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15–101. The 10-lb block rests on a surface for which mk = 0.5. It is acted upon by a radial force of 2 lb and a horizontal force of 7 lb, always directed at 30° from the tangent to the path as shown. If the block is initially moving in a circular path with a speed v1 = 2 ft>s at the instant the forces are applied, determine the time required before the tension in cord AB becomes 20 lb. Neglect the size of the block for the calculation.

A 4 ft B 7 lb

30

2 lb

SOLUTION ©Fn = man ; 20 - 7 sin 30° - 2 =

10 v 2 ( ) 32.2 4

v = 13.67 ft>s

(

L

MA dt = (HA)2

th an Th sa eir d i is w le co s p or w of a urs rov k is ill de ny es a ided pro st pa nd s te ro rt o c y of as lel ted th t se y e his s fo by s r in te wo ing the Uni gr rk s u te ity ( tu s d of inc de e o Sta th lud nt f in te e in lea s s c w g r tru o or o ni c p k n ng to yri an th . rs gh d e W Dis in t l is t a no orl sem eac ws t p d W in hi er id ati ng m e on itt W o ed e r . b)

(HA) t + ©

10 10 )(2)(4) + (7 cos 30°)(4)(t) - 0.5(10)(4) t = (13.67)(4) 32.2 32.2

t = 3.41 s

Ans. Ans A

© 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.

15–102. The 10-lb block is originally at rest on the smooth surface. It is acted upon by a radial force of 2 lb and a horizontal force of 7 lb, always directed at 30° from the tangent to the path as shown. Determine the time required to break the cord, which requires a tension T = 30 lb. What is the speed of the block when this occurs? Neglect the size of the block for the calculation.

A 4 ft B 7 lb

30

SOLUTION ©Fn = man ; 30 - 7 sin 30° - 2 =

10 v2 ( ) 32.2 4

v = 17.764 ft>s

L

MA dt = (HA)2

0 + (7 cos 30°)(4)(t) = t = 0.910 s

10 (17.764)(4) 32.2

th an Th sa eir d i is w le co s p or w of a urs rov k is ill de ny es a ided pro st pa nd s te ro rt o c y of as lel ted th t se y e his s fo by s r in te wo ing the Uni gr rk s u te ity ( tu s d of inc de e o Sta th lud nt f in te e in lea s s c w g r tru o or o ni c p k n ng to yri an th . rs gh d e W Dis in t l is t a no orl sem eac ws t p d W in hi er id ati ng m e on itt W o ed e r . b)

(HA)1 + ©

Ans. An s.

© 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.

2 lb

15–103. A 4-lb ball B is traveling around in a circle of radius r1 = 3 ft with a speed 1vB21 = 6 ft>s. If the attached cord is pulled down through the hole with a constant speed vr = 2 ft>s, determine the ball’s speed at the instant r2 = 2 ft. How much work has to be done to pull down the cord? Neglect friction and the size of the ball.

B

r1 = 3 ft

(v B )1 = 6 ft>s

SOLUTION H1 = H2

v r = 2 ft>s

4 4 v (2) (6)(3) = 32.2 32.2 u vu = 9 ft>s v2 = 292 + 22 = 9.22 ft>s

Ans.

T1 + ©U1 - 2 = T2

©U1 - 2 = 3.04 ft # lb

th an Th sa eir d i is w le co s p or w of a urs rov k is ill de ny es a ided pro st pa nd s te ro rt o c y of as lel ted th t se y e his s fo by s r in te wo ing the Uni gr rk s u te ity ( tu s d of inc de e o Sta th lud nt f in te e in lea s s c w g r tru o or o ni c p k n ng to yri an th . rs gh d e W Dis in t l is t a no orl sem eac ws t p d W in hi er id ati ng m e on itt W o ed e r . b)

1 4 1 4 ( )(6)2 + ©U1 - 2 = ( )(9.22)2 2 32.2 2 32.2

Ans. An s.

© 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.

*15–104. A 4-lb ball B is traveling around in a circle of radius r1 = 3 ft with a speed 1vB21 = 6 ft>s. If the attached cord is pulled down through the hole with a constant speed vr = 2 ft>s, determine how much time is required for the ball to reach a speed of 12 ft/s. How far r2 is the ball from the hole when this occurs? Neglect friction and the size of the ball.

B

r1 = 3 ft

(v B )1 = 6 ft>s

SOLUTION v = 2(vu)2 + (2)2

v r = 2 ft>s

12 = 2(vu)2 + (2)2 vu = 11.832 ft>s H1 = H2 4 4 (6)(3) = (11.832)(r2) 32.2 32.2

¢r = v rt (3 - 1.5213) = 2t t = 0.739 s

Ans.

th an sa eir d i w le co s p or w of a urs rov k is ill de ny es a ided pro st pa nd s te ro rt o c y of as lel ted th t se y e his s fo by s r in te wo ing the Uni gr rk s u te ity ( tu s d of inc de e o Sta th lud nt f in te e in lea s s c w g r tru o or o ni c p k n ng to yri an th . rs gh d e W Dis in t l is t a no orl sem eac ws t p d W in hi er id ati ng m e on itt W o ed e r . b)

r2 = 1.5213 = 1.52 ft

Ans. An s.

© 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.

15–107. The ball B has a weight of 5 lb and is originally rotating in a circle. As shown, the cord AB has a length of 3 ft and passes through the hole A, which is 2 ft above the plane of motion. If 1.5 ft of cord is pulled through the hole, determine the speed of the ball when it moves in a circular path at C.

1.5 ft C 2 ft 3 ft

SOLUTION

©Fn = man;

vC

B

Equation of Motion: When the ball is travelling around the first circular path, 2 u = sin - 1 = 41.81° and r1 = 3 cos 41.81° = 2.236. Applying Eq. 13–8, we have 3 ©Fb = 0;

T

A

2 T1 a b -5 = 0 3

vB

T1 = 7.50 lb

7.50 cos 41.81° =

v2t 5 a b 32.2 2.236

v1 = 8.972 ft>s

©Fb = 0; ©Fn = man;

T2 sin f - 5 = 0 T2 cos f =

th an Th sa eir d i is w le co s p or w of a urs rov k is ill de ny es a ided pro st pa nd s te ro rt o c y of as lel ted th t se y e his s fo by s r in te wo ing the Uni gr rk s t ity ( tu s ed e i d of nc e o Sta th lud nt f in te e in lea s s c w g r tru o or o ni c p k n ng to yri an th . rs gh d e W Dis in t l is t a no orl sem eac ws t p d W in hi er id ati ng m e on itt W o ed e r . b)

When the ball is traveling around the second circular path, r2 = 1.5 cos f. Applying ng Eq. 13–8, we have (1) (1)

v22 5 a b 32.2 1.5 cos f

(2)

Conservation of Angular Momentum: Since no force actss on the e ball ball along allong the th tangent of the circular, path the angular momentum is cconserved con nserved about abou ut z axis. Applying Eq. 15–23, we have (Ho)1 = (Ho)2

r1mv1 = r2mv2 2.236a

5 5 b (8.972) = 1.5 coss f a 5 co bbv v2 32.2 32.2 2

Solving Eqs. (1),(2) and (3) yields f = 13.8678°

(3)

T2 = 20.85 lb

v2 = 13.8 ft>s

Ans.

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

15–110. An amusement park ride consists of a car which is attached to the cable OA. The car rotates in a horizontal circular path and is brought to a speed v1 = 4 ft>s when r = 12 ft. The cable is then pulled in at the constant rate of 0.5 ft>s. Determine the speed of the car in 3 s.

O

A r

SOLUTION Conservation of Angular Momentum: Cable OA is shorten by ¢r = 0.5(3) = 1.50 ft. Thus, at this instant r2 = 12 - 1.50 = 10.5 ft. Since no force acts on the car along the tangent of the moving path, the angular momentum is conserved about point O. Applying Eq. 15–23, we have (HO)1 = (HO)2 r1 mv1 = r2 mv¿ 12(m)(4) = 10.5(m) v¿

The speed of car after 3 s is

th an Th sa eir d i is w le co s p or w of a urs rov k is ill de ny es a ided pro st pa nd s te ro rt o c y of as lel ted th t se y e his s fo by s r in te wo ing the Uni gr rk s u te ity ( tu s d of inc de e o Sta th lud nt f in te e in lea s s c w g r tru o or o ni c p k n ng to yri an th . rs gh d e W Dis in t l is t a no orl sem eac ws t p d W in hi er id ati ng m e on itt W o ed e r . b)

v¿ = 4.571 ft>s

v2 = 20.52 + 4.5712 = 4.60 ft>s

Ans. Ans. A

© 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.