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.
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19. Fluid Mechanics
Pascal's Law & Hydraulic Lift
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