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Block one is on top of block two, with blow two on a frictionless table. The system of blocks is accelerating to the right due to a force applied strictly to block one. What direction is the frictional force being exerted on the lower block by the upper block, and is it kinetic or static friction. 

                 ___
               _|_1_|_-------->F applied
               |  2  |
               |     |       ------> direction of acceleration of system

|-----------(Frictionless Table)----------------|

I know so far that the frictional force being exerted by block 2 onto block 1 is directed to the right in the direction of the acceleration. But would it be considered as kinetic friction or static? I'm not sure if block 2 would move backwards or not.

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    $\begingroup$ "I know so far that the frictional force being exerted by block 2 onto block 1 is directed to the right"... No, try again. $\endgroup$
    – Bill N
    Commented Jun 21, 2018 at 20:09
  • $\begingroup$ Have you drawn a free body diagram for each of the blocks showing the forces acting on each of them, or do you feel that you have advanced to the point where you no longer have the need to use free body diagrams? $\endgroup$
    – Chet Miller
    Commented Jun 21, 2018 at 21:15

3 Answers 3

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Since the blocks move together as a system, block 2 is stationary wrt block 1. This makes the frictional force between 1 and 2 static. If the force applied is greater than this frictional force, then block 1 would be moving relative to block 2. The fact they are moving relative to one another would make this friction kinetic.

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  • $\begingroup$ Your answer is self-contradictory: " Since the blocks move together ... block 2 is stationary wrt block 1. " "...block 1 would be moving relative to block 2." I think you need to clarify your answer. $\endgroup$
    – Bill N
    Commented Jun 21, 2018 at 20:08
  • $\begingroup$ Are you sure they are self-contradictory? I meant that 1 and 2 moved as one object. so if you were standing on 1, 2 would not be moving. i thought that meant that 2 was stationary relative to 1, Relative to the frictionless plane it is moving. Maybe its semantics. $\endgroup$
    – Natsfan
    Commented Jun 21, 2018 at 21:57
  • $\begingroup$ They can both be moving but at different velocities relative to the table, or they can move at the same velocity. Your first sentence semantically implies they have the same velocity, but that's not a given. Then you say "the fact they are moving relative to one another." That's not a fact, either. Those are two possible situations, not definite facts. That's why I say the answer is self-contradictory. $\endgroup$
    – Bill N
    Commented Jun 22, 2018 at 3:15
  • $\begingroup$ I stated if force is less than the static friction force the blocks move together. Otherwise, the blocks move at different velocities. Does that help or confuse the issue? Thanks for you comments. $\endgroup$
    – Natsfan
    Commented Jun 22, 2018 at 14:25
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As jmh, pointed out, this is a static friction case since there's no relative movement between the blocks. There is a force being applied from the upper block to the lower block in the same direction of the movement.

"not sure if block 2 would move backwards or no" Since there's no relative moment between the blocks that means that block two is not moving backwards.

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As stated by jmh, as long as block $1$ and block $2$ are not moving relative to each other, the friction between them, by definition, is static.

This static friction slows down block $1$ by pulling it back or to the left and accelerates block $2$ by pulling it forward or to the right. So the direction of the static friction, as any force between two bodies, will depend on which of the two bodies we are looking at.

For the block $1$ to accelerate, the applied force $F$ will always be greater than the friction force pulling it back - otherwise, there would be no net force acting on block $1$ and, therefore, no acceleration.

If we want to figure out whether the blocks will move together or block $2$ will fall behind, we can write Newton's second law equations for each block, noting, that, for the blocks to move together, their acceleration has to be the same.

$F-f_{st}=ma$

$f_{st}=Ma$

From here we can find the relationship between force $F$ and friction force $f_{st}$ required for the block to move together:

$F=f_{st}(\frac m M +1)$

If $F$ increases, for the equations to hold, $f_{st}$ has to increase as well, but, at some point, $f_{st}$ will reach its maximum value, $f_{st-max}$, and, from this point on, block $2$ won't be able to keep up and will start slipping back relative to block $1$, while the static friction will be replaced by the kinetic friction.

This will happen, when force $F$ exceeds $f_{st-max}(\frac m M +1)$.

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