# Frictional force on stacked objects

1. Feb 9, 2012

### moonbase

1. The problem statement, all variables and given/known data
Three 10 kg blocks are stacked on top of each other on top of a frictionless table. A hand touching the bottom block applies a force of 90 N to the right. The coefficient of friction between the blocks is sufficient to keep the blocks from moving with respect to each other. What is the total force exerted on the middle block by the bottom block?

2. Relevant equations
Fn=Fg=mg
Fs≤usFn

3. The attempt at a solution
I drew a diagram and have determined that the forces acting on the middle block are the normal and frictional forces from both the top and bottom block. But since it's only asking for the bottom block, I think I can what the top block is doing. I know that the normal force must be 98 N (10 kg x 9.8 m/s2) but I can't figure out what the frictional force would be. I know it must be greater than 90 N if it's resisting the push of the hand, but I don't know how I would find the "upward" force it is is giving to the block above.

2. Feb 9, 2012

### PhanthomJay

This reasoning is incorrect. Draw a Free Body Diagram of the entire system of 3 blocks and identify the weight and external forces acting on the system, and then use Newton's Law's in both directions to solve for the unknown force and acceleration of the system. Then look at a free body diagram perhaps of the bottom block alone, and continue using Newton's Laws again for that block.

3. Feb 9, 2012

### moonbase

I did that and know that the only forces acting on the middle block from the lower block are the normal force pointing up, and the frictional force pointing right, so the sum of these would be the answer. I know that the normal force is equal to the downward force on the lower block (which is mtopg + mmiddleg), but I don't understand how Newton's laws would help me find the frictional force since I'm not given a constant.

4. Feb 9, 2012

### PhanthomJay

The normal force and friction force act in different directions, so thet have to be added vectorially, not algebraically. But you can just state that the total or net force acting on the middle block by the bottom block is so many Newtons up (the normal force) and so many Newtons to the right (the friction force).
You don't need to find the friction coefficient, just the friction force, which you can get by using Newton 2 in the horizontal direction, once you solve for 'a' in the earlier step.

5. Feb 9, 2012

### moonbase

Thanks, I got it now!