A box of mass `8 kg ` placed on a rough inclined plane of inclination `theta` its downward motion can be prevented by applying an upward pull `F` and it can be made to slide upward appliying a force `2F` .The coefficient of friction between the box and the inclined plane is

Options

A. `(tan theta)//3`
B. `3 tan theta`
C. `(tan theta)//2`
D. `2 tan theta`

Explanation

To solve the problem, we need to analyze the forces acting on the box on the inclined plane and set up equations based on the information provided.

### Step-by-Step Solution:

1. **Identify the Forces Acting on the Box**:
- The weight of the box (W) is given by \( W = mg = 8 \, \text{kg} \times 9.8 \, \text{m/s}^2 = 80 \, \text{N} \).
- The weight can be resolved into two components:
- Parallel to the incline: \( W_{\parallel} = mg \sin \theta = 80 \sin \theta \)
- Perpendicular to the incline: \( W_{\perpendicular} = mg \cos \theta = 80 \cos \theta \)

2. **Normal Force (N)**:
- The normal force acting on the box is equal to the perpendicular component of the weight:
\[ N = mg \cos \theta = 80 \cos \theta \]

3. **Frictional Force**:
- The limiting frictional force (F_f) can be expressed as:
\[ F_f = \mu N = \mu (80 \cos \theta) \]

4. **First Condition (Preventing Downward Motion)**:
- When the upward force \( F \) is applied to prevent downward motion, we have:
\[ F = mg \sin \theta - F_f \]
- Substituting the expression for friction:
\[ F = 80 \sin \theta - \mu (80 \cos \theta) \]
- Rearranging gives:
\[ F = 80 \sin \theta - 80 \mu \cos \theta \]
- Dividing through by 80:
\[ \frac{F}{80} = \sin \theta - \mu \cos \theta \]
- This is our **Equation (1)**.

5. **Second Condition (Sliding Upward)**:
- When the force \( 2F \) is applied to make the box slide upward, we have:
\[ 2F = mg \sin \theta + F_f \]
- Substituting for the frictional force:
\[ 2F = 80 \sin \theta + \mu (80 \cos \theta) \]
- Rearranging gives:
\[ 2F = 80 \sin \theta + 80 \mu \cos \theta \]
- Dividing through by 80:
\[ \frac{2F}{80} = \sin \theta + \mu \cos \theta \]
- This is our **Equation (2)**.

6. **Equating the Two Conditions**:
- From Equation (1):
\[ \frac{F}{80} = \sin \theta - \mu \cos \theta \]
- From Equation (2):
\[ \frac{2F}{80} = \sin \theta + \mu \cos \theta \]
- Multiply Equation (1) by 2:
\[ \frac{2F}{80} = 2\sin \theta - 2\mu \cos \theta \]
- Now equate this to Equation (2):
\[ 2\sin \theta - 2\mu \cos \theta = \sin \theta + \mu \cos \theta \]

7. **Solving for the Coefficient of Friction (μ)**:
- Rearranging gives:
\[ 2\sin \theta - \sin \theta = \mu \cos \theta + 2\mu \cos \theta \]
\[ \sin \theta = 3\mu \cos \theta \]
- Therefore:
\[ \mu = \frac{\sin \theta}{3 \cos \theta} = \frac{1}{3} \tan \theta \]

8. **Final Result**:
- The coefficient of friction \( \mu \) is:
\[ \mu = \frac{1}{3} \tan \theta \]