Dynamics Block

Shatzkinator

1. The problem statement, all variables and given/known data
A 2.5 kg block is initially at rest on a horizontal surface. A horizontal force F of magnitude 6.0 N and a vertical force P are then applied to the block (the vertical force is upwards). The coefficients of friction for the block and surface are Ms = 0.40 and Mk = 0.25. Determine th emagnitude of the frictional force acting on the block if the magnitude of P is a) 8.0 N, b) 10 N, and c) 12 N


2. Relevant equations
fs = Mk * Fn
Fnet = ma
Fg - Fn - P = 0
3. The attempt at a solution
I tried using the third equation to get Fn and plug this value into the first equation to get force of friction. I didn't get the right answer... for P = 8.0 N the answer should be 6.0 N. How?

Doc Al

Show exactly what you did. Hint: First you must determine whether the relevant friction is static or kinetic.

Shatzkinator

Fg - Fn - P = 0
24.5 = Fn + 8
Fn = 16.5

fs = 16.5 * 0.25 = 4.1 N

Doc Al

Good.

This assumes that the block moves. Does it?

Shatzkinator

well Yea if the force applied is 6.0 N and the friction is 4.1 then yes it should move.

Shatzkinator

Oh wait no it doesn't because Fn * Ms = 8.25 which is greater than 6 N so it doesn't move. Now what?

Doc Al

How does static friction work? Note that for static friction, μN is the maximum possible static friction that the surfaces can generate before being forced to move. But static friction will be only what it needs to be to prevent slipping (up to the maximum). How much does static friction need to be in this case to prevent slipping?

Shatzkinator

6 to prevent slipping?

Doc Al

Right!