The direction of three forces 1N, 2N and 3N acting at a point,are parallel to the sides of an equilateral triangle taken in order. The magnitude of their resultant is:

Options

A. `sqrt(3) N`
B. `sqrt(3)/(2)N`
C. `(3)/(2)N`
D. Zero

Answer

To find the resultant of the three forces acting at a point, we will follow these steps:

### Step 1: Identify the Forces and Their Directions
The forces acting at a point are:
- \( F_1 = 1 \, \text{N} \)
- \( F_2 = 2 \, \text{N} \)
- \( F_3 = 3 \, \text{N} \)

These forces are directed along the sides of an equilateral triangle. Let’s assume:
- \( F_1 \) acts along the first side.
- \( F_2 \) acts along the second side.
- \( F_3 \) acts along the third side.

### Step 2: Resolve the Forces into Components
Since the angles in an equilateral triangle are \( 60^\circ \), we can resolve each force into its horizontal (x-axis) and vertical (y-axis) components.

1. **Force \( F_1 = 1 \, \text{N} \)**:
- \( F_{1x} = 1 \cos(0^\circ) = 1 \)
- \( F_{1y} = 1 \sin(0^\circ) = 0 \)

2. **Force \( F_2 = 2 \, \text{N} \)** (acting at \( 60^\circ \)):
- \( F_{2x} = 2 \cos(60^\circ) = 2 \times \frac{1}{2} = 1 \)
- \( F_{2y} = 2 \sin(60^\circ) = 2 \times \frac{\sqrt{3}}{2} = \sqrt{3} \)

3. **Force \( F_3 = 3 \, \text{N} \)** (acting at \( 120^\circ \)):
- \( F_{3x} = 3 \cos(120^\circ) = 3 \times \left(-\frac{1}{2}\right) = -\frac{3}{2} \)
- \( F_{3y} = 3 \sin(120^\circ) = 3 \times \frac{\sqrt{3}}{2} = \frac{3\sqrt{3}}{2} \)

### Step 3: Sum the Components
Now, we can find the resultant components by summing the x and y components:

- **Resultant in x-direction**:
\[ R_x = F_{1x} + F_{2x} + F_{3x} = 1 + 1 - \frac{3}{2} = 2 - 1.5 = 0.5 \]

- **Resultant in y-direction**:
\[ R_y = F_{1y} + F_{2y} + F_{3y} = 0 + \sqrt{3} + \frac{3\sqrt{3}}{2} = \frac{2\sqrt{3}}{2} + \frac{3\sqrt{3}}{2} = \frac{5\sqrt{3}}{2} \]

### Step 4: Calculate the Magnitude of the Resultant Force
The magnitude of the resultant force \( R \) can be calculated using the Pythagorean theorem:
\[ R = \sqrt{R_x^2 + R_y^2} = \sqrt{(0.5)^2 + \left(\frac{5\sqrt{3}}{2}\right)^2} \]
Calculating this:
\[ R = \sqrt{0.25 + \frac{75}{4}} = \sqrt{\frac{0.25 + 75}{4}} = \sqrt{\frac{75.25}{4}} = \frac{\sqrt{75.25}}{2} \]

### Step 5: Final Result
The resultant force is approximately \( R \approx 3.5 \, \text{N} \) (after calculating \( \sqrt{75.25} \)).