Altering the shape of a flasks walls

[aspodkfpo]

Homework Statement:

13 b) i) https://www.asi.edu.au/wp-content/uploads/2015/03/2011-Physics-exam-solutions.pdf

On page 13 of pdf

Relevant Equations:

n/a

" For the conical vessel, we note that pressure is a local force, and so must be independent of conditions far away. Hence altering the shape of the flasks walls while keeping local conditions constant will not change the pressure locally, and so F = (ρgh+Pa)πr2 also holds for this flask "

Can someone expand on this and the theory behind it? My intuition is lower mass, lower force, but this is countered in the next part of this problem.

 

[haruspex]

Homework Statement:: 13 b) i) https://www.asi.edu.au/wp-content/uploads/2015/03/2011-Physics-exam-solutions.pdf

On page 13 of pdf
Relevant Equations:: n/a

" For the conical vessel, we note that pressure is a local force, and so must be independent of conditions far away. Hence altering the shape of the flasks walls while keeping local conditions constant will not change the pressure locally, and so F = (ρgh+Pa)πr2 also holds for this flask "

Can someone expand on this and the theory behind it? My intuition is lower mass, lower force, but this is countered in the next part of this problem.

Consider first a rectangular tank.
If you increase the weight by increasing the width then the force per unit area stays the same. If you increase the depth instead then the force per unit area increases in proportion.
Now let the tank have sides that lean outwards. Some of the weight is taken by those sides, and it turns out that the force per unit area at the base still only depends on depth.
If we make the sides lean in, as in the example, the fluid exerts pressure on those sides, and the reaction force has a downward component. This makes up for the fact that there is less water near the top.