Problem source

A uniform ladder of mass $M$ rests with its upper end against a smooth vertical wall, and with its lower end on a rough slope which rises upwards towards the wall and makes an angle of $\phi$ with the horizontal.
The acute angle between the ladder and the wall is $\theta$. If the ladder is in equilibrium, show that $N$ and $F$, the normal reaction and frictional force at the foot of the ladder are given by 
\[
N=Mg\left(\cos\phi-\frac{\tan\theta\sin\phi}{2}\right),
\]
\[
F=Mg\left(\sin\phi+\frac{\tan\theta\cos\phi}{2}\right).
\]
If the coefficient of friction between the ladder and the slope is $2$, and $\phi=45^{\circ}$, what is the largest value of $\theta$ for which the ladder can rest in equilibrium?

Solution source

\begin{align*}
\overset{\curvearrowleft}{X}: && 0&= \frac{l}{2} Mg\sin \theta - l R_1 \cos \theta  \\
\Rightarrow && R_1 &= \frac12 \tan \theta Mg \\

\text{N2}(\uparrow): && 0 &= R\cos \phi +F \sin \phi - Mg \\
\text{N2}(\rightarrow):&& 0&=R_1-F \cos \phi + R \sin \phi \\ 
\Rightarrow && \frac12 \tan \theta Mg &= F \cos \phi- R \sin \phi \\
&& Mg &= F \sin \phi +R \cos \phi \\
\Rightarrow && F &= Mg \left ( \sin \phi + \frac12 \tan \theta \cos \phi  \right) \\
&& N &= Mg \left (\cos \phi - \frac12 \tan \theta \sin \phi \right )
\end{align*}

If $\mu = 2$ and $\phi = 45^{\circ}$, we must have $F \leq 2 N$, so:

\begin{align*}
&&  Mg \left ( \sin \phi + \frac12 \tan \theta \cos \phi  \right)  &\leq 2 Mg \left (\cos \phi - \frac12 \tan \theta \sin \phi \right ) \\
\Rightarrow && 1 + \frac12 \tan \theta \leq 2-\tan \theta \\
\Rightarrow && \frac 32 \tan \theta \leq 1 \\
\Rightarrow && \tan \theta \leq \frac23 \\
\Rightarrow && \theta \leq \tan^{-1} \frac23
\end{align*}