Question

A ladder of length \(l\) and mass \(m\) is placed against a smooth vertical wall, but the ground is rough. The coefficient of friction between the ground and the ladder is \(\mu\). The angle \(\theta\) at which the ladder will stay in equilibrium is:

• A. \(\theta = \tan^{-1}(\mu)\)
• B. \(\theta = \tan^{-1}(2\mu)\)
• C. \(\theta = \tan^{-1}\!\left(\dfrac{\mu}{2}\right)\)
• D. \(\theta = \tan^{-1}\!\left(\dfrac{1}{2\mu}\right)\)

Hint:

For ladder problems:
Always take moments about the point where friction acts
Smooth wall \(\Rightarrow\) no friction at the wall
Limiting friction \(f=\mu N\)

Answer 1

The Correct Option is C

Step 1: Identify forces acting on the ladder.
Weight \(mg\) acting downward at the centre of the ladder.
Normal reaction \(N_1\) from the ground (vertical).
Friction \(f\) at the ground (horizontal).
Normal reaction \(N_2\) from the smooth wall (horizontal).
Step 2: Apply equilibrium of forces. Vertical equilibrium: \[ N_1 = mg \] Horizontal equilibrium: \[ f = N_2 \] Since the ladder is on the verge of slipping, \[ f = \mu N_1 = \mu mg \] Hence, \[ N_2 = \mu mg \]
Step 3: Take moments about the point of contact with the ground. Moment of weight about ground: \[ mg \left(\frac{l}{2}\right)\cos\theta \] Moment of wall reaction about ground: \[ N_2 \cdot l \sin\theta \] For rotational equilibrium: \[ mg\left(\frac{l}{2}\right)\cos\theta = N_2 l \sin\theta \] Substitute \(N_2 = \mu mg\): \[ \frac{1}{2}\cos\theta = \mu \sin\theta \]
Step 4: Solve for \(\theta\). \[ \tan\theta = \frac{1}{2\mu} \] \[ \theta = \tan^{-1}\!\left(\frac{\mu}{2}\right) \] Final Answer: \[ \boxed{\theta = \tan^{-1}\!\left(\dfrac{\mu}{2}\right)} \]