Learn about Fluids - part 2 Video

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Learn about Fluids - part 2

Welcome back, so just review on what I was I doing in 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 there really the same thing is equal to the work that I get out of the system or the energy that I got out 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 such as the definition of work. And let me just rewrite this equation here. Well if I can just rewrite this exact equation, I could say so the force input force and let me just divide it by just for kicks—Let me divide it by this area because what—the input here I'm pressing down this piston that’s pressing down on this area of water. So it’s input port force and let me say that times the input area. Oh let me call this input, let’s call it the input one and let’s call the output two for simplicity. So this is going to be some; let say I have a piston on the top here. Let me do this in ink color, brown. Brown is good color. So I have another piston here right and there's going to be some output force, F2 and these are the general. The general notion is that I'm pushing on this water. The water can't be compress. So the water going to push up on this ends 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 conversation energy and everything we did at the work etcetera. So I just rewrote, I'm rewriting this equation, so if I take the input force and divide it by the input area. Let me switch back to green and then I'm, let say I'm multiply by the area then and then I just multiply times D1. You see what I did here, I just took and multiply and divided by A1 which can you do. You can multiply and divide it by any number right. These two cancel out is equal to the same thing at inner sight, F2. I'm not good at managing my space on my whiteboard. F2/A2 × A2 × D2, hopefully that makes sense. So what's this quantity right here? This F1 divided by A1. Well force divided by area if you haven’t been familiar with already and actually if you’re just watching my videos. There’s no reason for you to be is to find us pressure. Pressure is force in a given area. So this is pressure, so well call this the pressure that I'm in putting into the system. Pressure one times and what's area one times distance one? So that’s the area of the two but this point, the cross section 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 V1, pressure times V1 is equal to the output pressure right. Force two divided by area two, right that’s the output pressure that the water is exerting on this piston. So that’s the output pressure, P2 and what's area 2 times D2? So this area, the cross sectional area times the height in which how much the water is displays upward that is equal to volume two, right. But what do we know about these two volumes and I went over it probably redundantly in the previous video. Those two volumes are equal, right V1 = V2. So we could just divide both sides by that equation and you get the pressure input is equal to the pressure output. So P1 = P2, so P1 = P2 coming out 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 the section of the cross sectional area and we have this notion of pressure. So where that does help us? Well this actually tells us and you can do this example in multiples situations but I like to think of it wold we if we didn’t have gravity first because gravity tends to confuse things but well introduce gravity in the video or two. It’s that when you have any external pressure onto a liquid, sort onto in compressible fluid that pressure is distributed evenly throughout the fluid and that’s what we essentially just prove just using the law of conservation of energy and everything we know about work. And I 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 prove it with this little drawing here. Another way to think about it is let’s say that I'll have a tube and at the end of the tube is a balloon, Let say I'm doing this on the space shuttle and its saying that if I increase so you know, let say that I have some piston here, right. And let say I were to and I knew this stable right and let me say I have water throughout this whole thing. Let say I have water throughout the whole piston and let me see if I can use that fill function again. Oh no there's must have been holding my drawing. Let me just draw the water. So I have 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 that, that net pressure. So if I going to know and I would have and that pressure is going to; that extra pressure I'm applying is going to compress this little bit. That extra compression is going to be distributed through the whole balloon. Let say that this right here is rigid. Let say that some kind of metal structure. The rest of the balloon is going to expand uniformly. So that pressure that increase pressure I'm doing is going to the whole thing. Its not like the balloon would 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 just up here. The balloon is going to get wider and just it stays the same like that. Hopefully that gives you a little bit of intuition. But going back to what I have 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 define it as a simple machine when I initially drew right. Let’s draw that weird thing again where you know it looks like this. We have water in it where there's bunch of water. Make sure I fill it so that when I do the filling it’s completely filled. Doesn’t filled other things and there you go. So this is cool because this is now another simple machine because we know that 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 pressures is just divided as 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 say; lets give you an example. Let say that I were to apply with the pressure of; lets say the pressure in is equal to 10-pascals well that’s a new word and its name after pascals principle or Blaise Pascal. And so and what is the pascal? Well that is just equal to 10-newtons per meter squared that’s all a pascal is. It’s a newton per meter squares. It’s a very natural unit. So let say my pressure in is 10-pascals and let say that my input area is 2-square meters. So you know if I work to the surface of law there. There’ll be two square meters and let say that my output area is equal to—I don’t know—four meters 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-Pascals is equal to my input force divided by area. So I'll multiply both sides by two, 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-Pascals, my output pressure would also be 10-Pascals. So I also have 10-Pascals is equal to my out force over my out cross sectional area, right. So I'll have like a piston here goes up like that and so that’s four meters. So I'll do 4 x 10, so I got 40 newtons is equal to my output force. So what just happened here? I input it some my input force is equal to 20-newtons and my output force is equal to 40-newtons. So I just doubled my force or essentially I had a mechanical advantage of two. So this is an example of a simple machine. This is or; it’s a hydraulic machine. Anyway I've just run out of time, I'll see you in the next video.

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Welcome back, so just review on what I was I doing in 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 there really the same thing is equal to the work that I get out of the system or the energy that I got out 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 such as the definition of work.

And let me just rewrite this equation here. Well if I can just rewrite this exact equation, I could say so the force input force and let me just divide it by just for kicks—Let me divide it by this area because what—the input here I'm pressing down this piston that’s pressing down on this area of water. So it’s input port force and let me say that times the input area. Oh let me call this input, let’s call it the input one and let’s call the output two for simplicity.

So this is going to be some; let say I have a piston on the top here. Let me do this in ink color, brown. Brown is good color. So I have another piston here right and there's going to be some output force, F2 and these are the general. The general notion is that I'm pushing on this water. The water can't be compress. So the water going to push up on this ends 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 conversation energy and everything we did at the work etcetera.

So I just rewrote, I'm rewriting this equation, so if I take the input force and divide it by the input area. Let me switch back to green and then I'm, let say I'm multiply by the area then and then I just multiply times D1. You see what I did here, I just took and multiply and divided by A1 which can you do. You can multiply and divide it by any number right. These two cancel out is equal to the same thing at inner sight, F2.

I'm not good at managing my space on my whiteboard. F2/A2 × A2 × D2, hopefully that makes sense. So what's this quantity right here? This F1 divided by A1. Well force divided by area if you haven’t been familiar with already and actually if you’re just watching my videos. There’s no reason for you to be is to find us pressure. Pressure is force in a given area. So this is pressure, so well call this the pressure that I'm in putting into the system. Pressure one times and what's area one times distance one? So that’s the area of the two but this point, the cross section 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 V1, pressure times V1 is equal to the output pressure right. Force two divided by area two, right that’s the output pressure that the water is exerting on this piston. So that’s the output pressure, P2 and what's area 2 times D2? So this area, the cross sectional area times the height in which how much the water is displays upward that is equal to volume two, right.

But what do we know about these two volumes and I went over it probably redundantly in the previous video. Those two volumes are equal, right V1 = V2. So we could just divide both sides by that equation and you get the pressure input is equal to the pressure output. So P1 = P2, so P1 = P2 coming out 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 the section of the cross sectional area and we have this notion of pressure. So where that does help us?

Well this actually tells us and you can do this example in multiples situations but I like to think of it wold we if we didn’t have gravity first because gravity tends to confuse things but well introduce gravity in the video or two. It’s that when you have any external pressure onto a liquid, sort onto in compressible fluid that pressure is distributed evenly throughout the fluid and that’s what we essentially just prove just using the law of conservation of energy and everything we know about work. And I 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 prove it with this little drawing here. Another way to think about it is let’s say that I'll have a tube and at the end of the tube is a balloon, Let say I'm doing this on the space shuttle and its saying that if I increase so you know, let say that I have some piston here, right. And let say I were to and I knew this stable right and let me say I have water throughout this whole thing. Let say I have water throughout the whole piston and let me see if I can use that fill function again. Oh no there's must have been holding my drawing.

Let me just draw the water. So I have 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 that, that net pressure. So if I going to know and I would have and that pressure is going to; that extra pressure I'm applying is going to compress this little bit. That extra compression is going to be distributed through the whole balloon. Let say that this right here is rigid. Let say that some kind of metal structure.

The rest of the balloon is going to expand uniformly. So that pressure that increase pressure I'm doing is going to the whole thing. Its not like the balloon would 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 just up here. The balloon is going to get wider and just it stays the same like that. Hopefully that gives you a little bit of intuition.

But going back to what I have 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 define it as a simple machine when I initially drew right. Let’s draw that weird thing again where you know it looks like this. We have water in it where there's bunch of water. Make sure I fill it so that when I do the filling it’s completely filled. Doesn’t filled other things and there you go. So this is cool because this is now another simple machine because we know that 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 pressures is just divided as 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 say; lets give you an example. Let say that I were to apply with the pressure of; lets say the pressure in is equal to 10-pascals well that’s a new word and its name after pascals principle or Blaise Pascal.

And so and what is the pascal? Well that is just equal to 10-newtons per meter squared that’s all a pascal is. It’s a newton per meter squares. It’s a very natural unit. So let say my pressure in is 10-pascals and let say that my input area is 2-square meters. So you know if I work to the surface of law there. There’ll be two square meters and let say that my output area is equal to—I don’t know—four meters 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-Pascals is equal to my input force divided by area. So I'll multiply both sides by two, 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-Pascals, my output pressure would also be 10-Pascals. So I also have 10-Pascals is equal to my out force over my out cross sectional area, right. So I'll have like a piston here goes up like that and so that’s four meters.

So I'll do 4 x 10, so I got 40 newtons is equal to my output force. So what just happened here? I input it some my input force is equal to 20-newtons and my output force is equal to 40-newtons. So I just doubled my force or essentially I had a mechanical advantage of two. So this is an example of a simple machine. This is or; it’s a hydraulic machine. Anyway I've just run out of time, I'll see you in the next video.

Transcription by: Scribe4you Transcription Services