100 coplanar forces each equal to 10 n act on a body. Each force makes angle `pi//50` with the preceding force. What is the resultant of the forces.

Options

A. 1000 N
B. 500 N
C. 250 N
D. Zero

Answer

To find the resultant of 100 coplanar forces, each of 10 N, acting at an angle of \(\frac{\pi}{50}\) radians with respect to the preceding force, we can use the concept of vector addition.

### Step-by-Step Solution:

1. **Identify the Forces**:
Each force \( F \) is given as 10 N, and there are 100 such forces.

2. **Determine the Angle Between Forces**:
The angle between each consecutive force is \(\theta = \frac{\pi}{50}\) radians.

3. **Calculate the Resultant Force**:
The forces can be represented as vectors in a plane. The resultant \( R \) of these forces can be calculated using the formula for the resultant of forces acting at an angle:
\[ R = n \cdot F \cdot \cos\left(\frac{(n-1) \theta}{2}\right) \]
where \( n \) is the number of forces, \( F \) is the magnitude of each force, and \( \theta \) is the angle between the forces.

4. **Substituting Values**:
Here, \( n = 100 \), \( F = 10 \, \text{N} \), and \( \theta = \frac{\pi}{50} \):
\[ R = 100 \cdot 10 \cdot \cos\left(\frac{(100-1) \cdot \frac{\pi}{50}}{2}\right) \]
Simplifying the angle:
\[ R = 1000 \cdot \cos\left(\frac{99 \cdot \frac{\pi}{50}}{2}\right) \]
\[ R = 1000 \cdot \cos\left(\frac{99\pi}{100}\right) \]

5. **Calculate the Cosine**:
The cosine of \(\frac{99\pi}{100}\) can be calculated:
\[ \cos\left(\frac{99\pi}{100}\right) = -\cos\left(\frac{\pi}{100}\right) \]
Thus:
\[ R = 1000 \cdot -\cos\left(\frac{\pi}{100}\right) \]

6. **Final Result**:
The resultant force \( R \) is negative, indicating that the resultant vector points in the opposite direction of the initial forces. The magnitude can be calculated using a calculator:
\[ R \approx 1000 \cdot -0.99985 \approx -999.85 \, \text{N} \]
Therefore, the magnitude of the resultant force is approximately \( 999.85 \, \text{N} \).

### Final Answer:
The resultant of the 100 coplanar forces is approximately \( 999.85 \, \text{N} \) in the direction opposite to the initial forces.