A block of mass 4kg rests on a rough horizontal plane. The plane is gradually inclined until at an angle `theta=60^(@)` , with the vertical, the mass just begins to side. What is the coefficient of static friction between the block and the plane?

Options

A. `sqrt(3)`
B. `1/2`
C. `(1)/sqrt(2)`
D. `(1)/sqrt(3)`

Explanation

To solve the problem, we need to determine the coefficient of static friction (μ) between the block and the inclined plane when the block just begins to slide at an angle of θ = 60° with the vertical.

### Step-by-Step Solution:

1. **Identify the Forces Acting on the Block:**
- The weight of the block (W) acts downwards, which can be calculated as:
\[ W = mg = 4 \, \text{kg} \times g \]
- The gravitational force can be resolved into two components:
- Perpendicular to the inclined plane: \( W_{\perp} = mg \cos(30^\circ) \)
- Parallel to the inclined plane: \( W_{\parallel} = mg \sin(30^\circ) \)

2. **Calculate the Components of the Weight:**
- Using \( g \approx 9.81 \, \text{m/s}^2 \):
\[ W_{\perp} = 4g \cos(30^\circ) = 4g \left(\frac{\sqrt{3}}{2}\right) = 2\sqrt{3}g \]
\[ W_{\parallel} = 4g \sin(30^\circ) = 4g \left(\frac{1}{2}\right) = 2g \]

3. **Determine the Normal Force (N):**
- The normal force (N) is equal to the perpendicular component of the weight:
\[ N = W_{\perp} = 2\sqrt{3}g \]

4. **Determine the Maximum Static Friction (F_max):**
- The maximum static friction force (F_max) can be expressed as:
\[ F_{\text{max}} = \mu N = \mu (2\sqrt{3}g) \]

5. **Set Up the Equation for Forces:**
- At the point of impending motion, the maximum static friction equals the parallel component of the weight:
\[ F_{\text{max}} = W_{\parallel} = 2g \]
- Therefore, we can equate the two expressions:
\[ \mu (2\sqrt{3}g) = 2g \]

6. **Solve for the Coefficient of Static Friction (μ):**
- Cancel \( g \) from both sides (assuming \( g \neq 0 \)):
\[ \mu (2\sqrt{3}) = 2 \]
- Rearranging gives:
\[ \mu = \frac{2}{2\sqrt{3}} = \frac{1}{\sqrt{3}} \]

### Final Answer:
The coefficient of static friction (μ) between the block and the plane is:
\[ \mu = \frac{1}{\sqrt{3}} \]