Consider two different shaped containers having same area: one is cylinder, the other is like an inverted pyramid (roughly). Both have the same level of water, the weight of the inverted pyramidal container will therefore be greater than that of the cylinder. But I know from Pascal's law that the pressure should be same in both containers. If the base areas are the same then the force (weight as measured by a scale) should also be the same. Why are the weights different?
Fluid Statics – Understanding Pressure and Weight in Hydrostatic Paradox
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No, the entire weight will not directly rest on the base of the slanted container (although it does indirectly). There are a number of ways to approach this, but the easiest way is to observe that the total force acting on the bottom of the container is equal to the sum of the hydrostatic pressure force (the pressure at the bottom of the container multiplied by its area) and the shear force around the edge of the base (draw a free-body diagram to convince yourself of this). The hydrostatic pressure depends only on the height of the column of fluid above the given location ($P=\rho gh$), and the shear force is equal to the weight of the fluid outside of the base (supported by the walls). Since both vessels contain the same volume, but the slanted container has a wider cross section above the base, the total height of the body of fluid will be less for the container with slanted walls than the other container. Thus, the hydrostatic pressure at the base will be less. But remember, any reduction in the pressure force will be compensated by an increase in the shear force around the edge of the base. The vertical walls support no vertical force, and so they exert no shear force on the base. The slanted walls do bear some of the weight, and so they transfer this force to the base via shear. In both cases, the sum of the hydrostatic pressure force plus the shear force (if there be any) is equal to the weight of the fluid, and this is what the scale measures.
Now what I think is that the pressure on the base × base area = normal force
If you mean the internal pressure in the vessel, then no. The vessel has rigid walls and the pressure inside is not necessarily related to the pressure outside.
Imagine a gas cylinder with a movable piston. We can depress the piston to increase the pressure inside without adding mass. But the normal force from the ground is unaltered.
If the fluid pressure on the bottom of the vessel increases, but the normal force does not, then some other force must be changing so that the bottom does not accelerate. This force comes from the walls of the cylinder.
The extra area of the bottom cylinder that is not under the upper cylinder is under internal pressure. Before the mixing the fluid is pushing up on this area with pressure $2\rho g H$, but after the mixing the pressure is $3 \rho g H$. This extra pressure creates an upward force that counteracts the force from the increased downward pressure on the base.
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The flaw in your reasoning is that in the inverted cylinder, the hydrostatic pressure is also applied to the "diagonal" walls of the container. This causes a net downward force that, if left unbalanced, would cause a downward acceleration. However, the force is transmitted down through the (rigid) walls of the container to the base, and from there to the scale, which registers a higher weight.