---
title: "Hydrostatic force on sector gate (4 m long)"
subject: "Hydro Power"
chapter: 1
key: "semester_6/hydro_power:sector-gate-4m"
source: https://www.bctengineer.com/q/semester_6/hydro_power/sector-gate-4m
site: BCT Engineering
---

# Hydrostatic force on sector gate (4 m long)

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

- Asked in: 2071 Magh, 7 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/sector-gate-4m
- Whole chapter: https://www.bctengineer.com/md/semester_6/hydro_power/chapter-1

## Question

The sector shown in figure below consists of a cylindrical surface of which AB is the trace, supported by a structural frame hinged at C. The length of the gate is 4 m. Determine the magnitude and location of the horizontal and vertical component of the total hydrostatic pressure on the gate. [Figure: circular-arc gate AB of radius 6 m centred at hinge C; A is at the water surface and B at the bed; radius CA is inclined at 60° to the horizontal, and an angle of 30° is also marked at C; water is held on the convex side of the arc.]

## Short answer

Horizontal force FH = 529.74 kN acts 1.732 m above bed. Vertical force FV = 177.88 kN downward acts 5.158 m from hinge. Resultant force = 558.81 kN at 18.56° to horizontal passing through hinge C.

## Answer

Resolve the hydrostatic force into horizontal and vertical components. Gate length $L = 4$ m.

**Reading of the figure:** radius $CA$ (A at the water surface) makes 60° with the horizontal, and $CB$ (B at the bed) is horizontal, so the gate subtends 60° at C (the 30° mark is the angle of CA from the vertical). Water is on the convex side.

#### Geometry
Take the hinge as origin. Radius $R = 6$ m, angle between the two radii $= 60^\circ$, one radius horizontal at bed level. The other end of the arc is at the water surface, so
$$
H = R\sin 60^\circ = 6 \times 0.8660 = 5.196\ \text{m (water depth)}
$$
Horizontal distance of the upper end from the hinge $= R\cos 60^\circ = 3$ m.

```text
   water surface ======= upper end of arc
                    (3, 5.196)
       hinge  .   |\  
        |   .     | \ water on the
        | .  60deg|  \  convex side
        o---------+---* lower end (6, 0)
        <--- R = 6 m --->        bed
```

#### Horizontal component
Equal to the force on the vertical projection ($H \times L$):
$$
F_H = \rho g \frac{H^2}{2} L = 9.81 \times \frac{5.196^2}{2} \times L = 132.435\,L\ \text{kN}
$$
It acts at $H/3 = 1.732$ m above the bed (3.464 m below the surface).

#### Vertical component
Equal to the weight of water lying vertically above the arc, up to the surface. Area of that region:
$$
\begin{aligned}
A_v &= \underbrace{3 \times 5.196}_{\text{rectangle}} - \left(\underbrace{\tfrac{60}{360}\pi 6^2}_{\text{sector}} - \underbrace{\tfrac12 \times 3 \times 5.196}_{\text{triangle}}\right) \\
&= 15.588 - (18.850 - 7.794) = 4.533\ \text{m}^2
\end{aligned}
$$
$$
F_V = \rho g A_v L = 9.81 \times 4.533 \times L = 44.470\,L\ \text{kN (downward)}
$$
Its line of action passes through the centroid of this area, found by integration: 5.158 m horizontally from the hinge, i.e. 0.842 m from the lower end of the arc.

**Check:** every pressure force on a circular arc passes through the centre, so the resultant must pass through the hinge. Moment about the hinge: $132.435 \times 1.732 = 229.4$ and $44.470 \times 5.158 = 229.4$ (per m), which balance.

#### Results for $L = 4$ m
$$
\begin{aligned}
F_H &= 132.435 \times 4 = 529.74\ \text{kN} \\
F_V &= 44.470 \times 4 = 177.88\ \text{kN} \\
F &= \sqrt{529.74^2 + 177.88^2} = 558.81\ \text{kN at } 18.56^\circ \text{ to the horizontal}
\end{aligned}
$$

**Answer:** $F_H = 529.74$ kN, acting 1.732 m above the bed (3.464 m below the surface); $F_V = 177.88$ kN downward, acting 5.158 m horizontally from hinge C. The resultant passes through C.

## Check yourself

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

1. How is the horizontal component of force on a curved surface obtained?
   A. Weight of water above the surface
   B. Force on the horizontal projection
   C. Force on the vertical projection of the surface
   D. Pressure at the centroid times arc length only
   Answer: C. F_H equals the force on the vertical projection (H x L), i.e. rho g H^2/2 x L, acting at H/3 above the bed. (Key point: Horizontal component rule)

2. How is the vertical component computed for this gate?
   A. Force on the vertical projection
   B. Weight of the gate
   C. Pressure at the bed times the base length
   D. Weight of water vertically above the curved surface up to the free surface
   Answer: D. F_V equals the weight of the liquid above the curved surface (real or imaginary) up to the free surface. (Key point: Vertical component rule)

3. Why must the resultant force pass through the hinge?
   A. Pressure acts normal to a circular arc, so every elemental force passes through the centre
   B. The hinge is at the centre of gravity of the water
   C. The gate is weightless
   D. Because F_H equals F_V
   Answer: A. Elemental pressure forces are radial for a circular arc, so their resultant passes through the centre. This was verified by equal moments of 229.4 kN m per metre. (Key point: Resultant passes through centre)

4. Per metre values are 132.435 kN (horizontal) and 44.47 kN (vertical). What are the components for L = 4 m?
   A. 529.74 kN and 177.88 kN
   B. 1324 kN and 445 kN
   C. 177.88 kN and 529.74 kN
   D. 132.4 kN and 44.5 kN
   Answer: A. Multiply the per-metre forces by L = 4 m. (Key point: Scaling with gate length)

## Close variants set in other exams

- [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×