Problem source

\begin{center}
\begin{tikzpicture}[scale=0.5]
    % Circle
    \def\a{5}
    \def\b{10}
    \def\k{\b/sqrt(1 + (\b*\b - \a*\a)/(2*\a*\b)*(\b*\b - \a*\a)/(2*\a*\b))}
    
    \coordinate (O) at (0,0);
    \coordinate (CO) at (-\a, \b);
    \coordinate (Q1) at (0, \b);
    \coordinate (W1) at (0, 1.5*\b);
    \coordinate (W2) at ({-1.5*\k}, {1.5*\k * (\b*\b - \a*\a)/(2*\a*\b)});
    \coordinate (Q2) at ({-\k}, {\k * (\b*\b - \a*\a)/(2*\a*\b)});
    \pgfmathsetmacro{\Px}{\a*cos(150)}
    \pgfmathsetmacro{\Py}{\a*sin(150)}
    \coordinate (P) at ($(CO) + (\Px, \Py)$);
    % \coordinate (P) at ($(CO) + (\a*cos(120), \a*sin(120))$);
    
    \draw (CO) circle (\a);
    
    % Line from (4,0) to (-0.85,3.65)
    \draw (O) -- (W1);
    \draw (O) -- (W2);
    
    \node at (Q1) [right] {$Q_1$};
    \node at (W1) [right] {$W_1$};
    \node at (W2) [left] {$W_2$};
    \filldraw (Q1) circle (2pt);
    \filldraw (Q2) circle (2pt);
    \node at (Q2) [left] {$Q_2$};
    
    
    \node at (P) [left] {$P$};
    \filldraw (P) circle (2pt);
    \pic [draw, angle radius=1cm, "$\alpha$"] {angle = Q1--O--Q2};
\end{tikzpicture}\end{center}
A uniform sphere of mass $M$ and radius $r$ rests between a vertical wall $W_{1}$ and an inclined plane $W_{2}$ that meets $W_{1}$ at an angle $\alpha.$ $Q_{1}$ and $Q_{2}$ are the points of contact of the sphere with $W_{1}$ and $W_{2}$ resectively, as shown in the diagram. A particle of mass $m$ is attached to the sphere at $P$, where $PQ_{1}$ is a diameter, and the system is released. The sphere is on the point of slipping at $Q_{1}$ and at $Q_{2}.$ Show that if the coefficients of friction between the sphere and $W_{1}$ and $W_{2}$ are $\mu_{1}$ and $\mu_{2}$ respectively, then 
\[
m=\frac{\mu_{2}+\mu_{1}\cos\alpha-\mu_{1}\mu_{2}\sin\alpha}{(2\mu_{1}\mu_{2}+1)\sin\alpha+(\mu_{2}-2\mu_{1})\cos\alpha-\mu_{2}}M.
\]
If the sphere is on the point of rolling about $Q_{2}$ instead of slipping, show that 
\[
m=\frac{M}{\sec\alpha-1}.
\]

Solution source

\begin{center}
\begin{tikzpicture}[scale=0.5]
    % Circle
    \def\a{5}
    \def\b{10}

    \def\k{\b/sqrt(1 + (\b*\b - \a*\a)/(2*\a*\b)*(\b*\b - \a*\a)/(2*\a*\b))}
    
    \coordinate (O) at (0,0);
    \coordinate (CO) at (-\a, \b);
    \coordinate (Q1) at (0, \b);
    \coordinate (W1) at (0, 1.5*\b);
    \coordinate (W2) at ({-1.5*\k}, {1.5*\k * (\b*\b - \a*\a)/(2*\a*\b)});
    \coordinate (Q2) at ({-\k}, {\k * (\b*\b - \a*\a)/(2*\a*\b)});

    \pgfmathsetmacro{\Px}{\a*cos(180)}
    \pgfmathsetmacro{\Py}{\a*sin(180)}
    \coordinate (P) at ($(CO) + (\Px, \Py)$);
    % \coordinate (P) at ($(CO) + (\a*cos(120), \a*sin(120))$);
    
    \draw (CO) circle (\a);
    
    % Line from (4,0) to (-0.85,3.65)
    \draw (O) -- (W1);
    \draw (O) -- (W2);
    
    \node at (Q1) [right] {$Q_1$};
    \node at (W1) [right] {$W_1$};
    \node at (W2) [left] {$W_2$};
    \filldraw (Q1) circle (2pt);
    \filldraw (Q2) circle (2pt);
    \node at (Q2) [left] {$Q_2$};
    
    
    \node at (P) [left] {$P$};
    \filldraw (P) circle (2pt);

    \node at (CO) [above] {$O$};

    \pic [draw, angle radius=1cm, "$\alpha$"] {angle = Q1--O--Q2};

    \draw[-latex, blue, ultra thick] (Q1) -- ++(-2,0) node[below] {$R_1$};
    \draw[-latex, blue, ultra thick] (Q1) -- ++(0,-2) node[right] {$F_{r1}$};
    \draw[-latex, blue, ultra thick] (P) -- ++(0,-2) node[right] {$mg$};
    \draw[-latex, blue, ultra thick] (CO) -- ++(0,-2) node[right] {$Mg$};

    \draw[-latex, blue, ultra thick] (Q2) -- ($(CO)!0.6!(Q2)$) node[below] {$R_2$};
    \draw[-latex, blue, ultra thick] (Q2) -- ($(W2)!0.6!(Q2)$) node[below] {$F_{r2}$};
\end{tikzpicture}
\end{center}

Since the sphere is on the point of slipping at both $Q_1$ and $Q_2$, $F_{r1} = \mu_1 R_1$ and $F_{r2} = \mu_2 R_2$

\begin{align*}
    \text{N2}(\uparrow): && -mg-Mg-\mu_1 R_1 + R_2 \sin \alpha + \mu_2 R_2 \cos \alpha &= 0 \\
    \text{N2}(\rightarrow): && -R_1 + R_2 \cos \alpha - \mu_2 R_2 \sin \alpha &= 0 \\
    \\
    \Rightarrow &&  R_2 \cos \alpha - \mu_2 R_2 \sin \alpha &= R_1 \\
    % && -mg-Mg+\mu_1 (R_2 \cos \alpha - \mu_2 R_2 \sin \alpha) + R_2 \sin \alpha + \mu_2 R_2 \cos \alpha &= 0 \\
    % \\
    \overset{\curvearrowleft}{O}: && mg - \mu_1 R_1 - \mu_2R_2 &= 0 \\
    \Rightarrow && \mu_1 R_2 \l \cos \alpha - \mu_2 \sin \alpha \r - \mu_2 R_2 &= -mg \\
    && \mu_1 (R_2 \cos \alpha - \mu_2 R_2 \sin \alpha) + R_2 \sin \alpha + \\ && \quad \quad \mu_2 R_2 \cos \alpha - \mu_1 R_2 \l \cos \alpha - \mu_2 \sin \alpha \r - \mu_2 R_2 &= Mg \\
    \Rightarrow && \frac{\mu_2+\mu_1  \l \cos \alpha - \mu_2 \sin \alpha \r }{\mu_1 ( \cos \alpha - \mu_2  \sin \alpha) +  \sin \alpha + \mu_2  \cos \alpha - \mu_1  \l \cos \alpha - \mu_2 \sin \alpha \r - \mu_2 } &= \frac{m}{M} \\
    && \frac{\mu_2+\mu_1 \cos \alpha - \mu_1\mu_2 \sin \alpha }{\cos \alpha (-2\mu_1+\mu_2) + \sin \alpha (1 +2\mu_1\mu_2) -\mu_2} &= \frac{m}{M}
\end{align*}

If instead the sphere is about to roll about $Q_2$, then the forces at $Q_1$ will be $0$, we can then take moments about $Q_2$.

\begin{center}
\begin{tikzpicture}[scale=0.5]
    % Circle
    \def\a{5}
    \def\b{10}

    \def\k{\b/sqrt(1 + (\b*\b - \a*\a)/(2*\a*\b)*(\b*\b - \a*\a)/(2*\a*\b))}
    
    \coordinate (O) at (0,0);
    \coordinate (CO) at (-\a, \b);
    \coordinate (Q1) at (0, \b);
    \coordinate (W1) at (0, 1.5*\b);
    \coordinate (W2) at ({-1.5*\k}, {1.5*\k * (\b*\b - \a*\a)/(2*\a*\b)});
    \coordinate (Q2) at ({-\k}, {\k * (\b*\b - \a*\a)/(2*\a*\b)});

    \pgfmathsetmacro{\Px}{\a*cos(180)}
    \pgfmathsetmacro{\Py}{\a*sin(180)}
    \coordinate (P) at ($(CO) + (\Px, \Py)$);
    % \coordinate (P) at ($(CO) + (\a*cos(120), \a*sin(120))$);
    
    \draw (CO) circle (\a);
    
    % Line from (4,0) to (-0.85,3.65)
    \draw (O) -- (W1);
    \draw (O) -- (W2);
    
    \node at (Q1) [right] {$Q_1$};
    \node at (W1) [right] {$W_1$};
    \node at (W2) [left] {$W_2$};
    \filldraw (Q1) circle (2pt);
    \filldraw (Q2) circle (2pt);
    \node at (Q2) [left] {$Q_2$};
    
    
    \node at (P) [left] {$P$};
    \filldraw (P) circle (2pt);

    \node at (CO) [above] {$O$};

    \pic [draw, angle radius=1cm, "$\alpha$"] {angle = Q1--O--Q2};

    \draw[-latex, blue, ultra thick] (P) -- ++(0,-2) node[right] {$mg$};
    \draw[-latex, blue, ultra thick] (CO) -- ++(0,-2) node[right] {$Mg$};

    \draw[-latex, blue, ultra thick] (Q2) -- ($(CO)!0.6!(Q2)$) node[below] {$R_2$};
    \draw[-latex, blue, ultra thick] (Q2) -- ($(W2)!0.6!(Q2)$) node[below] {$F_{r2}$};

    \draw[dashed] (Q2) -- (CO) -- (P) -- cycle;

    \pic [draw, angle radius=1cm, "$\alpha$"] {angle = P--CO--Q2};
    
\end{tikzpicture}
\end{center}


Looking at perpendicular distances from $Q_2$ to $O$ and $P$ we have $r \cos \alpha$ and $r(1-\cos \alpha)$

\begin{align*}
    \overset{\curvearrowleft}{Q_2}: && mg (1 - \cos \alpha) - Mg \cos \alpha &= 0 \\
    \Rightarrow && \frac{1}{\sec \alpha-1} &= \frac{m}{M}
\end{align*}