Maximum force on bottom block before top block slides

[mewmew]

Hi, I was giving a problem by my teacher that will most likely be on the exam. A block, m1, is ontop of another block, m2 and there is a constant friction, mu, between the two blocks. Between block m2 and the ground there is no friction. What is the largest force, P, that can be put on m2 and have m1 stay in its same position. I have F=m1a and F=m2a, F = (mu)N1 = (mu)m1g. It seems really easy but for some reason I am having trouble with it. Thanks

Edit: My exam is in an hour so I was typing fast without thinking.

 

[arildno]

F is NOT N1!
It is the frictional force between m1 and m2!
Hence, you must solve the system when F equals MAXIMAL STATIC FRICTION,
F= (mu)N1.

 

[Pyrrhus]

I also don't understand the P-F ? isn't the ground with block 2 frictionless?

 

[mewmew]

Sorry, I was typing faster than I was thinking so I got a few things mixed up, I think it should be correct now.

 

[Pyrrhus]

Ok draw a freebody diagram on each block,

On the block on the ground you have the contact force exerted by the block on top, and the force p, plus its weight.

On the block on the top, you got the friction force, plus the same magnitude for the contact force exerted by block on the ground, and its weight.

You know if the friction force of the block on the top is equal to the acceleration caused by the P force, then the block won't fall.

Try to solve from there.

 

[mewmew]

So, is the following correct?

F = mu(m1g)

P=m2a
F=m1a

I could then solve one for a, and get for example a = f/m1 and then plug that into P = m1a and get P = m2(mu(m1g))/m1 ? It doesn't really seem like a good answer, but that's all I can figure out.

 

[arildno]

No, you have:
P-F=m2a (1)
F=m1a (2)
So, adding (1) and (2):
P=(m1+m2)a(3)
Since F is maximal static friction, we get a=(mu)g, or:
P=(m1+m2)(mu)g