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There is a block $A$ on the table; the friction between the table and the block is not taken into account. A cube $B$ lies on the block, the coefficient of friction between the cube and the block is $u$. If we pull $B$ with a force $\vec F_\text{ext}$ (external) such that $B$ does not move relative to $A$, then $\vec F_\text{friction} = -\vec F_\text{ext}$, as the force of static friction.
In this case, $\vec F_\text{friction}$ will act on body $A$ according to Newton's third law. Body $A$ will acquire some acceleration, which means that body $B$ will move relative to the Earth with the same acceleration as body $A$.
However, if we write Newton's second law for body $B$, we get: $m \vec a =\vec F_\text{ext} -\vec F_\text{friction} = 0$, but body $B$ is moving with acceleration, which means that $m \vec a$ is not equal to $0$.

This results in a very strange situation. Can you explain this absurd situation?

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    $\begingroup$ “then F (friction) = -F (external), as the force of static friction.” Why? $\endgroup$ Commented Oct 26 at 13:40

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Your problem is in this assumption:

If we pull B with a force F (external) such that B does not move relative to A, then F (friction) = -F (external), as the force of static friction.

The key error here is that the force of static friction is not always equal to the external applied force. Rather, it will be equal to whatever value it needs to be to keep the contact surfaces at rest relative to each other. The only constraint is that its magnitude has to be less than some critical value (typically taken to be proportional to the normal force.)

For a block sitting on a horizontal table, with only friction and one external force acting on it, we know that we must have $a = 0$; and therefore we must have $F_\text{fr} = - F_\text{ext}$. But if the block is accelerating (as it is in this case), then $a \neq 0$ and $F_\text{fr} \neq - F_\text{ext}$.

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  • $\begingroup$ You should be able to confirm this experimentally if you have a bunjee cord and a sled and some snow, or a big rubber band and a toy car, etc! $\endgroup$ Commented Oct 26 at 14:45
  • $\begingroup$ It turns out that in the reference frame associated with bar (A), ma = F (external) - F (friction) - F (inertial) = 0. That is, F (external) > F (friction). However, F (inertial) depends on acceleration (i.e., on F (friction)), and F (friction) depends on F (inertial). This results in a circular dependency. Have I understood you correctly? $\endgroup$ Commented Oct 26 at 17:03
  • $\begingroup$ @Marmajuck: Your analysis in a non-inertial frame is also valid. It's not a circular dependency, though, since $F_\text{inertial}$ depends on the acceleration of the frame, not on the acceleration of the block. You've picked a frame in which these accelerations happen to be the same, which leads to a self-consistent and unique solution for all of the variables; if you picked a different non-inertial frame (or an inertial frame) you'd still get the same result for $F_\text{fric}$. $\endgroup$ Commented Oct 26 at 17:21
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You have great understanding, but you missed one idea: friction is only equal to $F_{\text{ext}}$ as long as $F_{\text{ext}}$ is not > $\mu N$, after that it will cross it in this case $$F_{\text{ext}}-\mu N = ma.$$

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