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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.

15–107. 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.

15–109. The 150-lb car of an amusement park ride is connected to a rotating telescopic boom. When r = 15 ft, the car is moving on a horizontal circular path with a speed of 30 ft>s. If the boom is shortened at a rate of 3 ft>s, determine the speed of the car when r = 10 ft. Also, find the work done by the axial force F along the boom. Neglect the size of the car and the mass of the boom. F

SOLUTION

r

Conservation of Angular Momentum: By referring to Fig. a, we notice that the angular momentum of the car is conserved about an axis perpendicular to the page passing through point O, since no angular impulse acts on the car about this axis. Thus,

A HO B 1 = A HO B 2 r1mv1 = r2m A v2 B u 15(30) r1 v1 = 45 ft>s = r2 10

Thus, the magnitude of v2 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)

Av2Bu =

v 2 = 4A v 2 B r 2 - A v2 B u 2 = 232 + 452 = 45.10 ft>s = 45.1 ft>s

Ans. Ans.

Principle of Work and Energy: Using the result of v2, T1 + ©U1 - 2 = T2 1 1 mv1 2 + UF = mv 2 2 2 2

1 150 1 150 a b (302) + UF = a b (45.102) 2 32.2 2 32.2 UF = 2641 ft # lb

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.

16–10. The vertical-axis windmill consists of two blades that have a parabolic shape. If the blades are originally at rest and begin to turn with a constant angular acceleration of ac = 0.5 rad>s2, determine the magnitude of the velocity and acceleration of points A and B on the blade when t=4 s.

ac

0.5 rad/s2

B

SOLUTION

10 ft

Angular Motion: The angular velocity of the blade at t = 4 s can be obtained by applying Eq. 16–5.

20 ft

A

v = v0 + ac t = 0 + 0.5(4) = 2.00 rad>s Motion of A and B: The magnitude of the velocity of points A and B on the blade can be determined using Eq. 16–8. Ans.

vB = vrB = 2.00(10) = 20.0 ft>s

Ans.

th an sa eir d i s 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)

vA = vrA = 2.00(20) = 40.0 ft>s

The tangential and normal components of the acceleration of points A and B can n be determined using Eqs. 16–11 and 16–12 respectively. (at)A = arA = 0.5(20) = 10.0 ft>s2

(an)A = v2 rA = A 2.002 B (20) = 80.0 ft>s2 (at)B = arB = 0.5(10) = 5.00 ft>s2

(an)B = v2 rB = A 2.002 B (10) = 40.0 ft>ss2 The magnitude of the acceleration of points A and d B are

(a)A = 2(at)2A + (an)2A = 210.02 + 80.0 80.02 = 80 80.6 0.6 fft>s >s2

Ans.

(a)B = 2(at)2B + (an)2B = 25.002 + 40.0 0.02 = 40.3 4 fft>s2

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.

16–19. The vacuum cleaner’s armature shaft S rotates with an angular acceleration of a = 4v3>4 rad>s2, where v is in rad>s. Determine the brush’s angular velocity when t = 4 s, starting from v0 = 1 rad>s, at u = 0. The radii of the shaft and the brush are 0.25 in. and 1 in., respectively. Neglect the thickness of the drive belt.

SOLUTION Motion of the Shaft: The angular velocity of the shaft can be determined from dt =

L t

L0 2

t

dt =

A

dvS L aS vs

S

A

S

dvS

L1 4vS 3>4 2

vs

t 0 = vS 1>4 1 t = vS 1>4 – 1

When t = 4 s

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)

vS = (t+1) 4

vs = 54 = 625 rad>s

Motion of the Beater Brush: Since the brush is connected to thee shaft sh ft by by a non-slip non-sslip p belt, then v B r B = vs r s vB = ¢

rs 0.25 b (625) = 156 1 6 rad>s rad>>s ≤v = a rB s 1

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.

16–22. For a short time the motor turns gear A with an angular acceleration of aA = (30t1>2) rad>s2, where t is in seconds. Determine the angular velocity of gear D when t = 5 s, starting from rest. Gear A is initially at rest. The radii of gears A, B, C, and D are rA = 25 mm, rB = 100 mm, rC = 40 mm, and rD = 100 mm, respectively.

A C

B D

SOLUTION Motion of the Gear A: The angular velocity of gear A can be determined from dvA =

L

L

vA

vA

t

dvA =

L0

vA 0

L0

= 20t3>2 2

30t1>2dt t 0

vA = A 20t

B rad>s

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)

3>2

When t = 5 s

adt

vA = 20 A 53>2 B = 223.61 rad>s

Motion of Gears B, C, and D: Gears B and C which are mounted on the same the sa m axle me will have the same angular velocity. Since gear B is in mesh with gea gear ge arr A, A then vB rB = vA rA vC = v B = ¢

rA 25 b (223.61) = 55.90 55. 0 ra rad>s d>s ≤v = a rB A 100

Also, gear D is in mesh with gear C. Then vD rD = vC rC vD = ¢

rC 40 55.9 = 22.4 rad>s b (55.90) ≤v = a rD C 100

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.

*16–116. At a given instant, the gear racks have the velocities and accelerations shown. Determine the acceleration of point A.

a v

2 ft/s2 6 ft/s

SOLUTION Velocity Analysis: The angular velocity of the gear can be obtained by using the method of instantaneous center of zero velocity. From similar triangles, yD yC v = = rD>IC rC>IC 6 rD>IC

2

=

(1)

rC>IC

A 0.25 ft B

Where rD>IC + rC>IC = 0.5

(2)

Solving Eqs.(1) and (2) yields rD>IC = 0.375 ft v =

3 ft/s2 2 ft/s

rC>IC = 0.125 ft

yD 6 = 16.0 rad>s = rD>IC 0.375

th an Th sa eir d i is w le co s p or w of a urs rov k is ill de ny e a ided pro st pa nd s te ro r 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)

Thus,

a v

Acceleration Equation: The angular acceleration of the gear can be obtained ained d byy analyzing the angular motion of points C and D. Applying Eq. 16–18 6–1 with w th rD>C = { -0.5i} ft, we have aD = aC + a * rD>C - v2 rD>C

( -0.5i) 0.5i) (aD)ni + 2j = -(aC)ni - 3j + (- ak) * ( -0.5i) - 16.02 ((aD)ni + 2j = -(aC)ni + (0.5a - 3)j + 128i 128i Equating the j components, we have

2 = 0.5 a - 3

10.0 ra rrad>s ad>> 2 a = 10.0

The acceleration of point A can be obtained btain d by y aanalyzing aly yzi zing g tthe angular motion of h rA { 0.25i} 0.25i}} ft, we have points A and C. Applying Eq. 16–18 with A>C >C = {aA = aC + a * rA>C - v2 rA>C

aAj = - (aC)ni - 3j + (- 10.0k) k) * ( - 0.25i) - 16.02 ( -0.25i) Equating the i and j components, we have aA = 0.500 ft>s2 T

Ans.

(aC)n = 64 m>s2 d

© 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.

16–119. The wheel rolls without slipping such that at the instant shown it has an angular velocity V and angular acceleration A. Determine the velocity and acceleration of point B on the rod at this instant.

A 2a

O

V, A

a

B

SOLUTION vB = vA + vB/A (Pin) + v = ; B

1 Qv 22aR + 2av¿ a b 2 22

+c O = -

1 22

Qv22aR + 2av¿ a

23 b 2

v 23

vB = 1.58 va

Ans.

a A = aO + aA/O (Pin) (a A)x + (aA)y = aa + a(a) + v2a ;

T

;

T

:

(a A)x = aa - v2a (a A)y = aa a B = aA + aB/A (Pin)

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¿ =

1

v 2 23 1 a B = aa - v2a + 2a(a¿) a b - 2a a b 2 2 23 O = - aa + 2aa¿ a

2 23

b + 2a a

a¿ = 0.577a - 0.1925v2 a B = 1.58aa - 1.77v2a

v

2 1 b a b 2 23

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.

16–127. The slider block moves with a velocity of vB = 5 ft>s and an acceleration of aB = 3 ft>s2. Determine the angular acceleration of rod AB at the instant shown. 1.5 ft A vB aB

30

5 ft/s 3 ft/s2

2 ft B

SOLUTION Angular Velocity: The velocity of point A is directed along the tangent of the circular slot. Thus, the location of the IC for rod AB is indicated in Fig. a. From the geometry of this figure, rB>IC = 2 sin 30° = 1 ft

rA>IC = 2 cos 30° = 1.732 ft

Thus, vB rB>IC

Then

=

5 = 5 rad>s 1

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 s 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)

vAB =

vA = vAB rA>IC = 5(1.732) = 8.660 ft>s

Acceleration and Angular Acceleration: Since point A travels along circular long g the circ ular

slot, the normal component of its acceleration hass a m magnitude agnittud de of vA 2 8.6602 = 50 ft>s2 and is directed towards the (aA)n = = t center cen er of of the circular cirrc r 1.5 slot. The tangential component is directed along the tangent Applying e tan geentt of the he slot sslot. A the relative acceleration equation and referring to Fi Fig. F g. b, b aA = aB + aAB * rA>B - vAB 2 rA>B

0 i + 2 sin sin 30°j) 30 j) - 52( -2 cos 30° i + 2 sin 30°j) 50i - (aA)t j = 3i + (aAB k) * ( - 2 cos 30°i 50i - (aA)t j = (46.30 - aAB)i + (1.732a aAB + 25 25)j 5)jj Equating the i components, 50 = 46.30 - aAB aAB = -3.70 rad>s2 = 3.70 rad>s2 b

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.

*17–52. at

The pipe has a mass of 800 kg and is being towed behind a truck. If the angle u = 30°, determine the acceleration of the truck and the tension in the cable. The coefficient of kinetic friction between the pipe and the ground is mk = 0.1.

B

A G 0.4 m

SOLUTION + ©F = ma ; : x x

T cos 45° - 0.1NC = 800a

+ c ©Fy = may ;

NC - 800(9.81) + T sin 45° = 0

a + ©MG = 0;

45

u

C

T sin 15°(0.4) - 0.1NC(0.4) = 0

NC = 6161 N Ans.

a = 1.33 m>s2

Ans.

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)

T = 2382 N = 2.38 kN

© 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.

17–71. The pendulum consists of a 10-kg uniform slender rod and a 15-kg sphere. If the pendulum is subjected to a torque of M = 50 N # m, and has an angular velocity of 3 rad>s when u = 45°, determine the magnitude of the reactive force pin O exerts on the pendulum at this instant.

O

u

M ⫽ 50 N⭈m

600 mm A

100 mm

SOLUTION Equations of Motion: Since the pendulum rotates about a fixed axis passing through point O, [(aG)OA]t = a(rG)OA = a(0.3), [(aG)B]t = a(rG)B = a(0.7), [(aG)OA]n = v2(rG)OA = (32)(0.3) = 2.7 m>s2, and [(aG)B]n = v2(rG)B = (32)(0.7) = 6.3 m>s2. The mass moment of inertia of the rod and sphere about their respective 1 1 (IG)OA = ml2 = (10)(0.62) = 0.3 kg # m2 mass centers are and 12 12 2 2 (IG)B = mr2 = (15)(0.12) = 0.06 kg # m2. Writing the moment equation of 5 5 motion about point O, we have

B

+ ©MO = ©(Mk)O; - 10(9.81) cos 45°(0.3) - 15(9.81) cos 45°(0.7) - 50 =

a = 16.68 rad>s2

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 he th se y f b in is w ssin or t y U te o g he ni 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)

- 10[a(0.3)](0.3) - 0.3a - 15[a(0.7)](0.7) - 0.06a

This result can also be obtained by applying ©MO = IOa, where IO = © ©I IG + md2 = 2 1 Thu us, (10)(0.62) + 10(0.32) + (15)(0.12) + 15(0.72) = 8.61 kg # m2. Thus, 12 5 + ©MO = IOa;

- 10(9.81) cos 45°(0.3) - 15(9.81) cos 45°(0.7) 5°(0.7) - 50 5 = - 8.61a 8.6 61 a = 16.68 rad>s2

Using this result to write the force equations of motion motio m on along alo al lon ongg the t n and and t axes, ©Ft = m(aG)t;

(9.81 1) co coss 45 45°° + Ot 10(9.81) cos 45° + 15(9.81)

= 10[1 10 0[1 116.6 10[16.68(0.3)] + 15[16.68(0.7)]

Ot = 51.81 N ©Fn = m(aG)n;

5°° - 15(9.81) 15( sin 45° = 10(2.7) + 15(6.3) On - 10(9.81) sin 45° On = 294.92 N

Thus, FO = 4Ot 2 + On2 = 451.812 + 294.922 = 299.43 N = 299 N

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.

17–103. If the truck accelerates at a constant rate of 6 m>s2, starting from rest, determine the initial angular acceleration of the 20-kg ladder. The ladder can be considered as a uniform slender rod. The support at B is smooth.

C 1.5 m

B 2.5 m

60

A

SOLUTION Equations of Motion: We must first show that the ladder will rotate when the acceleration of the truck is 6 m>s2. This can be done by determining the minimum acceleration of the truck that will cause the ladder to lose contact at B, NB = 0. Writing the moment equation of motion about point A using Fig. a, a + ©MA = ©(Mk)A;

20(9.81) cos 60°(2) = 20amin (2 sin 60°) amin = 5.664 m>s2

a + ©MA = ©(Mk)A;

th an Th sa eir d i is w le co s p or w of a urs rov k s 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 n p k n ing cto 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)

Since a min 6 6 m>s2, the ladder will in the fact rotate.The mass moment of inertia about bout 1 1 2 2 2 # its mass center is IG = g. b, ml = (20) A 4 B = 26.67 kg m . Referring to Fig. 12 12 20(9.81) cos 60°(2) = -20(a G)x (2 sin 60°) - 20(a G)y (2 cos 60°) - 26.67a

(1) (1 1)

Kinematics: The acceleration of A is equal to thatt off the Thus, the truck. truck. Th h and refe referring attion n an erring g to o Fig. c, a A = 6 m>s2 ; . Applying the relative acceleration equation aG = aA + a * rG>A - v2 rG>A >A

oss 60° 60° i + 2 si sin n 60° 60 j) - 0 (aG)x i + (aG)y j = - 6i + ( - ak) * ( -2 cos (aG)x i + (aG)y j = (2 sin n 60 60° 0° a - 6)i 6))i + aj Equating the i and j components,

(a G)x = 2 sin 60° a - 6

(2)

(a G)y = a

(3)

Substituting Eqs. (2) and (3) into Eq. (1), a = 0.1092 rad>s2 = 0.109 rad>s2

Ans.

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

13–67. 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 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.

u

SOLUTION Free-Body Diagram: The free-body diagram of the passenger is shown in Fig. (a). Here, an must be directed towards the center of the circular path (positive n axis). 1h km 1000 m ba ba b h 1 km 3600 s = 22.22 m>s. Thus, the normal component of the passenger’s acceleration is given by 22.222 v2 = 4.938 m>s2. By referring to Fig. (a), = an = r 100 + c ©Fb = 0; + ©F = ma ; ; n n

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)

Equations of Motion: The speed of the passenger is v = a80

N cos u - m(9.81) = 0

N =

9.81m 99.81 1m cos u

9.81m sin u = m(4.938) cos u u = 26.7°

Ans.

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

13–95. The ball has a mass of 2 kg and a negligible size. It is originally traveling around the horizontal circular path of radius r# 0 = 0.5 m such 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 au = ru + 2ru = 11>r21d1r2u2>dt2 = 0. # When integrated, r2u = c, where the constant c is determined from the problem data.

A r u

B

r0

· u0

0.2 m/s C F

SOLUTION a Fu = mau;

$ 1 d 2# ## (r u) d = 0 0 = m[ru + 2ru] = m c r dt

# d(r2u) = 0 # r2u = C # (0.5)2(1) = C = (0.25)2u # u = 4.00 rad>s # Since r = - 0.2 m>s,

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)

Thus,

Ans. An ns.

$ r = 0

$ # ar = r - r(u)2 = 0 - 0.25(4.00)2 = - 4 m>s2 a Fr = mar;

- T = 2( - 4) T = 8N

Ans.

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

13–107. 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.

2 ft

r u · u

SOLUTION r = 2 + cos u # r = - sin uu # $ $ r = -cos uu2 - sin uu

3 ft

u = 0.5t2 # u = t $ u = 1 rad>s2

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

tan c =

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)

$ r = - cos 0.5(1)2 - sin 0.5(1) = -1.357 ft>s2 # $ ar = r - ru2 = -1.375 - 2.8776(1)2 = - 4.2346 ft>s2 $ ## au = ru + 2ru = 2.8776(1) + 2( -0.4794)(1) = 1.9187 ft>s2 r 2 + cos u 2 = = - 6.002 dr>du - sin u u = 0.5 rad

c = - 80.54°

2 ( - 4.2346) 32.2

+Q©Fr = mar;

- N cos 9.46° =

+a©Fu = mau;

F - 0.2666 sin 9.46° =

F = 0.163 63 lb

N = 0.2666 0 2666 6 lb

2 ((1.9187) (1.9 1 18 87) 32.2 2

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.

14–14. If the cord is subjected to a constant force of F = 300 N and the 15-kg smooth collar starts from rest at A, determine the velocity of the collar when it reaches point B. Neglect the size of the pulley.

200 mm B

C 30⬚ 200 mm F ⫽ 300 N

200 mm

SOLUTION

300 mm

Free-Body Diagram: The free-body diagram of the collar and cord system at an arbitrary position is shown in Fig. a.

A

Principle of Work and Energy: Referring to Fig. a, only N does no work since it always acts perpendicular to the motion. When the collar moves from position A to position B, W displaces vertically upward a distance h = (0.3 + 0.2) m = 0.5 m, while

force

F

displaces

a

distance

of

s = AC - BC = 20.72 + 0.42 -

20.2 + 0.2 = 0.5234 m . Here, the work of F is positive, whereas W does negative work. 2

2

TA + gUA - B = TB vB = 3.335 m>s = 3.34 m>s

1 (15)vB2 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 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)

0 + 300(0.5234) + [- 15(9.81)(0.5)] =

Ans. Ans.. Ans A

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14–21. The steel ingot has a mass of 1800 kg. It travels along the conveyor at a speed v = 0.5 m>s when it collides with the “nested” spring assembly. If the stiffness of the outer spring is kA = 5 kN>m, determine the required stiffness kB of the inner spring so that the motion of the ingot is stopped at the moment the front, C, of the ingot is 0.3 m from the wall.

0.5 m 0.45 m kB

kA A

SOLUTION

C

B

T1 + ©U1 - 2 = T2 1 1 1 (1800)(0.5)2 - (5000)(0.5 - 0.3)2 - (kB)(0.45 - 0.3)2 = 0 2 2 2 Ans.

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)

kB = 11.1 kN m

© 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.

*14–28. The cyclist travels to point A, pedaling until he reaches a speed vA = 4 m>s. He then coasts freely up the curved surface. Determine how high he reaches up the surface before he comes to a stop. Also, what are the resultant normal force on the surface at this point and his acceleration? The total mass of the bike and man is 75 kg. Neglect friction, the mass of the wheels, and the size of the bicycle.

y

C

SOLUTION 1 2

x1/2

y1/2

B y

x

2

4m

45

A

1 2

x + y = 2

4m

1 -1 1 1 dy x 2 + y-2 = 0 2 2 dx 1

dy -x-2 = 1 dx y-2 T1 + ©U1 - 2 = T2

y = 0.81549 m = 0.815 m x1>2 + (0.81549)1>2 = 2 x = 1.2033 m tan u =

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 (75)(4)2 - 75(9.81)(y) = 0 2 Ans. An s.

-(1.2033) - 1>2 dy = = - 0.82323 dx (0.81549) - 1>2

u = -39.46° Q+ ©Fn = m an ;

Nb - 9.81(75) cos 39.46° = 0 Nb = 568 N

+ R©Ft = m at ;

Ans.

75(9.81) sin 39.46°° = 75 at a = at = 6.23 m>s2

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.

x

*15–8. If the jets exert a vertical thrust of T = (500t3>2)N, where t is in seconds, determine the man’s speed when t = 3 s. The total mass of the man and the jet suit is 100 kg. Neglect the loss of mass due to the fuel consumed during the lift which begins from rest on the ground. T

SOLUTION Free-Body Diagram: The thrust T must overcome the weight of the man and jet before they move. Considering the equilibrium of the free-body diagram of the man and jet shown in Fig. a, + c ©Fy = 0;

500t3>2 - 100(9.81) = 0

t = 1.567 s

Principle of Impulse and Momentum: Only the impulse generated by thrust T after t = 1.567 s contributes to the motion. Referring to Fig. a, (+ c)

t2

m(v1)y + ©

Lt1

Fy dt = m(v2)y

3s

L1.567 s

a200t5>2 b 2

3s 1.567 s

v = 11.0 m>s

500 t3>2dt - 100(9.81)(3 - 1.567) = 100v

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)

100(0) +

- 1405.55 = 100v

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.

15–22. T (lb)

If the force T exerted on the cable by the motor M is indicated by the graph, determine the speed of the 500-lb crate when t = 4 s, starting from rest. The coefficients of static and kinetic friction are ms = 0.3 and mk = 0.25, respectively.

60 30 t (s) 2

T

M

SOLUTION Free-Body Diagram: Here, force 3T must overcome the friction Ff before the crate T - 30 60 - 30 moves. For 0 … t … 2 s, or T = A 15t + 30 B lb. Considering the = t - 0 2 - 0 free-body diagram of the crate shown in Fig. a, where Ff = mk N = 0.3N, + c ©Fy = 0;

N - 500 = 0

N = 500 lb

+ : ©Fx = 0;

3(15t + 30) - 0.3(500) = 0

t = 1.333 s

I =

L

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)

Principle of Impulse and Momentum: Only the impulse of 3T after t = 1.333 s contributes to the motion. The impulse of T is equal to the area under the T vs. t graph. At t = 1.333 s, T = 50 lb. Thus, 1 3Tdt = 3c (50 + 60)(2 - 1.333) + 60(4 - 2) d = 470 lb # s 2

Since the crate moves, Ff = mkN = 0.25(500) = 125 lb. Referring to Fig. Figg a,, Fi + ) (:

t2

m(v1)x + ©

Lt1

Fx dt = m(v2)x

500 500 bbv v (0) + 470 - 125(4 - 1.333) = a 32.2 32.2 v = 8.80 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.

*15–36. The 50-kg boy jumps on the 5-kg skateboard with a horizontal velocity of 5 m>s. Determine the distance s the boy reaches up the inclined plane before momentarily coming to rest. Neglect the skateboard’s rolling resistance.

s 30⬚

SOLUTION Free-Body Diagram: The free-body diagram of the boy and skateboard system is shown in Fig. a. Here, Wb,Wsb, and N are nonimpulsive forces. The pair of impulsive forces F resulting from the impact during landing cancel each other out since they are internal to the system. Conservation of Linear Momentum: Since the resultant of the impulsive force along the x axis is zero, the linear momentum of the system is conserved along the x axis. + ) (;

mb(vb)1 + msb(vsb)1 = (mb + msb)v

v = 4.545 m>s

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)

50(5) + 5(0) = (50 + 5)v

Conservation of Energy: With reference to the datum set in Fig. b, b, the tth he gravitational potential energy of the boy and skateboard at positions A and B are re (9.81) 1)( )(ss sin sin in 30 30°) 0) 0°) A Vg B A = (mb + msb)ghA = 0 and A Vg B B = (mb + msb)ghB = (50 + 5)(9.81)(s = 269.775s. TA + VA = TB + VB

1 1 (mb + msb)vA 2 + A Vg B A = (mb + msbb))v vB 2 + A Vg B B 2 2 1 (50 + 5) A 4.5452 B + 0 = 0 + 269.775s 69.7 77 75 5s 2 s = 2.11 m

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.