Problem source

One end of a light inextrnsible string of length $l$ is fixed to a point on the upper surface of a thin, smooth, horizontal table-top, at a distance $(l-a)$ from one edge of the table-top. A particle of mass $m$ is fixed to the other end of the string, and held a distance $a$ away from this edge of the table-top, so that the string is horizontal and taut. The particle is then released. Find the tension in the string after the string has rotated through an angle $\theta,$ and show that the largest magnitude of the force on the edge of the table top is $8mg/\sqrt{3}.$

Solution source

\begin{center}
    \begin{tikzpicture}[scale=2]
        \draw (0,5) -- (2,5) -- (2,4.5) -- (0,4.5);

        \coordinate (P) at ({2.05+3*cos(45)}, {5.05-3*sin(45)});
        \coordinate (E) at (2.05, 5.05);
        \coordinate (S) at (5,5.05);

        \draw[dashed] (0,5.05) -- (S);
        \draw (0,5.05) -- (E) -- (P);

        \pic [draw, angle radius=1.5cm, "$\theta$"] {angle = P--E--S};

        \filldraw (P) circle (1pt);
        \filldraw (S) circle (1pt);

        \draw[-latex, blue, ultra thick] (P) -- ++(0, -0.5) node[below] {$mg$};
        \draw[-latex, blue, ultra thick] (P) -- ($(P)!0.2!(E)$) node[above left] {$T$};
        \draw[-latex, blue, ultra thick] (E) -- ($(P)!0.8!(E)$) node[below] {$T$};
        \draw[-latex, blue, ultra thick] (E) -- ++(-0.5,0) node[left] {$T$};
        \draw[-latex, blue, ultra thick] (E) -- ++(0.2,0.5) node[above] {$R$};
        
    \end{tikzpicture}
\end{center}

\begin{align*}
\text{N2}(\nwarrow): && T - mg \sin \theta &= m \left ( \frac{v^2}{r}\right) \\
&&&= \frac{m v^2}{a} \\
\text{COE}:&& \underbrace{0}_{\text{assume initial GPE level is }0} &= \frac12 m v^2 - mga\sin \theta \\
\Rightarrow && v^2 &= 2ag \sin \theta \\
\Rightarrow && T &= \frac{m}{a} \cdot 2 ag \sin \theta + mg \sin \theta \\
&&&= 3mg \sin\theta
\end{align*}

Considering the force on the edge of the table will be:

\begin{align*}
&& \mathbf{R} &= \binom{-T}{0} + \binom{T \cos \theta}{-T \sin \theta} \\
&&&= \binom{T(1-\cos \theta)}{-T \sin \theta} \\
&&&= 3mg \sin \theta \binom{1-\cos \theta}{-\sin \theta} \\
\Rightarrow && |\mathbf{R}| &= 3mg \sin \theta \sqrt{(1-\cos \theta)^2 + \sin ^2 \theta} \\
&&&= 3mg \sin \theta \sqrt{2 - 2 \cos \theta} \\
&&&= 3mg \sin \theta\sqrt{4 \sin^2 \tfrac{\theta} {2}} \\
&&&= 6mg \sin \theta |\sin  \tfrac{\theta} {2} | \\
s = \sin \tfrac \theta2:&&&= 12mg s^2 \sqrt{1-s^2} 
\end{align*}

We can maximise $V = x\sqrt{1-x}$ by differentiating: 
\begin{align*}
&& \frac{\d V}{\d x} &= \sqrt{1-x} - \frac{x}{2\sqrt{1-x}} \\
&&&= \sqrt{1-x} \left ( 1 - \frac{x}{2-2x}\right) \\
&&&= \sqrt{1-x} \frac{2-3x}{2-2x} \\
\Rightarrow && x &= \frac23
\end{align*}

Therefore the maximum for will be:

\begin{align*}
|\mathbf{R}| &= 12 mg \frac 23 \sqrt{\frac13} \\
&= 8mg/\sqrt{3}
\end{align*}

as required.