Forces proportional to AB, BC and 2CA act along the sides of triangle ABC in order, their resultants represented in magnitude and direction is

Options

A. CA
B. AC
C. BC
D. CB

Answer

To solve the problem, we need to find the resultant of the forces acting along the sides of triangle ABC, which are proportional to the lengths of the sides AB, BC, and 2CA.

### Step-by-Step Solution:

1. **Identify the Forces**:
- Let the length of side AB = c, BC = a, and CA = b.
- The forces acting along the sides are:
- F1 (along AB) = k * c (where k is a proportionality constant)
- F2 (along BC) = k * a
- F3 (along CA) = 2k * b

2. **Determine the Directions of Forces**:
- F1 acts along AB.
- F2 acts along BC.
- F3 acts along CA but in the opposite direction (since it is acting along CA).

3. **Resolve the Forces into Components**:
- We can resolve these forces into their horizontal (x) and vertical (y) components.
- Assuming angle A is at vertex A, angle B at vertex B, and angle C at vertex C, we can use trigonometric functions to resolve the forces:
- F1:
- F1x = F1 * cos(A)
- F1y = F1 * sin(A)
- F2:
- F2x = F2 * cos(B)
- F2y = F2 * sin(B)
- F3 (acting in the opposite direction):
- F3x = -F3 * cos(C)
- F3y = -F3 * sin(C)

4. **Calculate the Resultant Components**:
- Sum the x-components:
- Rx = F1x + F2x + F3x
- Sum the y-components:
- Ry = F1y + F2y + F3y

5. **Calculate the Magnitude of the Resultant Force**:
- The magnitude of the resultant force R can be found using Pythagoras' theorem:
- R = √(Rx² + Ry²)

6. **Determine the Direction of the Resultant Force**:
- The direction (angle θ) of the resultant can be calculated using:
- θ = tan⁻¹(Ry/Rx)

### Final Result:
The resultant force R has a magnitude of √(Rx² + Ry²) and an angle θ with respect to the horizontal axis.