A block of mass m is placed over rough inclined plane having inclination `30^(@)` .The co-efficient of friction between the block and inclined plane is 0.75.The contact force on the block is `x/4mg`. Find the value of x

Options

A. `3/4 mg`
B. 2 mg
C. `5/4 mg`
D. mg

Answer

To find the value of \( x \) in the given problem, we will follow these steps:

### Step 1: Identify the forces acting on the block
The forces acting on the block on the inclined plane are:
- The gravitational force \( mg \) acting vertically downwards.
- The normal force \( N \) acting perpendicular to the inclined plane.
- The frictional force \( F_f \) acting parallel to the inclined plane, opposing the motion.

### Step 2: Resolve the gravitational force
The gravitational force can be resolved into two components:
- Perpendicular to the inclined plane: \( mg \cos(30^\circ) \)
- Parallel to the inclined plane: \( mg \sin(30^\circ) \)

Calculating these components:
- \( mg \cos(30^\circ) = mg \cdot \frac{\sqrt{3}}{2} \)
- \( mg \sin(30^\circ) = mg \cdot \frac{1}{2} \)

### Step 3: Calculate the normal force
Since there is no motion in the vertical direction, the normal force \( N \) is equal to the perpendicular component of the weight:
\[ N = mg \cos(30^\circ) = mg \cdot \frac{\sqrt{3}}{2} \]

### Step 4: Calculate the maximum static friction force
The maximum static friction force \( F_{f_{\text{max}}} \) can be calculated using the coefficient of friction \( \mu \):
\[ F_{f_{\text{max}}} = \mu N = 0.75 \cdot N = 0.75 \cdot \left( mg \cdot \frac{\sqrt{3}}{2} \right) \]
\[ F_{f_{\text{max}}} = \frac{3}{4} \cdot \left( mg \cdot \frac{\sqrt{3}}{2} \right) = \frac{3\sqrt{3}}{8} mg \]

### Step 5: Determine the actual friction force
Since the block is not moving, the actual friction force \( F_f \) will be equal to the component of the gravitational force acting down the incline:
\[ F_f = mg \sin(30^\circ) = mg \cdot \frac{1}{2} \]

### Step 6: Compare the actual friction force with the maximum friction force
We find that:
\[ mg \cdot \frac{1}{2} < \frac{3\sqrt{3}}{8} mg \]
This means the actual friction force \( F_f \) is less than the maximum static friction force, confirming that the block is at rest.

### Step 7: Calculate the resultant contact force
The resultant contact force \( R \) is the vector sum of the normal force and the frictional force:
\[ R = \sqrt{N^2 + F_f^2} \]
Substituting the values:
\[ R = \sqrt{\left( mg \cdot \frac{\sqrt{3}}{2} \right)^2 + \left( mg \cdot \frac{1}{2} \right)^2} \]
\[ R = mg \sqrt{\left( \frac{\sqrt{3}}{2} \right)^2 + \left( \frac{1}{2} \right)^2} \]
\[ R = mg \sqrt{\frac{3}{4} + \frac{1}{4}} = mg \sqrt{1} = mg \]

### Step 8: Relate the resultant contact force to the given expression
According to the problem, the contact force is given as:
\[ R = \frac{x}{4} mg \]
Setting the two expressions for \( R \) equal gives:
\[ mg = \frac{x}{4} mg \]
Dividing both sides by \( mg \) (assuming \( mg \neq 0 \)):
\[ 1 = \frac{x}{4} \]
Thus, multiplying both sides by 4:
\[ x = 4 \]

### Final Answer
The value of \( x \) is \( 4 \).