A box of mass 4 kg is placed on a rough inclined plane of inclination `60^(@)`. Its downward motion can be prevented by applying an upward pull is F and it can be made to slide upwards by applying a force 3F. The coefficient of friction between the box and inclined plane is

Options

A. `(2)/(sqrt3)`
B. `(sqrt3)/(2)`
C. `(1)/(sqrt2)`
D. `(1)/(2)`

Explanation

To solve the problem of finding the coefficient of friction between the box and the inclined plane, we will follow these steps:

### Step 1: Identify the forces acting on the box
The forces acting on the box on the inclined plane include:
- The gravitational force \( mg \) acting downward.
- The normal force \( N \) acting perpendicular to the inclined plane.
- The frictional force \( f \) 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 incline: \( mg \cos \theta \)
- Parallel to the incline: \( mg \sin \theta \)

Given that \( m = 4 \, \text{kg} \) and \( \theta = 60^\circ \):
- \( mg = 4 \times 9.8 = 39.2 \, \text{N} \)
- \( mg \cos 60^\circ = 39.2 \times \frac{1}{2} = 19.6 \, \text{N} \)
- \( mg \sin 60^\circ = 39.2 \times \frac{\sqrt{3}}{2} = 33.94 \, \text{N} \)

### Step 3: Set up the equations for the forces
1. **For the downward motion prevented by force \( F \)**:
\[ F + f = mg \sin 60^\circ \]
The frictional force \( f \) can be expressed as \( f = \mu N \), where \( \mu \) is the coefficient of friction.

The normal force \( N \) can be expressed as:
\[ N = mg \cos 60^\circ = 19.6 \, \text{N} \]
Thus, the equation becomes:
\[ F + \mu N = mg \sin 60^\circ \]
Substituting \( N \):
\[ F + \mu (19.6) = 33.94 \]

2. **For the upward motion with force \( 3F \)**:
\[ 3F - f = mg \sin 60^\circ \]
Again substituting \( f = \mu N \):
\[ 3F - \mu (19.6) = 33.94 \]

### Step 4: Solve the equations
Now we have two equations:
1. \( F + 19.6\mu = 33.94 \) (1)
2. \( 3F - 19.6\mu = 33.94 \) (2)

From equation (1):
\[ F = 33.94 - 19.6\mu \]

Substituting \( F \) into equation (2):
\[ 3(33.94 - 19.6\mu) - 19.6\mu = 33.94 \]
Expanding this:
\[ 101.82 - 58.8\mu - 19.6\mu = 33.94 \]
Combining terms:
\[ 101.82 - 78.4\mu = 33.94 \]
Rearranging gives:
\[ 78.4\mu = 101.82 - 33.94 \]
\[ 78.4\mu = 67.88 \]
\[ \mu = \frac{67.88}{78.4} \approx 0.866 \]

### Step 5: Final answer
The coefficient of friction \( \mu \) between the box and the inclined plane is approximately \( 0.866 \).

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