Direction of Friction on two stacked boxes

Question

I have box 2 stacked on box 1. Box 1 is on a frictionless table and accelerated by a force. However, there is friction between box 1 and box 2.

Now, I was told that this is the diagram for that problem:

I thought that since m1 and m2 are moving to the right, then the frictional force between box 1 and 2 will be to the left, opposite to the direction of motion. Can someone confirm what is the actual direction of friction for the frictional force between box 1 and box 2?

Also, is the action-reaction pair of forces between box 1 and box 2 the normal force of box 1 and the negative normal force of box 1?

It is the same issue with another problem I saw online:

In one of the free-body-diagrams, the friction seems to be going to the right (which makes sense since the movement is to the left), but in the lower free-body diagram the friction appears to be going to the right (in the same direction of movement). How is this possible?

Answer 1

Consider Newton's Third Law; commonly stated "every action has an equal and opposite reaction".

You should expect the surfaces to have friction in the opposite direction. Friction doesn't help one of them move, it resists the relative movement of both, causing forces in opposite directions.

You can sense this yourself by taking two rough surfaces and trying to move them one with each hand. The friction will resist movement with both hands, not just one.

Answer 2

Fictional forces do one of two things.
Static friction tries to prevent relative movement between two surfaces.
Dynamic friction tries to reduce the relative movement between two surfaces.

In the diagrams below $f_{12}$ is the frictional force on box $1$ due to box $2$ and $f_{21}$ is the frictional force on box $2$ due to box $1$.
Newton's third law tells us that these two frictional forces are equal in magnitude and opposite in direction.

Force $F$ accelerates both boxes to the right.

Or you could say frictional force $f_{12}$ accelerates box $1$ to the right. and the net force $F-f_{21}$ on box $2$ accelerates box $2$ to the right the frictional forces having adjusted themselves so that there is no relative movement between the two boxes.

As an aside suppose that there is relative movement between the two boxes and box $2$ is moving faster to the right than box $1$, then $f_{21}$, the dynamic friction force on box $2$, you can think of as trying to reduce the rate of increase of speed of box $2$ thus trying to make the speed of box $2$ closer to the speed of box $1$.