Problem source

A horizontal rail is fixed parallel to a vertical wall and at a distance $d$ from the wall. A uniform rod $AB$ of length $2a$ rests in equilibrium on the rail with the end $A$ in contact with the wall. The rod lies in a vertical plane perpendicular to the wall. It is inclined at an angle $\theta$ to the vertical (where $0 < \theta < \frac12\pi$) and  $a\sin\theta < d$, as shown in the diagram. 
\begin{center}
\begin{tikzpicture}
    % 1. Wall on the left 
    % A thick line with a dash pattern replicates the grey segmented wall.
    % The right edge of this line naturally touches x=0 due to its thickness.
    \draw[line width=6.2pt, dash pattern=on 11pt off 1pt, black!60] (-0.12,0) -- (-0.12,5);
    % 2. Main rod from A to B
    \draw[line width=2pt] (0,1.86) -- (5.86,3.34);
    % 3. Vertical dotted line passing through the pivot
    \draw[thick, dotted] (4,0.58) -- (4,4.48);
    % 4. Horizontal dotted line representing distance d
    \draw[thick, dotted] (0,4) -- (4,4);
    % 5. Angle arc (theta)
    % The start angle is arctan((3.34-1.86) / 5.86) ≈ 14.17 degrees.
    % We use a scope shifted to point A (0, 1.86) to easily draw the arc.
    \begin{scope}[shift={(0,1.86)}]
        \draw[thick] (14.17:0.8) arc (14.17:90:0.8);
    \end{scope}
    % 6. Support / pivot dot
    % Original PSTricks size 5pt diameter translates to a 2.5pt radius in TikZ
    \fill (4,2.78) circle (2.5pt);
    % 7. Labels
    % Using anchors to mimic the original top-left (\rput[tl]) positioning
    \node[anchor=north west, inner sep=1pt] at (0.15,2.4) {$\theta$};
    \node[anchor=south] at (2,4) {$d$}; % Centered above the line for better typography
    \node[anchor=north west, inner sep=1pt] at (0.12,1.78) {$A$};
    \node[anchor=north west, inner sep=1pt] at (5.66,3.14) {$B$};
\end{tikzpicture}
\end{center}
The coefficient of friction between the rod and the wall  is $\mu$, and the coefficient of friction between the rod and the rail is $\lambda$.
Show that in limiting equilibrium, with the rod on the point of slipping at both the wall and the rail, the angle $\theta$ satisfies
\[
d\cosec^2\theta = a\big( (\lambda+\mu)\cos\theta + (1-\lambda \mu)\sin\theta
\big)
\,.
\]
Derive the corresponding result  if, instead, $ a\sin\theta > d $.

Solution source

\begin{center}
\begin{tikzpicture}

    % 1. Wall on the left 
    % A thick line with a dash pattern replicates the grey segmented wall.
    % The right edge of this line naturally touches x=0 due to its thickness.
    \draw (-0,0) -- (-0,5);

    % 2. Main rod from A to B
    \draw[line width=2pt] (0,1.86) -- (5.86,3.34);

    \draw[-latex, blue, ultra thick] (0,1.86) -- ++(2, 0) node[right] {$R_W$};
    \draw[-latex, blue, ultra thick] (0,1.86) -- ++(0, 1.5) node[right] {$F_W$};

    % 5. Angle arc (theta)
    % The start angle is arctan((3.34-1.86) / 5.86) ≈ 14.17 degrees.
    % We use a scope shifted to point A (0, 1.86) to easily draw the arc.
    \begin{scope}[shift={(0,1.86)}]
        \draw[thick] (14.17:0.8) arc (14.17:90:0.8);
    \end{scope}

    % 6. Support / pivot dot
    % Original PSTricks size 5pt diameter translates to a 2.5pt radius in TikZ
    \fill (4,2.78) circle (2.5pt);

    \draw[-latex, blue, ultra thick] ({5.86/2}, {(3.34+1.86)/2}) -- ++(0,-2) node[below] {$W$};

    \draw[-latex, blue, ultra thick] (4,2.78) -- ++({5.86/5}, {(3.34-1.86)/5}) node[below] {$F_R$};
    \draw[-latex, blue, ultra thick] (4,2.78) -- ++({-(3.34-1.86)/4},{5.86/4}) node[above] {$R_R$};

    % 7. Labels
    % Using anchors to mimic the original top-left (\rput[tl]) positioning
    \node[anchor=north west, inner sep=1pt] at (0.15,2.4) {$\theta$};
    \node[anchor=north west, inner sep=1pt] at (0.12,1.78) {$A$};
    \node[anchor=north west, inner sep=1pt] at (5.66,3.14) {$B$};

\end{tikzpicture}
\end{center}

Notice everything is at limiting equilibrium, so $F_W = \mu R_W$ and $F_R = \lambda R_R$.

\begin{align*}
    \text{N2}(\nearrow): && \lambda R_R  - W \cos \theta+ R_W \sin \theta+\mu R_W \cos \theta &= 0 \\
    \text{N2}(\nwarrow): && R_R -W \sin \theta -R_W \cos \theta+\mu R_W \sin \theta &= 0 \\
    \overset{\curvearrowright}{A}: && a W \sin \theta -R_R \frac{d}{\sin \theta} &= 0 \\
    \overset{\curvearrowright}{\text{rod}}: && -W\left (d-a\sin \theta \right)+\mu R_W d-R_W d \cot \theta &= 0 \\
\end{align*}

So 

\begin{align*}
    && R_W d(\mu - \cot \theta) &= W (d - a \sin \theta) \\
    && a W  &= R_Rd \textrm{cosec}^2 \theta \\
    \Rightarrow && d \textrm{cosec}^2 \theta &=\frac{aW}{R_R} \\
    && \lambda R_R &= W \cos \theta -R_W(\sin \theta + \mu \cos \theta) \\
    &&&= W\cos \theta - W \frac{d - a \sin \theta}{d(\mu - \cot \theta)} ( \sin \theta + \mu \cos \theta) \\
    &&&= W { \left ( \frac{d\mu \cos \theta - d\cos \theta \cot \theta - d \sin \theta - d \mu \cos \theta+a \sin^2 \theta + a \mu \sin \theta \cos \theta}{d(\mu - \cot \theta)} \right) }\\
    &&&=  W \left ( \frac{ - d \textrm{cosec} \theta +a \sin^2 \theta + a \mu \sin \theta \cos \theta}{d(\mu - \cot \theta)} \right) \\
    \Rightarrow && d \textrm{cosec}^2 \theta &=\frac{aW}{R_R} \\
    &&&= \frac{ad\lambda(\mu - \cot \theta)}{- d \textrm{cosec} \theta +a \sin^2 \theta + a \mu \sin \theta \cos \theta} \\
    &&&= \frac{ad\lambda(\mu \sin \theta  - \cos  \theta)}{-d + a \sin^2 \theta (\sin \theta + \mu \cos \theta)} \\
    \Rightarrow && -d^2 \textrm{cosec}^2 \theta &+ a(\sin \theta + \mu \cos \theta) = ad\lambda(\mu \sin \theta  - \cos  \theta) \\
    \Rightarrow && d \textrm{cosec}^2 \theta &= a(\sin \theta + \mu \cos \theta)-a\lambda(\mu \sin \theta  - \cos  \theta) \\
    &&&= a( (\mu+\lambda)\cos \theta + (1-\mu \lambda)\sin \theta)
\end{align*}

If the rod is before the midpoint, the directions of both frictions will be reversed, ie we should obtain the same result, but with $\mu \to -\mu, \lambda \to -\lambda$ ie $d \textrm{cosec}^2 \theta = a( -(\mu+\lambda)\cos \theta + (1-\mu \lambda)\sin \theta)$