# How do you find the resultant hydrostatic force on a curved surface which is not exactly semicircle

Assuming this is a pressurized chamber with hydro-static force acting on a semi circle stuck in a surface: To find the resultant force, I would first divide the semicircle into half. Pressure on curve surface is equivalent to pressure on projected surface form the curve. Therefore, I can find the resultant force (F1) from multiplying water pressure (P) with radius of the circle (R): Where $$F_1 = P \times R$$

Considering the other half of the semicircle, my free body diagram should look like this also: Where $$F_2 = P \times (R-r)$$

Therefore, my resultant force should equals to $$F_1 - F_2$$.

I would like to know if this is the correct way of calculating the resultant force for such scenario.

• Yes, the only caveat is that there is probably some third hydrostatic force acting on the region bounded by little r unless the pressure is gauge pressure relative to this region. – Phil Sweet May 2 '20 at 14:17

Hello I would point out here Pressure is defined over an area so you would need to decide between a cylinder or a sphere (instead of circle) to fully define your problem.

Onwards I would change the perspective of the problem. I make use of the geometric symmetry: The surface AB and BC are identical and have indentical pressures therefore cancelling each other out. .

But below A there is hydrostatic force component though the pressure is not constant as one proceeds down. So I find the average pressure acting on the center of the surface found by $$P_{av} = (P_A + P_0)/2$$ ; where $$P_0$$ is pressure at rock bottom in tank.

If you have a massive tank i.e. Height of tank >> R then you may take $$P_{av} = \rho g H$$. Secondly, the area you are concerned with can be the cross section (AA') normal to the pressure (refer fig). This area needs to be visualised from the side view. It is a rectangle of length L and breadth R-r in case of a cyinder of length L Hence area $$L(R-r)$$\ It is a semicircle of radius r in case of a sphere hence area $$\pi r^2/2$$

Edit:
As pointed out in the comments, there is a vertical downward acting component on surface ABC. This can be calculated in a similar way as AA' was analysed except here the same pressure act on projected area AC.

• Welcome to Engineering! Very good answer, though, for the record, the pressures from AB and AC don't cancel out. Their horizontal components cancel out, but their vertical components both point down and are therefore added together. Also, I'm fairly certain we can assume a constant pressure $P$, as shown in the image itself. As you put it, this would be a simplification where $\text{depth} \gg R$. – Wasabi May 2 '20 at 19:45

You are partially correct in as far as you recognize that you have to subtract the right side forces from the left side.

But the pressure distribution is not even and is sloped increasing proportionally to the depth.

If you look at the blue trapezoid in my diagram below, all we need to do is to find out the force which is the difference between the area of the large triangle of the pressure ACD and small one ABE and I let you calculate it's the height from the vessel bottom.

Let's call your vessel's depth AC and breadth W. Then we call B and E the level at which your cylinder (not a circle!) cuts onto the wall AC.

Assuming $$\rho=1 \$$ for simplicity. The hydrostatic force is just the blue shaded area and we can find it by subtracting the area of triangle ACD by triangle ABE (sorry for terrible annotation)

$$F= 9.8 W \frac{( AC)^2 -(AB)^2 }{2}$$

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