A minimum force F is applied to a block of mass 102 kg to prevent it from sliding on a plane with an inclination angle `30^(@)` with the horizontal. If the coefficients of static and kinetic friction between the block and the plane are 0.4 and 0.3 respectively, then the force F is

Options

A. 157 N
B. 224 N
C. 315 N
D. zero

Explanation

To solve the problem step by step, we will analyze the forces acting on the block on the inclined plane and calculate the minimum force \( F \) required to prevent the block from sliding down.

### Step 1: Identify the forces acting on the block
The forces acting on the block are:
1. The weight of the block \( W = mg \) acting vertically downward.
2. The normal force \( N \) acting perpendicular to the inclined plane.
3. The frictional force \( f \) acting parallel to the inclined plane, opposing the motion.

### Step 2: Break down the weight into components
The weight of the block can be resolved into two components:
- The component parallel to the incline:
\[ W_{\parallel} = mg \sin \theta \]
- The component perpendicular to the incline:
\[ W_{\perpendicular} = mg \cos \theta \]

Given:
- Mass \( m = 102 \, \text{kg} \)
- Gravitational acceleration \( g = 10 \, \text{m/s}^2 \)
- Inclination angle \( \theta = 30^\circ \)

Calculating the components:
- \( W_{\parallel} = 102 \times 10 \times \sin(30^\circ) = 1020 \times 0.5 = 510 \, \text{N} \)
- \( W_{\perpendicular} = 102 \times 10 \times \cos(30^\circ) = 1020 \times \frac{\sqrt{3}}{2} \approx 882.6 \, \text{N} \)

### Step 3: Calculate the maximum static friction
The maximum static friction \( f_{\text{max}} \) can be calculated using:
\[ f_{\text{max}} = \mu_s N \]
Where \( \mu_s = 0.4 \) is the coefficient of static friction.

Substituting for \( N \):
\[ f_{\text{max}} = \mu_s (mg \cos \theta) = 0.4 \times 882.6 \approx 353.04 \, \text{N} \]

### Step 4: Set up the equilibrium condition
To prevent the block from sliding down, the applied force \( F \) plus the frictional force must balance the component of weight acting down the incline:
\[ F + f_{\text{max}} = W_{\parallel} \]
Substituting the values we calculated:
\[ F + 353.04 = 510 \]

### Step 5: Solve for the applied force \( F \)
Rearranging the equation gives:
\[ F = 510 - 353.04 \approx 156.96 \, \text{N} \]

### Final Answer
Thus, the minimum force \( F \) required to prevent the block from sliding is approximately:
\[ F \approx 157 \, \text{N} \]