# Three Blocks Contact force

1. Sep 27, 2008

### robvba

1. The problem statement, all variables and given/known data
Three blocks on a frictionless horizontal surface are in contact with each other. A force F is applied to block A (mass mA ). If mA=mB=mC=10.0 kg and F = 96.0 N, determine the force of contact that each block exerts on its neighbor.

2. Relevant equations

3. The attempt at a solution

a=96/(10+10+10)=3.2N

contact force between mA and mB: 96-3.2*10=64
contact force between mB and mC: 3.2*10=32 or 64-3.2*10=32

did i solve this right?

thank you

2. Sep 27, 2008

### tiny-tim

Hi robvba!

Correct result … and it looks like you got it the right way, but I'm not sure.

This is a Newton's second law problem … you really shouldn't talk about the "force between" two bodies … each equation should only concern the forces on a particular body.

3. Sep 27, 2008

### robvba

thank you. you're right. it's force of contact that each block exerts on it's neighbor. which would mean that block mA exerts a force of 64N on block mB while block mB exerts an equal and opposite force back; so mA = -mB. is that a fair statement?

4. Sep 27, 2008

### tiny-tim

erm … I don't think you mean mA = -mB, do you?

I was thinking more of mass times acceleration = sum of forces …

you know the acceleration of block A, so the sum of forces on block A is … , and so the reaction force from block B on block A is … ?

5. Sep 27, 2008

### robvba

right. i meant Fba = - Fab (Force of A on B = negative Force of B on A)...?