A fireman want to slide down a rope. The breaking load the rope is `3//4^(th)` of the weight of the man. With what minimum acceleration should the fireman slide down ?

Options

A. `g//6`
B. `g//4`
C. `g//3`
D. `g//2`

Answer

To solve the problem, we need to determine the minimum acceleration with which the fireman can slide down the rope without exceeding the breaking load of the rope.

### Step-by-Step Solution:

1. **Identify the Forces Acting on the Fireman**:
- The weight of the fireman \( W = mg \) (where \( m \) is the mass of the fireman and \( g \) is the acceleration due to gravity).
- The tension in the rope \( T \) acts upwards against the weight of the fireman.

2. **Set Up the Equation of Motion**:
- When the fireman slides down with acceleration \( a \), the net force acting on him can be expressed using Newton's second law:
\[ mg - T = ma \]
- Rearranging gives:
\[ T = mg - ma \]

3. **Determine the Maximum Tension**:
- According to the problem, the maximum tension \( T_{\text{max}} \) that the rope can withstand is \( \frac{3}{4} mg \).

4. **Set the Tension Equal to Maximum Tension**:
- To find the minimum acceleration, we set the tension equal to the maximum tension:
\[ mg - ma = \frac{3}{4} mg \]

5. **Solve for Acceleration \( a \)**:
- Rearranging the equation gives:
\[ mg - \frac{3}{4} mg = ma \]
\[ \frac{1}{4} mg = ma \]
- Dividing both sides by \( m \) (assuming \( m \neq 0 \)):
\[ \frac{1}{4} g = a \]

6. **Final Result**:
- Therefore, the minimum acceleration \( a \) with which the fireman should slide down is:
\[ a = \frac{g}{4} \]