Download 14 Fluids Chapters 14_-_fluids_combined

AP Physics – Fluid Stuff
Matter:
Matter is your basic stuff. You
probably agonized over what it was in chemistry.
Pretty simple though. Matter is just stuff –
anything with mass that takes up space. We often
keep track of matter via its mass and the volume it
occupies.
Density:
An important property of matter is
density. Density is defined as mass per unit
volume. The symbol for density is .
The equation for density is:

m
V
Where m is the mass, V is the volume and  is the
density.

A rectangular chunk of granite measures 1.5 m by 0.75 m by 2.5 m. How much does the thing
weigh?
First, we can get the density of granite for the table that the Physics Kahuna has kindly placed in
this very handout. Granite’s density is 2.7 x 103 kg/m3. Then let’s find the volume of the stone. V =
lwh
V = 2.8125 m3
V = 1.5 m (0.75 m)(2.5 m)
m
m  V
V
kg
m  2.7 x 103 3 2.8125 m 3 
m
Solve for the mass:



7.59 x 103 kg
Find the weight of the stone:
w  mg
m

 7.59 x 103 kg  9.8 2 
s 


7.44 x 104 N
The Physics Kahuna is relatively certain that you probably dabbled around a bit with good old
density in your chemistry studies.
Fluids:
A fluid is any material that flows and offers little resistance to changing its shape.
Essentially, what we’re talking about here, is a gas or liquid. A gas is a collection of very small
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particles that are rapidly moving around, independently of each other. Gases have an indefinite
shape and an indefinite volume.
In a liquid the particles are much closer together and exert attractive forces on each other. The
attraction isn't great enough to make the collection rigid, like we would see in a solid, but loose
enough so that the particles can move around fairly freely, but not loose enough that they can easily
separate. Thus a liquid can flow, but it all stays together. Liquids have a definite volume and an
indefinite shape.
Gases can be compressed or expanded - the volume can be easily changed. This is not true for
liquids. Liquids are basically incompressible. This is because they are about as close to one
another as they can get so squeezing them together doesn’t make much of a difference.
Pressure:
There’s a good chance that you have studied pressure in a chemistry class, so we’ll
quickly review the important points about it. Firstly, pressure is a scalar quantity.
Pressure  force per unit area.
Mathematically:
Pressure 
Force
Area
P
or
F
A
The unit for pressure in the United States is the psi, which stands for pounds per square inch or
lb/in2. Other common units are the atm, which stands for atmosphere, and inches of mercury.
These are the ones that you probably dealt with in chemistry.
The metric system uses the pascal which is abbreviated as Pa.
1 Pa  1
N
m2
Naturally we will use the pascal.

What pressure does a force of 1125 N exert on a surface that measures 2.0 cm by 1.1 cm?
p
F
A

125 N
0.02 m  0.011 m 
 570 000 Pa 
570 kPa
A pascal is one Newton of force acting upon a 1 square meter surface. Turns out that these here
pascals is most definitely small – kinda like they’re tiny or something - so the kilopascal (kPa) is
commonly used.
Atmospheric pressure is caused by the weight of the air pressing down on the earth's surface.
Imagine a column of air that measures one inch by one inch, this means it has a cross sectional area
of one square inch. The column soars upward to where the atmosphere ends and the vacuum of
space takes over. (This would make it be around 100 000 feet high.) So, imagine weighing this 100
000 foot high by one inch by one inch column on a handy bathroom scale. It would, at sea level,
weigh around 14.7 lb.
Its pressure would be:
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P
F
A
P
14.7 lb
1 in 2
 14.7 psi
Table of Density Values for Various Substances
Substance
kg/m3
Air
1.20
Carbon dioxide 1.84
Helium
0.17
Hydrogen
0.084
Methane
0.67
Nitrogen
1.16
Oxygen
1.33
Steam (100C) 1.99
Gases
Alcohol, ethyl
Aluminum
Copper
Gold
Granite
Ice
Brass
Iron
Silver
Lead
Mercury
Marble
Oil
Quartz
Rubber
Seawater
Styrofoam
Water
Wood
Density
kg/m3
0.791 x 103
2.70 x 103
8.9 x 103
19.3 x 103
2.7 x 103
0.917 x 103
4.70 x 103
7.8 x 103
10.5 x 103
11.3 x 103
13.6 x 103
2.7 x 103
0.85 x 103
2.65 x 103
1.15 x 103
1.025 x 103
0.10 x 103
1.000 x 103
0.50 x 103
In metric units, a column of air with an area of one square meter weighs 1.013 x 10 5 N (at sea
level). Therefore, atmospheric pressure would be 1.013 x 105 Pa or 1.013 x 102 kPa (or 101.3 kPa).

A 115 lb woman wearing high heel shoes is at a dance. Also attending the dance is a rather
large 325 lb man m (maybe a football player or a professional wrestler or some other large man
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type of profession). You, clumsy as always, trip and fall. The woman and the man both step on
one of your hands, placing all their weight on a heel of their respective shoe (which is on your
hand). Which would cause the greatest damage? The man's heel measures 3.0 in by 3.5 in and
the woman's heel measures 0.50 in by 0.50 in (she’s into those high fashion stiletto high heel
type shoes).
The heel that exerts the greatest pressure is the one that will cause the most damage to your hand.
First, we calculate the area of each heel.
Aw = 0.50 in(0.50 in) = 0.25 in2
A=lw
Am = 3.0 in (3.5 in)
P
115 lb
0.25 in 2
285 lb
Pm 
10.5 in 2
F
A
Pw 
= 10.5 in2

460 psi

27 psi
The woman exerts a pressure that is seventeen times greater than that of the man. So avoid high
heel wearing women when you are lying about on the deck at fancy dances!

The atmospheric pressure is 1.013 x 102 kPa. What force does it exert on the top of a desk that
measures 152 cm by 76 cm?
A =lw
P
A = 1.155 m2
A = 1.52 m (0.76 m)
F
A
F  PA  1.013

N
1.155 m 2
2
m


1.17 N
The study of liquids at rest is called hydrostatics.
If you dive under water, as you get deeper, the weight of the water above you exerts pressure on
you. As you go deeper, the pressure increases - more water above you, right? The pressure
increases by roughly 1 atm for every 32 feet of depth.
How does the pressure act on you?
The way a force acts on a solid is different than the way it acts on a fluid. Since a solid is a rigid
body, the force does not change its shape. The force mostly tries to move the object. A liquid
cannot sustain a force in this way. Push on the water in a wading pool and you makes a splash –
you make the water flow. If the fluid is restrained so that it can’t flow, and a force is exerted on it,
the force will increase the internal pressure of the fluid. The pressure exerted on a fluid in a closed
vessel is transmitted throughout the fluid and pushes at right angles to all surfaces that it touches.
This is called Pascal's Principle.
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Pascal’s principle  The force exerted by a fluid
on the walls of its container always acts
perpendicular to the walls.
When you are under water the water’s pressure pushes in
on you from all sides. The force is perpendicular to your
body. The clever drawing to the right shows you some of
the force vectors acting on the intrepid snorkel diver.
(Although she appears to have lost her snorkel.)
Forces exerted by fluid on wall of container are
perpendicular at every point.
The Barometer:
The barometer is a device used to measure air pressure.
If you fill a glass with water in a tub and then invert the glass and partially pull it out, the water will
stay in the glass. Why?
Force from
atmosphere
The weight of the atmosphere pushes down on the surface of the water – the old atmospheric
pressure. The water in the tub is confined so the pressure exerted on the surface is transmitted
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throughout the liquid. The pressure exerts a force perpendicular to the surfaces in the tub and in the
glass. So the water in the glass is pushed upward.
Weight of water
in the glass
The water in the glass wants to run out because of its weight, so it exerts a force throughout the
water that acts perpendicular to the various surfaces. It acts on the water surface, pushing it up.
This can be seen in the lovely drawing above.
The effect of this is that the two sets of forces cancel out. The water
wants to run out of the glass and raise the surface in the tub but the
weight of the air pushes down and that force is greater, so the water
is pushed up the glass. We end up with a static column of water in
the glass.
Forces cancel out!
This is how a barometer works. The atmosphere can support a tall
column of water. If we have a column of air that has a cross sectional area of one square meter, it
weighs 1.013 x 105 N. It turns out that it can support a column of water of the same cross sectional
area so long as the water weighs the same or less than the air. For sea level, this works out to about
a ten-meter tall column. This is about 32 feet.
Minor variations in the atmospheric pressure cause the water column height to vary slightly. The
height of the column could then tell you what the atmospheric pressure is.
Well, barometers don’t use no water, instead they use mercury. Why?
Mercury is very dense – about 13.5 times denser than water. So a barometer
using mercury as the working fluid doesn’t have to be as tall as a water
barometer. It turns out that at sea level, the column of mercury will be around
760 mm high (which is around 30 inches). 30 inches high is a lot more handy
and practical than 32 feet.
So how do you make a barometer? You takes you a glass tube, closed at one end,
that is around 80 cm in length or so and fill the thing with mercury. Plug up the
open end and turn the whole thing upside down. The open end, now on the
bottom is placed into a reservoir of mercury and the plug is removed. The
mercury will run out of the tube until the weight of the mercury is equal to the
weight of the air column. The area above the mercury in the tube is essentially a vacuum (it will
have a small amount of mercury vapor that has evaporated, but there isn’t much vapor in there).
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A common pressure unit when using barometers is the height of the mercury column. At sea level,
this is around 760 mm or 76 cm. In America we use inches. The height of the mercury column at
sea level is around 29 inches. Thus, when you hear the weather report the weather person might say
that the barometric pressure is 29.2 inches of mercury. In Gillette the atmospheric pressure is less
because of the altitude. The meteorologists give us a relative air pressure. They pretend that the
normal mercury column height is 30 inches and give us readings based on that.
Dear Doctor Science,
Why is barometric pressure given in inches of mercury?
-- Hugh Grant from Ardmore, PA
Dr. Science responds:
Back in the 1840's, when barometric pressure was first discovered, it was
considered vulgar to make a direct statement about the laws of nature. The phases
of the moon were called "Lunar Melancholic Waning" and even rainfall was
referred to as "The Lamentations of the Firmament." Everyone was a long-winded
poet, including Gilbert Shelton, the English amateur meteorologist, who coined the
metaphor "Mercuric Altitude" to describe his mental condition just before a storm.
Today, even when half the nation is on anti-depressants, we continue this proud
tradition.
Back to Barometers:
Barometers are useful in forecasting the weather. The rule of thumb is
that falling air pressure means bad weather, and rising air pressure means good weather. A steady
air pressure also means good weather. This is because weather is caused by huge, moving masses
of air which have different pressures. So pressure changes signal the movement of air masses that
have different temperatures which means weather. A low pressure means that weather is coming
towards you. A high pressure means that the weather is somewhere else, but not where you are.
Today, any device that measures air pressure is called a barometer, so not all instruments use
mercury columns. Aneroid barometers use little bellows (like accordions have) to measure the air
pressure.
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Pressure and Density: As mentioned before, when one is swimming under water, the
deeper you go, the greater the pressure you are exposed to.
The weight of the water above you is:
w  mg
Recall that density, , is equal to:

so,
m  V
w  mg
m
V
Plug this in for mass:
  V  g
 Vg
so
Volume is equal to the area multiplied by the height, or:
w  Vg
V  Ah
Plug in A h for V and you get:
w  Vg
Pressure is:
P
P
   Ah  g
F
A
w   Ahg
We can substitute Agh for F since the weight is the force
 Agh
A
so
P   gh
Thus the pressure in a fluid column can be found using this equation:
P   gh
This equation is provided you in a slightly different form for the AP Physics Test. It looks like this:
p  p0   gh
Here the only new thing is p0 - this simply stands for the initial pressure.
The pressure is proportional only to the depth and density of the fluid. The shape of the container
or the object has no effect on pressure.

What is the pressure exerted by water at a depth of 45.0 m?
p  p0   gh   gh  1.00 x 103
kg 
m
9.8

  45 m 
s2 
m32 
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Unit wise, you have a kg m/s2 (which is a N) over a m2, which makes the thing a Pascal.
P  441 000 Pa 
441 kPa
Dear Doctor Science,
Why does nature hate a vacuum?
-- ? from The Internet
Dr. Science responds:
Nature hates a lot of things and, actually, a vacuum is far down the list -- behind
perpetual motion machines and lemon-lime flavorings. The older Creation gets,
the crankier Nature becomes. Some say that Nature has simply lost her joie de vivre
and is looking to spend her autumn years railing against a long list of modern
developments and voting against school bond issues. Nature's first vacuum cleaner
was an old upright Hoover. It lasted 50 years. Nowadays, it's tough to find one of
these new plastic machines that'll make it five. Nature hates that, and come to think
of it, I'm not too happy about that, either.
Dear Doctor Science,
What is the given length for a Mercury-filled Torricelli tube?
-- Gern Sabourin from Sherwood Park, Alberta, Canada
Dr. Science responds:
It varies, depending on nose shape and hair color. Black haired people with small,
upturned noses have very short Torricelli tubes. Fortunately, the amount of
mercury in each tube remains constant,
independent of tube length. Blondes with sharp noses have the longest Torricelli
tubes of all, but compensate with decreased general libido and affect a cowering,
groveling demeanor. Many of them dye their eyebrows to appear more interesting,
and disguise their essential blandness by adopting an artificial manner of speech,
punctuated by loud belching.
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Dear Cecil:
How does one suck in a piece of spaghetti? Think about it. How one sucks milk
through a straw is easy. The lowered pressure in the mouth due to sucking causes the
air pressure over the milk to force the liquid up. But if one pushes on the end of a
piece of spaghetti it just buckles. The mouth is closed and sealed over the sides of the
spaghetti, so passing air doesn't drag it along. Somehow the air very close to the
mouth must obliquely communicate a force along the length, and it's far from clear
how it's possible. --Berg [I guess; kind of scrawled], El Cerrito, California
Cecil replies:
You're thinking, this is the lamest question Cecil has ever answered. However, this is
because you lack an appreciation of the scientific issues. I blame myself. It took me a
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while to get the regulars on the Usenet newsgroup sci.physics to focus on this too. Apart
from the one lamer who said the partial vacuum inside your mouth exerted a positive force
that pulled the spaghetti in, most reasoned as follows:
(1) Air pressure is customarily conceived of as acting perpendicularly to the surface on
which it bears. In other words, it presses straight down.
(2) Air pressure at any point on the side of a strand of spaghetti is exactly counteracted by
the air pressure on the opposite side.
(3) The one place where the air pressure is not counteracted is on the end of the spaghetti.
The pressure on the outside end is much greater than the pressure on the inside (mouth)
end.
(4) Therefore, the force on the spaghetti is equal to outside air pressure minus the pressure
inside your mouth times the cross section of the spaghetti.
You're not getting this, I said. I know how much pressure is exerted. What I want to know is
where it's exerted, since it seems pretty obvious that literally pressing on the end of a strand
of limp spaghetti doesn't do jack.
What do we care where it's exerted? said the sci.physics regulars. We are scientists. We
deal in the world of quantifiable effects. It is enough to know that the air bears somewhere,
and that the pressure differential in aggregate is some mathematically determinable amount,
as a consequence whereof the spaghetti is sucked, or rather forced, into your mouth.
Freaking gearheads, I said. Screw the mathematics. I want to know, what is actually
happening at the level of individual particles?
"Heisenberg tells us . . . ," the sci.physics types began.
Screw Heisenberg, I replied.
Finally a few of the scientific types conceded that the question had a certain practical
interest. After some discussion we concluded that whereas it is customary to think of air as
pressing straight down, most individual air particles, in fact, strike the surface of the
spaghetti obliquely. Those particles striking the spaghetti close to the point where it entered
the mouth, and whose vector had some inward-pushing component, would force it in.
Exactly how close to the mouth the particles would have to hit would of course depend on
whether the spaghetti was al dente or boiled to within an inch of its life. You want to take it
up with the sci.physics crowd, be my guest.
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Dear Cecil:
Mama mia, Cecil! The sci.physics crowd's pathetic shot at explaining how anyone can suck up
spaghetti makes one wonder why our tax dollars are still subsidizing hopeless efforts at science
education. We might as well be funding pictures of naked thermocouples. These wonks have
missed the point entirely. Spaghetti sucking is not merely a function of the difference in
pressure between the outside of the mouth and the inside. You can easily recognize this using a
simple thought experiment. Imagine that a physicist--Erwin Schroedinger will do as an
example--places the end of a strand of cooked spaghetti in his mouth. Then, instead of
sucking, imagine that he turns on a pump that rapidly increases the air pressure surrounding
his head. The air-pressure differential will, according to probability, crush Schroedinger's
skull like an eggshell long before it neatly forces the limp spaghetti through his pursed lips. (I
encourage any sci.physics wonks who doubt this to try the experiment at home.) Ergo, it's not
just air pressure.
A closer approximation to the right answer is that the spaghetti strand has its own density and
cohesion. When you suck, the difference in density between the spaghetti strand and the air
inside your mouth pulls the molecules on the surface of the spaghetti into your mouth; because
the spaghetti's cohesion holds it together, the rest of the spaghetti is pulled in too. But then,
I'm sure you knew that all along. --Harry Doakes, Portland, Oregon
Dear Harry:
When will I learn? I thought this question, while not without interest, was too esoteric to get much
reaction, forgetting that giving the Teeming Millions a question that's esoteric is like throwing the
piranhas raw meat. I got rants about spaghetti sucking from every unemployed Ph.D. in North
America, plus a few who aren't unemployed but who--how shall I put this?--probably ought to keep
their resumes up-to-date. A sampling of alternative theories, starting with yours:
"When you suck, there is a difference in density between the spaghetti strand and the air inside your
mouth."
I never suck, although I have my off days. What does "a difference in density" have to do with it?
There is a difference in density if the stuff just lies there on the freaking table. We're talking about
spaghetti, not an ideal gas.
"Spaghetti isn't a solid (although one can use air pressure differential to pull a round solid into the
mouth, but at nowhere near the spectacular rate at which spaghetti can be sucked, as it were) but a
starchy gel and so subject to physical laws that govern both solids and liquids. When subjected to a
pressure differential, the spaghetti strand necks down slightly and . . ." (Brian).
OK, the spaghetti isn't solid but mushy. So what? You can suck solid things too. Cecil has been . . .
well, I was about to say Cecil has been sucking various cylindrical objects, but I recognize that a
certain element will find this comical. Let us pause while the lads get it out of their systems. Very
well. By process of experiment, we learn that the speed at which something can be sucked depends
very little on whether it's solid or spongy--mostly it's a
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matter of its weight relative to its cross section. I therefore feel entitled to ignore the fluidic
aspects of the situation.
"The air pressure around the spaghetti causes an internal pressure in the spaghetti. At the
interface between outside and inside your lips, the internal pressure of the spaghetti drops;
the spaghetti flows down this pressure gradient" (Bruce Kline).
Bruce. It's "spaghetti." Quite a few people persuaded themselves that spaghetti exhibits
characteristics of flow and that it somehow extrudes into your mouth. But this could not
result in motion of the spaghetti without loss of structural integrity or at least permanent
deformation, which does not occur, and in any case would be slow. This is spaghetti, not
lava.
"I'm going with the dragging-it-along theory, wherein the seal of the lips around the strand
is not perfect but allows some air or, more probably, a thin layer of spaghetti sauce to be
sucked in, dragging the strand along with it" (J. Ebert).
First thing I thought of. First thing I rejected. The sauce acts as a lubricant, sure. However,
from our experiments with nonpasta above, we conclude that an extremely thin film of spit
is sufficient to facilitate sucking. Given the minute amount of fluid and the minimal cross
section it presents, it's implausible to suggest that the sauce or whatever is the sole or even
the primary medium of propulsion.
Cecil must be really desperate if he's consulting the noodleheads on Usenet.
Not at all. Cecil loves the noodleheads on Usenet. It's like consulting the regulars at the
local bar. The quality of advice is easily as good, and you don't have to stand for the price
of drinks.
--CECIL ADAMS
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AP Physics - Buoyancy
Why do your basic massively heavy humungous ships
float? An aircraft carrier weighs almost a hundred
thousand tons. So why don’t it sink like a stone?
Aristotle thought it was due to the shape of a body. If
the body had sharp edges, it would sink. If the body
was smooth and rounded – like the hull of a ship – it
would float. (So why do you have all those round
rocks on the bottom of a river?)
Like most of the teachings of Aristotle, this is pure
nonsense. The guy who actually figured out why
things float was another Greek genius, Archimedes
(ca. 287 – 212 B.C.). He discovered how buoyancy
works. Buoyancy is an upward force that a fluid
exerts on an object that is immersed in it. It causes
things to float or else results in an apparent loss of
weight of a body when it sinks in the fluid. This is
called the buoyant force.
The important physics law dealing with buoyancy is
called Archimedes' Principle.
Archimedes’ Principle  An object is buoyed up by a force equal to
the weight of the fluid it displaces.
This rule is true both for gases and liquids, since they are both fluids.
Archimedes' lost principle  When a body is immersed in water, the phone rings.
Floating and Sinking: You have 2
tons of steel. Steel is 9 times denser than
water, so the 2 tons of steel will only
displace about 2/9 or 0.22 tons of water.
The buoyant force is not nearly enough to
float the block, so it will sink.
We now take the same 2 tons of steel and
make a box from it. This hollow box has a
much greater volume than did the cube.
When you set the box in the water, is
displaces a larger volume of water than the
cube did, it is the displaced volume of fluid
which creates the buoyant force. As the box
sinks deeper into the water, it displaces
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more and more water, which causes the buoyant force to increase. When the buoyant force is equal
to the weight of the box, it stops sinking and floats. The weight of the box is equal to the buoyant
force so the net force on the system is zero.
So if the box is big enough, it will float.
As cargo ships are loaded with freight, they sink deeper into the water. As they sink deeper, they
displace more water. This increases the buoyant force. The boat’s weight and the buoyant force or
at equilibrium. The upward buoyant force is equal to the weight of the ship. As long as the total
weight of the ship does not exceed the weight of the water displaced, the ship will float.
Empty Ship
Buoyancy In Air:
Loaded Ship
Air is a fluid and Archimedes' principle applies to it just as it does to
liquids.
An object surrounded by air is buoyed up by a force equal to the
weight of the air displaced.
Balloons and blimps float in air just as boats float on water. A cubic meter of air has a mass of
around 1.2 kg, so its weight is around 12 N. Thus any object that has a volume of one cubic meter
and a weight less than 12 N would float, i.e., rise. If it weighs more than 12 N, it would sink, i.e.,
fall.
When you built a hot air balloon in chemistry, you took advantage of Archimedes' principle, even
though you were thinking in terms of volumes and numbers of moles and all that ever so exciting
chemistry stuff.
Imagine the blimp taking pictures of the superbowl as a large fish swimming in water. They both
maintain their 'altitude' by controlling their density and taking advantage
of Archimedes' principle.
Cause of the Buoyant Force:
You toss a rock into the water
and it sinks. The forces exerted by the pressure of the water act
perpendicularly to the outer surfaces of the rock (Pascal’s principle).
Let’s look at the forces.
The horizontal forces pushing in on the sides cancel each other out.
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This does not happen with the vertical forces. As the depth increases, the pressure increases. This
means that the forces pushing up, on the bottom of the rock, are greater than the forces pushing
down on the top of the rock. This is because the bottom of the rock is at a greater depth than the top
of the rock, so the pressure is greater. Recall that P   gh , so the greater the depth the greater the
pressure.
The difference in the forces on the top and bottom are responsible for the buoyant force.
Finding the Buoyant Force: A cube with area A on all its sides is immersed in a fluid.
We can find the force pushing down on the top. We can also find the force
pushing up on the bottom surface of the cube:
F
P
A
so
F  PA
We also know that P   gh
h
l
We can plug that into the force equation we just developed:
Force on top:
FTop  PTop A   ghA
Force on Bottom:
FBot  PBot A   g  h  l  A
g is the acceleration of gravity. l is the length of each side of the cube. h is the depth of the
top of the cube.
The force on the bottom is greater than the force on top, the difference is the buoyant force. We can
find this difference:
F  FBot  FTop   g  h  l  A   ghA
F   gA   h  l   h    gAl
The area of the cube A multiplied by length of the cube’s side l is the volume of the cube, so:
F  Vg
But the difference in the force on the top and bottom, F, is the buoyant force. Let's call it FB.
Fbuoy   Vg
Buoyant force equation
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FBuoy is the buoyant force,  is the density of the fluid, g is the acceleration of gravity, and V
is the volume of fluid that is displaced.
This is the exact same equation we worked out to find the weight of an object using density and
volume. All we’re really doing is finding the weight of the fluid that was displaced.
Apparent Weight: If the object’s weight is less than the buoyant force, the object will float.
If the object’s weight is greater than the buoyant force, the object will sink, but it will weigh less.
We call this the apparent weight, FA. The apparent weight is simply the difference between its
actual weight and the buoyant force.
FA  w  FB
Practice Problems:

A cube of steel that measures 5.0 cm on each side is immersed in water. The density of steel is
9.0 x 103 kg/m3. The density of water is 1.0 x 103 kg/m3. What is the (a) buoyant force acting
on the cube and what is (b) its apparent weight?
(a) Find the buoyant force.
V   0.50 m 
The volume of the cube is:
FB  Vg
 1.0 x 104
3
 0.000125 m3


kg
m
4
3 
x
m
1.25
10
9.8

m3
s2


 

 1.25 x 104 m3
1.2 N
(b) Find the apparent weight.
The weight of the cube is:
w  Vg
Now we can plug and chug:

kg 
m

FW   9.0 x 104 3  1.25 x 104 m 3  9.8 2   110 N
m 
s 




To find the apparent weight we use this equation:
FA  FW  FB

 110 N  12 N

98 N
A cork has a volume of 4.25 cm3. The density of cork is 207 kg/m3. [a] What volume of the
cork is beneath the surface when the cork floats in water? [b] What downward force is needed
to completely submerge the cork?
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[a]. The floating cork will displace a volume of water equal to its own weight. We can use the
density of the cork to find its weight. With the weight, we can then calculate the volume of water
displaced.
First let’s convert the volume of the cork to cubic meters:
 1m 
4.25 cm 

 100 cm 
3
 0.00000425 m3
3
 4.25 x 106 m3
Now we find the weight of the cork:
w  Vg

kg 
m

  207 3  4.25 x 106 m 3  9.8 2   8620 x 106 N
m 
s 




 8.62 x 103 N
Now we can find the volume of water this represents. We set the buoyant force equal to this value:
FB  Vg
V
FB
g




1
3 kg m 

 8.62 x 10
2
kg


m 
s
3
1.00
10
9.8
x



m3 
s2 

V  0.880 x 106 m3
8.80 x 107 m3

Or
2
3  10 cm 
7
8.80 x 10 m 

 1m 
3
 8.80 x 101 cm3

0.880 cm3
(b) To find the force needed to sink the cork, we have to analyze the forces. First we will draw a
FBD:
FB
The forces are: FB the upward buoyant force, w the downward weight of the cork, and
F is the force we must push down with to make it sink.
The sum of the forces must be zero:
FB  F  w  0
F  FB  w
We’ve already calculated the weight of the cork, w.
w
F
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All we have to do is calculate the buoyant force, this is different than when the cork was floating
because the entire cork is under water. So we use the cork’s entire volume to find FB.
FB  Vg
 1.0 x 104
FB  41.6 x 102 N


kg
m

4.25 x 106 m 3  9.8 2
3
m
s

 4.16 x 102 N



Now we can find the force
F  FB  w  4.16 x 102 N  8.62 x 103 N
F  4.16 x 102 N  0.862 x 102 N

3.30 x 102 N

A large 9.0 m radius spherical balloon is filled with helium. The mass of the balloon bag and
the little basket thingee that hangs underneath is 168 kg. How much additional weight can the
balloon carry (assume it is at sea level)?
FB
The fluid that is displaced is air, so we can easily find the buoyant force.
Let’s look at a FBD:
The sum of the forces must equal zero, this is the point where the upward force
(the buoyant force) is equal to the total downward force – the weight of the bag,
basket, and helium:
The forces are: FB (buoyant force), w (weight of bag and basket), FHe (weight
of helium), and the lift (lifting force) FL (weight that is to be lifted).
FB  FW  FL  FHe  0
Volume of balloon:
FHe
FL
FL  FB  FW  FHe
4
V   r3
3
4
3
   9.0 m 
3
 3 054 m3
FW
 3.054 x 103 m3
We already know w (weight of bag and basket). We can now calculate the buoyant force.
We’ll have to look up the density of the air and use that to find the buoyant force.
FB  Vg

kg 
m

 1.2 3  3.054 x 103 m 3  9.8 2   3.592 x 104 N
m 
s 




Next we have to find the weight of the helium. We have to look up its density as well.
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FHe  Vg

kg 
m

  0.17 3  3.054 x 103 m 3  9.8 2   5.088 x 103 N
m 
s 




Now we can find the additional weight the balloon can lift.
FL  FB  FW  FHe
 3.592 x 104 N  168 N  5.088 x 103 N
FL  35 920 N  168 N  5 088 N
 30 660 N

31 000 N
The Famous Archimedes Story:
Archimedes is quite a famous fellow. He is credited
with many inventions. There is a wonderful story told about him that involves the idea of buoyancy
and how he discovered his principal. Archimedes lived in Syracuse, a Greek city state in the third
century BC. He was a natural philosopher and worked as a sort of technology troubleshooter for
the king, King Hiero. It is said that Archimedes built some sort of lens that enabled him to set an
invading fleet of ships on fire from the walls of the city. He also is supposed to have built some sort
of lever device that could reach out over the walls of the city, grasp, pick up ships, shake them, and
then drop them back into the water so they sank. Quite a deal. He also invented the Archimedes
screw, a hollow tube that had a long screw inside it. One end could be placed in water, the screw
would be rotated, and water could be pumped up the tube. The screw is still in use in the middle
east.
So here goes the story. One day King Hiero asked Archimedes to determine if his new gold crown
contained the actual quantity of gold given the crown maker. The king suspected that it might have
been adulterated with some other less dear metal (like brass or silver). However, just in case the
goldsmith had not cheated the king, the crown could not be harmed in any way (so no core
samples).
Archimedes worried over the problem. He had to come up
with a form of what today we call "nondestructive" testing.
The king was not noted for being a patient man, so
Archimedes was under a lot of pressure. Bad things could
happen to people who failed to deliver for his majesty.
Finally, unable to solve the problem, Archimedes decided to
take a break and refresh himself with a nice bath. Well, the
bathtub ended up being his salvation for it was in the tub that
the solution finally came to him. As he dunked his body, he
observed that the water level rose when he got in the tub. He
also seemed to weigh less. The solution was at hand! He
realized he could measure the volume of the crown by
immersing it in water and could then determine if it was gold
by finding its density. He, supposedly, was so excited that he leapt naked from the tub and ran
through the streets of Syracuse shouting, "Eureka! Eureka! !" (I have found it).
The state motto of California is “Eureka”.
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Dear Doctor Science,
Humans often times swallow air while eating.
Later they release that air in the form of a burp.
Do fish swallow water and burp water?
-- Rob King from Harrodsburg, KY
Dr. Science responds:
Only those fish who suffer from an eating
disorder. Many carp seem to have this problem,
and early studies suggest that being kept captive
in garden ponds and fish tanks and stared at by
uncaring humans is the root cause of this
affliction. Some fish can be trained to burp air, at
varying pitches, so that they form a sort of
burping fish choir. In fact, a school of them, the
"Tuneful Tunas" was a staple on the old Lawrence
Welk show, filling in for the Lennon sisters while
they were in Moscow, visiting their grandfather's
tomb.
Dear Doctor Science,
Humans often times swallow air while eating. Later they release that air in the form of a burp. Do fish
swallow water and burp water?
-- Rob King from Harrodsburg, KY
Dr. Science responds:
Only those fish who suffer from an eating disorder. Many carp seem to have this problem, and early studies
suggest that being kept captive in garden ponds and fish tanks and stared at by uncaring humans is the root
cause of this affliction. Some fish can be trained to burp air, at varying pitches, so that they form a sort of
burping fish choir. In fact, a school of them, the "Tuneful Tunas" was a staple on the old Lawrence Welk
show, filling in for the Lennon sisters while they were in Moscow, visiting their grandfather's tomb.
Dear Doctor Science,
Since helium is lighter than air, does a helium tank get lighter or heavier as you empty it?
-- Carey Smith Moorman from Carlinville, IL
Dr. Science responds:
A helium tank is actually a vacuum of sorts, one that sucks the heaviness out of things. So when a helium
tank is connected to a balloon, it's merely removing the heaviness from the balloon. The law of Conservation
of Energy and Mass demands that heaviness go somewhere, and it goes into the tank. When you return the
tank to the chemical supply shop, they sell that heaviness to record companies where it is used to give
substance to popular music. People who handle such heaviness must wear rubber gloves, because absorbing
heaviness through the skin can result in depression, a natural consequence of working with popular culture
in any form.
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Dear Cecil:
For a long time I've heard stories about a man who tied a bunch of balloons to a lawn chair
and went soaring into the heavens. I even spent an afternoon searching at the library to see if
it was true, but no luck. I gave up, thinking it must be someone's wild imagination. Then the
other day a story in the paper made mention of a mad balloonist named Larry Walters. Can
you tell me more? -- Roger K., Dallas, Texas
Dear Roger:
How fleeting is fame. It's been a mere 16 years since Larry Walters made his legendary flight, and
already people are starting to think he's a mythical being. Au contraire. Larry, an authentic
working-class hero (at the time he was driving a truck), went aloft July 2, 1982, from his
girlfriend's backyard in suburban Los Angeles. His craft: an aluminum lawn chair borne by 42
helium-filled weather balloons.
Larry's original idea was that he would fly east to the Mojave desert, but it didn't quite work out
that way. As his girlfriend and buddy were feeding out the tether, the line broke and he shot
skyward. Eventually he reached 16,000 feet, where the pilots of at least two airliners saw him. Not
wanting to cause a fuss, he began putting out calls on his portable CB radio. After a while his feet
got cold, so he pulled out a pellet pistol and began shooting out balloons.
The descent was uneventful except for the fact that the balloons wrapped around some power lines
at the end, knocking out the electricity in a Long Beach residential neighborhood for about 20
minutes. But Larry and his chair stayed clear--he simply dropped a few feet to the ground, having
spent about 90 minutes in the air.
Most people thought the whole thing was pretty funny, and Larry got to appear on Letterman and
the Today show. But the FAA was not amused. "We know he broke some part of the Federal
Aviation Act, and as soon as we decide which part it is, some type of charge will be filed," a
spokesman said.
Sure enough, Walters was charged with reckless operation of an aircraft, failure to stay in
communication with the tower, and flying a "civil aircraft for which there is not currently in effect
an airworthiness certificate." He wound up paying a $1,500 fine.
Well worth it, you think? Maybe. But Larry's life did not end well. He quit his job and made the
rounds on the motivational speaking circuit for a while, but eventually this petered out. He and his
longtime girlfriend broke up, and he could find only part-time work. In despair, he hiked into the
Angeles National Forest near LA in 1993 and shot himself to death. He was 44 years old.
--CECIL ADAMS
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AP Physics - Moving Fluids
There are two types of flow that fluids can undergo; laminar flow and turbulent flow. Laminar
flow is also known as streamline flow.
Laminar flow is the motion of a fluid in which every particle in the fluid follows the same path as
that followed by previous particles. Basically it means that the particles in the fluid are traveling in
smooth lines, one right after the other. We can represent laminar flow by drawing streamlines,
which are example paths that the fluid particles are traveling along.
Turbulent flow
occurs at high flow rates and when the fluid is moving past irregular shapes. In turbulent flow, the
motion of the fluid becomes chaotic, and it forms eddies and whirlpools. Turbulent flow absorbs
energy and increases the frictional drag throughout the fluid.
Turbulent flow is very complicated and the actual motion of the fluid cannot be precisely
calculated. The mathematics needed to even approximate the flow is pretty hairy (and not all that
accurate, models are used which come fairly close to describing the behavior, but the models can be
off).
We will deal mainly with laminar flow and ignore turbulent flow.
In dealing with flow through pipes and tubes, we will assume that the fluid is incompressible (pretty
close for liquids, they pretty much are incompressible, but not exactly perfect for gas flow, gases
being subject, as you remember from chemistry, being compressible). We will also assume that
they encounter no internal friction as they flow through the pipe.
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Rate of Flow:
The rate of flow is defined as the volume of fluid that passes a certain crosssection in a given time. In mathematical terms, this rate of flow is:
R  vA
Where R is the flow rate and v is the velocity of the fluid. A is the cross-sectional area of the pipe.
Common units for rate of flow are cubic feet per second, cubic meters per second, gallons per
second, liters per second, etc. Almost any volume unit and almost any time unit can be used to
express flow rate.
The flow rate must be a constant throughout the length of the pipe, as we are ignoring friction and
assuming that the fluid cannot be compressed. What goes in has got to be what comes out.
Imagine water entering a hose at one end, traveling through the hose, and then coming out of the
other end of the hose. The water that enters the hose in a given time has to equal the water that
leaves the hose in the same time. So R, the flow rate, remains constant no matter what happens
inside the hose.
A2
v1
A1
v2
The flow rate will be constant even if the radius of the pipe changes.
Upstream, the cross-sectional area, A1, is larger than the cross-sectional area downstream, A2. The
flow rate at both of these points must be the same. The flow rate is:
R  vA
so
R  v1 A1  v2 A2
v1 is the velocity of the fluid upstream, v2 is the flow rate downstream, A1 is the upstream
cross-sectional area, and A2 is the downstream cross-sectional area.
So
v1 A1  v2 A2
This is the equation that you’ll get to play with when you take the AP Physics
Test.

Water flows through a rubber hose 2.0 cm in diameter at a velocity of 4.0 m/s. If the hose is
coupled into a hose that has a diameter of 3.5 cm, what is the new speed of the fluid?
v1 A1  v2 A2
v1 r12
 v2 r2
2
v1r12
 v2 r2
2
v2 
v1r12
r2 2
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v2 
m  2.0 cm 
 4.0
s  3.5 cm 2
2
v1r12
r2
2

1.3
m
s
The Nozzle Effect: When a fluid flows through a narrow opening, a nozzle, its velocity must
increase. We can see this in the following problem.

Water flows through a rubber hose 3.0 cm in diameter at a velocity of 5.0 m/s. If the hose is
coupled into a nozzle that has a diameter of 0.50 cm, what is the new speed of the fluid?
v1 A1  v2 A2
v2 
v1r12
r2
2
v1 r12
 v2 r2
2
m  3.0 cm 
 5.0
s  0.50 cm 2
v1r12
 v2 r2
2

180
That’s a pretty good velocity increase!
Bernoulli's Principle: A fluid's
velocity increases, you have seen, when it
flows through a constriction - the diameter
of the pipe decreases. To accelerate the
fluid as it goes into the constriction, the
pushing force in the large diameter area
must be greater than the pushing force in
the constriction.
v2 
2
v1r12
r2 2
m
s
y
B
This is shown in the drawing. We have a
horizontal pipe that narrows and then
resumes its original diameter. Attached at
A, B, and C are small tubes filled with fluid. The height of the liquid in these tubes indicates their
relative pressures. At A and C the pressure is greater than it is at B. If y is the difference in height
between the liquid columns, then the pressure difference is given by:
A
C
PA  PB   g y
This change in pressure that takes place in a constriction is called the Venturi effect. The Venturi
effect says that pressure changes are accompanied by changes in velocity.
In the 1700’s Daniel Bernoulli (1700 – 1782), a Swiss scientist, experimented with water flowing
through pipes. He found that the pressure exerted by a liquid on its walls decreased as its velocity
increased. He found it to be true for both liquids and gases. Today we call this the Bernoulli
principle. In simple form, Bernoulli's principle says this:
When the speed of a liquid increases, its internal pressure
decreases.
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This is a consequence of the conservation of energy. Bernoulli developed an equation which relates
pressure and velocity in a fluid system which is called Bernoulli's equation.
1
P   g y   v 2  const.
2
Where const. is some constant, P is the pressure,  is the fluid density, y is the height
of the fluid, and v is the fluid velocity.
The
 g y term is the potential energy per unit volume of the flowing fluid.
The
1 2
 v is the kinetic energy per unit volume of the flowing fluid.
2
We can analyze the flow of a fluid through a system using this equation.
P1
v1
P2
y
1
v2
y
2
If we look at two locations in the system, we know that the sum of the pressure, kinetic energy, and
potential energy have to equal a constant, i.e., they have to be the same, so we can write:
P1   g y1 
1
1
 v12  P2   g y2   v2 2
2
2
Water flows through a pipe that has a constriction in it as shown.
P1
v1
y
P2
v2
y
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P1   g y1 
1
1
 v12  P2   g y2   v2 2
2
2
Analyzing the terms in Bernoulli’s equation, we see that the potential energy remains the same and
cancels out (no change in height, y). Therefore:
P1 
1
1
 v12  P2   v2 2
2
2
P2  P1 
1
1
 v12   v2 2
2
2
 P1 

1
 v12  v2 2
2

The initial velocity is smaller than the final velocity, so P2 has to be less than P1.
Let’s use Bernoulli’s equation to examine a static system.

An oddly shaped tank is filled with water to a depth of 1.20 m. Use Bernoulli’s equation to
calculate the pressure at point B at the bottom of
the tank.
A
P1   g y1 
1
1
 v12  P2   g y2   v2 2
2
2
The system is static, so the fluid has zero velocity.
The equation becomes:
P1   g y1  P2   g y2
1.20 m
Assume the pressure at point A on the surface is zero
and that the y value is also zero. The equation
becomes:
0  P2   g y2
P2    g y2
B
If we chose the downward direction to be negative, we get:
P2   g y2
Which is the equation we used before for hydrostatic pressure: P   gh
P2   g y2
 1.0 x 103
kg
m 32
m

 9.8 2  1.20 m 
s 

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P2 

1.18 x 104 Pa
A container of water, diameter 12 cm, has a small opening
near the bottom that can be unplugged so that the water can
run out. If the top of the tank is open to the atmosphere, what
is the exit speed of the water leaving through the hole. The
water level is 15 cm above the bottom of the container. The
center of the 3.0 diameter hole is 4.0 cm from the bottom.
y2
Let’s figure what is going on with the flow out the hole at the
bottom. We have the area of the hole A1 and the area of the
container A2.
The water spurts out of the hole with a speed of v1. The flow of water in the container, which
makes the surface level drop is very slow by comparison. So slow that we can say that it is  zero.
So v2 = 0.
The pressure on the top of the surface is the atmospheric
pressure. The surface acting on the water at the opening on
the bottom is also the atmospheric pressure (actually it is a
tiny bit bigger because it is slightly lower, but the difference
is insignificant). So we can let the two pressures equal each
other. So P1 = P2.
A2 P0 = P2
y
15 cm
Therefore, Bernouli’s equation,
4.0 cm
P1   g y1 
A1
P0
v1
1
1
 v12  P2   g y2   v2 2
2
2
Becomes:
1
2
 g y1   v12   g y2
This simplifies to:
1
g y1  v12  g y2
2
1
g y1  v12  g y2
2
We solve for the velocity, v1:
v12  2  g y2  g y1  v1  2 g  y2  y1 
We can plug in the data:
m

v1  2  9.8 2   0.15 m 0.040 m 
s 


1.5
m
s
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Applications of Bernoulli's Principle:
The Venturi effect is well illustrated in the
classic hose nozzle. Water, if allowed to pour out of the end of a one-inch hose, is not traveling at
much of a speed. Attach a nozzle to the thing. The nozzle makes the water squeeze through a very
small opening. As it does this, its velocity increases greatly, and the stream shoots out like crazy.
Its pressure decreases as it goes through the nozzle. This pressure drop does not immediately make
sense, because you know that the stream of water blasting out of the nozzle can put a hurt on stuff
you go squirting at. Do not confuse the pressure that a liquid has within it with the pressure it can
exert when something interferes with its flow. The pressure within a shooting stream of water is
low, but the pressure it can exert on something in its path can be quite large. The force it exerts on
things in its path is a result of the kinetic energy it has and its momentum.
Bernoulli and Flight:
Bernoulli's principle is often used to explain why birds and airplanes
can fly. One of the reasons that their flight is possible is because of the shape of their wings and the
way that air flows over and under these wings. Birds evolved the proper shape for a wing at least
135 million years ago. Human beings didn't catch on until fairly recently.
In the 1880s a British scientist named George Cayley developed the cambered wing. It shared a
cross-sectional shape with the wing of birds. It looks like this:
Here follows the standard explanation. As the wing passes through the air, the flow rate in front of
the wing on the top and bottom has to equal the flow rate at the end of the wing on the top and
bottom. The air passing over the top of the wing has to travel a longer distance than the air passing
under the wing. This is because of the curve on the top of the wing (the camber). Since it is
covering a longer distance in the same time, the velocity of the air on the top of the wing is greater
than the velocity of the air on the bottom of the wing. From Bernoulli's principle, we know that the
pressure decreases as the velocity increases. So the pressure above the wing is lower than the
pressure under the wing. Air pressure pushes the wing up to try and equalize the pressures. This
upward force is called lift.
For something to fly, of course, the lift must be greater than the weight. Birds are engineered to be
extremely light in weight. They have hollow bones, no teeth, no bony tail, female birds have only
one working ovary, etc. This helps them fly as the amount of lift their wings must develop is very
small. The light weight of birds can be deceptive. We think of large birds as being something like
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a turkey (which, at least for the domesticated ones) do not fly and are therefore quite heavy. But the
actual large flying bird is quite different. The Physics Kahuna fondly recalls the wonderful Laysan
Albatross (the fabled "gooney bird" of Midway Island). These are large birds - they probably stand
2 - 3 feet tall and have a wingspread of around 8 to 9 feet. On occasion, they would get in the way
(like when you were mowing your lawn and you didn't want to run one over). So you'd grab the
critter by the neck and tote it to safety. The first time the Physics Kahuna did this, he expected a
turkey sized mass for the bird, and applied what he thought was an appropriate amount of force. It
wasn't appropriate. The bird ended up being hauled into the sky with a terrific acceleration. The
bird was incredibly light. It felt like it was made of a few sticks of balsa wood, some tissue paper,
and stuffed with Styrofoam.
Other Bernoulli Applications:
Another application of the principle includes the
chimney. The chimney had a tremendous effect upon the history of Europe. Up till the 10th
century, people in many parts of Europe lived in large buildings in a huge room called a great hall.
In the winter, a fire would be kept in the center of the room. Everyone - peasants, the nobles, etc.
would sleep in the room to keep warm. In the center of the ceiling was a hole that was supposed to
allow the smoke to escape. The smoke, of course, did not escape all that well, but would fill the
room. Nasty business - breathing smoke is most unpleasant. The chimney changed all that. A
chimney is a pipe that extends from the room, through the ceiling, pierces the roof, and then up into
the air. The fire is semi-enclosed beneath it. Wind on the outside of the house moves across the
roof and across the opening of the chimney. The air in the house at the opening of the chimney is
not moving, so there is a pressure difference as a result of Bernoulli's
principle. The pressure above the chimney is less, so air is drawn from
the room into the chimney and then out the chimney above the house.
This carries the smoke to the outside. The chimney is said to "draw".
The air does this for two reasons, Bernoulli's principle and the fact that
the fire produces hot air which, being less dense, is buoyed upward.
Bernoulli's principle is responsible for making sure that none of the
smoke leaks into the room. Some of the heat is lost in this way, but
enough radiates into the room to keep it warm.
After the invention of the chimney, houses were broken up into
individual rooms, which was practical with the chimney. The nobles removed themselves from the
peasants – kicked them out. Who wants to have a bunch of peasants lying about! Eventually the
nobles built their own separate houses. The peasants were left to get by on their own, they ended up
living in small, mean, little huts. The rest is history.
Other animals, besides birds, have made use of
Bernoulli's principle. Prairie dogs build their burrows
with multiple entrances. One of the entrances is
always higher than the others. The wind near the
ground is usually less than it is above the ground. The
entrance that is higher than the others usually has
more wind blowing over it. This makes the pressure
above this opening less. Because of this, air is drawn into the other openings and comes out of the
high one. This gives the little critters a sort of natural air conditioning.
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Baseball pitchers and tennis players also make use of Bernoulli's
principle. The curve ball is a pitch that actually does curve. When the
ball is released, it is given a spin. As it travels through the air, one side
of the ball is going in the same direction as the air. So the ball's spin
adds to the air velocity. On the other side of the ball, the spin is in the
opposite direction of the air motion So the air is traveling slower on
this side of the ball. The pressure is lower on the side with the highest
air speed, so the ball is pushed to that side by air pressure; therefore
curving.
Tennis players learn the technique
of giving the ball a spin when they
serve. This causes the ball to curve in the same way a curve
ball curves -- makes it harder to return.
Houses have had their roofs ripped off when strong winds
passed over the roofs, creating very low pressures. Air pressure
then pushes the roof up and off.
Another example of Bernoulli's principle in action is the 'Bernoulli blower' at the Adventurarium.
The device has a big fan that blows a strong stream of air straight up. A ball is then thrown into the
air stream. It stays in the stream and is not blown out. How come? Well
the velocity in the center of the wind stream is greatest. As the ball moves
to the outside, one side is in the strong air stream in the center and the other
side is in air that is moving slower. The pressure in the center is less
because the air is moving faster, so the ball is pushed to the center by air
pressure and stays in the stream.
The Physics Kahuna will have shown you all sorts of marvelous Bernoulli
demonstrations. Make sure that you understand how these wonders
worked.
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AP Physics – Fluids Wrap Up
Here are the equations that you get to play around with:
p  p0   gh
This is the equation for the pressure of something as a function of depth in a fluid. You
would use it to figure out the pressure acting at a depth of 25.0 m in a lake for example.
FBuoy  Vg
This is the equation for the buoyant force. It is also an equation that calculates the
weight of an object as a function of its density, volume, and the acceleration of gravity.
A1v1  A2v2
This equation represents flow rate, which is the cross sectional area, A, multiplied by
the velocity of the fluid, v. This is set up for two locations in a flow system. The
flow rate for a fluid that is incompressible must stay constant, so this equation allows
you to calculate the linear speed of the fluid as a function of the cross sectional area
of the system.
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p   gy   v 2  const.
2
This is a Bernoulli’s equation. This allows you to calculate pressure, linear speed, &tc.
For a system at different places within the system.
Here is the stuff you need to be able to do.
A. Fluid Mechanics
1. Hydrostatic pressure
a) You should understand that a fluid exerts pressure in all directions.
This is basic. For example, atmospheric pressure goes in all directions about an object – under
it, over it, on the sides, &tc. Good old Pascal’s Principle.
b) You should understand that a fluid at rest exerts pressure perpendicular to any surface
that it contacts.
This is also an application of Pascal’s principle. The pressure is everywhere throughout the
liquid. The direction of the force acting on a surface is always perpendicular to the surface.
c) You should understand and be able to use the relationship between pressure and depth
in a liquid, p   g h .
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The actual equation that is provided you is p  p0   gh
where p0 would be
some initial pressure. We did a bunch of these problems. Guage pressure is based on the idea that the
atmospheric pressure is zero pressure. Absolute pressure uses a perfect vacuum – zero pascals – as its
zero pressure. So guage pressure differs from absolute pressure by one atmosphere. The pressure at a
certain depth would be give by p  p0   gh . For an absolute pressure you would set p0 equal to the
atmospheric pressure. For a gage pressure you would drop the
p0
term.
2. Buoyancy
a) You should understand that the difference in the pressure on the upper and lower
surfaces of an object immersed in a liquid results in an upward force on the object.
We went through this when the Physics Kahuna derived the buoyancy equation for you.
Because the pressure depends on depth, the pressure increases with the depth. So if the top of
a regular object is 10 m below the surface and the bottom of it is 15 m below – five meters
deeper, the force, which is pressure times area, must be greater. Thus there is a larger force
pushing up on the bottom of the body than the pressure pushing down on the top of the body.
The net force is upward and is given the name of ‘buoyant’ force.
b) You should understand and be able to apply Archimedes’ principle; the buoyant force on
a submersed object is equal to the weight of the liquid it displaces.
Well, the statement gives you Archimedes’ principle and tells you to understand it. So do that.
3. Fluid flow continuity
a) You should understand that for laminar flow, the flow rate of a liquid through its cross
section is the same at any point along its path.
So okay, do that too.
b) You understand and be able to apply the equation of continuity,
1 A1v1   2 A2v2 .
Actually the equation that you are given is: A1v1  A2v2 the density part isn’t in the equation.
This is because in the type of problem that you’ll be doing, the density won’t change and will remain
constant. Because of that, it cancels out of the equation. The Physics Kahuna is not at all sure why
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statement b) above had a different form of the equation. Probably some miscommunication at the
College Board. Anyway, we did a bunch of problems where you got to use the equation. It is
all pie.
4. Bernoulli’s equation
a) You should understand that the pressure of a flowing liquid is low where the velocity is
high, and vice versa.
Simple principle, simple stuff. Hey you can do it!
b) You should understand and be able to apply Bernoulli’s equation,
1
p   gy   v 2  const.
2
Well. That’s it for what you need to know.
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From 2002:
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In the laboratory, your are given a cylindrical beaker containing a fluid and you are asked to
determine the density  of the fluid. You are to use a spring of negligible mass and unknown
spring constant k attached to a stand. An irregularly shaped object of known mass m and
density D (D >>  ) hangs from the spring. You may also choose from among the following
items to complete the task.
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A metric ruler
A stopwatch
String
(a) Explain how you could experimentally determine the spring constant k.
The mass of the weight is known, suspend the mass from the spring in air, measure the
displacement of the spring and calculate k from the equation Fs   kx where Fs is the
mg, the weight of the thing.
(b) The spring-object system is now arranged so that the object (but not the spring) is immersed
in the unknown fluid, as shown above. Describe any changes that are observed in the
spring-object system and explain why they occur.
The mass will have less weight in the fluid because of the buoyant force. It will decrease
by the amount of the force which is Fbuoy  Vg
(c) Explain how you could experimentally determine the density of the fluid.
Knowing k we can calculate the apparent weight of the object in the fluid. The difference
between its weight in the air and in the fluid will equal the buoyant force.
The volume of the object could be calculated using the equation for density:

m
V
The volume of fluid displaced will be the same as the volume of the object.
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Knowing the buoyant force, we can use the buoyant force to calculate the density of the
fluid.
Fbuoy  Vg

FBuoy
Vg
(d) Show explicitly, using equations, how you will use your measurements to calculate the fluid
density  . Start by identifying any symbols you use in your equations.
Symbol
Physical quantity
Fs
This is the force that stretches the spring. In air, it will be the
weight of the object.
k
The spring constant
x
The spring displacement
Fg
m
The weight of the object
g
The acceleration of gravity
Fbuoy
The buoyant force

The density of the fluid
V
Volume of fluid displaced by the object
g
Acceleration of gravity
The mass of the object
Measure the displacement of the spring by the object in air. Calculate the weight of the object
using Fg  mg . Using this weight, calculate the spring constant from Fs   kx . Calculate
the volume of the object using the equation for density. This will be the same as the volume of
the fluid displaced.
Note the spring displacement. From this calculate the weight of the object in the fluid using the
buoyant force equation. The difference in the two forces is the buoyant force. Using the volume
displaced, the buoyant force, and the acceleration of gravity, calculate the density of the fluid.
(Using the buoyant force equation.)
So there it is. Your’re all set fluid mechanicswise.
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FLUID CONCEPTS
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Matter is made up of atoms. Every atom of an element is alike.
We can model atoms as tiny, hard “billiard balls.”
In a substance, atoms are combined to form molecules.
When atoms or molecules collide, they collide elastically (no loss of kinetic energy).
There are three forms of matter that we encounter in this class: solids, liquids, and gases. In all
cases, the atoms or molecules are in constant motion in a material.
In a solid, the average position of the atoms or molecules remains constant. That is why solids
tend to maintain their shape, in general.
In a liquid, the atoms or molecules are free to move around, but the distance between atoms or
molecules remains relatively small. For this reason, a liquid takes the shape of its container, but
does not expand to fill it.
The molecules of a liquid feel a mutual attraction. This creates surface tension, which acts
something like a “skin” on the surface of the liquid.
In a gas, the molecules are widely separated and feel little mutual attraction. This is why a gas
tends to expand to take the shape of its container.
We will treat liquids as incompressible. It is true that the compressibility for liquids in the
situations we consider is negligible.
Pressure is force per unit measure of area. The SI unit of pressure is the Pascal (Pa) or N/m2.
Since a fluid is not rigid, it can only exert a force perpendicular to a surface. Picture the
particles as little balls colliding with the surface, as in when you bounce a ball off a wall. Can
you see that the direction of the velocity change is always perpendicular to the wall? (Try
drawing it).
A fluid has weight. This is the reason why it exerts a pressure.
The pressure of a fluid is proportional to the density of the fluid, the gravitation constant (for
earth, average g=9.8 N/kg), and the depth of the fluid. Thus the pressure exerted by a fluid only
varies with the depth, not with the volume, the shape of the container, or any other factors.
The atmosphere exerts considerable pressure on objects at the surface of the earth.
Many simple devices depend on atmospheric pressure: drinking straws, suction cups, mercury
barometers. Be sure you can explain how these and others work using the concepts in this
section.
Gauges measure atmospheric pressure as 0 Pa. Atmospheric pressure must be added to gauge
pressure to give the actual pressure (called “absolute pressure”).
Since a liquid is incompressible, a pressure exerted on it is transmitted throughout the liquid
(Pascal’s Principle).
An object floats b/c it displaces a volume of water whose weight equals the object’s weight
(Archimedes’ Principle). The sum of the vertical forces is zero. Alternate version: an object
floats b/c its average density is less than the surrounding fluid. Second alternate version: an
object floats b/c the pressure at the bottom of the object is higher than the pressure at the top
(the resulting force must be equal to the weight of the object).
We will treat only fluids that flow in thin sheets, i.e. laminar flow. The thin sheets can be
modelled in two dimensions as streamlines.
Fluids flow from high pressure to low pressure.
Because a fluid is incompressible, the volume flow rate is constant, even if the diameter of the
pipe changes (Continuity Principle). The product of Area and velocity of a fluid is a constant
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for a particular pipe. This means the fluid speeds up in a constriction and slows down in a
wider part of the pipe.
Bernoulli made a conservation of energy statement for a fluid in laminar flow: the energy per
volume is constant in a pipe (Bernoulli’s Principle). This is true even if the diameter of the pipe
and therefore the velocity change.
Fast-moving fluids are at lower pressure than slow-moving fluids (Venturi Effect). This is a
consequence of Bernoulli’s Principle (i.e., conservation of energy).
Airplanes fly b/c of Bernoulli’s Principle and the momentum change of the air (due to the
“attack angle” of the wing). Can you explain this?
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