1. fluids Chapter 4 References:
1- Physics in Biology and Medicine, 3rd E, Paul Davidovits.
2- College Physics, 6th E, Serway
3- Web Sites
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2. 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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3. Introduction Fluids are liquids and gases
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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4. 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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5. 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.
P2 −P1 = ρgh
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6. 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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7. 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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8. Barometer
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9. 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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10. 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

11. 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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12. 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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13. 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 g/cm3. Find the pressure at the bottom of the tank. (Take 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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14. 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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15. 4-3 The human brain
The human brain is immersed in a fluid (the cerebrospinal fluid) of density kg/m3, which is slightly less than the average density of the brain, kg/m3.
Most of the weight of the brain is supported by the buoyant force of the surrounding fluid.
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16. 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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17. 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 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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18. 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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19. 4-5 Surface Tension FT = TL
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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20. 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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21. capillary action. h =2T cos θ / Rρg rgh= 2T/R if q=0 Air
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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22. 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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23. 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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