# On hydrostatic pressure and weight

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?

• Have a look at this article: usuario.cicese.mx/~ovelasco/archivos/cursos/Wilson1995.pdf Commented Jul 17, 2023 at 8:51
• @MarkoGulin: The second sentence says: "If the sides of the vessel slope generally inwards from bottom to top, than the force on the bottom is greater than the weight of the fluid contained." So if you can put such a setup on a scale and eliminate the 'weight' of the vessel itself (by supporting it with clamsp for example), the scale would actually register a weight more than due to the mass of the water itself. Correct? If that's correct, my only remaining question is whether such a setup can be achieved. Can you effectively balance 2L against 1L by playing with shapes and clamps? Commented Jul 18, 2023 at 6:52
• You can never balance $2\ell$ against $1\ell$ on a correctly constructed and operated mass balance. Commented Jul 18, 2023 at 8:11

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 $$m''g, the container exerts forces, the $$w$$'s in the diagram, on the right-hand column of water which compensate for the "missing liquid.

• Thank you for answering, I'll look at the link as quickly as possible. One remark, I was actually considering a conical vessel inverted to yours (so wider at the bottom than top). Here I think it's somewhat obvious the container is supporting some weight of the water as it is 'above' the wall. With an inverted cone, even that dissappears which makes the setup less intuitive. Commented Jul 17, 2023 at 11:18
• Okay, I've read your linked answer as well. I'm pretty sure I get that. Basically your answer here agrees with what I say. If you somehow make a setup where your scale only register the force due to pressure at the bottom, then of course setups with equal heights of water and the same base area will exert the same force (with the clamps I aim to eliminate the weight of the vessel). For an inverted cylinder (wider at the bottom than top), the force applied is still the same, despite there being less water being supported by the scale as opposed to the cylindrical setup. Is that correct? Commented Jul 18, 2023 at 6:48
• You can engineer a mass balance and clamped vessel to be balanced, but you cannot support different amounts of water and yet get balance. It is very clear that you did not get what it is you are reading. Commented Jul 18, 2023 at 8:09
• @naturallyInconsistent : With all due respect, your answers are not very enlightening and basically you're just telling me I don't understand anything. You might be right, but you're doing a horrible job at explaining. In Farcher's drawing above, the scale register the same force as the pressures at the bottoms are equal with equal areas. The sloping walls of the vessel on the right support the extra water and if the vessel is not being supported by the scale, this wouldn't register. Already here there is a setup with different amounts of water that can be balanced. Commented Jul 19, 2023 at 7:19
• You can argue that this is not supporting 2L against 1L for example as the extra water is supported by the vessel itself. I agree with that. So now imagine the slope to be inward as you go up. The article you referred me to as well says: "If the sides of the vessel slope generally inwards from bottom to top, than the force on the bottom is greater than the weight of the fluid contained." Wouldn't that setup still balance as the only force acting on the scale is simply due to the pressure? Commented Jul 19, 2023 at 7:21

As is well known, hydrostatic pressure depends only on the depth of the water, not the shape of the container.

This is correct, but for a reason that you do not yet know.

Therefore, the scale should register the same 'weight' even though there is clearly less water in the second setup than the first.

This is very wrong. The scale will register the weight of water and the flask holding them.

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.

This is a very big issue. To make sure that there would be no leakages, the bottom must be something like rubber, and firmly pressed against the weighing scale, to provide enough traction to resist the water's pressure attempting to leak through the sides. It can be done, but the scale weighing will be difficult to do.

Where you actually went wrong, is that you assumed that the walls of the container do nothing. Instead, the water presses against the walls at the pressure for each height, and if the walls are slanted, then there will be a vertical component of pressing, either upwards or downwards, depending upon how the slanting is. In your case, it is downwards, i.e. the walls increase the pressure of the water just to agree with the height distribution, and in so doing, is being lifted off the scale by the water, increasing the likelihood of leakage. Similarly, a conical flask that is slanted the other way, will be lifting the water, and thereby be pressed downwards onto the scale by the water.

This is how the scale will always measure the weight of the water and the flask (and the excess securing force to prevent leakage), regardless of the shape of the walls.

• You write: "The scale will register the weight of water and the flask holding them." In this experiment, the bottomless flask is being supported by a clamp near the bottom of the flask. Does the scale still register the flask in that case? I'm well aware that balancing water without leakage is extremely difficult. The practical difficulties is not what concerns me here. I'm also aware the reaction force of the water pushing on the walls is what makes the hydrostatic paradox work. Commented Jul 17, 2023 at 9:30
• My question really is, if you do this setup where the container is not supported by the scale but by clamps, is the resulting force on the scale in both setups the same? If the answer is yes, you can essentially balance 2L of water against 1L by varying the shape of the water, the instrument used to force the water in different shapes must be supported by something other than the scale to make it work. Also, thank you in advance for taking time to answer my questions! I do appreciate that. Commented Jul 17, 2023 at 9:32
• My answer pointed out why it is that you are wrong and it is definitely not possible to balance $2\ell$ of water against $1\ell$, with reasoning why. Marko Gulin also deleted his original wrong answer and linked to a paper that talks around the correct answer I just gave. The scale must register the flask for this to work. If you lift the flask, then nothing can prevent the leakage. Commented Jul 17, 2023 at 9:42
• @Mathematician42, Suppose some magic prevents water from leaking through the interface between the walls and the plate on the top of the scale. What you're still missing is that the clamp cannot support the weight of a conical vessel that opens upward without also supporting some of the weight of the water. And, if the area of the opening where it touches the scale approaches zero, then the fraction of the water supported by the clamp approaches 100%. Water pressure is normal to the surface of its container, and a vector normal to the surface of the cone has a downward component Commented Jul 17, 2023 at 10:43
• @naturallyInconsistent: Couldn't a counterweight on the other side of the balance counter the force applied by the water (and prevent leakage)? Commented Jul 17, 2023 at 11:09