For a vertical retaining wall of height h supporting a hydrostatic or earth-pressure distribution that varies linearly with depth, the resultant total pressure acts parallel to the free surface and at what distance above the base?

Introduction / Context:
Retaining walls resist lateral pressures from fluids or soils. When the pressure distribution is triangular (zero at the top and maximum at the base), determining the line of action of the resultant is essential for stability checks against overturning and sliding.

Given Data / Assumptions:

• Wall height = h.
• Pressure varies linearly from zero at the top to a maximum at the base.
• Resultant acts parallel to the free surface (i.e., horizontal for hydrostatic/active pressure).

Concept / Approach:
A linearly varying pressure distribution forms a triangle. The resultant of a triangular load acts through its centroid, located at one-third of the height from the base (i.e., h/3 above the base). The magnitude equals the area of the triangle (0.5 * base_pressure * h), and its line of action is crucial for overturning moment calculations.

Step-by-Step Solution:
Identify pressure diagram: triangular with zero at top and maximum at base.Compute resultant location: centroid of triangle is at h/3 from the base.Direction: parallel to the free surface (typically horizontal).

Verification / Alternative check:
For hydrostatics, an identical conclusion arises from integrating pressure p = gamma * z over depth. For earth pressure with linear distribution, the same centroidal property holds, confirming h/3 from the base.

Why Other Options Are Wrong:

• h/4 and h/2: not the centroid of a triangle.
• 2h/3: corresponds to the location from the top, not from the base.

Common Pitfalls:
Mixing the location measured from the top versus from the base; forgetting that the direction is parallel to the free surface, not inclined like the wall backface in some cases.

Final Answer:
h/3.