Normal force on a fluid container 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?
 A: Your approach makes some sense, but is missing an important factor that should explain the discrepancy.
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.
There is also a pressure acting on the top face of the rectangular prism that is greater than the atmospheric pressure pushing down.
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 $$F_{net weight} = F_{down} - F_{up} = 4A \rho g H - 1A \rho g H = 3A \rho g H$$
As long as all the net forces due to pressure are accounted for, the two methods will agree with each other.  You just forgot about the extra area being pushed up on that makes the total weight less than the pressure at the bottom times the bottom area.  This concept is the general reason why these two methods give the same result if all factors are considered.
