A 5.410 kg block of wood rests on a steel desk. The coefficient of static friction between the block and the desk is 𝜇s=0.455 and the coefficient of kinetic friction is 𝜇k=0.205. At time 𝑡=0, a For

Question

A 5.410 kg block of wood rests on a steel desk. The coefficient of static friction between the block and the desk is 𝜇s=0.455 and the coefficient of kinetic friction is 𝜇k=0.205. At time 𝑡=0, a force 𝐹=14.9  is applied horizontally to the block. State the force of friction applied to the block by the table at times 𝑡=0 and 𝑡>0.

Consider the same situation, but this time the external force 𝐹 is 29.9 N. Again, state the force of friction acting on the block at times 𝑡=0 and 𝑡>0.

Answer 1

Draw a free body diagram:

Since there is no acceleration (or motion at all) in the y-direction then the sum of the forces in the y-direction is zero:

∑Fy = 0 = F_N - W or F_N = W or F_N = mg = (5.410)(9.81) = 53.0721 N

Given that the static coefficient of friction (μ_s) is 0.455, we can use F_F = (μ_s)(F_N) to calculate the maximum possible force of friction prior to the object beginning to move.

F_F = (μ_s)(F_N) = (0.455)(53.0721) = 24.1 N.

Since the applied force is only 14.9 N, there will not be enough applied force to break the static friction so there will be NO MOTION at t = 0 or at any other time "t"

However, an applied force of 29.9 N is greater than the "break free" force of friction (24.1 N) therefore the object will begin to move. Once moving, the kinetic force of friction takes over which is calculated at:

F_F = (u_k)(F_N)

F_F = (0.205)(53.0721) = 10.9 N

So with an applied force of 29.9 N, at t = 0 the force of friction is 24.1 N and at t > 0, the force of friction is 10.9 N