Hydrostatic Force on a Hemispherical Wall Projection: 04-BS-7, December 2014, Q1

Question 1 — Hydrostatic Force on a Hemispherical Wall Projection (5 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given.

Quantity	Value
Hemisphere radius, R	1 m
Depth of centre A below free surface, h_A	2.5 m
Density of water, ρ	1000 kg/m³

Find. The horizontal force F_H, vertical force F_V, and the resultant total force F_R on the hemispherical projection.

Fig. Q1 — solid hemisphere (radius R = 1 m) bulging horizontally into the tank from the wall, centre A at depth 2.5 m; the resultant hydrostatic force F_R combines a horizontal thrust and an upward "buoyant" component.

Approach. The horizontal force equals the pressure at the centroid of the vertical-plane projection (a full circle of radius R) times that area; the vertical force equals the weight of the (real or virtual) water that would fill the hemisphere's own volume — the "buoyant effect" the hint points to. The resultant combines the two vectorially.

Horizontal force (projected-area method). The vertical-plane projection of the curved hemisphere is a full circle of radius R centred at the same depth as A: $$F_H = \rho g h_A (\pi R^2) = 1000 \times 9.81 \times 2.5 \times \pi(1)^2 = \boxed{77.05\text{ kN}}$$
Vertical force (buoyant-effect hint). Integrating the pressure's vertical component over the curved surface reduces exactly to the weight of a hemisphere-shaped volume of water, acting upward: $$F_V = \rho g \left(\tfrac{2}{3}\pi R^3\right) = 1000\times 9.81\times \tfrac{2}{3}\pi(1)^3 = \boxed{20.55\text{ kN (upward)}}$$
Resultant total force. Combining the perpendicular horizontal and vertical components: $$F_R = \sqrt{F_H^2+F_V^2} = \sqrt{77.05^2+20.55^2} = \boxed{79.74\text{ kN}}, \qquad \theta=\tan^{-1}\!\left(\frac{20.55}{77.05}\right)=14.9^\circ \text{ above horizontal}$$

Quantity	Result
Horizontal force, F_H	77.05 kN
Vertical force, F_V	20.55 kN (up)
Resultant force, F_R	79.74 kN at 14.9° above horizontal

Topic: Hydrostatic force on a curved (3-D) submerged surface.

Key relations: F_H = ρg h_c A_proj (vertical-plane projected area); F_V = ρg × (volume of real or virtual fluid enclosed above/by the surface).

Why this works: pressure acts normal to every surface element; resolving into horizontal and vertical components and integrating collapses the curvature — any curved surface sharing the same silhouette gives the same F_H, while the vertical component always reduces to the weight of the associated fluid volume by the same reasoning used to derive Archimedes' principle.

Common pitfall: assuming F_V is zero by "up-down symmetry" of the hemisphere — pressure increases with depth, so the lower cap and upper cap do NOT cancel; that residual asymmetry is exactly what survives integration and produces the hint's buoyant-volume result.

Source: White, Fluid Mechanics, 8th ed., §2.6 (hydrostatic forces on curved surfaces).

Question 1 of 13: Hydrostatic Force on a Hemispherical Wall Projection

Question 1 — Hydrostatic Force on a Hemispherical Wall Projection (5 marks)