# Does pressure really varies with depth in a fluid?

In the derivation of pressure variation with depth in a fluid, we consider a hypothetical cylinder (or any convenient shape) and make a fbd for that shape. In the fbd , the forces included are

1. $$mg$$ (where $$m$$ is the mass of the cylinder)
2. The pressure difference force on the cross sections and finally we equalise them.

But from Pascal's law, we also know that pressure of a force in a fluid is transmitted to all the points without being diminished.

So my question is:

why don't we include the Normal force acting on the fluid due to the bottom surface of the container keeping the fluid in the derivation of pressure variation with increasing depth in a fluid?

Hope the question is clear.

• Comments are not for extended discussion; this conversation has been moved to chat. Nov 22, 2020 at 11:53
• @ACuriousMind I apologise for that. I will not repeat this from Next time 🙂. Nov 22, 2020 at 12:13
• I have posted a meta question on this particular question here Nov 28, 2020 at 13:15

The normal force acting on the fluid in the bottom of the container is a consequence of pressure increasing with depth. This normal force is part of what allows the vessel to actually contain the pressure, along with the normal force on the walls. The deeper the water, the greater those pressures at the bottom. A common example is Pascal's barrel where by filling a thin tube with water with a barrel below it, the pressure eventually becomes too great and breaks the barrel.

Without those normal forces, hydrostatic pressure could not really develop at all (as JustJohan mentioned in a comment). If you tried to put more water on top of some other water, you need a container for it to actually stay above the other water, or else you basically just get a spreading pool of water at approximately atmospheric pressure.

So you could possibly say without something to contain the fluid, you can't really experience pressure increasing with depth, and therefore buoyancy. If the fluid isn't somehow contained, the fluid above that pushes on the fluid below will just move the fluid away, instead of having the force of it's weight contained by a vessel so that it generates pressure in the fluid.

I wouldn't really call it the "ultimate source of buoyant force" though. I probably wouldn't call anything that really, since buoyancy takes several factors; but if anything I would consider the weight of the fluid and gravity to be the ultimate source of that force. The container just facilitates the process by providing a reaction force to contain the pressure.

• Comments are not for extended discussion; this conversation has been moved to chat. Nov 22, 2020 at 11:56

why don't we include the Normal force acting on the fluid due to the bottom surface of the container keeping the fluid in the derivation of pressure variation with increasing depth in a fluid ?

We do.

Pascal’s law for a hydrostatic fluid can be written $$\Delta P = \rho g \Delta h$$. This applies to all points in the fluid, including those in contact with the bottom of the container. Whatever that pressure is, it defines the pressure at one place. Then the pressure at all other points in the fluid can be calculated by applying Pascal’s law. If the normal force increases, then the pressure will increase throughout the fluid, per Pascal’s law.

EDIT: the question was edited to invalidate my previous answer, but I have left it here. Below is an answer to the edited question.

But from Pascal's law, we also know that pressure of a force in a fluid is transmitted to all the points without being diminished.

Sometimes the natural-language wording of a physical law is confusing or ambiguous. In those cases it is best to rely on the mathematical expression. Pascal's law for a hydrostatic fluid is $$\Delta P = \rho g \Delta h$$, where $$\Delta P$$ is the difference in pressure between two points, $$\Delta h$$ is the difference in the height of fluid above the same two points (i.e. the change in depth which increases as you go down deeper in the fluid), $$\rho$$ is the fluid density, and $$g$$ is the local gravitational acceleration. Using the mathematical expression will avoid confusion due to the English wording.

Suppose we have a large cylindrical container whose bottom area is $$1m^2$$ and it is holding water of say $$5m^3$$ volume. So the total mass of water will be $$5000kg$$. Since the center of mass of the fluid is at rest , this means that the normal force from the bottom of the container balances its weight (using the same procedure as we do for solids). So normal force from the bottom is equal to $$50000N$$.

Therefore the pressure on the bottom of the container is $$50\text{ kPa}$$.

Now let us consider a hypothetical cylinderical portion of water with base area $$\frac{1}{4}m^2$$ and height $$1m$$. So it's weight is $$1000 ×\frac{1}{4}m^3× 10 = 2500N$$.

The pressure of the Normal force at the bottom surface is $$\frac{50000N}{1m^2}=50000Pa$$ and thus from Pascal's law, the force on the bottom area of the hypothetical cylinder is $$50000×\frac{1}{4} N > 2500 N$$

This is an incorrect use of Pascal's law. You have not specified where the cylinder is located in the vertical direction. Suppose that the bottom of the cylinder is $$2\text{ m}$$ above the bottom of the container. So by Pascal's law

$$\Delta P = \rho g \Delta h = 1000 \text{ kg/m}^3 \ 10 \text{ m/s}^2 \ (-2 \text{ m}) = -20 \text{ kPa}$$

So the pressure on the bottom of the cylinder is $$50\text{ kPa}-20\text{ kPa} = 30 \text{ kPa}$$ which provides an upwards force of $$7.5 \text{ kN}$$.

This means that the hypothetical cylinder can be at rest only if the pressure difference force is in downward direction and not in the upward direction i.e. pressure should decrease with increasing depth in the water or any fluid.

To determine this we will need to include the force from the pressure on the top. To determine that we will apply Pascal's law again.

$$\Delta P = \rho g \Delta h = 1000 \text{ kg/m}^3 \ 10 \text{ m/s}^2 \ (-3 \text{ m}) = -30 \text{ kPa}$$

So the pressure on the top of the cylinder is $$50\text{ kPa}-30\text{ kPa} = 20 \text{ kPa}$$ which provides a downward force of $$5 \text{ kN}$$.

So the sum of the forces on the top and the bottom are $$2.5 \text{ kN}$$ upwards which balances the weight force of $$2.5 \text{ kN}$$ downwards.

• I think something is unclear here (may be my question to you ). Let me rephrase : Suppose we have a container having some liquid. Now since it is at rest in the container and not doing vertical motion , we say that normal force from the bottom equals the total weight of the fluid. Now from Pascal's law , we also know that if a force is a applied to one point , it is transmitted to all the points . So the same normal force is transmitted to all the points in the liquid. Nov 20, 2020 at 4:48
• Now if we consider a hypothetical cylinder I'm the fluid , we show its weight , and force due to pressure difference but not the normal force . So my question was why don't we show this normal force even if it is transmitted to all the points in the liquid ? Nov 20, 2020 at 4:52
• @Ankit “Now from Pascal's law, we also know that if a force is a applied to one point, it is transmitted to all the points”. That is not a correct statement of Pascal’s law. Pascal‘s law is $\Delta P=\rho g \Delta h$. In that form it is clear that the pressure from the normal force is accounted for. Notice, it is the pressure, not the force. That should be clear since force has a direction and pressure does not. Hydraulic forces do not retain the source’s direction
– Dale
Nov 20, 2020 at 5:11
• en.m.wikipedia.org/wiki/Pascal%27s_law Nov 20, 2020 at 5:13
• @Ankit I really wish you would stop editing your questions to invalidate my answers. I have already told you about this and so this was not done out of ignorance. Now you have received an answer to the modified question so I cannot even revert it without invalidating their answer (and unlike you I will not do that to a respondent). Your behavior is very inconsiderate as well as violating the forum rules.
– Dale
Nov 20, 2020 at 17:50

In your example, you have the bottom of the cylinder being pushed up with $$50000\cdot\frac{1}{4}=12,500 \text N$$ of force, which is greater than the force of gravity pulling on the cylinder. Were this cylinder to be subject only to the forces of gravity and the forces of the base, you are correct that this would cause upward motion. However, there are more forces at play. There are the forces of the water above the cylinder, pushing down on it. In your example the smaller cylinder you are looking at is 1 meter high, and it is "supporting" 4 meters of water above it. The sum of gravity, the force of the base, and the force from above will equal zero. Indeed, given that it is 1 unit of mass supporting 4 units of mass above it, it should be no surprise that the forces on the bottom of that cylinder are exactly 5 times the weight of the cylinder itself!

Some of the confusion with Pascal's principle is that it speaks to perturbations around the equilibrium point. We then develop the mathematical expression $$\Delta P = mg\Delta h$$ from exploring what must be true about pressure in order for those perturbation statements to be true.

• how did you assumed that the considered cylinder is at the bottom and is having 4 m of water above it ? Nov 20, 2020 at 17:41
• The area of the lager tank's bottom was 1 square meter, and it held 5 cubic meters of water. For a prism such as a cylinder, we can divide volume by surface area to get height, so we see it must be 5m tall. Then the test slug of water (also a cylinder) is stated to be 1m tall. Nov 20, 2020 at 17:48

If the cross sectional area of a container is constant with height, you are correct that the normal force at the bottom of the container is the weight of the fluid. Actually it is the weight plus atmospheric pressure which is the pressure at the top of the fluid. For simplicity we well ignore the atmospheric pressure by using gauge pressure.

But the normal force at any point above the bottom is just the weight of the fluid above that point, which is less than the normal force at the bottom. In that way the normal force, and therefore the pressure decreases linearly from a maximum at the bottom to zero at the top.