Four particles each of mass 'm' are placed at the four vertices of a square 'a' .Find net force on any one the particle.

Explanation

To find the net gravitational force acting on one of the particles located at the vertices of a square, we can follow these steps:

### Step 1: Identify the Configuration
Consider a square with vertices labeled A, B, C, and D, where each vertex has a particle of mass \( m \). We will calculate the net force acting on particle D due to particles A, B, and C.

### Step 2: Calculate Forces Due to Each Particle
1. **Force on D due to A (F_DA)**:
- The distance between particles D and A is equal to the side of the square, \( a \).
- The gravitational force is given by:
\[ F_{DA} = \frac{G m^2}{a^2} \]

2. **Force on D due to B (F_DB)**:
- The distance between particles D and B is also equal to the side of the square, \( a \).
- The gravitational force is:
\[ F_{DB} = \frac{G m^2}{a^2} \]

3. **Force on D due to C (F_DC)**:
- The distance between particles D and C is the diagonal of the square, which can be calculated using the Pythagorean theorem:
\[ \text{Distance} = \sqrt{a^2 + a^2} = a\sqrt{2} \]
- The gravitational force is:
\[ F_{DC} = \frac{G m^2}{(a\sqrt{2})^2} = \frac{G m^2}{2a^2} \]

### Step 3: Resolve Forces into Components
Next, we need to resolve the forces \( F_{DA} \), \( F_{DB} \), and \( F_{DC} \) into their x and y components.

1. **Components of F_DA**:
- \( F_{DA,x} = F_{DA} \cdot \cos(45^\circ) = \frac{G m^2}{a^2} \cdot \frac{1}{\sqrt{2}} = \frac{G m^2}{a^2\sqrt{2}} \)
- \( F_{DA,y} = F_{DA} \cdot \sin(45^\circ) = \frac{G m^2}{a^2} \cdot \frac{1}{\sqrt{2}} = \frac{G m^2}{a^2\sqrt{2}} \)

2. **Components of F_DB**:
- \( F_{DB,x} = 0 \) (since it acts vertically downwards)
- \( F_{DB,y} = -F_{DB} = -\frac{G m^2}{a^2} \)

3. **Components of F_DC**:
- \( F_{DC,x} = -F_{DC} \cdot \cos(45^\circ) = -\frac{G m^2}{2a^2} \cdot \frac{1}{\sqrt{2}} = -\frac{G m^2}{2a^2\sqrt{2}} \)
- \( F_{DC,y} = -F_{DC} \cdot \sin(45^\circ) = -\frac{G m^2}{2a^2} \cdot \frac{1}{\sqrt{2}} = -\frac{G m^2}{2a^2\sqrt{2}} \)

### Step 4: Sum the Components
Now, we sum the x and y components to find the net force components acting on particle D.

1. **Net Force in the x-direction (F_net,x)**:
\[ F_{\text{net},x} = F_{DA,x} + F_{DB,x} + F_{DC,x} = \frac{G m^2}{a^2\sqrt{2}} + 0 - \frac{G m^2}{2a^2\sqrt{2}} = \frac{G m^2}{a^2\sqrt{2}} - \frac{G m^2}{2a^2\sqrt{2}} = \frac{G m^2}{2a^2\sqrt{2}} \]

2. **Net Force in the y-direction (F_net,y)**:
\[ F_{\text{net},y} = F_{DA,y} + F_{DB,y} + F_{DC,y} = \frac{G m^2}{a^2\sqrt{2}} - \frac{G m^2}{a^2} - \frac{G m^2}{2a^2\sqrt{2}} \]
\[ = \frac{G m^2}{a^2\sqrt{2}} - \frac{G m^2\sqrt{2}}{2a^2\sqrt{2}} = \frac{G m^2}{a^2\sqrt{2}} - \frac{G m^2}{2a^2} = \frac{G m^2}{2a^2\sqrt{2}} \]

### Step 5: Calculate the Magnitude of the Net Force
Finally, we can calculate the magnitude of the net force using the Pythagorean theorem:
\[ F_{\text{net}} = \sqrt{(F_{\text{net},x})^2 + (F_{\text{net},y})^2} \]
\[ = \sqrt{\left(\frac{G m^2}{2a^2\sqrt{2}}\right)^2 + \left(\frac{G m^2}{2a^2\sqrt{2}}\right)^2} \]
\[ = \sqrt{2 \left(\frac{G m^2}{2a^2\sqrt{2}}\right)^2} = \frac{G m^2}{2a^2\sqrt{2}} \sqrt{2} = \frac{G m^2}{2a^2} \]

### Final Result
Thus, the net force acting on any one of the particles is:
\[ F_{\text{net}} = \frac{G m^2}{2a^2} \]