Problem source

A thin non-uniform rod $PQ$ of length $2a$ has its centre of gravity a distance $a+d$ from $P$. It hangs (not vertically) in equilibrium suspended from a small smooth peg $O$ by means of a light inextensible string of length $2b$ which passes over the peg and is attached at its ends to $P$ and $Q$. Express $OP$ and $OQ$ in terms of $a,b$ and $d$. By considering the angle $POQ$, or otherwise, show that $d < a^{2}/b$.

Solution source

\begin{center}
    \begin{tikzpicture}[scale=2]
        \begin{scope}[rotate=20]
            \coordinate (P) at (-1,0);
            \coordinate (Q) at (1,0);
            \coordinate (M) at (0.2,0);
        \end{scope}

        \coordinate (O) at ($(M)+(0,1)$); 
        
        \filldraw (M) circle (1.2pt) node [below] {$G$};
        \draw (O) circle (0.8pt) node [above] {$O$};

        \node[left] at (P) {$P$}; 
        \node[right] at (Q) {$Q$}; 

        \draw[dashed] (O) -- (M);

        \node[above left] at ($(P)!0.5!(O)$) {$2b-x$};
        \node[above right] at ($(Q)!0.5!(O)$) {$x$};
        \node[below] at ($(P)!0.5!(M)$) {$a+d$};
        \node[below] at ($(Q)!0.5!(M)$) {$a-d$};

        \draw[thick] (P)--(Q);
        \draw (P) -- (O) -- (Q);

        \draw[-latex, ultra thick, blue] (O) -- ($(O)!0.3!(P)$) node[below] {$T$};
        \draw[-latex, ultra thick, blue] (O) -- ($(O)!0.5!(Q)$) node[below] {$T$};
        
    \end{tikzpicture}
\end{center}

Resolving horizontally, it's clear that $\angle POG = \angle GOQ$, in particular applying the sine rule:

\begin{align*}
&& \sin \angle POG &= \frac{a+d}{2b-x} \sin \angle PGO \\
&& \sin \angle GOP &= \frac{a-d}{x} \sin \angle OGQ \\
\Rightarrow && \frac{a+d}{2b-x} &= \frac{a-d}{x} \\
\Rightarrow && x(a+d) &= (2b-x)(a-d) \\
\Rightarrow && 2ax &= 2b(a-d) \\
\Rightarrow && x &= b - \frac{db}{a} \\
\Rightarrow && PO &= b+\frac{db}{a} \\
&& OQ &= b - \frac{d}{a}
\end{align*}

Applying the cosine rule:

\begin{align*}
&& \cos POQ &= \frac{(b + \frac{db}{a})^2 + (b - \frac{db}{a})^2 -4a^2}{2(b^2 - \frac{d^2b^2}{a^2})} \\
&&&= \frac{2b^2 + \frac{2d^2b^2}{a^2}-4a^2}{2(b^2 - \frac{d^2b^2}{a^2})} \\
&&&= \frac{2a^2b^2 + 2d^2b^2-4a^4}{2b^2(a^2 - d^2)}  \\
&&&< 1  \\
\Leftrightarrow && 2a^2b^2 + 2d^2b^2-4a^4 &< 2b^2(a^2-d^2) \\
\Leftrightarrow &&  2d^2b^2-4a^4 &< -2b^2d^2 \\
\Leftrightarrow &&  4d^2b^2&< 4a^4 \\
\Leftrightarrow &&  d^2&< \frac{a^4}{b^2} \\
\Leftrightarrow &&  d&< \frac{a^2}{b} \\


\end{align*}