The Scenario
Suppose I have a cylindrical beaker and a conical flask (which cross-sectional area is bigger at the bottom), each located on an identical digital weight scale. Both of them have the same area, $A$, at the bottom and both are weightless. I fill both of them with identical fluids so that the height of the fluid in each container is the same. Obviously, the conical flask will require less to do so. Assume equal atmospheric pressure and gravity.
The Inconsistency
According to hydrostatics, the pressure at depth $h$, independent of the container shape, is: $$P(h)=\rho gh + P_0$$ where $\rho$ is the density of the fluid and $P_0$ is the atmospheric pressure at the fluid surface. Both the cylindrical beaker and the conical flask have the same fluid height; therefore, both should have the same pressure at the bottom, regardless of their shape. Since they have the same area at the bottom, the force exerted on the weight scale should be equal ($F=PA$). Therefore, the weight scales should register the same weight for both the cylindrical beaker and the conical flask.
This is impossible, since the amount of fluid in each container is significantly different (approximately by a factor of $3$).
The answers to this related question says that the additional weight of an inverted conical flask is due to the forces acting on the walls. However, I don't believe that it is possible for the water to act an upward force on the walls and thus lift the container upwards.
How does one explain this inconsistency?
Answer
Pressure acts in all directions. When the pressurized fluid touches the tilted side of the container, the force due to that pressure is outward and perpendicular to the surface.
A similar question (Pressure,Pascals Law) was asked the other day, and this was part of the answer.
Is this counter-intuitive? It shouldn't be. It's easy to think of cases when pressurized water is forced to spray upwards. If that spray is covered up, the force of the water on the plug is upward as well.
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