At what angle should the two force vector 2F and `sqrt(2)F` act so that the resultant force is `sqrt(10)F`?

Options

A. `45^(@)`
B. `60^(@)`
C. `90^(@)`
D. `120^(@)`

Explanation

To solve the problem of finding the angle at which the two force vectors \(2F\) and \(\sqrt{2}F\) should act so that their resultant is \(\sqrt{10}F\), we can follow these steps:

### Step 1: Understand the Problem
We have two force vectors:
- \( \vec{A} = 2F \)
- \( \vec{B} = \sqrt{2}F \)

We need to find the angle \( \theta \) between these two vectors such that the magnitude of their resultant \( R \) is \( \sqrt{10}F \).

### Step 2: Use the Formula for Resultant of Two Vectors
The magnitude of the resultant \( R \) of two vectors can be calculated using the formula:
\[ R = \sqrt{A^2 + B^2 + 2AB \cos \theta} \]
Substituting the values of \( A \) and \( B \):
\[ R = \sqrt{(2F)^2 + (\sqrt{2}F)^2 + 2(2F)(\sqrt{2}F) \cos \theta} \]

### Step 3: Substitute the Values
Calculating \( A^2 \) and \( B^2 \):
\[ (2F)^2 = 4F^2 \]
\[ (\sqrt{2}F)^2 = 2F^2 \]
Thus, we have:
\[ R = \sqrt{4F^2 + 2F^2 + 2(2F)(\sqrt{2}F) \cos \theta} \]
This simplifies to:
\[ R = \sqrt{6F^2 + 4\sqrt{2}F^2 \cos \theta} \]

### Step 4: Set Up the Equation
We know that the resultant \( R \) is equal to \( \sqrt{10}F \):
\[ \sqrt{6F^2 + 4\sqrt{2}F^2 \cos \theta} = \sqrt{10}F \]

### Step 5: Square Both Sides
Squaring both sides to eliminate the square root gives:
\[ 6F^2 + 4\sqrt{2}F^2 \cos \theta = 10F^2 \]

### Step 6: Rearrange the Equation
Rearranging the equation:
\[ 4\sqrt{2}F^2 \cos \theta = 10F^2 - 6F^2 \]
\[ 4\sqrt{2}F^2 \cos \theta = 4F^2 \]

### Step 7: Simplify
Dividing both sides by \( 4F^2 \) (assuming \( F \neq 0 \)):
\[ \sqrt{2} \cos \theta = 1 \]

### Step 8: Solve for \( \cos \theta \)
\[ \cos \theta = \frac{1}{\sqrt{2}} \]

### Step 9: Find the Angle
The angle \( \theta \) can be found using the inverse cosine function:
\[ \theta = \cos^{-1}\left(\frac{1}{\sqrt{2}}\right) = 45^\circ \]

### Conclusion
Thus, the angle at which the two force vectors \( 2F \) and \( \sqrt{2}F \) should act to achieve a resultant of \( \sqrt{10}F \) is \( 45^\circ \).

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