An inclined rectangular sluice gate AB, 1.2 m by 5 m size, is installed to control the discharge of water. The end A is hinged. Determine the force normal to the gate applied at B to open it.

$$ \text{Area, } A = 1.2 \, \text{m} \times 5.0 \, \text{m} = 6.0 \, \text{m}^2 $$ $$ \bar{h} = (5.0 - 1.2 \sin 45^\circ) + \frac{1.2}{2} \sin 45^\circ $$ $$ \bar{h} = 5.0 - 0.6 \sin 45^\circ $$ $$ \bar{h} = 5.0 - 0.6 \times \frac{1}{\sqrt{2}} \approx 4.576 \, \text{m} $$

$$ F = \rho g A \bar{h} $$ $$ F = 1000 \times 9.81 \times 6.0 \times 4.576 $$ $$ F \approx 269343 \, \text{N} $$

$$ I_G = \frac{\text{width} \times \text{length}^3}{12} = \frac{5.0 \times (1.2)^3}{12} = 0.72 \, \text{m}^4 $$ $$ h^* = \frac{I_G \sin^2\theta}{A\bar{h}} + \bar{h} $$ $$ h^* = \frac{0.72 \times (\sin 45^\circ)^2}{6.0 \times 4.576} + 4.576 $$ $$ h^* = \frac{0.72 \times 0.5}{27.456} + 4.576 $$ $$ h^* \approx 0.0131 + 4.576 \approx 4.589 \, \text{m} $$

$$ \text{Distance from surface to Centroid, } y_{cg} = \frac{\bar{h}}{\sin 45^\circ} = \frac{4.576}{1/\sqrt{2}} \approx 6.471 \, \text{m} $$ $$ \text{Distance from surface to Center of Pressure, } y_{p} = \frac{h^*}{\sin 45^\circ} = \frac{4.589}{1/\sqrt{2}} \approx 6.489 \, \text{m} $$ $$ \text{Moment Arm about A, } AH = y_{p} - y_{cg} + \frac{\text{length}}{2} $$ $$ \text{Wait, the distance from CG to CP is } \frac{I_G}{A y_{cg}} = \frac{0.72}{6 \times 6.471} \approx 0.0185 \text{m} $$ $$ \text{The force F acts at this distance below the centroid. Hinge A is at the top.} $$ $$ \text{Moment Arm } AH = (\text{distance from A to centroid}) + (\text{distance from centroid to CP}) $$ $$ AH = \frac{1.2}{2} + 0.0185 = 0.6185 \, \text{m} $$

$$ P \times (\text{Length of Gate}) = F \times (AH) $$ $$ P \times 1.2 = 269343 \times 0.6185 $$ $$ P = \frac{269343 \times 0.6185}{1.2} $$ $$ P \approx 138800 \, \text{N} $$