Find the vector sum of `N` coplanar forces, each of the magnitude `F`,when each force makes an angle of `2pi//N` with that preceding it.

Explanation

To find the vector sum of \( N \) coplanar forces, each of magnitude \( F \), where each force makes an angle of \( \frac{2\pi}{N} \) with the preceding one, we can follow these steps:

### Step-by-Step Solution:

1. **Understanding the Arrangement of Forces**:
Each force \( F \) is represented as a vector in a plane, and they are arranged such that the angle between each consecutive force is \( \frac{2\pi}{N} \). This means that if you visualize these forces, they will form a regular polygon (specifically, a regular \( N \)-gon) when drawn from a common point.

2. **Expressing Forces in Component Form**:
We can express each force in terms of its components. The \( i^{th} \) force \( \vec{F_i} \) can be expressed as:
\[ \vec{F_i} = F \cos\left(\frac{2\pi(i-1)}{N}\right) \hat{i} + F \sin\left(\frac{2\pi(i-1)}{N}\right) \hat{j} \]
where \( i = 1, 2, \ldots, N \).

3. **Calculating the Resultant Force**:
To find the resultant force, we need to sum up all the components in the \( x \) and \( y \) directions separately.
- **Sum of \( x \)-components**:
\[ R_x = \sum_{i=1}^{N} F \cos\left(\frac{2\pi(i-1)}{N}\right) \]
- **Sum of \( y \)-components**:
\[ R_y = \sum_{i=1}^{N} F \sin\left(\frac{2\pi(i-1)}{N}\right) \]

4. **Using Symmetry**:
Due to the symmetry of the arrangement (a regular polygon), the sum of the \( x \)-components and the sum of the \( y \)-components will both equal zero:
\[ R_x = 0 \quad \text{and} \quad R_y = 0 \]

5. **Conclusion**:
Since both components of the resultant force are zero, the vector sum of all \( N \) coplanar forces is:
\[ \vec{R} = R_x \hat{i} + R_y \hat{j} = 0 \hat{i} + 0 \hat{j} = \vec{0} \]

### Final Answer:
The vector sum of \( N \) coplanar forces, each of magnitude \( F \), when each force makes an angle of \( \frac{2\pi}{N} \) with the preceding one, is \( \vec{0} \).

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