1. The problem statement, all variables and given/known data (I'm going to let the coefficient for friction be "u" since the actual greek letter that's generally used doesn't seem to be a a forum option; also, I'm going to use a dash since subscripts aren't an option) A 40 kg slab rests on a frictionless floor. A 10 kg block rests on top of the slab. The coefficient of static friction (u-s) between the block and the slab is 0.60, whereas their kinematic friction coefficient (u-k) is 0.40. The 10 kg block is pulled by a horizantal force with a magnitude of 100N. What are the resulting accelerations of (a) the block and (b) the slab? 2. Relevant equations mass of the block = 10kg mass of the slab = 40kg u-s = 0.60 u-k = 0.40 F = 100N The diagram shown with the problem has the large rectangular slab resting on a frictionless surface, with the smaller cubed block on top of it. The force acting on the the block is going left. The specific equations I use are dependent on how I set up my free body diagrams, which is what I'm sort of needing help with; but in general, friction is equal to the Normal force multiplied by the coefficient of friction; also, force equals mass multiplied by acceleration. 3. The attempt at a solution All I need help with is setting up the equations; I'll know how to work it from there. Also, I already know what the answers to the problem are. The forces acting on the block: There's a leftward force of 100 newtons; a frictional force going right; a normal force exerted by the slab going up; and mg going down The forces acting on the slab: So far, I have a normal force the the floor exerts on the slab, mg going down, and a normal force exerted by the box going downward. Which way does friction go? I first tried it with the same frictional force going in the same direction as it did with the diagram for the block, using u-s when I calculated the slab's acceleration and u-k when I calculated the box's acceleration, but I didn't get the right answers. So, I'm thinking I set up my free body diagram for the slab wrong.