Two forces with equal magnitudes F act on a body and the magnitude of the resultant force is F /3. The angle between the two forces is

Options

A. `cos^-1(-17/18)`
B. `cos^-1(-1/3)`
C. `cos^-1(2/3)`
D. `cos^-1(8/9)`

Answer

To solve the problem, we need to find the angle between two equal forces \( F \) when the resultant force is given as \( \frac{F}{3} \).

### Step-by-Step Solution:

1. **Identify the Given Information:**
- Two forces \( F_1 \) and \( F_2 \) with equal magnitudes \( F \).
- The magnitude of the resultant force \( R = \frac{F}{3} \).

2. **Use the Formula for Resultant of Two Forces:**
The formula for the resultant \( R \) of two forces \( F_1 \) and \( F_2 \) acting at an angle \( \theta \) is given by:
\[ R = \sqrt{F_1^2 + F_2^2 + 2 F_1 F_2 \cos \theta} \]
Since \( F_1 = F_2 = F \), we can substitute:
\[ R = \sqrt{F^2 + F^2 + 2 F \cdot F \cos \theta} \]
This simplifies to:
\[ R = \sqrt{2F^2 + 2F^2 \cos \theta} \]

3. **Substituting the Resultant Value:**
We know that \( R = \frac{F}{3} \), so we can set up the equation:
\[ \frac{F}{3} = \sqrt{2F^2 + 2F^2 \cos \theta} \]

4. **Square Both Sides:**
Squaring both sides to eliminate the square root gives:
\[ \left(\frac{F}{3}\right)^2 = 2F^2 + 2F^2 \cos \theta \]
This simplifies to:
\[ \frac{F^2}{9} = 2F^2 + 2F^2 \cos \theta \]

5. **Rearranging the Equation:**
Rearranging the equation, we get:
\[ \frac{F^2}{9} - 2F^2 = 2F^2 \cos \theta \]
To combine the terms, convert \( 2F^2 \) to a fraction:
\[ \frac{F^2}{9} - \frac{18F^2}{9} = 2F^2 \cos \theta \]
This simplifies to:
\[ \frac{-17F^2}{9} = 2F^2 \cos \theta \]

6. **Dividing by \( F^2 \):**
Dividing both sides by \( F^2 \) (assuming \( F \neq 0 \)):
\[ \frac{-17}{9} = 2 \cos \theta \]

7. **Solving for \( \cos \theta \):**
Dividing both sides by 2 gives:
\[ \cos \theta = \frac{-17}{18} \]

8. **Finding the Angle \( \theta \):**
To find the angle \( \theta \), we take the inverse cosine:
\[ \theta = \cos^{-1}\left(\frac{-17}{18}\right) \]

### Final Answer:
The angle between the two forces is \( \theta = \cos^{-1}\left(\frac{-17}{18}\right) \).