Equilibrium of a Ladder against a Wall – Rankers Physics
Topic: Rotational Motion
Subtopic: Torque

Equilibrium of a Ladder against a Wall

A ladder of length \(\ell\) and mass m is placed against a smooth vertical wall but the ground is not smooth. Coefficient of friction between the ground and the ladder is \(\mu\). The minimum angle \(\theta\) with ground at which the ladder will stay in equilibrium is :
\(tan^{-1}(\mu)\)
\(tan^{-1}(2\mu)\)
\(tan^{-1}(\mu/2)\)
\(tan^{-1}(1/2\mu)\)

Solution:

For translational and rotational equilibrium of the ladder, taking torque about the base gives \(N_{\text{wall}} \ell \sin\theta = mg \frac{\ell}{2} \cos\theta\). With \(N_{\text{wall}} = f \le \mu mg\), we get the minimum angle for equilibrium to be \(\tan\theta = \frac{1}{2\mu}\).

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