Three masses of 1 kg each are kept at the vertices of an equilateral triangle of side 3 m. What is the force acting on a point mass of 5 kg placed at its centroid ?

Explanation

To solve the problem of finding the net gravitational force acting on a point mass of 5 kg placed at the centroid of an equilateral triangle with three 1 kg masses at its vertices, we can follow these steps:

### Step 1: Understand the Setup
We have three masses (m1, m2, m3) of 1 kg each located at the vertices of an equilateral triangle with a side length of 3 m. The point mass (M) of 5 kg is placed at the centroid of the triangle.

### Step 2: Calculate the Distance from the Centroid to the Vertices
For an equilateral triangle, the distance (r) from the centroid to any vertex can be calculated using the formula:
\[ r = \frac{a}{\sqrt{3}} \]
where \( a \) is the side length of the triangle. Here, \( a = 3 \, \text{m} \).
\[ r = \frac{3}{\sqrt{3}} = \sqrt{3} \, \text{m} \approx 1.732 \, \text{m} \]

### Step 3: Calculate the Gravitational Force Exerted by Each Mass
The gravitational force \( F \) exerted by each mass on the point mass can be calculated using Newton's law of gravitation:
\[ F = \frac{G \cdot m \cdot M}{r^2} \]
where:
- \( G \) is the gravitational constant \( 6.674 \times 10^{-11} \, \text{N m}^2/\text{kg}^2 \)
- \( m = 1 \, \text{kg} \) (mass at the vertex)
- \( M = 5 \, \text{kg} \) (mass at the centroid)
- \( r = \sqrt{3} \, \text{m} \)

Substituting the values:
\[ F = \frac{6.674 \times 10^{-11} \cdot 1 \cdot 5}{(\sqrt{3})^2} \]
\[ F = \frac{6.674 \times 10^{-11} \cdot 5}{3} \]
\[ F = \frac{33.37 \times 10^{-11}}{3} \approx 11.12 \times 10^{-11} \, \text{N} \]

### Step 4: Analyze the Forces Acting on the Point Mass
Each of the three forces \( F_1, F_2, F_3 \) acts along the lines connecting the centroid to each vertex. Since the triangle is equilateral, the angle between any two forces is \( 120^\circ \).

### Step 5: Calculate the Resultant Force
To find the resultant force \( F_R \) acting on the point mass, we can use the vector addition of the forces. The forces have equal magnitudes and are separated by \( 120^\circ \).

Using the formula for the resultant of two vectors:
\[ F_R = \sqrt{F^2 + F^2 + 2F \cdot F \cdot \cos(120^\circ)} \]
Since \( \cos(120^\circ) = -\frac{1}{2} \):
\[ F_R = \sqrt{F^2 + F^2 - F^2} = \sqrt{F^2} = F \]

### Step 6: Determine the Net Force
Since the resultant of any two forces will be equal in magnitude to the third force and they are symmetrically arranged, the net force acting on the point mass at the centroid will be zero.

### Final Answer
The net force acting on the point mass of 5 kg placed at the centroid of the triangle is:
\[ \text{Net Force} = 0 \, \text{N} \]