# What is the physical interpretation of this limiting case of a block on a wedge?

I have a question about the physical interpretation of a particular limiting case in a problem. Please note my question is not about how to actually solve the problem, but rather an entirely conceptual question.

Consider a block on a wedge inclined at angle $$\theta$$. The coefficient of friction between the block and plane is $$\mu$$. What is the maximum acceleration, $$A$$, that we can accelerate the wedge at so that the block remains on the wedge without sliding?

At large accelerations, we can expect the friction on the block to point down the slope. From this, I was able to derive that the maximum possible acceleration has magnitude $$A=\frac{\sin{\theta}+\mu\cos{\theta}}{\cos{\theta}-\mu\sin{\theta}}g.$$ Here is my question. For the magnitude above to be positive, we require $$\cos{\theta}-\mu\sin{\theta}>0$$, or $$\tan{\theta}<\frac{1}{\mu}$$.

But what if we are given that $$\theta$$ is large enough so that $$\tan{\theta}>\mu$$? Combining the two conditions, we must have $$\mu<\tan{\theta}<\frac{1}{\mu}.$$ In particular, we have that $$\mu<\frac{1}{\mu}$$, which is only true if $$\mu<1$$ for positive $$\mu$$.

This means that if $$1<\mu<\tan{\theta}$$, the magnitude of the maximum acceleration is negative, which doesn't make any sense. It seems that such a maximum acceleration does not exist in this case, and I can make the acceleration arbitrarily large without the block moving. How can I physically interpret this?

The case under consideration is $$\tan\theta > \mu>1$$.
Let's do this step-by-step:

### Minimum Acceleration

It is clear that because $$\tan \theta>\mu$$, the block on the unaccelerated wedge will slide down.
Let's slowly accelerate the wedge (with acceleration $$a$$) and notice the effect of the value of acceleration on the forces acting on the block. At one point, we observe that the net force on the object is zero: $$ma_{\text{min}}\cos\theta + \mu (ma_{\text{min}}\sin\theta + mg\cos\theta)=mg\sin\theta \tag{1-1}$$ $$\Rightarrow a_{\text{min}}=\frac{g\sin\theta-\mu g\cos\theta}{\cos\theta+\mu\sin\theta}\tag{1-2}$$ Alright, so $$a_{\text{min}}$$ is the minimum value of acceleration needed to keep the block stationary with respect to the wedge.

### Maximum Acceleration?

Let's keep increasing the value of $$a$$. We observe that the frictional force keeps reducing (still pointing up the slope) and at one point, it becomes zero.

$$ma_0\cos\theta = mg\sin\theta \Rightarrow a_0=g\tan\theta \tag{2-1}$$

Let's not stop here and keep increasing the value of $$a$$. From this moment, the frictional force starts to point down the slope.

Here, we need to be careful about an important point:
The frictional force necessary to keep the object in equilibrium might be less than the maximum value it can take, no matter how much you increase $$a$$ further. If this is true, then there is no meaning to "maximum acceleration" and we can conclude that the object will stay in equilibrium for any $$a \geq a_{\text{min}}$$.

$$ma\cos\theta -mg\sin\theta = f \tag{2-2}$$ $$f \leq f_{\text{max}}=\mu m(a\sin\theta + g \cos\theta) \tag{2-3}$$ $$\stackrel{\text{Comparing (2-2) and (2-3)}}\Rightarrow (a-g\tan\theta) < (\underbrace{\mu \tan\theta}_{\text{Slope>1}} a + g) \;\forall \;a>a_0 \tag{2-4}$$

In short, the notion of maximum acceleration is not defined for the case $$\tan\theta > \mu >1$$.