# Hydrostatic force on inclined immersed surface

Help me to find out the hydrostatic force acting on an inclined immersed surface that is actually the diagonal of a 4mx3m rectangular tank filled with water. Given that the tank is 2m deep.

I know, F = Specific weight x Area x Depth of the center of gravity of the immersed surface from the free water surface.

But the problem is that I have no idea how to calculate that depth of the center of gravity of the immersed surface from the free water surface. Any help will be appreciated.

First of all a drawing would really help. Possibility #1

Make a sketch and add the forces to the inclined area. You will either get a triangle or a trapezoidal, depending on the exact setup.

You can look up the center of gravity or calculate it yourself (Parallel axis theorem). Then you need to calculate the pressure at that point, let's call it $p_\text{res}$ for resulting Force.

$F = p_\text{res} \cdot A$

Possibility #2

Just integrate

$\vec{F} = \int p \vec{n} dA$

with $\vec{n}$ being the normal vector. I'd prefer this way if you don't really need to know where the resulting force is attacking.

• In my case the tank is rectangular shaped and the panel is exactly along the diagonal, which has divided the tank into two equal shaped triangular bodies. Since the tank is a 4mx3mx2m one, I believe the length of the diagonal is 5m. I am not sure whether the Area should be computed as 5mx2m in this case or not. Moreover in normal case for triangle the centre of gravity is h/3 (above the base) and 2h/3 (from the top). I do not know how it should be calculated here. Nov 1 '17 at 15:06
• In fact I wonder the centre of gravity for what needs to be computed. The triangular water volume or the rectangular panel on which water is pressurizing. Nov 1 '17 at 15:16
• Maybe I wasn't really clear with terms: When you calculate the centroid of the triangular shape you find the center of pressure. This is the point where the resulting pressure force acts on the plane surface. As I usually go with the second method of just integrating I'm not too familiar with the parallel axis theorem but I think you would need centroid of the area for that, in order to calculate how much and in which direction the center of pressure deviates from the area centroid. Although in case of the triangular shape you already know where the center of pressure is. Nov 1 '17 at 16:42

For the general case of an inclined plane, given the angle $\alpha$, the heights $h_1$ and $h_2$, the water density $\rho$, and the (constant) width $w$.

The pressure at a certain depth $h$ is equal to $p=\rho\cdot g\cdot h$, so the force on the incline is the integral of the green area. For a trapeze: $$F=\int p\cdot dx dw = \frac{h_1+h_2}{2}xw \rho$$ where $x$ is the length of the base of the trapze, in this case $x=\frac{h_2}{\sin\alpha}$. To calculate the $x_c$-centroid line of a trapeze you can use the following formula: $$x_c=\frac{x}{3}\cdot\frac{h_2+2h_1}{h_1+h_2}$$ The height above base of the centroid $h_c=x_c\cdot \sin\alpha$

I would not find out the sum of average force acting on the glass. But a more appropriate way would be to find the most critical stress which happens near the bottom of the your panel.

I would pick the pressur at the bottom and use that multiplied by vertical projection of you surface to get the force downward.

$P= \rho.d =\rho.3 \space$

$Assuming \ rho=1$

you get $P =3000 kg/m^2$

And you can plug this pressure into the equation for plate tress for a horizontal plat of 4by 2 meters.

one such formula is b is the width=2, a is the length=4, and t is thickness. Off course there are finite element softwares that are more precise, but in this case you will be fine!

Source of the formule:http://www.roymech.co.uk/Useful_Tables/Mechanics/Plates.html