If two forces each of magnitude 20N, are acting on a particle such that the first force `(vecE_(1))` is acting along east direction and the second force `(vercF_(2))` is acting along the direction `60^(@)` north of west, then the magnitude of resultant force on particle is

Options

A. `20N`
B. `20sqrt(3)N`
C. `(20)/(sqrt(3))N`
D. None of these

Explanation

To find the magnitude of the resultant force acting on the particle, we will follow these steps:

### Step 1: Identify the Forces
We have two forces:
- Force \( \vec{F_1} = 20 \, \text{N} \) acting towards the east.
- Force \( \vec{F_2} = 20 \, \text{N} \) acting at an angle of \( 60^\circ \) north of west.

### Step 2: Resolve the Forces into Components
1. **Force \( \vec{F_1} \)**:
- \( F_{1x} = 20 \, \text{N} \) (east direction)
- \( F_{1y} = 0 \, \text{N} \)

2. **Force \( \vec{F_2} \)**:
- Since \( \vec{F_2} \) is acting \( 60^\circ \) north of west, we can resolve it into components:
- The angle with respect to the negative x-axis (west) is \( 60^\circ \).
- \( F_{2x} = -20 \cos(60^\circ) = -20 \times \frac{1}{2} = -10 \, \text{N} \) (west direction)
- \( F_{2y} = 20 \sin(60^\circ) = 20 \times \frac{\sqrt{3}}{2} = 10\sqrt{3} \, \text{N} \) (north direction)

### Step 3: Calculate the Resultant Components
Now, we can add the components of the two forces:
- **Resultant in the x-direction**:
\[ R_x = F_{1x} + F_{2x} = 20 + (-10) = 10 \, \text{N} \]

- **Resultant in the y-direction**:
\[ R_y = F_{1y} + F_{2y} = 0 + 10\sqrt{3} = 10\sqrt{3} \, \text{N} \]

### Step 4: Calculate the Magnitude of the Resultant Force
Using the Pythagorean theorem, we find the magnitude of the resultant force \( R \):
\[ R = \sqrt{R_x^2 + R_y^2} = \sqrt{(10)^2 + (10\sqrt{3})^2} \]
\[ R = \sqrt{100 + 300} = \sqrt{400} = 20 \, \text{N} \]

### Conclusion
The magnitude of the resultant force on the particle is \( 20 \, \text{N} \).

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