If a ladder weighting `250 N` in placed a smooth vartical wall having coefficient of friction between it and floor 03, then what is the maximum force of fraction available at the point of contact between the ladder and the floor?

Options

A. `75 N`
B. `50 N`
C. `35 N`
D. `25 N`

Explanation

To solve the problem of finding the maximum force of friction available at the point of contact between the ladder and the floor, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Given Data:**
- Weight of the ladder (W) = 250 N
- Coefficient of friction (μ) = 0.3

2. **Understand the Forces Acting on the Ladder:**
- The ladder is in equilibrium, meaning that the sum of the vertical forces and the sum of the horizontal forces must both be zero.
- The weight of the ladder acts downward at its center of mass.
- The normal force (N2) acts upward from the floor.
- There is no friction from the wall since it is smooth.

3. **Balance the Vertical Forces:**
- In the vertical direction, the normal force (N2) must balance the weight of the ladder:
\[ N2 = W = 250 \, \text{N} \]

4. **Calculate the Maximum Force of Friction:**
- The maximum force of friction (f_max) can be calculated using the formula:
\[ f_{\text{max}} = \mu \cdot N2 \]
- Substitute the values:
\[ f_{\text{max}} = 0.3 \cdot 250 \, \text{N} = 75 \, \text{N} \]

5. **Conclusion:**
- The maximum force of friction available at the point of contact between the ladder and the floor is:
\[ \text{Maximum Force of Friction} = 75 \, \text{N} \]

### Final Answer:
The maximum force of friction available at the point of contact between the ladder and the floor is **75 N**.

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