In Figure 6-63, a block weighing 22 N is held at rest against a vertical wall by

Question

In Figure 6-63, a block weighing 22 N is held at rest against a vertical wall by a horizontal force of magnitude 60 N. The coefficient of static friction between the wall and the block is 0.53, and the coefficient of kinetic friction between them is 0.36. In six experiments, a second force is applied to the block and directed parallel to the wall with these magnitudes and directions: (a) 35 N, up, (b) 12 N, up, (c) 47 N, up, (d) 63 N, up, (e) 4.1 N, down, and (f) 21 N, down. In each experiment, what is the frictional force on the block, including sign? Take the direction up the wall as positive, and down the wall as negative. Next, calculate the acceleration, including sign, of the block in each case. Note that acceleration is zero if the block does not move.
(g) What is the acceleration in (a)?
(h) What is the acceleration in (b)?
(i) What is the acceleration in (c)?
(j) What is the acceleration in (d)?
(k) What is the acceleration in (e)?
(l) What is the acceleration in (f)?

Explanation / Answer

w = 22 N, F = 60 N. us = 0.53, uk = 0.36 normal force N = F = 60 N maximum friction force fs = us*N = 31.8 N, kinetic friction force fk = uk*N = 21.6 N the 2nd force = P (a) P = 35 N, up, vertical resultant force (except friction force) = 35 - 22 = 13 N (up) fs, so it will move friction force f = fk = 21.6 N (up), net force = 43 - 21.6 = 21.4 N (down), acceleration a = 21.4*9.8/22 = 9.53 m/s^2 (down)

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