Force on the gate due to air pressure

$F_{air} = p_{air}A = 45\left[ \frac{1}{4}\pi (3^2) \right]$
$F_{air} = 318.09 ~ \text{kN}$

Force on the gate due to water

$F_w = \gamma_w \bar{h} A = 9.81(1.5)\left[ \frac{1}{4}\pi (3^2) \right]$
$F_w = 728.10 ~ \text{kN}$

Horizontal force at the hinge support

$\Sigma F_x = 0$
$R_O = 728.10 - 318.09$

$R_O = 410.01 ~ \text{kN}$ ← [ C ] *answer for part 1*

Location of *F*_{w} from the invert

$e = \dfrac{I_g}{A\bar{y}} = \dfrac{\frac{1}{64}\pi (3^4)}{\frac{1}{4}\pi (3^2) \times 10.5}$
$e = 0.0536 ~ \text{m}$

$y = 1.5 - e = 1.5 - 0.0536$

$y = 1.4464 ~ \text{m}$ ← [ A ] *answer for part 2*

Location of hinge support

$\Sigma M_O = 0$
$(y - z)F_w = (1.5 - z)F_{air}$

$(1.4464 - z)728.10 = (1.5 - z)318.09$

$z = 1.4048 ~ \text{m}$ ← [ D ] *answer for part 3*