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 of finding the coefficient of static friction between a block of mass 4 kg resting on a rough horizontal plane that begins to slide when the plane is inclined at an angle of θ = 60° with the vertical, we can follow these steps:

### Step 1: Understanding the Angle
The angle θ is given as 60° with the vertical. To find the angle with the horizontal, we can use the relationship:
\[ \text{Angle with horizontal} = 90° - \theta = 90° - 60° = 30° \]

### Step 2: Drawing the Free Body Diagram (FBD)
We draw the block on the inclined plane. The forces acting on the block are:
- The gravitational force (weight) \( W = mg \) acting downwards.
- The normal force \( N \) acting perpendicular to the surface of the incline.
- The frictional force \( F_f \) acting parallel to the incline, opposing the motion.

### Step 3: Resolving the Weight into Components
The weight of the block can be resolved into two components:
- Perpendicular to the incline: \( W_{\perp} = mg \cos(30°) \)
- Parallel to the incline: \( W_{\parallel} = mg \sin(30°) \)

Given \( m = 4 \, \text{kg} \) and \( g = 10 \, \text{m/s}^2 \):
\[ W = mg = 4 \times 10 = 40 \, \text{N} \]
Now, we calculate the components:
\[ W_{\perp} = 40 \cos(30°) = 40 \times \frac{\sqrt{3}}{2} = 20\sqrt{3} \, \text{N} \]
\[ W_{\parallel} = 40 \sin(30°) = 40 \times \frac{1}{2} = 20 \, \text{N} \]

### Step 4: Normal Force
The normal force \( N \) is equal to the perpendicular component of the weight:
\[ N = W_{\perp} = 20\sqrt{3} \, \text{N} \]

### Step 5: Frictional Force
The frictional force \( F_f \) must equal the parallel component of the weight when the block is about to slide:
\[ F_f = W_{\parallel} = 20 \, \text{N} \]

### Step 6: Coefficient of Static Friction
The frictional force can also be expressed in terms of the coefficient of static friction \( \mu \) and the normal force:
\[ F_f = \mu N \]
Substituting the known values:
\[ 20 = \mu (20\sqrt{3}) \]
Solving for \( \mu \):
\[ \mu = \frac{20}{20\sqrt{3}} = \frac{1}{\sqrt{3}} \]

### Final Answer
The coefficient of static friction \( \mu \) is:
\[ \mu = \frac{1}{\sqrt{3}} \approx 0.577 \]