Question

A uniform ladder rests in limiting equilibrium with its lower end on a rough horizontal plane with coefficient of friction $\mu$ and its upper end against a smooth vertical wall. If $\theta$ is the inclination of the ladder with the wall, then $\theta$ is equal to

• A. ${\tan }^{-1}\mu$
• B. ${\cot }^{-1}\mu$
• C. ${\cot }^{-1}(2\mu )$
• D. ${\tan }^{-1}(2\mu )$

Answer 1

Answer: (D)

In limiting equilibrium

$R=W...(i)$
$S=\mu R...(ii)$
$\Rightarrow S=\mu W$
Taking moment about $A$,
$-W\cdot AD+S\cdot BC=0$
$\Rightarrow -W\cdot AG\cos \left(90^{\circ }-\theta \right)+\mu W\cdot AB\sin \left(90^{\circ }-\theta \right)=0$
$\Rightarrow -W\cdot AG\sin \theta +\mu W\cdot 2AG\cdot \cos \theta =0$
$\Rightarrow -\sin \theta +2\mu \cos \theta =0$
$\Rightarrow \sin \theta =2\mu \cos \theta$
$\Rightarrow \theta ={\tan }^{-1}(2\mu )$