Problem source

Two uniform ladders $AB$ and $BC$ of equal length are hinged smoothly at $B$. The weight of $AB$ is $W$ and the weight of $BC$ is $4W $.
The ladders stand on rough horizontal ground with  $\angle ABC=60^\circ\,$.  The coefficient of friction between each ladder  and the 
ground is  $\mu$.
A decorator of weight $7W$ begins to climb the ladder $AB$ slowly. When she has climbed up $\frac13$ of the ladder, one of the ladders slips.
Which ladder  slips, and what is the value of $\mu$?

Solution source

\begin{center}    
\begin{tikzpicture}[scale=2]
    \coordinate (A) at (1,0);
    \coordinate (B) at (2, {sqrt(3)});
    \coordinate (C) at (3,0);
    \coordinate (D) at ({4/3},{sqrt(3)/3});
    \draw (A) -- (B) -- (C);
    \draw (0,0) -- (4, 0);
    \filldraw (B) circle (1pt);
    \filldraw (D) circle (1pt);
    \node[above] at (B) {$B$}; 
    \node[above right] at (C) {$C$}; 
    \node[above left] at (A) {$A$}; 
    % \draw[thick] (0,1) circle (1);
    \draw[-latex, blue, ultra thick] (D) -- ++(0,-0.3) node[below] {$7W$};
    \draw[-latex, blue, ultra thick] ($(A)!0.5!(B)$) -- ++(0,-0.3) node[below] {$W$};
    \draw[-latex, blue, ultra thick] ($(C)!0.5!(B)$) -- ++(0,-0.3) node[below] {$4W$};

    \draw[-latex, blue, ultra thick] (A) -- ++(0,0.3) node[above] {$R_A$};
    \draw[-latex, blue, ultra thick] (A) -- ++(0.3,0) node[below] {$F_A$};
    \draw[-latex, blue, ultra thick] (C) -- ++(0,0.3) node[above] {$R_C$};
    \draw[-latex, blue, ultra thick] (C) -- ++(-0.3,0) node[below] {$F_C$};

    % \draw[-latex, blue, ultra thick] (B) -- ++(-0.3,0) node[below] {$F_C$};
    
\end{tikzpicture}
\end{center}

\begin{align*}
     \text{N2}(\rightarrow): && F_A - F_C &= 0\\
     && F_A &= F_C \\
     \text{N2}(\uparrow): && R_A + R_C - 7W - W - 4W &= 0\\
     && R_A + R_C &= 12W \\
     \overset{\curvearrowright}{A}: && \frac{1}{6}7W + \frac{1}{4}W + \frac{3}{4}4W - R_C &= 0 \\
     \Rightarrow && R_C &= \frac{53}{12}W\\
     \Rightarrow && R_A = 12W - \frac{53}{12}W &= \frac{91}{12}W \\
     \overset{\curvearrowleft}{B}(AB): && \frac{1}{2}W + \frac{2}{3}7W - R_A+\sqrt{3}F_A &= 0 \\
     \Rightarrow && F_A = \frac{1}{\sqrt{3}} \l \frac{91}{12}-\frac12-\frac{14}3\r W &= \frac{29}{12\sqrt{3}}W
\end{align*}

We know that the system is about to slip, so equality holds in one of $F_A \leq \mu R_A$ or $F_C \leq \mu R_C$. Since $F_A = F_C$, we know it must occur for whichever of $\mu R_A$ and $\mu R_C$ is smaller. Since $R_C$ is much smaller, this must be the ladder about to slip $BC$ and \[ \mu = \frac{F_C}{R_C} = \frac{\frac{29}{12\sqrt{3}}W}{\frac{53}{12}W} = \boxed{\frac{29}{53\sqrt{3}}}\]