Problem source

A uniform rectangular lamina $ABCD$ rests in equilibrium in a vertical plane with the $A$ in contact with a rough vertical wall. The plane of the lamina is perpendicular to the wall.
It is supported by a light inextensible string attached to the side $AB$ at a distance $d$ from $A$. The other end of the string is attached to a point on the wall above $A$ where it makes an acute angle  $\theta$ with the downwards vertical. The side $AB$ makes an acute angle $\phi$ with the upwards vertical at $A$. The sides $BC$ and $AB$ have lengths $2a$ and $2b$ respectively. The coefficient of friction between the lamina and the wall is $\mu$.
  \begin{questionparts}
  \item Show that, when the lamina is in limiting equilibrium with the frictional force acting upwards,
    \begin{equation}
      d\sin(\theta +\phi) = (\cos\theta +\mu \sin\theta)(a\cos\phi
      +b\sin\phi)\,.  \tag{$*$}
    \end{equation}
  \item How should $(*)$ be modified if the lamina is in limiting equilibrium with the frictional force acting downwards?
  \item Find a condition on $d$, in terms of $a$, $b$, $\tan\theta$ and $\tan\phi$, which is necessary and sufficient for the frictional force to act upwards. Show that this condition cannot be satisfied if $b(2\tan\theta+ \tan \phi) < a$.
  \end{questionparts}

Solution source

\begin{center}
\begin{tikzpicture}
        \begin{scope}[rotate=60]
            \coordinate (D) at (0,0);
            \coordinate (C) at (2,0);
            \coordinate (B) at (2,6);
            \coordinate (A) at (0,6);

            \coordinate (G) at (1,3);
            \coordinate (Gx) at (0,3);

            \coordinate (T) at (1.2,6);
            
        \end{scope}

        \coordinate (X) at (-1,0);
        \coordinate (Y) at ({6*cos(150)},6);

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

        % \draw (-7, 0) -- (0.5,0);
        \draw ({6*cos(150)},0) -- ({6*cos(150)},8);

        \draw (T) -- (Y);

        \node[left] at (A) {$A$};
        \node[above] at (B) {$B$};
        \node[right] at (C) {$C$};
        \node[below] at (D) {$D$};

        \node[right] at ($(A)!0.5!(B)$) {$2b$};
        \node[above] at ($(B)!0.5!(C)$) {$2a$};

        \filldraw (G) circle (1pt) node [right] {$G$};
        \filldraw (Gx) circle (1pt) node[below] {$X$};
        \draw[dashed] (Gx) -- (G);
        \draw[dashed] ({6*cos(150)},{3*sin(150)}) -- (Gx);



        \pic [draw, angle radius=1cm, angle eccentricity=1.25, "$\theta$"] {angle = A--Y--T};
        \pic [draw, angle radius=0.8cm, angle eccentricity=1.25, "$\phi$"] {angle = T--A--Y};

        \draw[-latex, ultra thick, blue] (T) -- ($(Y)!0.5!(T)$) node[above] {$T$};
        \draw[-latex, ultra thick, blue] (A) -- ++(0,1) node[above] {$F$};
        \draw[-latex, ultra thick, blue] (A) -- ++(1,0) node[above] {$R$};
        % \draw[-latex, ultra thick, blue] (O) -- ++(-1,0) node[left] {$F_G$};
        % \draw[-latex, ultra thick, blue] (O) -- ++(0,1) node[above] {$R_G$};

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

\begin{questionparts}
\item 
\begin{align*}
\text{N2}(\uparrow): && T \cos \theta + F -W &= 0 \\
&& W &= T\cos \theta + \mu R \tag{1} \\
\text{N2}(\rightarrow): && R-T\sin \theta &= 0 \\
&& R &= T \sin \theta  \tag{2}\\
\\
(1)+(2): && W&=(\cos \theta + \mu \sin \theta)T \tag{3} \\
\overset{\curvearrowright}{A}: && 0 &= W(b\sin \phi + a \cos \phi) - Td\sin(\phi+\theta)  \tag{4} \\
\\
(3)+(4): && 0 &= (\cos \theta + \mu \sin \theta)(b\sin \phi + a \cos \phi)-d\sin(\phi+\theta) \\
\Rightarrow && d\sin(\phi+\theta) &= (\cos \theta + \mu \sin \theta)(b\sin \phi + a \cos \phi)
\end{align*}

as required.

\item If $F$ is operating downwards, it's equivalent to $-\mu$, ie:
\[d\sin(\phi+\theta) = (\cos \theta - \mu \sin \theta)(b\sin \phi + a \cos \phi)\]

\item For the frictional force to be acting upwards, we need
\begin{align*}
&& d\sin(\phi+\theta) &\geq  \cos \theta(b\sin \phi + a \cos \phi) \\
\Rightarrow && d &\geq \frac{\cos \theta(b\sin \phi + a \cos \phi)}{\sin(\phi + \theta)} \\
&&&= \frac{\cos \theta(b\sin \phi + a \cos \phi)}{\sin\phi \cos\theta+\cos\phi\sin \theta)}\\
&&&= \frac{(b\sin \phi + a \cos \phi)}{\sin\phi+\cos \phi \tan \theta)}\\
&&&= \frac{a+b\tan \phi}{\tan\theta+\tan\phi }\\
\end{align*}

We know that $d < 2b$, so

\begin{align*}
&& 2b &>\frac{a+b\tan \phi}{\tan\theta+\tan\phi }\\
\Rightarrow && 2b \tan \theta + 2b \tan \phi &> a + b \tan \phi \\
\Rightarrow &&b(2 \tan \theta +  \tan \phi) &> a\\
\end{align*}

Therefore we will have problems if the inequality is reversed!
\end{questionparts}