The forces each of `20N` act on a body at `120^(@)` The magnitude and direction of resultant is

Options

A. `20N,phi=60^(@)`
B. `20sqrt2N,phi=60^(@)`
C. `10sqrt2N,phi=0^(@)`
D. `10sqrt2N,phi=120^(@)`

Explanation

To solve the problem of finding the magnitude and direction of the resultant of two forces each of 20 N acting at an angle of 120 degrees to each other, we can follow these steps:

### Step 1: Identify the Forces and Angle
Let \( F_1 = 20 \, \text{N} \) and \( F_2 = 20 \, \text{N} \). The angle between the two forces is given as \( \theta = 120^\circ \).

### Step 2: Use the Law of Cosines to Find the Magnitude of the Resultant
The formula for the magnitude of the resultant \( R \) of two vectors is given by:
\[ R = \sqrt{F_1^2 + F_2^2 + 2F_1F_2 \cos \theta} \]
Substituting the values into the formula:
\[ R = \sqrt{20^2 + 20^2 + 2 \cdot 20 \cdot 20 \cdot \cos(120^\circ)} \]

### Step 3: Calculate Each Component
Calculating \( 20^2 \):
\[ 20^2 = 400 \]
Thus, we have:
\[ R = \sqrt{400 + 400 + 2 \cdot 20 \cdot 20 \cdot \cos(120^\circ)} \]
Now, calculate \( \cos(120^\circ) \):
\[ \cos(120^\circ) = -\frac{1}{2} \]
Substituting this value:
\[ R = \sqrt{400 + 400 + 2 \cdot 20 \cdot 20 \cdot \left(-\frac{1}{2}\right)} \]
Calculating the last term:
\[ 2 \cdot 20 \cdot 20 \cdot \left(-\frac{1}{2}\right) = -400 \]
So, we have:
\[ R = \sqrt{400 + 400 - 400} \]
\[ R = \sqrt{400} = 20 \, \text{N} \]

### Step 4: Determine the Direction of the Resultant
To find the direction of the resultant, we can use the sine rule or the tangent rule. The angle \( \phi \) of the resultant with respect to one of the forces can be calculated using:
\[ \tan \phi = \frac{F_2 \sin \theta}{F_1 + F_2 \cos \theta} \]
Substituting the values:
\[ \tan \phi = \frac{20 \sin(120^\circ)}{20 + 20 \cos(120^\circ)} \]
Calculating \( \sin(120^\circ) \):
\[ \sin(120^\circ) = \frac{\sqrt{3}}{2} \]
And substituting \( \cos(120^\circ) \):
\[ \tan \phi = \frac{20 \cdot \frac{\sqrt{3}}{2}}{20 + 20 \cdot \left(-\frac{1}{2}\right)} = \frac{10\sqrt{3}}{20 - 10} = \frac{10\sqrt{3}}{10} = \sqrt{3} \]
Thus, \( \phi = 60^\circ \).

### Step 5: Final Result
The magnitude of the resultant force is \( 20 \, \text{N} \) and the direction is \( 60^\circ \) from the direction of one of the forces.

### Summary
- Magnitude of Resultant: \( 20 \, \text{N} \)
- Direction of Resultant: \( 60^\circ \)