The force `F_(1)` required to just moving a body up an inclined plane is double the force `F_(2)` required to just prevent the body from sliding down the plane. The coefficient of friction is `mu`. The inclination `theta` of the plane is :

Options

A. `tan^(-1) mu`
B. `tan^(-1) ( mu//2)`
C. `tan^(-1) ( 2mu)`
D. `tan^(-1)( 3mu)`

Answer

To solve the problem, we will analyze the forces acting on a body on an inclined plane and derive the relationship between the forces required to move the body up and to prevent it from sliding down.

### Step-by-Step Solution:

1. **Identify the Forces Acting on the Body**:
- When the body is on the inclined plane, the gravitational force acting on it can be resolved into two components:
- Parallel to the incline: \( mg \sin \theta \)
- Perpendicular to the incline: \( mg \cos \theta \)

2. **Normal Force**:
- The normal force \( N \) acting on the body is equal to the perpendicular component of the weight:
\[ N = mg \cos \theta \]

3. **Frictional Force**:
- The frictional force \( f \) can be expressed as:
\[ f = \mu N = \mu (mg \cos \theta) \]

4. **Force to Move the Body Up the Incline (F1)**:
- When the body is just about to move up the incline, the force \( F_1 \) required is given by the sum of the gravitational component and the frictional force acting down the incline:
\[ F_1 = mg \sin \theta + f = mg \sin \theta + \mu mg \cos \theta \]
- Thus, we can write:
\[ F_1 = mg (\sin \theta + \mu \cos \theta) \]

5. **Force to Prevent Sliding Down the Incline (F2)**:
- When the body is just about to slide down, the force \( F_2 \) required is given by the difference between the gravitational component and the frictional force acting up the incline:
\[ F_2 = mg \sin \theta - f = mg \sin \theta - \mu mg \cos \theta \]
- Thus, we can write:
\[ F_2 = mg (\sin \theta - \mu \cos \theta) \]

6. **Relationship Between F1 and F2**:
- According to the problem, \( F_1 \) is double \( F_2 \):
\[ F_1 = 2 F_2 \]
- Substituting the expressions for \( F_1 \) and \( F_2 \):
\[ mg (\sin \theta + \mu \cos \theta) = 2 \cdot mg (\sin \theta - \mu \cos \theta) \]
- We can cancel \( mg \) from both sides (assuming \( mg \neq 0 \)):
\[ \sin \theta + \mu \cos \theta = 2 (\sin \theta - \mu \cos \theta) \]

7. **Simplifying the Equation**:
- Expanding the right side:
\[ \sin \theta + \mu \cos \theta = 2 \sin \theta - 2 \mu \cos \theta \]
- Rearranging the terms:
\[ \sin \theta + \mu \cos \theta - 2 \sin \theta + 2 \mu \cos \theta = 0 \]
\[ -\sin \theta + 3\mu \cos \theta = 0 \]
- This leads to:
\[ \sin \theta = 3\mu \cos \theta \]

8. **Dividing by Cosine**:
- Dividing both sides by \( \cos \theta \) (assuming \( \cos \theta \neq 0 \)):
\[ \tan \theta = 3\mu \]

9. **Finding the Angle**:
- Finally, we can express the angle \( \theta \) as:
\[ \theta = \tan^{-1}(3\mu) \]

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