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Welcome back.

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

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

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

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

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

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

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That means that the input work
is equal to the output work,

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or that the input force times
the input distance is equal to

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the output force times the
output distance-- that's just

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the definition of work.

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Let me just rewrite this
equation here.

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

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

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

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pressing down on this
area of water.

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So this input force-- times
the input area.

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

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simplicity.

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

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

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

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outward force F2.

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The general notion is that I'm
pushing on this water, the

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water can't be compressed, so
the water's going to push up

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on this end.

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The input force times the input
distance is going to be

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equal to the output force times
the output distance

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

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everything we did with
work, et cetera.

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

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

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then I multiply by the
area, and then I just

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multiply times D1.

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

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A1, which you can do.

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

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cancel out.

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

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

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over A2 times A2 times D2.

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Hopefully that makes sense.

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

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

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already, and if you're just
watching my videos there's no

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reason for you to be, is
defined as pressure.

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Pressure is force in a given
area, so this is pressure--

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we'll call this the
pressure that I'm

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inputting into the system.

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What's area 1 times
distance 1?

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

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cross-sectional area,
times this distance.

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

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previous video-- we could
say that's the

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input volume, or V1.

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

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divided by area 2 is the output
pressure that the water

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is exerting on this piston.

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So that's the output
pressure, P2.

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And what's area 2 times D2?

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The cross sectional area, times
the height at which how

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much the water's being displaced
upward, that is

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equal to volume 2.

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But what do we know about
these two volumes?

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I went over it probably
redundantly in the previous

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

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we could just divide both
sides by that equation.

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

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so P1 is equal to P2.

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I did all of that just to show
you that this isn't a new

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concept: this is just the
conservation of energy.

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The only new thing I did is I
divided-- we have this notion

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of the cross-sectional area,
and we have this notion of

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pressure-- so where
does that help us?

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

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multiple situations, but I like
to think of if we didn't

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have gravity first, because
gravity tends to confuse

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things, but we'll introduce
gravity in a video or two-- is

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that when you have any external
pressure onto a

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liquid, onto an incompressible
fluid, that pressure is

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distributed evenly throughout
the fluid.

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

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

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

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external pressure is applied to
a fluid, that pressure is

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distributed throughout
the fluid equally.

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

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

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and at the end of the
tube is a balloon.

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

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

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piston here.

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This is stable, and
I have water

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throughout this whole thing.

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

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there must have been a
hole in my drawing.

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Let me just draw the water.

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

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principle is telling us that
if I were to apply some

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pressure here, that that net
pressure, that extra pressure

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I'm applying, is going to
compress this little bit.

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That extra compression is
going to be distributed

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through the whole balloon.

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Let's say that this right here
is rigid-- it's some kind of

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middle structure.

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

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that increased pressure I'm
doing is going through the

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whole thing.

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

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pressure is just translated
down here, or that just up

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

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to stay the same length there.

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Hopefully, that gives you a
little bit of intuition.

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

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interesting, because that's
actually another simple or

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maybe not so simple machine
that we've constructed.

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I almost defined it as a
simple machine when I

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initially drew it.

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

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this, where I have
water in it.

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

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

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

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

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And pressure is force divided
by area, so the force in,

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divided by the area in, is equal
to the force out divided

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by the area out.

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

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with a pressure in equal
to 10 pascals.

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

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principle, for Blaise Pascal.

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What is a pascal?

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That is just equal to 10 newtons
per meter squared.

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

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it's a very natural unit.

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

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input area is 2 square meters.

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

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

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to 4 meters squared.

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

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

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

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my input force?

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Input pressure is equal to input
force divided by input

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

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my area, so I multiply
both sides by 2.

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I get input force is equal
to 20 newtons.

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My question to you is what
is the output force?

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How much force is the
system going to push

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upwards at this end?

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We know that must if my input
pressure was 10 pascals, my

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output pressure would
also be 10 pascals.

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So I also have 10 pascals is
equal to my out force over my

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out cross-sectional area.

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

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

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newtons is equal to
my output force.

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So what just happened here?

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I inputted-- so my input force
is equal to 20 newtons, and my

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output force is equal to 40
newtons, so I just doubled my

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force, or essentially I had a
mechanical advantage of 2.

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This is an example of a
simple machine, and

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it's a hydraulic machine.

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Anyway, I've just
run out of time.

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I'll see you in the next video.

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