A mass of 200 kg is placed on a rough inclined plane of angle `30^(@)` . If coefficient of limiting friction is `1 // sqrt 3` , find the least forces in newton , acting parallel to the plane (i) to keep the mass from sliding down (ii) to move the mass up the plane.

Explanation

To solve the problem, we need to determine the forces acting on a mass of 200 kg on a rough inclined plane at an angle of 30 degrees, given that the coefficient of limiting friction is \( \frac{1}{\sqrt{3}} \). We will find the least force required to (i) keep the mass from sliding down and (ii) to move the mass up the plane.

### Step-by-Step Solution:

**Given:**
- Mass, \( m = 200 \, \text{kg} \)
- Angle of incline, \( \theta = 30^\circ \)
- Coefficient of limiting friction, \( \mu = \frac{1}{\sqrt{3}} \)
...