Hydrostatic force on sector gate (10 m long)

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.

   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.