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Good afternoon, ladies and gentlemen. The next unit we are going to learn is fluids.
Before we can actually talk about fluids there are two main types of fluid categories, hydrostatics,
and hydrodynamics. The first category we are dealing with, of course, is hydrostatics which
simply means about fluids that are staying still. Hydrodynamics, on the other hand, of
course means fluids that are moving but we are going to deal with fluids that are not
moving first. But before we can actually talk about some hydrostatic formulas I want to
talk about an old chemistry equation, which is, of course, density. If you guys remember
back in chemistry you probably had to do a density lab. So let’s just recap the density
lab quickly. In the density lab you probably had some sort of random object. You then had
to find the mass and then afterwards the volume. If it was an easy object to measure, you could
use length times width times height over the circle. If it was an irregular object the
easiest way to measure the volume of the object was to place it, of course, in fluid. You
could then measure exactly how much the fluid went up. In this case 9 mL which we will convert
to our units in a minute and based on knowing the mass and knowing the volume we are able
to solve for the density. So if we are to look at that again one more time, the mass
of the object was 11 grams. We convert that into kilograms, which is 0.011 kg. The volume of the object was 9 mL, which if I
convert that into meters3 you will get 9•10-6 m3. If you were to divide those two numbers
of course density ultimately gives you mass over volume. If the object was more dense
than the fluid, the object would sink. If it was less dense than the fluid the object
would float. In this case the object is more dense than the water because it sank. Now
the symbol we actually use for the object is not d. We will use the Greek letter ρ
(rho). It kind of looks like a curvy p. So basically density equals mass divided by volume.
We will be using that a lot later on so remember to keep that formula in mind. If I want to
do one quick practice question with that "what is the density of a 200 g box with a side
length of 10 cm?" So the mass would be 0.2 kg. Volume would equal length times width
times height. Now make sure you convert centimeters to meters before you plug it into the formula.
Now each side equals 0.1 m. So 0.2 kg would be the mass, the volume is 0.001 m3 and your
density would equal 200 kg/m³. The only density you need to memorize however is that of water.
In chemistry you learned it to be 1 g/cm³. For physics however we don't use grams or
centimeters. So it's going to be 1000 kg/m³, which means that this box above would, of
course, float on the water. So the next part we should talk about is pressure. What exactly
is pressure? Now when you think about forces, we've dealt with forces before but a little
bit different from the idea of pressure itself. Pressure is defined as the force applied to
a specific area; meaning to take this pen as an example. The pen has two points, one
point has a larger surface area and the other point has a smaller surface area. So if I
use this pen and apply a force to my hand I'll feel a certain pressure. Now if I turn
my pen to the other side and apply the same force it hurts a lot more. There's a lot more
pressure because there is a much smaller area. That force is concentrated over a much smaller
area. The formula for pressure is P=F/A. Based on the formula itself the unit would be newtons
per meter squared (N/m2). So keeping this formula in mind let’s take look at this
example of a person wearing snow shoes. This person is able to walk on top of snow. How
is that possible? Well basically because the snow might be soft but his weight is now distributed
over a larger area than before. Since it’s distributed over a larger area the pressure
is less so it’s not enough to actually penetrate through the snow. So when we talk about P=F/A,
with a larger area you have less pressure. Force of course being the person’s weight.
If you look at this next example there is a physics teacher who is brave enough to be
laying on a bed of nails. But he is not punctured. Why is that? Well individually one nail would
have very little surface area so if you put your entire body weight then there is going
to be enough pressure to actually penetrate through your skin and go in. But over an entire
bed of nails where there's a lot of them and a lot more surface area the force is evenly
distributed so therefore there is a lot less pressure. If you actually take a look at this
problem over here "estimate the pressure exerted on a floor by a 50 kg Jamie Schiffer, standing
momentarily on a spiked heel, and compare it to the pressure exerted by 1500 kg elephant."
I gave you Jamie Schiffer and her weight as she is standing on one foot. So let's keep
one foot up and one foot down and she is standing on 1 foot. Her weight, of course, is 500 Newtons.
The area was given as 0.05 cm². Of course we don't use centimeters we use meters. To
convert to m2, like we convert from cm to m we divide by 100, to convert from cm2 to
m2 you dived by 100 twice. If you divide by 100 twice you get 5×10-6 m². So if we now
take the time to get pressure, P= 500N / 5×10-6 m² which would equal to 1×108 N/m2, or 100
million N/m2. If you compare that to an elephant now, if an elephant were to stand on your
foot you might think it might hurt a lot more compared to if Jamie stood on your foot. But
if you calculate it out the pressure exerted by the elephant was a mass of 1500 kg so that’s
a weight of 15,000N divided by area of 800 cm² so that's 0.08 m2, which would equal
to 187,500 N/m2. A large amount but nearly as large as Jamie Schiffer. If Jamie Schiffer
were to actually stand on a single heeled foot and stand on your foot it would be enough
pressure to actually penetrate through you. I mean why do you think it's so hard for a
girl to walk across the grass in high heels? It's because as she's walking she's literally
poking a hole down into the grass as she's walking across. Anyways so if we go back to
over here, so what does pressure exactly have to do with fluids then? We have defined density,
we have defined pressure itself. What exactly is so special about fluid pressure? Now for
fluid pressure there are a couple things that are special about it. First of all it's that
the fluid surrounds an object on all sides. So for fluid pressure, pressure is exerted
on all sides and the pressure amount is exactly the same if you are at the exact same depth,
okay? So pressure is constant at a specific depth. The reason the second point is so important
is because we use hydraulics to do the most amazing things. Keeping in mind that pressure
always stays the same at a specific depth. Our first example here is a car lift. The
way a car lift works is you have a force being applied at one side and a much larger force
is applied on the other side automatically. That simply is because according to fluid
pressure we know that that pressure here and the pressure there must be exactly equal.
P1 is equal to P2 but they have different area sizes. Different piston sizes. So if
I actually wrote it out, F1 over A1 is equal to F2 over A2. I can see that there's a lot,
if I increase the area of the lift I automatically increase how much force is being applied because
of a lot more water being over here is applying that upward force due to the pressure that
exists over there. If we're actually to solve this as a problem, in a car lift, compressed
fluid exerts a force on a piston with a radius of 5.00 cm. If I were to draw this out just
like the picture we saw before, it has a first piston and a second piston. The first piston
and a radius of 5 cm which is equal to 0.05 m and the second one has a radius of 15 cm
or 0.15 m. So how much force must it exert to lift a car that weights this amount? So
there's a car over here. How much force must be exerted downward on one side to exert the
other force on the other side? So as long as we understand that the pressure are equal,
P1 will equal to P2. Now we know that F1 over A1 is equal to F2 over A2. So if I write this
all out, F1 over π times 0.052 is equal to 1.33 x 104 N over π times 0.152. Notice that
there is a π on both sides, I'll cross those out first and if I do a little bit of quick
multiplication, you get 1,477.78 N. A lot less force to lift a heavy car due to the
power of hydraulics. We see hydraulics whenever you talk about brakes in a car. Have you ever
wondered or even brakes in your bicycle, have you ever wondered how your brake pedal in
your car works? Well when you push down on the brake pedal what it's really doing is
that it's pushing on a tube of a liquid. So therefore, the force being applied when you
push down automatically transfers the force to where the tires are and the clamp closes
down on the tires, forcing the tire to slow down. Well let’s talk a little bit more
about fluid pressure. When we talk about fluid pressure, as I said before, pressure is being
exerted from all sides. Literally it's as if the weight of the water above the box were
all pushing down and compressing onto that box. Not only of all that weight of that water
above but there is also air outside the box that's adding on to that pressure as well.
In fact if we are to quickly look at this model over here, we have a pressure sensor.
We can see the deeper and deeper we go the more the pressure increases. Now one quick
note over here. Notice that they say there pressure is in atm. If you guys know the standard
pressure when we talk about STP, standard pressure is 1 atm but we don't use 1 atm in
physics. We use Newtons per meter squared. So the quick conversion is that 1 atm the
same thing as 1•105 N/m2. 1 atm is 100,000 N acting on one m2. When the sensor goes deeper,
deeper, deeper there's literally more water on top forcing down onto that sensor, which
is why the pressure increases as you go down. If I go out of the water, even if I go out
of the water my pressure never reaches zero. It reaches 1. What exactly is that 1? Well
that is the pressure of the outside air. So even when you are outside there's still air
right now pushing on that sensor but when you put that sensor into the fluid you have
the fluid pushing onto the sensor plus the air pushing down onto the water. All squeezing
down onto that sensor. So the formula you should be writing down for pressure is P=
Po+ ρgh. Po stands for atmospheric pressure. Which of course you guys know is 1 atm or
100,000 N/m2 and ρgh is the fluid pressure. Now I'm going to rewrite this later but all
together you create a total pressure which as you can see in this example over here.
Now before I actually write out the formula itself, let’s take a look at a couple different
examples of how strong this pressure really is. I'm sure you guys have may have seen many
submarine movies where somehow the submarine loses power and it starts sinking. In this
YouTube example over here they're just showing you the different compartments inside the
submarine. The submarine is designed to withstand high pressure. While it is designed to go
deep into water but even a submarine has its limits. So if then the power goes down, in
this case it's sinking down, down, down. I'll fast forward a little bit. Eventually there's
going to be a point where the compartment can no longer withstand that pressure and
the entire thing caves in and centrally breaks apart. Even with us being in air, air itself
exerts a huge amount of force on us from all sides. We just don't really realize it. In
this example we have a steel drum that's heated and then cooled. Now the reason it's being
cooled is to lower the pressure inside as much as possible. But when you lower the pressure
inside, I'm just going to pause this, well just watch. What happens is that air pressure
is still on the outside while it will actually crush the steel drum in. Pretty crazy but
normally you might ask why didn't the steel drum crush before? I mean everything is being
surrounded by this 100,000 N/m2. Why exactly aren't they being crushed? Well if you take
a look at the steel drum, the fact is yes, there is air pressure pushing on the still
drum from outside with 1 atm or 1•105 N/m2. There is also air inside the steel drum which
is pushing out of the same pressure of 1•105 N/m2. Ultimately those two pressures are going
to cancel each other out. So now let’s talk about the actual formula for fluid pressure.
You have P=PO + ρgh. P means absolute pressure. Which is just a fancy way of saying the total
pressure and when I say total it's just the total pressure of the atmospheric, which is
PO, which we also know to be 1•105 N/m2. That's the constant that will be given to
you plus ρgh. Rho, well this is still, rho is equal to the density of the fluid, g of
course is gravity and h is depth. Meaning measured from the top of the water itself.
So ρgh is the fluid pressure also sometimes called the gauge pressure. So if you take a look at our animation from
before, so outside we have air pressure. All the way at top we have air pressure. Inside
we air pressure plus the fluid pressure. So right now this value over here is the absolute
pressure. So P=PO + ρgh, now since this is in atm I'll leave it in atm for now. That
means outside we have 1 atm pushing down. So that's 15.51 atm = 1 atm + ρgh. That means
that the total fluid pressure has to be 14.51 atm. Notice that as we go deeper and deeper
and deeper the depth increases, the pressure is increasing and increasing and increasing.
As I said before the fluid pressure is ultimately caused by having the entire weight of the
water above pushing down onto the sensor over here. So it’s all pushing down onto this.
So if you think about it, ρgh, that makes perfect sense. If the fluid is more dense
it will be heavier than before and will increase the pressure acting on it. If you were from
a different planet and the gravity were increased, that would also increase how much pressure
is acting on it. If you increase the depth there would simply be more water on top pushing
down. So the formula is P=PO + ρgh. Let’s do a quick practice problem with that. What
is the absolute pressure in the bottom of a swimming pool 22 m by 12 m whose uniform
depth is 2 m? So if we're to sketch this out, let's see, let's make this into a cube. So
if we are looking for the absolute pressure inside the pool, that would simply be P=PO
+ ρgh. PO is, of course, like I said before is 1 atm or 1•105 N/m2 plus rho. The density
of water is 1000 kg/m3, the value that you will eventually memorize. Times gravity, times
the depth. So we're talking about the depth measured from the surface. The depth of course
is 2 m and if I take all the values and I combine them all together, we'll end up with
100,000+20,000. So it'll become 1.2•105 N/m2. That's the total absolute pressure.
Now the thing is, if for some reason if this scenario had the pool covered with a lid of
some sort that was tight. If you put a cover on top of this pool that completely blocked
off the air from hitting it, then we would have not included Po. But because it was open
in the air we are going to include air pressure in our calculation. As a quick follow up question
I also want to know how much force is acting on the bottom of the pool. Now if we go back
to the pool and take a quick look at the diagram, they say the pool has bottom which is 22 m
by 12 m. So the area is 22 m by 12 m. The area is 264 m2. Now we know the pressure already
which we calculated already. So therefore, we use the formula P=F/A. F = (1.2•105 N/m2)
• (264 m2), which equals to 3.17•107 N. A very, very large number�