---
title: "Resultant force on curved surface BCDEF"
subject: "Fluid Mechanics"
chapter: 3
key: "semester_3/fluid_mechanics:curved-surface-bcdef-force"
source: https://www.bctengineer.com/q/semester_3/fluid_mechanics/curved-surface-bcdef-force
site: BCT Engineering
---

# Resultant force on curved surface BCDEF

Fluid Mechanics · Chapter 3: Hydrostatics

- Asked in: 2074 Asoj, 8 marks
- Page: https://www.bctengineer.com/q/semester_3/fluid_mechanics/curved-surface-bcdef-force
- Whole chapter: https://www.bctengineer.com/md/semester_3/fluid_mechanics/chapter-3

## Question

Find the resultant pressure force due to water on a curved surface BCDEF of 10 m length as shown in figure below. [Figure: tank of water (s = 1) with gauge reading -19 kN/m² at the top; curved surface from B through C, D, E to F with portions of 2 m, 2 m, 2 m, 2 m, 3 m dimensions marked]

## Short answer

Horizontal component, from pressure on vertical projection, is 625.8 kN; vertical component equals weight of water above surface up to equivalent free surface, 1812.2 kN. Resultant 1917 kN at 70.9° below horizontal.

## Answer

The resultant force on a curved surface has a horizontal component (force on the vertical projection) and a vertical component (weight of the liquid vertically above the surface up to the equivalent free surface). The figure is not fully readable, so the geometry below is assumed.

**Assumed geometry** (water on the concave side, length $L = 10$ m, gauge at level of B reads $-19$ kN/m²):

```text
   B |  gauge level, p = -19 kPa
     | BC = 2 m (vertical wall)
   C |
      \  CD, DE = two quarter circles,
       ) D  R = 2 m (semicircle C-D-E)
      /
   E |________ F   EF = 3 m (horizontal)
```

Vertical extent of B to E = 2 + 2 + 2 = 6 m, and EF lies at 6 m below B.

**Pressure at B and equivalent free surface**

$$
\frac{p_B}{\gamma} = \frac{-19}{9.81} = -1.937\ \text{m}
$$

The equivalent free surface is 1.937 m below B. Pressure at depth $z$ below B is $p = -19 + 9.81z$ kPa.

**Horizontal component** (vertical projection BE, height 6 m):

$$
\begin{aligned}
F_H &= L\int_0^6 (-19 + 9.81z)\,dz = L\left(-19\times 6 + 9.81\times\frac{6^2}{2}\right)\\
&= 10(-114 + 176.58) = 625.8\ \text{kN}
\end{aligned}
$$

**Vertical component**

- On the semicircle CDE: the upper quarter is pushed up and the lower quarter is pushed down by the liquid. The net force equals the weight of liquid in the half-disc enclosed between the two arcs:
$$
F_{V1} = \gamma\left(\frac{\pi R^2}{2}\right)L = 9.81\times 6.283\times 10 = 616.4\ \text{kN (downward)}
$$
- On the flat EF at 6 m depth: $p = -19 + 9.81\times 6 = 39.86$ kPa
$$
F_{V2} = 39.86\times 3\times 10 = 1195.8\ \text{kN (downward)}
$$

$$
F_V = 616.4 + 1195.8 = 1812.2\ \text{kN}
$$

**Resultant**

$$
\begin{aligned}
F &= \sqrt{625.8^2 + 1812.2^2} = 1917.2\ \text{kN}\\
\theta &= \tan^{-1}\frac{1812.2}{625.8} = 70.9^\circ \text{ below the horizontal}
\end{aligned}
$$

**Answer: $F \approx 1917$ kN (about 1.92 MN), inclined at $70.9^\circ$ to the horizontal, with $F_H = 625.8$ kN and $F_V = 1812.2$ kN.** If the actual figure differs, use the same method: $F_H = \int p\,dA_{proj}$ and $F_V$ = weight of liquid up to the equivalent free surface.

## Check yourself

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

1. How are the components of the force on a curved surface found?
   A. F_H on the vertical projection; F_V as weight of liquid up to the equivalent free surface
   B. F_H as weight of liquid; F_V on the vertical projection
   C. Both from the weight of liquid
   D. Both from the pressure at the centroid
   Answer: A. This is the standard curved-surface method. (Key point: Components)

2. The gauge reads -19 kPa at B. Where is the equivalent free surface?
   A. 1.937 m above B
   B. 1.937 m below B
   C. At B
   D. 19 m below B
   Answer: B. p/γ = -19/9.81 = -1.937 m, so the zero-pressure level is below B. (Key point: Negative gauge pressure)

3. What is F_H on the vertical projection BE (6 m, L = 10 m)?
   A. 10(19 x 6 + 176.58) = 2906 kN
   B. 9.81 x 36/2 x 10 = 1766 kN
   C. L(-19 x 6 + 9.81 x 6^2/2) = 625.8 kN
   D. -19 x 6 x 10 = -1140 kN
   Answer: C. Integrating p = -19 + 9.81 z over 0 to 6 m gives -114 + 176.58 per metre. (Key point: Horizontal component)

4. With F_H = 625.8 kN and F_V = 1812.2 kN, what is the resultant?
   A. F_H + F_V ≈ 2438 kN
   B. F_V - F_H ≈ 1186 kN
   C. F_V ≈ 1812 kN
   D. √(F_H^2 + F_V^2) ≈ 1917 kN
   Answer: D. Components are perpendicular so combine by Pythagoras; angle tan^-1(F_V/F_H) = 70.9°. (Key point: Resultant)

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