A box is placed on an inclined plane and has to be pushed down.The angle of inclination is

Options

A. equal to angle of repose
B. more than angle of repose
C. less than the angle of repose
D. equal to angle of friction

Explanation

To solve the problem of determining the relationship between the angle of inclination of a box on an inclined plane and the angle of repose, we can follow these steps:

### Step-by-Step Solution:

1. **Understanding Forces on the Box**:
- When a box is placed on an inclined plane, it experiences gravitational force (weight) acting downwards. This weight can be resolved into two components:
- One component acting parallel to the incline: \( F_{\text{parallel}} = mg \sin \theta \)
- Another component acting perpendicular to the incline: \( F_{\text{perpendicular}} = mg \cos \theta \)

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

3. **Frictional Force**:
- The frictional force \( F_{\text{friction}} \) opposing the motion of the box is given by:
\[ F_{\text{friction}} = \mu N = \mu (mg \cos \theta) \]
- Here, \( \mu \) is the coefficient of friction.

4. **Condition for Downward Movement**:
- For the box to move downward, the component of gravitational force along the incline must be greater than or equal to the frictional force:
\[ mg \sin \theta \geq \mu (mg \cos \theta) \]
- Dividing both sides by \( mg \) (assuming \( m \neq 0 \)):
\[ \sin \theta \geq \mu \cos \theta \]

5. **Using the Angle of Repose**:
- The angle of repose \( \alpha \) is defined as the angle at which the box just begins to slide down the incline. This is given by:
\[ \tan \alpha = \mu \]
- Therefore, we can relate \( \sin \theta \) and \( \cos \theta \) to \( \tan \alpha \):
\[ \sin \theta \geq \tan \alpha \cos \theta \]
- This can be rewritten as:
\[ \tan \theta \geq \tan \alpha \]
- Thus, we conclude that:
\[ \theta \geq \alpha \]

6. **Conclusion**:
- Based on the analysis, we find that the angle of inclination \( \theta \) must be greater than or equal to the angle of repose \( \alpha \) for the box to slide down the incline. Therefore, the correct option is that the angle of inclination is **more than the angle of repose**.

### Final Answer:
The angle of inclination is **more than the angle of repose**.