A weightless rod is acted upon by upward parallel forces of 2 newton and 4 newton magnitudes at ends A and B respectively. The total length of the rod AB = 3metere. TO keep the rod in equilibrium a force of 6 newton act in the following manner.

Options

A. downward at any point between A and B
B. downward at the mid-point of AB
C. downward at a point C such that AC = 1 metre
D. downward at a point D such that BD = 1 metre

Explanation

To solve the problem step by step, we need to analyze the forces acting on the rod and apply the conditions for equilibrium.

### Step 1: Identify the Forces Acting on the Rod
- At point A, there is an upward force of 2 N.
- At point B, there is an upward force of 4 N.
- To maintain equilibrium, we need to introduce a downward force of 6 N.

### Step 2: Set Up the Equilibrium Conditions
For the rod to be in equilibrium, the net force and net torque acting on it must both be zero.

1. **Net Force Condition**:
\[ F_{\text{up}} - F_{\text{down}} = 0 \]
Here, \( F_{\text{up}} = 2 \, \text{N} + 4 \, \text{N} = 6 \, \text{N} \) and \( F_{\text{down}} = 6 \, \text{N} \). Thus, the net force is zero.

2. **Net Torque Condition**:
We can take moments about point A or point B. Here, we will take moments about point A.

### Step 3: Calculate the Torque About Point A
- The torque due to the 4 N force at point B (which is 3 m from A):
\[ \tau_{B} = 4 \, \text{N} \times 3 \, \text{m} = 12 \, \text{N m} \, (\text{clockwise}) \]
- The torque due to the 6 N force (let's say it is at a distance \( x \) from point B, which means it is \( 3 - x \) from point A):
\[ \tau_{6N} = 6 \, \text{N} \times (3 - x) \, \text{m} \, (\text{counterclockwise}) \]

### Step 4: Set Up the Torque Equation
For equilibrium, the sum of the torques about point A must be zero:
\[ \tau_{6N} - \tau_{B} = 0 \]
Substituting the values:
\[ 6(3 - x) - 12 = 0 \]

### Step 5: Solve for \( x \)
Expanding the equation:
\[ 18 - 6x - 12 = 0 \]
\[ 6 - 6x = 0 \]
\[ 6x = 6 \implies x = 1 \, \text{m} \]

### Step 6: Determine the Position of the 6 N Force
Since \( x = 1 \, \text{m} \) is the distance from point B, the distance from point A is:
\[ 3 - x = 3 - 1 = 2 \, \text{m} \]

### Conclusion
The 6 N force acts downward at a point that is 1 meter from B and 2 meters from A. Therefore, the correct answer is that the force acts downward at point D such that BD = 1 meter.