solution manual for engineering mechanics dynamics 13th ... · 13–7. if the 50-kg crate starts...

Solution Manual for Engineering Mechanics

Dynamics 13th Edition by Hibbeler

link full downoad:

https://testbankservice.com/download/solution

-manual-for-engineering-mechanics-

dynamics-13th-edition-by-hibbeler

13–1.

The 6-lb particle is subjected to the action of

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

5t2

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.

z

F 2 y

F 3

x F1

SOLUTION

©F = ma; (2i + 6j - 2tk) + (t2

i - 4tj - 1k) - 2ti - 6k = ¢326.2 ≤(axi + ay j + azk)

Equating components:

6

≤ax = t2

6 6

¢ 32.2 - 2t + 2 ¢ 32.2 ≤ay = - 4t + 6 ¢ 32.2 ≤az = - 2t - 7

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

6 t3 6 6

¢ 32.2 ≤vx = 3 - t2

+ 2t ¢ 32.2 ≤vy = - 2t2 + 6t ¢ 32.2 ≤vz = - t

2 - 7t laws or

s =

s

¢

teaching Web)

3 + t -

3 + 3t

- 3 2.

¢32.2

≤ x 12 -

¢32.2

≤ y = 32.2 ≤sz = - Dissemination

Since ds = v dt, integrating from s = 0, t = 0 yields copyright Wide .

instr u ctors

permitted 6 t4 t

3 6 2t

3 6

2 2

When t = 2 s then,

States learning

World

= -

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

United on and use

for student work s = Ans.

Thus, by the

protected of the

is assess ing

work

solely

work provided and of integrity

this

This is the and courses part

https://testbankservice.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

v = 4.025t2

+ 10

When t = 3 s,

v = 46.2 ft>s laws or

teaching Web)

ds = v dt .Dissemination

copyright Wide A s. .

s t

L0 ds = L0 (4.025t2

+ 10) dt States instructors World permitted

s = 1.3417t3

+ 10t United

of learning the is not on

and

use When t = 3 s,

s = 66.2 ft

the student for (including work by

protected of the is assessing

solely work

work provided and of

this

Ans.

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, 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

a = 1.121 m>s2

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 the +

s = s0 + v0t +

act

learning is

:

2

Uniteduse

A B 1 on

and

(1.121) A32

B = 5.04 m by

s = 0 + 0 + 2 the student work Ans. for (including

protected of the is solely work

assessing

this

workprovided and of integrity This is the

and courses part any

of their destroy

sale will

P

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.

( : )+

v 2

= v 2

+ 2a c

(s - s ) 0 0

42

= 02

+ 2a(5 - 0) a =

1.60 m>s2

:

Free-Body Diagram: Here, the kinetic friction Ff = mkN = 0.3N is required to be

directed to the left to oppose the motion of the crate which is to the right, Fig. a.

Equations of Motion:

+ c ©Fy = may; N + P sin 30° - 50(9.81) = 50(0) or laws

N = 490.5 - 0.5P

teaching Web)

.

Using the results of N and a, copyright Wide

Dissemination

States

.

World permitted + = max; P cos 30° - 0.3(490.5 - 0.5P) = 50(1.60)

instructors

: ©Fx not

United of learning the is

P = 224 N on andAns. u se

by

the student

for (including work protected of the is solely work

assessing

this



of their destroy

sale will

P

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 sin 45° - 30 =

800

a 32.2

a = 21.561 ft>s2

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

1 0 c 0

2

v = 0 + 2(21.561)(100 2 2 - 0)) 1 v1 = = 78.093 ft>s

800 -80 = 32.2a

a = -3.22 ft>s2

2 2

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

v22

= v12

+ 2ac(s2 - s1)

100 ft

s

laws teaching Web)

. Dissemination or

copyright Wide . permitted

instructors

States World

United of learning

not on the v 2 = 7 7 . 9 ft >s Aisns.

use and


protected assess ing

of the

is solely

work


this

This is the

and courses part

of any sale will



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.

13–6. 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

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

30°

laws Web) teaching

or

P

30°

Using the result of N, . Dissemination a = 0.8993 m>s

2 copyright Wide

©Fx¿

= may¿;

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

States instructors World permitted

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

use on not

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

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,

13–7.

If the 50-kg crate starts from rest and travels a distance of 6 m P

up the plane in 4 s, determine the magnitude of force P acting

on the crate. The coefficient of kinetic friction between the

30°

crate and the ground is mk = 0.25.

SOLUTION 30°

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

motion is known.

s = s + v t + 1 a t2

0 0 c

2

6 = 0 + 0 + 1 a(42

) 2

a = 0.75 m>s2

Free-Body Diagram: Here, the kinetic friction Ff = mkN = 0.25N is required to be laws Web) directed down the plane to oppose the motion of the crate which is directed up the

teaching

or

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

plane, Fig. a. .

©Fy¿ = may¿; N + P sin 30° - 50(9.81) cos 30° = 50(0) instructors permitted

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

This is the

and courses part

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

60 dv

Kinematics: For 0 … t 6 10 s. v = 10 t = {6t} ft>s. Applying equation a = dt ,

we have

dv

= 6 ft>s2

a = dt 10

t (s)

v - 60

80 - 60 30

For 10 6 t … 30 s, = , v = {t + 50} ft>s. Applying equation

dv t - 10 30 - 10

v

a = dt , we have

dv

a = dt =

1 ft>s 2

laws or

Equation of Motion: teaching Web) Dissemination

copyright Wide .

For 0 … t 6 10 s

+ ¢

3500 ≤(6) =

instructors permitted ; aFx = max ; F = 32.2 learning

States . World

of the is not

For 10 6 t … 30 s United on

652 lb use and

+ 3500

Ans.

¢

≤(1) =

the student

; aFx = max ; F = 32.2

for (including work

by


of the

109 lb is solely work Ans.


this

This is the

and courses part

of any sale will


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

a = 1.66 m>s

2 Aisns

of the . United on

and

use


protected of the

is solely assess ing

work


this

This is the

and courses part

of any sale will


13–10.

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

p

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

Equations of Equilibrium: At t = 2 s, P = 90 A 22

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

a = 1.75 m>s2

States World

A s.

United


and

use


protected of the


work


this

This is the

and courses part

of any sale will


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 aB (1)

From FBD(b),

+ c ©Fy = may ; 2T - 200 = - a 200 b aS

(2)

32.2

laws or

teaching Web) Dissemination

Kinematic: Establish the position-coordinate equation , we have copyright Wide . Taking time derivative twice yields

= l

instructors

permitted

2sS +sB

States . World learning

United of

on the 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

T = 96.43 lb

assessing

the aS = 1.15 ft>s T

is work work provided and of integrity

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,


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

speed of v = 14t2

2 ft>s, where t is in seconds.

SOLUTION (a) T = 40 lb Ans.

(b) v = 4t2

a = 8t

+ c aFy = may ; 2T - 80 =

80

18t2

32.2

At t = 2 s.

T = 59.9 lb Ans.

laws or

teaching Web) Dissemination copyright Wide .

instructors permitted States . World

United


and

use


protected of the


work


this

This is the

and courses part

of any

sale will



13–13. The bullet of mass m is given a velocity due to gas pressure

caused by the burning of powder within the chamber of the gun.

Assuming this pressure creates a force of F = F0 sin 1pt>t02

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

it is in the barrel. What is the bullet’s maximum velocity? Also, F0 determine the position of the bullet in the barrel as a function of

time.

SOLUTION

+ pt : ©Fx = max ; F0 sin a t b = ma

0

dv F

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

0

v t F 0 pt F t 0 0 pt t

L dv = L m 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

2F0t0 instruct ors World perm itted

States

Ls L t

pm t learning

is not

ds = F0t0

b a 1 - cos a pt

United of the

a b b dt

0 0 0 use

on and

s = F0t0 -

t0 pt t

for the student

a pm b c t p sin a 0 b d 0 (including by

protected of the work is solely

assessing work

F0t0

t0

pt

provided and of integr ity

s =

b a t -

sin a

b b work

this

Ans.

a pm p t0


of any

sale will


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,

13–14.

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

C

before coming to rest. Determine the constant horizontal force

developed in the coupling C, and the frictional force developed

between the tires of the truck and the road during this time. The

total mass of the boat and trailer is 1 Mg.

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

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

a : b

0 = 152

+ 2a(10 - 0)

a = -11.25 m>s2

= 11.25 m>s2 ;

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), United of learning the is

not

on

+ c ©Fx = max ; 11 250 - F = 2000( -11.25)

use

and

for student work

(including

protected by theof the assessing

F = 33 750 N = 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 Ateachings. Web)

aE = 0.7 5 m>s

laws

1.5= 0 + aE (2) .Dissemination or

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

AWor

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

United


and

use



of the

is solely

work


this

This is the

and courses part

of any sale will


*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

a = 14.8 ft>s2

laws

Web) teaching

An . or Dissemination

copyright Wide .


United


and

use


protected of the


work


this

This is the

and courses part

of any

sale will

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

a y¿ = may¿; NA - 10 cos 60° = a

b (0)

NA = 5.00 lb

32.2

Q + F 10

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

b a

(1)

32.2

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


10

States World

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

. (2)

United and

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

on

a = 3.69 ft>s2

the student Ans. for (including work

by

protected of

T = 7.013 lb

assessing

is work the

work provided and of in tegri ty

this

This is the

and courses part of any

sale will


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

a = 16.1 ft>s2

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 .

1

United

of learningthe is

act2

on use

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

2 by the and

1 for student work

2 (including

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


work of the

solely


this

Total time = tAB + tBC

= 1.82 s This is part the

Ans.

andcourses any

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

+ R©Fx = max; 40

40 sin 30° - 6.928 =

a 32.2

a = 10.52 ft>s2

( + R) v2

= v2

0 + 2 ac (s - s0);

v2

B = (10)2

+ 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 Dissemination ( : ) sx = (sx )0 + (vx )0 t copyright Wide .

( + T ) sy = (sy)0 + (vy)0 t + ac t 2 instructors World permitted R = 0 + 22.82 cos 30° (tBC) learning . is

2 of the not

1 2 United on

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

use

by the student and (including

t BC = 0.2572 s protected for of the work Ans.

is solely

assess ing

R = 5.08 ft work

Total time = tAB + tBC = 1.48 s provi ded

and of integrity Ans. work this

This is the

and courses part

of any

sale will



*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 =

13200t2

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

Q + ©Fx¿ = max¿ ; 3200t2

- 40019.812 a

b = 400a a = 8t2

- 4.616 17

dv = adt

M

v1 2 m/s

17

8 15

v 2

L2 dv = L0 18t 2

-4.6162 dt

v = 14.1 m>s Ans.

laws or



United


and

use



of the

is solely

work


this

This is the

and courses part

of any sale will


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 =

13200t2

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

Q + ©Fx¿ = max¿;3200t2

8

a = 8t2

- 4.616

- 40019.812 a 17 b = 400a

dv = adt

v t

L2 dv = L0 18t2

- 4.6162 dt

ds 3 v = dt = 2.667t - 4.616t + 2

s 2

M

v1 2 m/s

17

8 15

L0 ds = L0 (2.667 t3 - 4.616t + 2)

dt

laws or

s = 5.43 m teaching Web)

.Dissemination copyright A s. Wide .


United


and

use



of the

is solely

work


this

This is the

and courses part

of any sale will



13–22. Determine the required mass of block A so that when it is

released from rest it moves the 5-kg block B a distance of 0.75

m up along the smooth inclined plane in t = 2 s. Neglect the

mass of the pulleys and cords.

SOLUTION 1

Kinematic: Applying equation s = s0 + v0 t + 2 ac t2

, we have 1

(a +) 0.75 = 0 + 0 + 2 aB A 22

B aB = 0.375 m>s2

Establishing the position - coordinate equation, we have

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

Taking time derivative twice yields

From Eq.(1), laws or

3aA - aB = 0 teaching Web)

(1) Dissemination

3aA - 0.375 = 0 aA = 0.125 m>s2

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), is

protected assessing of 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

rate of 5 m>s2

. 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

aB = -2.5 m>s2 = 2.5 m>s

2 c

+ c ©Fy = may ; 2T - 800(9.81) = 800(2.5)

T = 4924 N = 4.92 kN Ans.

laws or



United


and

use


protected of the


work


this


of any sale will


*13–24.

If the motor draws in the cable at a rate of v = (0.05s3>2

) m>s,

s

where s is in meters, determine the tension developed in the v 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 M

is mk = 0.2.

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

first.

dv

A0.05s3>2

Bc(0.05) a

3

a = vds = 2 bs1>2

d = 0.00375s2

m>s2

When s = 10 m,

a = 0.00375(102

) = 0.375 m>s2

:

Free-Body Diagram: The kinetic friction Ff = mkN = 0.2N must act to the left to

oppose the motion of the crate which is to the right, Fig. a.

Equations of Motion: Here, ay = 0. Thus, laws or

+ c ©Fy = may; N - 20(9.81) = 20(0)

teaching Web)

copyright Wide

.

N = 196.2 N

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

T = 46.7 N student work

for (including Ans. protected of the is solely work

assessing

this



of their destroy

sale will


13–25.

If the motor draws in the cable at a rate of v = (0.05t2

) 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,

a = 0.1(5) = 0.5 m>s2

:

Free-Body Diagram: The kinetic friction Ff = mkN = 0.2N must act to the left to

oppose the motion of the crate which is to the right, Fig. a.

Equations of Motion: Here, ay = 0. Thus, laws or

+ c ©Fy = may; N - 20(9.81) = 0

teaching Web)

copyright Wide

.

N = 196.2 N

Dissemination


instructors

Using the results of N and a, United

of learning the is not + on

use

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

T = 49.2 N student work

for (including Ans. protected of the is solely work

assessing

this



of their destroy

sale will


13–26.

The 2-kg shaft CA passes through a smooth journal bearing at s¿ s B. Initially, the springs, which are coiled loosely around the

F 5 kN shaft, are unstretched when no force is applied to the shaft. In C B

A 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 k

the shaft at the instant s = 50 mm, s¿ = 450 mm. The ends of kCB 3 kN/m

2 kN/m the springs are attached to the bearing at B and the caps at C AB

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

2500(0.2) 2 v2

2500(0.2) - ¢ 2 ≤ = 2 laws or

teaching Web)

v = 30 m>s .Dissemination . copyright Wide

instructors

States World

United


and

use



of the

is solely

work


this


of any sale will



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

acceleration of 6 ft>s2

. 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



United


and

use


protected of the

is assess ing

work

solely


this


of any sale will



*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. laws or

teaching Web)

+ c ©Fy = may; N + 200 sin 30° - 500 = 0 N = 400 lb . copyright Wide

Dissemination + 500 . permitted

: ©Fx = max ; 200 cos 30° - 0.3(400) = 32.2

a States World instructors

a = 3.43 ft>s

United

of learning the is not 2 on Ans. use

by and

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

assessing

this



of their destroy

sale will


13–29.

The force exerted by the motor on the cable is shown in the F (lb)

graph. Determine the velocity of the 200-lb crate when t = 2.5

s. 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

a = (16.1t - 32.2) ft>s2

Kinematics: The velocity of the crate can be obtained by integrating the kinematic equation, dv = adt. For 2 s … t 6 2.5 s, v = 0 at t = integration limit. Thus,

( + c ) dv = adt L L

v t

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

t

laws or

teaching Web) 2 s wil l be used as the lower .

copyright Wide

Dissemination .

States instructors World permitted not

Uniteduse of learningon the is by and

for the student work (including

v = A8.05t2

- 32.2tB 2 sis protectedsolely work of the

- 32.2t +work2 assessing

= A8.05t2 32.2providedB

ftand>s of this integrity

When t = 2.5 s, This is the

v =

2) part

Ans. and coursesany

their destroy

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.

13–30.

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

F 5 625 t2

M t (s)

2

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

time required to move the crate is given by

+ c ©Fy = 0; 2(625t2

) - 400(9.81) = 0 A

t = 1.772 s

Equations of Motion: F = A625t2

B N. By referring to Fig. a,

+ c ©F y

= ma ; 2 A625t2

B - 400(9.81) = 400a

y

a = (3.125t2

- 9.81) m>s2

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 World permitted instructors not

L L of the is

v

t

United learning

on

1.772 s A3.125t2

- 9.81 Bdt by

use

L0 dv = the and L forstudent work

t

(including

A1.0417t3

v = - 9.81tB 1.772 protectedsolely work of the is

3

assessing m sthis

work+2provide

= A1.0417t - 9.81t d11.587andB of> integrity This is the When t = 2 s,

and courses part

any

v = 1.0417(23

) -

9.81(2)t of+ 11.587destroy= 0.301

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.

12 m

SOLUTION s

B

12 - s B

+

2s 2

+ (12)2

= 24 B

A

#

- sB + As

2 - 12

A + 144 B asAs Ab = 0 sA

- s$

B

- As 2

+ 144 B-

23 asAs

#Ab2 + As 2

+ 144 B

- 21 a #A

2 b

+ A A 2

+ 144 B

- 2

1 a A $ Ab

= 0

A A s s s s

$ 2

s # 2 # 2 $

sA A s A + sAs A

s B

= -

-

3

1

C

AsA2 + 144 B 2

AsA 2 + 144 B 2 S

a (5)

2(4)

2 1.0487 m s 2

laws

3 (4)2

+ 0 or

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


protected of the is

assess ing

work

solely


this

This is the

and courses part

of any

sale will


*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 the right with an acceleration of 3 m>s

2 and has a velocity of

4 m>s at the instant sA = 5 m, determine the tension in the

rope at this instant. When sA = 0, sB = 0. 12 m

SOLUTION s

B

+ 2s B

12 = s 2 + (12)2

= 24 B A

#

1

2

- 3

a

#

sA

- s B +

2s s Ab = 0

2 A

As + 144 B 2 A

$ 2 - 3 #

Ab

2 2

+ 144 B - 1

- s B - AsA + 144 B

2 asAs + AsA 2

as#

A2 b + AsA

2 +

144 B- 2

1

$ asAs Ab = 0

$ sA

2 # 2 # 2 $

s

B = - s A

- s A + sAs A

3 1

C

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

aB = - C (5)2

(4)2

(4) 2 + (5)(3) laws or

-

S

teaching .

((5)2 3 ((5) + 144)

+ 144)2

. Dissemination Web) 2

2 1 = 2.2025 m>s

2 copyright Wide

+ c ©Fy = may ; T - T =


1.80 kN learning Ans.not

United on the is

of use and



of the

is solely

work


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

learning . World

of the not United on

and is

use

for student work protected by the of assessing

= 20 ax solely work

the is


this

This coursesany part the and

is

: ©Fx = m ax; F - 18 = 10(2.138)

their of destroy

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

ac = 4.11 m>s2

:

Plate D:

+

= m ax; 19.62 - 18 = 10 aD

: ©Fx

aD = 0.162m>s2 :

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) ; ©Fx = max ; P - 3mmg = m aA P

aA = m - 3mg Ans.

(b) sB + sA = l

aA = - aB (1)

Block A:

(a)

B P A

(b)

laws

+ teachingWeb) ; ©Fx = max ; P - T - 3mmg = m aA (2) or

Dissemination

+ copyright Wide

Block B: .

; ©Fx = max ; instructors permitted

mmg - T = maB States World .

Subtract Eq.(3) from Eq.(2): of learning the is United on not

P - 4mmg = m (aA - aB) use

and

for student work

by the

(including

protected of P

assess ing

the

Use Eq.(1); A

2m - 2mg is work Ans.


this

This is the

and courses part

of any sale will

13–35.

The conveyor belt is moving at 4 m>s. If the coefficient of static friction between the conveyor and the 10-kg package B is

B

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; 0.2198.12 = 10 a

a = 1.962 m>s2

+

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

t = 2.04 s Ans.

laws or



United


and

use



of the

is solely

work


this

This is the

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 s = 4A 2 1 + s2

- 1B

s

+ s 2

dv

2

- 1B ¢ 2

≤ = a 32.2 b a v ds b

: ©Fx = max ; - 4A 21 + s 1 + s2

1 4s ds v 2

-

21 + s2 ≤ =

L

0 ¢4s ds - L15 a 32.2 b v dv

1

- C 2s2

- 43 1 + s2 D 0

1 =

A v2 - 15

2 B

32.2

v = 14.6 ft>s Ans.

laws or


copyright Wide .


United


and

use


protected of the

is assess ing

work

solely


this


of any

sale will


13–37. Cylinder B has a mass m and is hoisted using the cord and pulley system shown. Determine the magnitude of force F as a function of the cylinder’s vertical position y so that when F is

applied the cylinder rises with a constant acceleration aB.

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

S O L U T I O N

+ c ©Fy = may ; 2F cos u - mg = maB where cos u =

y

2F£ 2 y2 + A

d2 B

2 ≥ - mg = maB

d/2 d/2

F

y

y aB

B

2 y2

+ A d

2 B 2

F = m(aB + g)2 4y

2 + d

2

Ans.

4y

laws or



United


and

use


protected of the


work


this

This is the

and courses part

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

aC = 2.356 m>s2

(+ R) v 2 = v 2

+ 2a 1s - s A

2 B A c B

vB 2 = 12.522

+ 212.356213 - 02

vB = 4.5152 = 4.52 m>s

vA 2.5 m/s

A

Ans.

laws or

3 m

u B



United


and

use


protected of the


work


this

This is the

and courses part

of any sale will


13–39.

An electron of mass m is discharged with an initial y

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 v0 x

+

= max;

F0 = max

: ©Fx

+ c ©Fy = may; 0.3 F0 = may Thus,

vx dv =

tF0

Lv0

x L0 m dt

vx = F0 m t + v0

vy dv = 0.3F 0.3F 0 0

L0 y L0 m dt vy = m t laws or F

0.3F0 0 2

v = C

t + v + t a m 0 b

1 a m b

2 2

= + 2F0tmv0 + m v0

F

m

L0

x dx =

L0 m

t

F0 t

x = 2m + v0 t

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


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.

*13–40. 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


a = g sin u Ans. Wide

copyright .


United


and

use



of the

is solely

work


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

(b) From part (a) aC>A = 6.94 m>s2

aC = aA + aC>A Where aA = 0

B

45 C

A

= 6.94 m>s2

(c) a

C = a

A + a

C>A

= 2 + a

√ C>Ab

laws (1) teaching Web)

T

.Dissemination or 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

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

T b ; T

of learning is

not

United use the

aC = 23.905 + 5.905 = 7.08 m>s

on andAns.

5.905

u = tan-1 = 56.5° ud

for student work Ans. by the

protected of is

assess ing

work

sol e l y

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

an upward acceleration of 4 m>s2

.

SOLUTION +

; ©Fx = max ;N sin 45° = 2aC>AB sin 45°

N = 2 aC>AB

B

458 C

A

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

aC>AB = 9.76514

a = a + a C AB C>AB

1aC)x = 0 + 9.76514 sin 45° = 6.905 ;

(aC)y = 4 - 9.76514 cos 45° = 2.905T Ans.la

aC = 2 16.9052 2 +

12.9052 2

= 7.49 m>s2

ws or teaching Web) Dissemination -1 2.905


United


and

use


protected of the

is assess ing

work

solely


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

for the truck to reachs 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

+

= max; - 0.3(1962) = 200( - a)

: ©Fx

a = 2.943 m>s2

;

km

1000 m

1 laws

Kinematics: The final velocity of the truck is v = a60 b a b a bteaching= .Dissemination copyright

h 1 km 3600

in

or Web)

.

+ instructors permitted

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

( ; ) v = v0 + ac t States World

16.67 = 0 + 2.943t

United


and

use

for student protected by the of the

is solely work work provided and of integrity

this


of any sale will


*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


not

on

2aA + aB = 0 use

and(3)

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

> =

>

by the

a A

= - B protected T of the

a

> = assess ing>

Ans.

3.773 m s 2 3.77 m s c 2 7.546 mis s 2 sole ly

7.55 mworks 2

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

This is the

and courses part

of any

sale will


A

10 kg

B

30 kg

13–45. If the force exerted on cable AB by the motor is A B F = (100t

3>2) N, where t is in seconds, determine the 50-kg

crate’s velocity when t = 5 s. The coefficients of static and

kinetic friction between the crate and the ground are ms =

0.4 and mk = 0.3, respectively. Initially the crate is at rest.

SOLUTION

Free-Body Diagram: The frictional force Ff is required to act to the left to oppose

the motion of the crate which is to the right.

Equations of Motion: Here, ay = 0. Thus,

+ 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 x

= ma ; 100t3>2

- 147.15 = 50a x

a = A2t3>2

- 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 Uniteduse

of learning the is not t = 1.567 s on

by

and

the student Kinematics: Using the result of a and integrating the kinemati cfor eq u at io n wo r kd v = a d t

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 ( : ) assessing

L L work

this

t and of

integrity

v 2t

provided

L0 dv = L1.567 s

3>2

- This is the

2.943 Bdt part and courses t any

v = A0.8t5>2

- 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,

v = 0.8(5)5>2

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



13–46. Blocks A and B each have a mass m. Determine the largest

horizontal force P which can be applied to B so that A will not

move relative to B. All surfaces are smooth.

A

B P

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: +

laws or

; ©Fx = max;

teaching Web)

P = 2mg tan u

.AnDissemination.

P - N sin u = ma copyright

Wide .

P - mg tan u = mg tan u instructors permitted

States World

United


and

use



of the

is solely work


this

This is the

and courses part

of any sale will



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

B P

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 coslearningu

P - mga b = mga Uniteduse of

b andthe is not

on

sin u + ms cos u

for student work

bycos

cos u - ms sin u u - ms s the

cos u - ms sin u protected of P = 2mga

is b assessing the Ans.

solely work


this

This is the

and courses part

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

resistance is FD = kv2

, 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

mg - kv2

dv

+ T ©Fz = m a z ; = m dt

v m dv t

m

L0 1mg - kv2

2 = L0 dt v

m L dv = t k

0

mg

k

mg m

2A

k ¢ k k copyright

mt¢2A k

F D

v

- v2

1 Ak

¢ A k

mg ln D mg

¢

= ln

mg

A mg

A k

mg

+ v

2t2 mg

A k

e k

A k assessing

=

mg

mg

A

e - v e

k

mg e2t2 k

mg

A k

e2t2

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

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

S O L U T I O N

Q + ©Fx = m ax; 0 = m aB sin f

a = a + a B AC B>AC

a = a + a B 0 B>AC

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

Thus,

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

aB>AC = a0 sin

u t

B>AC

a

0 θ

C

A B

s

laws

. teaching

or

Dissemination Web)

L0 L0

v

sin u dt

copyright Wide

States instructors permitted

. learning

of the is not

United on = s = a0sin u t dt use and

t

L0 for student work by (including the

protected

of

2

a0 sin u t2

assessing

s = is solely work the Ans.


this

This is the

and courses part

of any sale will

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

dx d2

x

- k dt = m dt2

- m g - k

dy

= m

d2

y

dt dt 2

# - k

Integrating yields mg k In x = m t +

C1

In (y + ) = t + C2

y

v 0

θ0 O x

laws or teaching .

Dissemination Web)

copyright Wide permitted

instructors

# 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 # -(k>m)t protected for of the

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

x =

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

any

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 =

cos u0(1 - e

( )

) k

y = - m g t

+ m (v0 sin u0 +

mg )(1 - e-(k>m)t)

k k k As t : q

Ans.

Ans.

m v cos

x =

max 0

u Ans.

0 k

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 Fs

(mA + mB)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 or

laws

teaching Web)

Thus, .

copyright Wide - (mA + mB)g + kc a

Require y = d, N = 0

Since d is downward,

mA + mB - mAg + N . Dissemination

k b g + yd = (m

A + m

B )a States b World permitted

mA instructors

of learning the is not Uniteduse on

by the and kd = - (mA + mB)g for student work

(including

protected of the is solely work

assessing

(mA + m this

d = Bwork)g and of

Ans. k is

provided integrity

This the

and courses part

of any

their

destroy

sale will


*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

v2

©Fn = man ;

0.2(147.15) = 15a 5 b

v = 3.13 m>s Ans.

laws or



United


and

use



of the

is solely

work


this

This is the

and courses part

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).

v2 10

2

Equations of Motion: Realizing that an = r = r and referring to Fig. (a),

102

©Fn = man;

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

laws or



United


and

use



of the

is solely

work


this


of any sale will


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

v2

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

laws or



United


and

use


protected of the


work


this

This is the

and courses part

of any sale will


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,

which is increasing at 3 m>s 2

. 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.

N + 5 A9.81 B - 17.32 sin 30° = 5 a

42

+ T ©Fn = man; 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


and

use



of the

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

8 m/s

ρ

+ c ©Fb = m ab;

+

; ©Fn = m an;

N - W = 0 N = W

Fx = 0.7W

W 82

0.7W = 9.81 ( r ) r = 9.32 m

Ans.

laws or



United


and

use



of the

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 ;

4 (0.5)2 3

+ Q©Fy = may; T - 0.2(9.81)a 5 b = B0.2¢ 0.120 ≤ Ra 5 b T = 1.82 N

(0.5)2

Ans.

3 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;

copyright Wide

Ta 5 b - NB a 5 b 2

(0.5)


4

3

0.120

+ 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

protected of the

is

solely assess ing

work


this

This is the

and courses part

of any

sale will

z

A

200 mm

B 400 mm

300 mm


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 b

r = 0.25a 5

= 0.2 m +

N a 3 4 v2

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

; ©Fn = m an; s 5 4 3

b

N a b 0.2N a

+ 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


instructors

United


and

by

use

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

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

z

S

0.25 m

A

5

3 4

4

r = 0.25( 5 ) = 0.2 m

+ 3 4 v2

; ©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 m s

Ans. laws or



United


and

use



of the

is solely

work


this

This is the

and courses part

of any sale will


*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

= 16.1 ft>s2

+ R©Ft = mat ;

at 60 cos 60° = 32.2 at

60 152

Q + ©Fn = man ; 2T - 60 sin 60° =

T = 46.9 lb 32.2 a 10 b

Ans.

Ans.

laws or

u

10 ft

G



United


and

use



of the

is solely

work


this

This is the

and courses part

of any sale will



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 + R© = ma ; 60 cos u = 32.2 ta

t = 32.2 cos u

t t 60 v

2

Q + ©Fn = man;

2T - 60 sin u = 32.2 a 10 b (1)

v dn = a ds however ds = 10du

v 90° v dn =

L0 L

60° 322 cos u du v = 9.289 ft>s Ans.

From Eq. (1) laws or

60

9.289

2 teaching Web) Dissemination

2T - 60 sin 90° =

T = 38.0 lb

copyright Ans. Wide . 32.2a 10 b


United


and

use


protected of the


work


this

This is the

and courses part

of any

sale will

u

10 ft

G


13–62. The 10-lb suitcase slides down the curved ramp for which the 1 2 y

y = –– x

coefficient of kinetic friction is mk = 0.2. If at the instant it

reaches point A it has a speed of 5 ft>s, determine the normal force on the suitcase and the rate of increase of its speed.

S O L U T I O N

1 2

n = 8 x

dy 1

= tan u =

= - 1.5

u = - 56.31° dx 4 x 2 x = - 6

d2

y 1

dx2

= 4

B1 + a

dy 2

R23 C 1 + (- 1.5)

2 D 23

dx b r =

=

= 23.436 ft d y 1

2 dx2

2 2 4 2

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

2

N = 5.8783 = 5.88 lb

8

6 ft

A

x

laws or teaching .

Dissemination Web)

copyright Wide

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

This is the

and courses part

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

G

S O L U T I O N

202 u

+ aaFy = m1an2y ; N - 150 cos 60° =

150

b sin 60°

a

32.2 8

N = 277 lb Ans.

+ b aF = m1a 2 x

; - F + 150 sin 60° = 150 a 20

2 b cos 60°

x n 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


of any sale will



*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

+ 150 13022

; ©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

10.5 cos u + sin u2150 150 13022

1cos u - 0.5 sin u2 = 32.2 a 8 b

0.5 cos u + sin u = 3.49378 cos u - 1.74689 sin u

u = 47.5°

z

8 ft

laws Web) teaching

Dissemination or Ans.

G

u

copyright Wide .


United


and

use



of the

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

+ v2

; ©Fn = m an ;

T sin 30° = 80( 4 + 6 sin 30° )

+ c ©Fb = 0;T cos 30° - 8019.812 = 0 T =

906.2 N

4 m

b 6 m

u

t n

©Fn = m an ;

©Ft = m at;

©Fb = m ab ;

v = 6.30 m>s Ans. 16.302

) = 283 N

Ans.

Fn = 50(

7

Ft = 0 Ans. F

b - 490.5 = 0 laws or

Fb = 490 N Ans.teach ing Web)



United


and

use

the student for (including work

by

protected of the


work


this

This is the

and courses part

of any sale will


13–66. The man has a mass of 80 kg and sits 3 m from the center of the

rotating platform. Due# to the rotation his speed is increased from rest

by v = 0.4 m>s2

. 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

3 m

10 m

©Ft = m at ; Ft = 80(0.4)

Ft = 32 N

v2

©Fn = m an ; Fn = (80) 3

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

2

2 v2 2

)

0.3(80)(9.81) = A(32) + ((80) 3

v4

55 432 = 1024 + (6400)(

9 ) laws or teaching Web)

v = 2.9575 m>s

Dissemination

copyright Wide .

dv

at

= 0.4

= dt instructors permitted

v t States . World

L0 dv = L0 0.4 dt United


and

v = 0.4 t use

the student

2.9575 = 0.4 t for (including work by

t = 7.39 s protected of the is solely

assess ing

work

provi ded

and of integrity Ans.

this

work

This is the

and courses part

of any

sale will


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

an = = = 4.938 m>s

2. By referring to Fig. (a),

laws

r 100

Web) v 22.22 teaching

cos . Dissemination

copyright Wide .

9.81m nstructors i not permitted

9.81m World

States

+ 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


of the

is solely work


this


of any sale will


*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

Geometry: Here, dy = - 0.00625x and d2

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

2

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

[1 + (dy>dx)2

]3>2

[1 + (- 0.00625x)2

]3>2

r = |d2

y>dx2

| = |- 0.00625| 2 x = 80 m = 223.61 m

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

with u = 26.57° andla

ws Web) teaching

Disseminationor © Ft = 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

learning

. 92 is

©Fn = man;

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

of thenot b on

and

N = 6729.67 N = 6.73 kN

for student work Ans.

by the 223.61

protected of the

is assess ing

work

solely


this

This is the

and courses part

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 point A, it is traveling at 9 m>s

and increasing its speed at 3 m>s2

. 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

Geometry: Here, dy = - 0.00625x and d2

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

dx dx

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

r = |d2

y>dx2

| = |- 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

N = 6729.67 N = protected by the of the

Ans.

6.73 kN


this

This is the

and courses part

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

+ a©Fn = man ; 5 v

N - 5 sin f =

32.2 (20 )

v dv = at ds v f

45

8 ft/s

θ = 30 45

20 ft A

B C

Lg v dv = L45°32.2 cos f (20 df)

1 2 1 2 laws Web) teachi ng

Dissemination or v - (8) = 644(sin f - sin 45°) 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

the

is work


this

This is the

and courses part

of any sale will


13–71.

If the ball has a mass of 30 kg and a speed v = 4 m>s at the

instant it is at its lowest point, u = 0°, determine the tension in

the cord at this instant. Also, determine the angle u to which the

ball swings and momentarily stops. Neglect the size of the ball.

u

4 m

S O L U T I O N

(4)2

+ c ©Fn = man;

T - 30(9.81) = 30a 4 b

T = 414 N Ans.

+ Q©Ft = mat;- 30(9.81) sin u = 30at

at = - 9.81 sin u

at ds = v dv Since ds = 4 du, then u 0

- 9.81 L L4 v dv laws or 0 s in u(4 d u) =

1

teaching Web) C 9.81(4)cos uD

u0 = - 2 (4)

2

Dissemination . copyright Wide

39.24(cos u - 1) = - 8

u = 37.2° instructors not permitted States . World

of learning the is United on Ans.

use and


protected of the


work


this

This is the

and courses part

of any sale will


*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

9.81(4) cos u 20 = 2 (v)2

- 2 (4)2

1 2

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

At u = 20° v = 3.357 m>s

at = - 3.36 m>s2

= 3.36 m>s2

b T

= 361 N

laws or



United


and

use

the student by (including

Ans.

for work

protected of Ans. assessing

the

is solely work


this

This is the

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

A r r

B r r

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

Realizing that an =


r = r

v2

+ T ©Fn = man; mg = m¢ r ≤ v = 2 gr Ans. teaching Web)

+ c © Fn = man; N - mg = mg component of .Dissemination or Using the result 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

instructors not permitted Stat e s

learning World

N = 2mg

of is

the Ans.

United on and

use



of the

is solely

work


this

This is the

and courses part

of any sale will


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,

v

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

r

200 ft

weight of 3500 lb.

S O L U T I O N

3500 v2

T ©Fn = man; 3500 = 32.2 a 200 b

v = 80.2 ft>s Ans.

laws or



United


and

use


protected of the


work


this

This is the

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.

v2

v

B 2

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

+ c ©Fb = 0; mAg cos u - mBg = 0

mB

u = cos- 1 a

m

A b Ans.

+ vB 2 or = ma ;

m g sin u B

(l - h) sin u

; ©F laws n n A

teaching Web)

m

g (l - h)

.

vB =

A

sin u

copyright

(1) Wide

B

mB

Dissemination 2 2

States .

World permitted From Fig. b, sin u = 2mA - mB . Substituting this value into Eq. (1),instructors not

A of the is

United learning

on

3m 2

2

use

m g(l - h) - m by and

vB = A A B

the student

work

for

B

mB

mA

(including

£ protected the

is ≥ solely assessing work of

2 2

= B

g(l - h)(mA - mB ) this Ans. m m work and of integrity

A B provided


any

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

v2

2 = gh

v = 2 2gh Q.E.D.

laws or



United


and

use



of the

is solely

work


this

This is the

and courses part

of any sale will


13–77. The skier starts from rest at A(10 m, 0) and descends the

smooth slope, which may be approximated by a parabola. If she

has a mass of 52 kg, determine the normal force the ground

exerts on the skier at the instant she arrives at point B. Neglect

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

y

10 m 1 2

y –– x

5 20

A

x 5 m

B

S O L U T I O N

dy 1 d2

y 1

Geometry: Here, dx = 10x and dx2

= 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

a

10

r = |d2

y>dx2

| =

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 positive

Kinematics: The speed of the skier can be determined usingby

1

x

is

102 1 + Here, tan u = 10x. Then, sin u = v 0 This

is

+ L0

= -

L10 m

2

10 1 +

v 100 xof

2

= 9.81 m sale> 2

Substituting v2

= 98.1 m2

>s2

, 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 6 in. 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.

A

u

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.

v2 6

2

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

+ c ©Fb = 0; 20(l - 2) cos u - 10 = 0 (1)

+ 10 62

; ©Fn = man; 20(l - 2) sin u = laws

Web)

a

b teaching

Solving Eqs. (1) and (2) yields (2) or Ans.Dissemi

32.2

0.5 + l sin u

nation

u = 31.26° = 31.3° copyright Wide .

l = 2.585 ft instructors permitted

States . World

of learning the is

Ans.no t

United on use and



of the

is solely

work


this


of any sale will

k 20 lb>ft

13–79.

The airplane, traveling at a constant speed of 50 m>s,

is executing a horizontal turn. If the plane is banked at u = 15°, when the pilot experiences only a normal force on

the seat of the plane, determine the radius of curvature r of

the turn. Also, what is the normal force of the seat on the

pilot if he has a mass of 70 kg.

S O L U T I O N

+c aFb = mab; NP sin 15° - 7019.812 = 0

NP = 2.65 kN

+ 502

F

; a n = man; NP cos 15° = 70a r b

r = 68.3 m

u

r

Ans. Ans.

laws or



United


and

use



of the

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

= 97.22 m>s. Realizing that an = v2

= 97.222

= 3.151 m>s2


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 copyright Wide

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


United


and

use


protected of the


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

b a

b a

b

h 1 km 3600 s

v2

97.22

= 97.22 m>s. Realizing that an =

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

+ c ©Fb = 0; L cos 15° - 5000(9.81) = 0

L = 50780.30 N = 50.8 kN Ans.

+ 97.222

; ©Fn = man; 50780.30 sin 15° = 5000¢

≤

r laws or


r = 3595.92 m = 3.60 km copyright Wide .

Ans.


United


and

use



of the

is solely

work


this


of any sale will

L u


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.

A

2

y 2x

100 m

S O L U T I O N

dy

d2y

22 2 2

Geometry: Here, y = 22x1>2

. Thus,

= and dx2 = - . The angle that

dx 1>2 3>2

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

dy

2 2

u = tan- 1

a

= tan- 1 ¢

dx b 2 x = 100 m 2x1>2 ≤

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

dy 2 3>2

B1 +

rA = a dxb R d y 5

2

2

dx x = 100 m

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

Here, an must be directed towards the center of curvature (positive n axis). Equations of Motion: The speed of the motorcycle is

v = a km 1000 m 1 h

80 h

b a b a b

1 km 3600 s

Thus, v2 22.222

an= =

rA 2849.67 R +

©Fn = man; 800(9.81)cos 4.045° - N = 7689.82 N =

This and

sale


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

satisfy the equation tan u sin u = v20>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

l sin2

u

mv 2 cos u

0

a l b a sin2

u b = mg

tan u sin u = v 2

Q.E.D. gl

0

laws or



United


and

use


protected of the


work


this

This is the

and courses part

of any

v

0

sale will



*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

y = 8 - 1 x2

2

dy

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

d2

y

dx 2 = - 1

y

10 ft/s

A

1 2

y 8 –– x

2

O x

2 ft

dy

2

2

3

(1 + (- 2) )2

3

B1 + a dx b

r = R2

=

= 11.18 ft

d2

y

|- 1|

dx2

2 laws or

teaching Web)

Dissemination

co p yr i g h t Wide .

OA = 2 (2)2

+ (6)2

= 6.3246 instructors permitted

2 States . World

tan f = 6

; f = 71.565°

2

of the is not United on use and

the student work by (including

protect ed f or of 2

assessing32.2 the 11.18

is solely work5 (10)

+ b©Fn = man;


5 cos 63.435° - N + 33.246 cos 45.0° = b a b

N = 24.4 lb This is the Ans.

+ R©Ft = mat; and courses part 5 b at any

their 32.2

at = 180.2 ft>s2

sale will

an = v2 =

(10)2 = 8.9443 ft>s

2

r 11.18

a = 2 (180.2)2

+ (8.9443)2

a = 180 ft>s2

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

z =

ma

z; FA - 12(z + 0.3) = mz

0.75

FA - 12(0.1 sin 2u + 0.3) = 32.2(- 14.4 sin 2u) For u = 45°,

FA - 12(0.4) = 0.75

(- 14.4) 32.2 FA = 4.46 lb

laws or teaching Web) Dissemination

copyright Wide .

permitted

States

instructors . World

United of learningthe is not on

by use and

for student work Ans.

protected the of the is

assess ing

work

solely work provided and of integrity

this

This is the

and courses part

of any

sale will


13–86. Determine the magnitude of the resultant force acting on a 5-kg

particle at the instant t = 2 s, if the particle is moving along a horizontal path defined by the equations r = (2t + 10) m

2

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

SOLUTION

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

# r = 2

$ r = 0

u = 1.5t2 - 6t

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

$ u = 3

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

$ # #

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

laws or teaching Web)


©Fr = mar; Fr = 5(0) = 0

©Fu = mau; Fu = 5(42) = 210 N instructors permitted

States . World

F = 2(Fr)2

+ (Fu)2

= 210 N United

of learning the is not on use andAns.



of the

is solely work


this


of any sale will



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

= (0.5t2

- 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

u = 0.5t2

- t|t= 2 s = 0 rad

#

= t - 1|t= 2 s = 1 rad>s u $ # 2

= 2

= - 5 ft>s 2

ar = r - ru 0 - 5(1)

$ # # 2

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

©Fr = mar; Fr = 32.2 (- 5) = - 0.7764 lb

5

©Fu = ma u; Fu = 32.2 (9) = 1.398 lb

$ 2

u = 1 rad>s

2 2 2 2 laws Web) teaching

+ (1.398)

= 1.60 lb .Dissemination or

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


United


and

use



of the

is solely

work


this


of any sale will


*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 # $

r = 3 m>s r = 0 # $

u = t2

+ 2 u = 2t| t = 2 s = 4 rad>s u = 2 rad>s2

3 # 2 # # 2

z = 6 - t z = - 3t z = - 6t| t = 2 s = - 12 m>s

ar = # # # 2

= 0- 10(4) 2

= - 2

r - r u 160 m>s

$ # 2

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

# # 2

az = z = - 12 m>s = 1.5(- 160) = - 240 N

Ans.

©F

r = ma ; Fr

r Anslaw

©F = ma ; Fu = 1.5(44) = 66 N s. Web) u u teaching

.Dissemination or

©Fz = maz; copyright Wide

.

Fz - 1.5(9.81) = 1.5(- 12) Fz = - 3.28 N

A .


United


and

use


protected of the

is assess ing

work

solely


this

This is the

and courses part

of any sale will



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 )| 2

r = 1.5(sin uu + cos uu u =

120°

= - 18.75 ft>s Applying Eqs. 12–29, we have

A

B

r

r = 1.5 (2 – cos ) ft.

O

= 5 rad/s

$ #

2 2 2

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 = d r > 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

United


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 assessing

32.2 is solely work the

N = 2.773 lb = 2.77 lb

work provide d and of this integrity 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

z = - 0.5t # $ # #

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

$ $

u = 0 z = 0

ar = 2

2 =

r$

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

au = ru + 2ru = 0

az = $

z = 0

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

©Fu = mau; Fu = 0 Ans.la

ws

teaching Web) Dissemination or

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

Fz = 392 N

instructors permitted States World

Ans. .

United


and

use



of the

is solely work


this

This is the

and courses part

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$ angular velocity u = 4 rad>s

and an angular acceleration u = 8 rad>s2

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

SOLUTION

r

u

0.5 ft

0.5 ft

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

# # r = - sin uu # 2 $

r = - cos uu - sin uu # $

At u = 30°, u = 4 rad>s and u = 8 rad>s2

r = 1 cos 30° = 0.8660 ft #

r = - sin 30° (4) = - 2 ft>s ..

= - cos 30° (4)2

- sin 30° (8) = - 17.856 ft>s2

r

ar = $ # 2

= - 17.856 - 0.8660(4) 2

= - 31.713 ft>s 2

r - ru

$ # #

au = ru + 2ru = 0.8660(8) + 2(- 2)(4) = - 9.072 ft>s Q + ©Fr = mar; - N cos 30° = 0.5 (- 31.713)

32.2

0.5

laws or

2

teaching

Web) .

copyright Wide

N = 0.5686 lb Dissemination

. permitted

States instructors

World

United of learning the is not

on use

+ a©Fu = mau; F - 0.5686 sin 30° = 32.2 (- 9.072) by the and student

for (including work

F = 0.143 lb protected of the Ans. is solely work

assessing

this


and courses part

any

of their destroy

sale will



*13–92. Using a forked rod, a smooth cylinder C having a mass of

0.5 kg is forced to move along the vertical slotted path r = 10.5u2 m, where u is in radians. If the angular position

of the arm is u = 10.5t2

2 rad, where t is in seconds, determine the force of the rod on the cylinder and the

normal force of the slot on the cylinder at the instant

t = 2 s. The cylinder is in contact with only one edge of the

rod and slot at any instant.

S O L U T I O N

r = 0.5u

r = # $ $

0.5u r = 0.5u

u = 0.5t2

# $

u = t u = 1

At t = 2 s, #

$

u = 2 rad = 114.59° u =2 rad>2 u = 1 rad>s2

# $ 2

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

r 0.5(2) tan c =

c = 63.43°

dr>du = 0.5

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

# # 2 2 . Dissemination or

#

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

copyright Wide

au = ru + 2r u =


States World NC = 3.030 = 3.03 N of learning Ans.not United on the is use

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


C

u r 0.5 u

13–93.

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

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 # #

r =

4 sec u(tan u)u| = 4 sec 45° tan 45°(1.5) = 8.485 ft>s u= 45°

#

$ $

# #

+ uAsec u sec2

uu + tan u sec u tan uuB D

r =

4 Csec u(tan u)u $

# 2

= 4 Csec u(tan u)u + sec3

uu2

+ sec u tan2

uu D u= 45° laws

Using the above time derivatives, C

2 2 in teaching .Disseminationpermitted

A

B

or

.

r

u

u O

4 ft

= 38.18 ft>s + sec

3 45°(1.5)

2 copyright Wide Web)

=4 sec 45° tan 45°(0) + sec 45° tan 45°(1.5)

instructors not

ar = r - ru = 38.18 - 5.657 1.5 States World = 25.46 ft>Uni t e ds learningon

$ # 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>

Fig. a, assessing

is solely work shown in Equations of Motion: 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

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# = 90°, if the force

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

SOLUTION

r = eu

#

u

= e u

r

$ u 2 u

r = e (u) + e u At u = 90°

#

u = 2 rad>s $

u = 0 r = 4.8105

F

r r eu

u

# a = r - r u

r = 9.6210()

$ r = 19.242

$ #

#

$ # 2

au = ru + 2ru =

r tan c = dr

AduB

c = 45°

laws Web) teaching

= 19.242 - 4.8105(2) = 0 . Dissemination

or

copyright Wide World permitted 2 States instructors

2 learning

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

of the is

United on use

and

= eu

>eu

= 1

by the student for (including work


is solely work the

+ c a Fr = mar; - NC cos 45° + work provided and of integrity

F cos 45° = 2(0) this

This is the

Ans.

+

NC = 54.4 N

and course s part

; aF

u an y destroy

their of

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

direction yields #2au = $ # 2 #

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

When integrated, r u = c, where the constant c is determined from the problem data.

SOLUTION

$ # 1d 2 #

F =

0 = m[ru + 2r#

u]

= mc

r dt (r u ) d = 0

a u mau; Thus,

A

r

Bu

r0 u·0

0.2 m/s

C

F

2

d(r u) = 0

2

r u = C #

(0.5)2

(1) = C = (0.25)2

u

#

u = 4.00 rad>s #

r$

= 0 Since r = - 0.2 m>s,

$ # ar = r - r (u)

2 = 0 - 0.25(4.00)

2 =

- 4 m>s

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

T = 8 N

laws or

teaching Web) Dissemination copyright

Ans. Wide

instructors not permitted States . World

Uniteduse of learning the is on and

the student

for (including work by

protected of Ans. assessing

is solely work the


this

This is the

and courses part

of any sale will


*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

O

r =

cos u = 0.5 sec u

# #

r = 0.5 sec u tan uu

$ =

# + sec u(sec

2 # $

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

# # $

= 0.5C sec u tan2

uu + sec3

uu2

+ sec u tan uuD # $

When u = 30°, u = 2 rad>s and u = 0 r = 0.5 sec 30° = 0.5774 m $ 2 r = 0.5 sec 30° tan 30°(2)

r = 0.5C sec 30° tan 30°(2)

ar =

$ # 2

r - ru = 3.849 -

$ #

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

2 3 2 laws Web)

= 0.6667 m>s Dissemination or + sec 30°(2) + sec 30° t an 30°(0)D copyright Wide .

2 2 States World permitted

0.5774(2) = 1.540 m>s instructors not

= 3.849 m>s2

. learning

United

of the is

2.667 m>s 2

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

+ R©Fu = mau; F + 0.5(9.81) sin 30° -

protectedsolely of Ans.

assessing is work the

and of integrity

p r o v id ed

=this0.5(2.667) work5.79 s n 30° 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

$ #

of u = 3 rad/s2

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

# #

r = 0.5 sec u tan uu $

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

sec u(sec

2 ## $ r uu)D u + sec u tan uuF

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

When u = 30°, u = 2 rad>s and u = 3 rad>s2

r = 0.5 sec 30° = 0.5774 m

#

= 0.5 sec 30° tan 30°(2) = 0.6667 m>s

r

r

$ = 4.849 m>s

2

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

2 + sec

3 30°(2)

2

A

· u 2 rad/s

r

0.5 m

u

O

laws teaching Web)

Dissemination or copyright Wide

.

ar =

$ # 2

= 4.849 - 0.5774(2) 2 = 2.5396 m>s

2 States World permitted r - ru instructors

$

# .

learning

# of the is not

u = ru + 2r u = 0.5774(3) + 2(0.6667)(2) = 4.3987 m> s2

Uniteduse on and

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

the (including

by

N = 6.3712 = 6.37 N protected of Ans. assessing

is 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

r = eu

#

= e u #

r u

$ u #2 u

r = e (u) + e u At u = 45°

F

r r = eθ

θ

#

u = 2 rad>s $ u = 0 r = 2.1933 r# = 4.38656

$ r = 8.7731

$ # r = r - r(u)

laws Web) teaching

Dissemination or

copyright Wide .

# # 2 instructors not permitted

2 2 States

World

= 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

and by

use

= eu

>eu

= 1

tan c = dr the student work

a b (including

c = u = 45° r

assessing

du

the

is solelywork

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.


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 z = 0 Applying Eqs. 12–29, we have

ar = $ $2

= 0 - 8(0.100 2

) = -0.0800 m>s 2

r - ru

au = $ # #

r u + 2 ru = 8(0) + 2(0)(0.200) = 0

$

az = z = 0 laws or

Equation of Motion: teaching Web)

Fr = 250(- 0.0800) = - 20.0 N .

DisseminationAns.

©Fr = mar ; copyright Wide .

©Fu = mau ; instructors World permitted

Stat e s

©Fz = maz ; of learning the is

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


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

# $

At u = 30°, u = 0.4 rad>s, and u = 0.8 rad>s2

r = 0.8 cos 30° = 0.6928 ft

r# = - 0.8 sin 30°(0.4) = - 0.16 ft>s

$

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

2

r = $ # 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

0.5Uni

ted of learningthe is not

a + ©Fu = mau;

FOA + 0.2790 sin 30° - 0.5 cos 30° = use on

by (0.4263) and student

for (including work

32.2t he


FOA = 0.300 lb is solely work


this

This is the

and courses part

the Ans.

of any

sale will


13–101. The ball of mass m is guided along the vertical circular path

r = 2rc cos u usin#g the arm OA. If the arm has a constant

angular velocity u0, determine the angle u … 45° at which

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

P A

r

rc

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

0 )

# c #

4r c u

2 4r u 2

0

u = tan- 1

¢

c 0

≤

Anlaws

.

tan u = g g or teaching Web) Dissemination copyright Wide .


United


and

use



of the

is

solely work


this

This is the

and courses part

of any sale will


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 =

0.6u vu = ru = 0.6uu 2

2 #

2

v = r+ a rub # 2 2 # 2

22

= a 0.6ub

+ a 0.6uub

u =

uu 2

laws Web)

# $

# $

$

teaching

0.621 + u 2 Dissemination or # copyright Wide

u = -

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

2

.

At u = p rad, # = = 1.011 rad>s instructors permitted 1 + u2 States . World

learning

0.62 1 + p2

2

of the is not

$

(p)(1.011) United on

use and

= - 0.2954 rad>s2

u = - 1 + p2

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 ing

work

$ # 2 = - 0.1772 m>s

2 work and of integrity r = 0.6(- 0.2954) this

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

mpart>s the

$ # # and coursesany 2

their of destroy

r 0.6u sale will

tan c =

c = 72.34°

dr>du = 0.6 = u = p

+

; ©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

11>r2 d1r2

u2>dt = 0. When integrated, r2

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 # #

From Eq. (2), ¢ r ≤ dt ¢r2

u≤ = 0 r 2

u = c

# laws

Web) teaching

Dissemination

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

.

u = ¢ World permitted

(4)2 ≤

Statesinstructors not

Hence, when r = 4 ft, . of learning the is

# 51.2 $ Uniteduse

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

by and

for student work

the (including

- T = protected of

32.2 a 0 - (4)(3.2) b assessing the

400 2 solely is work

T = 509 lb work provided and of integrity

this Ans.

This is the

and courses part

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.

#

The$ arm is rotating at a rate of u = 5 rad>s when u = 2 rad>s2

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 $

u = 2 rad>s2

r = 0.2>u = 0.12732 m

# - 2 # r = - 0.2 u u = - 0.40528 m>s

$ - 3 #2 - 2

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

ar = r - r(u) = 2.41801 - 0.12732(5) = - 0.7651 m>s laws . au = r u + 2 r u = 0.12732(2) + 2(- 0.40528)(5) = - 3.7982 m>s

2

2

2 teaching

$ # 2 or

$ Dissemination Web)

copyright Wide

tan c =

=

World permitted

( dr ) - 0.2u - 2 instructors

r 0.2>u . p learning

of the is not

c = tan

- 1

= - 57.5184°

United on

(- 2 ) by use and

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

+


N 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

F = - 1.66 N any

Ans.

of

sale will


r 0.2 —

u

r · ·· 2

u 5 rad/s, u 2 rad/s

u 90

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 $

- cos 90°(0.5)2

- sin 90°(0) = 0

r = $ # 2 2 2 laws or

. $ # 2 teaching

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

Dissemination Web) a

r

c = - 63.43° copyright Wide

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

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


2 on

use

and

32.2 student for (including work

- N cos 26.57° = ( - 0.5)

by

+ c ©Fr = mar; 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

2 ft r path contact the particle on only one side.

SOLUTION r = 2 + cos u

# #

r = - sin uu

$ #

2 $ r = - cos uu - sin uu # $ At u = 60°, u = 0.5 rad>s, and u = 0 r = 2 + cos 60° = 2.5 ft

r# = - sin 60°(0.5) = - 0.4330 ft>s

$ - cos 60°(0.5)

2 - 0.125 ft>s

2

r = - sin 60°(0) =

a $ # 2

= - 0.125 - 2.5(0.5) 2

2

laws

or r =

r - ru = - 0.75 ft>s

$ #

2

teaching

Web) .

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

tan c =

=

= - 2.887

c = - 70.89° .

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


of

their destroy

u ·

u 3 ft

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.

If u = (0.5t2

) 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.

SOLUTION

r = 2 + cos u u = 0.5t2

r# = - sin uu

#

u = t $ # 2 $ $ 2

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

At t = 1 s, u = 0.5 rad, u = 1 rad>s, and u = 1 rad>s2

r

= 2 + cos 0.5 = 2.8776 ft

2 ft r

u

u·

3 ft

r# = - sin 0.5(1) = - 0.4974 ft>s2

$

- cos 0.5(1)2

- sin 0.5(1) = - 1.357 ft>s2

r = $ # 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

- N cos 9.46° =

( - 4.2346)

by

+ Q©Fr = mar; N =the0.2666 lb

protected of assessing the + a©Fu = mau; F - 0.2666 sin 9.46° = is (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

$

(1 - cos u)2

#

#

$ - 4 sin u u

- 4 cos u (u) 8 sin u u

+

+

r = (1 - cos u) 2

(1 - cos u)2

(1 - cos u)3

# $

At u = 90°, u = 4, u = 0

r = 4

r = - 16 $

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

# r = 128

laws or teaching .

. Dissemination Web)

copyright Wide instructors

P

r

u

$ 2 2 $ # States World

learning is

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

provided integrity

1 - cos u)

4 work and of this

tan c =

=This-

dr

= - 4 sin u

1is and

the

( )

2

2 u = 90° = - 4 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 m. If the cord has a stiffness k = 30 N>m and an unstretched length of 0.25 m, determine the force of the guide on the

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

velocity u = 5 rad>s.

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 = 0

At u = 60°, r = 0.6928

r#

= 2

$

r = - 17.321

P

r

·

0.4 m u 5 rad/s

u

O

laws or teaching Web)

Dissemination

$ # 2 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

of learning the is not on 0.08(-

Q + ©Fr = m ar; - 13.284 + NP cos 30° = 34.641)by

use

and

for student work (including

protected the of the assess ing

a + ©Fu = mau; F - NP sin 30° = 0.08(20) solely F = 7.67 N is work Ans.

NP = 12.1 N and of integrity provi ded

this

work

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, determine# the force of the guide on the

particle when u = 2 rad>s2

, u = 5 rad>s, and u = 60°.

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 $

= - 16.521 r

$ 2 2 laws Web) teaching

au= r u + 2 ru =

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

Dissemination or

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

copyright Wide

ar = r = - 33.841 .

$ # #

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. assessing

is solely work the

NP = 12.2 N work provided and of integrity

this

This is the

and courses part

of any

sale will

P

r

·

0.4 m u 5 rad/s

u

O



13–111.

A 0.2-kg spool slides down along a smooth rod. If the rod has a constant angular rate of rotation #

= 2 rad>s in the vertical plane, show that the equations of u

motion for the spool 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

ar = $ # 2

= $ 2 $

- 4r

r - ru r - r(2 ) = r

$ # # # #

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

r - 4r - 9.81 sin u = 0 (Q.E.D.)

Disseminationor

copyright (1) Wi de

©Fu = mau; 1.962 cos u -

Since u.

= 2 rad>s, then 0.8r + Ns u = 2

L0 # L0 u equation (Eq.(1)) is given by

- 2 t

r = C1 e + C2

Thus,

# - 2t

r = - 2 C1 e + 2C2

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

At t = 0, r

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

Solving Eqs. (5) and (6) yields

C1

= - 16

Thus,

9.81 - 2t 9.81 2t

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

= 9.81 a - e 2t

+ e2t

- sin 2tb

8

9.81

(sin h 2t

- sin 2t)

= 8

p 9.81 p

At u = 2t

a sin h

= 4 , r = 8 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

vp2 # 2

+ a rub

= r # 2 #

(80)2

= 0 + a 1200ub u = 0.06667

$ # #

$

2vpvp = 2rr + 2a rub a ru + rub

0 = 0 +

0 + 2r 2 $ $

= 0 uu u

ar $

# 2 2

=r - ru = - 600(0.06667)

$ # #

0 au = ru + 2ru = 0 + 0 =

+ c ©F r = ma r

; N - 150 =

or laws

2 teaching Web)

2 .

- 1200(0.06667) = - 8 ft>s copyright Wide

Dissemination .

States World permitted

150 instructors not

of the is a b (

- 8) N

= 113 lb learningAns.

32.2

by United use on

and

the student

for (including work protected of the is solely work

assessing

this



their

of 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

Ch2

1 GMS r

2 ≤ and h = r

e = GM where C =0 ¢1 - r v v S 0 0 0 0

1 GMS r0 v02

r0v02

e =

¢1 -

≤(r0v0)2

GMS r0 r0 v02

e = ¢ GMS - 1≤ GMS GMS (e + 1) y = r

0B 0

= B66.73(10- 12

)(1.99)(1030

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

km>s 151.3(109

)

= e + 1

Ans.

laws Web) or

1 = 1 GMS GMS . Dissemination r

r0

r v

r0

v0

.

not permitted 1 1 66.73(10

- 12

)(1.99)(1030

) 66.73(10 instructors

States World

= 0.502(10

9

2

b cos u +

C 151.3(10 )D

92 (30818)2

) cos u + 6.11(10 ) learning Ans.

- 12 - 12 United of on the is

1 use

and


protected of the

is assess ing

solely work


this


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 around the circular orbit of radius

r 0 = h + r e = C h + 6.378(106

)D m is given by

T =

2pr0 v s

2pC h + 6.378(106

)D

24(3600) =

vs

vs =

2pC h + 6.378(106

) (1)

86.4(10 3

)

r 0 = h + r e = C h + 6.378(106

)D m is given by laws Web) teaching

The velocity of the satellite orbiting around the circular orbit of radius or Dissemination copyright Wide

66.73(10-12

)(5.976)(1024

) .

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.


this

This is the

and courses part

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

66.73(10 )(5.976)(10 ) v0 = 3

B (800 + 6378)(10 )

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

laws or



United


and

use



of the

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

vC = A r

0 = B 4000(103

) + 6378(103

) = 6198.8 m>s Parabolic Orbit:

2(66.73)(10)5.976(10

)

2GM e

ve = A r0 = B 4000(103

) + 6378(103

) = 8766.4 m>s

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

m>s ¢v = 2.57 km>s

h = 4 Mm

Ans.

laws or



United


and

use



of the

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 + h2

1 GMs

ra = - C + h2

teaching Web)

Dissemination or

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

2a

2GMs copyright Wide .

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

States . World

p Uniteduse of

learningthe is not on

forstudent

and

T = h (2a)(b) by the

T2h2 protected of

is assessing

Thus, work the

4p a provided integrity

This is part the and courses their destroy

of any

4p2

sale will

T2 = a GMs ba3

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

Ch2

e = GMe

1 GMe

where C = r0 ¢1 - r0v0 2 ≤ and h = r0 v0.

1 GMe

e = GMe r0 ¢1 - r0 v02 ≤ (r0 v0)

2

r v 2 0 0

e = ¢ GMe - 1≤

laws

r0 v0

2 GMe (e + 1)

teachingWeb) or

2 Mm

B

GMe

= e + 1 v

0 = B

vB = v0 = 66.73(10 )(5.976)(10 )(0.25 + 1) C 8.378(10

6)

where r0 = rp= 2A 106 B + 6378A 10

3 B = 8.378A 10

6 B

r 8.378(106

) 0

ra = =

r0

. Dissemination

copyright Wide .

= 7713 m>s = 7.States71 Ans.World

km>

permitted m.


on use

and 13by.96A 106 B

= m

student - 1 - 1 for wor k

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

of the

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

rp 8.378(106

) is solely work


ra 13.96(106

)

this

This is the

and courses part

of any sale will

13–119. The elliptical orbit of a satellite orbiting the earth has an eccentricity of e = 0.45. If the satellite has an altitude of 6 Mm at perigee, determine the velocity of the satellite at apogee and the period.

P A

SOLUTION 6 Mm

6 3 6

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

h = 12.378(106

)vP (1)

and

1 GMe

C = r P

a1 - r v 2 b P P

66.73(10- 12

)(5.976)(1024

)

1

C = 12.378(106

) c1 - 12.378(106

)vP 2 d

C = 80.788(10- 9

) - 2.6027 laws(

or

2)

teaching Web)

2 Dissemination

vP

2 copyright Wide .

permitted

GMe instructorsnot

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

e =

of learningthe is Unitedon

and

use

c80.788(10- 9

) - d c12.378(106)vforP d

student work

v 2

P by the (including

- 12 2.6027

protected24 2

assessing the

0.45 =

66.73(10 is

)solely

work

)(5.976)(10 of

vP = 6834.78 m>s work and ofintegrity provided this

This is part the

a 2GM and courses their of des troy

rP any Using the result of vP,

e

- 1

r v 2 sale will P P

12.378(106

) =

2(66.73)(10- 12

)(5.976)(1024

) 6 2 - 1

12.378(10 )(6834.78 )

6

= 32.633(10 ) m

Since h = rPvP = 12.378(106

)(6834.78) = 84.601(109

) m2

>s is

constant, rava = h

32.633(106

)va = 84.601(109

)

va = 2592.50 m>s = 2.59 km>s Ans.

Using the result of h,

T = p (rP + ra) 2

rPra

p = 84.601(10 ) C 12.378(10

6) + 32.633(10

6) D312.378(10

6)(32.633)(10

6)


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

y2 S

e

s s

F = G r2 AlsoF = ms a r b Hence y 2 m s m e 0

ms a r b = G r2

5.976(1024

)

me

B66.73(10 - 12

) a

y = AG

r = 15(106

) b = 5156 m>s = 5.16 km>s Ans.

laws or



United


and

use



of the

is solely

work


this

This is the

and courses part

of any sale will


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

Central-Force Motion: Use ra =

r

0 , with r0 = rp = 6A 10

6

B m and

(2 GM>r y 2 B - 1

M = 0.70Me, we have 0 0

9A 106

B = 6(10)6

2(66.73) (10-12

) (0.7) [5.976(1024

)]

¢

6(10

6 2 ≤ - 1

laws or )yP teaching Web)

Dissemination

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


United


and

use



of the

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 - GMe

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 2v 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 GMe

1 + e = r0 v02

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

v0 = D r0

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

1va21 =

a ra b 1vp21 = c

31.89

1201106

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.

v = GMe 66.73110

- 12

215.976211024

2 r

101106

2

cD 0= D = 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



United


and

use



of the

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

rp = r0 = 9.301109

2 km

ch2

1 GMs r0v02

e = GM = r a 1 - r v b a GM b s 0 0 0 s

r v 2

0 0

e = a GMs - 1b

r v 2

0 0

GMs = e + 1 (1) GMs 1

r

b

laws Web)

r v 2

+ 1 teaching

e . Dissemination or 0 0

- 1 A B - 1 copyr ight Wide .

r

0

States instructors World

permitted = r0 1e + 12 ra = 2GMs =

2 (2)

0 0 e + 1

of learning the is

11 - e2 0.927 United

not

r 9

on

use

a = 10.8110 2 km

andAns.


protected a s se ss in g

of the

is solely work


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?

S O L U T I O N

e = 0.130

vp = v0 = 15 Mm>h = 4.167 km>s

Ch2

1 GMe r02

v02

e = GMe = r0 ¢1 - r0v02 ≤a GMe b

r v2

0 0

e = ¢ GMe - 1≤

2

r0 v0 GMe = e+ 1

r0 =

v02

= 1.130166.732110

- 12

215.976211024

2

C 4.167110 32 D 2

A P

laws Web) teaching

copyright Wide .

instructors permitted States . W orl d

learning is

= 25.96 Mm of thenot

1

United on

GMe use

and

r0

v0 2 = e + 1 the student

for (including work by

protected of 2

e - 1 A B - 1 assessing the

rA=

2GM = r

is solely work

r01e + 12 + 1

work and of thisintegrity 0 0 e

rA =

This is part the

25.96110 211.1302 and courses destroy

1 - e

6 their of any 0.870 sale

will

=

= 33.7111062 m = 33.7 Mm

v r r

vA = 0 0

A

= 15125.9621106

2 33.71110

62

= 11.5 Mm>h Ans.

d = 33.711106

2 - 6.37811062

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

C2

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

= GMe 1

r0 v2

0

GMe = 34.4(10- 9

)(409)(1021

)

= 14.07(1015

) r0 = e =

= 1.046(109

) ft laws or 2 13 2

GM 14.07(1015

) teaching Web)

1.047(109

)

- 3960 = 194(103) mi

Dissemination r =

v

)] copyrightAns. Wide

0 [3.67(10

.

instructors permitted 5280

States . World C

2h learning is

(b) e = GM = 1 Uniteduse of

the not

on and

for student work e by the (including

v 2 ≤ = 1

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

assessing

1

2GM

1

2(14.07)(1015

)

2 2 GMe is solely work v0 2 [3.67(10 )] provided ofintegrity this

r0 = = 9 3 ) m

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

of any sale will

(c) e 6 1

194(103

) mi 6 r 6 392(103

) 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

B

u R

A

nR

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 va = kvO =

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

planet, r = rB = R. 1 1 GM

=

a1 -

r r r v P P P

1 1 GM R = a r

P - r

2v 2

P P

Since h = rava = nRakA nR b = k GM

rPvP = h copyright

vP =

rP Also,

k2nGMRlearning r

P

2

ra =

rP nR =

2nGMR

vP2

=

Solving Eqs.(2) and (3),

2GM

r

Pv

P 2GM

- 1 2 is

of P P

rp(rp + nR)

k2

n

rp = 2 - k2

R Substituting the result of rp and vp into Eq. (1),

1 2 - k2

R = a k2

nR -

u = cos

- 1 a

k2

n -

1 - k

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

then k2

n - 1

u = p - cos- 1

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

r0 = 3(106

) m, its velocity is

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

B m

c 0 = B 3(10 )

= 1278.67 m>s

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

v = 2GM 2(66.73)(10 )(0.0123)(5.976)(10 ) m

e B r0 = B 3(10 )

= 1808.31 m>s

laws or


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

States . World

United

of learningthe is not on

and

use


protected of the

is assess ing

solely work


this

This is the

and courses part

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. A B

S O L U T I O N 9 Mm

1 GMe h = r0 v 0

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

1 A 1 - GMe B 1 r0

2 v0

22

2

2

ch 2 2

r0 0 0 r0v0

e = GMe

= GMe

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

1 GMe laws( 1 + e = r

0 v

2 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 ing

work


Rearrange Eq.(4), we have this

1 + e 1 + 0.76 This is 6 part the 6 1 - e 1 - 0.76 and coursesany their destroy

2 m

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

of

Substitute r

= r

= 9A 106

B m into

0 p Eq.(3)saleyieldswill

(1 + 0.76)(66.73)(10 12

)(5.976) (1024

)

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

Applying Eq. 13–20, we have

rp 91106

2

v a = a ra b np = B 66.01106

2 R18830.822 = 1204.2 m>s = 1.20 km>s Ans. If the rocket travels in a circular free-flight trajectory, its speed is given by Eq. 13–25.

GM 66.73110- 12

215.976211024

2 e

v e = D r0 =

D 91106

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,


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.

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

vC = B r

0 = B 4000(103

) + 6378(103

) = 6198.8 m>s Parabolic orbit: 12

)5.976(1024

) 2GM 2(66.73)(10 e

ve = B r0 = B 4000(103

) + 6378(103

) = 8766.4 m>s

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

m>s ¢v = 2.57 km>s

h 4 Mm

laws or

Ans.



United


and

use



of the

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

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

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

C = rP ¢1 - rP vP 2

≤

1 66.73(10-12

)(5.976)(1024

)

C = rp B1 - rp(4166.672

) R

15 Mm/h

P A

B1 -

R laws(

C = rP rP 2)

1 22.97(106

) teaching Web)

GMe DisseminationWide copyright .

22.97(10 )

instructors permitted 1 Ch

2

6

States . World

r B1 -

r R(4166.67 rP)

2 of learningthe is not

P P

United

0.15 =

66.73(10 12

)(5.976)(1024

) on

by use and

rP = 26.415(106

) m for student work

protected theof the

is assess ing

work

solely

rP

work and of thisintegrity

rA =

This is provided

the

2 p a r t

2GMe

- 1 their

any destroy

rP vP

26.415(106

)

of

= sale will

2(66.73)(10-12

)(5.976)(1024

) - 1

26.415(106

)(4166.672

)

= 35.738(106

) m

Since h = rP vP = 26.415(106

)(4166.672

) = 110.06(109

) m2

>s is constant, rA

vA = h

35.738(106

)v = 110.06(109

) A

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

Using the results of h, rA, and rP,

T =

p A rP + rA B 2

P A

6

p

=

C 26.415(106

) + 35.738(106

)D 226.415(106

)(35.738)(106

)

110.06(10 9 )

= 54 508.43 s = 15.1 hr 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,

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

Elliptical Orbit:

Ch2

e = GMe

1 GMe r0

e = a

r0 y02

e =

GMe

r0 y0 2

= e +

GMe

ye = A r0

1 GMe where C = r ¢ 1 - r y 2 0 0 0

¢ GMe ≤

- 1b

1 - 2 (r0 y0)2

GMe (e + 1) 1 y0 =

B r0

≤ and h = r0 y0

laws . teaching

or

. DisseminationWeb)

copyright Wide

instructors not permitted States

learning World

of the is

United on use

and

student work

2GMe

0

(includingebthe

GMe (e + 1) GMepro tecte d by

for the

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?

S O L U T I O N

r0

Central-Force Motion: Substitute Eq. 13–27, ra = (2GM>r0v0 2

) - 1 , with

ra = (14 + 6.378)(106

) = 20.378(106

) m and r0 = rp = (6 + 6.378)(106

)

= 12.378(106

) m, we have

A A'

O

6 Mm

14 Mm

¢

2(66.73)(10- 12

)[5.976(1024

)] ≤ -

1 laws Web)

20.378(106

) =

12.378(10 )vp teaching .

12.378(106

) 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 (including work

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

protected of 6

is assessing 2 the

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

= solely work

78.368(10 ) m >s.Thus, applying

p

provided of integrity

Eq. 13–31, we have work and this

h This courses part the

ra)2rp ra

and

is

T = (rp + destroy =

p 9 [(12.378 + 20.378)(10 any )

6

)]their212.378(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) GM 66.73(10)(5.976)(10 ) vc = A

e

A

= 4423.69 m>s = 4.42 km>s r

20.378(10 6

0 ) 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.

Ans.

solution manual for engineering mechanics dynamics 13th ... · 13–7. if the 50-kg crate starts...

Documents