Problem source

A sphere of radius $a$ and weight $W$ rests  on horizontal ground.  A thin uniform beam of weight $3\sqrt3\,W$  and length $2a$ is freely hinged to the ground at $X$, which is a distance ${\sqrt 3} \, a$  from the point of contact of the sphere with the ground.  
    The beam rests on the sphere, lying in the same vertical plane as the centre of the  sphere. The coefficients of friction between the beam and the sphere and between the sphere  and the ground are $\mu_1$ and $\mu_2$ respectively. 
 
    Given that the sphere is on the point of slipping at its contacts  with both the ground and the beam, find the values of $\mu_1$ and $\mu_2$.

Solution source

\begin{center}    
\begin{tikzpicture}[scale=2]

    \draw (-1,0) -- (3,0);
    \draw[thick] (0,1) circle (1);
    \draw ({sqrt(3)}, 0) -- ({1/sqrt(3)}, {2});
    % \draw ({sqrt(3)}, 0) -- ({0}, {3});

    \draw[-latex, blue, ultra thick] (0,1) -- ++(0,-0.3) node[left] {$W$};
    \draw[-latex, blue, ultra thick] (0,0) -- ++(0,.3) node[right] {$R_2$};
    \draw[-latex, blue, ultra thick] (0,0) -- ++(.3,.0) node[right] {$\mu_2 R_2$};

    \draw[-latex, blue, ultra thick] ({2/sqrt(3)},1) -- ++(0,-0.3) node[right] {$3\sqrt{3}W$};

    \draw[-latex, blue, ultra thick] ({2/sqrt(3)},1) -- ++(0,-0.3) node[right] {$3\sqrt{3}W$};

    \draw[-latex, blue, ultra thick] ({sqrt(3)/2},{3/2}) -- ++({sqrt(3)/8},{1/8}) node[right] {$R_1$};
    \draw[-latex, blue, ultra thick] ({sqrt(3)/2},{3/2}) -- ++({1/8},{-sqrt(3)/8}) node[right] {$\mu_1 R_1$};
    
    \node[below] at ({sqrt(3)}, 0) {$X$};
    

\end{tikzpicture}
\end{center}


The first important thing to observe is the angle at $X$ is $60^{\circ}$.

Now we can start resolving:

\begin{align*}
     \overset{\curvearrowleft}{X}: && 3\sqrt{3} W \cos 60^{\circ} a - R_1\sqrt{3}a &= 0  \tag{$1$}\\
     \overset{\curvearrowleft}{O}: && \mu_2 R_2 a - \mu_1R_1a &= 0 \tag{$2$} \\
     \text{N2}(\rightarrow): && \mu_2 R_2 + \mu_1R_1 \cos 60^{\circ} - R_1 \cos 30^{\circ} &= 0 \tag{$3$} \\
     \text{N2}(\uparrow): && R_2 - W - \mu_1 R_1 \cos 30^{\circ} - R_1 \cos 60^{\circ} &= 0  \tag{$4$} \\
     \Rightarrow && \frac{3}{2}W &= R_1 \tag{$(5)$ from $(1)$} \\
      && \mu_1 R_1 &= \mu_2 R_2  \tag{$2$}\\
      && \mu_1 R_1 \l 1 + \frac{1}{2} \r - R_1 \frac{\sqrt{3}}2 &= 0 \tag{$(3)$ and $(2)$} \\
      && \mu_1 &= \frac{1}{\sqrt3} \\
      \\
      && R_2 - W - \frac{1}{\sqrt3} \frac{3}{2}W \frac{\sqrt3}{2} - \frac{3}2W \frac12 &= 0 \\
      \Rightarrow && R_2 &= W \l 1 + \frac{3}{2}\r \tag{$6$} \\
      \Rightarrow && \mu_2 &= \frac{\mu_1 R_1}{R_2} = \frac{1}{\sqrt{3}} \frac{3}{5} = \frac{\sqrt3}{5} \tag{$(5)$ and $(6)$}
\end{align*}