### Video Transcript

In this video, we’re going to learn
about hydraulics. We’ll learn what this term means,
the discovery behind its usefulness, and we’ll also see how to apply it in practical
situations. To get started, imagine that you
are in preparation for this year’s upcoming World’s Strongest Human weight lifting
competition. One day while reading through the
rule book for the competition, you notice a loophole. It turns out that aside from using
their own bodies to supply the power to move weight in the competition, competitors
are in no other way restricted. Your goal for this year’s
competition is to win the prize for the greatest deadlift, that is, lifting the
maximum amount of weight a small distance off the ground.

This loophole in the rules gives
you an idea. You start collecting materials,
some lengths of tubing, a welded metal box, and some fittings to go between the
tubing and the box. To understand the idea you might
have in mind, we’ll want to learn something about the field of hydraulics. The word hydraulics has to do with
channeling or directing liquids, typically incompressible liquids, through pipes in
order to exert mechanical force and control. Here’s the basic idea behind
hydraulics and why it’s useful.

Say that we have a pipe system that
connects two endpoints of different areas. We further imagine that this pipe
is filled with a liquid of some sort which can’t be compressed. It’s incompressible and, moreover,
isn’t able to leak out around the edges where the endpoints are. So this liquid fills a fixed volume
and it can’t be compressed. Looking at the left side of our
system, say that we apply a force to this end cap. We’ll call the force 𝐹. And we’ll call the area of the end
cap 𝐴. This force 𝐹 spread over the area
𝐴 creates a pressure at this end cap. And this pressure is transmitted
all through the fluid, all throughout the entire system.

Now, here’s the really neat thing
about this system. If we were to look at different
cross sections all along the lengths of our tubing, even though these
cross-sectional areas are clearly different, the pressure we would measure at each
point is the same. This discovery that pressure is
constant across a hydraulic system is so useful that it became called a
principle. It’s named after its discoverer
Blaise Pascal. And in recognition of this
contribution, the unit of pressure 𝑃 is called the pascal. We’ve said that hydraulics is about
exerting mechanical force and control. And here’s how Pascal’s principle
helps us do that.

Consider the second end cap of our
hydraulic system, in particular its area which we can call 𝐴 sub two. On inspection, we can see that 𝐴
sub two is much bigger than 𝐴, our original end cap. And yet, the pressure at each of
these two points is said to be the same. This means that if we write our
pressure ratio 𝐹 over 𝐴 of our first end cap, that’s equal to the pressure at the
second end cap, where we have a force we can call 𝐹 two over area 𝐴 two. If we rearrange this equation to
solve for 𝐹 two, we see it’s equal to the ratio of the areas 𝐴 two to 𝐴
multiplied by the original force 𝐹. It’s this equation that explains
the mechanical force and control that hydraulics lets us exert.

Just think about it. For a given force 𝐹, if 𝐴 two is
10 times bigger than 𝐴, then that means our output force 𝐹 two is 10 times bigger
than the input force. And really, we can make that output
force 𝐹 two as large as we want. And we do it by varying the ratio
𝐴 two to 𝐴. There are many practical
applications for this principle. For example, when a dump truck is
full of heavy materials and it wants to empty out its load, a hydraulic press is
used to raise up one end of the dump trucks back, applying a very powerful force to
lift a lot of weight. A hydraulic system is also used for
digger machines in order to operate the large digging arm that carries masses of
earth or heavy stone. Let’s get some practice applying
Pascal’s principle through a few examples.

What magnitude force must be
exerted on the master cylinder of a hydraulic lift to support the weight of a car
with a mass of 2500 kilograms resting on a second cylinder? The master cylinder’s diameter is
2.50 centimeters and the second cylinder’s diameter is 25.0 centimeters.

From the information given, we can
label the mass of the car 2500 kilograms as 𝑚, the diameter of the master cylinder
2.50 centimeters as 𝑑 sub 𝑚, and the diameter of the second cylinder 25.0
centimeters we’ll call 𝑑 sub two. We want to solve for the force
magnitude that must be exerted on the master cylinder to accomplish this lift. We’ll call this force capital 𝐹
and start on our solution by drawing a diagram of the situation. In this scenario, we have a master
cylinder where we apply a force and a secondary cylinder that does the lifting of
the object we’re interested in elevating. In our case, that object is a car
of mass 𝑚 of 2500 kilograms. We apply a force of magnitude 𝐹 to
our master cylinder creating a pressure in our hydraulic system. That pressure is transmitted
through this system and then acts to keep the car elevated.

To solve for the magnitude of that
applied force 𝐹, we can recall that pressure is equal to force spread over an area
𝐴. And we also recall that in a
hydraulic system such as we have here pressure is constant. That means if we choose any two
points in our system, we’ll choose the input master cylinder and the output or
secondary cylinder. The pressure at these two points,
we can call them 𝑃 one and 𝑃 two, are the same. They’re equal to one another. This means that if we look at the
area, where 𝑃 one is applied and we call that area 𝐴 sub 𝑚, then we look at the
area, where 𝑃 two is applied and we call that area 𝐴 sub two, by the equality of
𝑃 one and 𝑃 two, we can write 𝐹 over 𝐴 sub 𝑚 equals the force applied by the
car, its mass times gravity, divided by 𝐴 sub two.

Rearranging this expression to
solve for the force 𝐹, we see it’s equal to 𝐴 sub 𝑚, the area of our master
cylinder in cross section, divided by 𝐴 sub two, the cross-sectional area of our
secondary lift, all multiplied by the weight force of the car 𝑚 times 𝑔. We can treat 𝑔, the constant, as
exactly 9.8 meters per second squared. And to solve for 𝐴 sub 𝑚 and 𝐴
sub two, we remember that the cross sections of these cylinders are circles. We recall that the area of the
circle is equal to 𝜋 times its diameter squared divided by four. The diameter of the master cylinder
and the diameter of the secondary cylinder are given.

So we can rewrite our expression
for force as 𝜋 over four 𝑑 sub 𝑚 squared over 𝜋 over four 𝑑 sub two squared all
multiplied by 𝑚𝑔. And we see the factors of 𝜋 over
four cancel out. Knowing 𝑑 sub 𝑚, 𝑑 sub two, 𝑚,
and the acceleration due to gravity 𝑔, we’re ready to plug in and solve for 𝐹. When we do, entering these values
on our calculator, we find that 𝐹 is 245 newtons. That’s the force we would need to
apply to lift up a car of mass 2500 kilograms.

Let’s get a bit more practice with
hydraulics through another example.

A certain hydraulic system is
designed to exert an output force 50.0 times as large as its input force. The system consists of two
cylinders, the master cylinder and the second cylinder. What must the ratio of the areas of
the second cylinder and master cylinder be? What must the ratio of the
diameters of the second cylinder and master cylinder be? By what factor is the distance
through which the input force moves relative to the distance through which the
output force moves? Assume no losses due to
friction.

In this scenario, we have a
hydraulic system with an input or master cylinder and an output or secondary
cylinder. The system is designed to meet a
particular specification that the output force — we can call it 𝐹 sub two — is 50.0
times greater than the input force — we can call it 𝐹 sub one. If we call the location in the
system where the first force is applied, location one, and the place in the system
where the second is applied, location two, we want to solve for three pieces of
information. First, we want to find the ratio of
𝐴 sub two, the area at location two, to 𝐴 sub one, the area at location one.

Next, we want to find the ratio of
the diameters of the two circular cylinders at those same two spots. And finally, if we call the
distance that the first or master cylinder descends 𝑠 one and the distance that the
secondary cylinder ascends 𝑠 two, then we want to solve for the ratio of those
distances 𝑠 one to 𝑠 two. We can start solving for these
three ratios by recalling the fact that pressure 𝑃 is equal to force divided by
area and that all throughout this closed hydraulic system the pressure 𝑃 is a
constant, which means we can write that the pressure at point one is equal to the
pressure at point two or 𝐹 one over 𝐴 one equals 𝐹 two over 𝐴 two.

We can rearrange this expression to
read 𝐴 two over 𝐴 one equals 𝐹 two over 𝐹 one. And by our given design constraint,
we know that 𝐹 two is equal to 50 times 𝐹 one. So in this ratio, 𝐹 one cancels
out. And we find that 𝐴 two over 𝐴 one
is 50.0. That’s the ratio of the area of the
second cylinder to that of the primary or master cylinder. Next, we want to solve for the
ratios, not the areas of these cylinders, but their diameters. And to solve for this, we can build
off of our result for part one. Recalling that the area of a circle
is equal to 𝜋 times its diameter squared divided by four, we can rewrite our area
ratio in terms of the diameters. And we see that the factor of 𝜋
over four in numerator and denominator cancel.

From our earlier work, we know that
this ratio 𝑑 two over 𝑑 one quantity squared is equal to 50.0. So if we take the square root of
both sides, at that point then, we have an expression for the ratio 𝑑 two to 𝑑
one. To three significant figures, this
is 7.07. That’s the ratio of the diameter of
the secondary cylinder to that of the master cylinder. Finally, we move on to solving for
𝑠 one over 𝑠 two, the ratio of the distance the master cylinder descends to the
distance the secondary cylinder ascends. To solve for this ratio, what we’ll
consider is the volume of fluid that’s being moved at each of these points one and
two.

Since the fluid in this system is
incompressible, that means that the volume 𝑉 one, which is the amount of fluid
moved by 𝐹 one pressing down on the master cylinder, is equal to the volume 𝑉
two. Since this hydraulic system is a
closed system, no fluid can escape and the fluid is incompressible. Since 𝑠 one and 𝑠 two are
distances, if we multiply those distances by the cross-sectional areas of their
respective points, writing 𝑉 one equals 𝑉 two is equivalent to writing 𝑠 one 𝐴
one equals 𝑠 two 𝐴 two. When we cross-multiply, we find
that 𝑠 one over 𝑠 two is equal to 𝐴 two over 𝐴 one, which we solved for
earlier. It’s equal to 50.0. That’s the ratio of the master
cylinder’s distance moved to the secondary cylinder’s distance moved.

Let’s summarize now what we’ve
learnt about hydraulics. We’ve seen that hydraulic systems
use liquids to amplify or multiply forces for the sake of creating mechanical
advantage. This mechanical advantage is
created thanks to Pascal’s principle, which says that within a hydraulic system, a
pressure is constant or the same everywhere. The equality of pressure at one
location to pressure at any other location is useful for figuring out how much
forces are amplified. And finally, the equation for
pressure connects force and area. It’s these two quantities, which
are the typical variables within a hydraulic system.