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?
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.
Best Answer
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.