Three equal masses of 1.5 kg each are fixed at the vertices of an equilateral triangle of side 1 m. What is the force acting on a particle of mass 1 kg placed at its centroid ?

Explanation

To solve the problem of finding the force acting on a particle of mass 1 kg placed at the centroid of an equilateral triangle with three equal masses of 1.5 kg each at its vertices, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Setup**:
We have three masses, each of 1.5 kg, located at the vertices of an equilateral triangle with a side length of 1 m. A particle of mass 1 kg is placed at the centroid of this triangle.

2. **Calculate the Distance from the Centroid to the Vertices**:
The distance from the centroid of an equilateral triangle to any of its vertices can be calculated using the formula:
\[ d = \frac{a}{\sqrt{3}} \]
where \(a\) is the side length of the triangle. For our triangle:
\[ d = \frac{1 \, \text{m}}{\sqrt{3}} \approx 0.577 \, \text{m} \]

3. **Calculate the Gravitational Force from Each Mass**:
The gravitational force \(F\) exerted by each mass on the 1 kg mass 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 \((6.674 \times 10^{-11} \, \text{N m}^2/\text{kg}^2)\),
- \(m_1 = 1.5 \, \text{kg}\) (mass at the vertex),
- \(m_2 = 1 \, \text{kg}\) (mass at the centroid),
- \(r = 0.577 \, \text{m}\) (distance from the centroid to the vertex).

Plugging in the values:
\[ F = \frac{(6.674 \times 10^{-11}) \cdot (1.5) \cdot (1)}{(0.577)^2} \approx \frac{(6.674 \times 10^{-11}) \cdot 1.5}{0.333} \approx 3.0 \times 10^{-10} \, \text{N} \]

4. **Determine the Direction of Forces**:
Each of the three forces acts along the line connecting the centroid to each vertex. The angles between the forces from different vertices are 120 degrees.

5. **Calculate the Net Force**:
Since the forces are equal in magnitude and symmetrically placed, we can analyze the components of these forces. The angle between any two forces is 120 degrees, which means the net force can be calculated using vector addition.

The resultant of three equal forces \(F\) at angles of 120 degrees can be shown to be zero due to symmetry. This can be confirmed by resolving the forces into components along the x and y axes. The components in one direction will cancel out the components in the opposite direction.

Thus, the net force acting on the mass at the centroid is:
\[ F_{\text{net}} = 0 \, \text{N} \]

### Final Answer:
The force acting on the particle of mass 1 kg placed at the centroid is **0 N**.