Two non-collinear forces, one of 10 N and another of 6 N act upon a body. The directions of the forces are unknown. The resultant force on the body is :

Options

A. between 6 and 10 N
B. between 4 and 16 N
C. more than 6 N
D. more than 10 N

Explanation

To find the resultant force acting on a body due to two non-collinear forces of magnitudes 10 N and 6 N, we can use the concept of vector addition. The resultant force will depend on the angle between the two forces. Here’s how we can calculate it step by step:

### Step-by-Step Solution:

1. **Identify the Forces**:
- Let \( F_1 = 10 \, \text{N} \)
- Let \( F_2 = 6 \, \text{N} \)

2. **Use the Formula for Resultant Force**:
The magnitude of the resultant force \( R \) when two forces are acting at an angle \( \theta \) is given by the formula:
\[ R = \sqrt{F_1^2 + F_2^2 + 2F_1F_2 \cos \theta} \]

3. **Calculate the Minimum Resultant**:
- The minimum resultant occurs when \( \theta = 180^\circ \) (forces act in opposite directions). In this case, \( \cos(180^\circ) = -1 \):
\[ R_{\text{min}} = \sqrt{10^2 + 6^2 + 2 \cdot 10 \cdot 6 \cdot (-1)} \]
\[ R_{\text{min}} = \sqrt{100 + 36 - 120} = \sqrt{16} = 4 \, \text{N} \]

4. **Calculate the Maximum Resultant**:
- The maximum resultant occurs when \( \theta = 0^\circ \) (forces act in the same direction). In this case, \( \cos(0^\circ) = 1 \):
\[ R_{\text{max}} = \sqrt{10^2 + 6^2 + 2 \cdot 10 \cdot 6 \cdot 1} \]
\[ R_{\text{max}} = \sqrt{100 + 36 + 120} = \sqrt{256} = 16 \, \text{N} \]

5. **Conclusion**:
The resultant force \( R \) can vary between the minimum and maximum values calculated:
\[ R \text{ varies from } 4 \, \text{N to } 16 \, \text{N} \]

### Final Answer:
The resultant force on the body varies from **4 N to 16 N**.