A mass of 200kg is resting on a rough inclined plane of angle `30^(@)`. If the coefficient of friction is `1//sqrt(3)` , find the greatest and the least force acting parallel to the plane which can just maintain the mass in equilibrium.

Explanation

To solve the problem, we need to analyze the forces acting on the mass resting on the inclined plane. We will find the greatest and least force acting parallel to the plane that can maintain the mass in equilibrium.

### Step 1: Identify the Forces
1. The weight of the mass (W) acts vertically downwards and is given by:
\[ W = mg \]
where \( m = 200 \, \text{kg} \) and \( g = 10 \, \text{m/s}^2 \).
\[ W = 200 \times 10 = 2000 \, \text{N} \]

2. The weight can be resolved into two components:
- Perpendicular to the incline: \( W_{\perp} = mg \cos \theta \)
- Parallel to the incline: \( W_{\parallel} = mg \sin \theta \)

### Step 2: Calculate the Components
1. Calculate \( W_{\perp} \) and \( W_{\parallel} \):
- For \( \theta = 30^\circ \):
\[ W_{\perp} = 2000 \cos(30^\circ) = 2000 \times \frac{\sqrt{3}}{2} = 1000\sqrt{3} \, \text{N} \]
\[ W_{\parallel} = 2000 \sin(30^\circ) = 2000 \times \frac{1}{2} = 1000 \, \text{N} \]

### Step 3: Calculate the Normal Force
1. The normal force (N) is equal to the perpendicular component of the weight:
\[ N = W_{\perp} = 1000\sqrt{3} \, \text{N} \]

### Step 4: Calculate the Frictional Force
1. The maximum frictional force (f_max) can be calculated using the coefficient of friction (\( \mu \)):
\[ f_{\text{max}} = \mu N = \frac{1}{\sqrt{3}} \times 1000\sqrt{3} = 1000 \, \text{N} \]

### Step 5: Determine the Forces for Equilibrium
1. For the mass to be in equilibrium, the net force acting parallel to the incline must be zero. Thus, we can write:
\[ F_{\text{applied}} - W_{\parallel} - f_{\text{max}} = 0 \]
Rearranging gives:
\[ F_{\text{applied}} = W_{\parallel} + f_{\text{max}} = 1000 + 1000 = 2000 \, \text{N} \]

2. For the least force (F_min), we consider the friction acting in the opposite direction:
\[ F_{\text{applied}} - W_{\parallel} + f_{\text{max}} = 0 \]
Rearranging gives:
\[ F_{\text{applied}} = W_{\parallel} - f_{\text{max}} = 1000 - 1000 = 0 \, \text{N} \]

### Conclusion
- The greatest force acting parallel to the plane that can maintain the mass in equilibrium is **2000 N**.
- The least force acting parallel to the plane that can maintain the mass in equilibrium is **0 N**.