Self lock: conditions to make it work

[Karol]

For F outside moments around A:
$$\mbox{and}\left\{\begin{array}{ll}F\left(\frac{x}{2}+l\right)=Ry+\mu Rx\\2\mu R<F\end{array}\right.\rightarrow F\left(\frac{x}{2}+l\right)<\frac{F}{2\mu}(y-\mu x)\rightarrow y>2\mu l$$
And for F inside its $\mu x < y$

 

[haruspex]

For F outside moments around A:
$$\mbox{and}\left\{\begin{array}{ll}F\left(\frac{x}{2}+l\right)=Ry+\mu Rx\\2\mu R<F\end{array}\right.\rightarrow F\left(\frac{x}{2}+l\right)<\frac{F}{2\mu}(y-\mu x)\rightarrow y>2\mu l$$
And for F inside its $\mu x < y$

I agree with your algebra for the F outside case, but the condition seems to be inverted. I thought we wanted the non-slip condition, so it would be $2\mu R > F$, leading to $\mu > y/(2l)$.
For F inside, perhaps you're doing the same, since you keep asserting $\mu x < y$ when I have shown the opposite must be true. Anyway, the inside case is more complicated because we cannot assume the top right corner is on the point of slipping. The frictional force there is anything from zero to $\mu R$. It will take a bit of thought to decide which of those extremes is the right one for the purpose of answering the question.