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  • Welcome back.

  • To just review what I was doing on the last video before

  • I ran out of time, I said that conservation of energy tells

  • us that the work I've put into the system or the energy that

  • I've put into the system-- because they're really the

  • same thing-- is equal to the work that I get out of the

  • system, or the energy that I get out of the system.

  • 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.

  • Let me just rewrite this equation here.

  • If I could just rewrite this exact equation, I could say--

  • the input force, and let me just divide it by this area.

  • The input here-- I'm pressing down this piston that's

  • pressing down on this area of water.

  • So this input force-- times the input area.

  • Let's call the input 1, and call the output 2 for

  • simplicity.

  • Let's say I have a piston on the top here.

  • Let me do this in a good color-- brown is good color.

  • I have another piston here, and there's going to be some

  • outward force F2.

  • 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.

  • The input force times the input distance is going to be

  • equal to the output force times the output distance

  • right-- this is just the law of conservation of energy and

  • everything we did with work, et cetera.

  • I'm rewriting this equation, so if I take the input force

  • and divide by the input area-- let me switch back to green--

  • then I multiply by the area, and then I just

  • multiply times D1.

  • You see what I did here-- I just multiplied and divided by

  • A1, which you can do.

  • You can multiply and divide by any number, and these two

  • cancel out.

  • It's equal to the same thing on the other side, which is

  • F2-- I'm not good at managing my space on my whiteboard--

  • over A2 times A2 times D2.

  • Hopefully that makes sense.

  • What's this quantity right here, this F1 divided by A1?

  • Force divided by area, if you haven't been familiar with it

  • already, and 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--

  • we'll call this the pressure that I'm

  • inputting into the system.

  • What's area 1 times distance 1?

  • That's the area of the tube at this point, the

  • cross-sectional area, times this distance.

  • That's equal to this volume that I calculated in the

  • previous video-- we could say that's the

  • input volume, or V1.

  • Pressure times V1 is equal to the output pressure-- force 2

  • divided by area 2 is 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?

  • The cross sectional area, times the height at which how

  • much the water's being displaced upward, that is

  • equal to volume 2.

  • But what do we know about these two volumes?

  • I went over it probably redundantly in the previous

  • video-- those two volumes are equal, V1 is equal to V2, so

  • we could just divide both sides by that equation.

  • You get the pressure input is equal to the pressure output,

  • so P1 is equal to P2.

  • 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 we have this notion of

  • pressure-- so where does that help us?

  • This actually tells us-- and you can do this example in

  • multiple situations, but I like to think of 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 pressure onto a

  • liquid, onto an incompressible fluid, that pressure is

  • distributed evenly throughout the fluid.

  • That's what we essentially just proved just using the law

  • of conservation of energy, and everything we know about work.

  • What I just said is called Pascal's principle: if any

  • external pressure is applied to a fluid, that pressure is

  • distributed throughout the fluid equally.

  • Another way to think about it-- we proved it with this

  • little drawing here-- is, let's say that I have a tube,

  • and at the end of the tube is a balloon.

  • Let's say I'm doing this on the Space Shuttle.

  • It's saying that if I increase-- say I have some

  • piston here.

  • This is stable, and I have water

  • throughout this whole thing.

  • Let me see if I can use that field function again-- oh no,

  • there must have been a hole in my drawing.

  • Let me just draw the water.

  • I have water throughout this whole thing, and all Pascal's

  • principle is telling us that if I were to apply some

  • pressure here, that that net pressure, 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's say that this right here is rigid-- it's some kind of

  • middle structure.

  • The rest of the balloon is going to expand uniformly, so

  • that increased pressure I'm doing is going through the

  • whole thing.

  • It's not like the balloon will get longer, or that the

  • pressure is just translated down here, or that just up

  • here the balloon's going to get wider and it's just going

  • to stay the same length there.

  • Hopefully, that gives you a little bit of intuition.

  • Going back to what I had drawn before, that's actually

  • interesting, because that's actually another simple or

  • maybe not so simple machine that we've constructed.

  • I almost defined it as a simple machine when I

  • initially drew it.

  • Let's draw that weird thing again, where it looks like

  • this, where I have water in it.

  • Let's make sure I fill it, so that when I do the fill, it

  • will completely fill, and doesn't fill other things.

  • This is cool, because this is now another simple machine.

  • We know that the pressure in is equal to the pressure out.

  • And pressure 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.

  • Let me give you an example: let's say that I were to apply

  • with a pressure in equal to 10 pascals.

  • That's a new word, and it's named after Pascal's

  • principle, for Blaise Pascal.

  • What is a pascal?

  • That is just equal to 10 newtons per meter squared.

  • That's all a pascal is-- it's a newton per meter squared,

  • it's a very natural unit.

  • Let's say my pressure in is 10 pascals, and let's say that my

  • input area is 2 square meters.

  • If I looked the surface of the water there it would be 2

  • square meters, and let's say that my output area is equal

  • to 4 meters squared.

  • What I'm saying is that I can push on a piston here, and

  • that the water's going to push up with some piston here.

  • First of all, I told you what my input pressure is-- what's

  • my input force?

  • Input pressure is equal to input force divided by input

  • area, so 10 pascals is equal to my input force divided by

  • my area, so I multiply both sides by 2.

  • I get input force is equal to 20 newtons.

  • My question to you is what is the output force?

  • How much force is the system going to push

  • upwards at this end?

  • We know that must 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.

  • So I'll have a piston here, and it goes up like that.

  • That's 4 meters, so I do 4 times 10, and so I get 40

  • newtons 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 essentially I had a mechanical advantage of 2.

  • This is an example of a simple machine, and

  • it's a hydraulic machine.

  • Anyway, I've just run out of time.

  • I'll see you in the next video.

Welcome back.

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    簡宇謙 に公開 2021 年 01 月 14 日
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