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The following problem is taken from an MIT homework assignment:

Hold a yardstick horizontally on your index fingers and slide your fingers together smoothly. The stick slides first on one finger, then on the other and it keeps alternating. This was demonstrated in lectures. Try it for yourself, it’s great fun! Why does this happen?

This question has already been asked elsewhere, but I have a question specifically about the MIT solution:

Let $N_1$ and $F_1$ be the normal and frictional forces for one finger, and let $N_2$ and $F_2$ be the normal and frictional forces for the other finger. The key to understanding this problem is to realize that the fraction of weight supported by each finger can be different. Clearly, the finger closest to the center of the yardstick will bear a larger fraction of the weight and hence will exert a larger normal force on the yardstick.

Imagine starting each finger under a separate end of the yardstick. Initially, each finger shares the weight equally, but as you attempt to move your fingers one of them, say finger 1, starts to slide. (To avoid sliding you would have to start with your fingers exactly the same distance from each end and move with exactly the same speed. Clearly human fingers are not capable of this. And the yardstick itself is too irregular for that precision). Immediately after finger 1 slides, both fingers will share the same weight equally ($N_1 = N_2$) but because the kinetic coefficient of friction is less than the static coefficient the friction on finger 2 is greater than the friction on finger 1. As finger 1 continues to slide in, it will bear more of the weight of the yardstick until $N_1$ is large enough that $F_1 = F_2$. As finger 1 moves in just a bit more, finger 2 will no longer be able to sustain the frictional force from 1, and hence finger 2 will move and finger 1 will stop. The whole procedure will begin again.

I do not understand this last part that I made bold. How does the frictional force on 1 relate in any way to what 2 is experiencing? What do they mean by "it cannot sustain"?

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  • $\begingroup$ Related: physics.stackexchange.com/questions/282160/… $\endgroup$ – dmckee Jul 13 at 14:05
  • $\begingroup$ @dmckee Thank you for your answer. I did come across that post, but I was wondering if I could get a clarification on this explanation $\endgroup$ – user10796158 Jul 13 at 14:18
  • $\begingroup$ Usually when people comment related post links they are not for the OP's benefit but more for future readers' benefit. If @dmckee thought that answered your question they would have marked it as a possible duplicate instead. $\endgroup$ – Aaron Stevens Jul 13 at 16:13
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As one finger slides it exerts a kinetic frictional force in the direction of the other, stationary finger. Since the rod is at rest with respect to the stationary finger, it must be that the magnitude of the static friction force between the rod and the stationary finger is equal to and opposite to the kinetic friction force between the rod and the moving finger.

You have to keep in mind that static friction has an upper limit to what it can "sustain" until it fails. This is what the solution means. The weight supported by the stationary finger decreases, and hence that maximum static friction force becomes less sand less until it finally can't match the kinetic friction force. Then sliding occurs on the other finger. This is the key. The maximum static friction force decreases as the sliding finger moves inward.

One small thing is that once one finger moves the other finger doesn't technically need to stop at the same time. It depends on the magnitude of the force you are applying to your fingers (which it seems like it's assumed to be constant and equal for both fingers).

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  • $\begingroup$ Thank you for your answer. Wait but that seems to contradict what the solution is saying though. You say "it must be that the magnitude of the static friction force between the rod and the stationary finger is equal to and opposite to the kinetic friction force between the rod and the moving finger." But the answer says "Immediately after finger 1 slides, both fingers will share the same weight equally (N1=N2) but because the kinetic coefficient of friction is less than the static coefficient the friction on finger 2 is greater than the friction on finger 1." Could you clarify? $\endgroup$ – user10796158 Jul 13 at 16:52
  • $\begingroup$ @user10796158 Coefficients of friction and friction forces are different things. Keep in mind for static friction $F\neq\mu_s N$ in general. $\mu_sN$ is just the maximum value it can have. Think of a book sitting on a table. The static friction force acting on the book is $0$, even though $\mu_sN\neq0$ $\endgroup$ – Aaron Stevens Jul 13 at 17:24

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