You've seen this example of a hydraulic jack in one of your reality check videos. We're going to look a little bit here at the pressure acting on these two pistons versus the forces acting on these pistons. We know from practice that we can push with a small force on an hydraulic jack and wind up lifting up something that's exerting a large force on the output side with this larger piston. So let's just check it and see how it works. If we simplify to make sure these pistons are exactly the same elevation, which is not terribly bad simplification, then we can look at locations one and two at the same elevation and see that P1 is equal to P2 because we're at the same elevation. But our observations tell us that F1 is not the same as F2. We've seen this in our application and the reason for that is because we got the same pressure but it's acting over different areas. The area of this piston is larger than the area of that piston. It becomes a little clearer if we draw some free body diagrams. Let's take the small piston first. It's got a pressure acting over this surface area on the bottom like that. So that F2 we can get as the integral over the area of our piston at 2 times the pressure times the incremental area that we're integrating over. And since the pressure here is constant and it's just equal to P2, then we can take that outside the integration and we wind up with F2 equal to P2 A2. If we do the same thing over here, we've got the same pressure but it's acting over this larger area on this larger piston. So the result is for exactly the same reasons. We've got F1 equal to the integral over area 1 of pressure at 1 over DA and we'll wind up with F1 equal to P1 A1. But both of those pressures are the same, P2 and P1. So the ratio F1 divided by F2, well F1 is equal to P1 A1 over F2 is equal to P2 A2 is equal to just the area ratio between location 1 and location 2. Or because area is proportional to diameter squared, if these are round pistons it'll be D1 squared over D2 squared. So this gives us a mechanical advantage. It can be quite a big mechanical advantage but it's not magic. It doesn't actually give us something for nothing. So suppose this piston is twice as big in diameter as this one, which is about how I've drawn it, I'll get four times as much force over here as I've applied over here. So I'm amplifying my force to move this other piston. But I'm not getting something for nothing because when I move this piston down, that piston will only move up one quarter as much. So the two must move down four times as much as one moves up. So I'm doing the same amount of work over here as I'm getting out over there. So we can use the area ratio to amplify forces by having the same pressure producing different forces. But we gain a mechanical advantage only in the force ratio. We still have to do just as much work. We still have conservation of energy in terms of what we accomplished by applying this force through a distance. So fluid statics can help us deal with practical problems such as jacking up your car. And we'll deal with a bunch of practical examples a little later.