### Video Transcript

In this video, we’re talking about
the pressure produced by fluids. And it’s true! All fluids — and those include
gases as well as liquids — produce pressure. In this video, we’ll learn what
causes fluid pressure. And we’ll also learn how to
calculate fluid pressure for different types of fluids at different depths.

Now here’s a bit of an interesting
thought. Whenever a person steps outside and
walks around, whether they realize it or not, they’re under the influence of fluid
pressure. A fluid we can recall is any form
of matter that can flow. This includes gases and
liquids. This means that whenever we’re
outside, we’re actually underneath many layers of gases in the atmosphere. Altogether the weight of all these
layers of gases in the atmosphere exert a force. And this force spread over the area
of our body as we walk around creates a pressure. We call this atmospheric
pressure. And this is a pressure experienced
by any object at sea level.

In general, in order to create
pressure, we need to have two ingredients. The first is a force and the second
is an area that that force is spread over. For example, if we have some
surface with an area that we can call capital 𝐴 and then we exert a force, we’ll
call it capital 𝐹 on that surface, then the average pressure — we’ll call it 𝑃
acting on the surface — is equal to the force divided by the area. So that’s what pressure is. It’s a force spread out over an
area. And just like stacks of layers of
air in the atmosphere all create a downward-acting force through their weight. And when that force is spread out
over the area of our body, we experience a pressure. Just like that, pressure is a force
spread out over an area. And just briefly so it doesn’t
catch us by surprise when we see it in an example, let’s say a quick word about the
units of pressure.

Since pressure is a force divided
by an area, we know that those units will be the units of newtons, the SI unit of
force, divided by meters squared, the unit of area. But we almost never see pressure
quoted this way. We almost never see a pressure
given as some number of newtons per square meter. Instead, the typical way to quote a
pressure is in a unit called a pascal abbreviated Pa. Here is what a pascal of pressure
means. Say we had some surface and the
area of that surface was exactly one meter squared. And then say that we apply a force
to that surface. We apply exactly one newton of
force. Well, one newton of force spread
out over one square meter of area is equal to one pascal of pressure. And just as a sidenote, one pascal
of pressure is a very, very small amount.

For example, the atmospheric
pressure we talked about experiencing anytime we step outside involves over 100000
pascals of pressure. But in any case, this is what one
pascal is, a newton of force spread over a meter of area. We’ve said here that any fluid
whether a gas or a liquid will produce pressure thanks to its weight. And when we consider the pressure
produced by liquids specifically, this is something lots of us have experience
with. Anyone who’s ever been inside the
diving well of a swimming pool has probably experienced that as they go deeper into
the wheel, the pressure that they feel increases. Often, the first place we feel this
is in our ears, one of the more sensitive spots on our body to pressure. Practically then, we get this
overall sense that as our depth in the water increases, the pressure grows along
with it.

And if we think about the water in
this tank in terms of layers just like we thought of air in the atmosphere that way,
this makes sense. Say that we consider each one of
these separate layers of water in the water tank. We know that the water in each
layer will have some weight thanks to its mass and the force of gravity on it. And if we start out by just
considering the water in this topmost layer here, we could say that the weight of
that water could be modelled this way. It creates a downward-acting force
below that layer. This means if we were underneath
that one layer of water, then we would feel the weight from it by itself.

Well, strictly speaking, there is
also the weight of the air in the atmosphere acting on us at this point. But let’s forget about that for the
moment. So we’re going to say that at this
position in the water tank, we’re only experiencing the downward force from this one
layer of water. But then, say that we drop down a
bit. At our new location, we’ll now
under two layers of water. So now, the force of the top layer
and the force of the second layer are both acting down on us. And then, if we continue to
descend, more layers of water are now above us. And therefore, the additional
weight of those layers is felt. And that weight, which is a force,
when it’s spread over the area of our body is experienced as a pressure.

Now, we’ve said that pressure in
general is a force spread over some area. It turns out though that the
pressure created by a fluid — a gas or a liquid — can be expressed a different
way. We can write that pressure as
capital 𝑃 sub 𝑓 to signify that it applies specifically to fluids. This pressure is equal to the
product of three different values. The first value is the density of
the fluid. We use the Greek letter 𝜌 to
symbolize density. Now density if we recall is equal
to the ratio of the mass of an object divided by the volume that object takes
up. So for example, air has a fairly
low density not much mass per unit volume. While a solid material, like
concrete or wood, will have a much higher density much more mass over some amount of
volume. So fluid pressure has to do with
density. And it also has to do with the
acceleration due to gravity, lowercase 𝑔.

One way to understand why this
acceleration should be here in this formula for pressure is to reconsider our diving
well example. In that case, we saw that it was
the weight of these layers of water which created pressure on the diver. And the reason those layers have
any weight at all is because of the attractional gravitational force on them. The Earth is pulling downward on
these layers. And it does so through the
gravitational field created by the Earth. So gravity’s acceleration is a
vital part of calculating fluid pressure.

The last piece of this equation is
the height ℎ of our fluid. And when we talk about the height
of a fluid, we have to be a little bit careful. Let’s say that looking at our
diving well, we want to calculate the pressure created by the water at this
location. We might think that the height of
that location will be measured from the bottom of the diving well. After all, if we had a cup of water
and we’re asked to measure the height of the water, we would probably measure it
that way. But when we’re calculating
pressure, things are different.

Recall that the pressure created on
our diver was thanks to the water that was above the diver rather than below
them. It’s only the layers of water above
a certain point that can contribute pressure to that point. And this means if we would have
measured the height of this point we’ve worked out, that height will be measured
actually from the surface of the water. That’s how tall the stack of fluid
is that’s responsible for creating pressure at that point.

Although this idea of measuring
height starting at the top rather than at the bottom of a liquid may at first seem
strange, we can at least see that measuring the height this way makes physical
sense. Given our equation for fluid
pressure, we know that the deeper an object is below a fluid surface, the more
pressure it will experience. If we have a small height, a small
distance below the surface, then our pressure will be relatively smaller. But as we drop deeper and deeper
down into a fluid, the pressure will grow. In fact, we could draw in arrows
representing the pressure exerted on the wall of the diving well as the depth of the
well increases.

At the top of the diving well, the
pressure in the wall is very small. But that pressure grows as our
depth increases until at the bottom of the well the pressure on the wall is fairly
significant. This is one reason why if you ever
see a concrete dam used to hold a lot of water in place, the bottom of the dam will
be much thicker than the top part. It’s because the pressure of the
fluid acting on the dam lower down is much greater and therefore requires more
reinforcement. Knowing all this about fluid
pressure, let’s get some practice now through a couple of examples.

Which of the following formulas
correctly shows the relationship between the pressure exerted by a fluid, the height
ℎ of the fluid above the point at which pressure is measured, the density 𝜌 of the
fluid, and the gravitational acceleration rate 𝑔? A) 𝑃 is equal to 𝑔 times ℎ
divided by 𝜌. B) 𝑃 is equal to 𝜌 times 𝑔
divided by ℎ. C) 𝑃 is equal to 𝑔 divided by 𝜌
times ℎ. D) 𝑃 is equal to 𝜌 times 𝑔 times
ℎ. E) 𝑃 is equal to ℎ divided by 𝜌
times 𝑔.

So in this exercise, we’re trying
to find the correct mathematical relationship between four quantities: pressure,
density, height, and gravity. We’re told that one of our five
answer options shows this correct relationship. To figure out which one it is,
there are two ways we can go about it. The first way — and certainly the
quickest way — is to recall off the top of our head the equation for the pressure
produced by a fluid. But let’s say that we’re not able
to recall that equation. Well, there is still a way we can
solve for the correct answer. The way we can do it is by looking
at the units on either side of each one of these equations.

Here’s what we mean by that. We can see that each one of these
five equations has four terms in it. Each one has pressure, it has
height, ℎ, it has density, 𝜌, and it has acceleration due to gravity, 𝑔. Now we can take each one of these
terms and we can look at the units that that term involves. For example, the units of pressure
are pascals, abbreviated Pa. And a pascal we can recall is equal
to a newton of force spread over a square meter of area. And we can further recall that
because a Newton is equal to a kilogram meter per second squared, then that means
the units of pressure are kilograms meters per second squared per meter squared,
which simplifies in the most basic of base units to a kilogram per meter second
squared.

The base unit of height on the
other hand is meters. The base unit of density is
kilograms per cubic meter. And we can recall that the units of
the acceleration due to gravity are meters per second squared. Now here’s the reason we’ve gone
through all this. If we look at the left-hand side of
each one of our five candidate equations, we see they all involve the pressure 𝑃 by
itself. We’ve seen from our analysis that
the units of pressure are kilograms per meter second squared. This means that these are the units
on the left-hand side of all of our candidate equations.

And for an equation to be true,
that is for it to be accurate, that means that the units on the right-hand side must
match these. They must also be kilograms per
meter second squared. If they’re not, then that means the
two sides of the equation can’t be equal. And that means that that particular
equation doesn’t correctly represent the relationship between these four values. So here is what we’ll do. We’ll go through these equations
one by one, starting with equation A. And we’ll analyse the units on the
right-hand side. If those units match up to the
units of pressure, then we’ll mark that equation as a possibly correct solution.

Starting out with equation A, the
right-hand side here has 𝑔 times ℎ divided by 𝜌. Looking over at the units of each
one of these terms, we see that the units of 𝑔 are meters per second squared, the
units of ℎ are meters, and the units of density are kilograms per cubic meter. In the numerator of this fraction,
these two factors of meters combine. And then if we multiply this
fraction by meters cubed divided by meters cubed, then that term meters cubed
cancels out in our denominator. And we have meters to the fifth
divided by second squared divided by kilograms or simply meters to the fifth divided
by second squared kilograms. When we compare these units to the
units we have for pressure, we see that there’s not a match, which means that option
A is not a reasonable candidate for the correct mathematical relationship.

Moving on to the right-hand side of
option B. Here, we have units of kilograms
per cubic meter for 𝜌 multiplied by meters per second squared for 𝑔 divided by
meters for ℎ. This numerator simplifies to
kilograms per meter squared second squared. And then if we divide both
numerator and denominator by meters, we have a final result of kilograms per cubic
meter second squared. This is close to the units for
pressure, but it’s not quite right. Therefore, option B isn’t our
answer either.

Moving to the right-hand side of
option C. Here we have meters per second
squared, the units of 𝑔, divided by kilograms per cubic meter, the units of 𝜌,
multiplied by meters, the units of height. A meter divided by meters cubed is
equal to one over meters squared. And then, if we multiply top and
bottom of this fraction by the inverse of the denominator, in other words multiply
it by meter squared per kilogram, then the meter squared per kilogram, the
denominator, cancels out with kilograms per meter squared. We’re left with a final result of
meters cubed per kilogram second squared, also not equal to the units of
pressure. So we cross option C off our
list.

Looking at the right side of
equation D. Here we have units of kilograms per
cubic meter multiplied by meters per second squared multiplied by meters. Combining all these factors of
meters, we get a result of kilograms per meter second squared. And we see that this is a match for
the units of pressure. So let’s put a star by option
D. And we’ll keep that in mind as a
possible answer choice.

Looking finally at the right-hand
side of option E. Here we have units of meters
divided by kilograms per cubic meter multiplied by meters per second squared. This denominator simplifies to
kilograms per meter squared second squared. And if we multiply both top and
bottom by meters squared second squared, this term cancels out in our denominator,
leaving us with meters cubed second squared per kilogram. We see this is not a match for the
units of pressure. And therefore, option E is off the
table. And this means that option D is our
answer. The pressure exerted by a fluid is
equal to the density of that fluid multiplied by 𝑔 multiplied by ℎ.

Let’s look now at a second example
involving fluid pressure.

What is the pressure exerted by
water at a depth of 2.5 meters? Use a value of 1000 kilograms per
cubic meter for the density of water.

So in this example, say that we
have a column of water. And we’re interested in the
pressure exerted by that water at a depth of 2.5 meters below the surface. So let’s say that’s a point here in
our water column. The pressure at this point we’ve
marked is created by the weight of all the water that’s above that point in our
water column. And by the way, it doesn’t make a
difference how wide the column is. However wide or narrow it is, the
pressure will be the same as long as we have this certain depth, 2.5 meters.

To answer this question of what is
the pressure exerted by the water at that point, we can recall that the pressure
created by a fluid is equal to the density of that fluid multiplied by its height
below the surface of the fluid times 𝑔, the acceleration due to gravity. Recalling that 𝑔 is 9.8 meters per
second squared, when it comes to the density of our fluid, we’re given that in our
problem statement, 1000 kilograms per cubic meter. And we’re also given the height, ℎ,
2.5 meters. And this means we can get right to
calculating this pressure. It’s equal to the density of the
water multiplied by the height below the surface of the water times the acceleration
due to gravity.

Now before we multiply these
numbers together, notice the units involved. That all the units are in base unit
form. We see that in the numerator of our
units, we have these two factors of 𝑚, the distance in meters. While in the denominator, we have
meters cubed. This means if we were to multiply
all the units involved together, we would get an overall result of kilograms per
meter second squared. This is equivalent to a newton per
meter squared. And we can recall that a newton per
meter squared is equal to the unit pascal, which is a pressure unit. This means that the units we’ll end
up with after we do our calculation are pascals. When we multiply these three
numbers together, we find a result of 24500 pascals. That’s the pressure exerted by the
water at this depth.

Let’s summarise now what we’ve
learned in this lesson.

We started out talking about
pressure. And we saw that pressure is equal
to a force spread out over an area. Written as an equation, we say that
𝑃, the pressure, is equal to 𝐹, the force, divided by 𝐴, the area. We saw further that pressure is
measured in units of pascals, where one pascal is equal to a newton per meter
squared. We learned in this lesson that
fluids — liquids and gases — exert pressure due to their weight. And this pressure is experienced by
any object under the fluid. We saw that the pressure created by
fluids — sometimes called 𝑃 sub 𝐹 — is equal to the density of the fluid
multiplied by the acceleration due to gravity multiplied by the height of that
fluid. And lastly, we saw that the height
in this equation is measured not from the bottom of a fluid but rather from the top
down. And this is because pressure comes
from the weight of a fluid layer acting down on some point.