1 problem found
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: \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*}