# Friction between two boxes

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1. Mar 1, 2015

### x86

1. The problem statement, all variables and given/known data

2. Relevant equations
F=ma
Ff = N u

3. The attempt at a solution
Pretty much, I did the same thing in the solution.

For box A:
N = mg
Ffs = mg*uk = 9.81*20*.4=78.48N

I note that Ffs > 60 N, so the boxes will move with the same acceleration

Now there are two ways I can approach this problem, each gives me a different answer??

1) Take two boxes as the same system, then Fx = 70*a = 60 ==> a = 0.857 m/s^2
2) Look at the FBD of box 2, note that Fx = 50*a = 60 ==> a = 1.2 m/s^2

However, both of these answers are different than the solution.

Where did I mess up (and why do 1 &2 give different solutions?)

2. Mar 1, 2015

### PhanthomJay

Solution messed up the math, although equation is correct. . When you look at the lower Block , you did not properly address the force acting on the lower block. It isn't 60. Draw an FBD.

3. Mar 1, 2015

### UVCatastrophe

Check for errata. Looks like they did 60 = 70a --> a = 70/60, which is wrong.

4. Mar 1, 2015

### x86

Isn't this the FBD? Ffs on the top block is -60N, and since its an action-reaction pair it should be +60N on the 50kg block

EDIT: Actually, now that I think about it, the friction on the bottom/top block should be different, because its making the top block accelerate.

So the I have more equations to solve:

20a = -Ff + 60
50a = Ff

20a = -50a + 60
70a = 60
a = 60/70

Thank you everyone for your help

Last edited: Mar 1, 2015