479 Horizontal Motion: The horizontal component of velocity ... - Wuala

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91962_05_R1_p0479-0512 6/5/09 3:53 PM Page 484 © 2010 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. R1–7. The man A has a weight of 100 lb and jumps from rest onto the platform P that has a weight of 60 lb. The platform is mounted on a spring, which has a stiffness k = 200 lb>ft. If the coefficient of restitution between the man and the platform is e = 0.6, and the man holds himself rigid during the motion, determine the required height h of the jump if the maximum compression of the spring is 2 ft. h A Conservation of Energy: The datum is set at the initial position of platform P. When the man falls from a height of h above the datum, his initial gravitational potential energy is 100h. Applying Eq. 14–21, we have P Conservation of Momentum: T 1 + V 1 = T 2 + V 2 0 + 100h = 1 2 a 100 32.2 b(y M) 2 1 + 0 (y H ) 1 = 264.4h m M (y M ) 1 + m P (y P ) 1 = m M (y M ) 2 + m P (y P ) 2 ( +T) a 100 100 b(264.4h) + 0 = a 32.2 32.2 b(y M) 2 + a 60 32.2 b(y P) 2 [1] Coefficient of Restitution: e = (y p) 2 - (y M ) 2 (y M ) 1 - (y p ) 1 ( +T) 0.6 = (y p) 2 - (y M ) 2 264.4h - 0 [2] Solving Eqs. [1] and [2] yields (y p ) 2 = 264.4h T (y M ) 2 = 0.4264.4h T Conservation of Energy: The datum is set at the spring’s compressed position. 60 Initially, the spring has been compressed = 0.3 ft and the elastic potential 200 1 energy is . Here, the compression of the spring caused by 2 (200) A0.32 B = 9.00 ft # lb impact is (2 - 0.3) ft = 1.7 ft. When platform P is at a height of 1.7 ft above the datum, its initial gravitational potential energy is 60(1.7) = 102 ft # lb . When platform P stops momentary, the spring has been compressed to its maximum and 1 the elastic potential energy at this instant is . Applying 2 (200)A22 B = 400 ft # lb Eq. 14–21, we have T 1 + V 1 = T 2 + V 2 1 2 a 60 32.2 b A 264.4hB2 + 102 + 9.00 = 400 h = 4.82 ft Ans. 484
91962_05_R1_p0479-0512 6/5/09 3:53 PM Page 485 © 2010 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. *R1–8. The baggage truck A has a mass of 800 kg and is used to pull each of the 300-kg cars. Determine the tension in the couplings at B and C if the tractive force F on the truck is F = 480 N. What is the speed of the truck when t = 2 s, starting from rest? The car wheels are free to roll. Neglect the mass of the wheels. C B F A : + ©F x = ma x ; 480 = [800 + 2(300)]a a = 0.3429 m>s 2 ( : + ) v = v 0 + a c t v = 0 + 0.3429(2) = 0.686 m>s Ans. : + ©F x = ma x ; T B = 2(300)(0.3429) T B = 205.71 = 206 N Ans. : + ©F x = ma x ; T C = (300)(0.3429) T C = 102.86 = 103 N Ans. R1–9. The baggage truck A has a mass of 800 kg and is used to pull each of the 300-kg cars. If the tractive force F on the truck is F = 480 N, determine the acceleration of the truck. What is the acceleration of the truck if the coupling at C suddenly fails? The car wheels are free to roll. Neglect the mass of the wheels. C B F A : + ©F x = ma x ; 480 = [800 + 2(300)]a a = 0.3429 = 0.343 m>s 2 Ans. : + ©F x = ma x ; 480 = (800 + 300)a a = 0.436 m>s 2 Ans. 485
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479 Horizontal Motion: The horizontal component of velocity ... - Wuala

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