--- title: "Hydrostatic force on sector gate (10 m long)" subject: "Hydro Power" chapter: 1 key: "semester_6/hydro_power:sector-gate-10m" source: https://www.bctengineer.com/q/semester_6/hydro_power/sector-gate-10m site: BCT Engineering --- # Hydrostatic force on sector gate (10 m long) Hydro Power · Chapter 1: Introduction to Basic Fluid Mechanics and Hydraulics - Asked in: 2078 Chaitra, 10 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-10m - Whole chapter: https://www.bctengineer.com/md/semester_6/hydro_power/chapter-1 ## Question The sector gate shown in figure below consists of a cylindrical surface of which AB is the trace, supported by a structural frame hinged at O. The length of the gate is 10 m. Determine the magnitude and location of the horizontal and vertical component of the total hydrostatic force on the gate. [Figure: circular-arc gate AB of radius 6 m centred at hinge O; OA is horizontal (radius = 6 m) with A at the bed level of O, OB makes θ = 60° with OA, and B is at the water surface; water lies on the convex side of the arc.] ## Short answer Sector gate (radius 6 m, 60° arc, length 10 m) on convex side: horizontal force 1324.35 kN at 1.732 m above bed from vertical projection; vertical force 444.70 kN downward at 5.158 m from hinge from water weight above arc. Resultant 1397 kN at 18.56° passes through hinge. ## Answer Resolve the hydrostatic force on the curved gate into a horizontal component (force on the vertical projection) and a vertical component (weight of water above the curved surface). Gate length $L = 10$ m. #### 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 $L = 10$ m $$ \begin{aligned} F_H &= 132.435 \times 10 = 1324.35\ \text{kN} \\ F_V &= 44.470 \times 10 = 444.70\ \text{kN} \\ F &= \sqrt{1324.35^2 + 444.70^2} = 1397.0\ \text{kN}, \quad \tan\phi = \frac{444.70}{1324.35} \Rightarrow \phi = 18.56^\circ \end{aligned} $$ **Answer:** $F_H = 1324.35$ kN, acting 1.732 m above the bed; $F_V = 444.70$ kN downward, acting 5.158 m horizontally from hinge O. The resultant (≈ 1397 kN at 18.56° below the horizontal) passes through O. ## 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. Pressure at the centroid times arc length only B. Force on the vertical projection of the surface C. Weight of water above the surface D. Force on the horizontal projection Answer: B. 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. Weight of the gate B. Pressure at the bed times the base length C. Weight of water vertically above the curved surface up to the free surface D. Force on the vertical projection Answer: C. 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. With R = 6 m and the radii 60 deg apart (one horizontal at the bed), what is the water depth? A. R cos60 = 3 m B. R = 6 m C. R sin60 = 5.196 m D. R tan60 = 10.39 m Answer: C. The upper end of the arc is R sin60 deg above the bed, so H = 5.196 m. (Key point: Water depth from geometry) 4. What are F_H and F_V for the 10 m long gate? A. 132.4 kN and 44.5 kN B. 1397 kN and 444.70 kN C. 1324.35 kN and 444.70 kN D. 444.70 kN and 1324.35 kN Answer: C. F_H = 132.435 x 10 = 1324.35 kN and F_V = 44.47 x 10 = 444.70 kN; the resultant is about 1397 kN. (Key point: Resultant for L = 10 m) ## Close variants set in other exams - [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× ## Related in Hydro Power - [Force to hold hinged rectangular gate](https://www.bctengineer.com/question.md?key=semester_6%2Fhydro_power%3Ahinged-rectangular-gate) — Hydro Power, chapter 1 ## Related in other subjects - [Force method: frame with hinged support](https://www.bctengineer.com/question.md?key=semester_5%2Ftheory_of_structures_ii%3Aforce-method-frame-30knm-50kn-hinged) — Theory of Structures II, chapter 2 - [Force method: frame reactions at hinged support](https://www.bctengineer.com/question.md?key=semester_5%2Ftheory_of_structures_ii%3Aforce-method-frame-50knm-hinged-d) — Theory of Structures II, chapter 2