Where must a `800 N` weight be hung on a uniform horizontal `100 N` pole of length `L` so that a body at one end supports one-third as much as a man at the other end ?

Options

A. at a distance of `0.22 L` from man
B. at a distance of `0.22 L` from boy
C. at a distance of `0.33 L` from man
D. in the middle

Explanation

To solve the problem, we need to find the position \( x \) where an \( 800 \, \text{N} \) weight should be hung on a uniform horizontal \( 100 \, \text{N} \) pole of length \( L \) such that a body at one end supports one-third as much as a man at the other end.

### Step-by-Step Solution:

1. **Identify Forces and Setup the Problem:**
- Let \( R_M \) be the force exerted by the man at one end of the pole.
- Let \( R_B \) be the force exerted by the boy at the other end of the pole.
- The weight of the pole is \( 100 \, \text{N} \) and acts at the midpoint of the pole (i.e., at \( \frac{L}{2} \)).
- The weight \( 800 \, \text{N} \) is hung at a distance \( x \) from the man.

2. **Translate the Given Condition:**
- According to the problem, the force exerted by the boy is one-third of that exerted by the man:
\[ R_B = \frac{1}{3} R_M \]

3. **Apply the Equilibrium Condition:**
- For vertical equilibrium, the sum of the upward forces must equal the sum of the downward forces:
\[ R_M + R_B = 800 + 100 \]
- Substituting \( R_B \) in terms of \( R_M \):
\[ R_M + \frac{1}{3} R_M = 900 \]
- This simplifies to:
\[ \frac{4}{3} R_M = 900 \implies R_M = 900 \times \frac{3}{4} = 675 \, \text{N} \]
- Therefore, substituting back to find \( R_B \):
\[ R_B = \frac{1}{3} R_M = \frac{1}{3} \times 675 = 225 \, \text{N} \]

4. **Set Up the Torque Equation:**
- To find \( x \), we will take moments about the position of the man (taking counterclockwise moments as positive):
\[ \text{Torque due to } 800 \, \text{N weight} + \text{Torque due to } 100 \, \text{N weight} = \text{Torque due to } R_B \]
- The equation becomes:
\[ 800 \cdot x + 100 \cdot \frac{L}{2} = R_B \cdot L \]
- Substituting \( R_B \):
\[ 800x + 50L = 225L \]
- Rearranging gives:
\[ 800x = 225L - 50L = 175L \]
- Therefore:
\[ x = \frac{175L}{800} = \frac{7L}{32} \]

5. **Calculate the Position:**
- To express \( x \) in terms of \( L \):
\[ x = 0.21875L \approx 0.22L \]

### Final Answer:
The weight must be hung at a distance of \( \frac{7L}{32} \) or approximately \( 0.22L \) from the man.