forcesfree-body-diagramfrictionnewtonian-mechanics
In this situation:
Enough tension is applied towards the right on $m_2$ so that $m_1$ starts sliding backwards.
I am confused about what the free body diagram for $m_1$ looks like.
Here is my attempt:
The net force in the $x$ direction on $m_1$ is negative, right? Otherwise the box wouldn't be sliding backwards.
I think that this is a very interesting problem which is conceptually difficult.
You do not need to worry about the FBD for the truck. The box should be your main focus.
Diagram 1 is the FBD as long as the box does not slide relative to the truck.
With the aid of diagram 1 work out the maximum acceleration $a$ the box can have as a result of the static frictional force $\mu_sN_{bt}$ acting on it.
Hopefully this will lead you swiftly onto phase two of the problem and the FBD diagram 2.
Now this is where you might think that the kinetic friction direction is incorrect because it is actually going to make the box move faster as you might have heard the statement "friction opposes motion"? In this case the reason for the kinetic friction acting in the direction shown is that the kinetic frictional force is trying to reduce the relative velocity between the box and the truck. So it is relative motion that kinetic friction opposes and sometimes, as in this case, it has to make something go faster in order to try an achieve this.
Your conceptual problems are possibly not over yet as you now have the difficult task of having a truck accelerating to the right and a box accelerating to the right and you have to figure out how long it will take the box to move a given distance on the truck and fall off.
You might a good way of doing this is to draw, using the same axes, graphs of velocity against time for the truck and the box and remember that the area under a $v-t$ graph is the displacement (distance + direction).
If the top block is sliding to the left relative to the bottom block, or if it would slide to the left if there was zero friction, then the friction on the top block due to its interaction with the bottom block would be to the right, as you have shown it. Newton's third law says that there must also be a force on the bottom block to the left. This also agrees with the frictional interaction which actually produces that force. diagram 2 is correct, conditional on the beginning "if"s in my first sentence.
The reason 3 is wrong is because forces always come in pairs, and you don't double the pairs. Force on top because of bottom is the partner of force on bottom because of top. They are a simultaneous pair. Calling them action-reaction is misleading at best, because it sounds like one is causing the other. That is a bad conclusion.
Yes, the direction of friction is determined by going opposite the relative sliding of two objects across each other.
Best Answer
This is a very good question that addresses a common misconception.
Short answer: The box $m_1$ does NOT slide backward. The friction force in the $+x$ direction causes the box 1 to slide forward, but not fast enough to keep up with the larger box 2. There are no backward force in the x-direction - there is only the forward friction force in the +x direction.
Be careful with your free body diagram. You should only show the forces applied directly on box 1. The tension force is applied on a different body (e.g. Box 2), so it plays no role in describing the motion of Box 1.