Pascal's Law & Hydraulic Lift - Online Tutor, Practice Problems & Exam Prep

Pascal's Law and Hydraulic Lift

Hey, everyone! So in this video, we (re) gonna talk about Pascal's law and probably the most important application of Pascal's law, which is called the hydraulic lift. It's the basis for a lot of, like, our different hydraulic tool that we use in industrial purposes and stuff like that. Right? So let's go ahead and get started here. What Pascal's law basically says is that pressure in a confined fluid is transmitted equally throughout the fluid. Now there's a lot of words here. Basically, what this just means is if you have like a liquid that's sort of like in this weird shape like this, if you push down on one area of the tube or you apply some force here, that's gonna generate a pressure and that pressure basically gets transmitted through all of the points throughout the fluid. Alright, now how does this help us solve problems? It doesn't really, at least not yet. We got a couple more conceptual things to cover. All right. So, remember that pressure in a liquid, does or does not depend on the shape of the container. What do you guys think? Hopefully, you guys realize that it does not depend on the shape of the container. We can actually tell straight from our P bottom equation. Alright, Pbottomtop+ρgh. Now this sort of like the second point here has to do with this image. Notice how we have a bunch of different shapes, different shapes containers. And let's say I wanted to ca...

Hydraulic Lift / Proportional Reasoning

Hey guys, so let's check out this hydraulic lift example, and this is a very straightforward example. I really want to nail this point that you can solve some of these very easily if you just know the ratio between the areas. Let's check it out. So it says here, the hydraulic lift is designed with cylindrical columns, one having double the radius of the other. So, let's draw this real quick. So we've got little cylindrical columns there and one is double the radius. I'm going to call this r1 is just r, and then r2 will be 2r because it's double the radius. It says both columns are capped with pistons of the same density and thickness. That's just standard language so that you know that the pistons basically don't have an impact on anything. They cancel each other out. So it looks like that. And then it says if you push on the thinner column with the force f, so if you push with F, I'm going to say that your force F1 has a magnitude of F. How much force will act on the other piston? So how much force do you get here? Cool. And I hope you remember that the way to start this is to say, hey, the pressure on both sides is the same. So to say, p1 equals p2. Therefore, F1 equals, f1 over a1 is f2 over a2. This is because, of course, pressure is force over area. So then I can write that F2 must be F1 times (a2/a1). Now the areas here are the areas of a circle because the surface area of a cylindrical column is going to be the area of a circle, pi r squared because it is cylindrical. Cool? So this means you're going to rewrite this as πr₂πr₁, and I can cancel out the πs. And if you want, you can even rewrite, you can factor out the square and it's going to look like this. So it's proportional, your new force is proportional to the square of the ratio of the radii. That sounds like a mouthful, but if you have F here, which is the original force, and then the second radius is double the first, so it looks like this. 2nd radius is double the first. So the r's cancel and you're left with 22, which is 4. And we're done. The answer is 4F. Now there's a lot of math here. You could have solved this more simply by knowing that, hey, if the radius is double and the area is pi r squared, if you double the radius and the radius is squared, that means that the area is going to be 4 times greater. If the radius is doubled, the area is quadrupled and if the area is quadrupled, that means that the new force is also going to be quadrupled, or 4 times greater than the original force. Okay. Double the radius means you quadruple the area, which means you quadruple the force. You could have just done that as well. Hopefully, this serves as a little bit of a review on how to do the full solution, but you could have been that quick also. And if you remember from a previous video, I told you that if the force becomes 4 times larger, then the height difference or the height gain on the right side is going to become 4 times smaller. It's just the opposite of what happens. So if the force becomes 4 times bigger, then the height becomes 4 times smaller. That's it. Okay? That's all you need to know. Now I'm going to show how to solve this but at this point, you could have already known this but by just knowing that the amount that you're going to gain inside here is going to be smaller by the same factor that the force gets multiplied by. But let's solve this real quickly. And we solve this by knowing that the change in volume on the left, the amount of volume that goes down here is the same amount of volume that goes up here. I didn't draw that properly because I'm trying to move this quickly, but the volume on the left is the same as the volume on the right. And volume, remember, is area times height. Right? So I can write a1h1 or Δh1 because it's the change in height, equals a2Δh2. And we're looking for Δh2. So Δh2 is Δh1a1 over a2. All I've done is move the a2 to the other side of the equation and I now know that this area here this area here is 4 times larger. We know this from over here. Right? So I can just say this is Δh1, which we call just big H. And I have an area, and then I have an area that is 4 times that size. These can so and you see how you end up with H/4. Okay? That's it for this one. Let's keep going.

Force to Lift a Car

Hey guys, so in this hydraulic lift example, we're being asked to find the minimum force needed to lift a car. Let's check it out. Alright. So, it says the hydraulic lift is designed with cylindrical columns having these radiators here. So I'm going to draw a standard hydraulic lift. It looks something like this. The radius of the first one here, we're going to call this r₁, is 20 centimeters or 0.2 meters and then this one is radius r₂ is 2 meters and notice that radius r₂ is 10 times larger than radius r₁. Cool. And you are going to push down here with a force F₁ and that's what we want to know. So that you lift a car. I'm going to draw an ugly car here. So that you can lift the car, lift car, with it. Okay. And we want to know how much force that is. Now remember, every hydraulic lift question or almost all of them are going to start with either the fact that the pressure on the left equals the pressure on the right or it's going to start with the fact that the change in volume on the left has to equal the change in volume on the right side. Okay. And because we're looking for force and force has to do with pressure, this is the equation we're going to use. Remember that pressure is force over area. So we're going to immediately, as soon as we write this, the next step is going to be to write that P₁ becomes F₁ over a₁ and P₂ becomes F₂ over a₂. And here we're looking for F₁, the input force. I can calculate the areas because I know the radii. Right? Remember area, when you have a cylindrical column, it's going to be πr². So if I have R, I have A, I can calculate that. What about f₂? Well, f₂ is the force required to lift a car. How much force do you need to lift something? The amount of force you need to lift something, you should remember this. The amount of force to lift something is the weight of that something. The weight of the object. Now you might be thinking, if I, if mg is 100 and you push with 100, isn't that just going to cancel itself out? It is. Technically, you should push with 100.0001. Right? So you have to be just barely enough. But this is kind of silly so we just set them equal to each other. Okay? But don't get confused. Don't think that just because they're equal to each other, it's not going to move. The idea is that it's slightly more than that, but we use that amount. So hopefully that makes sense. F Lift will be replaced with MG. So this right here will be MG. Okay. So now I can solve for F₁ by moving a1 to the other side. So it's going to be F₁ equals F₂ which is mg times a1, a1 goes to the top, divided by a₂. Okay. And the mass of the car is 100. Gravity, I'm going to use 10 just to make this easier. The first area, let's calculate the first area. It's going to be πr squared. I'm going to do this slowly here. And then here, πr squared. R₁, r₂. We're going to cancel these 2. 800 times 10 is 8,000. The first radius is 0.2 and the second radius is 2.0. Alright. So if you do this, if you do this, the best way to do 0.2 divided by 20 by the way, oop, by 2 is to multiply both sides by 10. So this becomes 2 over 20 and now you can easily see that this is just 1 over 10. Okay. Hopefully, you get that or you could just do it in a calculator. But this is going to be 8,001 divided by 10 squared. If you square the top and the bottom, you get 8,000 divided by 100. So now, 20s are going to cancel, and you're left with 80. So this is the final answer. This means that you need a force of 80 newtons. Now, check out what's happening here. If you apply a force of 80 newtons, you're gonna be able to lift this car even though it takes 8,000 newtons to lift this car. Notice that this force here is 100 times greater than your input force. And that should make sense because your radius was 10 times greater. Let me write this out. The second radius was 10 times greater than the first radius. Remember, area is πr². So if the radius is 10 times larger, then the area is 10 square times larger. So the area is going to be 100 times larger, and therefore the force will be magnified by a factor of 100. Cool. That's it for this one. Classic example of a hydraulic lift. Let's keep going.