Why is the force of Friction less than the force applied on the block in this situation?

Question

This is a seemingly basic mechanics problem but I'm having a dilemma in understanding what happens. You start with two blocks, one on top of the other (the bottom block has a larger mass than the top block). There is friction between the blocks, so they stick together. These two blocks rest on a frictionless surface. If I apply a force F to the top block, what happens?

Drawing the free body diagram, the force F I apply is counter acted by a frictional force f because of the friction between the two blocks. Additionally, this frictional force will be opposed by another frictional force on the bottom block that forces it in the direction I applied the force. Thus, we see that the blocks will move in the direction I have pushed.

HOWEVER, I can't understand why the top block will move forward as well. The force I have applied should be negated by the frictional force, but this is not the case. Clearly, the block accelerates forward despite the friction. Why is this paradox created???

Answer 1

I'm guessing your FBD looked something like this, where $F_1$ is the external force you apply. I'm assuming here that the top and bottom block don't slide relative to each other, so the forces at their junction ($F_2$) are equal and opposite.

The net force on the top block is the force you apply, $F_1$, minus the frictional force the bottom block applies to the top block, $F_2$:

$$ F_{top} = F_1 - F_2 $$

Because $F_1 > F_2$ the net force $F_{top} > 0$ and the top block accelerates.

Response to comment:

If the two blocks don't slide relative to each other then their accelerations must be the same so:

$$ \frac{F_{top}}{m_{top}} = \frac{F_{bottom}}{m_{bottom}} $$

We know that $F_{top} = F_1 - F_2$ and $F_{bottom} = F_2$, so:

$$ \frac{F_1 - F_2}{m_{top}} = \frac{F_2}{m_{bottom}} $$

and a quick rearrangement gives:

$$ F_1 = F_2 \frac{m_{top} + m_{bottom}}{m_{bottom}} $$

and since $m_{top} + m_{bottom} > m_{bottom}$ this means $F_1 > F_2$.

Basically $F_2$ is only accelerating the bottom block while $F_1$ is accelerating both blocks, so $F_1$ has to be greater than $F_2$.

Answer 2

The reason you give does not work merely for the fact that the force F and the friction balance each-other, thus creating no relative motion between the top block, and the bottom one. But, as the surface on which the system rests is friction-less, a net-force F acts on the lower block----- the frictional-force, which acts on the lower-block in the direction, same as that in which, the force F is directed. This direction is just the opposite to the direction of frictional-force, acting on the top block. Thus, there's a net acceleration of the system, in the direction, in which F has been applied. As our system contains both top, and lower blocks, the top-block moves along with the lower block, without slipping on the latter. Thus, the apparent paradox is resolved.

Response to comment:

Your confusion is then why the top block skids forward relative to the bottom. There's two type of friction: static, and kinetic. If the force F you apply exceeds the static friction, the top block will move relative to the bottom, and its acceleration relative to the lower-block is then due to force (F-kinetic frictional force). Thus, its cumulative motion is due to the relative motion, as well as motion of the system. The frictional-force is a self-adjusting force, and when the force F is lower than the static-friction, then there's no relative motion (and hence, no relative acceleration) between the blocks, and its motion is entirely due to system's motion.

In my previous post, I considered that F does not exceed the static-frictional force between the blocks. Hope it is a bit more clear to you now.

P.S., to be precise and to the point:

• The top block skids on the lower only when the force F applied on the top block exceeds the static-frictional force acting between the lower-surface of the top-block, and the upper-surface of the lower-block.

• We know, that kinetic-frictional force between any two surfaces is always smaller than the static-frictional force between them.

• Once the top-block starts skidding on the lower block, the friction that acts between the two surfaces in contact, is the kinetic-frictional force, which is lower than the static-frictional force between them.

• But, the force F applied is greater than the static-frictional force.

• So, force F applied on top-block is greater than the force of friction (kinetic-frictional force)

*Or, "force of friction is less than the force applied on the block in this situation".

Thus proved.

Hope you follow the arguments well.