# How can we show that pressure is exerted sideways too?

We can show that pressure upward and downward in a fluid is caused by weight of fluid column or volume. A simple derivation of this:

$$\text{Pressure}=\frac{\text{Force}}{\text{Area}}=\frac{\text{Weight of the fluid column}}{\text{Area}}$$ $$~~~~~=\frac{\text{mg}}{\text{A}}=\frac{\rho Vg}{\text{A}}=\frac{\rho g\times Ah}{\text{A}}=\rho g h$$

I am able to show that pressure in a fluid at a height is $\rho gh$ upwards and downwards, but how can I show that it acts sideways too?

I found a pretty reasonable answer here. So is this the correct reason?

Fluids are made of a large number of very small particles, much too small to see. These particles are in constant, rapid motion. They bump into one another. They bump into the walls of any container that holds them. They bump into objects in the fluid.

As the particles of a fluid bump into an object in the fluid, they apply forces to the object. The forces, acting over the object’s surface, exert pressure on the object. When the pressure in a fluid increases, the particles bump together more frequently. This increases the pressure on objects in the fluid.

The pressure a fluid exerts on an object in the fluid is applied in all directions. That is because the particles that make up the fluid can move in any direction. These particles exert forces as they bump into objects in the fluid.

This is essentially Pascal's law. I would like to add that it depends on the fluid being in equilibrium, as well as incompressible. Otherwise local pressure differences can build up and the fluid can behave rather differently.

I like to look at this problem from a thermodynamics point of view. One way to think of this is that the pressure of a fluid is a measure of it's potential energy (ability to do work), and in the static case, you can almost take this to be the definition of pressure. So $U = P V$ for some element of fluid with volume $V$. Then it is a matter of mechanics to show that there exists a force $\mathbf{f} = -\boldsymbol{\nabla} U$ wherever there is a potential energy gradient. This force must be cancelled by other forces (e.g. gravity), else the fluid will move, and not be in equilibrium any more. It follows that $U$, and thus $P$, is constant in a direction perpendicular to the external force (e.g. gravity).

Now, to answer your question, if you can mentally delete the vertical walls of your fluid, then outside the fluid the potential energy of a fluid particle would be zero, so there exists a sharp energy gradient where the wall used to be. This sharp energy gradient would result in a large horizontal force. The way to balance this force, of course, is to put the walls back. I should mention that the potential energy gradient is not strictly infinite because pressures are transmitted via electrostatic forces, which are long range forces.

• Actually I am not familiar with Thermodynamics yet. But Thanks!
– user49111
Commented Nov 28, 2014 at 16:45

It is not true that the pression of a fluid is due to gravity. Do you think fluids inside a bottle in the space station have no preassure? The right answer is the one you have quoted. The first link only refers to buoyancy, that does require gravity.

• -1. Pressure in a fluid is due to gravity. Pressure at certain is $\rho g h$, which is essentially derived from $\frac{Weight}{area}$.
– user49111
Commented Nov 28, 2014 at 11:22
• @ThePragmatick afaik, the pressure is due to the molecules hitting the walls of a domain. The gravity makes just the pressure increase with deepth. I wonder why among all the quantities playing a role in pressure (temperature, flow...), you chose gravity ?
– TZDZ
Commented Nov 28, 2014 at 12:59
• @ThePragmatick no, increase of preassure with depth is due to gravity. gravity at the top is not zero, is $p_0$, the atmospheric preassure.$p=p_0+\rho g h$
– user65081
Commented Nov 28, 2014 at 13:54
• @wolprhramjonny Why are you complicating such a simple thing? Pressure of a fluid at $h=0$ is 0. But the atmospheric pressure is not 0 because atmosphere forms another fluid. $h=0$ for one fluid doesn't mean $h$ is 0 for the other too.
– user49111
Commented Nov 28, 2014 at 14:29
• @ThePragmatick if you refuse to understand is up to you, I explained as much as I could.
– user65081
Commented Nov 28, 2014 at 16:23