Can you explain the two-block problem discussed in lecture? There was friction between the two blocks and the ground and the block has zero friction. What is the maximum force on the bottom so block $A$ wouldn't move? How to analyze this problem?

I guess this is on p. 37 of lecture 6. This is an example which can be done by the Force How-To. First, decide what system to consider. Take the top block and draw a FBD. What are the forces on it? Answer: gravity $mg$ down, normal force $N_1$ up, and friction force to the right (the bottom block is ``trying to'' drag the top block along to the right). The maximum friction force before slipping is $f_{s {\rm max}} = \mu_s N_1$. Since there's no vertical acceleration, $N_1 = mg$. So $f_{s {\rm max}} = \mu_s m g$.

Now take the bottom block and draw the FBD. In the vertical direction, we have $Mg$ down, $N_2$ up, and also $N_1$ down (force on $M$ from $m$ is equal and opposite to force $N_1$ on $m$ from $M$)-- but it turns out this particular equation isn't needed. In the horizontal direction, there's $F$ to the right and $f_s$ to the left-- remember by the third law, frictional force due to $M$ on $m$ is equal and opposite to frictional force due to $m$ on $M$. So, just at slipping, in the horizontal direction we write Newton's second law as $F -f_{s {\rm max}} = m a$, where $a$ is the horizontal acceleration. Although we can substitute for $f_{s {\rm max}}$, we still don't know $a$.

But we can still consider one more system: the two blocks together (at the time they are moving together and the top block is just about to slip) can be considered a system. The only horizontal force on the system is $F$. (We don't consider friction, because it's an internal force to the two-block system.) The total mass is $m+M$. So we can write $F= (m+M)a$ for the second law in the horizontal direction. This equation together with the one above can be solved for $F$.

The key to this problem to carefully consider what the ``system'' is, and only the forces on the system, for each case.