# Help interpreting a force problem

lizette
A 4kg black is put on top of a 5 kg block. To cause the top block to slip on the bottom one, while the bottom one is held fixed, a horizontal force of at least 12N must be applied to the top block. The assembly of the blocks is now placed on a horizontal, frictionless table. Find the magnitudes of the a) max horizontal force that can be applied to the lower block so that the blocks will move together and b) the resulting acceleration of the blocks.

The figure looks like this:

A
-
B --> F
--------

where block A is on top of block B

My question is, what does this mean:
"To cause the top block to slip on the bottom one, while the bottom one is held fixed, a horizontal force of at least 12N must be applied to the top block."

Slip on it from where? What does this matter? Does it mean that for A to fall off of B it needs to be pushed off with a force of 12N? Is this in correlation to fs <= UsN? (Us = coefficient of static friction) Or does it mean that if block B is pushed with a F > 12N then A will slip off it?

Thanks!

Chiara
the bottom block is kept fixed and therefore is just like the surface of a floor on which block A could slide. however, there is obviously friction between the two surfaces (the one of block A and the one of block B) and in order to make A slide over B the force applied to A has to be greater than the static friction between the two blocks -(4)(9.8)( CSFcoefficient of static friction)
12 N now move the upper block and therefore
12>(4)(9.8)( CSF)
CSF = 0.306
(a)When a force is applied to block B, the reaction of this force will act on block A (Newton's third Law) in the opposite direction. we know that if this force is greater than 12N the upper block will start sliding. therefore in order to make the two blocks move together the force on B has to be less than or equal to 12N.
(b)If the force acting on B is 12 N the acceleration of the system will be
12= (4+5)a
a=1.3m/s2