A particle of small m is joined to a very heavy body by a light string passing over a light pulley. Both bodies are free to move. The total downward force on the pulley is

Options

A. `gt gtmg`
B. `4mg`
C. `2mg`
D. `mg`

Explanation

To solve the problem of finding the total downward force on the pulley when a small mass \( m \) is connected to a very heavy mass \( M \) by a light string over a light pulley, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the System**: We have two masses: a small mass \( m \) and a very heavy mass \( M \). The small mass is hanging on one side of the pulley, and the heavy mass is on the other side.

2. **Draw the Free Body Diagram**:
- For mass \( m \), the forces acting on it are:
- Downward force due to gravity: \( mg \)
- Upward tension in the string: \( T \)
- For mass \( M \), the forces acting on it are:
- Downward force due to gravity: \( Mg \)
- Upward tension in the string: \( T \)

3. **Set Up the Equations of Motion**:
- For mass \( m \):
\[ T - mg = ma \quad \text{(1)} \]
- For mass \( M \):
\[ Mg - T = Ma \quad \text{(2)} \]

4. **Add the Two Equations**:
- Adding equations (1) and (2):
\[ (T - mg) + (Mg - T) = ma + Ma \]
- This simplifies to:
\[ Mg - mg = (M + m)a \]
- Rearranging gives:
\[ a = \frac{Mg - mg}{M + m} \quad \text{(3)} \]

5. **Find the Tension \( T \)**:
- Substitute \( a \) from equation (3) into equation (1):
\[ T = mg + ma \]
- Substitute \( a \):
\[ T = mg + m \left(\frac{Mg - mg}{M + m}\right) \]
- Simplifying this:
\[ T = mg + \frac{m(Mg - mg)}{M + m} \]
- Factor out \( mg \):
\[ T = mg + \frac{mMg - m^2g}{M + m} \]
- Combine the terms:
\[ T = \frac{mg(M + m) + mMg - m^2g}{M + m} \]
- This simplifies to:
\[ T = \frac{2mgM}{M + m} \]

6. **Approximate for Very Heavy Mass**:
- Since \( M \) is very large compared to \( m \) (i.e., \( m \ll M \)), we can approximate:
\[ T \approx 2mg \]

7. **Calculate Total Downward Force on the Pulley**:
- The total downward force on the pulley is the sum of the tensions from both sides:
\[ F_{total} = 2T = 2 \times 2mg = 4mg \]

### Final Answer:
The total downward force on the pulley is \( 4mg \).