A weightless rod is acted on by upward parallel forces of `2 N` and `4N` ends `A` and `B` respectively. The total leng th of the rod `AB=3m`. To keep the rod in equilibrium a fo rce of `6 N` should act in the following manner :

Options

A. Downwards at any point between `A` and `B`.
B. Downward at mid point of `AB`.
C. Downward at a point `C` such that `AC = 1 m`.
D. Downward at a point `D` such that `BD = 1 m`.

Explanation

To solve the problem of keeping the weightless rod in equilibrium under the influence of the given forces, we need to analyze the forces and torques acting on the rod.

### Step-by-step Solution:

1. **Identify the Forces Acting on the Rod:**
- A force of `2 N` acts upward at point `A` (left end of the rod).
- A force of `4 N` acts upward at point `B` (right end of the rod).
- We need to apply a force of `6 N` at a distance `x` from point `B`.

2. **Determine the Length of the Rod:**
- The total length of the rod `AB` is given as `3 m`.

3. **Set Up the Torque Equation:**
- To keep the rod in rotational equilibrium, the sum of the torques about any point must be zero. We will take torques about point `B` for convenience.
- The torque due to the `2 N` force at point `A` (which is `3 m` from point `B`) is:
\[ \text{Torque from } A = 2 N \times 3 m = 6 N \cdot m \text{ (counterclockwise)} \]
- The torque due to the `6 N` force applied at a distance `x` from point `B` (which is `3 - x` meters from point `A`) is:
\[ \text{Torque from } 6 N = 6 N \times (3 - x) \text{ (clockwise)} \]

4. **Set the Torques Equal for Equilibrium:**
- For the rod to be in equilibrium, the torques must balance:
\[ 6 N \cdot m = 6 N \times (3 - x) \]

5. **Solve for `x`:**
- Simplifying the equation:
\[ 6 = 6 \times (3 - x) \]
\[ 1 = 3 - x \]
\[ x = 3 - 1 = 2 \text{ m} \]

6. **Conclusion:**
- The force of `6 N` should be applied `2 m` from point `B` to keep the rod in equilibrium.