A block of mass 5 kg on 25degree -inclined plane, is subjected to a horizontal f

Question

A block of mass 5 kg on 25degree -inclined plane, is subjected to a horizontal force F as shown in the figure. The coefficient of static friction is mu s = 0. 4 and the coefficient of kinetic friction is mu k = 0. 2 Find the minimum magnitude of the force F so that the block is about to slide down the inclined plane. Find the maximum magnitude of the force F so that the block is about to slide up the inclined plane. What value of F is required to move the block up the inclined plane at constant speed?

Explanation / Answer

a) about to slide , bothsides forces are equal upward force = uN + Fcos25 (N = normal force) downward force = mg sin25 N = mgcos25 + Fsin25 so mgsin25 = 0.4(mgcos25 + Fsin25) + F cos 25 (0.4 beacause they are at rest we should take static friction) so 20.73 = 17.78+ 0.17F + 0.91F so F = 2.73N b) slide up the plane so frictional force will act downward so force equation is mg sin25 + uN = F cos25 mgsin25 + 0.4(mgcos25 + Fsin25) = F cos 25 so 20.73+17.78+0.17F = 0.91 F so F = 52.04 N c) for moving with constant speed forces are equal but here kinetic friction is acting so mgsin25 + 0.2(mgcos25 + Fsin25) = F cos 25 20.73 + 8.89 + 0.08F = 0.91F so F = 35.69 N

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