A `4 m` long ladder weighing `25 kg` rests with its upper end against a smooth wall and lower end on rough ground.What should be the minimum coefficient of friction between the ground and the ladder for it to be inclined at `60^(@)` with the horizontal without slipping? `(Take g =10 m//s^(2))`

Options

A. a. 0.19
B. b. 0.29
C. c. 0.39
D. d. 0.49

Answer

To solve the problem of determining the minimum coefficient of friction required for a 4 m long ladder weighing 25 kg to remain stable while inclined at an angle of \(60^\circ\) with the horizontal, we will follow these steps:

### Step 1: Identify the forces acting on the ladder
1. **Weight of the ladder (W)**: This acts vertically downward at the center of the ladder.
\[ W = mg = 25 \, \text{kg} \times 10 \, \text{m/s}^2 = 250 \, \text{N} \]
2. **Normal forces**:
- \(N_1\) acts vertically upward at the base of the ladder (ground).
- \(N_2\) acts horizontally from the wall (since the wall is smooth, it only provides a horizontal reaction).
3. **Frictional force (F_r)**: This acts horizontally at the base of the ladder, opposing the tendency to slip.

### Step 2: Set up the equilibrium equations
For the ladder to be in equilibrium:
- The sum of vertical forces must be zero.
- The sum of horizontal forces must be zero.
- The sum of moments about any point must also be zero.

#### Vertical Forces:
\[ N_2 = W \quad \text{(1)} \]

#### Horizontal Forces:
\[ F_r = N_1 \quad \text{(2)} \]

### Step 3: Calculate moments about the base of the ladder
Taking moments about the base of the ladder (point O):
- The moment due to the weight of the ladder (acting at its center, 2 m from the base):
\[ \text{Moment due to } W = W \cdot \left(2 \cos(60^\circ)\right) = 250 \cdot (2 \cdot 0.5) = 250 \, \text{N} \cdot 1 = 250 \, \text{Nm} \]
- The moment due to the normal force from the wall:
\[ \text{Moment due to } N_1 = N_1 \cdot (4 \sin(60^\circ)) = N_1 \cdot (4 \cdot \frac{\sqrt{3}}{2}) = 2\sqrt{3} N_1 \]

Setting the moments equal for equilibrium:
\[ N_1 \cdot (2\sqrt{3}) = 250 \quad \text{(3)} \]

### Step 4: Substitute \(N_1\) from equation (2) into equation (3)
From equation (2):
\[ N_1 = F_r = \mu N_2 \]
Substituting \(N_2 = W\) from equation (1):
\[ N_1 = \mu W = \mu \cdot 250 \]
Substituting this into equation (3):
\[ \mu \cdot 250 \cdot (2\sqrt{3}) = 250 \]
Dividing both sides by 250:
\[ \mu \cdot 2\sqrt{3} = 1 \]
Solving for \(\mu\):
\[ \mu = \frac{1}{2\sqrt{3}} = \frac{\sqrt{3}}{6} \]

### Step 5: Calculate the numerical value of \(\mu\)
Using \(\sqrt{3} \approx 1.732\):
\[ \mu \approx \frac{1.732}{6} \approx 0.2887 \]

### Final Answer
The minimum coefficient of friction required is approximately:
\[ \mu \approx 0.29 \]