Torque due to friction on a series of disks 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. 
$\hskip2in$  
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.
$\hskip2in$ 
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!
 A: 
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.
A: 
For this situation, the applied force is exactly equal to the frictional force

Unfortunately this is impossible for the reason you give in your answer. If you were somehow able to apply a force equal to the static friction force$^*$, then you have no net force with a net torque about the center of the disk. Therefore you have sliding between the disk and the ground, which means there is no static friction. We have arrived at a contradiction. Therefore in this scenario the static friction force will never be able to be equal to our applied force. The static friction force will actually be less than the applied force (until the applied force becomes too large and slipping occurs).
If you want to apply a force equal to $\mu_sN$ then that is fine. But then for the disk, to have rolling without slipping, the friction force will be equal to $\frac13\mu_sN$ (Derivation here  showing that for an object with moment of inertia $I=\gamma mr^2$, in order to have rolling without slipping due to applying a force $F$ at the center of the object, the static friction force must be equal to $\frac{\gamma}{1+\gamma}F$ ). 
It turns out that for the disk as long as $F<3\mu_sN$ then you will have rolling without slipping in the direction of the applied force.

$^*$ Quick clarification, the static friction force is not always equal to $\mu_sN$. In fact, this is such a specific case that really you should write $f_s<\mu_sN$. Better yet, you should really never even think about $\mu_sN$ when static friction is involved unless the work you are doing requires you to think about when static friction is right up against this boundary. Essentially you should very rarely ever think that $f_s=\mu_sN$, which it seems like most of the confusion lies here.
A: 
Lets write the equations for a free body diagram and see if we can determined the force $F_r$ 
Left disk
$$-m\,a_1+F-F_c-\mu\,N_1=0\tag 1$$
$$-m\,g-F_r-N_1=0\tag 2$$
$$-m\,r^2\,\alpha_1+\mu\,N_1\,r+F_r\,r=0\tag3$$
right  disk
$$-m\,a_2+F_c-\mu\,N_2=0\tag 4$$
$$m\,g+F_r-N_2=0\tag 5$$
$$-m\,r^2\,\alpha_2+\mu\,N_2\,r-F_r\,r=0\tag 6$$
Constraint Equation
$x1-x2=0\quad \Rightarrow\quad $
$$a_1=a_2\tag 7$$
and
$F_r-\mu_r\,Fc=0\tag 8$
Where:
$F_c\quad N_1\quad N_2$ are constrain forces
$\mu$ and $\mu_r$ are the friction coefficients
$a_1$ and $a_2$ are the translation accelerations
$\alpha_1$ and $\alpha_2$ are the rotational  accelerations 
We have 8 equations for the 8 unknowns:
$a_1,\alpha_1,a_2,\alpha_2\,N_1,N_2,F_c,F_r$
Result
$$\boxed{F_r=\frac{1}{2}\,\frac{\mu_r\,F}{1-\mu_r\,\mu}}$$
