Chapter 4
fluids
References:
1- Physics in Biology and Medicine, 3rd E, Paul Davidovits.
2- College Physics, 6th E, Serway
3- Web Sites
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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.
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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.
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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.
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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.
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Pressure units
1 torr
=1mm Hg
=13.5mm water
=1.33×103 dyn/cm2
=1.32×10−3 atm
=1.93×10−2 psi
=1.33×102 Pa (N/m2)
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4-2-b PRESSURE MEASUREMENTS
• The open-tube manometer
• P = P0 - rgh.
• P is called the absolute pressure,
• P - P0 is called the gauge pressure.
• If P > P0 h is +ve
• If P > P0 h is -ve
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Barometer
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Pascal’s Principle
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.
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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=(Pa-Pb)A=(rfluid.g.h)A=rfluid g.V=Mfluid.g
4-2-d 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=(Pa-Pb)A=(rfluid.g.h)A
=rfluid g.V=Mfluid.g
Case 1: Totally Submerged Object
B= rfluid g.Vfluid
Fg=Mobject.g= Vobject robject.g
B-Fg= rfluid g.Vfluid- Vobject robject.g
, Vfluid= Vobject=V
B-Fg =g.V (rfluid – robject)
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Case 2 Floating Object
• rfluid g.Vfluid=Vobject robject.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.
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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/cm3. Find the pressure at the
bottom of the tank. (Take 1 025 kg/m3 as the density of salt water.)
Solution
• P1=P0 +roil gh1= 1.01x105 Pa+(7.00x102
kg/m3)(9.80m/s2)(8.00m)
•
= 1.56x105 Pa
• so P bott= P1+rwater gh2= 1.56x105 Pa+ (1.025x103 kg/m3)
)(9.80m/s2)(5.00m)
•
= 2.06x105 Pa
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Problem 2
• Estimate the net force exerted on your eardrum- A~1 cm2 - due to the
water above when you are swimming at the bottom of a pool that is 5.0
m deep.
• Solution
•
• DP=P-P0=rgh
•
=(1.00x103 kg/m3)(9.80m/s2)(5.00m)= 4.9x104 Pa
• Fnet = A DP=(1x10-4 m2)( 4.9x104 Pa)~ 5N
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4-3 The human brain
The human brain is immersed in a
fluid (the cerebrospinal fluid) of
density 1 007 kg/m3, which is
slightly less than the average
density of the brain, 1 040 kg/m3.
Most of the weight of the brain is supported
by the buoyant force of the surrounding fluid.
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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.
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4-4 Buoyancy of Fish
•
•
•
•
EX. Cuttlefish
contains a porous bone that has a density of 0.62 g/cm3
its body has a density of 1.067 g/cm3.
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/cm3) by using the following equation
-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.
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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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4-5 Surface Tension
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
FT = TL
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capillary action.
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
Fm = 2πRT
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capillary action.
W= πR2hrg
Fm = 2πRT
Fm = 2πRT cos q in Y-direction
Air
2πRT cos θ = πR2 hρg
h =2T cos θ / Rρg
rgh= 2T/R if q=0
Pin
Pout
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Problem 3
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 00 / rgR
=2(0.073 N/m) / (1.00x103 kg/m3)(9.80 m/s2)(5.0x10-5 m)
=0.30 m
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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
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