19. Fluid Mechanics

Pascal's Law & Hydraulic Lift

# Force to Lift a Car

Patrick Ford

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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. All right. So it says hydraulic lift is designed with cylindrical columns having these radio here, So I'm gonna draw a standard hydraulic lift. Looks something like this. Um, the radius of the first one here. We're gonna call this. Our one is 20 centimeters or 200.2 m. And then this one is radius two is 2 m and notice that radius two is 10 times larger than radius one. Cool. And you are going to push down here with a Force F one, and that's what we want to know. So that's you Lift a car. I'm gonna draw ugly car here so that you can lift the car lift car with it, okay? And we wanna know how much force that ISS 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 and 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 gonna immediately As soon as we write this, the next step is gonna be the right. That p one becomes F one over a one and p two becomes F two over a two. And here we're looking for F one. The input force. I can calculate the areas because I know the radio. I right. Remember area. When you have a cylindrical columns, it's going to be pi r Square. So if I have are I have a I can calculate that. What about F two? Well, F two is the force required to lift a car? How much force do you need to lift something? The amount of force that you need to lift something. You should remember this theme Mount of force to lift something is the weight of that. Something the weight of the object. Now, you might be thinking, if I if m g is 100 you push with 100 isn't that just gonna cancel itself out. It is. Technically, you should push with 100.1 Right. 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 each other, it's not going to move. The idea is that it's the slightly more then that's what we use that amount. So hopefully that makes sense. F lift will be replaced with M. G. So this right here will be m G. Okay, So now I can solve for F one by moving a one to the other side. So it's gonna be f one equals, um F two, which is mg times a one. A one goes to the top, divided by a to okay. And the mass of the car is 800 gravity. I'm going to use 10 just to make this easier. The first area Let's calculate the first area is gonna be pi r squared. I'm gonna do this slowly here and then here. Pi r square are one or two. We're gonna cancel these two 800 times. 10 is 8000. The first radius is 0. and the second radius is 2.0. All right, So if you do this, if you do this, the best way to do 0.2 divided by 20 by the way, by true is to multiply both sides by 10. So this becomes to over 20. And now you can easily see that this is just 1/10. Okay, hopefully you get that, or you could just doing a calculator. But this is gonna be 8000 one divided by 10 square. If you square the top in the bottom, you get 8000 divided by 100. So now 20 they're gonna 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 800. I'm sorry. 8000 Newtons toe, lift this car. Notice that this force here is 100 times greater then your input force. And that should make sense because your radius was times greater let me write this south. The second radius was 10 times greater than the first radius. Remember, area is Pi R Square, so if the radius is 10 times larger than the area is 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 in hydraulic lift, let's keep going.

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