**Given Information:**
- Mass (m) = 200 kg
- Angle of inclined plane (θ) = 30°
- Coefficient of limiting friction (μ) = (√3)/3

**To Find:**
- Greatest and least forces acting parallel to the plane to keep the mass in equilibrium.

**Assumptions:**
- The mass is on the verge of moving up or down the inclined plane.
- The force of gravity acting on the mass is 9.8 m/s².

**Analysis:**

1. Decompose Forces:
- Resolve the weight of the mass into two components: one parallel to the inclined plane (mg sinθ) and the other perpendicular to the inclined plane (mg cosθ).
- The perpendicular component (mg cosθ) is balanced by the normal force from the inclined plane.
- The parallel component (mg sinθ) is the force that needs to be balanced by the frictional force.

2. Maximum Force:
- The maximum force of friction (F_max) can be calculated using the formula: F_max = μN, where N is the normal force.
- The normal force (N) can be calculated as: N = mg cosθ.
- Substitute the values: N = (200 kg)(9.8 m/s²)(cos30°) = 1694.52 N.
- Substitute the value of N in the formula for F_max: F_max = (√3)/3 * 1694.52 N ≈ 977.75 N.
- Therefore, the greatest force acting parallel to the plane to keep the mass in equilibrium is approximately 977.75 N.

3. Minimum Force:
- The minimum force of friction (F_min) can be calculated using the formula: F_min = μN, where N is the normal force.
- Substitute the value of N calculated earlier: F_min = (√3)/3 * 1694.52 N ≈ 977.75 N.
- Therefore, the least force acting parallel to the plane to keep the mass in equilibrium is approximately 977.75 N.

**Conclusion:**
- The greatest force acting parallel to the plane to keep the mass in equilibrium is approximately 977.75 N.
- The least force acting parallel to the plane to keep the mass in equilibrium is approximately 977.75 N.

Maximum force is 1960 N and minimum force is 0