Force on Two Blocks: Find F, Acceleration

[mbrmbrg]

A block of mass 4.0 kg is put on top of a block of mass M = 5.5 kg. To cause the top block to slip on the bottom one, while the bottom one is held fixed, a horizontal force of at least 18 N must be applied to the top block. The assembly of blocks is now placed on a horizontal, frictionless table.

(a) Find the magnitude of the maximum horizontal force F that can be applied to the lower block so that the blocks will move together.

(b) Find the magnitude of the resulting acceleration of the blocks.

Once the blocks are on the frictionless table, I run into trouble.
Namely, I don't know where to place friction in my free-body-diagrams for the two blocks. Friction definitely acts on the upper block, and I think that it points in the same direction as the force applied to the bottom block. It will also probably be maximum frictional force, which was given as 18N.
But does that friction act on the bottom block? I want to say not, but it's a contact force, and the bottom block is after all in contact with the top block!

 

[mbrmbrg]

Something odd this way comes:
I worked with the assumption that in the x direction, \Sigma F_B=F=m_Ba (where F=applied force) and \Sigma F_A=f=m_Aa (where f=maximum static fictional force between the two blocks).
I subtracted the second equation from the first and got \frac{F}{m_B}-\frac{f}{m_A}=0 to get F=24.75N.
That is the wrong answer, but I didn't know that yet, so I plugged F=24.75 into my first equation to get a=4.5m/s^2. Lo and behold! The a is right even though the F is wrong!
Any ideas?

 

[mbrmbrg]

Going on the off chance that \Sigma F_B=F+f, I subtracted 18 from 24.75. That's still the wrong answer...

 

[OlderDan]

A block of mass 4.0 kg is put on top of a block of mass M = 5.5 kg. To cause the top block to slip on the bottom one, while the bottom one is held fixed, a horizontal force of at least 18 N must be applied to the top block. The assembly of blocks is now placed on a horizontal, frictionless table.

(a) Find the magnitude of the maximum horizontal force F that can be applied to the lower block so that the blocks will move together.

(b) Find the magnitude of the resulting acceleration of the blocks.

Once the blocks are on the frictionless table, I run into trouble.
Namely, I don't know where to place friction in my free-body-diagrams for the two blocks. Friction definitely acts on the upper block, and I think that it points in the same direction as the force applied to the bottom block. It will also probably be maximum frictional force, which was given as 18N.
But does that friction act on the bottom block? I want to say not, but it's a contact force, and the bottom block is after all in contact with the top block!

You are not far off. 18N is the maximum force that can be applied before moving the top block while holding the bottom block, so that is the maximum frictional force. When maximum F is applied to the lower block, the friction force between the two blocks will be 18N. Friction acts one way on the top block (direction of motion) and the opposite way on the bottom block. It will oppose the F applied to the lower block and it will result in accelerating the upper block. From the upper block you find the acceleration of both blocks. What force is strong enough to overcome the 18N it has to fight, and still accelerate the lower block the same as the upper block?

An easier way to look at it is that since both blocks are moving with the same acceleration they are acting as a single mass responding to the applied F. What force will give the total mass the acceleration of the upper block? Both points of view give the same result.

 

[mbrmbrg]

Neat-o!
Thanks OlderDan.