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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

• 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 posted on 2014/12/06
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