19. Fluid Mechanics

Pascal's Law & Hydraulic Lift

# Hydraulic Lift / Proportional Reasoning

Patrick Ford

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Hey, guys. So let's check out this hydraulic lift example, and this is a very straightforward example. I really wanna 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. Eso it says here hydraulically to design with cylindrical columns, one having double the radius of the other. So let's draw this real quick. So we got a little cylindrical columns there and one is double the radius. I'm gonna call this. Our one is just our and then our two will be to our because it's double the radius, says both columns, Air capital, 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. Um, 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 So I'm going to say that your force F one 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 P one equals p two. Therefore, F one equals F one over a one is F two over a To this is because, of course, pressure is force over area. So then I can write that f two must be f one times a two over a one. Now the areas here, the areas here are the areas of a of a circle because the surface area of a cylindrical, um, column is going to be the area of a circle pie are square because it is cylindrical quote. So this means you're gonna rewrite this as pi r two square the radius of the second one divided by pi r one square the radius of the first one and I can cancel out the pies and if you want, you can even rewrites. You can factor out the square, and it's gonna look like this. So it's proportional. Your your new force is proportional to the square of the ratio of the radio. I 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 second rate This is double the first. So the ours cancel and you left the two square, which is four. And we're done the answers for F Now there's a lot of Matthew you could have solved this were simply by knowing that. Hey, if the radius is double in in 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 four times greater. If the radius is double, the area is quadruple. And if the area is quadrupled, that means that the new forces also going to be quadruple or four times greater 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 served as a little bit of a review 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 you, um if the force becomes four times larger than the height difference or the height gain on the right side is going to become four times smaller. It's just the opposite of what happens. So if the force becomes four times bigger, then the height becomes four times smaller. That's it. Okay, that's all you need to know. Now I'm gonna show how to solve this, but at this point, you could have already known this. But you by just knowing that the amount of that you're gonna get on the side here is gonna be smaller by the same factor that the force gets multiplied by. But let's solve this real quick. 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 of the left is the same as the volume on the right and volume. Remember, Is volume, if you remember, is area times height, right. So I can write a one h one or Delta H one because it's the change in height equals H two Delta H two and we're looking for a Delta H two. So Delta H two is Delta H one A one over a two. All I've done is movie A to to the other side of the equation. And I now know that this area here, this area here is four times larger. We know this from over here, right? So I can just say this is Delta H one, which we call this Big H and I have an area, and then I have an area that is four times that size. So these can so and you see how you end up with HPE over four. Okay, that's it for this one. Let's keep going.

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