Problem source

A uniform rigid rod $BC$ is suspended from a fixed point $A$ by light stretched springs $AB,AC$. The springs are of different natural lengths but the ratio of tension to extension is the same constant $\kappa$ for each. The rod is \textit{not} hanging vertically. Show that the ratio of the lengths of the stretched springs is equal to the ratio of the natural lengths of the unstretched springs.

Solution source

\begin{center}
    \begin{tikzpicture}
        \def\t{30};
        \def\l{2};
        \coordinate (G) at (0,0);
        \coordinate (X) at (1,0);
        \coordinate (Y) at (-1,0);
        \coordinate (C) at ({\l*cos(\t)},{\l*sin(\t)});
        \coordinate (B) at ({-\l*cos(\t)},{-\l*sin(\t)});
        \coordinate (A) at ({0},{3});

        \draw[ultra thick] (B) -- (C);

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

        \draw[-latex, ultra thick, blue] (B) -- ($(B)!0.5!(A)$);
        \draw[-latex, ultra thick, blue] (C) -- ($(C)!0.5!(A)$);
        \draw[-latex, ultra thick, blue] (G) -- ++(0, -2);

        \filldraw (A) circle (1.5pt) node[above] {$A$};
        \filldraw (G) circle (1.5pt) node[right] {$G$};

        \draw[dashed] (A) -- (G);
        ]
        \pic [draw, angle radius=.9cm, angle eccentricity=1.5, "$\theta$"] {angle = X--G--C};
        \pic [draw, angle radius=.9cm, angle eccentricity=1.5, "$\alpha$"] {angle = B--A--G};
        \pic [draw, angle radius=.9cm, angle eccentricity=1.5, "$\beta$"] {angle = G--A--C};

        \node[right] at (C) {$C$};
        \node[left] at (B) {$B$};
    
    \end{tikzpicture}
\end{center}

By moments or "centre of mass" or whatever argument you choose, the centre of mass is directly below $A$.

\begin{align*}
N2:&& 0 &= \frac{1}{|AC|}\binom{-l\cos \theta}{h-l \sin \theta} T_{AC} + \frac{1}{|AB|} \binom{l \cos \theta}{h+l \sin \theta}T_{AB} + \binom{0}{-1}mg \\
\Rightarrow && \frac{T_{AC}}{AC} &= \frac{T_{AB}}{AB} \\
\Rightarrow && \frac{\kappa(AC-l_{AC})}{AC} &= \frac{\kappa(BC-l_{BC})}{BC} \\
\Rightarrow && \frac{l_{AC}}{AC} &= \frac{l_{BC}}{BC} \\
\Rightarrow && \frac{l_{AC}}{l_{BC}} &= \frac{AC}{BC}
\end{align*}