331

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13–87.

SOLUTION

Ans.F = 2Fr2 + F2

u = 2(-0.7764)2 + (1.398)2 = 1.60 lb ©Fu = mau; Fu =

532.2

(9) = 1.398 lb

©Fr = mar; Fr =5

32.2 (-5) = -0.7764 lb

au = ru$

+ 2r#u#

= 5(1) + 2(2)(1) = 9 ft>s2

ar = r$ - ru

#2 = 0 - 5(1)2 = -5 ft>s2

u = 0.5t2 - t|t = 2 s = 0 rad u#

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

= 1 rad>s2

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

$ = 0

The path of motion of a 5-lb particle in the horizontal planeis described in terms of polar coordinates as and rad, where t is in seconds. Determinethe magnitude of the unbalanced force acting on the particlewhen .t = 2 s

u = (0.5t2 - t)r = (2t + 1) ft

Ans:F = 1.60 lb

339

© 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

13–95.

SOLUTION

At ,

Ans. F = 17.0 N

+a©Fu = mau; F + 15.19 sin 30° - 3(9.81) cos 30° = 3(-0.3)

+Q©Fr = mar; N cos 30° - 3(9.81) sin 30° = 3(-0.5196) N = 15.19 N

au = ru$

+ 2r#u#

= 1.0392(0) + 2(-0.3)(0.5) = -0.3 m>s2

ar = r$ - ru

#2 = -0.2598 - 1.0392(0.5)2 = -0.5196 m>s2

r$ = -1.2 cos 30°(0.5)2 - 1.2 sin 30°(0) = -0.2598 m>s2

r# = -1.2 sin 30°(0.5) = -0.3 m>s

r = 1.2 cos 30° = 1.0392 m

u#

= 0.5 rad>s and u$

= 0u = 30°

r$ = -1.2 cos uu

#2 - 1.2 sin uu

$ r# = -1.2 sin uu

# r = 2(0.6 cos u) = 1.2 cos u

A smooth can C, having a mass of 3 kg, is lifted from a feedat A to a ramp at B by a rotating rod. If the rod maintains aconstant angular velocity of determine theforce which the rod exerts on the can at the instantNeglect the effects of friction in the calculation and the sizeof the can so that The ramp from A to Bis circular, having a radius of 600 mm.

r = 11.2 cos u2 m.

u = 30°.u#

= 0.5 rad>s,

600 mm

600 mm

B

A

C

u 0.5 rad/s · r

u

Ans:F = 17.0 N