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 incline, 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)`

Answer

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 information to find the angle between the inclined plane and the horizontal.

### Step 1: Identify the forces acting on the body
The forces acting on the body include:
- The weight of the body (mg), acting vertically downwards.
- The normal force (N), acting perpendicular to the inclined plane.
- The frictional force (f), acting opposite to the direction of motion, which can be calculated using the coefficient of friction (μ).

### Step 2: Break down the weight into components
The weight can be resolved into two components:
- The component parallel to the incline: \( mg \sin \theta \)
- The component perpendicular to the incline: \( mg \cos \theta \)

### Step 3: Write the expression for the normal force
The normal force (N) is equal to the perpendicular component of the weight:
\[ N = mg \cos \theta \]

### Step 4: Write the expression for the frictional force
The frictional force (f) can be expressed as:
\[ f = \mu N = \mu (mg \cos \theta) \]
Given that \( \mu = \frac{1}{3} \):
\[ f = \frac{1}{3} (mg \cos \theta) \]

### Step 5: Write the expression for the net downward force along the incline
The net downward force (F_net) along the incline is given by:
\[ F_{\text{net}} = mg \sin \theta - f \]
Substituting the expression for friction:
\[ F_{\text{net}} = mg \sin \theta - \frac{1}{3} (mg \cos \theta) \]
Factoring out \( mg \):
\[ F_{\text{net}} = mg \left( \sin \theta - \frac{1}{3} \cos \theta \right) \]

### Step 6: Use the given condition about the normal force
According to the problem, the normal force is three times the resultant downward force along the incline:
\[ N = 3 F_{\text{net}} \]
Substituting the expressions for N and F_net:
\[ mg \cos \theta = 3 \left( mg \left( \sin \theta - \frac{1}{3} \cos \theta \right) \right) \]

### Step 7: Simplify the equation
Cancelling \( mg \) from both sides (assuming \( mg \neq 0 \)):
\[ \cos \theta = 3 \left( \sin \theta - \frac{1}{3} \cos \theta \right) \]
Distributing the 3:
\[ \cos \theta = 3 \sin \theta - \cos \theta \]
Adding \( \cos \theta \) to both sides:
\[ 2 \cos \theta = 3 \sin \theta \]

### Step 8: Rearranging to find the relationship between sin and cos
Dividing both sides by \( \cos \theta \):
\[ 2 = 3 \tan \theta \]
Thus, we can express \( \tan \theta \):
\[ \tan \theta = \frac{2}{3} \]

### Step 9: Find the angle θ
To find the angle \( \theta \), we take the arctangent:
\[ \theta = \tan^{-1}\left(\frac{2}{3}\right) \]

### Conclusion
The angle \( \theta \) between the inclined plane and the horizontal can be calculated using a calculator or trigonometric tables.