A body is sliding down an inclined plane having coefficient of friction `1//3`. If the normal reaction is three times that of the resultant downward force along the inclined plane, the angle between the inclined plane and the horizontal is

Options

A. `tan^(-1)[(1)/(2)]`
B. `tan^(-1)(2)`
C. `tan^(-1)((2)/(3))`
D. `tan^(-1)((3)/(2))`

Explanation

To solve the problem step by step, we will analyze the forces acting on the body sliding down the inclined plane and use the given conditions to find the angle θ between the inclined plane and the horizontal.

### Step 1: Identify the forces acting on the body
The forces acting on the body are:
- Gravitational force (weight) acting downwards: \( mg \)
- Normal force \( N \) acting perpendicular to the inclined plane
- Frictional force \( F \) acting opposite to the direction of motion along the inclined plane

### Step 2: Resolve the gravitational force into components
The gravitational force can be resolved into two components:
- Perpendicular to the inclined plane: \( mg \cos \theta \)
- Parallel to the inclined plane: \( mg \sin \theta \)

### Step 3: Write the expression for the normal force
Since there is no acceleration perpendicular to the inclined plane, the normal force \( N \) balances the perpendicular component of the gravitational force:
\[ N = mg \cos \theta \]

### Step 4: Write the expression for the frictional force
The frictional force \( F \) can be expressed using the coefficient of friction \( \mu \):
\[ F = \mu N = \frac{1}{3} N = \frac{1}{3} (mg \cos \theta) \]

### Step 5: Write the expression for the net force along the inclined plane
The net force \( R \) acting down the inclined plane is given by:
\[ R = mg \sin \theta - F \]
Substituting the expression for \( F \):
\[ R = mg \sin \theta - \frac{1}{3} (mg \cos \theta) \]
\[ R = mg \sin \theta - \frac{1}{3} mg \cos \theta \]
Factoring out \( mg \):
\[ R = mg \left( \sin \theta - \frac{1}{3} \cos \theta \right) \]

### Step 6: Use the relationship between normal force and resultant force
According to the problem, the normal force is three times the resultant downward force:
\[ N = 3R \]
Substituting the expressions for \( N \) and \( R \):
\[ mg \cos \theta = 3 \left( mg \left( \sin \theta - \frac{1}{3} \cos \theta \right) \right) \]
Cancelling \( mg \) from both sides (assuming \( m \neq 0 \)):
\[ \cos \theta = 3 \left( \sin \theta - \frac{1}{3} \cos \theta \right) \]

### Step 7: Simplify the equation
Distributing the 3:
\[ \cos \theta = 3 \sin \theta - \cos \theta \]
Bringing all terms involving \( \cos \theta \) to one side:
\[ \cos \theta + \cos \theta = 3 \sin \theta \]
\[ 2 \cos \theta = 3 \sin \theta \]

### Step 8: Divide both sides by \( \cos \theta \)
\[ 2 = 3 \tan \theta \]
Thus,
\[ \tan \theta = \frac{2}{3} \]

### Step 9: Find the angle \( \theta \)
Taking the arctangent:
\[ \theta = \tan^{-1} \left( \frac{2}{3} \right) \]

### Final Answer
The angle between the inclined plane and the horizontal is:
\[ \theta = \tan^{-1} \left( \frac{2}{3} \right) \]

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