ch06-p031

# ch06-p031 - 31. (a) Free-body diagrams for the blocks A and...

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Since the weight of block A is 44 N, the least weight for C is (110 – 44) N = 66 N. (b) The second law equations become T f = ( W A / g ) a F N W A = 0 W B T = ( W B / g ) a . In addition, f = μ k F N . The second equation gives F N = W A , so f = k W A . The third gives T = W B – ( W B / g ) a . Substituting these two expressions into the first equation, we obtain W B – ( W B / g ) a k W A = ( W A / g ) a . Therefore, () ( ) 2 2 (9.8 m/s ) 22 N 0.15 44 N 2.3 m/s . 44 N + 22 N Bk A AB gW W a WW == = + 31. (a) Free-body diagrams for the blocks A and C , considered as a single object, and for the block B are shown below. T is the magnitude of the tension force of the rope, F N is the magnitude of the normal force of the table on block A , f is the magnitude of the force of friction, W AC is the combined weight of blocks A and C (the magnitude of force G F gAC shown in the figure), and W
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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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