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 A D+S \cdot B C=0 $
$ \Rightarrow -W \cdot A G \cos \left(90^{\circ}-\theta\right) +\mu W \cdot A B \sin \left(90^{\circ}-\theta\right)=0 $
$\Rightarrow -W \cdot A G \sin \theta+\mu W \cdot 2 A G \cdot \cos \theta=0 $
$\Rightarrow -\sin \theta+2 \mu \cos \theta=0 $
$ \Rightarrow \sin \theta=2 \mu \cos \theta $
$\Rightarrow \theta=\tan ^{-1}(2 \mu)$