Problem source

A step-ladder has two sections $AB$ and $AC,$ each of length $4a,$ smoothly hinged at $A$ and connected by a light elastic rope $DE,$ of natural length $a/4$ and modulus $W$, where $D$ is on $AB,$ $E$ is on $AC$ and $AD=AE=a.$ The section $AB,$ which contains the steps, is uniform and of weight $W$ and the weight of $AC$ is negligible. 
The step-ladder rests on a smooth horizontal floor and a man of weight $4W$ carefully ascends it to stand on a rung distant $\beta a$ from the end of the ladder resting on the floor. Find the height above the floor of the rung on which the man is standing when $\beta$ is the maximum value at which equilibrium is possible.

Solution source

\begin{center}
    \begin{tikzpicture}

        \def\t{50};
        \def\l{4};
        
        \coordinate (A) at (0,0);
        
        
        \coordinate (C) at ({\l*cos(-\t)},{\l*sin(-\t)});
        \coordinate (E) at ({\l*cos(-\t)/4},{\l*sin(-\t)/4});
        \coordinate (B) at ({-\l*cos(-\t)},{\l*sin(-\t)});
        \coordinate (D) at ({-\l*cos(-\t)/4},{\l*sin(-\t)/4});
        \coordinate (G) at ({-\l*cos(-\t)/2},{\l*sin(-\t)/2});

        \draw (C) -- (A) -- (B);

        \draw (D) -- (E);

        \filldraw (A) circle (1pt) node[above] {$A$};
        \filldraw (B) circle (1pt) node[left] {$B$};
        \filldraw (C) circle (1pt) node[right] {$C$};
        \filldraw (D) circle (1pt) node[left] {$D$};
        \filldraw (E) circle (1pt) node[right] {$E$};

        \draw[-latex, ultra thick, blue] (G) -- ++(0,-1) node[below] {$W$};
        \draw[-latex, ultra thick, blue] ($(A)!0.7!(B)$) -- ++(0,-2) node[below] {$4W$};
        \draw[-latex, ultra thick, blue] (B) -- ++(0,1) node[above] {$R_B$};
        \draw[-latex, ultra thick, blue] (C) -- ++(0,1) node[above] {$R_C$};
        \draw[-latex, ultra thick, blue] (D) -- ($(D)!0.5!(E)$) node[above] {$T$};
        \draw[-latex, ultra thick, blue] (E) -- ($(D)!0.5!(E)$) node[above] {$T$};
    
    \end{tikzpicture}
\end{center}

\begin{align*}
N2(\uparrow): && 0 &= R_B+R_C  - 5W \\
\Rightarrow && 5W &= R_B + R_C \\
\\
\overset{\curvearrowright}{A}: && 0 &= (R_B - R_C) \cdot 4a \cdot \cos \theta -W \cdot 2a \cdot \cos \theta - 4W \cdot (4 - \beta)a \cdot \cos \theta \\
\Rightarrow && R_B-R_C &= W \left ( \frac12 + (4-\beta)\right) \\
\Rightarrow && R_B &= \frac{W}2 \left ( 5+\frac12+(4-\beta)\right) = \frac{W}{2}\left(\frac{19}{2} - \beta\right) \\
&& R_C &= \frac{W}{2} \left (5 - \frac12 - 4 +\beta \right) = \frac{W}{2} \left (\frac12 + \beta \right) \\
\\
\overset{\curvearrowright}{(A, AC)}: && 0 &= T \cdot a \cdot \sin \theta - R_C \cdot 4a  \cdot \cos \theta \\
\Rightarrow && T &=4 \cot \theta \frac{W}{2} \left ( \frac12 + \beta\right) \\
&&&= 20W \cot \theta \\
\text{Hooke's Law}:&& T &= \frac{W(2a \cos \theta - \frac{a}{4})}{\frac{a}{4}} = W(8 \cos \theta - 1) \\
\Rightarrow && 8 \cos \theta -1 &= \cot \theta (2\beta+1)\\
\Rightarrow && 1+2\beta &=8\sin \theta-\tan \theta \\
\Rightarrow && \beta &=   4 \sin \theta - \frac12 \tan \theta - \frac12 \\
\Rightarrow && \frac{\d \beta}{\d \theta} &= 4 \cos \theta - \frac12 \sec^2 \theta \\
&&&= \frac{8\cos^3 \theta - 1}{\cos^2 \theta} \\
\Rightarrow && \cos \theta &= \frac12 \\
\Rightarrow && h &= \beta a \sin \theta \\
&&&= \left (4 \frac{\sqrt{3}}{2}-\frac12 \sqrt{3}-\frac12 \right) a \frac{\sqrt3}{2} \\
&&&= \left ( \frac{9-\sqrt{3}}{4}\right)a
\end{align*}