A block is at rest on an inclined plane making an angle `alpha` with the horizontal. As the angle of the incline is increased the block start slipping when the angle of inclination becomes `theta` then coefficient of friction is equal to

Options

A. `sintheta`
B. `costheta`
C. `tantheta`
D. Independent of `theta`

Answer

To find the coefficient of friction (μ) when a block starts slipping on an inclined plane at an angle θ, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Forces Acting on the Block:**
- The weight of the block (W = mg) acts vertically downward.
- The component of the weight acting parallel to the incline is \( W_{\parallel} = mg \sin \theta \).
- The component of the weight acting perpendicular to the incline is \( W_{\perpendicular} = mg \cos \theta \).

2. **Determine the Normal Force (N):**
- The normal force exerted by the inclined plane on the block is equal to the perpendicular component of the weight:
\[ N = mg \cos \theta \]

3. **Frictional Force (F_f):**
- The frictional force that opposes the motion of the block is given by:
\[ F_f = \mu N \]
- Substituting the expression for the normal force:
\[ F_f = \mu (mg \cos \theta) \]

4. **Condition for the Block to Start Slipping:**
- The block starts slipping when the gravitational force component down the incline equals the maximum static frictional force:
\[ mg \sin \theta = \mu (mg \cos \theta) \]

5. **Canceling the Mass (m):**
- Since mass (m) appears on both sides of the equation, we can cancel it out:
\[ g \sin \theta = \mu (g \cos \theta) \]

6. **Rearranging the Equation:**
- Dividing both sides by \( g \) (assuming \( g \neq 0 \)):
\[ \sin \theta = \mu \cos \theta \]

7. **Solving for the Coefficient of Friction (μ):**
- Rearranging gives:
\[ \mu = \frac{\sin \theta}{\cos \theta} \]
- This simplifies to:
\[ \mu = \tan \theta \]

### Final Answer:
The coefficient of friction (μ) when the block starts slipping is:
\[ \mu = \tan \theta \]

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