A rectangular sluice gate AB, 2 m wide and 3 m long, is hinged at A. It is kept closed by a weight fixed to the gate. The total weight of the gate and the fixed weight is 343,350 N, with a combined centre of gravity at G. Find the height of the water ‘h’ which will just cause the gate to open.

$$ \text{Vertical depth of hinge A} = h - (3 \times \sin 45^\circ) $$ $$ = h - 2.121 \, \text{m} $$

$$ \bar{h} = \text{Depth of centroid D} $$ $$ \bar{h} = h - (3 - 1.5) \times \sin 45^\circ $$ $$ \bar{h} = h - (1.5 \times \sin 45^\circ) $$ $$ \bar{h} = h - 1.061 \, \text{m} $$

$$ F = \rho g A \bar{h} $$ $$ F = 1000 \times 9.81 \times 6 \times (h - 1.061) $$ $$ F = 58860 (h - 1.061) \, \text{N} $$

$$ \text{Moment of Inertia, } I_G = \frac{b L^3}{12} $$ $$ I_G = \frac{2 \times 3^3}{12} $$ $$ I_G = 4.5 \, \text{m}^4 $$

$$ \text{Let } y_p \text{ be the distance from C.G. (D) to C.P. (H) along the incline.} $$ $$ y_p = \frac{I_G}{A y_{cg}} $$ $$ \text{where } y_{cg} \text{ is the distance to the centroid from the surface, along the incline.} $$

$$ y_{cg} = \frac{\bar{h}}{\sin 45^\circ} $$ $$ y_{cg} = \frac{h-1.061}{\sin 45^\circ} $$

$$ y_p = \text{Dist (D to H)} = \frac{4.5}{6 \times \left( \frac{h-1.061}{\sin 45^\circ} \right)} $$ $$ y_p = \frac{0.75 \sin 45^\circ}{h-1.061} $$ $$ y_p = \frac{0.530}{h-1.061} \, \text{m} $$

$$ \text{Moment Arm, } AH = (\text{Distance A to D}) + (\text{Distance D to H}) $$ $$ AH = 1.5 + \frac{0.530}{h-1.061} $$

$$ M_{\text{weight}} = M_{\text{hydrostatic}} $$ $$ W \times EG = F \times AH $$ $$ 343350 \times 0.6 = 58860 (h - 1.061) \times \left( 1.5 + \frac{0.530}{h-1.061} \right) $$ $$ 206010 = 58860(h-1.061) \times 1.5 + 58860(h-1.061) \times \frac{0.530}{h-1.061} $$ $$ 206010 = 88290(h - 1.061) + (58860 \times 0.530) $$ $$ 206010 = 88290h - 93675.69 + 31195.8 $$ $$ 206010 = 88290h - 62479.89 $$ $$ 206010 + 62479.89 = 88290h $$ $$ 268489.89 = 88290h $$ $$ h = \frac{268489.89}{88290} $$ $$ h \approx 3.04 \, \text{m} $$