The minimum force required to move a body up on 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/(2sqrt3)` the angle of the inclined plane is

Options

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

Explanation

To solve the problem, we need to find the angle of inclination (θ) of the inclined plane given that the minimum force required to move a body up the plane is three times the minimum force required to prevent it from sliding down. The coefficient of friction (μ) is given as \( \frac{1}{2\sqrt{3}} \).

### Step-by-Step Solution:

1. **Identify Forces Acting on the Body**:
- The weight of the body (W) acts downward: \( W = mg \).
- This weight can be resolved into two components:
- Perpendicular to the inclined plane: \( W_{\perp} = mg \cos \theta \)
- Parallel to the inclined plane: \( W_{\parallel} = mg \sin \theta \)

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

3. **Frictional Force (F_f)**:
- The frictional force opposing the motion when moving up is given by:
\[ F_f = \mu N = \mu (mg \cos \theta) = \frac{1}{2\sqrt{3}} (mg \cos \theta) \]

4. **Force Required to Move Up (F_u)**:
- The force required to move the body up the incline is:
\[ F_u = mg \sin \theta + F_f = mg \sin \theta + \frac{1}{2\sqrt{3}} (mg \cos \theta) \]

5. **Force Required to Prevent Sliding Down (F_d)**:
- The minimum force required to prevent the body from sliding down is:
\[ F_d = mg \sin \theta - F_f = mg \sin \theta - \frac{1}{2\sqrt{3}} (mg \cos \theta) \]

6. **Relationship Between Forces**:
- According to the problem, \( F_u = 3F_d \):
\[ mg \sin \theta + \frac{1}{2\sqrt{3}} (mg \cos \theta) = 3 \left( mg \sin \theta - \frac{1}{2\sqrt{3}} (mg \cos \theta) \right) \]

7. **Simplifying the Equation**:
- Cancel \( mg \) from both sides:
\[ \sin \theta + \frac{1}{2\sqrt{3}} \cos \theta = 3 \left( \sin \theta - \frac{1}{2\sqrt{3}} \cos \theta \right) \]
- Expanding the right side:
\[ \sin \theta + \frac{1}{2\sqrt{3}} \cos \theta = 3\sin \theta - \frac{3}{2\sqrt{3}} \cos \theta \]
- Rearranging gives:
\[ \sin \theta + \frac{1}{2\sqrt{3}} \cos \theta - 3\sin \theta + \frac{3}{2\sqrt{3}} \cos \theta = 0 \]
- Combine like terms:
\[ -2\sin \theta + 2\cos \theta = 0 \]
- This simplifies to:
\[ \tan \theta = 1 \]

8. **Finding the Angle**:
- Therefore, \( \theta = 45^\circ \).

9. **Substituting the Coefficient of Friction**:
- Since we have \( \mu = \frac{1}{2\sqrt{3}} \), we can substitute it back to find the specific angle:
- From the earlier derived equation \( \tan \theta = 2\mu \):
\[ \tan \theta = 2 \times \frac{1}{2\sqrt{3}} = \frac{1}{\sqrt{3}} \]
- Thus, \( \theta = 30^\circ \).

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