Two like parallel forces 5 Nand 15 N, act on a light rod at two points A and B respectively, 6 m apart. The resultant force and the distance of its point of action from the point A'are

Options

A. `10 N, 4.5 m`
B. `20 N, 4.5m`
C. `20 N, 1.5 m`
D. `10 N, 1.5 m `

Answer

To solve the problem step by step, we need to find the resultant force and the distance of its point of action from point A.

### Step 1: Identify the forces and their positions
We have two parallel forces acting on a light rod:
- A force of 5 N acting at point A.
- A force of 15 N acting at point B.
- The distance between points A and B is 6 m.

### Step 2: Calculate the resultant force
The resultant force (R) is the sum of the two forces acting in the same direction:
\[ R = F_1 + F_2 = 5 \, \text{N} + 15 \, \text{N} = 20 \, \text{N} \]

### Step 3: Set up the torque balance equation
To find the distance of the resultant force's point of action from point A, we can use the principle of moments (torque). We will take moments about point B. Let \( x \) be the distance from point A to the point of action of the resultant force.

The distance from point B to the point of action of the resultant force will then be \( 6 - x \).

The clockwise moment due to the 5 N force about point B is:
\[ \text{Clockwise Moment} = 5 \, \text{N} \times (6 - x) \]

The anticlockwise moment due to the 15 N force about point B is:
\[ \text{Anticlockwise Moment} = 15 \, \text{N} \times x \]

### Step 4: Set the moments equal to each other
For the system to be in equilibrium, the clockwise moment must equal the anticlockwise moment:
\[ 5(6 - x) = 15x \]

### Step 5: Solve for \( x \)
Expanding the equation:
\[ 30 - 5x = 15x \]
Combining like terms:
\[ 30 = 15x + 5x \]
\[ 30 = 20x \]
Dividing both sides by 20:
\[ x = \frac{30}{20} = 1.5 \, \text{m} \]

### Step 6: Calculate the distance from point A
Since \( x \) is the distance from point A to the point of action of the resultant force, we have:
\[ \text{Distance from A} = 1.5 \, \text{m} \]

### Final Result
The resultant force is 20 N, and the distance of its point of action from point A is 1.5 m.