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???


2 Answers 2



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 $F_2$.

  • $\begingroup$ Yes, that's what it looks like. I understand everything you wrote, except I don't understand why we take the force F1 to be greater than the frictional force when I've been told in textbooks/classes that the amount of frictional force is equivalent to the force that is applied. $\endgroup$
    – user50672
    Commented Jun 18, 2014 at 8:04
  • $\begingroup$ @user50672: I've edited my answer to respond to your comment $\endgroup$ Commented Jun 18, 2014 at 8:21
  • $\begingroup$ I understood what you wrote but it's not actually the answer to my question. I actually figured it out, it's because the definition "frictional force is equivalent to applied force" is incorrect. Frictional force is actually equal to Fapplied/((m1/m2) + 1), where m1 is the top mass and m2 is the bottom mass. This would explain why we are usually convinced into thinking that they are equivalent, because m2 is usually much larger than m1 when we are considering the traditional case of a smaller object on a surface. $\endgroup$
    – user50672
    Commented Jun 18, 2014 at 9:24
  • $\begingroup$ Your answer was comprehensive in its application of using Newton's 2nd law but didn't explain why the frictional force differed from the applied. $\endgroup$
    – user50672
    Commented Jun 18, 2014 at 9:28
  • $\begingroup$ @user50672: that's what my edit did. $F_2$ is the frictional force and I show why it's less than $F_1$. $\endgroup$ Commented Jun 18, 2014 at 9:39

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.

  • $\begingroup$ Can you expound upon the "relative motion"? I'm confused because when I do a FBD for the top block, there is no net horizontal force, yet it is clearly accelerating. I completely understand the force on the lower block due to the friction and the net force that is on the entire system. $\endgroup$
    – user50672
    Commented Jun 18, 2014 at 6:31
  • $\begingroup$ The top-block does not skid on the lower one. You can take two blocks, keep the bottom at rest by holding it tightly on the surface, and move the top over the lower. That is relative-motion between the blocks. But now, you keep the blocks attached with strong adhesive, and now move the system over a smooth surface, like a table. There is no relative motion between the blocks. $\endgroup$
    – abstract
    Commented Jun 18, 2014 at 6:34
  • $\begingroup$ I still don't quite understand what force is causing the top block to move. It's clearly moving relative to the surface, but why?!? $\endgroup$
    – user50672
    Commented Jun 18, 2014 at 6:42

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