m = rV
(33-2)
where r is the density, the mass-per-volume, of the water making up the slug and V is the
volume of the slug. The volume of a cylinder is its height times its face area so we can write
m = rh A
Substituting this expression for the mass of the slug into equation 33-1 yields
PA − rh Ag − P A = 0
o
P − rhg − P = 0
o
P = P +
(33-3)
o
rgh
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Chapter 33 Fluids: Pressure, Density, Archimedes’ Principle
While we have been writing specifically about water, the only thing in the analysis that depends
on the identity of the incompressible fluid is the density r. Hence, as long as we use the density
of the fluid in question, equation 33-3 ( P = P +
) applies to any incompressible fluid. It
o
rgh
says that the pressure at any depth h is the pressure at the surface plus rgh.
A few words on the units of pressure are in order. We have stated that the units of pressure are
N/m2. This combination of units is given a name. It is called the pascal, abbreviated Pa.
N
1 Pa = 1
2
m
Pressures are often quoted in terms of the non-SI unit of pressure, the atmosphere, abbreviated
atm and defined such that, on the average, the pressure of the earth’s atmosphere at sea level is
1 atm. In terms of the pascal,
1 atm =
5
1.013×10 Pa
The big mistake that folks make in applying equation 33-3 ( P = P +
) is to ignore the units.
o
rgh
They’ll use 1 atm for Po and without converting that to pascals, they’ll add the product rgh to it.
Of course, if one uses SI units for r, g, and h, the product rgh comes out in N/m2 which is a
pascal which is definitely not an atmosphere (but rather, about a hundred-thousandth of an
atmosphere). Of course one can’t add a value in pascals to a value in atmospheres. The way to
go is to convert the value of Po that was given to you in units of atmospheres, to pascals, and
then add the product rgh (in SI units) to your result so that your final answer comes out in
pascals.
Gauge Pressure
Remember the gauge we constructed for our thought experiment? That part about evacuating the
inside of the pipe presents quite the manufacturing challenge. The gauge would become
inaccurate as air leaked in by the disk. As regards function, the description is fairly realistic in
terms of actual pressure gauges in use, except for the pumping of the air out the pipe. To make it
more like an actual gauge that one might purchase, we would have to leave the interior open to
the atmosphere. In use then, the gauge reads zero when the pressure on the sensor end is
1 atmosphere, and in general, indicates the amount by which the pressure being measured
exceeds atmospheric pressure. This quantity, the amount by which a pressure exceeds
atmospheric pressure, is called gauge pressure (since it is the value registered by a typical
pressure gauge.) When it needs to be contrasted with gauge pressure, the actual pressure that we
2
43
Chapter 33 Fluids: Pressure, Density, Archimedes’ Principle
have been discussing up to this point is called absolute pressure. The absolute pressure and the
gauge pressure are related by:
P = P + P
(33-4)
G
O
where:
P is the absolute pressure,
P is the gauge pressure, and
G
P is atmospheric pressure.
O
When you hear a value of pressure (other than the so-called barometric pressure of the earth’s
atmosphere) in your everyday life, it is typically a gauge pressure (even though one does not use
the adjective “gauge” in discussing it.) For instance, if you hear that the recommended tire
pressure for your tires is 32 psi (pounds per square inch) what is being quoted is a gauge
pressure. Folks that work on ventilation systems often speak of negative air pressure. Again,
they are actually talking about gauge pressure, and a negative value of gauge pressure in a
ventilation line just means that the absolute pressure is less than atmospheric pressure.
Archimedes’ Principle
The net pressure-times-area force on an object submerged in a fluid, the vector sum of the forces
on all the infinite number of infinitesimal surface area elements making up the surface of an
object, is upward because of the fact that pressure increases with depth. The upward pressure-
times-area force on the bottom of an object is greater than the downward pressure-times-area
force on the top of the object. The result is a net upward force on any object that is either partly
or totally submerged in a fluid. The force is called the buoyant force on the object. The agent of
the buoyant force is the fluid.
If you take an object in your hand, submerge the object in still water, and release the object from
rest, one of three things will happen: The object will experience an upward acceleration and bob
to the surface, the object will remain at rest, or the object will experience a downward
acceleration and sink. We have emphasized that the buoyant force is always upward. So why on
earth would the object ever sink? The reason is, of course, that after you release the object, the
buoyant force is not the only force acting on the object. The gravitational force still acts on the
object when the object is submerged. Recall that the earth’s gravitational field permeates
everything. For an object that is touching nothing of substance but the fluid it is in, the free body
diagram (without the acceleration vector being included) is always the same (except for the
relative lengths of the arrows):
B
Fg
and the whole question as to whether the object (released from rest in the fluid) sinks, stays put,
or bobs to the surface, is determined by how the magnitude of the buoyant force compares with
2
44
Chapter 33 Fluids: Pressure, Density, Archimedes’ Principle
that of the gravitational force. If the buoyant force is greater, the net force is upward and the
object bobs toward the surface. If the buoyant force and the gravitational force are equal in
magnitude, the object stays put. And if the gravitational force is greater, the object sinks.
So how does one determine how big the buoyant force on an object is? First, the trivial case: If
the only forces on the object are the buoyant force and the gravitational force, and the object
remains at rest, then the buoyant force must be equal in magnitude to the gravitational force.
This is the case for an object such as a boat or a log which is floating on the surface of the fluid it
is in.
But suppose the object is not freely floating at rest. Consider an object that is submerged in a
fluid. We have no information on the acceleration of the object, but we cannot assume it to be
zero. Assume that a person has, while maintaining a firm grasp on the object, submerged the
object in fluid, and then, released it from rest. We don’t know which way it is going from there,
but we can not assume that it is going to stay put.
To derive our expression for the buoyant force, we do a little thought experiment. Imagine
replacing the object with a slug of fluid (the same kind of fluid as that in which the object is
submerged), where the slug of fluid has the exact same size and shape as the object.
From our experience with still water we know that the slug of fluid would indeed stay put,
meaning that it is in equilibrium.
Table of Forces
Symbol=?
Name
Agent
Victim
B
Buoyant
The Surrounding
The Slug
B
Force
Fluid
of Fluid
a = 0
Gravitational
F
= m
The Earth’s
The Slug
SF
g
SFg
Force on the
Gravitational Field
of Fluid
F
SF
Slug of Fluid
g
2
45
Chapter 33 Fluids: Pressure, Density, Archimedes’ Principle
Applying the equilibrium equation ∑ F = 0 to the slug of
↑
fluid yields:
B
∑F = 0
↑
a = 0
B − F
= 0
gSF
F
SF
g
B = F
SF
g
The last equation states that the buoyant force on the slug of fluid is equal to the gravitational
force on the slug of fluid. Now get this; this is the crux of the derivation: Because the slug of
fluid has the exact same size and shape as the original object, it presents the exact same surface
to the surrounding fluid, and hence, the surrounding fluid exerts the same buoyant force on the
slug of fluid as it does on the original object. Since the buoyant force on the slug of fluid is
equal in magnitude to the gravitational force acting on the slug of fluid, the buoyant force on the
original object is equal in magnitude to the gravitational force acting on the slug of fluid. This is
Archimedes’ principle.
B = the buoyant force, which is equal in magnitude to the
gravitational force that would be acting on that amount
of fluid that would fit in the space occupied by the
submerged part of the object.
F = m
(The gravitational force)
o
g
og
Archimedes’ Principle states that: The buoyant force on an object that is either partly or totally
submerged in a fluid is upward, and is equal in magnitude to the gravitational force that would be
acting on that amount of fluid that would be where the object is if the object wasn’t there. For an
object that is totally submerged, the volume of that amount of fluid that would be where the
object is if the object wasn’t there is equal to the volume of the object itself. But for an object
that is only partly submerged, the volume of that amount of fluid that would be where the object
is if the object wasn’t there is equal to the (typically unknown) volume of the submerged part of
the object. However, if the object is freely floating at rest, the equilibrium equation (instead of
Archimedes’ Principle) can be used to quickly establish that the buoyant force (of a freely
floating object such as a boat) is equal in magnitude to the gravitational force acting on the
object itself.
2
46
Chapter 34 Pascal’s Principle, the Continuity Equation, and Bernoulli’s Principle
34 Pascal’s Principle, the Continuity Equation, and
Bernoulli’s Principle
There are a couple of mistakes that tend to crop up with some regularity in the application of
the Bernoulli equation
1
2
P + rv + rgh = constant . Firstly, folks tend to forget to create a
2
diagram in order to identify point 1 and point 2 in the diagram so that they can write the
Bernoulli equation in its useful form:
2
2
1
1
P + rv + rgh = P + rv + rgh . Secondly,
1
2
1
1
2
2
2
2
when both the velocities in Bernoulli’s equation are unknown, they forget that there is another
equation that relates the velocities, namely, the continuity equation in the form A v = A v
1 1
2
2
which states that the flow rate at position 1 is equal to the flow rate at position 2.
Pascal’s Principle
Experimentally, we find that if you increase the pressure by some given amount at one location
in a fluid, the pressure increases by that same amount everywhere in the fluid. This experimental
result is known as Pascal’s Principle.
We take advantage of Pascal’s principle every time we step on the brakes of our cars and trucks.
The brake system is a hydraulic system. The fluid is oil that is called hydraulic fluid. When you
depress the brake pedal you increase the pressure everywhere in the fluid in the hydraulic line.
At the wheels, the increased pressure acting on pistons attached to the brake pads pushes them
against disks or drums connected to the wheels.
Example 34-1
A simple hydraulic lift consists of two pistons, one larger than the other, in
cylinders connected by a pipe. The cylinders and pipe are filled with water.
In use, a person pushes down upon the smaller piston and the water pushes
upward on the larger piston. The diameter of the smaller piston is
2.20 centimeters. The diameter of the larger piston is 21.0 centimeters. On
top of the larger piston is a metal support and on top of that is a car. The
combined mass of the support-plus-car is 998 kg. Find the force that the
person must exert on the smaller piston to raise the car at a constant velocity.
Neglect the masses of the pistons.
2
47
Chapter 34 Pascal’s Principle, the Continuity Equation, and Bernoulli’s Principle
Solution
We start our solution with a sketch.
D = 2.20 cm
S
D = 21.0 cm
L
Now, let’s find the force R exerted on the larger piston by the car support. By Newton’s 3rd
N
Law, it is the same as the normal force F exerted by the larger piston on the car support. We’ll
N
draw and analyze the free body diagram of the car-plus-support to get that.
F
N
Table of Forces
Symbol Name
Agent
Victim
Gravitational
The Earth’s
The
F =mg Force on
Gravitational
Support-
F = mg
g
g
Support-Plus-Car Field
Plus-Car
∑
The
F = 0
The Large
↑
¢
↑
F
Support-
N
Normal Force
Piston
Plus-Car
F −
g
m
= 0
N
F =
N
g
m
newtons
F = 998 kg
.
N
9 80
kg
F =
newtons
9780
N
2
48
Chapter 34 Pascal’s Principle, the Continuity Equation, and Bernoulli’s Principle
Now we analyze the equilibrium of the larger piston to determine what the pressure in the fluid
must be in order for the fluid to exert enough force on the piston (with the car-plus-support on it)
to keep it moving at constant velocity.
R = F = 9780
N
N
Table of Forces
a = 0
Symbol
Name
Agent
Victim
Interaction
R = F
Support (That part,
N
N
Partner to
Large
= 9780
of the hydraulic lift,
Normal Force
Piston
that the car is on.)
F = PA
newtons
(see above)
PL
L
Pressure-Related
F
Large
Force on Large
The Water
∑
PL
Piston
F = 0
Piston
↑
¢
↑
F − R = 0
PL
N
PA − R = 0
L
N
RN
P =
(34-1)
AL
A is the area of the face of the larger piston. We can use the given larger piston
L
diameter D = 0.210 m to determine the area of the face of the larger piston as
L
follows:
D
2
A = π r where
L
r =
is the radius of the larger piston.
L
L
L
2
2
L
=
D
A
π
L
2
2
0.210 m
A = π
L
2
2
A = 0.03464 m
L
Substituting this and the value R = F = 9780 newtons into equation 34-1 above yields
N
N
newtons
9780
P =
2
0.03464 m
N
P = 282333
(We intentionally keep 3 too many significant figures in this intermediate result.)
2
m
2
49
Chapter 34 Pascal’s Principle, the Continuity Equation, and Bernoulli’s Principle
Now we just have to analyze the equilibrium of the smaller piston to determine the force that the
person must exert on the smaller piston.
F
PERSON
a = 0
F = PA
PS
S
∑F = 0
↑
¢
↑
F − F
= 0
PS
PERSON
PA − F
= 0
S
PERSON
F
= PA
PERSON
S
The area A of the face of the smaller piston is just pi times the square of the
S
D
radius of the smaller piston where the radius is
S , half the diameter of the
2
smaller piston. So:
2
S
F
=
D
Pπ
PERSON
2
2
N
0.0220 m
F
= 282333
π
PERSON
2
m
2
F
= 107 N
PERSON
2
50
Chapter 34 Pascal’s Principle, the Continuity Equation, and Bernoulli’s Principle
Fluid in Motion—the Continuity Principle
The Continuity Principle is a fancy name for something that common sense will tell you has to
be the case. It is simply a statement of the fact that for any section of a single pipe, filled with an
incompressible fluid (an idealization approached by liquids), through which the fluid with which
the pipe is filled is flowing, the amount of fluid that goes in one end in any specified amount of
time is equal to the amount that comes out the other end in the same amount of time. If we