2 blocks, frictions, and no numbers

[musicfairy]

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

A. m_Ag
B. m_Bg
C. μ_sm_Ag
D. μ_sm_Bg
E. μ_s(m_A+m_B)g

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

f = μm_Bg
f = F - m_Aa_A


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?

 

[Mattowander]

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

 

[musicfairy]

It would be m_Bg, I guess.

 

[Mattowander]

Really? Draw a free body diagram.

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

 

[musicfairy]

I did.

I have force F acting on m_A.
m_Ag is directed down
m_Bg 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? =(

 

[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?

 

[musicfairy]

Ok, I see now. So E says that F = μN_A
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.

 

[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.

 

[musicfairy]

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

 

[Mattowander]

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