Slide

Photo to know the direction of reactions..

First start with conditions of stability $\displaystyle\sum F_{x}= \displaystyle\sum F_{y}=\displaystyle\sum M= 0$ Let's assume that reaction at end point is the rod in vertical surface is $R_{1}$ and has distance $X$ from horizontal surface , friction force equal to $R_{1} \times \mu$ we will also named it as $F$ . And the reaction at end point of the rod at horizontal surface is $R_{2}$ and has distance $S$ from the vertical surface , friction force equal to $R_{2}\times \mu$ . $\color{#D61F06}{Note}$ that $tan\theta=\frac{X}{S}$
So we get from equations of equilibrium $R_{1}=R_{2}\times \mu(1)$ $R_{2}+R_{1}\times \mu=M$ from $(1)$ we get $R_{2}= \frac {M}{1+\mu^2}(2)$ Now we will take the Moment around point $F$ we get $R_{2}\times \mu\times X+M\times \frac{S}{2}=R_{2}\times S$ using $(2)$ we have $\frac{\not{M} * X * \mu }{1+\mu ^{2}} +\frac{\not{M*} S}{2} = \frac{\not{M} * S}{1+\mu ^{2}}$ Then dividing by $S$ we get $\frac{X}{S} *\frac{\mu }{1+\mu ^{2}} =\frac{1}{1+\mu ^{2}} -\frac{1}{2}$ Then $\frac{X}{S} =\frac{1}{\mu } -\frac{1+\mu ^{2}}{2\mu } =\boxed {\frac{1-\mu ^{2}}{2\mu }}$