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
$F_w = \gamma_w \bar{h} A = 9.81(10.5)\left[ \frac{1}{4}\pi (3^2) \right]$
$F_w = 728.10 ~ \text{kN}$
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 Fw 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
