# In a system of stacked blocks with friction, how can the top block have no acceleration?

Suppose I have a block of mass $$m$$ on top of a block of mass $$M$$. Friction exists between them. I apply a force $$F$$ on the top block. The equations of motion are thus $$F-f_{frictional}=ma$$

$$f_{frictional}=Ma$$Consider the case when $$F=f_{frictional}$$. Then acceleration of the top block is 0, however the bottom block has an acceleration $$a=\frac {F} {M}$$. Doesn't it seem unlikely? What am I missing? P.S. No friction between bottom block and surface.

• Perhaps this will help Why is the tension on both sides of an Atwood machine identical?. Commented Mar 23, 2021 at 10:26
• Is there a friction force between the bottom block and the surface it contacts? That's an important detail that you haven't mentioned! Commented Mar 23, 2021 at 12:02
• Related post by OP: physics.stackexchange.com/q/622939/2451 Commented Mar 24, 2021 at 13:40
• @BillN NO friction!
– user290607
Commented Mar 24, 2021 at 13:44
• @Feynstein Then edit your question to include that important setup info! Commented Mar 24, 2021 at 14:11

If $$F=f_{friction}$$ then, as you say, we must have $$a=0$$ and so from $$f_{friction}=Ma$$ we can conclude that $$F=f_{friction}=0$$ i.e. there is no force acting on the upper block.

If, however, $$F>0$$ then we must have $$F>f_{friction}$$, and we need to find values of $$f_{friction}$$ and $$a$$ that satisfy the simultaneous equations

$$F-f_{friction} = ma \\ f_{friction}=Ma$$

These equations are both satisfied when

$$\displaystyle a = \frac {F}{M+m} \\ \displaystyle f_{friction} = F \frac {M}{M+m}$$

So, as expected, we have $$a>0$$ and

$$\displaystyle F = f_{friction} \frac{M+m}{M} = f_{friction} \left( 1+ \frac{m}{M} \right) > f_{friction}$$

• Why do we need to have $F>f_{friction}$?
– user290607
Commented Mar 24, 2021 at 12:03
• cannot $F=f_{friction}$?
– user290607
Commented Mar 24, 2021 at 12:07
• @Feynstein Remember that $F$ and $f_{friction}$ are acting on the upper block in opposite directions. If $F < f_{friction}$ then the upper block would accelerate in the direction of $f_{friction}$ i.e. in the opposite direction to $F$. So applying a force $F$ to the upper block would cause it to accelerate in the opposite direction to $F$ - this is unrealistic. And $F=f_{friction}$ only if $a=0$ in which case $F=0$. Commented Mar 24, 2021 at 12:08
• I understand the case $F<f_{frictional}$. But why does $F=f_{fric}$ imply $F=0$
– user290607
Commented Mar 24, 2021 at 12:12
• Cannot they have the same NONZERO value?
– user290607
Commented Mar 24, 2021 at 12:12

Consider the case when $$F=f_{frictional}$$. Then acceleration of the top block is 0, however the bottom block has an acceleration $$a=\frac {F} {M}$$. Doesn't it seem unlikely? What am I missing?

What you are missing is the applied force $$F$$ cannot equal the friction force. When approaching problems like this it is essential to draw a free body diagram (FBD). See the FBD below. It is based on your subsequent clarification that there is no friction between the bottom block and the surface underneath.

As long as the maximum possible static friction force, $$\mu_{s}mg$$, between the blocks is not exceeded the two blocks will move together. Ignoring all the vertical forces which sum to zero, there are two forces acting on the top block, the applied external force $$F$$ and the static friction force $$f_f$$ that the lower block applies to the top. There is only one external force acting on the bottom block, the friction force applied to it by the top block which is equal and opposite to the friction force applied to the top block by the bottom block.

The acceleration of the top block is

$$a_{m}=\frac {F-f_{f}}{m}$$

The acceleration of the bottom block is

$$a_{M}=\frac {f_f}{M}$$

As long as $$F$$ doesn't exceed the maximum static friction force between the blocks, the top and bottom blocks move together and therefore

$$a_{m}=a_{M}$$

Equating the two accelerations and doing some math gives us

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

The two forces $$F$$ and $$f_f$$ are therefore not equal.

Hope this helps.

• Actually gandall's answer answers my question. Anyways, +1 for the effort.
– user290607
Commented Mar 24, 2021 at 15:13

When $$F=f_\mathrm{friction}$$ or $$F, the friction cannot be overcome, i.e., the top block will not move (no acceleration from rest). But, the applied force $$F$$ will be "transferred" into the bottom block, which then experiences an acceleration $$a=F/(M+m)$$, since both masses act as one in this case. As an example, consider sitting on a bobsled with a "grippy" rubber layer on top, which in turn rests on an icy surface (so the friction between the ground and the sled is basically 0). If your friend pushes you, he will also move the sled underneath you by transferring the force onto both the sled and you. But in absence of a grippy rubber layer and instead a slippery bare metal surface, he would just push you off the sled (of course, there is still some friction in real life, consider it just for illustration purposes).

• I initially thought exactly like this, but the equations does not seem to say so. Are they correct for $F <f_frictional$?
– user290607
Commented Mar 23, 2021 at 10:37
• Ok I got it. The equation says it, it was me who didn't see it.
– user290607
Commented Mar 23, 2021 at 10:46
• If $f_{friction}$ means the actual force of friction between the blocks (rather than the maximum value of static friction) then $F$ can never be less than $f_{friction}$, otherwise the upper block would move in the opposite direction to $F$ ! And the only scenario when $F=f_{friction}$ is when $F=0$. Commented Mar 23, 2021 at 11:29
• If $F<f_\text{friction}$ then the top block accelerates in the opposite direction as $F$ Commented Mar 24, 2021 at 13:51
• @BioPhysicist What about the case when $F=f_{fric}$ and $F\neq0$?
– user290607
Commented Mar 24, 2021 at 13:59