In the given figure, find the force acting on a particle of mass 1 kg.

Explanation

To find the force acting on a particle of mass 1 kg in the given configuration, we will analyze the gravitational forces acting on it due to the other masses around it.

### Step-by-Step Solution:

1. **Identify the Masses and Distances**:
- We have a mass of 1 kg (let's call it \( m_1 \)) and it is surrounded by other masses: two 2 kg masses (let's call them \( m_2 \)) and another 1 kg mass (let's call it \( m_3 \)).
- The distances between the masses are given as follows: the distance between \( m_1 \) and \( m_2 \) is 1 meter, and the distance between \( m_1 \) and \( m_3 \) is also 1 meter.

2. **Calculate the Gravitational Force**:
- The gravitational force between two masses is given by Newton's law of gravitation:
\[ F = \frac{G \cdot m_1 \cdot m_2}{r^2} \]
- Here, \( G \) is the gravitational constant, \( m_1 \) is the mass of the particle (1 kg), \( m_2 \) is the mass of the other particles (2 kg), and \( r \) is the distance between them (1 meter).

3. **Calculate Forces in the X-Direction**:
- For the forces acting on \( m_1 \) due to the two 2 kg masses:
- The force from one 2 kg mass (let's say to the right) is:
\[ F_{x1} = \frac{G \cdot 1 \cdot 2}{1^2} = 2G \]
- The force from the other 2 kg mass (to the left) is:
\[ F_{x2} = \frac{G \cdot 1 \cdot 2}{1^2} = 2G \]
- Since these forces are equal and opposite, they cancel each other out:
\[ F_{x \text{ net}} = 2G - 2G = 0 \]

4. **Calculate Forces in the Y-Direction**:
- Now consider the forces acting in the y-direction:
- The force from the 1 kg mass (let's say above) is:
\[ F_{y1} = \frac{G \cdot 1 \cdot 1}{1^2} = G \]
- The force from the other 1 kg mass (below) is:
\[ F_{y2} = \frac{G \cdot 1 \cdot 1}{1^2} = G \]
- Again, these forces are equal and opposite, so they cancel each other out:
\[ F_{y \text{ net}} = G - G = 0 \]

5. **Calculate the Net Force**:
- Since both the net forces in the x-direction and y-direction are zero:
\[ F_{\text{net}} = F_{x \text{ net}} + F_{y \text{ net}} = 0 + 0 = 0 \]

### Conclusion:
The net force acting on the particle of mass 1 kg is **0 N**.