Three point masses 'm' each are placed at the three vertices of an equilateral traingle of side 'a'. Find net gravitational force on any point mass.

Explanation

To find the net gravitational force on any point mass located at one of the vertices of an equilateral triangle formed by three point masses, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Configuration**:
- We have three point masses, each of mass \( m \), located at the vertices of an equilateral triangle with side length \( a \).
- Let's denote the vertices as \( A \), \( B \), and \( C \).

2. **Choose a Point Mass**:
- We will calculate the net gravitational force acting on mass \( B \) due to masses \( A \) and \( C \).

3. **Calculate the Gravitational Force Between Masses**:
- The gravitational force \( F \) between two point masses is given by Newton's law of gravitation:
\[ F = \frac{G m_1 m_2}{r^2} \]
- For the force \( F_1 \) exerted on mass \( B \) by mass \( C \):
\[ F_1 = \frac{G m^2}{a^2} \]
- For the force \( F_2 \) exerted on mass \( B \) by mass \( A \):
\[ F_2 = \frac{G m^2}{a^2} \]

4. **Determine the Angle Between Forces**:
- Since the triangle is equilateral, the angle between the forces \( F_1 \) and \( F_2 \) is \( 60^\circ \).

5. **Calculate the Net Gravitational Force**:
- The net gravitational force \( F_{\text{net}} \) can be found using the law of cosines:
\[ F_{\text{net}} = \sqrt{F_1^2 + F_2^2 + 2 F_1 F_2 \cos(60^\circ)} \]
- Since \( F_1 = F_2 \):
\[ F_{\text{net}} = \sqrt{F_1^2 + F_1^2 + 2 F_1^2 \cdot \frac{1}{2}} = \sqrt{2F_1^2 + F_1^2} = \sqrt{3F_1^2} = \sqrt{3} F_1 \]
- Substituting \( F_1 \):
\[ F_{\text{net}} = \sqrt{3} \cdot \frac{G m^2}{a^2} \]

6. **Direction of the Net Force**:
- The direction of the net force will be along the angle bisector of the angle formed by the lines connecting \( B \) to \( A \) and \( C \). Since both forces are equal in magnitude and the angle between them is \( 60^\circ \), the resultant force will make an angle of \( 30^\circ \) with each of the lines \( BA \) and \( BC \).

### Final Result:
The net gravitational force on mass \( B \) due to masses \( A \) and \( C \) is:
\[ F_{\text{net}} = \sqrt{3} \cdot \frac{G m^2}{a^2} \]
and it acts along the angle bisector of \( \angle ABC \).