Normal force on a fluid container [closed]

I was solving some problems and came across the following situation.

Consider a container with one cylinder on top of the other. Area of cross section is $$2A$$ and $$A$$ for the bottom and top cylinder respectively. The height of both the cylinders is $$H$$. Something like,

Now calculating the normal force by the surface by two different methods two different results appear.

Method 1:

$$N = mg$$ $$N = \bigl(\rho(A * h + 2A*h)\bigr)g$$ $$N = 3Ah\rho g$$

Method 2:

                     N = Area of bottom * pressure at the bottom


$$N = 2A * \rho g(2h)$$ $$N = 4Ah\rho g$$

Which the same as that a cylinder of height 2H and base Radius 2A. Where did I go wrong?

• Second approach seems to be the incorrect one @Aaron Stevens is right here ... – Aditya Garg Mar 28 at 19:05
• @AaronStevens ... That's definitely correct. Pressure at the bottom will be due to the total height of fluid above (2 H in this case, as each shape has a height H, and they are stacked). There's a different issue with the approach. That's how hydrostatics work. – JMac Mar 28 at 19:13
• @AaronStevens It's a fluid container with a shape of a rectangular prism with a cylinder on top. The question didn't do a great job explaining that; but that clearly fits with "Normal force on a fluid container" and all the calculations done by OP. – JMac Mar 28 at 19:16
• @JMac Wow I completely misread everything. I definitely thought this was just two objects stacked on top of each other and the OP was looking at the force pushing on the ground – Aaron Stevens Mar 28 at 19:18
• @JMac I still believe that the second method has an issue. If we made the radius of the upper section smaller and smaller the final answer wouldn't change, which is odd to me – Aaron Stevens Mar 28 at 19:24

They are looking for the normal force at the bottom. Your calculation for pressure at the bottom is correct, and that face of the container would have a $$4A \rho g H$$ force but only on the bottom face. What you've missed is that it is not the only force acting on this container.
This pressure will be equal to the area being pushed up, multiplied by the pressure. Since the area of the rectangular prism is $$2A$$, and the cylinder is $$1A$$, this means that the upwards force on the prism acts on an area of $$1A$$. At this location, the pressure will be $$\rho g H$$, because the height of liquid column above is $$H$$. That leaves a force of $$F = 1A \rho g H$$ acting in the upwards direction on that face.
You can then look at the force balance $$F_{net weight} = F_{down} - F_{up} = 4A \rho g H - 1A \rho g H = 3A \rho g H$$