part destroy

their

-1

sale will

-12

)(5.976)(10

) -1

26.415(10 )(4166.67 )

= 35.738(10 ) m

Since h = rP vP = 26.415(10 )(4166.67 ) = 110.06(10 ) m >s is constant, rA vA = h

35.738(10 )v = 110.06(10 ) A

vA = 3079.71 m>s = 3.08 km>s

Ans.

Using the results of h, r A, and r P, p T = 6 A rP + rA B 2 p

P A

9 = 110.06(10 ) C 26.415(10 ) + 35.738(10 ) D 226.415(10 )(35.738)(10 ) = 54 508.43 s = 15.1 hr Ans.

photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

13–131. A rocket is in a free-flight elliptical orbit about the earth such that the eccentricity of its orbit is e and its perigee is r0. Determine the minimum increment of speed it should have in order to escape the earth’s gravitational field when it is at this point along its orbit.

SOLUTION To escape the earth’s gravitational field, the rocket has to make a parabolic trajectory. Parabolic Trajectory: 2GMe

ye = A r0 Elliptical Orbit:

Ch e = GMe 1 GMe r0

GMe 1 r ¢ 1 - r y 2 ≤ and h = r0 y0

where C =

- 1b e=a 2 r0 y0 1e= GMe

laws or

. DisseminationWeb)

instructors not permitted learning of the is on

States

United

y0 =

World

use

and

r0 student

2GMe

Wide

(r0 y0)

GMe

GMe (e + 1) 1

teaching

r0 y0 =e+

0 0

GMe ≤

pro

GMe (e + 1)

GMe

tecte d

work

(includingebthe

the

for

assessing 0

solely

work

work provided and of

inte grity

this This is and

the

courses part

of any sale will

*13–132. The rocket shown is originally in a circular orbit 6 Mm above the surface of the earth.It is required that it travel in another circular orbit having an altitude of 14 Mm.To do this,the rocket is given a short pulse of power at A so that it travels in free flight along the gray elliptical path from the first orbit to the second orbit.Determine the necessary speed it must have at A just after the power pulse, and at the time required to get to the outer orbit along the path AA ¿. What adjustment in speed must be made at A¿ to maintain the second circular orbit?

A' O 6 Mm

14 Mm

SOLUTION r0 Central-Force Motion: Substitute Eq. 13–27, ra

= (2GM>r 0v0 ) - 1 , with 6 = rp = (6 + 6.378)(10 )

ra = (14 + 6.378)(10 ) = 20.378(10 ) m and r0

= 12.378(10 ) m, we have 2(66.73)(10

- 12

)[5.976(10

)]

12.378(10 )vp

20.378(10 ) =

≤ -

laws Web) teaching .

12.378(10 )

Dissemination or copyright instructors States . of learning the United on

vp = 6331.27 m>s r

12.378(10 )

Wide

use

permitted World

is not

and

(including work

va = ¢ ≤ v p ra

= B 20.378(106) R (6331.27) = 3845.74 m> sby

the

protected 6 is

Eq. 13–20 gives h = rp vp = 12.378(10 )(6331.27)

p Eq. 13–31, we have h

This

and

2 the

78.368(10 )

work

78.368(10 ) m >s.Thus, applying

integrity this

the

destroy

any 9 [(12.378 + 20.378)(10 )]their212.378(20.378)(10

solely

courses part

and

provided work

ra )2rp ra

T = (rp +

of assessing

)

sale will

= 20854.54 s The time required for the rocket to go from A to A¿ (half the orbit) is given by T t = 2 = 10427.38 s = 2.90 hr

Ans.

In order for the satellite to stay in the second circular orbit, it must achieve a speed of (Eq. 13–25) GM vc = A

66.73(10)(5.976)(10 e

)

= 4423.69 m>s = 4.42 km>s

20.378(10

Ans.

)

The speed for which the rocket must be increased in order to enter the second circular orbit at A¿ is ¢v = vc - va = 4423.69 - 3845.74 = 578 m s

Ans.