A body of mass m is kept on an inclined plane . The force required to just move a body up the inclined plane is double the force required to just prevent the body from sliding down the plane. If the coefficient of friction is `mu`, what is the inclination of the plane ?

Explanation

To solve the problem, we need to analyze the forces acting on the body on the inclined plane and use the information given in the question.

### Step-by-Step Solution:

1. **Identify the Forces**:
- The weight of the body \( W = mg \) acts vertically downward.
- The component of weight acting parallel to the incline is \( W_{\parallel} = mg \sin \theta \).
- The component of weight acting perpendicular to the incline is \( W_{\perpendicular} = mg \cos \theta \).
- The frictional force \( F_f \) acting up the incline is \( F_f = \mu W_{\perpendicular} = \mu mg \cos \theta \).

2. **Force to Prevent Sliding Down**:
- To prevent the body from sliding down, the frictional force must balance the component of weight acting down the incline:
\[ F_2 = mg \sin \theta - \mu mg \cos \theta \]
Here, \( F_2 \) is the force required to just prevent the body from sliding down.

3. **Force to Move Up the Incline**:
- The force required to just move the body up the incline is given as double that required to prevent sliding down:
\[ F_1 = 2F_2 \]
Substituting for \( F_2 \):
\[ F_1 = 2(mg \sin \theta - \mu mg \cos \theta) \]

4. **Force Balance for Moving Up**:
- The force required to move the body up the incline must overcome both the component of weight acting down the incline and the frictional force:
\[ F_1 = mg \sin \theta + \mu mg \cos \theta \]

5. **Setting the Two Expressions for \( F_1 \) Equal**:
- We now have two expressions for \( F_1 \):
\[ 2(mg \sin \theta - \mu mg \cos \theta) = mg \sin \theta + \mu mg \cos \theta \]
Simplifying this equation:
\[ 2mg \sin \theta - 2\mu mg \cos \theta = mg \sin \theta + \mu mg \cos \theta \]
Rearranging gives:
\[ 2mg \sin \theta - mg \sin \theta = 2\mu mg \cos \theta + \mu mg \cos \theta \]
\[ mg \sin \theta = 3\mu mg \cos \theta \]

6. **Canceling \( mg \)**:
- Since \( mg \) is common on both sides, we can cancel it out (assuming \( m \neq 0 \)):
\[ \sin \theta = 3\mu \cos \theta \]

7. **Dividing by \( \cos \theta \)**:
- Dividing both sides by \( \cos \theta \):
\[ \tan \theta = 3\mu \]

8. **Finding the Angle**:
- The inclination \( \theta \) can be expressed as:
\[ \theta = \tan^{-1}(3\mu) \]

### Final Answer:
The inclination of the plane is given by:
\[ \theta = \tan^{-1}(3\mu) \]