enter image description here Suppose we have the following system, in which we know the masses of A and B, and we know the static friction coefficient between A and C(there's no friction between the table and A). What's the minimum mass of C that prevents A from moving? So, just writing the free body diagrams and writing the equations I get that $m_c$ must be greater than $\dfrac{m_b}{\mu_e}$, which is okay for me. Now, the problem comes when I analyze what happens to C. Since there's no other horizontal force on C besides the static friction, I assume C must move to the left. However, this goes against my intuition/life experience, in that it's impossible for C to move without A moving. Does C really move? Why/Why not?

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    $\begingroup$ Maybe I am missing something. Block A will always move no matter what the mass of C is. Are you sure you have the problem correct? $\endgroup$ Aug 12, 2021 at 0:21
  • $\begingroup$ Why would A always move? Isn't it possible that C is so heavy that it prevents A from moving? $\endgroup$
    – Diego
    Aug 12, 2021 at 0:28
  • $\begingroup$ No. Even if C is really heavy A will still move. $\endgroup$ Aug 12, 2021 at 0:30
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    $\begingroup$ "Since there's no horizontal force on C besides the static friction, I assume C must move to the left" so you have already shown that if there is any static friction force, C will move, which will allow A to move. Then how can C ever be heavy enough to prevent A from moving? $\endgroup$
    – Alwin
    Aug 12, 2021 at 1:07
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    $\begingroup$ Friction acting on A won't act in the direction of Tension. $\endgroup$ Aug 12, 2021 at 2:17

1 Answer 1


Imagine there is no C. Then masses A and B may accelerate, so that Bg = (A+B)a, where a is the acceleration of A (and hence the acceleration of B, since they are tied by a string). B has total force (Bg-T) and A has force T, so $(Bg-T) = Ba$ and $T = Aa$, leading to $Bg = (A+B)a$.

Now imagine there is C and they all move together. By similar logic, $Bg = (A+B+C)a$. As long as there is no friction between the table and A, what could possibly prevent the combined A+C from moving?

Now imagine there is C and somehow A and B slip away from C. This happens if the friction necessary to accelerate C to keep up with A and B is larger than the static friction. In that case, the A-C interface will no longer be static, and A will pull away from C.

$Cg\mu < Ca = \frac{CBg}{A+B+C}$


$\mu < \frac{B}{A+B+C}$

When B is too large, it's like doing the tablecloth (A) yank/pull trick and leaving the plates on top (C) behind.

  • $\begingroup$ The problem asks about A moving, not C moving. $\endgroup$ Aug 12, 2021 at 0:22
  • $\begingroup$ I agree with you. I chose to address the system and the title "understanding static friction in this problem" rather than the specifics about the problem, which you have also identified may be fallible. $\endgroup$
    – Alwin
    Aug 12, 2021 at 0:23

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