Will static friction be zero unless it's necessary for maintaining equilibrium? While solving a problem related to equilibrium, I assumed that friction would be zero (I wasn't sure if it was correct) unless friction was required to maintain the system in equilibrium. This got me thinking about forces, and I tried to think about easy scenarios.
Let's say you have two blocks placed next to each other on a rough floor. They do not exert any force on each other, and the frictional force exerted on them by the ground is zero. However, it is possible for them to stay in equilibrium if there's a normal force between them, and the ground exerts frictional force to balance the normal force between them. But of course, this does not happen (Unless you press them against each other).
Another scenario I could think of is a particle attached to a string, placed on a rough floor. If I were to pull the particle, would tension balance the force, or would static friction do it?
Here's the problem I'm talking about:

You have a plate of mass 2m bent at right angles (with both parts of the plate being equal) and placed on a fixed cylinder. My question is - if the coefficient of friction is greater than 1/2, it is possible for the system to be in equilibrium with f2 being zero. But it's also possible for the system to be in equilibrium if f2 is not zero. How do I decide what f2 is going to be? Will it be zero as long as it isn't required to maintain equilibrium?
(Note: The question I originally posted mentioned second law of thermodynamics, but I removed it as I felt it was misleading and could be very unrelated to my question.)
 A: 
I assumed that friction would be zero unless friction was required to maintain the system in equilibrium

This part of the question is circular. If static friction occurs, it is obviously "required" by nature. So, you cannot derive anything from a supposed "necessity" of friction.

How do I decide what f2 is going to be?

The problem is, you have 4 unknowns ($f_1,f_2,N_1,N_2$) but only 3 equations (2x force equilibrium for the dimensions of the sketch plane and 1x torque equilibrium perpendicular to the sketch plane). The inequalities for the static friction force do not count into that, because they only bound the inside of the solution set against the outside (i.e. the "forbidden" friction forces).
Hence, you will have one remaining degree of freedom in the algebraic solution. You are perfectly right in claiming that the equilibrium is undetermined, i.e. there is a continuum of possible equilibria (friction-wise), at least if we look at the situation macroscopically.
Suppose we consider $f_2$ as the remaining degree of freedom of the solution, i.e. we may choose $f_2$ almost freely within a certain interval to be determined. As you have already shown, $f_2$ then has to satisfy the upper bound
$$f_2\le \frac{\mu}{1+\mu}mg =: f_{2,max}(\mu)$$
which obviously depends on the friction coefficient. On the other hand, you have derived that $f_1$ has to satisfy the inequality
$$f_1=mg-f_2\le \frac{\mu}{1-\mu} mg$$
which results in the requirement that
$$f_2\ge \frac{1-2\mu}{1-\mu}mg =:f_{2,min}(\mu)$$
So, generally, depending on the value of $\mu$, the system will assume an equilibrium state for all friction forces $f_2$ that are in the interval
$$\frac{1-2\mu}{1-\mu}mg\le f_2\le \frac{\mu}{1+\mu}mg$$
However, this interval only exists if
$$\frac{1-2\mu}{1-\mu}\le\frac{\mu}{1+\mu}$$
If we check the limiting equals sign of that inequality, we get the boundary friction value $\mu_0$ according to
$$\mu_0^2+2\mu_0-1=0$$
which has only one positive solution
$$\mu_0=\sqrt{2}-1\approx 0.4142$$
So, finally, we can quickly check that the above inequality is satisfied, i.e. there is a continuum of solutions for the friction forces, if
$$\mu>\mu_0$$
while there is a unique solution for the friction forces only if $\mu=\mu_0$.
Obviously, your assumption $\mu>0.5$ (or, equivalently, $f_2=0$) is sufficient for an ambiguous solution, but it is also unnecessarily narrow.
Now, what does the ambiguity of the solution mean? Friction is a non-linear phenomenon, i.e. it is not at all surprising that it leads to ambiguous solutions, and moreover, the static solutions may even depend on the history of the system.
If we look at the microscopic level, however, static friction is a consequence of crystal boundaries or even individual atoms interlocking between the bodies in contact. You might imagine kind-of "gear teeth" at the surface of the bodies. If the bodies are in contact at more than one point or area, it may be undetermined which "gear teeth" win the battle for taking up the tangential forces. The different contact points may even take up various fractions of the forces.
In your example, you could first place the top plate so carefully on the cylinder, that it already takes up all the tangential forces (namely $mg$), so that the side plate will not need to take up anything more as soon as it also gets in contact with the cylinder (this is the case $f_2=0$). Or you could first bring the side plate into contact with the cylinder, so that it already "steals" much of the tangential force (again $mg$), and there is not so much left for the top plate (something close to the case $f_2=f_{2,max}$).
It gets more complicated, if we also consider shear deformations close to the surface (think of rubber), because this may cause one set of "gear teeth" to give up some fraction of the friction forces in favor of the other "gear teeth" (surface areas). You may even pre-load the elastic material against the friction forces (pulling the rubber contact areas against each other before settling the friction).
For simple configurations, like a body on an inclined plane, this is totally irrelevant macroscopically, because there is only one surface normal and we are only interested in the total friction force. But your specific example is obviously designed (by choosing two contact surface normals that are extremely different, i.e. right-angled) to cause the kind of ambiguity that is sketched above.
A: The assertion regarding the static friction force being dependent on the force it counteracts is correct. The only condition here is that the value of friction force shoudl not exceed the threshold
$$F_{fr}\leq \mu_s N.$$
Taking the language from Wikipedia article on friction:

The static friction force must be overcome by an applied force before an object can move. The maximum possible friction force between two surfaces before sliding begins is the product of the coefficient of static friction and the normal force: $F_{max}=\mu_s F_n$. When there is no sliding occurring, the friction force can have any value from zero up to $F_max$. Any force smaller than $F_max$ attempting to slide one surface over the other is opposed by a frictional force of equal magnitude and opposite direction. Any force larger than $F_max$ overcomes the force of static friction and causes sliding to occur. The instant sliding occurs, static friction is no longer applicable—the friction between the two surfaces is then called kinetic friction. However, an apparent static friction can be observed even in the case when the true static friction is zero.

This should be distinguished from the kinetic friction force which occurs when the object is sliding, and which is equal to $F_{fr}=\mu N$ (with the kinetic friction coefficient $\mu$ usually a bit smaller than the static friction coefficient $\mu_s$).
