Question related to maximum static friction and minimum $\mu_s$ I think I have the right formulas the answer the following question, but there is something conceptual that is throwing me off (see end of question).

A body with mass $m$ is located on a slope with angle $\alpha$ (see picture).
A. What should be the size of friction between the body and the inclined plane if it is known that the body does not move across the inclined plane, when the inclined plane is moving at acceleration $a_0$ rightwards?
B. Create an expression for the minimal size of $\mu_s$ in order for that to happen.

If $a_0$ is "small":
$$ma_x=ma_0\cos\alpha-mg\sin\alpha+F_s=0$$
$$ma_y=N-mg\cos\alpha-ma_0\sin\alpha=0$$
$$F_s=mg\sin\alpha-ma_0\cos\alpha$$
$$N = mg\cos\alpha + ma_0\sin\alpha$$
$$\mu_s = \frac {mg\sin\alpha-ma_0\cos\alpha}{mg\cos\alpha + ma_0\sin\alpha}=\frac {g\sin\alpha-a_0\cos\alpha}{g\cos\alpha + a_0\sin\alpha}$$
If $a_0$ is "big":
$$ma_x=ma_0\cos\alpha-mg\sin\alpha -F_s=0$$
$$ma_y=N-mg\cos\alpha-ma_0\sin\alpha=0$$
$$F_s=ma_0\cos\alpha-mg\sin\alpha$$
$$N = mg\cos\alpha + ma_0\sin\alpha$$
$$\mu_s = \frac {ma_0\cos\alpha-mg\sin\alpha}{mg\cos\alpha + ma_0\sin\alpha}=\frac {a_0\cos\alpha-g\sin\alpha}{g\cos\alpha + a_0\sin\alpha}$$
Here come the questions:

*

*I believe $F_s$ in my calculation inherently refers to the maximum static friction, but I am not sure of this.


*I have no idea what is meant by "the minimal size" of $\mu_s$. I thought it depended only on the materials, and therefore always has the same value. How should I interpret the word "minimal" here?

 A: The idea of maximum static friction is that the friction is never greater than the force it opposes. It's always less than or equal to the force it opposes. In the equation in itself, $F_s$ represents an instantaneous force, so any frictional force. However, in this case, there is no acceleration so the equation is for the maximum. But this does not change much in terms of the question.
As for the coefficient, yes it does depend on the surfaces but the question wants you to find what this value has to be at the very minimum to counteract the other forces (as $F_{net}=0$), as if you want to find this material to stop the motion. Your working has the right idea. They ask for minimum because if it is any bigger it does not make a difference since the force will be less than or equal to the force it opposes anyways.
A: 
I believe $F_s$ in my calculation inherently refers to the maximum static friction, but I am not sure of this.

No, it is not the maximum static friction, at least not until you replace it with $N\mu_s$. When you just have your expressions of the form $F_s=\pm mg\sin\alpha\mp ma_0\cos\alpha$ this is just the friction force needed to prevent the object from moving up or down the ramp. At this point you have not introduced any model of friction into your work. For all we care $F_s$ could represent the force a really sticky piece of gum would be exerting on the block given then acceleration $a_0$. In other words, $\mu_s$ could be infinite and your equations would look exactly the same. Therefore, those equations for $F_s$ do not tell you what the maximum static friction force is, at least on their own. This leads to your next question.

I have no idea what is meant by "the minimal size" of $\mu_s$. I thought it depended only on the materials, and therefore always has the same value. How should I interpret the word "minimal" here?

Once you set $F_s=N\mu_s$ you are assuming the static friction is at its maximum. You are right that $\mu_s$ only depends on the what the two surfaces are. However, you are not told what the surfaces are. The question is trying to get you to think in terms of the following: "If I wanted to accelerate the system with some given acceleration $a_0$, then what do I need $\mu_s$ to be at the very least?" Certainly, the smaller $\mu_s$ is then the smaller range of $a_0$ values the system can take before the block starts moving. 
For example, let's say $\mu_s=0$. Then we have no static friction, and we only have one possible acceleration $a_0=g\tan\alpha$. As we increase $\mu_s$ we start to get an interval of allowed accelerations centered around this $a_0=g\tan\alpha$ (each half of this interval is represented by your "small" and "large" cases). 
Therefore, each of the $\mu_s$ values you have for each case is the minimum value of $\mu_s$ you would need to keep the block from moving relative to the incline. If we lowered $\mu_s$ from this value for a given $a_0$ then the friction force would need to be larger than $N\mu_s$ and the block would move relative to the incline.
Therefore, to summarize,


*

*In your calculations before you implement $F_s=N\mu_s$ your equations are just looking at what force $F_s$ is required to keep the block from moving relative to the incline. What this force actually represents is irrelevant and has not come into play yet. Once you assume $F_s=N\mu_s$ you are now assuming $F_s$ represents the maximum friction force for a given value of $\mu_s$

*The expressions you obtain for $\mu_s$ in each case is the minimum coefficient needed to prevent slipping given a value of $a_0$. This is because if $F_s=N\mu_s$ and then you lowered $\mu_s$ the force needed to keep the block from moving relative to the incline would need to be larger than what $N\mu_s$ now is. How $a_0$ relates to $g\tan\alpha$ tells you which case to apply, or you can just generalize to all cases by using absolute value signs with what you already worked out:
$$\mu_s = \frac {\left|g\sin\alpha-a_0\cos\alpha\right|}{g\cos\alpha + a_0 \sin \alpha}$$
