A force `vec(F)` is applied to block `(m = 6 kg)` at rest on an inclined plane of inclination `30^(@)`. The force `vec(F)` is horizontal and parallel to surface of inclined plane. Maximum value of `F` so that block remains at rest is `40 N`. The coefficient of friction of the surface is.

Options

A. `(2)/(sqrt(3))`
B. `sqrt(3)`
C. `(5)/(3 sqrt(3))`
D. `(5 sqrt(3))/(2)`

Explanation

To find the coefficient of friction (μ) for the block on the inclined plane, we can follow these steps:

### Step 1: Identify the Forces Acting on the Block
The forces acting on the block include:
1. The gravitational force (weight) \( \vec{W} = m \cdot g \)
2. The normal force \( \vec{N} \)
3. The applied horizontal force \( \vec{F} \)
4. The frictional force \( \vec{F_f} \)

### Step 2: Calculate the Weight of the Block
Given:
- Mass \( m = 6 \, \text{kg} \)
- Acceleration due to gravity \( g = 10 \, \text{m/s}^2 \)

The weight \( W \) of the block is:
\[ W = m \cdot g = 6 \, \text{kg} \cdot 10 \, \text{m/s}^2 = 60 \, \text{N} \]

### Step 3: Resolve the Weight into Components
The weight can be resolved into two components:
1. Perpendicular to the inclined plane: \( W_{\perp} = W \cdot \cos(\theta) = 60 \cdot \cos(30^\circ) \)
2. Parallel to the inclined plane: \( W_{\parallel} = W \cdot \sin(\theta) = 60 \cdot \sin(30^\circ) \)

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

Thus,
\[ W_{\perp} = 60 \cdot \frac{\sqrt{3}}{2} = 30\sqrt{3} \, \text{N} \]
\[ W_{\parallel} = 60 \cdot \frac{1}{2} = 30 \, \text{N} \]

### Step 4: Apply the Condition for Equilibrium
For the block to remain at rest, the net force acting along the inclined plane must be zero. The forces acting along the incline are:
- The frictional force \( F_f = \mu \cdot N \)
- The component of the weight down the incline \( W_{\parallel} \)
- The applied force \( F \)

The equation for equilibrium along the incline is:
\[ F + F_f = W_{\parallel} \]

Substituting \( F_f \):
\[ F + \mu \cdot N = W_{\parallel} \]

### Step 5: Calculate the Normal Force
The normal force \( N \) is equal to the perpendicular component of the weight:
\[ N = W_{\perp} = 30\sqrt{3} \, \text{N} \]

### Step 6: Substitute Known Values
Given that the maximum applied force \( F = 40 \, \text{N} \) and substituting \( N \):
\[ 40 + \mu \cdot (30\sqrt{3}) = 30 \]

### Step 7: Solve for the Coefficient of Friction (μ)
Rearranging the equation:
\[ \mu \cdot (30\sqrt{3}) = 30 - 40 \]
\[ \mu \cdot (30\sqrt{3}) = -10 \]
\[ \mu = \frac{-10}{30\sqrt{3}} = -\frac{1}{3\sqrt{3}} \]

However, since the frictional force cannot be negative, we need to consider the absolute values and the correct direction of forces. The correct equation should be:
\[ \mu \cdot (30\sqrt{3}) = 40 - 30 \]
\[ \mu \cdot (30\sqrt{3}) = 10 \]
\[ \mu = \frac{10}{30\sqrt{3}} = \frac{1}{3\sqrt{3}} = \frac{\sqrt{3}}{9} \]

### Final Answer
The coefficient of friction \( \mu \) is:
\[ \mu = \frac{1}{3\sqrt{3}} \approx 0.192 \]