Problem source

The lower end of a rigid uniform rod of mass $m$ and length $a$ rests at point $M$ on rough horizontal ground. Each of two elastic strings, of natural length $\ell$ and modulus of elasticity $\lambda$, is attached at one end to the top of the rod. Their lower ends are attached to points $A$ and $B$ on the ground, which are a distance $2a$ apart. $M$ is the midpoint of $AB$.
$P$ is the point at the top of the rod and lies in the vertical plane through $AMB$.
Suppose that the rod is in equilibrium with angle $PMB = 2\theta$, where $\theta < 45°$ and $\theta$ is such that both strings are in tension.
\begin{questionparts}
\item Show that angle $APB$ is a right angle.
Show that the force exerted on the rod by the elastic strings can be written as the sum of
\begin{itemize}
\item a force of magnitude $\frac{2a\lambda}{\ell}$ parallel to the rod
\item and a force of magnitude $\sqrt{2}\lambda$ acting along the bisector of angle $APB$.
\end{itemize}
\item By taking moments about point $M$, or otherwise, show that $\cos\theta + \sin\theta = \frac{2\lambda}{mg}$.
Deduce that it is necessary that $\frac{1}{2}mg < \lambda < \frac{1}{2}\sqrt{2}mg$.
\item $N$ and $F$ are the magnitudes of the normal and frictional forces, respectively, exerted on the rod by the ground at $M$.
Show, by taking moments about an appropriate point, or otherwise, that
\[N - F\tan 2\theta = \frac{1}{2}mg.\]
\end{questionparts}

Solution source

\begin{center}
    \begin{tikzpicture}[scale=4]
        \coordinate (A) at (-1,0);
        \coordinate (M) at (0,0);
        \coordinate (B) at (1,0);
        \coordinate (P) at ({cos(60)}, {sin(60)});

        \draw[dashed] (B) arc (0:180:1);

        \draw ($(A)+(-0.2,0)$) -- ($(B)+(0.2,0)$);
        \draw (M) -- (P);
        \draw (A) -- (P) -- (B);
        
        \filldraw (M) circle (0.5pt) node[below] {$M$};
        \filldraw (A) circle (0.5pt) node[below] {$A$};
        \filldraw (B) circle (0.5pt) node[below] {$B$};
        \filldraw (P) circle (0.5pt) node[above] {$P$};

        \draw[-latex, blue, ultra thick] (P) -- ++($0.5*(B)-0.5*(P)$) node[right] {$T_B$};
        \draw[-latex, blue, ultra thick] (P) -- ++($0.5*(A)-0.5*(P)$) node[left] {$T_A$};

        \draw[-latex, blue, ultra thick] ($0.5*(P)$) -- ++(0, -0.2) node[below] {$mg$};
        \draw[-latex, blue, ultra thick] (M) -- ++(0, 0.4) node[above] {$N$};
        \draw[-latex, blue, ultra thick] (M) -- ++(0.4, 0) node[above] {$F$};

        \pic [draw, angle radius=0.8cm, "$2\theta$"] {angle = B--M--P};

        \pic [draw, angle radius=0.8cm, "$\theta$"] {angle = A--P--M};


        
    \end{tikzpicture}
\end{center}

\begin{questionparts}
\item Notice that $AM = MB = MP$ in particular $P$ lies on a semi-circle of radius $a$ and therefore by Thales' theorem $\angle APB = 90^{\circ}$. Notice that by angles in a triangle and the angles adding to $90^{\circ}$, $\angle APM = \theta$.

Therefore,

\begin{align*}
&& |PB| &= 2a \sin \theta \\
&& |PA| &= 2a \cos \theta \\
&& T_A &= \frac{\lambda}{l} \left (2a \cos \theta -l \right) \\
&& T_B &= \frac{\lambda}{l} \left (2a \sin\theta -l \right) \\
\end{align*}

Since $T_A$ and $T_B$ are perpendicular, we can consider the forces as having vector $\frac{\lambda}{l}\binom{2a\cos \theta-l}{2a\sin \theta - l}$ in this coordinate system, ie the sum of a vector $\frac{2\lambda a}{l}\binom{\cos \theta}{\sin \theta}$ and $\displaystyle -\sqrt{2}\lambda \binom{\frac1{\sqrt{2}}}{\frac1{\sqrt{2}}}$ which are unit vectors parallel to the rod and along the bisector of $APB$ respectively.

\item 
\begin{align*}
\overset{\curvearrowright}{M}: && 0 &= \frac{a}{2} \cdot mg \cos 2 \theta - a\cdot \sqrt{2}\lambda  \cos (90-(45-\theta))\\
\Rightarrow && \cos 2 \theta &= \frac{\lambda}{mg} 2 \sqrt{2} \cos (45 + \theta) \\
\Rightarrow && \cos^2 \theta - \sin^2 \theta &= \frac{2\lambda}{mg} (\cos \theta - \sin \theta) \\
\underbrace{\Rightarrow}_{\theta < 45^{\circ}} && \cos \theta + \sin \theta &= \frac{2\lambda}{mg}
\end{align*}

Over $(0, 45^{\circ})$, $\cos \theta + \sin \theta$ ranges from $1$ to $\sqrt{2}$, therefore

$1 < \frac{2 \lambda}{mg} < \sqrt{2} \Rightarrow \frac12 mg < \lambda < \frac12 \sqrt{2} mg$ as required.

\item 

\begin{align*}
\overset{\curvearrowright}{P}: && 0 &=- \frac{a}{2} \cdot \left ( mg \cos 2\theta \right) - a \cdot F \sin 2 \theta + a \cdot N \cos 2 \theta \\
\Rightarrow && \frac12 mg &= N - F \tan 2 \theta
\end{align*}
as required.
\end{questionparts}