# Classical mechanics problem for two boxes [closed]

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This question is truly annoying, and I have been stuck for an hour on part D, would greatly appreciate if anyone could shed a light on this problem. Why ans for part c and d are different?

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## closed as off-topic by Qmechanic♦Oct 26 '13 at 9:40

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The force applied to the table is wrong. – ja72 Oct 25 '13 at 20:24
Also where is the friction between the blocks on your FBD? In general start from a clean FBD with each body separated and every force included. A good start will lead you the way. – ja72 Oct 25 '13 at 20:27

Your problem has two degrees of freedom, acceleration of the top block $a_1$ and acceleration of the bottom block $a_2$. If the friction force between them is $F$ then the equations of motion are
$$m a_1 = F \\ M a_2 = P - F$$
If the blocks are stuck then $a=a_1=a_2$ and $F$ are unknown. When they slip then $F=\mu m g$ is known, but $a_1$, $a_2$ are different unknowns. In each case there are two unkowns and two equations. The tansition occurs when friction tries to be $F \ge \mu m g$ which you need to find first.