A mass of 2kg rests on a horizontal plane. The plane is gradually inclined until at an angle `theta=20^(@)` with the horizontal, the mass just begins to slides. What is the coefficient of static friction between the block and the surface ?

Explanation

To find the coefficient of static friction between the block and the surface, we can follow these steps:

### Step 1: Identify the forces acting on the block
When the inclined plane is at an angle \( \theta = 20^\circ \), the forces acting on the block are:
- The gravitational force \( mg \) acting downwards.
- The normal force \( N \) acting perpendicular to the surface of the incline.
- The frictional force \( f \) acting parallel to the surface of the incline, opposing the motion.

### Step 2: Break down the gravitational force into components
The gravitational force can be broken down into two components:
- The component parallel to the incline: \( mg \sin(\theta) \)
- The component perpendicular to the incline: \( mg \cos(\theta) \)

### Step 3: Write the equation for the normal force
The normal force \( N \) is equal to the perpendicular component of the gravitational force:
\[ N = mg \cos(\theta) \]

### Step 4: Write the equation for static friction
The static friction force \( f \) can be expressed as:
\[ f = \mu_s N \]
where \( \mu_s \) is the coefficient of static friction.

### Step 5: Set up the equilibrium condition
At the angle where the block just begins to slide, the static friction force is equal to the component of the gravitational force acting down the incline:
\[ f = mg \sin(\theta) \]

### Step 6: Substitute the expressions for friction and normal force
Substituting the expressions for \( f \) and \( N \):
\[ \mu_s (mg \cos(\theta)) = mg \sin(\theta) \]

### Step 7: Simplify the equation
We can cancel \( mg \) from both sides (since \( m \) and \( g \) are not zero):
\[ \mu_s \cos(\theta) = \sin(\theta) \]

### Step 8: Solve for the coefficient of static friction
Rearranging the equation gives:
\[ \mu_s = \frac{\sin(\theta)}{\cos(\theta)} \]
This is equivalent to:
\[ \mu_s = \tan(\theta) \]

### Step 9: Substitute the angle
Now, substituting \( \theta = 20^\circ \):
\[ \mu_s = \tan(20^\circ) \]

### Step 10: Calculate the value
Using a calculator or trigonometric tables, we find:
\[ \tan(20^\circ) \approx 0.364 \]

### Final Answer
Thus, the coefficient of static friction \( \mu_s \) is approximately:
\[ \mu_s \approx 0.364 \]