A ladder of length `l` 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 angle `theta` at which the ladder will stay in equilibrium is

Options

A. `theta=tan^(-1)(mu)`
B. `theta=tan^(-1)(2mu)`
C. `theta=tan^(-1)(mu/2)`
D. none of these

Explanation

To solve the problem of determining the angle \(\theta\) at which a ladder of length \(l\) and mass \(m\) stays in equilibrium against a smooth vertical wall, we will follow these steps:

### Step 1: Draw the Free Body Diagram (FBD)
1. Draw the ladder leaning against the wall, making an angle \(\theta\) with the ground.
2. Identify the forces acting on the ladder:
- Weight \(mg\) acting downward at the center of the ladder.
- Normal force \(N_2\) acting upward at the base of the ladder.
- Normal force \(N_1\) exerted by the wall acting horizontally towards the ladder.
- Friction force \(f\) acting horizontally at the base of the ladder, opposing the motion.

### Step 2: Apply Force Balance in Horizontal Direction
- The horizontal forces must balance:
\[ f = N_1 \]
- The friction force can also be expressed in terms of the normal force at the base:
\[ f = \mu N_2 \]
- Therefore, we can write:
\[ N_1 = \mu N_2 \quad \text{(Equation 1)} \]

### Step 3: Apply Force Balance in Vertical Direction
- The vertical forces must balance:
\[ N_2 = mg \quad \text{(Equation 2)} \]

### Step 4: Substitute Equation 2 into Equation 1
- From Equation 2, substitute \(N_2\) into Equation 1:
\[ N_1 = \mu mg \quad \text{(Equation 3)} \]

### Step 5: Apply Torque Balance
- To find the angle \(\theta\), we need to balance the torques about the base of the ladder (point O):
- The torque due to the weight \(mg\) acts at a distance of \(\frac{l}{2} \cos \theta\) (perpendicular distance).
- The torque due to the normal force \(N_1\) acts at a distance of \(l \sin \theta\) (perpendicular distance).

- Setting the torques equal:
\[ mg \cdot \left(\frac{l}{2} \cos \theta\right) = N_1 \cdot (l \sin \theta) \]

### Step 6: Simplify the Torque Equation
- Cancel \(l\) from both sides:
\[ mg \cdot \left(\frac{1}{2} \cos \theta\right) = N_1 \sin \theta \]
- Substitute \(N_1\) from Equation 3:
\[ mg \cdot \left(\frac{1}{2} \cos \theta\right) = \mu mg \sin \theta \]
- Cancel \(mg\) from both sides:
\[ \frac{1}{2} \cos \theta = \mu \sin \theta \]

### Step 7: Rearrange to Find \(\theta\)
- Rearranging gives:
\[ \tan \theta = \frac{1}{2\mu} \]
- Therefore, the angle \(\theta\) can be expressed as:
\[ \theta = \tan^{-1}\left(\frac{1}{2\mu}\right) \]

### Final Answer
The angle \(\theta\) at which the ladder will stay in equilibrium is:
\[ \theta = \tan^{-1}\left(\frac{1}{2\mu}\right) \]