Chapter 4
fluids
References:
1

Physics in Biology and Medicine, 3
rd
E, Paul
Davidovits
.
2

College Physics, 6
th
E,
Serway
3

Web Sites
30 October 2013
1
Objectives
•
1

Understand the static behavior of fluids
•
2

Illustrate the properties of fluid pressure,
buoyant force in liquids, and surface tension
•
3

Understand the behaviors of fluids in motion
•
4

Study some examples from biology and
zoology.
30 October 2013
2
Introduction
•
The differences in the physical properties of solids, liquids, and gases
are explained in terms of the forces that bind the molecules.
•
1
–
Solids: the molecules are rigidly bound; a solid therefore has a
definite
shape and volume.
•
2

Liquids: The molecules constituting a liquid are not bound together with
sufficient force to maintain a definite shape, but the binding is sufficiently
strong to maintain a
definite volume
.
•
3

Gases: the molecules are not bound to each other. Therefore a gas has
neither a definite shape nor a definite volume
•
Fluids are liquids and gases
•
Fluids and solids are governed by the same laws of mechanics.
30 October 2013
3
1

Static Fluids
Force and Pressure in a Fluid
•
When a force is applied to
•
1

A solid:
this force is transmitted to the other
parts of the solid with its direction unchanged.
•
2

A fluid:
Because of a fluid’s ability to flow, it
transmits a force uniformly in all directions.
Therefore, the pressure at any point in a fluid at
rest is the same in all directions.
30 October 2013
4
Pressure
•
A fluid in a container exerts a force on all parts of the container in
contact with the fluid.
•
Fluid also exerts a force on any object immersed in it.
P=
F/A
•
F is always perpendicular to A.
•
The pressure in a fluid increases
with depth because of the weight
of the fluid above
.
P
2
−P
1
=
ρ
gh
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5
Pressure units
30 October 2013
6
1 torr
=1mm Hg
=13
.5mm water
=1
.33
×
10
3
dyn/cm
2
=1
.32
×
10
−3
atm
=1
.93
×
10
−2
psi
=1
.33
×
10
2
Pa (N/m
2
)
4

2

b PRESSURE MEASUREMENTS
•
The open

tube manometer
•
P = P
0

rgh.
•
P
is called the
absolute
pressure
,
•
P

P
0
is called the
gauge pressure
.
•
If P > P
0
h
is +ve
•
If P > P
0
h
is

ve
•
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7
Barometer
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8
Pascal’s Principle
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9
In an incompressible liquid, the increase in the pressure at any
point is transmitted undiminished to all other points in the
liquid. This is known as Pascal’s principle.
Archimedes’ Principle
•
Archimedes’ principle states that
•
a body
partially
or
wholly
submerged in a fluid
is buoyed upward by a force that is equal in
magnitude to the weight of the displaced fluid.
B=(P
a

P
b
)A=(
r
fluid
.g.h)A=
r
fluid
g.V=M
fluid
.g
4

2

d Archimedes’ Principle
30 October 2013
11
B=(P
a

P
b
)A=(
r
fluid
.g.h)A
=
r
fluid
g.V=M
fluid
.g
Archimedes’ principle states that a body
partially or wholly submerged in a fluid is
buoyed upward by a force that is equal in
magnitude to the weight of the displaced fluid.
Case 1
: Totally Submerged Object
B=
r
fluid
g.V
fluid
F
g
=
M
object
.g
=
V
object
r
object
.g
B

F
g
=
r
fluid
g.V
fluid

V
object
r
object
.g
,
V
fluid
=
V
object
=V
B

F
g
=
g.V
(
r
fluid
–
r
object
)
Case 2
Floating Object
30 October 2013
12
•
r
fluid
g.V
fluid
=V
object
r
object.
g
the fraction of the volume of a floating object that is below the fluid surface is
equal to the ratio of the density of the object to that of the fluid.
Problem 1
•
In a huge oil tanker, salt water has flooded an oil tank to a depth of 5.00 m. On
top of the water is a layer of oil 8.00 m deep, as in the cross

sectional view of the
tank in Figure. The oil has a density of 0.700 g/cm
3
. Find the pressure at the
bottom of the tank. (Take 1 025 kg/m
3
as the density of salt water.)
Solution
•
P
1
=P
0
+
r
oil
gh
1
= 1.01x10
5
Pa+(7.00x10
2
kg/m
3
)(9.80m/s
2
)(8.00m)
•
= 1.56x10
5
Pa
•
so P
bott
= P1+
r
water
gh
2
= 1.56x10
5
Pa+ (1.025x10
3
kg/m
3
)
)(9.80m/s
2
)(5.00m)
•
= 2.06x10
5
Pa
30 October 2013
13
Problem 2
•
Estimate the net force exerted on your eardrum

A
~
1 cm
2

due to the
water above when you are swimming at the bottom of a pool that is 5.0
m deep.
•
Solution
•
•
DP=P

P
0
=
r
杨
•
=(
1.00x10
3
kg/m
3
)(9.80m/s
2
)(5.00m)= 4.9x10
4
Pa
•
F
net
= A
D
P=(1x10

4
m
2
)( 4.9x10
4
Pa)
~
5N
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14
4

3 The human brain
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15
The human brain is immersed in a
fluid (the
cerebrospinal fluid
) of
density
1 007 kg/m
3
, which is
slightly less than the average
density of the brain,
1 040 kg/m
3
.
Most of the weight of the brain is supported
by the buoyant force of the surrounding fluid.
4

4 Buoyancy of Fish
•
The bodies of some fish contain:
•
either
porous bones
•
or
air

filled swim bladders
•
that decrease their average density and
allow them to float in water without an
expenditure of energy.
30 October 2013
16
4

4 Buoyancy of Fish
•
EX.
Cuttlefish
•
contains a porous bone that has a density of 0
.
62 g/cm
3
•
its body has a density of 1
.
067 g/cm
3
.
•
the percentage of the body volume occupied by the porous bone that
makes the average density of the fish be the same as the density of sea
water (1
.
026 g/cm
3
) by using the following equation
30 October 2013
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The cuttlefish lives in the sea at a depth of about 150 m.
At this depth, the pressure is 15 atm .

The spaces in the porous bone are filled with gas at a pressure of about
1 atm.

Therefore, the porous bone must be able to withstand a pressure of
14 atm.
The cuttlefish alters its density by injecting or
withdrawing fluid from its porous bone.
In fish that possess swim bladders
•
The decrease in density is provided by the gas
in the bladder.
•
To achieve the density reduction calculated in
the preceding example, the volume of the
bladder is only about 4% of the total volume
of the fish
•
Fish with swim bladders alter their density by
changing the amount of gas in the bladder.
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18
4

5 Surface Tension
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19
The surface of a liquid contract and behave somewhat like a
stretched membrane.
This contracting tendency results in a surface
tension that resists an increase in the free surface
That surface tension is force acting tangential to
the surface, normal to a line of unit length on the
surface
F
T
=
TL
capillary action
.
30 October 2013
20
The surface molecules near the wall
are attracted to the wall. This attractive
force is called
adhesion
.
These molecules are also subject to
the
attractive cohesive
force exerted
by the liquid
So
If the adhesive force is greater than the
cohesive force, the liquid wets the container
wall, and the liquid surface near the wall is
curved upward.
If the opposite is the case, the liquid surface is curved downward
F
m
= 2
πRT
capillary action
.
30 October 2013
21
W=
πR
2
h
r
g
F
m
= 2
πRT
F
m
= 2
πRT cos
q
in Y

direction
2
πRT
cos
θ
=
πR
2
hρg
h
=2
T
cos
θ / Rρg
r
gh= 2T/R if
q
=0
P
out
P
in
Air
Problem
3
30 October 2013
22
Find the height to which water would rise in a capillary tube with a radius
equal to 5.0 x 10

5
m. Assume that the contact angle between the water and the
material of the tube is small enough to be considered zero.
Solution
h=2Tcos
0
0
/
r
gR
=2(0.073 N/m) / (1.00x10
3
kg/m
3
)(9.80 m/s
2
)(5.0x10

5
m)
=0.30 m
Assignment
•
Solve the following problems
•
1,3,5,6
•
Surfactants
are molecules that lower surface
tension of liquids. (The word is an abbreviation of
surface active agent
.)
•
Write a short account on the effect of surfactants
that lowers the surface tension
30 October 2013
23
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