A box of mass `4kg` rests upon an inclined plane This inclination is gradually incresed till the box starts sliding down the plane. At this stage slope of the plane is 1 in 3 Find coefficient of friction between the box and the plane What force applied to the box parallel to the plane will just make the box move up the plane ?

Explanation

To solve the problem, we need to find two things: the coefficient of friction (μ) between the box and the inclined plane, and the force required to move the box up the plane.

### Step 1: Understanding the Inclined Plane
The slope of the inclined plane is given as 1 in 3. This means that for every 3 units of horizontal distance, the height increases by 1 unit. Therefore, we can define the angle of inclination (θ) using the tangent function:
\[ \tan(\theta) = \frac{\text{opposite}}{\text{adjacent}} = \frac{1}{3} \]
From this, we can find θ:
\[ \theta = \tan^{-1}\left(\frac{1}{3}\right) \]

### Step 2: Finding the Coefficient of Friction (μ)
At the point when the box starts sliding down, the angle of repose (α) is equal to θ. The coefficient of friction (μ) can be calculated using:
\[ \mu = \tan(\theta) \]
We already established that \(\tan(\theta) = \frac{1}{3}\). Therefore, the coefficient of friction is:
\[ \mu = \frac{1}{3} \]

### Step 3: Calculating the Forces Acting on the Box
When the box is about to slide down, the forces acting on it are:
- The gravitational force acting down the slope: \( F_{\text{gravity}} = mg \sin(\theta) \)
- The frictional force acting up the slope: \( F_{\text{friction}} = \mu mg \cos(\theta) \)

Where:
- \( m = 4 \, \text{kg} \)
- \( g = 10 \, \text{m/s}^2 \) (for simplicity)

### Step 4: Finding the Normal Force (N)
The normal force (N) acting on the box can be calculated as:
\[ N = mg \cos(\theta) \]

### Step 5: Force Required to Move the Box Up the Plane
To find the force \( F \) required to just move the box up the plane, we need to overcome both the gravitational force pulling it down and the frictional force:
\[ F = mg \sin(\theta) + \mu mg \cos(\theta) \]

### Step 6: Substitute Values
Now substituting the known values:
1. Calculate \( mg \sin(\theta) \):
- \( \sin(\theta) = \frac{1}{\sqrt{10}} \) (from the triangle formed by the slope)
- \( mg \sin(\theta) = 4 \times 10 \times \frac{1}{3} = \frac{40}{3} \)

2. Calculate \( mg \cos(\theta) \):
- \( \cos(\theta) = \frac{3}{\sqrt{10}} \)
- \( mg \cos(\theta) = 4 \times 10 \times \frac{3}{\sqrt{10}} = \frac{120}{\sqrt{10}} \)

3. Calculate the frictional force:
- \( F_{\text{friction}} = \mu mg \cos(\theta) = \frac{1}{3} \times \frac{120}{\sqrt{10}} = \frac{40}{\sqrt{10}} \)

### Step 7: Total Force Calculation
Now, substituting these into the equation for \( F \):
\[ F = \frac{40}{3} + \frac{40}{\sqrt{10}} \]

### Step 8: Final Calculation
Calculating the total force:
- Convert \( \frac{40}{\sqrt{10}} \) to a decimal or a common denominator to add to \( \frac{40}{3} \).

After calculating, we find that the total force required to move the box up the plane is approximately:
\[ F \approx 26.13 \, \text{N} \]

### Summary of Results
1. Coefficient of friction (μ) = \( \frac{1}{3} \)
2. Force required to move the box up the plane = \( 26.13 \, \text{N} \)