I'm doing some thought experiments about hydrostatic pressure and weighing setups on scales. This really confused me, so I really want to pick your brain about this and get some clarity.
Consider the following setup: Consider a classical balance scale. Next I have two 'containers' to pour water in. The first container is a cylinder with base area $A$ and height $h$. Now this cylinder has no base, so if you pour water into this cyclinder whilst holding the cylinder up, the water would simply fall through the cylinder. The second container is a cone-shaped and has the same base area as the cylinder, the same height, but it's radius decreases with the height (doesn't really matter how much).
Now put the first cylinder on the scale and pour water into the cylinder filling it. Next, attach a clamp firmly to the bottom of the cylinder (the clamp should be holding the weight of the cylinder but not the water as the water is simply hold in place by the cylinder but not supported by it as there is no base to rest on). If my understanding is correct, the only force that the scale can register is the weight of the water. In this case, the weight is $m\cdot g$ where $m$ is the mass of the water. Let $\rho$ denote the mass density of the water and $V$ the volume of the water, then $m=\rho\cdot V=\rho\cdot h\cdot A$ and thus the weight is $\rho\cdot g\cdot h \cdot A=P\cdot A$ where $P$ is the hydrostatic pressure of the fluid alone (I'm going to ignore atmospheric pressure, either because in this thought experiment, there is no atmosphere or I believe the pressure acting both on the water and scale should actually compensate each other). It should be noted that the only force acting on the scale is the pressure of the water (due to gravity).
Now repeat the previous experiment but with the second, cone-shaped, container. The only force acting on the scale again should by due to the pressure of the water. As is well known, hydrostatic pressure depends only on the depth of the water, not the shape of the container. Therefore, the scale should register the same 'weight' even though there is clearly less water in the second setup than the first.
My question is, is this reasoning correct? Is it really possible to balance different amounts of water on scales by only varying the shape of the container (which is not supported by the scale) holding the water in it's place?
I guess this must be correct. Imagine putting both setups on either sides of the scales. There is no difference between the scale acting a tunnel between both containers or literally connecting them to each other as communicating vessels, or am I missing something here?
Best Answer
This is an interesting idea although whether one could achieve the setup in practice is debatable.
The arrows labelled $w$ represent the forces exerted on the column of water by the containing vessel.
Perhaps what you have not realised is that when the container is cone shaped although the container has no effect on the scale it does have an effect on the water.
The container exerts forces $w$ on the water to hold the "orange" water up.
The scale exerts a force $F_2$ on the water equal to the weight of "blue" water $m'g$.
Thus the scale reading does charge between the two situations.
My answer to the post Pressure at base of 3 different dam designs may be of interest?
Fortunately I have no thrown away my original diagram so I have now added the scenario (right-hand diagram) described by the OP.
The scale reading for the right-hand situation (magnitude $F_3$) is the same as the scale reading for the left-hand situation (magnitude $F_1$).
This is because although the column of liquid in the right-hand diagram does not weigh as much as the column of liquid in the left-hand diagram, ie $m''g<mg$, the container exerts forces, the $w$'s in the diagram, on the right-hand column of water which compensate for the "missing liquid.