13–34. each of the two blocks has a mass m. the coefficient of kinetic friction at all surfaces of...

Solution Manual for Engineering Mechanics

Dynamics 13th Edition by Hibbeler

z

F 2 y

F 3

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13–1.

The 6-lb particle is subjected to the action of

x F

1 its2 weight and forces F1 = 52i + 6j - 2tk6 lb, F2 = 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

2

©F = ma; (2i + 6j - 2tk) + (t i - 4tj - 1k) - 2ti - 6k = ¢32 .2 ≤(axi + a y j + azk)

Equating components:

6 6 6

¢ 32.2 ≤

a

x = t2

- 2t + 2 ¢ 32.2 ≤ay = - 4t + 6 ¢ 32.2 ≤az = - 2

t -

7 Since dv = a dt, integrating from n = 0, t = 0, yields

6 t3 6 6

2

2

2

¢ 32.2 ≤vx = 3 - t + 2t ¢ 32.2 ≤vy = -

2t + 6t ¢ 32.2 ≤vz = - t - 7t laws or

s = s - ¢

2. teaching Web)

¢ ≤

- 3 + t ¢ ≤

y 3

- 3

12

+ 3t

- Dissemination

32.2 x 32.2 = 32.2 ≤sz =

Since ds = v dt, integrating from s = 0, t = 0 yields copyright Wide . instr u ctors permitted

6 t t 6 2t3 6

4 3

2

2

When t = 2 s then,

States World

= - learning

sx = 14.31 ft, sy = 35.78 ft sz 89.44offt the is not

United on use and

for student work s =

Thus, by the

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https://www.testbankfire.com/download/solution-manual-for-engineering-mechanics-dynamics-13th-edition-by-hibbeler/

https://www.testbankfire.com/download/solution-manual-for-engineering-mechanics-dynamics-13th-edition-by-hibbeler/

of any sale will

© 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 v = 10 ft/s 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 F = (2.5t) lb

distance the block travels during this time.

SOLUTION

+ 10

: ©Fx = max; 2.5t = ¢ 32.2 ≤a a

= 8.05t

dv = a dt

n t dv = L10 L0 8.05t dt 2 v = 4.025t + 10

When t = 3 s,

v = 46.2 ft>s

ds = v dt

s t 2

L0 ds = L0 (4.025t + 10) dt

s = 1.3417t3

+ 10t

When t = 3 s,

s = 66.2 ft

laws or

teaching Web)

.Dissemination copyright Wide

. A s.

States instructors World permitted

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Ans.

© 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.

13–3. 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.

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

2 a = 1.121 m>s

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

+ laws or

A : B v = v0 + act teaching Web)

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

copyrightAns. Wide

.

Dissemination

and States

. World permitted

1 instructors not

2 of theis +

s = s + v t +

a t

learning

2

:

B 0 0 c

Uniteduse on

A 1 2 and by

s = 0 + 0 + 2 (1.121) A3 B = 5.04 m the student work Ans. for (including

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308


*13–4. 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

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.

( : ) + 2

v 2

2a (s -s ) v = +

2 20 c 0

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

2 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) laws or

N = 490.5 - 0.5P

teaching Web)

.

Using the results of N and a, copyright Wide

Dissemination

+ .

P cos 30° - 0.3(490.5 - 0.5P) = 50(1.60)


: ©Fx = max; of learning the is not

United

P = 224 N on andAns.

by u se

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308


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;

+

; aFx = max;

800 800 sin 45° - 30 = 32.2 a

2 a = 21.561 ft>s v 2 = v 2 + 2a (s - s )

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

1 v1 = = 78.093 ft>s

800 -80 = 32.2a

2 a = -3.22 ft>s

2 2

v2 = (78.093) + 2( -3.22)(5 - 0)

2 2

v2 = v1 s2 - s1)

v 2 = 7 7 . 9 ft >s

s 100 ft

laws teaching Web)

. Dissemination or copyright Wide .

permitted

instructors

States learning World

United of on the not Aisns.

use and

the student

for (including work by

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and courses part

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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 Ri ver, 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.

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

teaching Web)

laws

Using the result of N, or

. Dissemination a = 0.8993 m >s

2 copyright Wide

©Fx¿

= may¿;

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


2 2 United of learning the is Kinematics: Since the acceleration a of the crate is constant,

use not

on

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

by the (including

v = 0 + 2(0.8993)(6 - 0) protected for of the work assessing

v = 3.29 m>s is solely work Ans. work provided and of integrity

this

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P

30°

© 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.

13–7. 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

crate and the ground is mk = 0.25.

SOLUTION Kinematics: Here, the acceleration a of the crate will be determined first since

its motion is known.

2

s = s + v t + 1 a t 0 0

2

c

6 = 0 + 0 + 2

1 a(4 )

2

a = 0.75 m>s 2

Free-Body Diagram: Here, the kinetic friction Ff = mkN = 0.25N is required to be laws

teaching

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

30°

Web) or

P

30°

Equations of Motion: Here, ay¿ = 0. Thus, Dissemination copyright Wide

plane, Fig. a. .

©Fy¿ = may¿; instructors permitted

N + P sin 30° - 50(9.81) cos 30° = 50(0)

States . World N = 424.79 - 0.5P learning on

of the is not

United and

use

Using the results of N and a, for student by the

©Fx¿ = max¿; P cos 30° - 0.25(424.79 - 0.5protectedP) - 50(9.81) sinof30° =

50(0.75) assessing

is solely work the

P = 392 N

work provided and of integrity Ans. this


of any sale will


*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.

SOLUTION

v(ft/s) F

80

60

Kinematics: For 0 … t

we have

dv 2 a = dt

= 6 ft>s For 10 6 t … 30 s,

dv a = dt , we have

Equation of Motion:

For 0 … t 6 10 s

+ ; aFx = max ; F =

For 10 6 t … 30 s

+ ; aFx = max ;F =

60 dv

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

t (s) 10 30

v - 60 = 80 - 60

,v = {t + 50} ft>s. Applying equation

t - 10 30 - 10 v

dv

= 2

a = dt 1 ft>s laws or

teaching Web) Dissemination copyright Wide .

3500 instructors permitted ¢ 32.2 ≤(6) = learning

States . World

of the is not United on

652 lb use and

3500 Ans.

¢ 32.2 ≤(1) = the student


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13–9.

The crate has a mass of 80 kg and is being towed by a chain p

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

P = 353.29 N N = 663.97 N

Equations of Motion: The friction force developed between the crate and its or laws

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

teaching Web)

Dissemination

+ c ©Fy = may ; N - 80(9.81) + 353.29 sin 20° = 80(0) copyright Wide .

: ©Fx = max ; 353.29 cos 20° - 0.3(663.97) = 80a instructors not permitted

+ N = 663.97 N States . World

2 Aisns a = 1.66 m>s of the . United on

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13–10.

The crate has a mass of 80 kg and is being towed by a chain p 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 mk = 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 N = 661.67 N

+

: ©Fx = 0; 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) teaching

.Dissemination or

©Fx = max ; 360 cos 20° - 0.3(661.67) = 80a copyright Wide .

instructors permitted 2 States World

a = 1.75 m>s A s.

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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; T-90 = - a 32.2 b

a

B (1) From FBD(b),

+ c ©F y

= ma ; 2T - 200 = - a

200 b a S

(2)

y

32.2

laws or

teaching Web) Dissemination

Ki nematic: Es tablis h the positi on-coordinate equation , we have copyright Wide .

Taking time derivativ e twice yields permitted

2sS +sB = l instructors

States . World

of learning

on the

United not 1 + T2 2aS + aB = 0

use and is(3)

2 2 for student work by

= 2.30

aB = -2.30 ft>s ft>sprotectedsolely of the Ans. Solving Eqs.(1),(2), and (3) yields the

aS = 1.15 ft>s T T = 96.43 lb assessing

is work work provided and of integrity

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*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

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

SOLUTION (a) T = 40 lb Ans.

2 (b) v = 4t

a = 8t

+ c aFy = may ; 80

2T - 80 =

32.2 18t2

At t = 2 s.

T = 59.9 lb Ans.

laws or


instructors permitted States . World

United

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and

use



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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

on the bullet, determine the velocity of the bullet at any instant F

it is in the barrel. What is the bullet’s maximum velocity? Also, F determine the position of the bullet in the barrel as a function of 0

time.

SOLUTION

+ pt

: ©Fx = max ; F0 sin a t b = ma 0

F dv 0 pt

sin a t

a = dt = a m b b 0

v t F 0 pt F t 0 0 pt t

L dv = Lm b sin a t b dt v = - a pm b cos a t b d 0 0 0 0 0

t

t 0

F0t0 pt

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

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

pt

0

v =

Ans.

laws

teaching Web) .Dissemination

copyright .

max pm A s. Wide instruct ors World perm itted

2F0t0

Ls L t pm t States learning

ds = F0t0

b a 1 - cos a

United of the is not

a b b dt

on and

0 0 t 0 use

F0t0

pt

the student

s =

- 0 t

for

a pm b c t

a

b d 0

(including

p sin 0 by

protected of the work is assessing

solely work

s = F0t0

b a t - p

t0 pt provided and of i ntegr ity

sin a t0 b b work this

Ans.

a pm


of any sale will


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

2 0 = 15 + 2a(10 - 0)

2 2 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

or .Dissemination Web)

copyright Wide

: ©Fx = max ; -T = 1000( -11.25) instructors permitted States World

T = 11 250 N = 11.25 kN A s.

Using the results of a and T and referring to Fig. (b), of learning the is United on not

+ c ©Fx = max ; use

11 250 - F = 2000( -11.25) and for student work

(including

protected by theof the

F = 33 750 N = assessing

33.75 kN Ans. is solely work


this


of any sale will

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

4

A + c B v = v0 + ac t

2 c

E

1.5= 0 + aE (2)

Ateachings. Web)

laws .Dissemination or

+ c ©Fy = may ; 4T - 500(9.81) = 500(0.75)

copyright Wide .

AW or

T = 1320 N = 1.32 kN States instructors ldns. permitted

United of learning the is not

use on

and



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and courses part

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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 07 458.

*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 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,

+ 60 : ©Fx = max ; 63.60 cos 30° - 0.3(91.80) = a 32.2 ba

2

a = 14.8 ft>s laws

Web) teaching

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copyright Wide .


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F

308


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 B

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 = mk 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 10

= ma y¿ ; N - 10 cos 60° = a

NA = 5.00 lb

a y¿ 32.2 b (0)

A

Q + F 10

= max¿; T + 0.1(5.00) - 10 sin 60° = - a

a x¿ 32.2 b a (1) From FBD(b),

10

laws or

a y¿ = may¿; NB - 10 cos 30° = a

b (0)

B = 8.660 lb

teaching Web)

32.2 Dissemination

copyright Wide .

a + F

instructors permitted

10

States World

a x¿ = max¿; T - 0.1(8.660) - 10 sin 30° = a 32.2 b a of

. (2)

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

and 2 on

a = 3.69 ft>s the student Ans. for (including work

by

protected of

T = 7.013 lb

assessing

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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?

SOLUTION

40

+ R ©Fx = m ax ;40 sin 30° = 32.2a

2 a = 16.1 ft>s

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

A

20 ft

30

B 4 ft

C

R

laws Web) teaching

Dissemination or

tAB = 1.576 s copyright Wide .

R = 0 + 25.38 cos 30°(tBC) instructors not permitted +

States World

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

of learningthe is 1

2 United on use

( + T) sy = (sy)0 + (vy)0 t + 2 act by the and

1 forstudent work

2 (including

4 = 0 + 25.38 sin 30° tBC + 2(32.2)(tBC) is

protected assess in g of the solely work


this

C = 1.82 s

This is part the Total time = tAB + tB andcourses any Ans.

R = 5.30 ft Ans. their

of destroy

sale will


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.

SOLUTION 40

+ R©Fx = max;

40 sin 30° - 6.928 = 32.2a 2

2 2 a = 10.52 ft>s

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

2 2 v B = (10) + 2(10.52)(20) vB = 22.82 ft>s

( + R) v = v0 + ac t;

22.82 = 10 + 10.52 tAB

+

A

laws Web)

teaching

or

20 ft

308

B 4 ft

C

R

tAB = 1.219 s ( : ) sx = (sx )0 + (vx )0 t

( + T ) sy = (sy)0 + (vy)0 t + ac t 2

R = 0 + 22.82 cos 30° (tBC)

2 1 2

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

t BC = 0.2572 s

Dissemination

copyright Wide .

instructors World permitted

learning . is

of the not

United on use

by the student and

(including

protected for of the work

assess in g

R = 5.08 ft is solely work Ans.

Total time = tAB + tBC = 1.48 s provi ded

and of integrity Ans. work this

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*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 =

2 13200t 2 N, where t is in seconds. If the car has an initial velocity v1 = 2 m>s when t = 0, determine its velocity when t = 2 s.

SOLUTION

8

2

Q + ©Fx¿ = max¿ ; 3200t - 40019.812 a 17 b = 400a

dv = adt

v 2 2

L2 dv = L0 18t -4.6162 dt

v = 14.1 m>s

M

v1 2 m/s

17

8 15

2 a = 8t - 4.616

Ans.

laws or

teaching Web)

Dissemination

copyright Wide .


United


use and



is solely work


this

This is the

and courses part

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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 sys tem, or transmission in any form or by any means, electronic, mechanical,

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 =

2 13200t 2 N, where t is in seconds. If the car has an initial velocity v1 = 2 m>s at s = 0 and t = 0, determine the distance it moves up the plane when t = 2 s.

SOLUTION 2 8

Q + ©Fx¿ = max¿;3200t - 40019.812 a 17 b = 400a

dv = adt

v t 2

L2 dv = L0 18t - 4.6162 dt

ds 3

v = dt = 2.667t - 4.616t + 2

s 2 3

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

M

v1 2 m/s

17

8 15

2 a = 8t - 4.616

laws or

s = 5.43 m teaching Web)

.Dissemination copyright A s. Wide .

States

instructorsWorld permitted

United


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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 sys tem, or transmission in any form or by any means, electronic, mechanical,

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

2

Kinematic: Applying equation s = s0 + v0 t + 2 ac t , we have 1

0.75 = 0 + 0 + 2 aB A

22 2 (a +) B aB = 0.375 m>s Establishing the position - coordinate equation, we have

2sA + (sA - sB) = l 3sA - sB = l

Taking time derivative twice yields

From Eq.(1), laws or teaching Web)

3aA - aB = 0 (1)

2 Dissemination

3aA - 0.375 = 0 aA = 0.125 m>s copyright Wide .

Equation of Motion: The tension T developed in the cord is the sameinstructorsthroughout permitted States . World learning

United of on the is not

by use and

T = 44.35 N for student work

a+ ©Fy¿ = may¿ ; T - 5(9.81) sin 60° = 5(0.375)the From FBD(a), protected as ses sing of is solely work

workprovided and of thisintegrity mA This is part the Ans.

+ c ©Fy = may ; 3(44.35) -9.81mA = mA ( -0.125) their

of destroy

sale will

E

C

D

B

A

13–23.

The winding drum D is drawing in the cable at an accelerated

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

D

SOLUTION

sA + 2 sB = l

aA = -2 aB

5 = -2 aB 2 2

aB = -2.5 m>s = 2.5 m>s c

+ c ©Fy

= may

; 2T - 800(9.81) = 800(2.5)

T = 4924 N = 4.92 kN Ans.

laws or



United


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*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 is mk = 0.2.

s

v

M

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

first.

dv 3

1>2 22

a = vds = A0.05s3>2

Bc(0.05) a 2 bs d = 0.00375s m>s When s = 10 m,

2 2

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.


+ c ©Fy = may; N - 20(9.81) = 20(0)

N = 196.2 N

laws or

teaching Web)

copyright Wide .

Dissemination

States World permitted instructors

Using the results of N and a, United

of learning the is not

+ on use

: ©Fx = max; T - 0.2(196.2) = 20(0.375) by the and studentwork

T = 46.7 N for (including Ans. protected of the is solelywork

assessing

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13–25. 2

If the motor draws in the cable at a rate of v = (0.05t ) m>s, s

where t is in seconds, determine the tension developed in the

v

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 M

is mk = 0.2.

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

first.

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

When t = 5 s,

2 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.


+ c ©Fy = may; N - 20(9.81) = 0

N = 196.2 N

laws or

teaching Web)

copyright Wide .

Dissemination


Using the results of N and a, United


+ on use

: ©Fx = max; T - 0.2(196.2) = 20(0.5) by the and

T = 49.2 N studentwork

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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.

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 v

v dv 2

2500(0.2) 2 v

2500(0.2) - ¢ 2 ≤ = 2 v = 30 m>s

s¿ s

F 5 kN

C B A

kCB 3 kN/m k AB

2 kN/m

laws or

teaching Web)

.Dissemination copyright Wide .

instructors

States World


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13–27.

The 30-lb crate is being hoisted upward with a constant 2 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.

SOLUTION Crate:

30

y

5 ft

B

A x

6 ft/s2

+ c ©Fy = may ;T - 30 = a 32.2 b(6) T = 35.59 lb Beam:

+

: ©Fx = 0; - Ax + 35.59 = 0 Ax = 35.6 lb Ans.

a+ c ©Fy = 0; Ay - 200 - 35.59 = 0 Ay = 236 lb Ans.

+ ©MA = 0; MA - 200(2.5) - (35.59)(5) = 0 MA = 678 lb # ft Ans.

laws or


copyright Wide .


United


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*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 30

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 (2)

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

T = 187.6 lb 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 N = 400 lb

+ 200 cos 30° - 0.3(400) =

500 : ©Fx = max ; 32.2 a States

laws or

teaching Web) .

copyright Wide Dissemination

. permitted

World instructors

United


a = 3.43 ft>s 2 on Ans. use

by and

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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)

SOLUTION 2.5

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 = 2 s

A

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 2

a = (16.1t - 32.2) ft>s

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

equation, dv = adt. For 2 s … t 6 2.5 s, v = 0 at t = 2 s will be used as the lower .

teaching Web)

integration limit. Thus, copyright Wide

Dissemination

( + c ) dv = adt

L L v t

L0 dv = L 2 s(16.1t - 32.2)dt

2 t

v = A8.05t - 32.2tB 2 sis - 32.2t +work2

.

States instructors World permitted not

Uniteduse of learningon the is

by and

for the student work (including

protectedsolely work of the assessing

= A8.05t 2 32.2providedB

ftand>s of this integrity

When t = 2.5 s, This is the

v = 2) part

Ans. and coursesany

their destroy

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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.

13–30.

The force of the motor M on the cable is shown in the graph. F (N)

Determine the velocity of the 400-kg crate A when t = 2 s.

2500

2 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 2

+ c ©Fy = 0; 2(625t ) - 400(9.81) = 0 A

t = 1.772 s

2 B N. By referring to Fig. a,

Equations of Motion: F = A625t 2

+ c ©F = ma ; 2 A625t B - 400(9.81) = 400a

y y

a = (3.125t2 2

- 9.81) m>s

Kinematics: The velocity of the crate can be obtained by integrating the kinema ic laws or

teaching Web)

equation, dv = adt. For 1.772 s … t 6 2 s, v = 0 at t = 1.772 s will be used ainthe .

lower integration limit. Thus, copyright Wide

.Dissemination

( + c ) dv = adt States instructors

World permitted not

L L United

of learning the

is v t on

L0 dv = 1.772 s A3.125t 2

- 9.81 Bdt by

use

the and L

t forstudent work

v = A1.0417t3 (including

- 9.81tB 1.772 protectedsolely work of the

is

assessing 3

m sthis

work+2provide

= A1.0417t - 9.81t d11.587andB of> integrity

When t = 2 s,


any 3 9.81(2)t of+ 11.587destroy= 0.301

v = 1.0417(2 ) - heir m>s Ans.

sale will


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.

SOLUTION

12 - s 2

+ 2s 2 + (12) = 24 B A

# 2 12

- sB + As A + 144 B asAs Ab = 0

12 m

s B

B

s A

- s$

- As

2 + 144 B

- 23 asAs

#

- 21 a #A

2 B- 2 A1

Ab2

+ As 2 B b A A 2 $ A

b

a

B A 2 # 2 # 2 $ A +144 s + s + 144 s s = 0 $ sAs A

s A + sAs A

s B = -

-

C 1

AsA2 + 144 B AsA 2 + 144 B 2 S 2 2

a (5) 1.0487 m s

2 laws

(4)

3 2

+

or (4) 0

B = - C - 150(9.81) = 150(1.0487) . Dissemination Web)

((5)2 + 144)2 2 2 = > copyright Wide

+ c ©Fy = may ; T -

T = 1.63 kN

instructors World permitted States

of learningthe is

Anots.

United on use and




this

This is the

and courses part

of any

sale will

© 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,

*13–32. 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 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.

SOLUTION

12 = s + 2s

(12)2

= 24 2 +

B A

# 1 - 3

# 2 A

2 2s s Ab = 0

- s B + As + 144 B 2 a A

$ 2 - 3 # 2 2 # B

+ AsA + 144 B - 1 as A

- s B - AsA + 144 2 as sAb 2

A

$ 2 # 2 # 2 $

s B = - sA s A s A + sAs A

3 - 1

C A sA2 + 144 B 2 AsA 2 + 144 B 2 S

2 2 (4)

2 + (5)(3)

aB = - C (5) (4)

2

23 - S

((5 + 144)

((5) + 144)

2

= 2.2025 m>s2 2

1

12 m

s B

B

s A

2 b + sA + $

asAs Ab = 0

-

1

144 B

2

laws or teaching .

. Dissemination Web)

copyright Wide

+ c ©Fy = may ; T - T

= 1.80 kN


learning Ans.not

United on the is of

use and


protected of the

assess in g

is solely work

work provided and of integrity this

This is the

and courses part

of any

sale will

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

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.

18 N

D

C 100 N

15 N

B

A

S O L U T I O N 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 +

: ©Fx = 0; 100 - 18 - F = 0

F = 82 N Fmax = 0.3(196.2) = 58.86 N 6

Slipping between B and C.

+

Assume no slipping between D and C, ax = 2.138 m>s :

: ©Fx = max; 100 - 39.24 - 18

+ Check slipping between D and C.

laws Web) teaching

Dissemination or copyright Wide

.

82N

instructors permitted States . World learning

of the not United on use and is

for student work protected by the of assessing

= 20 ax the is solely work


this

This coursesany part the and

is

: ©Fx = m ax;

their of destroy

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 ax; 100 - 39.24 - 19.62 = 10 ac

2

ac = 4.11 m>s :

Plate D:

+ : ©Fx = m ax; 19.62 - 18 = 10 aD

2

aD = 0.162m>s :

Ans.

Ans.


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 B

of the bottom block in each case. P A

SOLUTION Block A:

(a)

B P A

+

(a) ; ©Fx = max ; P - 3mmg = m aA P

aA = m - 3mg (b) sB + sA = l

aA = - aB Block A:

; ©Fx = ma mmg - T = maB

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

P - 4mmg = m (aA - aB)

protected

P Use Eq.(1); A

2m - 2mg is

(b)

Ans.

(1)

instructors permitted States World .

of learning the is United on not

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and for student work (including by the

of

assess in g

work the Ans.


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13–35. The conveyor belt is moving at 4 m>s. If the coefficient of B 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.

S O L U T I O N

+ : ©Fx = m ax;

+

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

t = 2.04 s

0.2198.12 = 10 a 2

a = 1.962 m>s

Ans.

laws or


copyright Wide .


United


and

use



is solely work


this

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and courses part

of any sale will


*13–36.

The 2-lb collar C fits loosely on the smooth shaft. If the spring 15 ft/s

is unstretched when s = 0 and the collar is given a velocity of s

15 ft>s, determine the velocity of the collar when s = 1 ft. C

1 ft

S O L U T I O N k4 lb/ft

F = kx; F = 4A 2 1 + s2

- 1B

s s

+ s 2 dv

2

2

- 4A 21 + s

: ©Fx = max ; - 1B ¢ 2 1 + s ≤ = a 32.2 b a v ds b

1 4s ds v 2

-

21 + s 2

≤ = L15 a 32.2 b v dv

0 ¢4s ds -

2 1

2 2

2

1

- C 2s - 43 1 + s D

0 =

32.2 A v - 15 B v = 14.6 ft>s Ans.

laws or teaching Web) Dissemination

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United


and

use



is solely work


this

This is the and courses part of any

sale will


13–37.

Cylinder B has a mass m and is hoisted using the cord and d/2 d/2

pulley system shown. Determine the magnitude of force F as a

function of the cylinder’s vertical position y so that when F is F

applied the cylinder rises with a constant acceleration aB. y

Neglect the mass of the cord, pulleys, hook and chain.

S O L U T I O N

y

aB

B

+ c ©Fy = may ;2F cos u - mg = maB

where cos u =

2 d 2

2 y + A 2 B

y d 2

2F£ 2y 2

+ A

2B

≥ - mg = maB

2

F = m(aB + g)2 4y + d

Ans.

4y


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United


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use



of the

is solely work


this


of any sale will


13–38. The conveyor belt delivers each 12-kg crate to the ramp at A 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°.

S O L U T I O N

Q +©Fy = may; NC - 1219.812 cos 30° = 0

NC = 101.95 N

+ R©Fx = max ; 1219.812 sin 30° - 0.31101.952 = 12 aC 2 aC = 2.356 m>s

(+ R) v 2 = v 2

+ 2a 1s - s 2

BA 2 c B A

vB 2 = 12.52 + 212.356213 - 02

vB = 4.5152 = 4.52 m>s

vA 2.5 m/s

A

Ans.

laws or

3 m

u B


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this

This is the

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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 Fx = F0 and Fy = 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; 0.3 F0 = may Thus,

y

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

v0 x

vx dv = t F0

Lv0

L0

x m dt

F 0

vx =

m t + v0 vy dv = 0.3F

0 0.3F

0

L0 y L0 m dt vy = m t laws or

F 0 2

v = t + v + t

C a m0 b 1 a m b 2 2

= + 2F0tmv0 + m v0

F

m L0

dx = L0

m

t

x

0.3F 0

F

0 t

+ v0 t

x = 2m

L0

y t 0.3F0

y = 0.3F0 t 2m

2

t = a B

by

0.3F0

2m

2m

F0

+ v 2m

x = 2m a 0.3F0 by 0 a B 0.3F0 y 2m

+ v

x = 0.3 0 a B 0.3F0 by 2


*13–40. 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.

v

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 or teaching Web) Dissemination

a = g sin u Ans. copyright Wide .


United


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this


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> s2

. In all cases, the collar moves in the plane.

S O L U T I O N

(a) + b©Fx¿ = max¿ ; 219.812 sin 45° = 2aC aC = 6.94 m>s2

2

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

aC = aA + aC>A Where aA = 0

2 = 6.94 m>s

(c)

a a a C = A + C>A

= 2 + a

√ C>Ab laws (1)

teaching T

.Dissemination From Eq.(1) copyright Wide

+ b©Fx¿ = max¿ ; 219.812 sin 45° = 212 cos 45°+ aC>A2 aC>A = 5.5225 m> 2 b in

B

45 C

A Web) or

.

aC = 2 + 5.5225 = 3.905 + 5.905 States instructors World permitted

T b ; T of learningthe is not aC = 23.905 + 5.905 = 7.08 m>s United use

5.905 on andAns.

u = tan-1 = 56.5° ud for student work Ans. by the

protected of is sol el y assess in g work

work provided and of

this

This is the

and courses part

of any

sale will


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

2 an upward acceleration of 4 m>s .

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 + a

CAB C>AB

1aC)x = 0 + 9.76514 sin 45° = 6.905 ;

(aC)y = 4 - 9.76514 cos 45° = 2.905T

aC = 2

2 2

16.9052 2 + 12.9052 = 7.49 m>s

B

458 C

A Ans.la ws or


-1 2.905


United


use and



is solely work


this

This is the

and courses part

of any sale will


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 s 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 Ff

develops. Since the crate is required to be on the verge of slipping, Ff = 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 +

: ©Fx = max; - 0.3(1962) = 200( - a)

2 a = 2.943 m>s ;

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

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

( ; ) v = v0 + ac t

km 1000 m 1 laws or

b a b a bteaching= Web)

copyright

.Dissemination

h 1 km 3600 .

in


Wide States World

16.67 = 0 + 2.943t United of learning the is not on

use and

for student protected by the of the


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 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.

Equations of Motion: By referring to Figs. b and c,

+ c ©Fy = may; 2T - 10(9.81) = - 10aA (1)

and

+ c ©Fy = may; T - 30(9.81) = - 30aB (2) laws or

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.


The second derivative of the above equation gives of learning the is United on not

2aA + aB = 0 use

and(3) Solving Eqs. (1), (2), and (3) yields for student work

>

>

by the

A = - =

B protected

T of the

> = assess in g>

a 3.773 m s 2 3.77 m s c 2 a 7.546 mis s 2

7.55 mworks 2

Ans.

solely

T = 67.92 N = 67.9 N provided ofintegrity Ans. work and this

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and courses part

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sale will

A

10 kg

B

30 kg


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

A B 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.


+ c ©Fy = may; N - 50(9.81) = 50(0) N = 490.5 N

Realizing that Ff = mkN = 0.3(490.5) = 147.15 N,

+ c ©F 3>2

= ma ; 100t - 147.15 = 50a

x x

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)

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. copyright Wide

Dissemination .

+ 3>2 States instructors World permitted

: ©Fx = 0; 100t - 196.2 = 0 of learning the is not t = 1.567 s Uniteduse on

and

by the student Kinematics: Using the result o f a and integ ratin g the kinem ati cfor equatio n wo r kd v = a dt

with the initial condition v = 0 at t = 1.567 as the lower inte (including ration limit,the

+ protected of

dv=adt is solely work

( : )

L assessing

L this

v

t

work and of

2t 3>2 This is provided the integrity

L0 dv = L1.567 s - 2.943 Bdt part 5>2 and courses t any

v = A0.8t - 2.943tB 1.567 stheir of destroy

2 sale will

v = A0.8t5>2

- 2.943t + 2.152 B m>s When t = 5 s,

5>2 v = 0.8(5) - 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 P - N sin u = ma P - mg tan u = mg tan u

laws or

teaching Web) .AnDissemination.

copyright

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

A

P

B u

C

SOLUTION Require

aA = aB = a

Block A: + c ©Fy = 0;

+ ; ©Fx = max ;

Block B:

+ ; ©Fx = max;

N cos u - msN sin u - mg =

0 N sin u + msN cos u = ma

mg N = cos u - ms sin u

a = ga cos u - ms sin u b

sin u + ms cos u

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

laws teaching Web)

Dissemination copyright Wide .


States . World sin u + ms cos u sin u + m c slearningu

of

b andthe is not P - mga b = mga Uniteduse

on

sin u + ms cos u

for student work

bycos

cos u - ms sin u u the

- ms s

cos u - ms sin u protected of P = 2mga b assessing the Ans. is solely work


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of any sale will

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

drag 2 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 .

S O L U T I O N

F D

v

2

dv

+ T ©Fz = m a z ;

v m dv t

m L0 1mg - kv 2

2 = L0 dt

v

m L dv = t k

k

mg

m

mg - kv = m dt

mg

0

- v2

1 Ak

¢ 2A k

k k

copyright

¢

Ak

mg ln D mg

mt

¢2A k ¢ = ln

mg

A mg

A k

mg

+ v

2t2 mg

A k

ek

A k assessing

A

A k e2t2

=

mg

e

mg

- v e

k

mg e2t2 k

mg

mg k

When t : q v

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

a 0

θ

time t. The block starts from rest when s = 0, t = 0. C

A B

S O L U T I O N s

Q + ©Fx = m ax;0 = m aB sin f Q +

Thus,

a = a + a B AC B>AC

a = a + a B 0 B>AC

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

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

sin aB>AC = a0 u t

B>AC

laws

. teaching

or

Dissemination Web) copyright Wide L0 L0 States instructors permitted

v

sin u dt

. learning

of the is not

= s =

a0sin u t dt

United on use and

t

L0 for student work by (including

the

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2

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is solely work

s = a0 sin u t the Ans.


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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?

S O L U T I O N

+

: ©Fx = max ; - kv cos u = m ax + c ©Fy = m ay ; - m g - k v sin u = m ay

or

y

v 0

O θ0

x

2 dx d x

2

- k dt = m dt 2 dy d y

= m

- m g - k dt dt 2

#

- k laws or

teaching .

Integrating yields Web)

Dissemination

mg k copyright Wide permitted

In x = m t + C1 instructors In (y + ) = t + C2

# States . World

not

k m of the United on use

# # student work

and x = v0 cos u0 e

When t = 0, x = v0 cos u0, y = v0 sin u0 by the

of the

# -(k>m)t protected for

m g mg - (k> m)tassessing

is solely work

provided integrity

y = -

k work and of this

+ (v0 sin u0 + ) This is part the

- m v0 and courses any

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

of

k sale will

y = - mg

t - (v0 sin u0 + mg

)( m

)e-(k>m)t k k k

When t = 0, x = y = 0, thus

m v0 - k>m t ( )

x =

k cos u0(1 - e )

y = - m g t

+ m

(v0 sin u0 + mg )(1 - e-(k>m)t)

k k k

As t : q

x =

m v cos

max 0 u

0 k

Ans.

Ans.

Ans.

13–51.

The block A has a mass mA and rests on the pan B, which has a 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 k 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 = (mA + mB)g

y F s (m A + m B )g

eq = k = k Block:

+ c ©Fy = may ; - mA g + N = mA a

Block and pan

+ c ©Fy = may; - (mA + mB)g + k(yeq + y) =(mA + mB)a

Thus,

or

laws

teaching Web) .

copyright Wide m A + m B - mAg + N . Dissemination

- (mA + mB)g + kc a k

b g + yd = (m A

+ m B

)a States b World permitted

mA instructors

Require y = d, N = 0 of learning the is not Uniteduse on

by the and

kd = - (mA + mB)g for student work (including

Since d is downward, protected of the is solely work

(mA + m assessing

this

Bwork)g and of

d = k provided integrity Ans. This is the


of their destroy

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

*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.

S O L U T I O N

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 2

v

©Fn = man ;

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

laws or



United


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of any sale will

z

5 m


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.

S O L U T I O N Free-Body Diagram: The free-body diagram of block B is shown in Fig. (a). The tension

A

r

B

v

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).

2 2 v 10

Equations of Motion: Realizing that an =

r = r and referring to Fig. (a),

2

10

©Fn = man;

147.15 = 2a r b r = 1.36 m Ans.


copyright Wide .


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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.

S O L U T I O N Free-Body Diagram: The free-body diagram of block B is shown in Fig. (a). The tension

A

r

B

v

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).

v2 v2

Equations of Motion: Realizing that an =

and referring to Fig. (a),

r =1.5 2

v

©Fn = man; 147.15 = 2a 1.5 b v = 10.5 m>s Ans.

laws or



United


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use



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this

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13–55. 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,

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

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.

P

v 5 4 m/s 308

A

0.5 m

laws or



United


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is solely work


this

This is the

and courses part

of any sale will


*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 between a carton and the conveyor are ms = 0.7 and mk = 0.5, respectively.

S O L U T I O N

+ c ©Fb = m ab; N - W = 0

N = W

Fx = 0.7W

+ 2

W 8

; ©Fn = m an;

0.7W = 9.81 ( r )

r = 9.32 m

8 m/s

ρ

Ans.

laws or


copyright Wide .


United


use and



is solely work


this

This is the

and courses part

of any sale will


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.

S O L U T I O N

r 300

r = 120 mm = 0.120 m

200 =500 ; 2

4 (0. 5) 3

+ Q©Fy = may;

T - 0.2(9.81)a 5 b = B0.2¢ 0.120 ≤ Ra 5 b T = 1.82 N

2 Ans.

3 (0.5) 4

+ a©Fx = max;NB - 0.2(9.81)a 5 b = - B0.2¢ 0.120 ≤ Ra 5 b NB = 0.844 N Ans.

Also, laws or

+ 3

4 teaching Web)

Dissemination

: ©Fn = man; Ta

b

5 b - NB a 5 copyright Wide (0.5) instructors permitted

0.120

4

3

+ c ©Fb = 0; States . World

Ta 5 b + NB a 5 b - 0.2(9.81) = 0 of learning the is not

T = 1.82 N United on use and

NB = 0.844 N for student work Ans.

by the (including


is solelywork


this


of any sale will

z

A

200 mm

B 400 mm

300 mm

© 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 publis her prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical,

13–58. The 2-kg spool S fits loosely on the inclined rod for which the z 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.

S 0.25 m

S O L U T I O N A

4

5 b

+

r = 0.25a = 0.2 m

v2 N a 3

b 4 4 s 35 0.2

; ©Fn = m an; s 5 - 0.2N a b = 2a b

N a b 0.2N a b

+ c ©Fb = m ab; s 5 + s 5 - 2(9.81) = 0

Ns = 21.3 N v = 0.969 m>s Ans.

laws or

teaching Web)

. copyright Wide

Dissemination


United


by use and

the student


assessing

this



of their destroy

sale will

5

3 4


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.

S O L U T I O N

4

r = 0.25( 5 ) = 0.2 m 2

+ 3 4 v

; ©Fn = m an ; Ns( 5 ) + 0.2Ns( 5 ) = 2( 0.2 )

4 3

+ c ©Fb = m ab ; Ns( 5) - 0.2Ns( 5 ) - 2(9.81) = 0 Ns = 28.85 N

v = 1.48 s Ans.


copyright Wide .


United


use and



is solely work


this


of any sale will

z

S

0.25 m

A

5

3 4


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

SOLUTION

60 2

Ans.

+ R©Ft = mat ; 60 cos 60° = 32.2 at at = 16.1 ft >s

60 2

15

Q + ©Fn = man ;

Ans.

2T - 60 sin 60° = 32.2 a

10 b T = 46.9 lb

laws or


copyright Wide .


United


use and



is solely work


this

This is the

and courses part

of any sale will

u

10 ft

G


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.

S O L U T I O N

60 a = 32.2 cos u

+ R© = ma ;

60 cos u = 32.2 ta

t t t 2

60 v

Q + ©Fn = man;

2T - 60 sin u = 32.2 a 10 b (1) v dn = a ds however ds = 10du

v 90° v dn =

L

L0

60° 322 cos u du

Ans.

v = 9.289 ft>s

From Eq. (1) 2

laws or teaching Web)

60 9.289 Dissemination

2T - 60 sin 90° =

copyright Ans. Wide .

32.2a 10

b T = 38.0 lb


United


use and



is solely work


this


of any sale will

u

10 ft

G


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

S O L U T I O N

1 2

n = 8 x

1 2 y

y = –– x 8

6 ft

A

x

dy 1

dx = tan u = 4 x 2 x = - 6 = - 1.5 u = - 56.31°

2 d y 1

2

dx = 4 2

dy 2 3

23

B1 + a dx b R2 C 1 + (- 1.5) D r =

=

= 23.436 ft

d y2 1

2

2 4 2

2 dx laws or teaching .

+ Q©Fn = man ; N - 10 cos 56.31° = a 32 .2 b ¢23 .436 ≤ Dissemination

Web)

2 copyright Wide

N = 5.8783 = 5.88 lb Ans. permitted instructors

States . World learning

+ R©Ft = mat; - 0.2(5.8783) + 10 sin 56.31° = 32.2 atUniteduse of on the is not 10

the student by (including

a

2 protected for of the work

= 23.0 ft s

Ans.

t


this


of any sale will


13–63.

The 150-lb man lies against the cushion for which the z

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

S O L U T I O N 2

150 20

+ aaFy = m1an2y ;

N - 150 cos 60° =

8 b sin 60° 32.2 a N = 277 lb

2 Ans.

+ b aF = m1a 2 ; - F + 150 sin 60° = 150 a 20 b cos 60°

x n x 32.2 8

F = 13.4 lb Ans. Note: No slipping occurs

Since ms N = 138.4 lb 7 13.4 lb

laws or


copyright Wide .


United


and

use



of the

is solely work


this

This is the

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G

u


*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.

S O L U T I O N 2

+ 150 1302

; ©Fn = man; 0.5N cos u + N sin u =32.2a 8 b

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

150

N = cos u - 0.5 sin u 2 10.5 cos u + sin u2150 150 1302

1cos u - 0.5 sin u2

= 32.2 a 8 b

z

8 ft

G

u

0.5 cos u + sin u = 3.49378 cos u - 1.74689 sin u teaching Web)

u = 47.5° laws

Dissemination or Ans.

copyright Wide .


United


and

use



is solely work


this

This is the

and courses part

of any sale will


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?

S O L U T I O N

2 + v

; ©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 Ans.

F 16.302

©Fn = m an ; = 50(

) = 283 N

Ans.

n

7

©Ft = m at; Ft = 0 Ans.

©F = m a b

; F b - 490.5 = 0

b laws or

Fb = 490 N

Ans.teach ing Web) Dissemination copyright Wide .


United


use and



is solely work


this


of any sale will

4 m

b 6 m

u

t n


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 2 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.

S O L U T I O N

©Ft = m at ; Ft = 80(0.4)

Ft = 32 N

3 m

10 m

v2

©Fn = m an ; Fn = (80) 3 2 2

F = ms Nm = 2(Ft) + (Fn)

v2 2 2

3 )

0.3(80)(9.81) = A(32) + ((80) 4

v

55 432 = 1024 + (6400)(

9 )

v = 2.9575 m>s

dv

at = dt = 0.4 v t

L0 dv = L0 0.4 dt v = 0.4 t 2.9575 = 0.4 t

t = 7.39 s

laws or

teaching Web)



United


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of the

is solely work

provi ded

and of integrity Ans.

work this

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

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

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

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

S O L U T I O N 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 1 h 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 2 an = r =

100 = 4.938 m>s . By referring to Fig. (a), laws

v 22.22 teaching

cos . Dissemination

copyright Wide

Web)

.

9.81m nstructors i not permitted

9.81m States World

+ c ©Fb = 0; N cos u - m(9.81) = 0 N = learning

+ of the is

cos u

United on use and

u = 26.7° for student Ans. by the


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*13–68. The 0.8-Mg car travels over the hill having the shape of a y 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.

S O L U T I O N 2

Geometry: Here, dy = - 0.00625x and d y

= - 0.00625. The slope angle u at point

2

dx dx

A is given by

dy

tan u =

= - 0.00625(80)

u = - 26.57°

dx 2 x = 80 m and the radius of curvature at point A is

2 3>2 2 3>2 [1 + (dy>dx) ] [1 + (- 0.00625x) ] 2 2

r = |d y>dx | = |- 0.00625| 2 x = 80 m = 223.61 m andla

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

with u = 26.57°

ws Web) teaching

Disseminationor ©

F t = m

at ; 800(9.81) sin 26.57 ° - Ff = 800(0 ) copyright Wide

r = 223.61 m, we have .

Ff = 3509.73 N = 3.51 kN

instructors permitted States Ans.World

. 92 learning is

©Fn = man;

800(9.81) cos 26.57° - N = 800Uniteda

of thenot us e b on

and

N = 6729.67 N = 6.73 kN for student work Ans. by the 223.61


is solely work


this


of any sale will

x2 y 20 (1 6400 )

A

x 80 m


13–69. The 0.8-Mg car travels over the hill having the shape of a y

parabola. When the car is at point2A, 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.

S O L U T I O N 2

Geometry: Here, dy

= - 0.00625x and d y

= - 0.00625. The slope angle u at point dx dx 2

A is given by

dy tan u = dx 2 x = 80 m = - 0.00625(80) u = - 26.57° and

the radius of curvature at point A is

C 1 + (dy>dx) 2

D 3>2

C 1 + (- 0.00625x)2

D 3>2

2 2 r = |d y>dx | = |- 0.00625| 2 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

Ff = 1109.73 N = 1.11 kN . DisseminationAns. or

©Ft = mat;

800(9.81) sin 26.57° - Ff = 800(3)

copyright Wide .

9 2instructors not permitted

©Fn = man; States World

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

the is

of

United and

use

forstudent of the

N = 6729.67 N = protected by the

6.73 kN Ans.


this


of any

sale will

x2 y 20 (1 6400 )

A

x 80 m


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.

S O L U T I O N

+ b ©Ft = mat ; 5

5 cos f = 32.2 at

a t = 32.2 cos f

5 v

+ a©Fn = man ;N - 5 sin f =

32.2 (20 )

v dv = at ds v f

Lg v dv = L45°32.2 cos f (20 df)

1 2 1 2 laws Web)

teachi ng

(8)= 644(sin f - sin 45°)

Dissemination or v - copyright Wide

2 2 .

At f = 45° + 30° =

vC = 19.933 ft>s = 19.9 ft>s World permitted

75°, .

NC = 7.91 lb United

of learning the is not on Ans.

use and

At f = 45° + 45° = 90° for student work by the (including

vB = 21.0 ft s protectedsolely of Ans. assessing is work the


this


of any sale will

45

8 ft/s

θ = 30 45

2 0 ft A

B C


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.

S O L U T I O N 2

(4)

+ c ©Fn = man; T - 30(9.81) = 30a 4 b

T = 414 N

+ Q©Ft = mat;- 30(9.81) sin u = 30at

at = - 9.81 sin u

at ds = v dv Since ds = 4 du, then

u 0

- 9.81 L

1

L4 v dv

C 9.81(4)cos uD u

(4 2

= - 2 0 )

39.24(cos u - 1) = - 8 u = 37.2°

for by

protected

assess in g

is solely

u

4 m

Ans.

laws or

teaching Web)


instructors not permitted

States . World

of learning the is

United on Ans. use and

the student

(including work

of the

work


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and courses part

of any

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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

4 m

S O L U T I O N

v2 + a©Fn = man; T - 30(9.81) cos u = 30a 4 b

+ Q©Ft = mat;- 30(9.81) sin u = 30at at

= - 9.81 sin u

at ds = v dv Since ds = 4 du, then

u v

- 9.81

L0 sin u (4 du) = L4 v dv u

1

1

2

2 laws or

9.81(4) cos u 20 = 2 (v) - 2 (4) teaching Web)

1 2 Dissemination

39.24(cos u - 1) + 8 = 2 v copyright Wide .

At u = 20° instructors permitted States . World

v = 3.357 m>s 2

2


United

on and

at = - 3.36 m>s =

3.36 m>s

b T use

the student

= 361 N by (including

Ans.

for work

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solely work the


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and courses part

of any sale will


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?

S O L U T I O N 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 v

A r r

B r r

Realizing that an = r = r and referring to Fig. (a), 2

+ T ©Fn = man;

v

mg = m¢ r ≤ v = 2 gr Ans. teaching Web)

+ c © F = ma ; N - mg = mg component of .Dissemination or Using nthe resultn of v,the normal car accelerationlaws

v gr copyright Wide

an =

=

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

N = 2mg

S tat e s

instructors not permitted World

of learning is the Ans.

United on use and



of the

is solely work


this


of any sale will

© 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:

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.

S O L U T I O N 2

3500 v

T ©Fn = man; 3500 = 32.2 a 200 b

v = 80 2 ft>s

v

r 200 ft

Ans.

laws or



United


use and



is solely work


this

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and courses part

of any sale will


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.

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.

2 v v B 2

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

+ c ©Fb = 0; mAg cos u - mmBg = 0 B

-

a m

u = cos 1 A b vB 2

Ans. + or

; ©F = ma ; m g sin u (l - h) sin u laws

n n A

teaching Web)

.

m A

g (l - h)

(1)

=

vB

sin u

mB

B

copyright Wide

Dissemination

From Fig. b, sin u = 2 2 .

2mA - mB States World permitted . Substituting this value into Eq. (1),instructors not

A of learning the is

2 United on use

vB =

m A g(l - h) 3m A- m 2 B by

the student

and

for

work

B

mB

mA

(including protected the

£ is ≥ solely assessing work of

2 2

= B g(l - h)(mA - mB ) this

Ans.

work and of integrity m m

A B provided

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of their destroy

sale will

C

u h

A

B


*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.

B

h

A

S O L U T I O N

+ R©Ft = mat; mg sin u = mat at = g sin u

v dv = at ds = g sin u ds However dy = ds sin u

v h

L0 v dv = L0g dy

2 v 2 = gh

v = 2 2gh

Q.E.D.


copyright Wide .


United


and

use




this


of any sale will


13–77.

The skier starts from rest at A(10 m, 0) and descends the y smooth slope, which may be approximated by a parabola. If she

has a mass of 52 kg, determine the normal force the ground 10 m 1 2 exerts on the skier at the instant she arrives at point B. Neglect 20

y –– x 5

the size of the skier. Hint: Use the result of Prob. 13–76. A x

5 m B

S O L U T I O N 2

dy 1 d y 1

2

Geometry: Here, dx = 10x and dx = 10 . The slope angle u at point B is given by dy

= 0 tan u = dx x = 0 m u = 0°

and the radius of curvature at point B is c 1 + 12 3>2

C 1 + (dy>dx)2

D 3>2

2

2

r = |d y>dx | =

teaching

y2

Equations of Motion:

permitted

+ b©Ft = mat; 52(9.81) sin u = - 52at + a©Fn = man;

use

must be in the direction of Kinematics: The speed of the skier can be determined usingby

1

is

Here, tan u = 10x. Then, sin u = v 0

is

+ L0 = - L10 m 10

v = 9.81 m sale> 2

a 10

positive

x

102 1 +

This

2

1 +

100 xof

2

2 2 2 Substituting v = 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

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.

S O L U T I O N 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. 2 2

v 6

Since an = r = 0.5 + l sin u , by referring to Fig. (a),

+ c ©Fb = 0; 20(l - 2) cos u - 10 = 0

+ 10 62

; ©Fn = man;

20(l - 2) sin u = a b Solving Eqs. (1) and (2) yields

32.2 0.5 + l sin u u = 31.26° = 31.3°

l = 2.585 ft

(1)

laws Web) teaching

(2) or Ans.Dissemi nation

copyright Wide .

6 in.

A

u k 20 lb>ft


of learning the is

Ans.no

t

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the student


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is solely work


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sale will

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.

S O L U T I O N

u

r

+c aFb = mab;

+

F ; a n = man;

NP sin 15° - 7019.812 = 0

NP = 2.65 kN

2 50

NP cos 15° = 70a r b r = 68.3 m

Ans.

Ans.

laws or


copyright Wide .


United


use and



is solely work


this

This is the

and courses part

of any sale will


*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.

r

S O L U T I O N 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 1 h Equations of Motion: The speed of the airplane is v = ¢350 h ≤ ¢ 1 km ≤ ¢ 3600 s ≤

2 2 2

= 97.22 m>s. Realizing that an = v = 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 or Solving Eqs. (1) and (2) yields teaching Web) Dissemination copyrigh t Wide

u = 17.8° T = 51517.75 = 51.5 kN Ans. .


United


use and



of the

is solely work


this

This is the

and courses part

of any sale will

L

u


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. r

S O L U T I O N 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 1 h

Equations of Motion: The speed of the airplane is v = a 350

h b a 1 km b a 3600 s b 2

v 97.22

= 97.22 m

>s. Realizing that

an = r = r and referring to Fig. (a),

L

u

+ c ©Fb = 0;

+

; ©Fn = man;

L cos 15° - 5000(9.81) = 0

L = 50780.30 N = 50.8 kN 2

Ans.

97.22

≤

50780.30 sin 15° = 5000¢ r laws or

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


United

of learning the is not use on

and the student for (including work

by

protected of the

assess in g

is solely work


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and courses part

of any

sale will


13–82. The 800-kg motorbike travels with a constant speed of y 80 km> h up the hill. Determine the normal force the

surface exerts on its wheels when it reaches point A.

Neglect its size.

S O L U T I O N 2

1>2 dy

22 d y 2

= 2

Geometry: Here, y = 22x. Thus, dx andx

2 = -

3>2 . The angle that 1>2

2x 4x the hill slope at A makes with the horizontal is

dy - 1

u = tana dx b 2 x = 100 m

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

dy

B1 +

rA = a dxb d y

2

2

dx 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

v = a km 1000 m

80 h

b a b a

1 km

2

Thus, v2 22.22 an= =

R + rA 2849.67

©Fn = man; This and

sale

A 2

y 2x

100 m

2 2

- 1 2x1>2

= tan ¢

≤ 2 3>2

R

5

x = 100 m

1 h

b

3600 s

800(9.81)cos 4.045° - N = 7689.82 N =


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 2

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

S O L U T I O N

O

u

l

+

: ©Fn = man; + c ©Fb = 0;

Since r = l sin u

v 2

0

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

T = mv0 2

2 l sin2

u mv cos u 0

a l b a sin2

u b = mg 2

tan u sin u = v Q.E.D.

0


copyright Wide .


United


use and



is solely work


this


of any

sale will

v0


*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.

S O L U T I O N 2

y = 8 - 1x 2

dy

- dx = tan u= x 2 x = 2 = 2 u = 63.435°

2 d y

y

10 ft/s

A 1 2

y 8 –– x

2

O x

2 ft

dx 2 = - 1 dy

(1 + (- 2) 2 )2 3

B1 + a dx b 2 R 3 = = 11.18 ft

r = d2

y 2 |- 1|

2

dx 2 laws or teaching Web) Dissemination co p yri gh t Wide .

OA = 2 2 2

= 6.3246

(2) + (6) instructors permitted

2 States . World

of the is not United on

6

use and

tan f = ; f = 71.565°

the student work by (including prot ect ed of 2

2

assessing32.2 the 11.18

work5 (10)

is solely work provided and of integrity

+ b©Fn = man;

+ R©Ft = mat;

5 cos 63.435° - N + 33.246 cos 45.0° N = 24.4 lb This is


their

at = 180.2 ft>s

2

2 sale will

(10)

an = v2 = = 8.9443 ft>s

2

r 11.18 2 2

a = 2 (180.2) + (8.9443) 2

a = 180 ft>s

= b a b

the Ans.

5 b at 32.2

Ans.


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.

S O L U T I O N 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 z

laws or

teaching Web)

FA - 12(z + 0.3) = mz Dissemination 0.75 copyright Wide .

permitted

FA - 12(0.1 sin 2u + 0.3) = 32.2(- 14.4 sin 2u)

St ates

instructors. World For u = 45°,

United of learningthe is not 0.75 on

FA - 12(0.4) = 32.2 (- 14.4) by use and

FA = 4.46 lb for student work Ans.

protected the of the assess in g

is solely work


this

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and courses part of any

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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

2 - 6t

u = 1.5t

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

$

u = 3

$ # 2 ar = r - ru= 0 - 0 = 0 $ #

or laws au = ru + 2ru = 14(3) + 0 = 42 Hence,

©Fr = mar; Fr = 5(0) = 0

©Fu = mau; Fu = 5(42) = 210 N

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


copyright Wide .


United

of learning the is not on use andAns.


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this

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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

2 = (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 $ 2

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

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

©Fr = mar; F

r =

32.2 (- 5) = - 0.7764 lb 5

©Fu = ma u; F

u =

32.2 (9) = 1.398 lb laws

22 2 2 Web) teaching

.Dissemination or

F = 2Fr + Fu = 2(- 0.7764) + (1.398) = 1.60 lb copyright An . Wide

.

States instructors World

permitted

United


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*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

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

= 3 m>s r 2 #

u = t + 2 u = 2t| t = 2 s

3 # 2

z = 6 - t z = - 3t

ar = # # # 2 2 2

r - r u = 0- 10(4) = - 160 m>s

$ #

au = ru + 2ru = 10(2) + 2(3)(4) = 44 m>s

# # 2

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

r r

©F = ma ; Fu = 1.5(44) = 66 N u u

©Fz = maz;

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

$ r = 0

$ 2

= 4 rad>s u = 2 rad>s

# # 2

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

2

Ans. Anslaw

s. Web) teaching

.Dissemination or

copyright Wide .

Fz =

- 3.28 N A .



on

use and



is solely work


this

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and courses part

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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.

S O L U T I O N # $

Kinematic: Here, u = 5 rad>s and u = 0. Taking the required time derivatives

at u = 120°, we have

r = 1.5(2 - cos u)|u = 120° = 3.75 ft #

r# = 1.5 sin uu|u = 120° = 6.495 ft>s #

$ $ 2 )| 120° 2

r = 1.5(sin uu + cos uu = - 18.75 ft>s Applying Eqs. 12–29, we have

# $ 2 2

2

A

B

r

r = 1.5 (2 – cos ) ft.

O

= 5 rad/s

ar = r - ru = - 18.75 - 3.75(5 ) = - 112.5 ft>s

au = ru + 2ru = 3.75(0) + 2(6.495)(5) = 64.952 ft>s laws Web) $ # # 2 teaching

r 1.5(2 - cos u) . Dissemination

copyright Wide . tan c =

dr > d u = 1.5 sin u = 2.8867 c = 70.89° States World permitted

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

u = 120°

instructors

of learning the is not United on

and

aF

r = mar ; - N cos 19.11° = (- 112.5) the student Applying Eq. 13–9, we have by use

work

0.75 for

protected of

is assessing

32.2 solely work the

N = 2.773 lb = 2.77 lb

work provide d and of thisintegrity Ans.

F

FOA + 2.773 sin (64.952) a u = mau ; 19.11°This = 32.2 part the

and courses

is 0.75 of destroy

any

FOA = 0.605 lb their

sale will

13–90. 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 = 1-0.5t2 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 r = 1.5 u = 0.7t z = - 0.5t

z

# $ # # r = r = 0 u = 0.7 z = - 0.5

$ $ u = 0 z = 0

ar = r

2

- r(u) = 0 - 1.5(0.7) 2 = - 0.735 $ #

au = ru + 2ru = 0 $

az

= z = 0

©Fr = mar; Fr = 40(- 0.735) = - 29.4 N Ans. Ans.la

©Fu = ma u; Fu = 0 ws

Web) teaching Dissemination or

©Fz = maz; Fz - 40(9.81) = 0 copyright Wide .

Fz = 392 N instructors permitted States World

Ans. .

United


use and



is solely work


this


of any sale will

z

u

r 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$ 2angular 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.

SOLUTION

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

#

= - sin uu # 2 $

r = - cos uu - sin uu # $

2

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

= - 17.856 - 0.8660(4) 2

= - 31.713 ft>s 2 laws or

ar = r - ru Web) $ # #

teaching 2

.

au = ru + 2ru = 0.8660(8) + 2(- 2)(4) = - 9.072 ft>s copyright Wide

Q + ©Fr = mar; - N cos 30° = 0.5 (- 31.713) N = 0.5686 lb

Dissemination

. permitted

32.2 States World instructors

of learning the is not 0.5 United on use

+ a©Fu = mau; F - 0.5686 sin 30° = 32. 2 (- 9.072) by the and student

F = 0.143 lb

for (including work protected of the Ans. is solely work

assessing

this



of their destroy

sale will

r

u

0.5 ft

0.5 ft


*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 in2radians. If the angular position

is u = 10.5t 2 rad, where tis in seconds,ofthearm 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.

S O L U T I O N # $ $

r = 0.5u r = 0.5u r = 0.5u

2 # $

u = 0.5t u = t u = 1

At t = 2 s, #

$

2

u = 1 rad>s u = 2 rad = 114.59° u =2 rad>2 # $ 2

r = 1 m r = 1 m>s r = 0.5 m>s

r 0.5(2)

tan c = dr>du = 0.5 c = 63.43°

ar = r - ru = 0.5 - 1(2) = - 3.5 $ #

# # 2 2

#

au = ru + 2r u = 1(1) + 2(1)(2) = 5

NC = 3.030 = 3.03 N

laws Web) teaching

. Dissemination or

copyright Wide


States World

of learning Ans.not United on the is

use

C

u r 0.5 u

for student work

+ b© Fu = mau;

by the (including

F - 3.030 sin 26.57° + 4.905 sin 24.59° = 0.5(5) and

protected of the F = 1.81 N assessing Ans.

is solely work


this

This is the

and courses part

of any sale will


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

SOLUTION

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

r = 4 sec u |u= 45° = 4 sec 45° = 5.657 ft

# =

4 sec u(tan u)u|

# r = 4 sec 45° tan 45°(1.5) = 8.485 ft>s

r

u

u O

4 ft

u= 45°

#

$ = $

2 #

r 4 Csec u(tan u)u + uAsec u sec uu + tan u sec u tan

$ 3 # 2 2 2

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

D u= 45°

C

2 2 Using the a bove time de riva tive s,

= 38.18 ft>s 3 2 =4 sec 45° tan 45°(0) + sec 45°(1.5) + sec 45° tan 45°(1.5)

#

uuB D

laws or

in teaching . .Disseminationpermitted

copyright Wide Web)

instructors not

r = 38.18 - 5.657 1.5 = 25.46 ft>

U nit e d s learningon

a = r - ru States World

$ #2 2 2 of the is

$ # # A B by use 2 and

student work for (including

au = ru - 2ru = 5.657(0) + 2(8.485)(1.5) = 25.46the ft>

assessing

Fig. a, is solely work shown in Equatio ns of Mot ion: By referring to the free-body diagram of the cy linderthe

provide 3 2.2d integrity

4 this

N cos 45° - 4 cos 45° work = ( 2 5 . 4 6 )and of

This is part the

©Fu = mau; and courses (25.46) N = 8.472 lb their any

4

of

sale will 32.2

FOA = 12.0 lb Ans.

13–94. 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# = 90°, if the force F maintains a constant angular motion u = 2 rad>s.

SOLUTION u

r = e

u

r # = e u $ u 2 u

F

r r u

e

u

r = e (u) + e u At u = 90°

#

u = 2 rad>s $

u = 0 r = 4.8105 laws Web) teaching

# . Dissemination or a = r - r u

r = 9.6210() = 19.242 - 4.810 5(2 ) = 0

$ copyright Wide

r = 19.242

2

States

World permitted

$ # instructors

#

$ # 2 2 learning

au

= ru + 2ru =

0 + 2(9.6210)(2) = 38.4838 m>s not of the is

r United on

u u use and

tan c =

dr = e >e = 1 by the student

AduB

c = 45° for (including work

protected of

is assessing

solely work the


+ c a Fr = mar; - NC cos 45° + F cos 45° = 2(0) this

+

This is the andcourses part

F

NC = 54.4 N

destroy

Ans. ; a u

their of an y

F = 54.4 N

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

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

# $ # 2 #

ru + 2ru = 11>r21d1r u2>dt2 = 0.

direction yields 2au

=

When integrated, r u = c, where the constant c is determined

from the problem data.

SOLUTION

F = $ # 1d 2 #

= mc

r

0 = m[ru + 2r#

u]

dt (r u ) d = 0

a u mau; Thus,

2 d(r u) = 0

2

r u = C #

laws or 2 2

(0.5) (1) = C = (0.25) u teaching Web) # Dissemination

u = 4.00 rad>s copyright

Ans. Wide

#

$

Since r = - 0.2 m>s, r = 0 instructors not permitted

$ # States . World

ar = r 2 2

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

- 4 m>s

of learning the is

Uniteduse on a nd

a Fr = mar; - T = 2(- 4) the student

2 for (including work

T = 8 N by

protected of Ans.

is assessing

solely work the


This is the

and courses part

of any sale will

A

r

Bu

r0 u ·0

0.2 m/s C

F


*13–96.

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.

S O L U T I O N

A

· u 2 rad/s

r

0.5 m

u

0.5

r = cos u = 0.5 sec u # #

r = 0.5 sec u tan uu # 2 $ = E C (sec u tan uu + sec u(sec

#

r 0.5 ) tan u D

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

$ 2 2 3 2

r = 0.5 sec 30° tan 30°(2) = 0.6667 m>s r = 0.5C sec 30° tan 30° (2 ) + sec 30°(2 ) + sec 30° t an 30°(0)D

$ # 2 2 2

ar = r - ru = 3.849 - 0.5774(2) = 1.540 m>s 2

= 3.849 m>s

$ #

ru +au =

#

2.667 m>s

2

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

O

$

laws Web)

Dissemination or copyright Wide .

States World permitted

instructors not

. learning

United of the is

on

use

and

Q + ©Fr = mar; N cos 30° - 0.5(9.81) cos 30° = 0.5(1.for540) student work by the (including

N = 5.79 N protectedsolely of Ans. assessing is work the

p r o v id ed

and of integrity

+ R©Fu = mau; F + 0.5(9.81) sin 30° - work 5.79 s n 30° =this0.5(2.667)

F = 1.78ThisN courses part the Ans. and

is 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.

S O L U T I O N

0.5

r = cos u = 0.5 sec u

# #

= 0.5 sec u tan uu

r

$ 2 ## $

r = 0.5E C (sec u tan uu) tan u + sec u(sec uu)D u + sec u tan uuF

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

# $ 2

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

r = 0.5 sec 30° = 0.5774 m

#

= 0.6667 m>s

r = 0.5 sec 30° tan 30°(2)

$ = 4.849 m>s2 2 2 3 2

r = 0.5C sec 30° tan 30°(2) + sec 30°(2) + sec 30° tan 30°(3)D

A

· u 2 rad/s

r

0.5 m

u

O

laws teaching Web)

Dissemination or

copyright Wide .

a $ # 2 2 2

States World permitted

r = r - ru = 4.849 - 0.5774(2) = 2.5396 m>s instructors

$ #

.

2 learning

# of the is not

u = ru + 2r u = 0.5774(3) + 2(0.6667)(2) = 4.3987 m> s Uniteduse on and

Q + ©Fr = mar; N cos 30° - 0.5(9.81) cos 30° = 0.5(2.5396)student work for (including

by the

N = 6.3712 = 6.37 N protected of Ans.

is assessing

solely work the

+ R©Fu = mau; work provided and ofintegrity

F + 0.5(9.81) sin 30° - 6.3712 sin 30°this= 0.5(4.3987)

F =

2.93ThisNis part the Ans.

and courses of any

sale will


13–98. The collar has a mass of 2 kg and travels along the smooth

horizontal rod defined by the equiangular spiral r = 1eu

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.

S O L U T I O N u

r = e

# u #

r = e u

$ u # 2 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

$ copyright Wide .

r = 8.7731 instructors not permitted

$ # # #

2 2

2

States

World

r = r - r(u) = 8.7731 - 2.1933(2) = 0 . learning

au = r u + 2 r u = 0 + 2(4.38656)(2) = 17.5462 m>s of the is United on

u u by use and

tan c = dr = e >e = 1 the student work a b (including

c = u = 45°

r assessing

du is

solelywork

the

Q + F workprovided and of thisintegrity 2(0)

a r = mar ;- NC cos 45° + FThiscosis45° = part the and courses

any

sale will

+ a aFu = mau ;F sin 45° + NC sin 45° = 2(17.5462)

N = 24.8 N Ans.

F = 24.8 N Ans.

F

r θ

r = e θ


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 = 8 m 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.

S O L U T I O N

# $

Kinematic: Here, r = 8 m, r = r = 0. Taking the required time derivatives at t = 2s, we have

$ #

u = 0.1t + 0.5|t = 2s = 0.700 rad u = 0.100 rad>s u = 0 # $

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 $

az = z = 0 Equation of Motion:

Fr = 250(- 0.0800) = - 20.0 N

©Fr = mar ;

©Fu = mau ;

z = 0


.

DisseminationAns. copyright Wide .

instructors World permitted

Stat e s

of learning the is

©Fz = maz ; Ans. Fu = 250(0) = 0 United on not

use

student work

Fz - 250(9.81) = 250(0) by the (including and

protected of the = 2.45forkN Ans.

assessing


this

This is the

and courses part

of any

sale will


*13–100.

The 0.5-lb ball is guided along the vertical circular path P A

r = 2rc cos# u using the arm OA. If the arm has an angular v$elocity u = 0.4 rad>s and an angular acceleration

2

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. Set rc = 0.4 ft.

r

rc

u

O

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

# $ 2

0.4 rad>s, and u = 0.8 rad>s

At u = 30°, u = r = 0.8 cos 30° = 0.6928 ft

r#

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

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

$ # # 2 teaching .

= - 0.4309 - 0.6928(0.4)

= - 0.5417 ft>s

Dissemination Web) ar = r - ru copyright Wide

+ Q©Fr = mar; N cos 30° - 0.5 sin 30° = ( - 0.5417) N = 0.2790 lb permitted au = ru + 2ru = 0.6928(0.8) + 2( - 0.16)(0.4) = 0.4263 ft>s instructors

0.5 States . World

0.5 Uni

ted of learningthe is not

a + ©Fu = mau; use on

FOA + 0.2790 sin 30° - 0.5 cos 30° = by (0.4263) and student

for (including work

32.2t he

protected of assessing

FOA = 0.300 lb is solely work work provided and of integrity

this

the Ans.

This is 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 Neglect friction and the size of the ball.

P

A

r

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

2 2 2 ar = r - ru = - 2rc cos uu0 - 2rc cos uu0 = - 4rc cos uu0

+ Q©Fr = mar; #

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

4r u 2

4r u 2 An . c 0

u = tan - 1

¢

c 0

tan u = g g ≤ laws or teaching Web)

Dissemination

copyright Wide .


United


and

use



of the

is

solely work


this


of any sale will


leave the surface of the semicylinder.

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 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.

S O L U T I O N # $ $

r = 0.6 u r = 0.6 u r = 0.6 u

# = # # #

vr = r 0.6 u vu = ru = 0.6uu

# 2

2 2

v = r+ a rub

2 # 2 2 # 2

= a 0.6ub

2 + a 0.6uub

u =

uu 2

laws Web)

# $

# $

$

teaching

0.621 + u 2

# Disseminationor

copyright Wide

u = -

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

At u = p rad, # = 2 = 1.011 rad>s instructors permitted 1 + u2

States .World

2 learning

0.62 1 + p 2

of the is not

$

(p)(1.011) United on

and

2 use

2

u = - 1 + p = - 0.2954 rad>s by the student work for (including

r = 0.6(p) = 0.6 p m r# = 0.6(1.011) = 0.6066protectedm> of the $ 2

is

solely assess in g

work

$ # 2 2 work and of integrity r = 0.6(- 0.2954) = - 0.1772 m>s this

= - 0.1772 - 0.6 p(1.011)This= -is2.104

ar = r - ru mpart>s the

$ # # and coursesany 2

their of destroy

r 0.6u sale will

tan c = dr>du = 0.6 = u = p c = 72.34°

+

; ©Fr = mar ; - N cos 17.66° = 0.4(- 2.104) N = 0.883 N Ans. + T ©Fu = mau ; - F + 0.4(9.81) + 0.883 sin 17.66° = 0.4(0.6698)

F = 3.92 N Ans.

P

u p

r

r 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

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

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

= .. + . . = motion

in the u direction yields au ru 2ru 2 # 2 #

11>r2 d1r u2>dt = 0. When integrated, r u = c, where the

constant c is determined from the problem data.

S O L U T I O N

400 $ # 2

+ Q©Fr = mar ; - T = ¢ 32.2 ≤ ¢r - ru ≤ (1)

400 $ #

#

+ a©Fu = mau ; 0 = ¢ 32.2 ≤ ¢ru + 2r u ≤ (2)

1 d # #

2

2

From Eq. (2), ¢ r ≤ dt ¢r u≤ = 0 r u = c laws

# teaching Web) Dissemination

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

.

u = ¢ (4) 2 ≤

World permitted

Statesinstructors not

Hence, when r = 4 ft, .

#

51.2 of learning the is

$ Uniteduse

on

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

for student work

the

(including

- T = protected of 32.2 a 0 - (4)(3.2) b assessing the

00 2 solely is work


T = 509 lb this Ans.


of any sale will

r C

· u u 0

O

© 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 reproduct ion, 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.

*13–104.

# 2

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

S O L U T I O N

p u = 2 = 90°

# u = 5 rad>s $

2 u = 2 rad>s r = 0.2>u = 0.12732 m

# - 2 # r = - 0.2 uu = - 0.40528 m>s $ - 3 #2 - 2

r = - 0.2[- 2u (u) + u u] = 2.41801

a = r - r(u) = 2.41801 - 0.12732(5) = - 0.7651 m>s r 2

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

$ # 2 2 2

$

r 0.2 —

u

r · ·· 2

u 5 rad/s, u 2 rad/s

u 90

laws

. teaching

or

Dissemination Web)

copyright Wide tan c = (dr )=

instructors World permitted - 0.2u - 2

r 0.2>u . p learning

of the is not - 1 United on

c = tan (- 2 ) = - 57.5184° by use and

+ c ©Fr = m ar ; Np cos 32.4816° = 0.5(- 0.7651) for student work

protected theof the

+ N is solely ass ess in g work

P

= - 0.453 N provided of integrity ; ©F u

= m a u

; F + Np sin 32.4816° = 0.5( - 3.7982) work and this

This is part the and courses any F = - 1.66 N

of Ans.

sale will


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.

SOLUTION r = 2 + cos u

r# #

= - sin uu

$ # 2 $

r = - cos uu - sin uu

# $

2 ft r

· u

u 3 ft

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 $ 2

- sin 90°(0) = 0

r = - cos 90°(0.5) $ # 2 2 2 laws or $ # 2 teaching .

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

Dissemination Web)

copyright Wide

tan c = r = 2 + cos u = - 2 c = - 63.43° . World permitted

dr>du - sin u 2 u= 90° Statesinstructors not

United of learning the is

2 on

use

and

32.2 student for (including work

by

+ c ©Fr = mar; - N cos 26.57° =( - 0.5) N = 0.0the3472 lb

+ protectedsolely of assessing the

; ©Fu = mau; F - 0.03472 sin 26.57° =is ( - 0.5) work

work and of thisintegrity F = - 0.0155 lbprovided Ans.

32.2

This is the

and courses part

of any sale will

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 path contact the particle on only one side.

SOLUTION r = 2 + cos u # #

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

2 ft r

u ·

u 3 ft

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

r#

= - sin 60°(0.5) = - 0.4330 ft>s $ 2 2

r = - cos 60°(0.5) - sin 60°(0) = - 0.125 ft

>s

a $ # 2 2 laws or

r = r - ru 2

= - 0.125 - 2.5(0.5) = - 0.75 ft>s

teaching Web)

$ # 2 .

au = ru + 2ru = 2.5(0) + 2( - 0.4330)(0.5) = - 0.4330 ft>s copyright Wide

tan c = r

= 2 + cos u

= - 2.887

c = - 70.89° Dissemination

.World permitted

dr>du - sin u 2 u=

60° States instructors not

United

of learning the is

2 on

by use and + Q©Fr = mar; - N cos 19.11° =

32.2 ( - 0.75) N =0.04930the lb

student for (including work

2 protected of the -

F - 0.04930 sin 19.11° =

( 0.4330)sole

+a©Fu = mau; is ly work 32.2

assessing

F = - 0.0108 lbprovided

work and of this integrity Ans.

This is the

and courses part a n y

of

their destroy

sale will


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

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.

2

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

SOLUTION 3 ft

r = 2 + cos u 2 u = 0.5t

r

# #

= - sin uu u = t $ # 2 $ $ 2

r = - cos uu - sin uu u = 1 rad>s $

2 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

2 $ 2 2

r = - cos 0.5(1)

- sin 0.5(1) = - 1.357 ft>s

$ # 2 2 2 laws or $ # 2 teaching .

ar = r - ru

= - 1.375 - 2.8776(1) = - 4.2346 ft>s Dissemination Web)

copyright Wide

tan c = r = 2 + cos u = - 6.002 c = - 80.54° . World permitted

dr>du - sin u 2 u= 0.5 rad Statesinstructors not

United

of learning the is

2 on

use

and

32.2 student for (including work

by

+ Q©Fr = mar; - N cos 9.46° = ( - 4.2346) N =the0.2666 lb

protected of

F - 0.2666 sin 9.46° = is assessing the

+ a©Fu = mau; (1.9187)solely work

work and of thisintegrity F =0.163 lb provided Ans.

32.2

This is the

and courses part of any

sale will

*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

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°.

S O L U T I O N 4

r = 1 - cos u

r#

= #

- 4 sin u u 2

$ (1 - cos u)

# #

- 4 sin u u

$ - 4 cos u (u) 8 sin u u

2 2 3 r = (1 - cos u) + (1 - cos u) + (1 - cos u)

# $ At u = 90°, u = 4, u = 0

r = 4

r = - 16 $

r = r - r(u) = 128 - 4(4) = 64 r = 128

#

$ 2 2

$ #

laws or teaching .

. Dissemination Web)

copyright Wide instructors

States World

learning is

P

r

u

au = ru + 2 ru = 0 + 2(- 16)(4) = - 128 United of on the not

4 use and

r = 1 - cos u the student for (including work by

dr - 4 sin u is protected of

r

4

1 - cos u) provided integrity 4 work and of this

tan c = dr =

=This-

the

- 4 sin u

2 u = 90°= - 4 1is and

( ) courses part

du (1 - cos u) any

their

sale will

3

+ c ©Fr = m ar ; P sin 45° - N cos 45° = 32.2 (64) +3

; © Fu = mau ; - P cos 45° - N sin 45° = 32.2 (- 128) Solving,

P = 12.6 lb 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 P

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 r

parti#cle when u = 60°. The guide has a constant angular

velocity u = 5 rad>s.

·

S O L U T I O N 0.4 m u 5 rad/s

r = 0.8 sin u u

r#

# = 0.8 cos u u O

$ # 2 $ r = - 0.8 sin u (u) + 0.8 cos uu # $ u = 5, u = 0

At u = 60°, r = 0.6928

# r = 2 $

r = - 17.321


$ # 2

Dissemination

2

ar = r - r(u) = - 17.321 - 0.6928(5) = - 34.641 .

Fs = ks; Fs = 30(0.6928 - 0.25) = 13.284 N instructors permitted

au = ru + 2 ru = 0 + 2(2)(5) = 20 States . World

United


0.08(- use

Q + ©Fr = m ar; - 13.284 + NP cos 30° =

a + ©Fu = mau; F - NP sin 30° = 0.08(20)

F = 7.67 N

NP = 12.1 N

34.641)by and

for student work

(including

protected the of the solelyassess in g

is work Ans.

and of integrity provi ded

work this This is the

and courses part

of any

sale will

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$ m, determine2# the force of the guide on

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

P

r

S O L U T I O N r = 0.8 sin u # # r = 0.8 cos u u $ # 2 $ r = - 0.8 sin u (u) + 0.8 cos uu

# $

u = 5, u = 2

At u = 60°, r = 0.6928

r# = 2 $ r = - 16.521

$ 2 2

au= r u + 2 r u =

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

ar = r - r(u) = - 16.521 - 0.6928(5) = - 33.841

laws teaching Web)

Dissemination or copyright Wide

.

·

0.4 m u 5 rad/s

u

O

$ # #

Fs = ks; Fs = 30(0.6928 - 0.25) = 13.284 N instructors permitted States . World

Q + ©Fr = m ar; - 13.284 + NP cos 30° = 0.08(- 33.841)Uniteduse


and

+ a©Fu = mau; F - NP sin 30° = 0.08(21.386) for student work by the

F = 7.82 N protected of Ans.

is assessing

work

solely the


NP = 12.2 N this


of any sale will


13–111.

A 0.2-kg spool slides down along a smooth rod. If the rod has a constant angular rate of rotation u = 2 rad>s in the vertical plane, show that t he equations of

#

motion for the spool are r - 4r - 9.81 sin u = 0 and # 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

zero when t = 0, evaluate the constants C1 and C2 to determine r at the instant u = p>4 rad.

u u 2 rad/s

r

S O L U T I O N #

#

Kinematic: Here, u. = 2 rad>s and u = 0. Applying Eqs. 12–29, we have

$ # 2 $ 2 $

ar = r - ru = 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 Web)

teaching

Disseminationor

r - 4r - 9.81 sin u = 0 (Q.E.D.) copyright (1) Wi de

©Fu = mau; u.

1.962 cos u - Since

0.8r + Ns

= 2 rad>s, then u =

L0 # L0 u

equation (Eq.(1)) is given by

- 2 t

r = C1 e + C2

Thus,

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

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

9.81

r = - 16e

= 9.81a

9.81

= 8(sin h 2t

At u = 2t

# - 2t

r = - 2 C1 e + 2C2

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

C1 = -

- 2t 9.81 2t

+ 16 e - 8 sin 2t 2t 2t

- e + e - sin 2tb

8

- sin 2t) p 9.81 p

= 4 , r = 8 a sin h 4


*13–112.

The pilot of an airplane executes a vertical A

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

S O L U T I O N

r = 600(1 + cos u)|u = 0° = 1200 ft #

r = - 600 sin uu u = 0° = 0 $ $ #2

2

r = - 600 sin uu - 600 cos uu u = 0° = -600u # 2

2 # 2 vp = r + a rub

#

#

2

2

(80) = 0 + a 1200ub u = 0.06667 $ #

# $

2vpvp = 2rr + 2a rub a ru + rub or 2 $ $ laws

0 = 0 +

= 0

0 + 2r uu u teaching Web)

$

2

2

2

2

ar .

=r - ru = - 600(0.06667) - 1200(0.06667) = - 8 ft>s copyright Wide $ # # Dissemination

0 .

au = ru + 2ru = 0 + 0 = States World permitted

150 instructors not

of the is

+c ©F

= m a r ;

- 150 = a b (

- 8)N

= 113 lb learningAns.

r N 32.2 United use on by and

the student


assessing

this


and courses part

of any

their destroy

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

S O L U T I O N 2

Ch e = GM where C =0

1 GMS

2

e = GM

S r0 ¢1 - r0 v0

1 GMS

r

- r v 2 ≤ and h = r v ¢1 2

2 r 0 v0

≤(r0v0)

e = ¢ GMS - 1≤

2

r0v0 = e + 1

GMS GMS (e + 1)

y = r 0B 0

- 12 30

= B66.73(10 )(1.99)(10 )(0.0821 + 91) = 30818 m>s = 30.8 km>s 151.3(10 )

1 = 1 GMS GMS r v r v

r r0 0 0

66.73(10

- 12

)(1.99)(10 30)

66.73(10 1

1

9 2

b cos u + C 151.3 (10 ) D

= 0.502(10 ) cos u + 6.11(10 )

- 12 - 12 United

Ans.

laws Web)

or

. Dissemination

.

instructors not permitted

States World

92 (30818)2

learning Ans.

of on the is 1

use and


protected of the

assess in g

is solely work


This is the

and courses part

of any

sale will


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.

S O L U T I O N The period

of

the

satellite

6

around

the

circular orbit

of

radius

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

T =

2pr0 v

s 6

24(3600) = 2pC h + 6.378(10 )D vs 6

vs =

2pC h + 6.378(10 )

86.4(10 3

) (1) 6

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

The velocity of the satellite orbiting around the circularorbit of radius or Dissemination

24 copyright Wide

66.73(10

-12 )(5.976)(10

. ) States World permitted

yS =

GMe

instructors

C r0

.

of learning the is

yS =

(2)no

United on t

and

use

6 for student work Solving Eqs.(1) and (2), protected by the of

assessing =

h = 35.87(10 ) m = 35.9 Mm yS = 3072.32 m> 3.07thekm>s Ans. is solely work work provided and of integrity

this


of any sale will

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.

S O L U T I O N For a 800-km orbit

- 12 24

v0 = 66.73(10 )(5.976)(10 )

3

B (800 + 6378)(10 )

= 7453.6 m>s = 7.45 km>s Ans.

laws or



United


use and


protected of the assess ing

is solely work


this

This is the

and courses part

of any sale will


*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.

S O L U T I O N Circular Orbit:

66.73(10)5.976(10 )

GM

e

r 3 3 vC = A 0 = B 4000(10 ) + 6378(10 ) = 6198.8 m>s Parabolic Orbit:

2GM e 2(66.73)(10)5.976(10 )

r0 = B 3 3

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

¢v = ve - vC = 8766.4 - 6198.8 = 2567.6 m>s ¢v = 2.57 km>s

h = 4 Mm

Ans.

laws or


copyright Wide .


United


use and



is solely work


this

This is the

and courses part

of any sale will


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,

1 GMs

r = C cos u + h2

For u = 0° and u = 180°,

1 GMs rp = C + h

2

1 GMs

2 ra = - C + h teaching Web) or

Dissemination

Eliminating C, from Eqs. 13–28 and 13–29, laws

2a 2GM s copyright Wide .

From Eq. 13–31,2 = 2 instructors permitted b h States . World

p

T = h (2a)(b) T2h2

Thus,

4p a

2 4p

3

T2 = a GMs ba

of Uniteduse learningthe is not on

forstudent

and

by the protected of

is assessing

work the

provided integrity

This is part the and courses their destroy

of any

sale will

Q.E.D.


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.

A

S O L U T I O N 2

Ch e = GMe

1 GMe

2 where C = r0 ¢1 - r0v0 ≤ and h = r 0 v .

1 0

GM e 2 2

e = GMe r0 ¢1 - r0

v0 ≤ (r0 v0)

r v 2 0 0

e = ¢ GMe - 1≤

laws

r v 2 GM (e + 1) teachingWeb)

0 0 e

or

GMe

= e + 1 0 = r

. Dissemination

B

0 copyright Wide

v = v

= 7713 m>s = 7.States71 Ans.W orld = 66.73(10 )(5.976)(10 )(0.25 + 1)

B 0 C 8.378(106

) km> permitted

where r0 =

rp= 2A 10 6

B 3

B

= 8.378A 6

B m.

+ 6378A 10 10

6 of learning the is

r 8.378(10 ) use

United on 0

13 by.96A 106 B

and

=

ra = = m student

-

1

- 1

for wor k

2GMe 2(66.73)(10 )(5.976)(10 ) protected the

of the

vA= vB = 6

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

rp 8.378(10 ) is solely work


ra

13.96(10 6

)

this This is the

and courses part

of any sale will

2 Mm

B

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.

SOLUTION

6 3 6

Here, rO = rP = 6(10 ) + 6378(10 ) = 12.378(10 ) m.

P A

6 Mm

6

h = 12.378(10 )vP (1) and

1 GMe

C = r

a1 - r v 2 b - 12

24

P P P

1 66.73(10 )(5.976)(10 )

6

C = 12.378(10

6

) c1 - 12.378(10 )vP 2 d

laws(

- 9 2.6027 or

C = 80.788(10 ) - 2) teaching Web) 2 Dissemination v

2 copyright

Wide . P

permitted

instructorsnot

Using Eqs. (1) and GMe

(2), States . World

e =

of learningthe is Unitedon

and

use - 9

c80.788(10 ) - 2 d c12.378(106)vforP d student work

v

P by the (including

- 12 2.6027

protected24 2

assessing the

66.73(10 i s

)solely

work

0.45 = )(5.976)(10 of

vP = 6834.78 m>s work and ofintegrity provided this

This is part the

a2GM and courses their of des troy

rP any Using the result of vP,

e

v 2 - 1

r sale will

P P 6 12.378(10 )

24

= - 12

2(66.73)(10 )(5.976)(10 ) - 1

6 2

12.378(10 )(6834.78 ) 6

= 32.633(10 ) m

6 9 2

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

9

6

32.633(10 )va = 84.601(10 )

va = 2592.50 m>s = 2.59 km>s Ans. Using the result of h,

T = p (r P + r ) 2 r r

p 6 6

6 6

= 84.601(10 )C 12.378(10 ) + 32.633(10 ) D312.378(10 )(32.633)(10 ) © 2013 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is prote cted 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, 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

m m 2 S

e y s

F = G r2 AlsoF = ms a r b Hence y

2 m

s m

e

0

2

ms a r b = G r 24

m e 5.976(10 )

G - 12 6

y = A r = B66.73(10 ) a 15(10 ) b = 5156 m>s = 5.16 km>s Ans.

laws or



United


and

use



is solely work


this


of any sale will

© 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,

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 3 Mm

A B O A¿

6 Mm 9 Mm

S O L U T I O N r

0 2 6

Central-Force Motion: Use ra = (2 GM>r y B - 1 , with r0 = rp = 6A 10 B m and 0 0

M = 0.70Me, we have

6(10) 6

9A 106

B = -12 24

2(66.73) (10 ) (0.7) [5.976(10 )]

6 2 ≤ - 1

laws or

¢ 6(10 )yP

teaching Web)

Dissemination

yA = 7471.89 m>s = 7.47 km>s copyrightAns. Wide .


United


use and



is solely work


this

This is 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.

S O L U T I O N

Central-ForceMotion: Here,

C = 1

11 - GM e

2 [Eq. 13–21] and r r v

0 0 0

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

1 GMe 2 2 ch r A 1 - r 2 v 2 B A r v B r v

0

0 0 0 0 0 0

e = GM = GM = GM - 1 e e e Rearrange Eq. (1) gives

Rearrange Eq. (2), we have

B 120 Mm

A

10 Mm S

h = r0 v0

(1)

1 GM e

1 +

2

e =r0 v0

Substitute Eq. (2) into Eq. 13–27, ra =

r v0 = D 0

A 2 GMe >r0 v0 B

r0 2A

B - 1 1 + e

ra = or

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

= r0 = a b C 120110 1 + 0.58

1 - 0.58

This

Substitute r0 = 1rp21 = 31.89911062 m intoandEqcourses. ( 1vp21 = D 11 + 0.582166.732110 6 215.976s

-

31.899110 2

Applying Eq. 13–20, we have

rp 31.89

1va21 = 6

a ra b 1vp21 = c

120110 2

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

6 1 - e

10110 2 =a 1 + e b C 120110

Substitute r0 = 1rp22

11 + 0.84622166.732110 215.9672110 2

v p

2

=

D 10110 6 )


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

1rp22 101106

2

(va22 = c 1ra22 d 1vp22 = c 1201106

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.

- 12 24

v = GM e 66.73110 215.9762110 2

r 6

cD

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

¢v = 1vp22 - ve = 8580.25 - 6314.89 = 2265.36 m>s = 2.27 km>s Ans.

laws or


copyright Wide .


United


and

use



is solely work


this

This is the

and courses part

of any sale will


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

S O L U T I O N 9

rp = r0 = 9.30110 2 km

2 2 ch 1 GM s r0v0

e = GM = r a 1 - r v b a GM b

s2 0 0 0 s

r v 0 0

e = a GMs - 1b 2

r v 0 0

GMs = e + 1 GM s 1

2 r b r v e + 1

(1)

laws

teaching

Web)

0 0

- 1 A

B - 1

r = r0 1e + 12 ra =

2GMs = 2

0 0 e + 1

. Dissemination or States instructorscopyright World Wide permitted.

(2)

of learning the is

11 - e2 0.927

r 9

United

use on

not

a = 10.8110 2 km andAns.

the student


protected of the

a s se ss in g


this

This is the

and courses part

of any

sale will


*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? A

P

S O L U T I O N

e = 0.130

vp = v0 = 15 Mm>h = 4.167 km>s 2 2 2 Ch 1 GMe r0 v0

2

e = GMe = r ¢1 - r v ≤a GMe b

2 0 0 0

r v

0

0

e = ¢ GMe - 1≤ 2

r 0 v 0

GMe = e+ 1 laws Web)

teaching

r0

= copyright Wide

. v0

2

instructors permitted St ates W orl d

learning .

is = 1.130166.73 21

- 12

215. 976 21 24

10 10 2

C 4.167110 32 D 2

= 25.96 Mm of thenot

1

United on GMe

use and r v

0 0 2 = e + 1 the student for (including work by

protected of 2 e - 1 A B - 1 ass essing the

rA = 2GM =

r is solely work

r01e + 12 e + 1

work and of thisintegrity 0 0

rA = This is part the

25.96110 211.1302 and courses destroy 1 - e

theirof any

6 0.870 sale

will

=

6

= 33.71110 2 m = 33.7 Mm v r r

vA = 0 0 A

6 =15125.962110 2

6 33.71110 2

= 11.5 Mm>h Ans. 6 6

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 2 2 21

Take G = 34.4110 21lb ft 2>slug , Me = 409110 2 slug,

S O L U T I O N

v 3

2 0 = 2500 mi>h = 3.67(10 ) ft>s

C h

(a) e =GMe = 0 or C = 0

1 = GMe2

r0 v 0 - 9 21

GMe = 34.4(10 )(409)(10 )

= 14.07(10 15

)

9 r0 = e = = 1.046(10

) ft 2 13 2

GM 14.07(1015

) 9 3

r = 1.047(10 ) - 3960 = 194(10 ) mi

v [3.67(10 )]

0

5280 2

C h

(b) e = GM = 1


Dissemination copyrightAns. Wide

.


States . World

of learning is

Uniteduse the not

on and

for student work e by the (including

GM e (r 0 v 0)a r b ¢1 - r v 2 ≤ = 1 protected of the

2GM 2(14.07)(1015

) assessing

1 2 2 1 GMe is solely work

v0 2 [3.67(10 )] provided ofintegrity

9

this r0 = = 3 ) m

This is part the r = 396(103) - 3960 = 392(103) mandi courses Ans.

of any

sale will

(c) e 6 1

3 3 194(10 ) mi 6 r 6 392(10 ) mi Ans.

(d) e 7 1

r 7 392(103

) 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 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.

SOLUTION

When the probe is orbiting the planet in a circular orbit of radius rO = nR, its

speed is given by

GM GM

v = r

O BO

= B nR

The probe will enter the elliptical trajectory with its apoapsis at point A if its speed is

GM decreased to v

a = kv

O =

kB nR at this point. When it lands on the surface of the planet, r = rB

= R.

1 1

=

a1 - r r

P

1 1

R = a r - P

Since h = rava = nRakA nR b = k GM

B

u R A

nR

GM r v

P P GM

2 r v 2 P P

rPvP = h

copyright

vP =

rP Also,

k2nGMRlearning r

P

2

ra = 2GM

rP r v P P

2GM

- 1

nR = 2 is

of

P P

2nGMR 2

vP = rp(rp + nR) Solving Eqs.(2) and (3), 2 k n

2

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

1

2 - k 2

2

R = a k nR - 2

- 1 a

k n - u = cos 1 - k Here u was measured from periapsis.When measured from apoapsis, as in the

figure 2 then k n - 1

- 1 u = p - cos a


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.

SOLUTION 3 Mm

When the command module is moving around the circular orbit of radius 6

r0 = 3(10 ) m, its velocity is

GM

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

c B 0= B 3(10 )

= 1278.67 m>s

The escape velocity of the command module entering into the parabolic trajectory is v =

m

2GM 2(66.73)(10 )(0.0123)(5.976)(10 )

r

e B 0 = B 3(10 ) = 1808.31 m>s

laws or

teaching Web)

Dissemination

Thus, the required increase in the command module is copyright Wide . instructors

States . World

United

of learningthe is not on

use and


protected of the

is

solely assess in g

work


this


of any sale will


*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.

S O L U T I O N 1 GMe

Central-Force Motion: Here C = r 0 a 1 - r 0 v 2 b [Eq. 13–21] and 0

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

A B

9 Mm

h = r0 v 0 1

A 1 - GMe 1 r

2 v 2 2

2 2 2 B 0 0 2

ch r0 0 0

r0v

0

e = GMe = GMe = GMe - 1 (1) Rearrange Eq.(1) gives

1 GMe

v 2

laws(

1 + e = r

0

2) or

Rearrange Eq.(2), we have 0 teaching .

Dissemination Web) copyright Wide


B

11 + r0 States World

v0 = e2 GMe learning

. (3)

r0 2 United of the is not

Substitute Eq.(2) into Eq. 13–27, ra

= 12GMe >r0 v0 2 - 1

, w use

work and

student

by the

2A 1 B - 1 protected of the 1 + e is solely assess in g work

Rearrange Eq.(4), we have work provided and of integrity

this

1 + e 1 + 0.76 This is 6 part the 6

1 - e 1 - 0.76 and coursesany their destroy

ra = a b r0 = a b C 9110 2 D = 66.0110 2 m of

6

Substitute r 0 = rp =

9A m into

10 B Eq.(3)saleyieldswill 12 24

(1 + 0.76)(66.73)(10 )(5.976) (10 )

p D 9(10 6) = 8830.82 m>s

Applying Eq. 13–20, we have 6 r p 9110 2

6

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.

- 12 24

GM 66.73110 215.9762110 2 e

=

6

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.


Copyright and written permission should be obtained from the publisher prior to any prohibited reproduct ion, 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.

13–129.

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.

h 4 Mm

S O L U T I O N Circular orbit: GM 66.73(10)5.976(10 ) e

r 3 3 vC = B0= B 4000(10 ) + 6378(10 ) = 6198.8 m>s Parabolic orbit:

12 24

2GM 2(66.73)(10 )5.976(10 ) e

3 3

ve = Br0 = B 4000(10 ) + 6378(10 )

= 8766.4 m>s

¢v = ve - vC = 8766.4 - 6198.8 = 2567.6 m>s ¢v = 2.57 km>s

laws or

Ans.


copyright Wide .


United


use and



is solely work


this

This is the

and courses part

of any sale will


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.

P A

S O L U T I O N

6 m 1h

Here, vP = c 15(10 ) h d a 3600 s b = 4166.67 m>s.

h = rPvP

h = rP (4166.67) = 4166.67rp (1)

and

1 GMe 2

C = rP ¢1 - rP vP ≤ 24

-12

1 66.73(10 )(5.976)(10 )

2

rp(4166.67

C = rp B1 - ) R laws(

C = rP B1 - rP 6 R 2) 1 22.97(10 ) teaching Web)

GMe DisseminationWide copyright .

1 22.97(10 ) instructors permitted Ch

2 6

States . World

r

B1 -

r

R(4166.67 rP)

2 of learningthe is not

P P

12

24

United

0.15 = on

66.73(10 )(5.976)(10 ) by use and

rP = 26.415(106

) m for student work

protected theof the assess in g

is solely work

rP work and of thisintegrity

provided

r A = 2 This is part

the

2GM e

- 1

their any

destroy

rP vP 6 of

= 26.415(10 ) sale will

- 12 24

2(66.73)(10 )(5.976)(10 ) - 1

6 2 26.415(10 )(4166.67 )

6 = 35.738(10 ) m

6 2 9 2

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

vA = h 6 9 35.738(10 )v = 110.06(10 )

A

vA = 3079.71 m>s = 3.08 km>s Ans.

Using the results of h, rA, and rP, p

T =6 A rP + rA B 2 P A p

6 6

6

6

= 110.06(10 9 ) C 26.415(10 ) + 35.738(10 )D 226.415(10 )(35.738)(10 )

= 54 508.43 s = 15.1 hr Ans. © 2013 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by

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.

S O L U T I O N To escape the earth’s gravitational field, the rocket has to make a parabolic trajectory.

Parabolic Trajectory:

2GMe

ye = A r0

Elliptical Orbit: 2

1

GM e

Ch

where C = r ¢ 1 - r y

2 ≤ and h = r 0

y e = GMe 0 0 0 0

1

¢ GMe r0

e = a 2 - 1b

e = r0 y0

1 -

GMe 2

r0 y0

= e + 1

GMe

2

GMe

0

GMe ≤ laws . teaching

or

. DisseminationWeb)

2 (r0 y0)2

copyright Wide

instructors not permitted States

learning World

GMe (e + 1) of the is

United on

y0 =

use

B r0 and

student work

by

(includingebthe

the

GMe (e + 1)

GMepro tecte d

for

of

assessing

0 is solely work 0


this

This is the

and 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 A' O

6 Mm

14 Mm

S O L U T I O N Central-Force Motion: Substitute Eq. 13–27, ra

6 6 ra = (14 + 6.378)(10 ) = 20.378(10 ) m and r0

6 = 12.378(10 ) m, we have

r0 2

= (2GM>r0v0 ) - 1 , with 6

= rp = (6 + 6.378)(10 )

- 12 24 2(66.73)(10 )[5.976(10 )] 1 laws Web)

20.378(106

) = ¢ 12.378(10 )vp ≤ - teaching . 6

12.378(10 ) Dissemination or

copyright Wide

vp = 6331.27 m>s


States . World

r of learning the is

United on not

p 12.378(10 ) use and

va = ¢ ≤ v p

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

(including work

ra

protected of 6

is assessing 2 the

solely work 78.368(10 ) m >s.Thus, applying

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

p provided of integrity Eq. 13–31, we have work and this

h This courses part the

ra)2rp ra

T = (r

p +

20.378)(10 is destroy

= p

9 [(12.378 + )]their2 any 6

)

12.378(20.378)(10 6 of

78.368(10 ) 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)

vc = A GM e A 66.73(10)(5.976)(10 )

r 6

= 4423.69 m>s = 4.42 km>s

Ans.

0 20.378(10 ) 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.

13–34. each of the two blocks has a mass m. the coefficient of kinetic friction at all surfaces of...

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