A body is placed on a rough inclined plane of inclination `theta`. As the angle `theta` is increased from `0^(@)` to `90^(@)` the contact force between the block and the plane

Options

A. remains constant
B. first remains constant then decreases
C. first decreases then increases
D. first increases then decreases

Answer

To analyze the contact force between a block and a rough inclined plane as the angle of inclination \(\theta\) increases from \(0^\circ\) to \(90^\circ\), we can follow these steps:

### Step 1: Understand the Forces Acting on the Block
When the block is placed on the inclined plane, the forces acting on it are:
- The gravitational force (\(mg\)) acting downwards.
- The normal force (\(N\)) acting perpendicular to the surface of the incline.
- The frictional force (\(f\)) acting parallel to the surface of the incline.

### Step 2: Resolve the Gravitational Force
The gravitational force can be resolved into two components:
- Perpendicular to the incline: \(mg \cos \theta\)
- Parallel to the incline: \(mg \sin \theta\)

### Step 3: Analyze the Normal Force
The normal force \(N\) is equal to the component of the gravitational force acting perpendicular to the incline:
\[ N = mg \cos \theta \]

### Step 4: Analyze the Frictional Force
The frictional force \(f\) can be expressed as:
\[ f = \mu N = \mu (mg \cos \theta) \]
where \(\mu\) is the coefficient of friction between the block and the inclined plane.

### Step 5: Consider the Motion of the Block
As \(\theta\) increases:
- At \(\theta = 0^\circ\): The block does not experience any inclination, and the contact force is simply the normal force \(N = mg\).
- As \(\theta\) increases, the normal force decreases because \(N = mg \cos \theta\).
- At \(\theta = 90^\circ\): The incline becomes vertical, and the normal force approaches zero, meaning the block would no longer be in contact with the incline.

### Step 6: Conclusion on Contact Force
As \(\theta\) increases from \(0^\circ\) to \(90^\circ\):
- The normal force \(N\) decreases from \(mg\) to \(0\).
- The contact force between the block and the inclined plane also decreases from \(mg\) to \(0\).

### Summary
The contact force between the block and the inclined plane decreases as the angle of inclination \(\theta\) increases from \(0^\circ\) to \(90^\circ\).