---
title: "Hydrostatic force on curved surface"
subject: "Hydro Power"
chapter: 1
key: "semester_6/hydro_power:hydrostatic-force-curved-surface"
source: https://www.bctengineer.com/q/semester_6/hydro_power/hydrostatic-force-curved-surface
site: BCT Engineering
---

# Hydrostatic force on curved surface

Hydro Power · Chapter 1: Introduction to Basic Fluid Mechanics and Hydraulics

- Asked in: 2081 Chaitra, 4 marks
- Video: [Hydrostatic pressure on submerged curved surfaces sample problem | Fluid Mechanics Lesson 10](https://www.youtube.com/watch?v=Z8KWoAl7nLY) — Tutor Jack Ph
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_6/hydro_power/hydrostatic-force-curved-surface
- Whole chapter: https://www.bctengineer.com/md/semester_6/hydro_power/chapter-1

## Question

How can the hydrostatic force on a curved surface be analyzed? Explain.

## Short answer

Resolve into horizontal force F_H = ρg A_v h̅_v on the vertical projection and vertical force F_V = ρg V, the weight of liquid above (real or imaginary). The resultant F = √(F_H²+F_V²) passes through the centre of curvature for cylindrical surfaces.

## Answer

On a curved surface the pressure acts normal to each element, so the elemental forces have different directions. The total force is found by resolving it into **horizontal** and **vertical** components.

```text
   free surface ==================
        :  water above   |
        :  (volume ABCD) |  F_V = weight of
        :                |  this water
        D________________A
                         \    <- curved surface AB
                          \
                           B
     F_H = force on vertical projection of AB
```

#### Horizontal component
$$
F_H = \rho g A_v \bar h_v
$$
- $A_v$ = area of the **vertical projection** of the curved surface, $\bar h_v$ = depth of its centroid.
- $F_H$ acts through the centre of pressure of this projected area, at depth $\bar h_v + \dfrac{I_G}{A_v \bar h_v}$.

#### Vertical component
$$
F_V = \rho g V
$$
- $V$ = volume of liquid (real or imaginary) lying vertically above the curved surface up to the free surface.
- $F_V$ acts through the centroid of that volume.
- If the liquid is above the surface, $F_V$ acts downward. If the liquid is below the surface, $F_V$ acts upward and equals the weight of the imaginary liquid above it.

#### Resultant
$$
F = \sqrt{F_H^2 + F_V^2}, \qquad \tan\theta = \frac{F_V}{F_H}
$$
For a circular (cylindrical) surface, every elemental force is normal to the surface and passes through the centre, so the resultant also passes through the centre of curvature.

Applications: radial (Tainter) gates, sector gates and drum gates on spillways, curved dam faces and pipe bends.

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. The hydrostatic force on a curved surface is found by:
   A. Treating it as a flat surface
   B. Resolving it into horizontal and vertical components
   C. Using only the surface area
   D. Using only the centroid depth
   Answer: B. Pressure acts normal to each element in changing directions, so horizontal and vertical components are found separately. (Key point: Components)

2. The horizontal component F_H equals:
   A. rho g V
   B. rho g A_v h-bar_v, force on the vertical projection
   C. rho g A h-bar of the curved area
   D. Weight of liquid above
   Answer: B. F_H is the total pressure on the vertical projection of the surface. (Key point: Horizontal component)

3. The vertical component F_V equals:
   A. Force on the vertical projection
   B. Weight of real or imaginary liquid vertically above the surface up to the free surface
   C. Weight of the surface
   D. Zero
   Answer: B. F_V = rho g V, with V the volume of liquid above the surface, acting through its centroid. (Key point: Vertical component)

4. The resultant force and its inclination are:
   A. F = F_H + F_V; tan(theta) = F_H/F_V
   B. F = sqrt(F_H^2 + F_V^2); tan(theta) = F_V/F_H
   C. F = F_H F_V; tan(theta) = F_V
   D. F = F_H - F_V; theta = 45 degrees
   Answer: B. The components are perpendicular, so they combine vectorially; for a circular surface the resultant passes through the centre of curvature. (Key point: Resultant)

## Related in Hydro Power

- [Hydrostatic force on sector gate (10 m long)](https://www.bctengineer.com/question.md?key=semester_6%2Fhydro_power%3Asector-gate-10m) — Hydro Power, chapter 1, asked 2×
- [Hydrostatic force on sector gate (4 m long)](https://www.bctengineer.com/question.md?key=semester_6%2Fhydro_power%3Asector-gate-4m) — Hydro Power, chapter 1, asked 2×
- [Hydrostatic thrust and uplift with drainage gallery](https://www.bctengineer.com/question.md?key=semester_6%2Fhydro_power%3Adam-thrust-uplift-gallery) — Hydro Power, chapter 4
- [Net force on panel with water and glycerin](https://www.bctengineer.com/question.md?key=semester_6%2Fhydro_power%3Apanel-water-glycerin) — Hydro Power, chapter 1

## Related in other subjects

- [Net hydrostatic force on conical surface](https://www.bctengineer.com/question.md?key=semester_3%2Ffluid_mechanics%3Anet-force-conical-surface) — Fluid Mechanics, chapter 3
- [Resultant force on curved surface BCDEF](https://www.bctengineer.com/question.md?key=semester_3%2Ffluid_mechanics%3Acurved-surface-bcdef-force) — Fluid Mechanics, chapter 3
- [Resultant force on curved surface ABCDE (S = 1.1)](https://www.bctengineer.com/question.md?key=semester_3%2Ffluid_mechanics%3Acurved-surface-abcde-s11-20m) — Fluid Mechanics, chapter 3
- [Resultant force on curved surface ABCDE (S = 1.25)](https://www.bctengineer.com/question.md?key=semester_3%2Ffluid_mechanics%3Acurved-surface-abcde-s125-10m) — Fluid Mechanics, chapter 3