A weightless rod is acted upon by two upward parallel forces of `2 N` and `4 N` at ends `A` and `B` respectively. The total length of the rod `AB = 3 m`. To keep the rod in equilibrium a force of `6 N` should act in the following manner.

Options

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

Answer

To solve the problem of keeping the weightless rod in equilibrium under the action of the given forces, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Forces and their Positions:**
- There are two upward forces acting on the rod:
- Force at point A (2 N)
- Force at point B (4 N)
- The total length of the rod AB is 3 m.

2. **Define the Position of the Applied Force:**
- Let the position where the 6 N force is applied be at a distance \( x \) from point A. Consequently, the distance from point B to the point where the force is applied will be \( 3 - x \).

3. **Set Up the Moment Equation:**
- To maintain equilibrium, the sum of the moments about any point must be zero. We can choose point C (the point where the 6 N force is applied) to calculate the moments.
- The moment due to the force at point B (4 N) about point C is \( 4 \times (3 - x) \).
- The moment due to the force at point A (2 N) about point C is \( 2 \times x \).
- Setting the sum of the moments to zero gives us the equation:
\[ 4(3 - x) - 2x = 0 \]

4. **Solve the Moment Equation:**
- Expanding the equation:
\[ 12 - 4x - 2x = 0 \]
- Combine like terms:
\[ 12 - 6x = 0 \]
- Rearranging gives:
\[ 6x = 12 \]
- Dividing both sides by 6:
\[ x = 2 \]

5. **Determine the Position of the Force:**
- The force of 6 N should be applied at a distance of \( x = 2 \) m from point A. Therefore, it is located 1 m from point B (since \( 3 - 2 = 1 \)).

### Conclusion:
To keep the rod in equilibrium, the 6 N force should be applied downward at a point 2 m from point A (or 1 m from point B).

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