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I have a hemispherical vessel held against the wall, which holds water inside it. I had to calculate the net force required to hold this arrangement in equilibrium. The diagram is:

Vessel Diagram

Whatever horizontal force $F_x$ I apply, the same will be applied by the wall on the water in it. So using calculus by taking elements of the wall of thickness ${\rm d}y$ and integrating the force due to pressure on the circular part of the wall, I got the answer $F_x = \rho g \pi R^3$, where $\rho=\,$density of water, $R=\,$radius of vessel.

However, when I checked the solution, they gave the answer as $\rho g R(\pi R^2)$, which although the same as my answer, is written in a format that seems to suggest the answer is just the product of pressure at half the total depth times the total area of the wall. Is there a general result like this that suggests the total horizontal force can be calculated just using the pressure at half the depth times the total area of the wall in contact with the fluid? How is it derived?

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Because you are in a uniform gravitational field, the pressure varies linearly with depth, which means that the average pressure in any vessel is equal to the pressure at half of the total depth.

From there, you simply multiply by the area of the fluid in contact with the wall to get the total force being applied.

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  • $\begingroup$ Is this still true if the top and bottom halves are not symmetric? $\endgroup$ – Tom B. Mar 19 '18 at 1:06
  • $\begingroup$ Does this rule follow as long as pressure increases linearly ..are there any restrictions for the container $\endgroup$ – Abhinav Apr 30 '18 at 5:26
  • $\begingroup$ Yes, there are certain symmetry requirements. Mirror symmetry would definitely work, but you could certainly loosen the requirements further than that. $\endgroup$ – J. Murray Apr 30 '18 at 5:37
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I divided the above in two parts as the imagining the the 3d figure the origin is the point the wall touching the line point P. From basic equation, $$F=PA$$ Here as the origin is point P,The pressure on it by the above portion is same as the lower one. $$\Rightarrow P=\rho gR$$ $$\Rightarrow A=\frac{\pi R^2}{2}$$ So this is on is for the upper porition for lower one $$P=\rho gR$$ $$A=\frac{\pi R^2}{2}$$ As the forces distributed at every point in the vessel is uniform and same below the divison. so Total force on the vessel is $$F_{u}=\frac{(\rho gR)(\pi R^2)}{2}$$ $$F_{l}=\frac{(\rho gR)(\pi R^2)}{2}$$ $$F_{total}=F_{u}+F_{l}$$ $$F_{total}=\rho gR(\pi R^2)$$.

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