# Understanding static friction in this problem 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?

• 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? Aug 12, 2021 at 0:21
• Why would A always move? Isn't it possible that C is so heavy that it prevents A from moving? Aug 12, 2021 at 0:28
• No. Even if C is really heavy A will still move. Aug 12, 2021 at 0:30
• "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? Aug 12, 2021 at 1:07
• Friction acting on A won't act in the direction of Tension. Aug 12, 2021 at 2:17

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.

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

so

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

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

• The problem asks about A moving, not C moving. Aug 12, 2021 at 0:22
• 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. Aug 12, 2021 at 0:23