If you're seeing this message, it means we're having trouble loading external resources on our website.

If you're behind a web filter, please make sure that the domains *.kastatic.org and *.kasandbox.org are unblocked.

Main content
Current time:0:00Total duration:10:12

Pressure and Pascal's principle (part 2)

Video transcript

welcome back so just to review what I was doing the last video before I run out of time I said that okay you know conservation of energy tells us that the work I put into the system or the energy that I put into the system because they're really the same thing is equal to the work then I the work that I get out of the system or the energy that are out of the system right and I said well that means that the input work is equal to the output work or that the input force times the input distance is equal to the output force times the output distance that's just the definition of work and let me just rewrite this equation here let me well if I I could just rewrite this this exact equation I could say so the force input force and let me just divide it by just just for for kicks let me divide it by this area because what I'm plotting the input here I'm pressing down this piston that's pressing down on this area of water so it's input port for us and let me say that times the input area well let me call this input let's call the input one let's call the output two for simplicity so there's going to be some let's say I have a piston on on the top here let me do this in a good color like brown brown is a good color so I have another piston here alright and there's going to be some outward force f-two and this is the general the general notion is that I'm pushing on this water the water can't be compressed so the water's going to push up on this end right so the input force times the input distance is going to be equal to the output force times the output distance right this is just law of conservation of energy and everything we did with work etc so I just rewrote I'm rewriting this equation so if I take the input force and divide by the input area let me just switch back to green and then I'm let's say I multiplied by the area then and then I just multiply times d1 you see what I did here I just multiplied and divided by a-1 which you can do you can multiply and divide by any number right these two cancel out is equal to the same thing on the other side f2 I'm not good at managing my space on my whiteboard F 2 / a2 times a2 times d2 hopefully that makes sense so what's what's this quantity right here this F 1 divided by a 1 well force divided by area if you haven't been familiar with it already and actually if you're just watching my videos there's no reason for you to be is defined as pressure pressure is force in a given area so this is pressure so we'll call this the pressure that I'm inputting into into the system pressure one times and what's area one times distance 1 so that's the area of the tube at this point the cross-sectional area times this distance well that's equal to this volume that I calculated in the previous video right we could say that's the input volume or V 1 pressure times V 1 is equal to the output pressure right force 2 divided by area 2 right that's the output pressure that the water is exerting on this piston so that's the output pressure P 2 and what's area 2 times D 2 so this area the cross-sectional area times the height at which how much the water is being displaced upwards that is equal to volume 2 right but what do we know about these two volumes and and I went over it probably redundantly in the previous video is that those two volumes are equal right V 1 is equal to V 2 so we could just divide both sides by that equation and you get the pressure input is equal to the pressure output so p1 is equal to p2 so p1 is equal to p2 coming out p2 and I did all of that just to show you that this isn't a new concept this is just the conservation of energy the only new thing I did is I divided we have this notion of the cross sectional area and and we have this notion of pressure so where does that help us well this actually tells us and and you can you can do this example in in multiple situations but I like to think of it with if we didn't have gravity first because gravity tends to confuse things but we'll introduce gravity in a video or two is that when you have any external a pressure onto a onto a a liquid so on to an incompressible fluid that pressure is distributed evenly throughout the fluid and that's what we essentially just proved just using the the law of conservation of energy and everything we know about work and what I just said that's called Pascal's principle that if any external pressure is applied to a fluid that pressure is distributed throughout the fluid equally and another way to think about it I mean we proved it with this little drawing here another way to think about it is let's say that I don't have a tube at the end of the tube is a balloon let's say I'm doing this on the space shuttle and it's saying that if I increase so let you know let's say that I have some some piston here right and let's say I were to I were to a tenia this is stable right and let me say I have I have water throughout this whole thing let's say I have water throughout the whole piston let me see if I can use that fill function again oh no there must have been a hole in my drawing edit let me just draw the water set of water throughout this whole thing and all Archimedes principle I mean sorry all Pascal's principle is telling us that if I were to apply some pressure if I were to apply some pressure here pressure in that that net pressure so if I you know and out of and that pressure is going to that extra pressure I'm applying is going to compress this a little bit that extra compression is going to be distributed through the whole balloon let's say that this this right here is rigid let's say that's some kind of metal structure the rest of the balloon is going to is going to expand uniformly so that pressure that increased pressure I'm doing is going through the whole thing it's not like the balloon will get you know it's not like the balloon is just going to get longer that the pressure is just translated down here or that you know just up here the balloon is going to get wider and it's just going to stay the same like there hopefully that gives you a little bit of intuition but going back to what I had drawn before that's actually interesting because that's actually another simple maybe or maybe not so simple machine that we've constructed and I almost defined it as a simple machine when I when I initially drew it right let's draw that weird thing again where you know it looks like this or I have water in it or there's a bunch of water let's say make sure I fill it so that when I do the fill it's completely filled doesn't fill other things there you go so this is cool because this is now another simple machine because we know that the the pressure in we know that the pressure in the pressure in is equal to the pressure out right pressure in is equal to the pressure out and pressure is just divided is force divided by area so the force in divided by the area in is equal to the force out divided by the area out right so let's say let's let's have it let's let me give you an example let's say that I were to apply with the pressure of so let's say the pressure in is equal to 10 Pascal's that's a new word and it's named after Pascal's principle or Blaise Pascal and so and what is a Pascal well that is just equal to 10 Newton's per meter squared that's all a Pascal is it's a Newton per meter squared it's very natural it's a very natural unit so let's say my pressure in is 10 Pascal's and let's say that my input area my input area is 2 square meters so you know if I looked at the surface of wall there would be 2 square meters and let's say that my output area my output area my output area is equal to my output area is equal to I don't know 4 meter squared so what would be the so what I'm saying is I can push on a piston here and that the water is going to push up with some piston here right so first of all I told you what my input pressure is what's my input force well input pressure is equal to input force divided by input area so 10 Pascal's is equal to my input force divided by my area so I'm multiply both sides by 2 I get input force is equal to 20 Newtons so my question to you is what is the output force how much force is the system going to push upwards at this end well we know that if my input pressure was 10 Pascal's my output pressure would also be 10 Pascal's so I also have 10 Pascal's is equal to my out force over my out cross-sectional area right so I'll have like a piston here and it goes up like that and so that's 4 meters so I do 4 times 10 so I get 40 Newton's is equal to my output force so what just happened here I inputted so my input force is equal to 20 Newtons and my output force is equal to 40 Newtons so I just doubled my force or I essentially add a mechanical advantage of 2 so this is an example of a simple machine this it's or and it's a hydraulic machine anyway I've just run out of time I'll see in the next video