SMANSA Science

gintable says:

I assume the gate is to the right of tank of water, and no water exists to its right.

Also assume variation of air pressure is negligible, and water is incompressible.

Pressure varies with depth in an incompressible fluid according to:
P = rho*g*y, where y is position below surface, rho = density, g=gravity

The gate extends at depths of the following:
its center elevation, c = 2.1 m
its height, h = 1.2 m
The gate will extend down to y2 = c+h/2 and from y1 = c-h/2. Other walls hold the water for y < y1.

Our integral for net force is the following:
F = integral(P(y)*L, y=y1..y2)

it manifests to:
F = L*rho*g*integral(y, y=c-h..c+h)

Which solves as
F = L*rho*g*(y^2/2)|y=c-h/2..c+h/2

Plug in limits:
F = L*rho*g*(1/2*(c+1/2*h)^2-1/2*(c-1/2*h)^2)

Simplify to get result:
F = L*rho*g*c*h

To find direction, we can just use intuition. All pressure acts perpendicular to gate, outward from water. Therefore the direction is rightward.

To find the location of the total force, we find the center of the load using the following integral:

ybar = integral(y*P(y)*L, y=y1..y2)/F

It manifests to:
ybar = rho*g*L*integral(y^2, y=c-h/2..c+h/2)/(L*rho*g*c*h)

Solution to integral:
ybar = rho*g*L*( 1/3*(c+1/2*h)^3 – 1/3*(c-1/2*h)^3 )/(L*rho*g*c*h)

Simplifying:
ybar = (12*c^2 + h^2)/(12*c)

Data:
rho = 1000 kg/m^3; g=9.8 N/kg; h = 1.2m; c = 2.1m; L = 1.8 m;

Conclusions (after plugging in numbers):
Force: F = 44,452.8 Newtons
Direction: Rightwards
Location of force: ybar = 2.157 meters below surface of water