Two forces, each of magnitude F have a resultant of the same magnitude F. The angle between the two forces is

Options

A. `45^@`
B. `120^@`
C. `150^@`
D. `60^@`

Explanation

To find the angle between two forces of equal magnitude \( F \) that have a resultant of the same magnitude \( F \), we can use the formula for the resultant of two vectors.

### Step-by-Step Solution:

1. **Understanding the Problem**:
We have two forces, \( \vec{F_1} \) and \( \vec{F_2} \), each with a magnitude \( F \). The resultant of these two forces is also \( F \). We need to find the angle \( \theta \) between the two forces.

2. **Using the Resultant Formula**:
The formula for the resultant \( R \) of two vectors \( A \) and \( B \) at an angle \( \theta \) is given by:
\[ R = \sqrt{A^2 + B^2 + 2AB \cos(\theta)} \]
In our case, both forces have the same magnitude \( F \), so we can substitute \( A = F \) and \( B = F \):
\[ R = \sqrt{F^2 + F^2 + 2F \cdot F \cos(\theta)} \]

3. **Substituting the Known Values**:
Since the resultant \( R \) is also \( F \), we can set up the equation:
\[ F = \sqrt{F^2 + F^2 + 2F^2 \cos(\theta)} \]
Simplifying the equation:
\[ F = \sqrt{2F^2 + 2F^2 \cos(\theta)} \]

4. **Squaring Both Sides**:
To eliminate the square root, we square both sides:
\[ F^2 = 2F^2 + 2F^2 \cos(\theta) \]

5. **Rearranging the Equation**:
Rearranging gives:
\[ F^2 - 2F^2 = 2F^2 \cos(\theta) \]
\[ -F^2 = 2F^2 \cos(\theta) \]

6. **Dividing by \( F^2 \)**:
Dividing both sides by \( F^2 \) (assuming \( F \neq 0 \)):
\[ -1 = 2 \cos(\theta) \]

7. **Solving for \( \cos(\theta) \)**:
From the above equation, we find:
\[ \cos(\theta) = -\frac{1}{2} \]

8. **Finding the Angle**:
The angle \( \theta \) for which \( \cos(\theta) = -\frac{1}{2} \) is:
\[ \theta = 120^\circ \quad \text{(or \( 240^\circ \), but we consider the acute angle)} \]

### Final Answer:
The angle between the two forces is \( 120^\circ \).