Three equal masses of 2kg each are placed at the vertices of an equilateral triangle and a mass of 4 kg is placed at the centroid of the triangle which is at a distance of `sqrt(2)` m from each of the vertices of the triangle. The force, (in newton) acting on the mass of 4 kg is

Options

A. 2
B. `sqrt(2)`
C. 1
D. zero

Explanation

To find the force acting on the mass of 4 kg placed at the centroid of the equilateral triangle formed by three 2 kg masses, we can follow these steps:

### Step 1: Understand the Configuration
We have three equal masses (2 kg each) located at the vertices of an equilateral triangle. A mass of 4 kg is placed at the centroid of the triangle. The distance from the centroid to each vertex is given as \( \sqrt{2} \) m.

### Step 2: Calculate the Gravitational Force Between Each Vertex Mass and the Centroid Mass
The gravitational force \( F \) between two masses can be calculated using Newton's law of gravitation:
\[ F = \frac{G \cdot m_1 \cdot m_2}{r^2} \]
where:
- \( G \) is the gravitational constant \( (6.674 \times 10^{-11} \, \text{N m}^2/\text{kg}^2) \)
- \( m_1 \) and \( m_2 \) are the masses (2 kg and 4 kg)
- \( r \) is the distance between the masses \( (\sqrt{2} \, \text{m}) \)

Substituting the values:
\[ F = \frac{(6.674 \times 10^{-11}) \cdot (2) \cdot (4)}{(\sqrt{2})^2} \]
\[ F = \frac{(6.674 \times 10^{-11}) \cdot 8}{2} \]
\[ F = \frac{53.392 \times 10^{-11}}{2} \]
\[ F = 26.696 \times 10^{-11} \, \text{N} \]

### Step 3: Analyze the Forces Acting on the 4 kg Mass
Each of the three 2 kg masses exerts a gravitational force on the 4 kg mass. Since the triangle is equilateral, the angles between the forces are 120 degrees.

### Step 4: Calculate the Resultant Force
To find the net force acting on the 4 kg mass, we can use vector addition. The forces from two of the masses will have components that can be resolved into horizontal and vertical directions.

1. The force from one mass can be represented as:
- \( F_1 = 26.696 \times 10^{-11} \, \text{N} \) at an angle of \( 0^\circ \)
- \( F_2 = 26.696 \times 10^{-11} \, \text{N} \) at an angle of \( 120^\circ \)
- \( F_3 = 26.696 \times 10^{-11} \, \text{N} \) at an angle of \( 240^\circ \)

2. The horizontal components of \( F_2 \) and \( F_3 \) will cancel out due to symmetry, and the vertical components will also cancel out.

### Step 5: Conclusion
Since the forces from the three masses are symmetrically arranged and cancel each other out, the net gravitational force acting on the mass of 4 kg at the centroid is:
\[ F_{\text{net}} = 0 \, \text{N} \]

### Final Answer
The force acting on the mass of 4 kg is \( 0 \, \text{N} \).

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