Consider the following example:

A mass ($m_1$) is stacked on top of another mass ($m_2$) resting on a frictionless table in a vacuum. There is a nonzero coefficient of static friction ($\mu_s$) between the masses. What is the maximum force $F$ that can be applied to the top mass such that no slipping occurs?

diagram of masses

Clearly, the normal force is $F_n = m_1 g$ and so the friction will resist a force up to $F_f = \mu_s m_1 g$ in magnitude. As far as I know, this is correct.

My (incorrect) intuition says that we should stop here, and $F_\text{max} = \mu_s m_1 g$ is the maximum force before $m_1$ starts to slip. However my textbook (and this answer) says that we should continue, and solve for the acceleration of the two masses: $$ \begin{align*} F_{\text{net, }1} &= F_\text{max} - F_f & F_{\text{net, }2} &= F_f \\\\ a_1 &= \frac{F_\text{max} - F_f}{m_1} & a_2 &= \frac{F_f}{m_2} \end{align*} $$ Now, since we want $m_1$ and $m_2$ to move as a single unit, these accelerations should match: $$ \begin{align*} a_1 &= a_2 \\\\ \frac{F_\text{max} - F_f}{m_1} &= \frac{F_f}{m_2} \\\\ F_\text{max} &= \frac{F_fm_1}{m_2} + F_f \\\\ F_\text{max} &= \boxed{\frac{\mu_s m_1^2 g}{m_2} + \mu_s m_1 g} \end{align*} $$

This result seems to imply that we can apply a force much larger than the friction (especially if $m_2$ is small) without causing any slipping. Is that true, and why is it possible, conceptually? I understand the algebra behind the calculation, but I'm failing to wrap my head around the reality of it; it seems impossible that we can apply an arbitrarily large force by making $m_2$ sufficiently small, even though the friction is constant.

  • 1
    $\begingroup$ Intuitively, it's a bit like pushing a box across a heavy table and seeing that the table does not slide with the box - even very little force will cause relative movement. But if you put a sheet of paper under the box, it's the box+paper may slide around together with no relative movement - you'd really have to jerk the box suddenly to accelerate it away from the paper. $\endgroup$ Commented Mar 21 at 13:56

3 Answers 3


To quickly review, friction arises because when two objects are being driven to slide past each other, any sliding generally incurs an energy penalty (because of dissipative interactions at the surfaces—microscale protrusions need to irreversibly deform during rubbing, for example). For sliding to proceed spontaneously, the driving force needs to be large enough that the input force–distance work pays for that energy penalty.

We often consider one object to be fixed in place, or sufficiently massive that it serves as an intuitive frame of reference. That object, acting as a base, never moves in those simple cases.

But what if that object—here, $m_2$—weighs almost nothing?

Any initial frictional force easily accelerates it, and now the driving force for sliding has disappeared, as both objects are moving in concert.

The only hope one has to detach from $m_2$ is essentially to jerk away so quickly that its own slight mass generates an inertial force comparable to the frictional threshold of sliding. In other words, we can't rely on the base staying fixed; we have to consider its mass as a relevant parameter—and we might first guess a simple scaling as $\sim 1/m_2$ from the above reasoning. Combination of the only other known parameters gives $\sim \mu_s m_1^2g/m_2$ as a term with dimensionality of force, and more rigorous analysis confirms the presence of this term.

It can be insightful to envision some practical cases of a very light $m_2$. Consider, say, trying to move a sealed ream of paper on a slick surface by brushing one's fingers on the top, in contrast to moving a single sheet of paper on the same surface.


The forces of static friction only need to accelerate $m_2$, the lower block. $F_{max}$ is the force to accelerate both blocks at the same rate. Thus it can be higher than the static friction -- its the static friction plus the force needed to accelerate the top block.

It may be helpful to rearrange the equation. You have $F_{max}=\frac{\mu_sm_1^2g}{m_2} + \mu_sm_1g$. If you factor out the $\mu_sm_1g$, you get $F_{max}=\mu_sm_1g(\frac{m_1}{m_2} + 1)$ This makes it a bit more clear that you've got the force you can apply to $m_2$ through $m_1$, $\mu_sm_1g\cdot 1$ and the force required to accelerate $m_1$ to match, $\mu_sm_1g\cdot\frac{m_1}{m_2}$


If you accept that the applied force is parallel to the table, the other answers are correct, but in the wording provided, there is no indication of that and the force is not explicitly marked parallel. While the diagram does show the applied force parallel to the table, I would do the algebra for an applied force at angle $\theta$ and then show that it reduces to the aforementioned answer when $\theta = 0$. The answer to the question, as worded, is "That's a a materials question." because $F(\theta = -\frac{\pi}{2}) = \infty$.

Physics teachers love pedantry.

  • $\begingroup$ Good point, I should have mentioned that the applied force is parallel to the table. Although, while we're being pedantic, I did note that the normal force does not depend on the applied force ($F_n = m_1 g$), which does imply that. $\endgroup$ Commented Mar 22 at 9:29
  • $\begingroup$ As I read it, that was your conclusion rather than part of the problem statement. $\endgroup$
    – user121330
    Commented Mar 22 at 22:42

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.