Square of the resultant of two forces of equal magnitude is equal to three times the product of their magnitude. The angle between them is

Options

A. `0^(@)`
B. `45^(@)`
C. `60^(@)`
D. `90^(@)`

Answer

To find the angle between two forces of equal magnitude, given that the square of the resultant of these forces is equal to three times the product of their magnitudes, we can follow these steps:

### Step-by-Step Solution:

1. **Define the Forces**: Let the magnitudes of the two forces be \( F_1 \) and \( F_2 \). Since they are of equal magnitude, we can say \( F_1 = F_2 = F \).

2. **Resultant of Two Forces**: The resultant \( R \) of two forces acting at an angle \( \theta \) can be calculated using the formula:
\[ R = \sqrt{F_1^2 + F_2^2 + 2F_1F_2 \cos \theta} \]
Substituting \( F_1 \) and \( F_2 \) with \( F \):
\[ R = \sqrt{F^2 + F^2 + 2F \cdot F \cos \theta} = \sqrt{2F^2 + 2F^2 \cos \theta} \]

3. **Square of the Resultant**: Now, squaring the resultant:
\[ R^2 = 2F^2 + 2F^2 \cos \theta \]

4. **Given Condition**: According to the problem, the square of the resultant is equal to three times the product of the magnitudes of the forces:
\[ R^2 = 3F_1F_2 = 3F \cdot F = 3F^2 \]

5. **Setting the Equations Equal**: Now, we can set the two expressions for \( R^2 \) equal to each other:
\[ 2F^2 + 2F^2 \cos \theta = 3F^2 \]

6. **Simplifying the Equation**: Rearranging the equation gives:
\[ 2F^2 \cos \theta = 3F^2 - 2F^2 \]
\[ 2F^2 \cos \theta = F^2 \]

7. **Dividing by \( F^2 \)**: Assuming \( F^2 \neq 0 \), we can divide both sides by \( F^2 \):
\[ 2 \cos \theta = 1 \]

8. **Solving for \( \cos \theta \)**: This simplifies to:
\[ \cos \theta = \frac{1}{2} \]

9. **Finding the Angle**: The angle \( \theta \) that corresponds to \( \cos \theta = \frac{1}{2} \) is:
\[ \theta = 60^\circ \]

### Final Answer:
The angle between the two forces is \( 60^\circ \).