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:
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?
Best Answer
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.