The Attempt at a Solution
Lets start with the easy part:
The pressure of the air is constant so the pressure will act as if it's a concentrated force on the center of gravity of the gate.
F = PA = (10,000Pa)(.3)(.6)=1800N
The moment about B = 1800N (.15m) = 270Nm.
Now we know that the water has to exert a 270Nm torque to make the gate move.
This is where I'm stuck. I'm not sure how to find the torque that the water exerts on the gate. Depending on the ratio of h to the height of the door, .3, it seems to me that the concentrated force of the water might land above the door. So it seems to me like I can't use the same method. I feel like I need to integrate and add up all the small moments across the whole door.
So the integral would be something like
integrating from B to .3
M = ∫ γ(volume of water)dy
but I don't know anything else except the height for the water....