A thin uniform rod of mass m moves translationally with acceleration a due to two antiparallel forces of level arm I. One force is of magnitude F and acts at one extreme end. The length of the rod is

Options

A. `(mal)/(ma+F)`
B. `(2(F+ma)l)/(ma)`
C. `l(l+F/(ma))`
D. `((F+ma)l)/(2ma)`

Explanation

To solve the problem, we need to find the length of the rod (L) given the forces acting on it and the acceleration. Let's break down the solution step by step.

### Step-by-Step Solution:

1. **Understanding the Forces**:
- We have a thin uniform rod of mass \( m \).
- Two forces are acting on the rod:
- Force \( F \) acting at one extreme end.
- A second force \( F_1 \) acting at the other end, which is greater than \( F \).
- The lever arm (distance between the points where the forces are applied) is denoted as \( l \).

2. **Net Force and Acceleration**:
- Since the rod is moving with a translational acceleration \( a \), the net force acting on the rod can be expressed as:
\[ F_1 - F = ma \]
- Rearranging gives us:
\[ F_1 = F + ma \]

3. **Torque Consideration**:
- For the rod to have only translational motion and no rotational motion, the torques about the center of mass must balance out.
- The torque due to force \( F \) about the center of mass (which is at \( L/2 \)) is:
\[ \tau_F = F \cdot \left(\frac{L}{2}\right) \]
- The torque due to force \( F_1 \) is:
\[ \tau_{F_1} = F_1 \cdot \left(\frac{L}{2}\right) \]
- Setting the torques equal gives us:
\[ F \cdot \left(\frac{L}{2}\right) = F_1 \cdot \left(\frac{L}{2} - l\right) \]

4. **Substituting \( F_1 \)**:
- Substitute \( F_1 = F + ma \) into the torque equation:
\[ F \cdot \left(\frac{L}{2}\right) = (F + ma) \cdot \left(\frac{L}{2} - l\right) \]

5. **Solving for \( L \)**:
- Expanding both sides:
\[ F \cdot \frac{L}{2} = (F + ma) \cdot \left(\frac{L}{2}\right) - (F + ma) \cdot l \]
- Rearranging gives:
\[ F \cdot \frac{L}{2} - (F + ma) \cdot \frac{L}{2} = -(F + ma) \cdot l \]
- Factor out \( L/2 \):
\[ \left(F - (F + ma)\right) \cdot \frac{L}{2} = -(F + ma) \cdot l \]
- Simplifying:
\[ -ma \cdot \frac{L}{2} = -(F + ma) \cdot l \]
- Thus:
\[ ma \cdot \frac{L}{2} = (F + ma) \cdot l \]
- Solving for \( L \):
\[ L = \frac{2(F + ma)l}{ma} \]

6. **Final Expression**:
- Therefore, the length of the rod \( L \) is:
\[ L = \frac{2(F + ma)l}{ma} \]