A block of mass m rests on a rough inclined plane. The coefficient of friction between the surface and the block is µ. At what angle of inclination `theta` of the plane to the horizontal will the block just start to slide down the plane?

Options

A. `theta=tan^(-1)mu`
B. `theta=cos^(-1)mu`
C. `theta=sin^(-1)mu`
D. `theta=sec^(-1)mu`

Explanation

To solve the problem of finding the angle of inclination \( \theta \) at which a block of mass \( m \) just starts to slide down a rough inclined plane, we can follow these steps:

### Step 1: Understand the forces acting on the block
When the block is on the inclined plane, two main forces act on it:
1. The gravitational force \( mg \) acting downwards.
2. The normal force \( N \) acting perpendicular to the surface of the incline.

### Step 2: Resolve the gravitational force
The gravitational force can be resolved into two components:
- **Parallel to the incline**: \( F_{\text{parallel}} = mg \sin \theta \)
- **Perpendicular to the incline**: \( F_{\text{perpendicular}} = mg \cos \theta \)

### Step 3: Determine 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: Calculate the frictional force
The frictional force \( F_{\text{friction}} \) that opposes the motion of the block is given by:
\[ F_{\text{friction}} = \mu N = \mu (mg \cos \theta) \]

### Step 5: Set up the equation for motion
For the block to just start sliding down, the force due to gravity parallel to the incline must equal the maximum static friction force:
\[ mg \sin \theta = \mu (mg \cos \theta) \]

### Step 6: Simplify the equation
We can cancel \( mg \) from both sides of the equation (assuming \( m \neq 0 \)):
\[ \sin \theta = \mu \cos \theta \]

### Step 7: Rearrange to find \( \tan \theta \)
Dividing both sides by \( \cos \theta \):
\[ \tan \theta = \mu \]

### Step 8: Solve for \( \theta \)
To find the angle \( \theta \), we take the inverse tangent (arctan) of both sides:
\[ \theta = \tan^{-1}(\mu) \]

### Final Answer
Thus, the angle of inclination \( \theta \) at which the block just starts to slide down the plane is:
\[ \theta = \tan^{-1}(\mu) \]

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