The force required to move a body up a rough inclined plane is double the force required to prevent the body from sliding down the plane. The coefficient of friction when the angle of inclination of the plane is `60^(@)` is .

Options

A. `1/3`
B. `1/sqrt2`
C. `1/sqrt3`
D. `1/2`

Answer

To solve the problem, we need to find the coefficient of friction (\( \mu \)) when the angle of inclination (\( \theta \)) is \( 60^\circ \). The force required to move a body up a rough inclined plane is given to be double the force required to prevent the body from sliding down the plane.

### Step-by-Step Solution:

1. **Identify Forces Acting on the Body:**
- When the body is moving up the incline, the forces acting on it include:
- The gravitational force (\( mg \)) acting downwards.
- The component of gravitational force parallel to the incline (\( mg \sin \theta \)).
- The normal force (\( N \)) acting perpendicular to the incline.
- The frictional force (\( f \)) acting down the incline (opposing the motion).

2. **Forces When Moving Up the Incline:**
- The net force equation when moving up the incline can be written as:
\[ F = mg \sin \theta + f \]
- The maximum frictional force is given by \( f = \mu N \), where \( N = mg \cos \theta \).
- Therefore, we can substitute \( f \):
\[ F = mg \sin \theta + \mu mg \cos \theta \]

3. **Forces When Preventing Sliding Down:**
- When preventing the body from sliding down, the forces acting on it include:
- The gravitational force component down the incline (\( mg \sin \theta \)).
- The frictional force acting up the incline (\( f \)).
- The equation can be written as:
\[ F' = mg \sin \theta - f \]
- Again substituting \( f \):
\[ F' = mg \sin \theta - \mu mg \cos \theta \]

4. **Relate the Two Forces:**
- According to the problem, the force required to move the body up the incline (\( F \)) is double the force required to prevent it from sliding down (\( F' \)):
\[ F = 2F' \]
- Substituting the expressions for \( F \) and \( F' \):
\[ mg \sin \theta + \mu mg \cos \theta = 2 \left( mg \sin \theta - \mu mg \cos \theta \right) \]

5. **Simplifying the Equation:**
- Expanding the right side:
\[ mg \sin \theta + \mu mg \cos \theta = 2mg \sin \theta - 2\mu mg \cos \theta \]
- Rearranging the terms:
\[ mg \sin \theta + \mu mg \cos \theta + 2\mu mg \cos \theta = 2mg \sin \theta \]
- This simplifies to:
\[ mg \sin \theta + 3\mu mg \cos \theta = 2mg \sin \theta \]
- Dividing through by \( mg \) (assuming \( mg \neq 0 \)):
\[ \sin \theta + 3\mu \cos \theta = 2\sin \theta \]
- Rearranging gives:
\[ 3\mu \cos \theta = 2\sin \theta - \sin \theta \]
- Thus:
\[ 3\mu \cos \theta = \sin \theta \]
- Therefore:
\[ \mu = \frac{\sin \theta}{3\cos \theta} = \frac{1}{3} \tan \theta \]

6. **Substituting the Angle:**
- For \( \theta = 60^\circ \):
\[ \tan 60^\circ = \sqrt{3} \]
- Thus:
\[ \mu = \frac{\sqrt{3}}{3} = \frac{1}{\sqrt{3}} \]

### Final Answer:
The coefficient of friction (\( \mu \)) is \( \frac{1}{\sqrt{3}} \).