Three particles each of mass m are placed at the three corners of an equilateral triangle of side L. A particle of mass M is kept at the midpoint of any one side. What is the force acting on M, due to this system of 3 particles ?

Options

A. `(GMm)/(4L^(2))`
B. `(3)/(4)(GMm)/(L^(2))`
C. `(4)/(3)(GMm)/(L^(2))`
D. `(4GMm)/(L^(2))`

Explanation

To solve the problem of finding the force acting on a particle of mass \( M \) placed at the midpoint of one side of an equilateral triangle formed by three particles each of mass \( m \), we will follow these steps:

### Step 1: Understand the Configuration
We have three particles of mass \( m \) located at the vertices of an equilateral triangle with side length \( L \). A fourth particle of mass \( M \) is placed at the midpoint of one of the sides of the triangle.

### Step 2: Identify the Positions
Let’s label the vertices of the triangle as \( A \), \( B \), and \( C \). The midpoint of side \( AB \) will be denoted as \( D \). The distances between the particles are as follows:
- Distance \( AD = BD = \frac{L}{2} \) (since \( D \) is the midpoint of \( AB \)).
- To find the distance \( CD \), we will use the Pythagorean theorem.

### Step 3: Calculate the Distance \( CD \)
Using the coordinates:
- \( A(0, 0) \)
- \( B(L, 0) \)
- \( C\left(\frac{L}{2}, \frac{\sqrt{3}}{2}L\right) \)
- \( D\left(\frac{L}{2}, 0\right) \)

The distance \( CD \) can be calculated as:
\[ CD = \sqrt{\left(\frac{L}{2} - \frac{L}{2}\right)^2 + \left(\frac{\sqrt{3}}{2}L - 0\right)^2} = \sqrt{0 + \left(\frac{\sqrt{3}}{2}L\right)^2} = \frac{\sqrt{3}}{2}L \]

### Step 4: Calculate the Forces Acting on \( M \)
The gravitational force between two masses is given by:
\[ F = \frac{G m_1 m_2}{r^2} \]
where \( G \) is the gravitational constant, \( m_1 \) and \( m_2 \) are the masses, and \( r \) is the distance between them.

1. **Force due to mass at \( A \)**:
\[ F_A = \frac{G M m}{\left(\frac{L}{2}\right)^2} = \frac{4GMm}{L^2} \]
This force acts vertically upwards towards \( A \).

2. **Force due to mass at \( B \)**:
\[ F_B = \frac{G M m}{\left(\frac{L}{2}\right)^2} = \frac{4GMm}{L^2} \]
This force acts vertically upwards towards \( B \).

3. **Force due to mass at \( C \)**:
\[ F_C = \frac{G M m}{\left(\frac{\sqrt{3}}{2}L\right)^2} = \frac{4GMm}{3L^2} \]
This force acts diagonally towards \( C \).

### Step 5: Resolve Forces
The forces \( F_A \) and \( F_B \) act in opposite directions and will cancel each other out since they are equal in magnitude. The net force acting on \( M \) will be due to the mass at \( C \).

### Step 6: Calculate the Resultant Force
The resultant force acting on \( M \) is simply the force due to mass \( C \):
\[ F_{net} = F_C = \frac{4GMm}{3L^2} \]

### Step 7: Direction of the Resultant Force
The direction of the resultant force is towards point \( C \).

### Final Answer
The force acting on the mass \( M \) due to the system of three particles is:
\[ F = \frac{4GMm}{3L^2} \quad \text{(towards point C)} \]