A body is acted upon by two forces of manitudes `F_(1)=sqrt2 Nand F_(2)=3 N` which are inclined at `45^(@)` to each other. There magnitude of resultant force acting on the body is

Options

A. 17 N
B. 11 N
C. `sqrt17 N`
D. `sqrt11 N`

Answer

To find the magnitude of the resultant force acting on a body when two forces are applied at an angle to each other, we can use the formula for the resultant of two vectors. Here are the steps to solve the problem:

### Step-by-Step Solution:

1. **Identify the Magnitudes of the Forces**:
- Let \( F_1 = \sqrt{2} \, \text{N} \)
- Let \( F_2 = 3 \, \text{N} \)

2. **Identify the Angle Between the Forces**:
- The angle between the two forces is given as \( \theta = 45^\circ \).

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

4. **Substitute the Values into the Formula**:
- First, calculate \( F_1^2 \) and \( F_2^2 \):
\[ F_1^2 = (\sqrt{2})^2 = 2 \]
\[ F_2^2 = 3^2 = 9 \]
- Now calculate \( 2 F_1 F_2 \cos(45^\circ) \):
- We know \( \cos(45^\circ) = \frac{1}{\sqrt{2}} \).
\[ 2 F_1 F_2 = 2 \times \sqrt{2} \times 3 = 6\sqrt{2} \]
\[ 2 F_1 F_2 \cos(45^\circ) = 6\sqrt{2} \times \frac{1}{\sqrt{2}} = 6 \]

5. **Combine All the Values**:
\[ F_R = \sqrt{F_1^2 + F_2^2 + 2 F_1 F_2 \cos(45^\circ)} = \sqrt{2 + 9 + 6} \]
\[ F_R = \sqrt{17} \]

6. **Final Result**:
The magnitude of the resultant force acting on the body is:
\[ F_R = \sqrt{17} \, \text{N} \]

### Conclusion:
The magnitude of the resultant force acting on the body is \( \sqrt{17} \, \text{N} \).

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