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 net gravitational force acting on the 4 kg mass placed at the centroid of an equilateral triangle formed by three 2 kg masses at its vertices, we can follow these steps:

### Step 1: Understand the Setup
We have three equal masses (m1 = m2 = m3 = 2 kg) located at the vertices of an equilateral triangle. The centroid of this triangle has a mass of 4 kg placed at it. The distance from each vertex to the centroid 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 is given by Newton's law of gravitation:
\[ F = \frac{G \cdot m_1 \cdot m_2}{r^2} \]
Where:
- \( G \) is the gravitational constant \( (6.674 \times 10^{-11} \, \text{N m}^2/\text{kg}^2) \)
- \( m_1 = 2 \, \text{kg} \) (mass at the vertex)
- \( m_2 = 4 \, \text{kg} \) (mass at the centroid)
- \( r = \sqrt{2} \, \text{m} \) (distance from vertex to centroid)

Substituting the values:
\[ F = \frac{(6.674 \times 10^{-11}) \cdot (2) \cdot (4)}{(\sqrt{2})^2} \]
\[ F = \frac{(6.674 \times 10^{-11}) \cdot 8}{2} = 2.6696 \times 10^{-10} \, \text{N} \]

### Step 3: Determine the Direction of the Forces
Each of the three forces \( F_{A}, F_{B}, F_{C} \) acting on the 4 kg mass from the three 2 kg masses will point towards the respective masses. Since the triangle is equilateral, the angles between these forces will be 120 degrees.

### Step 4: Resolve the Forces into Components
To find the net force, we can resolve these forces into x and y components. Since the forces are symmetrical, we can calculate the x and y components of one force and then multiply by three.

For one force \( F \):
- The x-component: \( F_x = F \cdot \cos(30^\circ) \)
- The y-component: \( F_y = F \cdot \sin(30^\circ) \)

Calculating these components:
\[ F_x = 2.6696 \times 10^{-10} \cdot \cos(30^\circ) = 2.6696 \times 10^{-10} \cdot \frac{\sqrt{3}}{2} \]
\[ F_y = 2.6696 \times 10^{-10} \cdot \sin(30^\circ) = 2.6696 \times 10^{-10} \cdot \frac{1}{2} \]

### Step 5: Calculate the Net Force Components
Since there are three forces, the net x-component \( F_{net,x} \) and net y-component \( F_{net,y} \) will be:
\[ F_{net,x} = 3 \cdot F_x \]
\[ F_{net,y} = 3 \cdot F_y \]

### Step 6: Calculate the Magnitude of the Net Force
The resultant force can be calculated using the Pythagorean theorem:
\[ F_{net} = \sqrt{(F_{net,x})^2 + (F_{net,y})^2} \]

### Step 7: Conclusion
After calculating the components and the resultant force, we find that the net force acting on the 4 kg mass at the centroid is zero due to the symmetry of the forces acting on it.

### Final Answer
The net force acting on the mass of 4 kg is **0 N**.

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