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

Options

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

Answer

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

### Step 1: Understanding the Forces Acting on the Block
When a block of mass \(m\) is placed on a rough inclined plane, the forces acting on it include:
- The gravitational force (\(mg\)) acting vertically downward.
- The normal force (\(N\)) acting perpendicular to the inclined surface.
- The frictional force (\(F_f\)) acting parallel to the inclined surface.

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

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

### Step 4: Calculating the Contact Force
The contact force (\(F_{\text{contact}}\)) between the block and the inclined plane can be expressed as:
\[ F_{\text{contact}} = \mu N \]
where \(\mu\) is the coefficient of friction. Substituting for \(N\):
\[ F_{\text{contact}} = \mu (mg \cos \theta) \]

### Step 5: Analyzing the Behavior of the Contact Force as \(\theta\) Increases
As \(\theta\) increases from \(0^\circ\) to \(90^\circ\):
- The term \(\cos \theta\) decreases from \(1\) (at \(\theta = 0^\circ\)) to \(0\) (at \(\theta = 90^\circ\)).
- Consequently, the contact force \(F_{\text{contact}} = \mu (mg \cos \theta)\) also decreases.

### Step 6: Conclusion
As the angle \(\theta\) increases from \(0^\circ\) to \(90^\circ\), the contact force between the block and the inclined plane decreases to zero.

### Summary
- At \(\theta = 0^\circ\), \(F_{\text{contact}} = \mu mg\).
- At \(\theta = 90^\circ\), \(F_{\text{contact}} = 0\).