Three identical particles force exerted on one body due to the other two.

Answer

To solve the problem of finding the net force exerted on one body due to the other two identical particles placed at the corners of an equilateral triangle, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Setup**:
We have three identical particles, each of mass \( m \), placed at the corners of an equilateral triangle with side length \( a \).

2. **Determine the Gravitational Force Between Two Particles**:
The gravitational force \( F \) between any two particles is given by Newton's law of gravitation:
\[ F = \frac{G m^2}{a^2} \]
where \( G \) is the gravitational constant.

3. **Analyze Forces Acting on One Particle**:
Let's focus on one particle (let's call it Particle 1) and determine the forces acting on it due to the other two particles (Particle 2 and Particle 3).

4. **Resolve Forces into Components**:
The forces \( F_{12} \) (from Particle 2) and \( F_{13} \) (from Particle 3) can be resolved into horizontal and vertical components. Since the triangle is equilateral, the angle between the line connecting any two particles and the horizontal is \( 60^\circ \).

- The horizontal components of the forces are:
\[ F_{12x} = F_{12} \cos(60^\circ) = \frac{G m^2}{a^2} \cdot \frac{1}{2} = \frac{G m^2}{2a^2} \]
\[ F_{13x} = F_{13} \cos(60^\circ) = \frac{G m^2}{a^2} \cdot \frac{1}{2} = \frac{G m^2}{2a^2} \]

- The vertical components of the forces are:
\[ F_{12y} = F_{12} \sin(60^\circ) = \frac{G m^2}{a^2} \cdot \frac{\sqrt{3}}{2} = \frac{\sqrt{3} G m^2}{2a^2} \]
\[ F_{13y} = F_{13} \sin(60^\circ) = \frac{G m^2}{a^2} \cdot \frac{\sqrt{3}}{2} = \frac{\sqrt{3} G m^2}{2a^2} \]

5. **Calculate Net Forces**:
- The net horizontal force \( F_{net,x} \) is:
\[ F_{net,x} = F_{12x} + F_{13x} = \frac{G m^2}{2a^2} + \frac{G m^2}{2a^2} = \frac{G m^2}{a^2} \]
- The net vertical force \( F_{net,y} \) is:
\[ F_{net,y} = F_{12y} + F_{13y} = \frac{\sqrt{3} G m^2}{2a^2} + \frac{\sqrt{3} G m^2}{2a^2} = \sqrt{3} \frac{G m^2}{a^2} \]

6. **Calculate the Resultant Force**:
The resultant force \( F_{net} \) can be calculated using the Pythagorean theorem:
\[ F_{net} = \sqrt{F_{net,x}^2 + F_{net,y}^2} = \sqrt{\left(\frac{G m^2}{a^2}\right)^2 + \left(\sqrt{3} \frac{G m^2}{a^2}\right)^2} \]
\[ = \sqrt{\frac{G^2 m^4}{a^4} + \frac{3 G^2 m^4}{a^4}} = \sqrt{\frac{4 G^2 m^4}{a^4}} = \frac{2 G m^2}{a^2} \]

7. **Final Result**:
The net force exerted on one particle due to the other two is:
\[ F_{net} = \frac{2 G m^2}{a^2} \]