Center of pressure on an inclined plane surface For a plane surface fully submerged and inclined, the center of pressure is located where relative to the centroid of the area?

Introduction / Context:
Designing gates, tanks, and dam faces requires knowing where the resultant hydrostatic force acts. For plane surfaces, the resultant acts at the center of pressure, generally not at the centroid due to the pressure variation with depth.

Given Data / Assumptions:

• Incompressible fluid at rest; hydrostatic pressure increases with depth.
• Plane surface fully submerged; surface is inclined with its centroid below the free surface.
• Pressure distribution is linear with depth.

Concept / Approach:

The resultant hydrostatic force equals pressure at the centroid times the area, but its line of action passes below the centroid. The depth of the center of pressure y_cp (measured vertically) is y_cp = y_c + I_G / (y_c A), where y_c is centroid depth, I_G is the second moment of area about the free-surface-parallel axis through the centroid, and A is area. Since I_G, y_c, A are positive, y_cp > y_c, i.e., deeper than the centroid.

Step-by-Step Solution:

Verification / Alternative check:

For a rectangular plate of height h with centroid depth y_c, the added term simplifies to h^2/(12 y_c) for the axis choice, which is positive, confirming the center of pressure is below the centroid for any finite y_c.

Why Other Options Are Wrong:

(a) Only true if pressure were uniform, which it is not in a fluid at rest with depth. (b) Above the centroid is impossible for the linear hydrostatic distribution. (d) Position depends on geometry and submergence. (e) Free surface is not a feasible location for a submerged plane’s resultant.

Common Pitfalls:

Confusing line of action with magnitude; forgetting the inclination angle does not change the conclusion about being deeper than the centroid.

Final Answer:

below the centroid (deeper than the centroid)