A block weighing 26 N is held at rest against a vertical wall by a horizontal fo

Question

A block weighing 26 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 0.65 and the coefficient of kinetic friction between them is 0.39. In six experiments, a second force is applied to the block and directed parallel to the wall with the following magnitudes and directions.


(a) 38 N, up
What is the magnitude of the frictional force on the block?
N

(b) N29 N, up
What is the magnitude of the frictional force on the block?


(c) N56 N, up
What is the magnitude of the frictional force on the block?
N

(d) 81 N, up
What is the magnitude of the frictional force on the block?
N

(e) 12 N, down
What is the magnitude of the frictional force on the block?
N

(f) 19 N, down
What is the magnitude of the frictional force on the block?

Explanation / Answer

For each force, you first need to determine whether or not the block is moving. There is a 26N force acting downwards (weight of the block). Counteracting this force is the resistant force of static friction, which is, at most, F(static friction) = 60N(.65), or 39N. In (d) the net force is F=81N-26N, or 55N up. Since 55N>39N, we expect the block to accelerat upwards, so now we must use the kinetic friction coefficient to calculate the opposing force of friction. F(kinetic) = 60N(.39)=23.4N.

The same thing happens in (f). Net force is now F=19N+26N=45N down (I added this time because the force of the weight and the applied force act in the same direction). 45N>39N, so now we expect acceleration downwards and thus must again use kinetic friction coefficient. F(kinetic)=60N(.39)=23.4N.

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