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

Options

A. `(2(F+ma)l)/(ma)`
B. `l(1+(F)/(ma))`
C. `((F+ma)l)/(2ma)`
D. `(mal)/(ma+F)`

Answer

To solve the problem of finding the length of a thin uniform rod subjected to two antiparallel forces, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Forces and Setup**:
- Let the rod have a mass \( m \) and a length \( D \).
- One force \( F \) acts at one end of the rod, and the other force \( F' \) acts at the opposite end.
- The lever arm (distance between the two forces) is given as \( l \).

2. **Apply Newton's Second Law**:
- The net force acting on the rod is given by:
\[ F' - F = m a \]
- Rearranging gives:
\[ F' = F + m a \]

3. **Calculate the Torque**:
- The torque due to force \( F' \) about the center of the rod (which is at \( D/2 \)) is:
\[ \tau' = F' \cdot \left(\frac{D}{2} - l\right) \]
- The torque due to force \( F \) about the center of the rod is:
\[ \tau = F \cdot \left(\frac{D}{2}\right) \]

4. **Set the Net Torque to Zero**:
- Since the rod is in translational motion and not rotating, the net torque must equal zero:
\[ \tau' - \tau = 0 \]
- This gives:
\[ F' \cdot \left(\frac{D}{2} - l\right) = F \cdot \left(\frac{D}{2}\right) \]

5. **Substituting for \( F' \)**:
- Substitute \( F' = F + m a \) into the torque equation:
\[ (F + ma) \cdot \left(\frac{D}{2} - l\right) = F \cdot \left(\frac{D}{2}\right) \]

6. **Expand and Rearrange**:
- Expand the left side:
\[ F \cdot \left(\frac{D}{2} - l\right) + ma \cdot \left(\frac{D}{2} - l\right) = F \cdot \left(\frac{D}{2}\right) \]
- Rearranging gives:
\[ ma \cdot \left(\frac{D}{2} - l\right) = F \cdot l \]

7. **Solve for the Length \( D \)**:
- Isolate \( D \):
\[ D = \frac{2Fl}{ma} + 2l \]

8. **Final Expression**:
- Thus, the length of the rod \( D \) can be expressed as:
\[ D = \frac{2(F + ma)l}{ma} \]

### Conclusion:
The length of the rod \( D \) is given by the formula:
\[ D = \frac{2(F + ma)l}{ma} \]