Hydrostatics – line of action of resultant on a vertical wall (single liquid) A vertical wall retains a single liquid to depth H. The resultant hydrostatic force acts at a distance __________ below the free surface (measured along the wall):

Introduction / Context:
Knowing where the resultant hydrostatic force acts is essential for designing retaining walls, gates, and tanks. For a vertical plane surface in a liquid, the pressure increases linearly with depth, producing a triangular distribution.

Given Data / Assumptions:

• Vertical wall of height H exposed to a single, homogeneous liquid.
• Free surface is open to atmosphere.
• Wall width is constant; gravity is uniform.

Concept / Approach:
For a vertical plane, gauge pressure p_g = w * h varies linearly from zero at the surface to w * H at the bottom, forming a triangular distribution. The line of action (center of pressure) of a triangle load lies at one-third the height from the base, i.e., two-thirds from the apex (free surface).

Step-by-Step Solution:
Step 1: Recognize triangular pressure diagram with zero at h = 0 and maximum at h = H.Step 2: For a triangle, centroid location from the base is H/3.Step 3: Thus, from the free surface, the distance is H − H/3 = 2H/3.Step 4: Therefore, the resultant hydrostatic force acts at 2H/3 below the surface.

Verification / Alternative check:
Formal integration for the center of pressure on a rectangle gives the same result for a vertical wall: y_cp = I_G / (A * ȳ) + ȳ, which simplifies to 2H/3 for the hydrostatic triangle.

Why Other Options Are Wrong:

• H/3: this is the location above the bottom, not from the surface.
• H/2, 3H/4, H/4: do not correspond to the centroid of a triangle with zero at top.

Common Pitfalls:
Confusing the center of pressure with the centroid of the plane area; due to pressure variation, the resultant is below the geometric centroid.

Final Answer:
2H/3