[Physics] Torque due to friction on a series of disks

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Recently I was looking at two situations involving friction and torque.

The first situation seemed pretty straightforward at first. A disk of mass $m$ and radius $r$, with a coefficient of static friction $\mu _s$ with the ground, is given a force $F$ originating at its center. I have drawn a free body diagram below.

$\hskip2in$ Free body diagram for the situation

For this situation, the applied force is exactly equal to the frictional force, that is to the coefficient of static friction is high enough, and that the applied force is low enough, such that $\mu_sN = F$, we can pretty trivially show that $\Sigma_x F=0$ and $\Sigma_y F=0$. However , when solving for the net torque, we arrive at $\Sigma \tau = \mu _srF$, which means a non-zero net torque is applied to the disk.

This paradoxically, at least to me, means that the ball is spinning in place without actually moving. How could this be?

The second situation is very similar to the first, except that a second disk, with the same mass and radius, has been placed directly next to the first disk. The coefficient of static friction between the two disks is $\mu_{sb}$. I have drawn another free body diagram below.

$\hskip2in$ Free body diagram for the next situation

The situation is very similar to the first, however I hypothesize that there is a torque ($f$) due to friction between the first and second disks. I think that the direction of the force points downward, as when the the first disk tries to rotate due to the friction with the table, the second disk resists this change, thus causing a force opposite the motion of the spin.

However, I am at a complete loss as to how to calculate this force new force.

Any help at all would be appreciated, thank you!

Best Answer

Assuming that the coefficient of static friction is high enough such that $μ_sN=F...$, we can pretty trivially show that $Σ_xF=0$ and $Σ_yF=0$.

Actually, I'm not sure you can. $\mu _sN$ gives you the maximum possible force of static friction, not necessarily the actual force. To find the actual force you'd need to relate the acceleration of the disk with the angular acceleration. Only a particular frictional force will provide the necessary linear and angular acceleration.

Your second force will be difficult to calculate. You don't know the normal force, and the second disk must either slip against the first or slip against the table.

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