Class exercise on Friction
 
 

The free-body diagram looks like this:


Since the object does not move vertically, the vertical forces must balance, so n = W = mg = (1 kg)(10 m/s2) = 10 N.

Case 1. The force F has magnitude 5 N. The friction force is limited and can be no greater than µsn = (0.75)(10 N) = 7.5 N. Thus the floor exerts a frictional force of magnitude 5 N, the forces balance, and the object remains at rest. The acceleration is zero.

Case 2. The force F has magnitude 8 N. This exceeds the maximum of 7.5 N that the floor can exert, so the object begins to accelerate. Once moving, the friction is kinetic, and we have f = µkn = (0.6)(10 N) = 6 N. There is a net force of 8 N - 6 N = 2 N, and the object's acceleration is a = Fnet/m = (2 N)/(1 kg) = 2 m/s2.