ch06-p068

# ch06-p068 - 68. The free-body diagrams for the two boxes...

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T mmg mm = + F H G I K J −= 21 12 105 ( N . μμ θ )cos . (b) These equations lead to an acceleration equal to ag =− + + F H G I K J F H G I K J = sin cos . . 22 11 362 m/s 2 (c) Reversing the blocks is equivalent to switching the labels. We see from our algebraic result in part (a) that this gives a negative value for T (equal in magnitude to the result we got before). Thus, the situation is as it was before except that the rod is now in a state of compression. 68. The free-body diagrams for the two boxes are shown below. T is the magnitude of the force in the rod (when T > 0 the rod is said to be in tension and when T < 0 the rod is under compression), 2 N F G is the normal force on box 2 (the uncle box), 1 N F G is the the normal force on the aunt box (box 1), G f 1 is kinetic friction force on the aunt box, and G f 2 is kinetic friction force on the uncle box. Also,
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## This note was uploaded on 05/17/2011 for the course PHY 2049 taught by Professor Any during the Spring '08 term at University of Florida.

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