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.