Question

A 4.770 kg block of wood rests on a steel desk. The coefficient of static friction between the block and the desk is ?s = 0.655 and the coefficient of kinetic friction is ?k = 0.155. At time t = 0, a force F = 18.8 N is applied horizontally to the block. State the force of friction applied to the block by the table at the following times: t=o and t>0

Consider the same situation, but this time the external force F is 38.0 N. Again state the force of friction acting on the block at the following times: t=0 and t>0

Answer 1

Answer 2

See, as long as the force applied on the block is less than the maximum static friction, the block remains at rest. And when at rest only static friction acts.
the static friction is given by f <= μ_sN, which means that the max value of static friction is μN. But if this was the value of frition all the time then would the net force on the block be zero.

Consider a block resting on table acted by 18 N as given. So if you draw its FBD and take f as friction force. So, f will come out to be 18 N.

18 - f = 0 [since block is in equilibrium, so net force zero].

Thus static friction has a self adjusting nature, which means that it adjusts itself according to the other forces on the block as long as the block is not moving. For example if a block is at and no external force is applied, what is the value of friction? It is ZERO, because net force is zero and since no other force is present so friction is 0.

Search google for self adjusting nature of static friction, you will get it. Just remember its not always μN but less than or equal to μN.

Answer 3

a)F = 18N

normal reaction on block = 4.77*9.8 N = 46.746 N

friction acting when block doesnt move = 0.655*46.746N = 30.6183 N

So F<30.6183 N:

so block doesnt move.

Force of friction applied on block = 18 N [since block is in equilibrium so static friction adjusts]

This remains for t=0 to t>0

b) Now F = 38 N which is greater than 30.62 N

so at t=0 max static friction is applied = 30.62 N

at t>0, block starts moving so kinetic friction takes over.

so friction force for t>0 = 0.155*46.746 N = 7.246 N