Block on a block problem, with friction

Question

Consider two blocks, one on top of the other on a frictionless table, with masses $m_1$ and $m_2$ respectively. There is appreciable friction between the blocks, with coefficients $\mu_s$ and $\mu_k$ for static and kinetic respectively. I'm considering the fairly routine problem of determining the maximum horizontal force $F$ (say, to the right) that can be applied to the top block so that the two blocks accelerate together.

The problem is not hard to solve symbolically. If the two blocks move together, their accelerations are the same, and the top block doesn't move with respect to the bottom block, so only static friction is in play. In a standard coordinate system (with $x$ oriented to the right), the sum of horizontal forces for the top block is

$$F-F_{sf}=m_1a$$

and for the bottom block

$$F_{sf}=m_2a$$

where $F_{sf}$ is the force of static friction. Solving for $a$ in these two expressions, and then equating them, gives

$$F=\frac{(m_1+m_2)F_{sf}}{m_2}$$

The maximum such force will therefore be achieved when $F_{sf}$ is maxed out at $\mu_s m_1g$, so

$$F_{max}=\frac{m_1}{m_2}\mu_s(m_1+m_2)g$$

I understand this solution, but conceptually I don't have a response to the following nagging question: $F_{max}$ is clearly larger than the max static friction force $\mu_sm_1g$ (because $\frac{m_1+m_2}{m_2}>1$), so why doesn't the application of a force of magnitude $F_{max}$ to the top block cause kinetic friction to take over? This line of reasoning would suggest that applying a force $F$ of magnitude greater than $\mu_sm_1g$ would cause the top block to start moving with respect to the bottom block (in which case the blocks no longer accelerate together, as in the above solution). I'm at a loss, conceptually, to say what's wrong here. I suspect it has something to do with being careful about reference frames, but a clear explanation would be much appreciated.

Answer 1

The key is that the bottom block is actually moving and is not held fixed like the ground typically is (here I am asusming $F$ is applied to the top block).

Elaborating on my comment: Your acceleration "$a$" is with respect to the ground. The equation $F-F_{sf}=m_1a$ shows that the reason you can accelerate is because $F_{max}>F_{sf}$, and this is accelerating with respect to ground, not with respect to the bottom block.

In particular, if you held the bottom block in place (treating it as the ground), then yes kinetic friction would kick in, but now your equations would be different (because $a_{bottom}=0$ and $a_{top}>0$).

Answer 2

You can also think of your problem in a slightly different way, which may (or may not) help you see things in a different light.

1. You apply your force $F$, and it accelerates the top block at a rate $a_1$.
2. You can consider the acceleration as an inertial force of $-m_1\cdot a_1$, so there is a surplus force $F-m_1\cdot a_1$ left over.
3. This surplus force will be used to accelerate the second block at a rate $a_2$.
4. Since the table is frictionless, there can be no surplus, and all the force must be consumed in accelerating the block, so $F-m_1\cdot a_1-m_2\cdot a_2 = 0$.

So far everything holds no matter what the nature of the interfacial contact between the blocks, which we must introduce to solve for $a_1$ and $a_2$:

1. If both blocks move together, then $a_1 = a_2 = a$, and the inequality for static friction applies, $F - m_1\cdot a \leq m_1 g \mu_s$. This, together with the equation of (4) above gives you an acceleration of $a=F/(m_1+m_2)$ for forces $F \leq \frac{m_1}{m_2} (m_1+m_2) g \mu_s$.
2. If $a_1 \neq a_2$ then we have relative motion, and then we know what the interfacial force is exactly, and we can calculate $a_1 = F / m_1 - g \mu_k$ and $a_2 = g \mu_k m_1 /m_2$.

But, to answer your question, why doesn't a force larger than the maximum static friction applied to the top block cause slipping, and thus dynamic friction? Because part of the force is used up in accelerating the top block, and it is only the surplus that is available to try to overcome friction between the blocks.

Answer 3

The reason you're finding this non-intuitive is because you're assuming that the applied force on the top block will cause the bottom block to experience the same force due to friction which is wrong.

For example, consider two blocks A and B , sitting side by side on a smooth table. If you apply a force F on A, then one may incorrectly say that the force that B experiences is also F. But that would be wrong. It's actually less than F.

The same is true for your example. The frictional force experienced by the bottom block is less than the actual force applied on the top block. Hence, for the bottom block to experience the maximum static frictional force, one has to apply a force greater than this. Note that this 'frictional force' is due to the top block itself and hence your example is analogous to the above example I gave.

But why does the bottom block experience lesser force in the first place? This is because the 'rest' of the force (or in more correct terms: the rest of the momentum) is used up to move the top block itself. If the top block is massive, then you will need to exert an even larger force to get the top block to accelerate at the same rate as that of the bottom block. Most of the imparted momentum (force) here is given to the top block while the rest of it is given to the bottom block. If the top block is light, then very little of the momentum imparted (force) is used up to move the top block and the rest of it is given to the bottom block. Hence in this case, the required force will be close to the maximum static frictional force.

You can see this in your expression: $F=\frac{(m_1+m_2)F_{sf}}{m_2}$

as $m_1 \to 0$, the force required approaches the frictional force.

However, in this expression: $F=\frac{m_1}{m_2}\mu_s(m_1+m_2)g$

as $m_1 \to 0$, the force also approaches $0$. This is because this expression incorporates the fact that the maximum frictional force decreases as the mass of the top block decreases (lighter the top block, less the normal force between the two blocks). Hence as the top block's mass approaches $0$, the maximum force you can apply also approaches $0$.

Answer 4

There is no kinetic friction. The system $F = (m1+m2)a$ The max force cannot be more than the static friction force for top block because then top block will start to slip of bottom block. Your condition is that top block should not slip. This is only possible if the fmax does not exceed the static friction of top block with bottom block.

Therefore, the top block is in equilibrium then $Fmax=Fstatic$ friction The bottom block experiences the same force $Fstatic$ friction due to newtons 3rd law. But there is no opposing friction force on bottom block. The block moves to the direction of force with $(m1+m2)a=Fmax=Fstatic$ friction.

Answer 5

I would approach the problem a bit differently. Firstly, there is no relative motion between the two blocks, and hence the force at play is the static force only. i.e. $F_{sf}$. This force accelerates the bottom mass, $m_2$. So, $F_{sf}=m_2.a$..... eq.1 Now, consider the top mass m1 to be in an accelerated frame of reference, which has an acceleration $a$. The forces acting on $m_1$ are the following 1. $F_{Max}$ - i.e. the pull force to the right (say) 2. $F_{sf}$ acting to the left 3. A pseudo force equal to m1.a acting to the left, due to the mass $m_1$ being in an accelerated frame. Since the body is at rest in this reference frame, these forces neutralize each other. Thus, $F_{max} = F_{sf} + m_1.a $ .....eq 2 equations 1 and 2 are the equations that you used to solve your problem

Answer 6

The force $F_{max}=\frac{m_1}{m_2}\mu_s(m_1+m_2)g$ accelerates both blocks with the bottom block taking up $\mu_{\rm s}m_1g$ of the force and the rest of the force $\frac{m_1}{m_2}\mu_s(m_1+m_2)g-\mu_{\rm s}m_1g = \mu_{\rm s}\frac{m_1^2}{m_2}g$ is used to accelerate the top block.

As a check, for the bottom block $m_2a_2 = \mu_{\rm s}m_1g\Rightarrow a_2 = \mu_{\rm s} \frac {m_1}{m_2}g$
and for the top block $m_1a_1 = \mu_{\rm s}\frac{m_1^2}{m_2}g \Rightarrow a_1 = \mu_{\rm s} \frac {m_1}{m_2}g$.

Both blocks have the same acceleration.