# Homework Help: 2 blocks, frictions, and no numbers

1. Sep 28, 2008

### musicfairy

Block A, with mass mA, is initially at rest on a horizontal floor. Block B, with mass mB, is initially at rest on the horizontal top surface of A. The coefficient of static friction between the two blocks is μs. Block A is pulled with a horizontal force. It begins to slide out from under B if the force is greater than:

A. mAg
B. mBg
C. μsmAg
D. μsmBg
E. μs(mA+mB)g

I came up with a couple of equations (that may not be right).

f = μmBg
f = F - mAaA

So I tried substitution and solving for F, but I can't get the right answer (E).
So what's the right way to do this problem?

2. Sep 28, 2008

### Mattowander

The equation for frictional force is the coefficient of friction times the normal force. What is the normal force on block A?

3. Sep 28, 2008

### musicfairy

It would be mBg, I guess.

4. Sep 28, 2008

### Mattowander

Really? Draw a free body diagram.

Don't forget that block B is sitting on top of block A.

5. Sep 28, 2008

### musicfairy

I did.

I have force F acting on mA.
mAg is directed down
mBg is directed down, and then up as a normal force
friction is in opposite direction of the 2 blocks

What am I doing wrong here??? =(

6. Sep 28, 2008

### Mattowander

The problem is that block A has block B on top of it. That means that the weight of block A is (Ma + Mb)g. What that means is that since block A is not accelerating down, it's normal force must be the same as the weight. Now do you understand?

7. Sep 28, 2008

### musicfairy

Ok, I see now. So E says that F = μNA
But what does that mean? I thought friction is supposed to be the same for both blocks in this problem. This is getting more confusing.

8. Sep 28, 2008

### Mattowander

The coefficient of friction is the same. The frictional force depends upon the normal force though. Because A has a greater normal force than B, it would produce a greater frictional force.

9. Sep 28, 2008

### musicfairy

I see. Thanks for the explanations. Now I really need to study hard for that test.

10. Sep 28, 2008

### Mattowander

No problem :) Glad I could help. Good luck!