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Problem 9.14 The box is stationary on the inclined surface. The coef f cient of static friction between the box and the surface is s . (a) If the mass of the box is 10 kg, D 20 , D 30 , and s D . 24, what force T is necessary to start the box sliding up the surface? (b) Show that the force T necessary to start the box sliding up the surface is a minimum when tan D s . T Solution: x y T N f mg D 20 s D . 24 m D 10 kg g D 9 . 81 m/s 2 X F x : T cos C f C mg sin D X F y : N C T sin mg cos D (a) D 30 , f D s N Substituting the known values and solving, we get T D 56 . 5 N , N D 64 . 0 N , f D 15 . 3 N Solving the 2nd equilibrium eqn for N and substituting for f.f D s N/ in the f rst eqn, we get T cos C s mg cos s T sin C mg sin D Differentiating with respect to , we get dT d D T. sin s cos / . cos C s sin / Setting dT d D 0, we get tan D s c 2008 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. 693 Problem 9.27 The ladder and the person weigh 30 lb and 180 lb, respectively. The center of mass of the 12-ft ladder is at its midpoint. The angle D 30 . Assume that the wall exerts a negligible friction force on the ladder. (a) If x D 4 ft, what is the magnitude of the friction force exerted on the ladder by the f oor? (b) What minimum coef F cient of static friction between the ladder and the f oor is necessary for the person to be able to climb to the top of the ladder without slipping? x a Solution: (a) Assume no slipping occurs D 30 , x D 4 ft X F x : f B N A D X F y : N B 210 lb D X M B : N A . 12 ft cos / 30 lb . 6 ft sin / 180 lb x D Solving f B D 77 . 9 lb , N B D 210 lb (b) At the top of the ladder D 30 , x D 6 ft X F x : f B N A D X F y : N B 210 lb D X M B : N A . 12 ft cos / 30 lb... View Full Document
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