Problem source

A non-uniform rod $AB$ has weight $W$ and length $3l$. When the rod is suspended horizontally in equilibrium by vertical strings attached to the ends $A$ and $B$, the tension in the string attached to $A$ is $T$.
When instead the rod is held in equilibrium in a horizontal position by means of a smooth pivot at a distance $l$ from $A$ and a vertical string attached to $B$, the tension in the string is $T$. Show that $5T = 2W$. 
When instead the end $B$ of the rod rests on rough horizontal ground and the rod is held in equilibrium at an angle $\theta$ to the horizontal by means of a string that is perpendicular to the rod and attached to $A$, the tension in the string is $\frac12 T$. Calculate $\theta$ and find the smallest value of the coefficient of friction between the rod and the ground that will prevent slipping.

Solution source

\begin{center}
    \begin{tikzpicture}
        \coordinate (A) at (0,0);
        \coordinate (B) at (3,0);
        \coordinate (G) at (1.2,0);
        \draw[thick] (A) -- (B);

        
        \node[left] at (A) {$A$};
        \node[right] at (B) {$B$};

        \draw (A) -- ++(0,2);
        \draw (B) -- ++(0,2);

        \draw[-latex, ultra thick, blue] (A) -- ++(0,1) node[above] {$T$};
        \draw[-latex, ultra thick, blue] (B) -- ++(0,1) node[above] {$W-T$};
        \draw[-latex, ultra thick, blue] (G) -- ++(0,-1) node[below] {$W$};
        
    \end{tikzpicture}
\end{center}



Suppose the centre of mass of the rod is $x$ away from $A$.

\begin{align*}
\overset{\curvearrowleft}{B}: && (3l-x)W - 3lT &= 0 \\
\Rightarrow && x &= \frac{3l(W-T)}{W} \tag{1}
\end{align*}

\begin{center}
    \begin{tikzpicture}
        \coordinate (A) at (0,0);
        \coordinate (B) at (3,0);
        \coordinate (G) at (1.2,0);
        \draw[thick] (A) -- (B);

        
        \node[left] at (A) {$A$};
        \node[right] at (B) {$B$};

        % \draw (A) -- ++(0,2);
        \draw (B) -- ++(0,2);

        % \draw[-latex, ultra thick, blue] (A) -- ++(0,1) node[above] {$T$};
        \filldraw (1,0) circle (0.1);
        \draw[-latex, ultra thick, blue] (B) -- ++(0,1) node[above] {$T$};
        \draw[-latex, ultra thick, blue] (G) -- ++(0,-1) node[below] {$W$};
        \draw[-latex, ultra thick, blue] (1,0) -- ++(0,0.4) node[above] {$?$};
        
    \end{tikzpicture}
\end{center}

In the second set up we have:

\begin{align*}
\overset{\curvearrowleft}{\text{pivot}}: && 2lT - (x-l)W &= 0 \\
\Rightarrow && x &= \frac{2lT + lW}{W} \tag{2} \\
\\
(1) \text{ & } (2): && 3l(W-T) &= l(2T+W) \\
\Rightarrow && 2W &= 5T
\end{align*}


\begin{center}
    \begin{tikzpicture}
        \coordinate (A) at ({3+3*cos(160)},{3*sin(160)});
        \coordinate (B) at (3,0);
        \coordinate (G) at ({3+1.8*cos(160)},{1.8*sin(160)});
        \draw[thick] (A) -- (B);

        
        \node[left] at (A) {$A$};
        \node[right] at (B) {$B$};

        \coordinate (X) at (0,0);
        \coordinate (Y) at (3.5,0);

        \draw[dashed] (X) -- (Y);

        % \draw (A) -- ++(0,2);

        % \draw[-latex, ultra thick, blue] (A) -- ++(0,1) node[above] {$T$};
        \draw[-latex, ultra thick, blue] (A) -- ++(1,0) node[right] {$\frac12T$};
        \draw[-latex, ultra thick, blue] (B) -- ++(-1,0) node[left] {$F$};
        \draw[-latex, ultra thick, blue] (B) -- ++(0,1) node[right] {$R$};
        \draw[-latex, ultra thick, blue] (G) -- ++(0,-1) node[right] {$W$};

        \pic [draw, angle radius=.8cm, angle eccentricity=1.25, "$\theta$"] {angle = A--B--X};
        
    \end{tikzpicture}
\end{center}



\begin{align*}
&& x&= \frac{3l(W-T)}{W}\\
&&&= \frac{3l(W - \frac25 W)}{W} \\
&&&= \frac{9}{5}l\\
\overset{\curvearrowleft}{B}: && -\frac12 T (3l \sin \theta) + W \frac{6}{5}l \cos \theta &= 0 \\
\Rightarrow && \tan \theta &= \frac{4}{5} \frac{W}{T} \\
&&&= \frac45 \frac52 \\
&&&= 2 \\
\Rightarrow && \theta &= \tan^{-1} 2 \\
\\
\text{N2}(\uparrow): && R &= W \\
\text{N2}(\rightarrow): && F &= \frac12 T \\
\Rightarrow && F & \leq \mu R \\
\Rightarrow && \frac12 T &\leq \mu W \\
\Rightarrow && \mu &\geq \frac12 \frac{T}{W}  = \frac12 \frac25 = \frac15

\end{align*}