PS3, Number 7

Miscellaneous correspondence with one of you:

On the top block, you have the normal force up, and its weight down, so the normal force is just the mass of the block times g.

That means the friction force is (mu)(mg). For motion to occur, you need (mu static)(mg) < (pulling force). This is true in this case, so the block is moving to the left, but also sliding against the bottom slab. Even though there is no friction between the slab and the ground, the block can pull away from the slab if the pulling force is big enough.

Since the block is moving, the *kinetic* friction force is (mu kinetic)(mg). The two horizontal forces are then this one and the pulling force, their difference gives mass times acceleration for the top block.

Now, if the top block has a friction force to the right due to the interaction with the bottom slab, the slab itself must feel the same force in the opposite direction by Newton's third law. Thus, if the top block is slipping off the bottom slab, the bottom slab has to feel the *same* friction force (but in the opposite direction) that the block feels. This is the only horizontal force on the slab, since it has no friction with the floor, nor is there a pulling force directly on it. So, (m slab)g = (friction force on top block) = (mu kinetic)(m block)g

I'll draw this out in class tomorrow and hopefully it will be clearer.