Four particles each of mass 'm' are placed at the four vertices of a square 'a' .Find net force on any one the particle.

Explanation

To find the net force on one of the particles placed at the vertices of a square, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Setup**:
We have four particles, each with mass \( m \), located at the vertices of a square with side length \( a \). Let's denote the vertices as A, B, C, and D.

2. **Choose a Particle**:
We will calculate the net force acting on particle C. The other particles (A, B, and D) will exert gravitational forces on C.

3. **Calculate Individual Forces**:
- **Force due to Particle D (F1)**:
The distance between particles C and D is \( a \).
\[ F_1 = \frac{G m^2}{a^2} \]
- **Force due to Particle B (F2)**:
The distance between particles C and B is also \( a \).
\[ F_2 = \frac{G m^2}{a^2} \]
- **Force due to Particle A (F3)**:
The distance between particles C and A is the diagonal of the square, which is \( \sqrt{2}a \).
\[ F_3 = \frac{G m^2}{( \sqrt{2}a )^2} = \frac{G m^2}{2a^2} \]

4. **Determine the Directions of Forces**:
- The forces \( F_1 \) and \( F_2 \) act along the sides of the square, while \( F_3 \) acts diagonally towards particle A.

5. **Calculate the Resultant of F1 and F2**:
Since \( F_1 \) and \( F_2 \) are equal and perpendicular to each other, we can find the resultant \( F_{12} \) using the Pythagorean theorem:
\[ F_{12} = \sqrt{F_1^2 + F_2^2} = \sqrt{F_1^2 + F_1^2} = \sqrt{2F_1^2} = F_1 \sqrt{2} \]

6. **Direction of the Resultant Force**:
The resultant \( F_{12} \) will be directed along the angle bisector of the angle between \( F_1 \) and \( F_2 \), which is also towards particle A.

7. **Calculate the Net Force on Particle C**:
The net force \( F_{net} \) on particle C is the vector sum of \( F_{12} \) and \( F_3 \). Since \( F_{12} \) is directed towards A and \( F_3 \) is also directed towards A, we can add them directly:
\[ F_{net} = F_{12} + F_3 = F_1 \sqrt{2} + \frac{G m^2}{2a^2} \]
Substituting \( F_1 \):
\[ F_{net} = \sqrt{2} \cdot \frac{G m^2}{a^2} + \frac{G m^2}{2a^2} \]
Factoring out \( \frac{G m^2}{a^2} \):
\[ F_{net} = \frac{G m^2}{a^2} \left( \sqrt{2} + \frac{1}{2} \right) \]

### Final Result:
The net force on particle C is:
\[ F_{net} = \frac{G m^2}{a^2} \left( \sqrt{2} + \frac{1}{2} \right) \]