The minimum force required to move a body up an inclined plane is three times the minimum force required to prevent it from sliding down the plane. If the coefficient of friction between the body and the inclined plane is `(1)/(2sqrt(3))`, the angle of the inclined plane is

Options

A. `60^(@)`
B. `45^(@)`
C. `30^(@)`
D. `15^(@)`

Answer

To solve the problem, we need to find the angle of inclination (θ) of the inclined plane given the relationship between the forces acting on a body on the plane. Here’s a step-by-step breakdown of the solution:

### Step 1: Understanding the Forces Acting on the Body
When a body is placed on an inclined plane, the gravitational force (mg) acting on it can be resolved into two components:
- Perpendicular to the incline: \( mg \cos \theta \)
- Parallel to the incline: \( mg \sin \theta \)

### Step 2: Setting Up the Equations
1. The limiting frictional force (F_L) that prevents the body from sliding down the incline can be expressed as:
\[ F_L = \mu N = \mu (mg \cos \theta) \]
where \( \mu = \frac{1}{2\sqrt{3}} \).

2. Therefore, substituting for N:
\[ F_L = \frac{1}{2\sqrt{3}} (mg \cos \theta) \]

### Step 3: Minimum Force to Prevent Sliding Down
To prevent the body from sliding down, the force of gravity acting down the incline must be balanced by the frictional force and the applied force (F_1):
\[ mg \sin \theta = F_1 + F_L \]
Substituting \( F_L \):
\[ mg \sin \theta = F_1 + \frac{1}{2\sqrt{3}} (mg \cos \theta) \]

### Step 4: Expressing F_1
Rearranging the equation gives:
\[ F_1 = mg \sin \theta - \frac{1}{2\sqrt{3}} (mg \cos \theta) \]

### Step 5: Minimum Force to Move Up the Incline
The minimum force required to move the body up the incline (F_2) is given as three times the minimum force to prevent sliding down:
\[ F_2 = 3F_1 \]

### Step 6: Setting Up the Equation for F_2
When moving up the incline, the friction acts in the opposite direction:
\[ F_2 = mg \sin \theta + F_L \]
Substituting for \( F_L \):
\[ F_2 = mg \sin \theta + \frac{1}{2\sqrt{3}} (mg \cos \theta) \]

### Step 7: Equating F_2 with 3F_1
Now, substituting \( F_2 \) and \( F_1 \) into the equation:
\[ mg \sin \theta + \frac{1}{2\sqrt{3}} (mg \cos \theta) = 3 \left( mg \sin \theta - \frac{1}{2\sqrt{3}} (mg \cos \theta) \right) \]

### Step 8: Simplifying the Equation
Dividing through by \( mg \) (assuming \( mg \neq 0 \)):
\[ \sin \theta + \frac{1}{2\sqrt{3}} \cos \theta = 3 \sin \theta - \frac{3}{2\sqrt{3}} \cos \theta \]

Rearranging gives:
\[ \sin \theta + \frac{1}{2\sqrt{3}} \cos \theta + \frac{3}{2\sqrt{3}} \cos \theta = 3 \sin \theta \]
\[ \sin \theta + \frac{2}{\sqrt{3}} \cos \theta = 3 \sin \theta \]
\[ \frac{2}{\sqrt{3}} \cos \theta = 2 \sin \theta \]

### Step 9: Finding the Angle θ
Dividing both sides by 2:
\[ \frac{1}{\sqrt{3}} \cos \theta = \sin \theta \]
This can be rewritten as:
\[ \tan \theta = \frac{1}{\sqrt{3}} \]
Thus, the angle θ is:
\[ \theta = 30^\circ \]

### Final Answer
The angle of the inclined plane is \( \theta = 30^\circ \).