# Vertical (upward) hydrostatic force created in a prismatic vessel with a slanted side?

I'm trying to work out the upward vertical force which will be generated (per linear m) when the prismatic form below is filled with a fluid, in order to counter it by applying a downward force equal or larger in magnitude.

As a rough indication of scale, H ≈ 1 m, a ≈ 0.1 m, b ≈ 0.5 H, making the slope close to 2:1.

I suspect I should use (Hρg) using the average depth (=H/2) and area of the slanted side to get the force acting on the slanted surface per linear m, then take this force as acting normal to the slanted surface and resolve its vertical component, and assume that's the only upward force generated on the prism. That would give a value of: (H/2)(ρg) . √(H2 + b2) . cos(tan-1(H/b)) = (Hb/2) . (ρg) (since cos(tan-1(H/b)) == b/√(H2 + b2) ) as the upward force to be countered per metre.

But while checking this, I've also seen calculations for dams and other submerged slanted surfaces using the centroid, and now I'm a bit unsure. Also hydrostatics was never my subject. Have I got this right in principle?

(The actual situation is that I'm casting a prismatic shape using a very dense low-viscosity fluid which sets after some time (imagine something similar to a very dense, very fluid, plaster of paris or cement, with a density ≈ 2700 kg/m^3). For practical reasons, while it's easy to clamp and brace the mould to counter the horizontal forces and any moments generated by the fluid before it sets, there's nothing available externally to hold/anchor the bottom of the mould to the floor or ground. So it has to be held down against uplift forces by adding equivalent weight on top. Most of the hydrostatic force *should* act horizontally, at a guess, but I need to be sure I've got enough weight on top when I start casting, or the fluid will just lift the mould off the ground and flow from under it ;-) )

• Don't forget that you don't just need sufficient weight to counter the total vertical force. You also may need to counter tipping torque as well. That prism may want to tip left to relieve pressure, and do so at a lower pressure than a "lift-off" would happen. Dec 10, 2016 at 0:48
• The frame and its bracing is robust and will handle the lateral forces and any moments - good catch though. Edited that into the Q. But that's why its mainly the upward force left to be countered. Dec 10, 2016 at 0:53

If the whole apparatus were a right prism (same area at top and bottom) there would be no upward force to counter. So, if you were to make a vertical wall at the right edge, and extend the mold to enclose the missing wedge-shaped region, you could pour liquid ballast into that wedge which is of the same density as the fluid of your casting. Then filling the mold would put zero net pressure on the downward-facing side, and would not produce any upward resultant force.

Because pressure varies with depth, this has the additional advantage of supporting the slanted surface with the correct pressure at all depths. So the slanted surface will not bend under the pressure.

The weight of the mold must be enough to maintain a seal at the base surface.

• Nice! That's a solution I hadn't considered! Lateral thinking! (And a suitable vertical face already exists in the place you described, as a bonus). My own lateral thinking afterwards was maybe to add a 'false floor' internally so the uplift force was also exerted by/on the frame itself (downwards) rather than by/on the ground or floor,and fill in under it when the main body of fluid is set. Will comment in a bit on yours. Dec 10, 2016 at 12:16

Yes your proposed method of calculation is correct. The only problem is that there appears to be nothing at the base of the prism, and the pressure $P_0$ at that level is unknown. What prevents the water from falling out? It is difficult to imagine how this structure could be supported.

The pressure at height $y$ above the base level is $P(y)=P_0-\rho gy$. When you have decided what value $P_0$ should have, you can calculate the average pressure on each face of the prism and the net upward force on it.

The horizontal forces from the fluid push outward on the walls of the structure but do not have any horizontal resultant. At each level the force per unit depth is the same on the vertical face on the left as on the slanted face on the right.

A shorter method of calculation is to realise thet the upward force $F=P_0A$ on the base of the prism (where $A$ is its area) supports the weight $W$ of everything above it. Any excess of $P_0A$ over $W$ is a net upward force on the structure, which must be balanced by some load or ballast.

Adding weight to the top does not sound like a good idea as this will raise the centre of gravity and cause the structure to become unstable.