Today we're going
to change topics. I'm going to talk to you
about fluids, hydrostatic pressure
and barometric pressure. If, for now, we forget gravity and I would have
a compartment closed off and filled with a fluid-- could be either a gas
or it could be a liquid-- this has area A, here-- and I apply a force on it
in this direction, then I apply a pressure. Pressure is defined
as the force divided by area-- has units newtons
per square meter which is also called pascal. One newton per square meter
is one pascal. Now, in the absence of gravity, the pressure is, everywhere
in this vessel, the same. And that is what's called
Pascal's principle. Pascal's principle says that the pressure applied
to an enclosed fluid is transmitted undiminished
to every point in the fluid and to the walls
of the container. Keep in mind, pressure is
a scalar, it has no direction. Force has a direction and the force exerted
by the fluid on anything-- therefore also on the wall-- must be everywhere
perpendicular to the wall, because if there were
any tangential component, then the fluid
would start to move. and we are talking here
about a static fluid. So if I take any element-- I take one here at the surface,
little element delta A and the force must be
perpendicular to that surface, delta F, and so delta F
divided by delta A-- in the limiting case
for delta A goes to zero-- is, then, that pressure P. This has some
truly amazing consequences which are by no means
so intuitive. This is the idea
of an hydraulic jack. I have here a vessel
which has a very peculiar shape. Ooh, ooh, an opening here. And let there be here
a piston on it with area A1 and here one with area A2. It's filled
with liquid everywhere and I apply here a force F1
and here a force F2. So the pressure that I apply
here is F1 divided by A1. So according to Pascal, everywhere in the fluid,
that pressure must be the same. For now, I just assume
that the effect of gravity, which I will discuss shortly, doesn't change the situation
very significantly. But I will address
the gravity very shortly. So the pressure, then,
will be the same everywhere, but the pressure due to this
side is F2 divided by A2... and so the two must be the same,
if the liquid is not moving. So what that means is
that if A2 over A1 were 100, it would mean
that this force could be a hundred times less
than that one. In other words, I could put
on here a weight, a mass of ten kilograms, and here I could put
1,000 kilograms and it would be
completely in equilibrium. That's not so intuitive. This is used in all garages. What they do is,
they put on top of this-- if I blow that up here,
so this is this platform, there's a rod here
and on top of it is a car. And someone pushes here
and then this goes up. The car goes up. If I push here with a force a little bit more
than ten kilograms-- so that would be 100 newtons--
this level would go up. And so your first thought
may be, "Gee, isn't that a violation
of the conservation of energy? Am I not getting
something for nothing?" Well, not really. Suppose I push this down
over a distance d1, then the amount of fluid
that I displace-- that is, the volume,
is A1 times d1. That fluid ends up here. So this one will go up
over a distance d2. But the same amount of fluid
that leaves here adds there. In other words, A1 d1
must be A2 times d2. Now, if the force here is a hundred times less
than the force there, the work that I am doing
on the left side is F1 times d1. If the force here is
a hundred times less than that, the distance that I move is
a hundred times larger than d2, because A2 over A1 is 100. And so F1 d1 will be
the same as F2 d2-- 100 times lower force but
over 100 times larger distance, and so the product is the same. So the work that I do
when I push this down I get back in terms of
gravitational potential energy by lifting the car. So if I wanted to move
the car up by one meter and if the ratio is 100 to one I would have to move
this down by 100 meters. That's a little bit impractical so these hydraulic presses
are designed in such a way that you can just jack it
like that and every time
that you bring it up, that liquid flows back in again into this side
of the hydraulic jack. But, indeed, you will have to go
effectively 100 meters, then, for that car to go up
by one meter if the ratio is 100 to one. Now, gravity, of course, has an effect
on the pressure in the fluid. If you go down into the oceans, we know that the pressure
will go up, and that is the result
of gravity. And I would like to derive
the pressure increase. Let this be the direction
of increasing y, and I choose a liquid element,
so this is in the liquid itself. I can choose it in any shape
that I want to. I just take
a nice horizontal slab. And this is area A
so the bottom is also A. And let this be
at height y plus delta y and this is at height y. And the pressure here
is P y plus delta y and the pressure here is P of y. And this object has
a mass, delta m, and the liquid has
a density, rho, which could be a function of y-- we will leave that open
for now. And so this mass--
the mass that I have here-- is the volume times the density. And the volume is A--
this area-- times delta y, and then times the density,
which may be a function of y. So if now I put in
all the forces at work here, there is gravity, which is delta
m times g in this direction. Then I have a force upwards due
to the pressure of the fluid. That's what
we want to evaluate. It's always perpendicular
to the surfaces. We talked about that earlier. So in this side,
it comes in like this and here it comes
in like this, the force. From the bottom,
it comes in like this and from the top,
I call this F2. I only consider
the vertical direction, because all forces
in the horizontal plane will cancel,
for obvious reasons. So, now, there has
to be equilibrium. This fluid element
is not going anywhere-- it's just sitting still
in the fluid. And so I now have that F1--
which is in this direction-- minus F2 minus delta mg
must be zero. Only then is the fluid element
in static equilibrium. But F1 is this pressure
times the area-- so that is P at level y
times the area, and F2 is P at level
y plus delta y times the area, minus delta m
is A delta y times rho, so I get minus A times delta y,
which could be a function... rho could be a function of y
times g, and that equals zero. Notice I lose my area. I'm going to rearrange this
slightly and divide by delta y. And so I get that P
at the level y plus delta y minus the pressure at level y
divided by delta y equals-- if I switch that around, so I
bring this to the other side-- equals minus rho y times g. And if I take
the limiting case of this-- for delta y goes to zero--
then we would call this dP/dy. And this tells you that when you
go to increasing values of y, that the pressure will go down,
it's a minus sign. Very natural. If you go with
decreasing values of y, then the pressure will go up. And we call this
hydrostatic pressure. hydrostatic pressure So it's due to the fact
that there is gravity; without gravity, there is
no hydrostatic pressure. Now, most fluids, most liquids
are practically incompressible. In other words, the density of
the liquid cannot really change. And so therefore,
you could remove this and simply always use
the same density. It's exceedingly difficult. It takes horrendous
forces and pressures to change the density of
a liquid, unlike that of a gas. A gas is compressible and you can very easily change
the density of gas. So liquid is incompressible. If I have here a piston
and I have here a liquid and I put a force on here, it would be impossible for me
to make that volume smaller-- even by a fraction of a percent,
it would be impossible. If this, however, were gas, then it would be very easy
for me to push that in and to change the volume,
make the volume smaller and thereby make
the density of the gas go up. If I took a sledgehammer and I would hit
a plastic pillow, just bang, and the pillow was filled with
air, it acts like a cushion and I could squeeze it. If I hit the sledgehammer
on the marble floor, I could not squeeze it and the pressure on the marble
floor and on the hammer would be way higher, because I
don't have this cushion action. If I take a paint can-- and we
have one here, we have two-- and this paint can is filled
to the brim with water and another one
is filled with air and I hit it
with a sledgehammer, then this acts like a cushion. This one, however, doesn't want
the volume to be decreased, so the force, like on the marble
floor, would be way higher. But remember that force divided
by area is pressure, and according to Pascal, that pressure propagates
undiminished in the whole fluid. And so if I would shoot
a bullet through here, then I get a huge force-- extremely small area
of the bullet. And so the pressure
inside the liquid would go up enormously,
and the can might explode, provided that it's really filled
to the brim with water, because if there is air left, then you have
this cushion action. Now, I don't remember
whether there's air in here or whether there's air in there. I'll leave you to decide. So we'll fire a bullet
from this side, and then we'll see which can
explodes and which does not. And the one that doesn't
is the one that has air in it, and the one that explodes...
has the water in it, provided that we
really filled it to the brim. Oh, boy, there's
still something in there. (blowing) Okay. I did something wrong,
but that's okay. All right,
there goes the bullet. Okay, are we ready for this? So you tell me
which can is filled with air and which can is filled
with water. Three, two, one, zero. (gunshot) (sound of gunshot in slowmotion) Okay, this one is closed. It has a nice hole here
and a nice hole there. And this one has a hole here
and a hole there, but you saw the top come off. So we know
which one had water in it and by the way,
it's still there. These things are
not so intuitive. I will assume from now on that liquids are
completely incompressible. In other words, I can now use
this law that we have there and do a very simple
integration. I have now dP, which I can integrate
from some value P1 to P2. This is y, level P2, level y2, level y1,
pressure P1 in the liquid, and that equals now minus rho g
dy, integrated from y1 to y2. So that's now a trivial integral
because rho is constant-- rho is not a function of y. With the atmosphere of the
Earth, that's more difficult, because rho is a function
of altitude with the atmosphere but not with liquids. And so we get that P2 minus P1 equals minus rho g
times y2 minus y1, and this is called Pascal's law. I prefer to write it
slightly differently, but it's the same thing. I write a plus sign here,
so I switch these around: rho g times y2 minus y1. So what it means
is I see immediately that if y2 minus y1
is positive-- this is higher than this-- that the pressure at P1 is
larger than the pressure at P2, but of course they are
completely identical, so this is
the hydrostatic pressure. This has
quite bizarre consequences. Suppose I had a vessel
that I filled with a liquid. It had a rather changed shape,
like so... a ratherstrange shape. So I would fill it
with liquid to this level, and the level here is y2. And let's take the bottom of
this vessel and call this y1. And so inside here
we have pressure P1 and right there,
we have pressure P2. Well, what Pascal is saying now is that the pressure here
is everywhere the same because y2 minus y1 is the same
for all these points here. And so you will say, "Well,
that is sort of intuitive." You will say, "Look,
if I take here a column-- "nicely cylindrical vertical
column, which has area A, "and I call this separation h,
for simplicity-- "then the weight
of that column-- "that's the weight
of the liquid-- "would be the area times h--
that's the volume-- times the density
of the liquid, rho, times g." And so you would say,
"That's a force." The weight... the bottom here
has to carry that weight and so the pressure
at the bottom is that weight
divided by the area, so that is rho hg. So you would say,
"That's very clear." Yeah, maybe, but how about here? The pressure is the same,
it couldn't be any different. If the pressure
were different here, then the liquid
would start to flow. But here you don't have
that column h over you. You only have it here. And how about here? The consequence
of Pascal's law is that if you had
a vessel like this and you filled it all the way
with liquid, that the pressure
here at the bottom would be exactly the same
as this vessel, which is filled with liquid
all the way to the bottom. And yet the weight of this is way more
than the weight of this. But yet, according
to Pascal's law, the pressure differential
is the same. It is not intuitive. We live at the bottom
of an ocean of air. So here's the Earth
and here's air. And when we go up in y, then we also expect
that the pressure will go down. It doesn't go down linearly,
like liquids do. Liquids are linear
because rho doesn't change. In the case of air, the density
does change with altitude. But if I can take
one square centimeter cylinder all the way to the top
of the atmosphere just like I did here,
in the liquids... I take a one square centimeter--
I could have taken area A-- and I weigh all this air, then I would get the right
answer for the pressure here, because I do that there
and that works, so that should work here. And what I find, then... then I find that
at the bottom... at sea level, I find roughly one kilogram
force, which is ten newtons, per square centimeter. It means, then,
that this whole column... If I take a one square
centimeter tubing all the way to the top
of the atmosphere-- a few hundred kilometers-- that that would weigh one
kilogram only, all that air. One kilogram
per square centimeter. If you convert that
to newtons per square meter, then you get roughly
ten to the fifth pascal. And that is called, generally,
one atmosphere. It's called
the atmospheric pressure. So the air pushes down on us-- that gives us
the atmospheric pressure-- not unlike the way
that liquid pushes down because of its weight and increased pressure
as you go down. This atmospheric pressure is also often called
barometric pressure. The idea of a barometer. Here's my hand,
and my hand has an area of roughly
150 square centimeters. Force is always perpendicular
to the surface. I discussed that several times. For each square centimeter, there is an equivalent weight
of one kilogram due to the air above me. That means 150 kilograms
is pushing down on my hand. Why is my hand not going down? Well, because there's
also 150 kilogram pushing up. And so, I feel very comfortable. I don't even notice it that there is this huge force
in this direction and this huge force
in that direction. So how can I measure
this atmospheric pressure if I can't feel it? The way that you can measure it
is by the following experiment. You take liquid, and you put
a hose in the liquid, as I will do very shortly. This is cranberry juice,
and this is the hose. And we're going to immerse the
hose completely into that liquid so that it's completely filled
with the liquid. And then we lift it out,
and as we lift it out, we will see that the liquid
will stay there. It's the barometric pressure
that's pushing it in. And I pull it out
and pull it out and pull it out and pull it out and there comes a time that it
will not stay in there anymore. So it's way down there now,
the vessel, and then it lets go. And this is now empty,
and here is the liquid. And this is a way that we can
measure the barometric pressure and I will show you shortly
how that works. But let me first convince you that if I let all
this cranberry juice inside the tubing that I have... I put my finger on top
of the tubing and I lift it out. Notice that the cranberry juice
stays there. It's not running down. It's only when I take my finger
off the top-- (whooshes)
then it goes down. But as long as I hold my finger
on the top, it isn't going down. If my hose were long enough-- and we will know shortly
how long-- it turns out to be more
than ten meters-- then we would see the cranberry
break loose from the top. And this is a way that you can
measure the atmospheric pressure and I will now be
more quantitative about that. I will leave this equation
because I like that equation. So imagine that
we have this experiment-- which in the old days
wasn't done with plastic hoses, which was done
with glass tubes-- and suppose I end up
here with liquid and here with such a tube and that the liquid had broken,
so it's empty here... and here is the liquid. This is y1, this is y2,
this is the pressure P1 and right inside here is the
pressure P2 which is zero, because it's empty,
there is nothing. A little bit of vapor pressure,
but that's very small. And so this distance here...
let that be h. And so now what is
the pressure here at P1 which, of course,
is the barometric pressure? It's just exposed
to the atmosphere. Well, P1 minus P2 equals
rho of the liquid times g, times y2 minus y1, which is h. But P2 is zero,
so P1 equals rho gh and that is
the barometric pressure. So all I have to do
is take a liquid, know the density of the liquid, measure how far I have
to pull that hose up before the liquid breaks loose, and then I know
what the barometric pressure is. Now, this was done
in the early days, in the 17th century
by Torricelli. He used mercury
and he found that... Mercury, by the way,
has a density of 13.6 times ten to the third
kilograms per cubic meter. So this is mercury. He found that h is about 76
centimeters-- 0.76 meters. This is a fact. It changes a little bit
from day to day. It could change
by a few centimeters-- a little down, a little up. If it's up, the barometric
pressure is higher than when it's down. And so the barometric
pressure, P1, is then 13.6 times ten
to the third times g-- for which I will take ten--
and the height is 0.76, and that is 1.03
times ten to the fifth pascal... which comes very close to the one-kilogram force
per square centimeter that I mentioned to you earlier. One atmosphere's pressure is
defined in a very special way in a very precise way-- namely, that it is
exactly the pressure when the column here
is 760 millimeters of mercury. Then we call the pressure here--
that's the definition-- one atmosphere. Now you can do the same
experiment with water, whereas we tried to do it
with cranberry juice. The density of water is 13.6
times lower than that of mercury, so the column has to
be 13.6 times higher than 76 centimeters,
which is about ten meters. So you would have
to raise this thing up to ten meters
before you would see the break. But you would have... then you've built yourself
a water barometer. If you do it with mercury,
you have a mercury barometer. You would see
this level go down. And if the pressure is high,
the weather is good; and if the pressure is low,
the weather is not so good. So you could build yourself
a water barometer-- has to be ten meters long. The story has it that Pascal,
who was French, did the whole thing
with red wine. So he had a red wine barometer. It's very good to remember that ten meters of water
produces a hydrostatic pressure of one atmosphere. So if you go down
into the oceans by 100 meters, then the hydrostatic pressure
increases by ten atmosphere. So every ten meters
is one atmosphere. Cornelis Van Drebbel-- and I know how to pronounce
that name because I'm Dutch; he was a Dutch inventor-- is usually credited with
building the first submarine in the very early 17th century,
around 1622. And he successfully operated
this submarine at a depth of about five meters. Imagine, five meters. The hydrostatic pressure
there is half an atmosphere. Ten meters, one atmosphere--
five meters, half an atmosphere. Nowadays, submarines go... It's a little secret
how far they go, but... they have gone up to 3,000
feet, which is 900 meters, where the hydrostatic pressure
is 90 atmospheres. On every square meter
of that submarine, if it is at 900 meters, there is a force of 900 tons--
900,000 kilograms. Now, Van Drebbel's submarine was an enormous accomplishment
for the 17th century, because how are you going
to seal a vessel whereby the inside of the vessel
is one atmosphere-- that's the air
that he was breathing-- is five meters below the level, and so the outside pressure
is one and a half atmosphere? Namely, one atmosphere
barometric pressure and half an atmosphere
from the hydrostatic pressure. So there's an overpressure
on the vessel of half an atmosphere. But that means
on every square centimeter, there is a force pointing
inwards of half a kilogram-- the equivalent
of half a kilogram weight. Force is always perpendicular
to the surface, and if you would take two square
meters of his submarine, that would be a force
of 10,000 kilograms. It's amazing
that he managed to do that and that he could
actually operate his submarine successfully. I can show you here in 26.100 what kinds of forces Van Drebbel
was dealing with. You see there in front of you
a paint can. And I'm going to
evacuate the paint can. I'm going to pump the air out. And so here is the paint can,
about 25 centimeters by 15. And so it has equilibrium-- there's one atmosphere outside,
one atmosphere inside. Paint can is happy. I'm going to suck the air out,
so I get an underpressure here. In other words, the pressure
outside is higher than inside-- exactly the problem
that Van Drebbel had. The pressure outside is higher
than inside. You get an implosion. He managed to counter that,
to build it strong enough. When we take out the air here,
you can argue, well, then, the overpressure
is really one atmosphere and he only dealt
with half an atmosphere. Well, before we reach
this to be a vacuum, believe me, it already implodes. So the forces that we're dealing
with are very comparable to what Van Drebbel
was dealing with when he built his submarine. And so this can
will start to crumble when we take the air out, and that's another way of really
seeing the atmospheric pressure. So I take the pressure out
of the inside and the can
will literally be squeezed because of the ocean of air
that is hanging on us and is pushing down on us. Okay. It has to be
properly sealed, which is always a bit
of a problem. And so I have here a vacuum
pump, and let's pump on it. (can buckling) You can already hear
the crushing. The force on the front cover
alone... this is 375 square centimeters. If the pressure inside
were zero, that would be a force
of 375 kilograms. But look--
it's not very happy, that can. And these are the kind
of forces very comparable to what Van Drebbel was dealing
with in the 17th century, and he was able
to even operate his submarine under these forces,
without collapse. Okay, I think we... you want to take this
as a souvenir? Oh, no, I can't give
that to you. We have to first take this off,
but you can pick it up later. That... that mouthpiece
is quite precious for us, because we have to use that
again, of course. So you see what tremendous
forces are at stake when you deal
with barometric pressure. If you go scuba diving, you go
to a depth of ten meters-- could you stick a tube
in your mouth, which could go all the way
to the surface, and could you breathe? Well, there's no way. If you were here
and you have a tube... here's the water level... and if this is ten meters,
then the overpressure here between your lungs and the water
is one atmosphere, overpressure. So here is one atmosphere
barometric pressure. Here is one atmosphere
hydrostatic pressure plus the barometric pressure, so here you have
two atmospheres. So there is ahuge force
on your chest. Inside your lungs
is one atmosphere, outside is two atmospheres and there is no way
that you could breathe. If the area of my chest is
some 30 by 30 centimeters-- which is a thousand
square centimeters-- it would be like having
a hundred-kilogram weight on my chest, and also
on my back, of course, because it's
in both directions-- it pushes like this
and it pushes like this and like this and like this. So you're really being
squeezed to death. So what do you do
when you go scuba diving? You need pressured air
with you in the tank and that you breathe, and so now,
with the pressured air, you can obviously counter the hydrostatic pressure
from the water. Now suppose we go snorkeling. That's different. Then we do have a little tube
in our mouth, and we snorkel. How deep do you think
we could snorkel-- that our lungs
could easily accommodate the hydrostatic pressure? Any idea? Do you think we could snorkel
maybe three meters? Who thinks we could
easily do three meters? Okay, who thinks
maybe only one meter? Who thinks
way less than one meter? Well, we know it's not way less, because snorkels
are this long, you know, so you know you can
at least do 30 centimeters, so it can't be
all that much less. Well, we can measure
how deep we could snorkel and we can measure
what the capacity... the capabilities
of our lungs are in order to counter
the hydrostatic pressure. If I'm underwater, there
is pressure on my chest, and so to let the air out
is easy. You just...
I'm squeezed in, right? That goes out. But in order to inhale,
to suck in the air... (inhales) I would have
to push out my chest with a force that counters
the force due to the water. And so the question is, what kind of pressure
could I generate with my lungs to overcome
the hydrostatic pressure? And we are going
to measure that today with an instrument
that we call a manometer. A manometer is a simple tube-- it could be made of anything,
but a plastic tube will do fine. And we have liquid in here. It's open here
and it's open here. I put my mouth here,
and I'm going to see how much overpressure
I can produce in my lungs by pushing, by blowing. And so I'm going to blow in here and then this level will go down
and this level will go up. And this height difference--
let's call that h-- and the density
of this fluid is rho. We will use water for that,
colored water. So the pressure here at y1
equals P1-- that's here-- and the pressure here
at y2 equals P2. And so P1 minus P2
equals rho times h. I apply the law that
we still have here-- P1 minus P2,
rho times h times g. I know that the P2 level is
one atmosphere, that's correct. This is one atmosphere--
that's open to the world. So P1 equals one atmosphere
plus rho hg. And so what my manometer
indicates is how much pressure
I can generate over and above
the one atmosphere, and we call that overpressure. We often work
with overpressure gauges. When you go to the gas station and you have
your tire pressures measured-- the pressure in your tires-- there's also a gauge which
measures the overpressure. And so right there...
we have such a manometer. I'm going to blow in here, and that's going to tell
us immediately how deep I can snorkel. If I was able to make this
height difference ten meters, then I could snorkel at a depth
of ten meters in the water, because it means
that I could generate an overpressure
of one atmosphere. If this would only be
five meters, then I could only snorkel
down to five meters. If this is only a sad one meter, then I could really
only comfortably breathe one meter below the surface. And that's what you see here
and I want you to... Already you know it's not going
to be so fantastically high. Otherwise, these hoses
would be longer. This is the level
that we have now. On this side is atmospheric
pressure, that's open there, and on this side is also
atmospheric pressure, it's open. That's why the levels
are the same. This mark is 50 centimeters
above here and this is 50 below. So if I can generate
an overpressure in my lungs of one-tenth of an atmosphere, then this one would come up
50 centimeters and this one would go down
50 centimeters. That's one meter of water. One meter of water
is equivalent to a tenth of an atmosphere,
remember? Ten meters of water is one
atmosphere hydrostatic pressure. So if I can manage that, then I can generate an
overpressure in my lungs of a tenth of an atmosphere and I could snorkel, then,
at a depth of one meter. Let me try it. (inhaling deeply) (exhaling) (inhaling) (exhaling noisily) Aarch! A meter is impossible. You can have it maybe
for a few seconds, but not for very long. This is very disappointing. So you cannot even snorkel
at one meter. If someone else wants to try, I will cut this off,
so it's very hygienic. Maybe some of you can do better. You are the strong guy,
remember? Now, when you blow,
don't make the liquid oscillate, because then
you can squirt it out. You should really try
to take a deep breath and then push
as hard as you can. STUDENT:
Okay. LEWIN:
Go ahead. Strong man! You were about one meter
and 20 centimeters. Great, terrific. (class applauds ) How about sucking air? How much underpressure
could I generate in my lungs? Well, we can measure it
with this instrument. (inhales) I can go like this
and, of course, the liquid will go
the other way around. Maybe I can do much better in terms of underpressure
than in overpressure. Let's try. (class laughs) About the same,
a lousy one meter. When you're underwater, it's
never a problem to let air out, because due
to the hydrostatic pressure, there is a force on your chest. So letting the air out is easy. The problem is to expand
your lungs again to raise your chest. That means the problem
is that you can't... (inhales deeply) suck in the air,
and so it's really the second experiment that I did that determines how deep
you can snorkel underwater, and we found that
it's about one meter. So it's not the blowing out,
but it is the... (inhales deeply) sucking in. And so, this weekend--
and this is a true story-- I said to myself, "Gee, suppose I see someone... "From the second floor, "I look down on someone
on the first floor "having a great glass of juice
or wine or whatever, beer... and I would like to steal that
by sucking it up with a straw." (class laughs) Could I do that? And the idea would then being... I would be standing here. The person being unaware
of this glass down here with some great stuff in it... This would be my straw,
and I would just suck it up. And I decided over the weekend that the straw could not
be much longer, then, than about that one meter
that you just saw-- that's the underpressure. In fact, when you go
to supermarkets and you buy yourself
this stuff for kids, you know that this can be done. You can suck up at least... this is maybe
40, 50 centimeters. That you should be able to do; otherwise
they wouldn't sell them. But I didn't think that I could
do much more than a meter. And so I went to the supermarket
and I bought myself a hose and I bought the hose
to be two meters and I stood on the...
in the kitchen, like this. And I had a glass there
and I managed to do it. And it surprised me. So I went back to the super...
to the hardware store, got myself a three-meter
hose... tube. I knew for sure that there
was no way I was going to do it, so I went to the second floor
of my home-- I can look down
on the first floor; that's the way
the house is built. And I can't believe it,
I can't believe it-- why I can suck so well. It looks like it's almost a
violation of what I showed you, and so I need some help
from someone. And I'm going to demonstrate how good I am
in stealing someone's drink. Could you assist me with that? Because you have to hold my
straw in that juice, you see, because I will go very high. And so you...
well, just stand here, and I will throw you
the hose in a minute. Stand a little bit on the side,
so that the class can see you. I'll see you shortly. Hello. (class laughs) Okay, here's my straw. Can you put it in there? Now, as I'm going to try
to get this liquid up to me, I want you to think
about why I can do that whereas there I could
only do one meter. There is something very special. There's no way
that in my lungs-- this is five meter almost-- there's no way that I could have half an atmosphere
underpressure in my lungs; that is not possible. So somehow I don't do it
with my lungs, and maybe I won't even make it
in the first place. I can't talk when I do it. I cannot talk. There we go. (class laughs) LEWIN:
Mmm. (class laughs) Okay, I drank cranberry juice,
believe it or not. Think about all this
and try it at home, it's fun. Buy yourself a hose
that is even longer. Now watch it...
watch it, hold it. If I take my finger off, what do you think will happen
with the cranberry juice? It will run down, there it goes. Okay, thank you. See you Wednesday.
