introductionmedicaphystudents38.pptx.pdf

A medical physics book for students of BSc. And MSc.

1. INTRODUCTION
TO MEDICAL
PHYSICS
1437-1438
UMM AL QURA UNIVERSITY
PREPARATORY YEAR

2. INTRODUCTION TO MEDICAL PHYSICS PREPARATORY YEAR 2015-2016 2
SYLLABUS
CHAPTER TITLE WEEK Modifications
1 Motion on a straightline 2+3
2 Motion in two dimensions 4
3 Newton’s law of motion 5
4 Statics 6
6 Work and energy 7
13 Mechanics of nonviscousfluids 8+10
14 Viscous fluids 11
24 Mirrors, lenses and optical systems 12+13
30 Nuclear Physics 14
31 Ionizing Radiation 15
Original textbook
"Physics" edited by Joseph W. KANE and Morton M. STERNHEIM.
Third Edition. JOHN WILEY & SONS, Inc. ISBN: 0-471-63845-5

3. INTRODUCTION TO MEDICAL PHYSICS PREPARATORY YEAR 2015-2016 3
What is medical physics?
Medicalphysicsistheapplicationofphysicstomedicine.
Allareasofphysicscanbeappliedtomedicine
(Mechanics,electromagnetism,thermodynamics,
nuclearphysics,optics,fluids,…..)
Medicalphysicsismainlyinvolvedinthedevelopmentof
newinstrumentationandtechnologyusedfordiagnosis
andalsofortreatments.
Thehumanbodyisaverycomplexsystem.Conceptsof
modelinginphysicscanbyappliedtosimulatedifferent
activitiesofthehumanbodysystems:
Forexamplethemodelingofthebloodflowinthestudy
ofthebody’scirculatorysystem.

4. The human body is made up of different systems working together to keep the body
in health. We can use analogies with physics to simulate the function of these
systems and to understandthe connections between them.
INTRODUCTION TO MEDICAL PHYSICS PREPARATORY YEAR 2015-2016 4

5. Application to medicaldiagnosis
Differenttechniquesofdiagnosisandmedicalinstrumentsarebasedonphysicalprinciples
suchas,themeasurementofthebodytemperature,themeasurementoftheblood
pressure,theeyepressure,theheartpulse,…
Medicalimaging(X-raysradiology,MagneticResonanceImagingMRI,ultra-soundscan,…)is
averyusefuldesciplineofmedicaldiagnosis.
The figures shows an old and a new instrument for breathing diagnosis
INTRODUCTION TO MEDICAL PHYSICS PREPARATORY YEAR 2015-2016 5

6. Medical instruments
INTRODUCTION TO MEDICAL PHYSICS PREPARATORY YEAR 2015-2016 6
Blood pressure measurement
Sphygmomanometer
Digital wrist tensiometer
Body temperature measurement
Medical mercury thermometer Infrared era thermometer

7. Medical instruments
INTRODUCTION TO MEDICAL PHYSICS PREPARATORY YEAR 2015-2016 7
Otoscope
An otoscope is a medical device
typically having a light and a set of
lenses, used for the visual
examination of the eardrum and
the canal of the outer ear.

8. Medical Imaging
Magnetic Resonance Imaging MRI
MRI uses the property of the nuclear magnetic resonance NMR to image the nuclei of atoms
inside the body, specially the hydrogen atom H since the body tissues contain lots of water. MRI is
used for pathologic diagnosis such as lesions in the brain
MRI device MRI Image
INTRODUCTION TO MEDICAL PHYSICS PREPARATORY YEAR 2015-2016 8

9. X-Ray Imaging
An X-ray is a painless medical test that helps
physicians diagnose. Radiography involves exposing
a part of the body to a small dose of ionizing
radiation to produce images inside of the body.X-
rays are the oldest and most frequently used form
of medical imaging.
INTRODUCTION TO MEDICAL PHYSICS PREPARATORY YEAR 2015-2016 9
Fluoroscopy
Oneofthemostimportantbenefitsofthis
procedureisthatitallowsthedoctortoviewthe
body’sinnersystemswhiletheyareactually
functioning.Forexample,adoctorcanwatcha
patient’sstomachasitdigestsfood,allowingthe
doctortoobtainvaluablediagnosticinformation.

10. Medical surgery
Lasersurgery(photocoagulation)
Retinaldetachmentoccurswhenpartofitisliftedfromitsnormalpositioninthebackoftheeyeball.
Duringphotocoagulationyoursurgeondirectsalaserbeamthroughacontactlensor
ophthalmoscopedesignedforthisprocedure.Thelasermakesburnsaroundtheretinaltear,and
thescarringthatresultsusually"welds"theretinatotheunderlyingtissue.
INTRODUCTION TO MEDICAL PHYSICS PREPARATORY YEAR 2015-2016 10

11. Endoscopy
Endoscopicsurgeryusesscopesgoingthroughsmallincisionsornaturalbodyopeningsinorderto
diagnoseandtreatdisease.Anotherpopulartermisminimallyinvasivesurgery(MIS),which
emphasizesthatdiagnosisandtreatmentscanbedonewithreducedbodycavityinvasion.
Anendoscopeisalong,thinandflexibletube
whichhasalightandavideocamera.
DiagnosisofstomachbyUpper-endoscopy
INTRODUCTION TO MEDICAL PHYSICS PREPARATORY YEAR 2015-2016 11

12. Prosthesis
Prosthesis is the replacement of a missing or a defect parts of the body by an other made
artificially and assuming the same function as the missing part.
Cartilage prosthesis of the knee
Artificial Leg
Hearing Aid
INTRODUCTION TO MEDICAL PHYSICS PREPARATORY YEAR 2015-2016 12

13. Artificial devices to replace vital organs continues to be a human dream for thousands of people
around the world who are waiting for a heart or a kidney grief.
INTRODUCTION TO MEDICAL PHYSICS PREPARATORY YEAR 2015-2016 13
Human heart
Total artificial heart

15. CHAPTER 1: MOTION ON A STRAIGHT LINE PREPARATORY YEAR 2015-2016 15
Chapter 1
MOTION ON A STRAIGHT LINE
COURSE TOPICS:
1.1 Measurements, Standards and Units
1.2 Displacements; Average Velocity
1.3 Instantaneous Velocity
1.4 Acceleration
1.5 Finding the Motion of an Object
1.6 The Acceleration of Gravity and Falling
Objects
Examples to be explained and solved:
1.2; 1.4; 1.14; 1.16 and 1.20
Homework Problems: 1.3-1.16 and 1.50

16. Introduction
The objective of this chapter is to learn how to describe the motion of an object on astraight line.
The description of motion is treated from a kinematics point of view, then the causes of motion
(Forces) are note considered and the object in motion is treated as a point particle.
In this case the motion can be described with the following three quantities:
The position as function of time �(�)
Thevelocity��
The acceleration ��
We define, also the basic quantities ( length, mass and time) playing a role in mechanics (science
of motion) and their standards.
1.1 Measurements standards and units
Physical quantities are classified into fundamental quantities such as mass,length,timeand
derived quantities such as velocity, acceleration, force, energy….
CHAPTER 1: MOTION ON A STRAIGHT LINE PREPARATORY YEAR 2015-2016 16

17. Internationalsystem British system C.G.S System
Physicalquantity unit symbol unit symbol unit symbol
Length Meter �
Foot ��centimeter ��
1��=0.3048�1��=10
−2
�
Mass kilogram ��
Pound ��gram �
1��=0.4536��1�=0.001��
Time second �Second �second s
TheinternationalsystemisalsoknownasthemetricsystemortheM.K.S.Asystem.Inthe
medicalareasomeunitsaremoreusedthanthoseoftheS.Iunits,suchastheuseofcalorie
asunitforenergythantheJoule(1���=4.2�),themillimeterofmercuryforthepressure
thanthePascal(1����=133.32��)ortheliterforthevolumethanthemetercube.
CHAPTER 1: MOTION ON A STRAIGHT LINE PREPARATORY YEAR 2015-2016 17

18. The scientificnotation
A number is said to be in scientific notation when it is written as a number between 1and 10,
times a power of 10. for example 521can be written as 5.21×10
2
, or a small number like
0.000000521 can be written as 5.21×10
−7
. The advantage of this notation is its compactness, it
also facilitates numerical calculations.
When a number is written with the powers of 10, we can use the following prefixes
Multiples Prefix symbol Sub-multiples
Prefix symbol
10 deca da0.1 10
−1
deci d
100 10
2
hecto h 0.01 10
−2
centi c
1000 10
3
kilo k 0.001 10
−3
milli m
1000 000 10
6
Mega M 0.000001 10
−6
micro �
1000 000 000 10
9
Giga G 0.000 000 001 10
−9
nano n
1000 000 000 000 10
12
Tera T 0.000000 000 001 10
−12
pico p
CHAPTER 1: MOTION ON A STRAIGHT LINE PREPARATORY YEAR 2015-2016 18

19. Conversion of units
Toconvertquantitiesfromaunitsystemtoanother,wecanusethefollowingsystematic
method:
Supposewewanttoconvertalength�=1.75�intofoot.Theconversionfactorbetween
thetwounitsisgivenby:1��=0.3048�
Toconvertfrommetertofootwehavetofollowthesesteps:
1-Multiplythequantitytoconvertbyone:
�=1.75�×1
2-Rearrangetheconversionfactorinquotientequalto�thatallowstheeliminationof
theunitfromwhichwewanttoconvert:
1��=0.3048�⇒
1��
0.3048�
=1
3-Replacethisforminthefirststep:
�=1.75�×1=1.75�×
1��
0.3048�
=5.74��
Example1.1:Convert100��intometers.
100��=100��×1=100��×(
0.3048�
1��
)=30.48�
CHAPTER 1: MOTION ON A STRAIGHT LINE PREPARATORY YEAR 2015-2016 19

20. Example 1.2:
Convert the velocity of 24Τ��into Τ��ℎ.
We have 1��=10
3
�and 1ℎ=3600�
Then 24��=24
�
�
×1×1{ 1is written twice because we have two units to convert}
24��=24
�
�
×
1��
10
3
�
×
3600�
1ℎ
=
24×3600
10
3
��
ℎ
=86.4Τ��ℎ
Example 1.3: The skin is the largest organ in the human body; for a human adult the average area of
the skin surface is about 1.8�
2
,how much squared foot is this area?
Solution:������ℎ��1��=0.3048�,�ℎ��1��
2
=(0.3048)
2
�
2
�=1.8�
2
=1.8�
2
×
1��
2
0.3048
2
�
2
=19��
2
Example 1.4: an ampoule contains a solution of drug of 300��/5��, convert this dose into �/�.
Solution:
300��
5��
=
300×10
−6
�
5×10
−3
�
=0.06�/�
CHAPTER 1: MOTION ON A STRAIGHT LINE PREPARATORY YEAR 2015-2016 20

21. 1.2 Displacement-Average velocity
Example:
In the figure above if the object moves from �
1���
2then its displacement is :
∆�=�
2−�
1=7�−4�=3�.
The displacement from �
2���
3is: ∆�=�
3−�
2=−5�−7�=−12�.
The displacement can be positive or negative; it is negative if the motion is in the negative
direction
To describe the motion of an object we should first set up a coordinate system to locate the position.
A coordinate system is made up of an origin, a positive direction and a unit of length.
Displacement: is the change in the position:
∆�=�
�??????���−�
??????�??????�??????��
CHAPTER 1: MOTION ON A STRAIGHT LINE PREPARATORY YEAR 2015-2016 21

22. Average velocity:the average velocity isthe displacement over an elapsed time ∆�:
Example: a car is at �
1=600�when �
1=5�and at �
2=500�when �
2=15�itsaverage
velocity is: ҧ�=
∆�
∆�
=
�2−�1
�
2−�
1
=
500−600�
(15−5)�
=−10�/�
1.3 Instantaneous velocity
The average velocity doesn’t give a description about the rate of change of the position at each
instant. We need often to know the velocity of an object at each second, referred to as
instantaneous velocity:
Instantaneous velocity: The instantaneous velocity is determined by computing the average
velocity for an extremely short time interval:
�=lim
∆�→�
∆�
∆�
=
��
��
CHAPTER 1: MOTION ON A STRAIGHT LINE PREPARATORY YEAR 2015-2016 22
ഥ�=
∆�
∆�
=
�
�−�
�
�
�−�
�
Mathematically this expression is the first derivative of �with respect to the time

23. Example:
the motion of an object is given by the equation : ��=2+3�−2�
2
.Where
����ℎ����������������and ����ℎ����������������.
(a) Find the velocity of the object at �=5�.
(b) What is its average velocity between the two instants �
1=3�and �
2=5�?
Answer:
(a) �=
��
��
=3−4�;then at the instant �=5��=3−4×5=−17�/�
(b) At �
1=3�:�
1=2+3×3−2×3
2
=−7�
At �
2=5�∶ �
2=2+3×5−2×5
2
=−33�
Then the average velocity is:
ҧ�=
(−33+7)�
5−3�
=−13�/�
CHAPTER 1: MOTION ON A STRAIGHT LINE PREPARATORY YEAR 2015-2016 23

24. 1.4 Acceleration
Like position, velocity can change with time. The rate at which velocity changes is the acceleration
. Again we can discuss the average and the instantaneous acceleration.
The Average accelerationis the change of velocity over an interval of time ∆�:
The Instantaneous acceleration is the rate change of velocity over an extremely short time
interval.
Example:the motion of an object is given by the equation : ��=27�−4�
2
.��������and
���������. (a) Find the acceleration of the object at �=5�.
Answer:
(a)�=
��
��
=
�
��
��
��
=
�
��
27−8�=−8m/�
2
;the motion is with constant acceleration.
A negative acceleration means that velocity is decreasing with time.
ഥ�=
∆�
∆�
�=���
∆�→�
∆�
∆�
=
��
��
CHAPTER 1: MOTION ON A STRAIGHT LINE PREPARATORY YEAR 2015-2016 24

25. 1.5 Finding the motion of an object
•Iftheinitialpositionandvelocityareknown,theirlatervaluescanthenbefoundfromthe
acceleration.
•Whentheaccelerationisconstant,wecanfindtheequationsofmotion.Inthiscasetheaverage
andtheinstantaneousaccelerationsareequalandthefollowingequationsareobtained:
Average velocity ҧ�=
1
2
�+�
�(1)
Relating the final velocity to the initial
velocity
Velocity equation ∆�=�∆�(2)
Relating the final velocity to the initial
velocity and the acceleration
Equation of motion
∆�=�
�∆�+
1
2
�∆�
2
(3)
Relating the final position to the initial
position, the initial velocity and the
acceleration
�
2
−�
�
2
=2�∆�(4)
Relating the final velocity to the initial
velocity, the acceleration and the position
change
Note that if the initial position �
�and velocity �
�are taken at the initial time �
�=0, then ∆�=�
−�
�will be simply �, equations (2) , (3) and (4) are written as follows: �=�
�+��(2’),
�=�
�+�
��+
1
2
��
2
(3’) and �
2
−�
�
2
=2��−�
�(4’)
CHAPTER 1: MOTION ON A STRAIGHT LINE PREPARATORY YEAR 2015-2016 25

26. Example 1.16 P 14:
A car, initially at rest at a traffic light, accelerates at 2�/�
2
when the light turns green. After 4
seconds what are its velocity and position?
Solution:
Since we know the acceleration �, the elapsed time �, and the initial velocity �
0=0, we can use
Equations (2) and (3) to find the velocity and the displacement.
Thus,�=�
0+�×�=0+2�.�
−2
×4�=8��
−1
And �=�
0�+
1
2
��
2
=0+
1
2
×2�.�
−2
×4�
2
=16�
After 4s the car has reached a velocity of 8m.s
-1
and is 16m far from the light.
Note that we could also have found �from Eq. (4) using our result for �
Exercise:Resolve example 1.16 with an initial velocity �
0=3��
−1
.
CHAPTER 1: MOTION ON A STRAIGHT LINE PREPARATORY YEAR 2015-2016 26

27. 1.6 Free falling objects
Falling objectsundergo an acceleration, which we attribute to gravity, the gravitational attraction
of the earth.
If gravity is the only factor affecting the motion of an object falling near the earth’s surface, and air
resistance is either absent or negligibly small. So long as the object’s distance from the surface of
the earth is small compared to the earth’s radius, it is found that:
1-The gravitational acceleration is the same for all falling objects, no matter what their size or
composition or mass.
2-The gravitational acceleration is constant. It does not change as the object falls.
•An object initially thrown upward has also the gravity acceleration . Its velocity steadily
decreases in magnitude until it becomes zero at the highest point reached.
•Free falling problems can be solved using the equations of motion in a straight line (which is on
the vertical direction) with a constant acceleration equal to 9.80�.�
−2
.
CHAPTER 1: MOTION ON A STRAIGHT LINE PREPARATORY YEAR 2015-2016 27

28. The equations of Free falling objects can be summarized in the following table:
Upwardpositive axis Downward positive axis
Δ�=�
�∆�+
1
2
(−9.80)∆�
2
�=�
0−9.80∆�
�
2
−�
�
2
=2(−9.80)∆�
∆�=�
�∆�+
1
2
(9.80)∆�
2
�=�
0+9.80∆�
�
2
−�
�
2
=2(9.80)∆�
Example 1.20 P 17:A ball is dropped from 84 m above the ground. when does the ball strike the
ground? (b) what is its velocity and its speed when it strikes the ground?
a)Let choose the positive axis is upward, ∆�=
1
2
�∆�
2
(the initial velocity is zero) then
∆�=
2(−84�)
−9.80��
−2
=4.14�
b) �=�∆�=−9.80��
−2
×4.14�=−40.6��
−1
then the ball hits the ground with a speed of
40.6��
−1
CHAPTER 1: MOTION ON A STRAIGHT LINE PREPARATORY YEAR 2015-2016 28

29. CHAPTER 2: MOTION IN TWO DIMENSIONS PREPARATORY YEAR 2015-2016 29
Chapter 2
MOTION IN TWO DIMEN SIONS
COURSE TOPICS:
An introductionto vectors
The velocity in two dimensions
The acceleration in two dimensions
Examples to be explained and solved:
2.1, 2.2, 2.3, 2.4 and 2.6
Homework Problems: 2.13, 2.15, 2.19 and 2.21

30. CHAPTER 2: MOTION IN TWO DIMENSIONS PREPARATORY YEAR 2015-2016 30
2.1 An introduction to vectors
Physical quantities can be classified as scalarsor vectors. A scalar quantity is simple number (with
unit) such as mass, distance, speed,…but a vector quantity is defined with both a magnitude
(which is a number with unit) and a direction; such as force, displacement, velocity,….
For example, to describe where the
school is located versus your home, it is
not sufficient to give the distance
between them. We have to give the
distanceand the direction.
Along this chapter we will see how to describe motion in two dimensions using vectors. That’s why
it’s first important to know what is a vector and how to add, subtract and multiply vectors.

31. CHAPTER 2: MOTION IN TWO DIMENSIONS PREPARATORY YEAR 2015-2016 31
Thecomponentsofavector
Figure 2.2: Components of vector
Definition of vector: A vector is defined with a magnitudeand a direction. The
magnitude is given by the length of the vector and the direction by a positive angle
(Fig. 2.2 a)
A vector Ԧ�is resolved in(�,�)plane into two components (Fig. 2.2 b) : �
�is the component
on the �axis and �
�is the component on the �axis, where:
�
�=���������
�=�sin�

32. CHAPTER 2: MOTION IN TWO DIMENSIONS PREPARATORY YEAR 2015-2016 32
�=�
�
2
+�
�
2
and �=tan
−1
��
��
+�
If the vector is defined in the (�,�)plane by its components �
�and �
�, then the
magnitude and the direction of the vector are:
�is a correction of the angle which depend on the quadrant where the vector is located
�=0if the vector is in the first quadrant
�=180°the vector is in the second quadrant
�=180°the vector is in the third quadrant
�=360°the vector is in the fourth quadrant

33. CHAPTER 2: MOTION IN TWO DIMENSIONS PREPARATORY YEAR 2015-2016 33
Example2.1page30:
Apersonwalks1��dueeast.Ifthepersonthenwalksasecondkilometer,whatisthefinaldistance
fromthestartingpointifthesecondkilometeriswalked:
(a)dueeast;(b)duewest;(c)duesouth?
WewillcallthefirstdisplacementԦ�andthesecond�.
Solution
Weconstructthesum�=�+�forthe
threecases(SeeFigurebelow).
(a)Since�and�areinthesamedirection,
�=�+�=2�=2��.
Thevector�isdirecteddueeast.
(b)Here,thevectorsareopposite,
so�=�–�=0.
(c)FromthePythagoreantheorem:
�
2
=�
2
+�
2
=2�
2
,
so�=2A=2��
�pointstowardthesoutheast(Fig.2.3)

34. CHAPTER 2: MOTION IN TWO DIMENSIONS PREPARATORY YEAR 2015-2016 34
Example 2.2 page 32:
Find the components of the vectors Ԧ�and �in Fig.2.7, if �=2and �=3.
Solution:
�
�and�
�are positive:
�
�=�cos=2cos30°=2(0.866)=1.73
�
�=�sin�=2sin30°=2(0.500)=1.00
From Fig.2.7b, �
�is positive and �
�is negative:
cos 45°=sin 45°=0.707,
�
�=3 cos45°=3 (0.707) =2.12
�
�=−3 sin 45°= −3 (0.707) =−2.12

35. CHAPTER 2: MOTION IN TWO DIMENSIONS PREPARATORY YEAR 2015-2016 35
Adding and subtracting vectors
Adding vectors using components:
1.Find the components of each vector to be added
2.Add the �−and �−components separately
3.Find the resultant vector.
�=�+�=�+�

36. CHAPTER 2: MOTION IN TWO DIMENSIONS PREPARATORY YEAR 2015-2016 36
Subtracting vectors:
The negative of a vector is a vector of the same magnitude pointing in the
opposite direction. Here �=�−�
�=�+−�=�−�

37. CHAPTER 2: MOTION IN TWO DIMENSIONS PREPARATORY YEAR 2015-2016
Example2.3page33:
Ԧ�=2ො�+ො�,�=4ො�+7ො�
(a)FindthecomponentsofԦ�=Ԧ�+�
(b)FindthemagnitudeofԦ�anditsangle�withrespecttothepositive
�axis
Solution:
a) Using the equation Ԧ�=�
�+�
�ො�+�
�+�
�ො�
We can write Ԧ�=2+4ො�+1+7ො�=6ො�+8ො�
Thus �
�=6, and �
�=8.
(b) From the Pythagorean theorem: �
2
=�
�
2
+�
�
2
=6
2
+8
2
=100
Then �=�
�
2
+�
�
2
=100=10
FromFig.2.9,weseethattheangle�satisfies:
����=
�
�
�
�
=
8
6
=1.33
Then �=���
−1
1.33=53.1°

38. CHAPTER 2: MOTION IN TWO DIMENSIONS PREPARATORY YEAR 2015-2016
Multiplying vector by scalar
Multiplying a vector by 3 increases its magnitude by a factor of 3, but does not change its
direction.
Ԧ�=�
�ො�+�
�ො�
3Ԧ�=3�
�ො�+3�
�ො�
−3Ԧ�=−3�
�ො�−3�
�ො�

39. CHAPTER 2: MOTION IN TWO DIMENSIONS PREPARATORY YEAR 2015-2016 39
The velocity in two dimensions
Intwodimensions,position,velocity,andaccelerationarepresentedbyvectors:motionina
plane.
Aprobleminvolvingmotioninaplaneisapairofone-dimensionalmotionproblems.
If the displacement in a time interval ∆tis denoted by the vector ∆�, then the average velocity of the
object is parallel to ∆�and is given by:
ഥ
�=
∆�
∆�
where ∆Ԧ�=∆�ො�+∆�ො�, then
ഥ
�=
∆�
∆�
ෝ�+
∆�
∆�
ෝ�

40. CHAPTER 2: MOTION IN TWO DIMENSIONS PREPARATORY YEAR 2015-2016
Example :
An ambulance travels from the hospital 10��due south in 7���, and then
5��due east in 3���. Find (a) the final position of the ambulance, (b) its
average velocity
Solution
(a) ∆�=5��ො�−10��ො�
(b) ∆�=7���+3���=10���=0.167ℎ���
then:
Ԧ�=
∆�
∆�
=
∆�
∆�
ො�+
∆�
∆�
ො�=
5��
0.167ℎ
ො�−
10��
0.167ℎ
ො�=30Τ��ℎො�−60Τ��ℎො�

41. CHAPTER 2: MOTION IN TWO DIMENSIONS PREPARATORY YEAR 2015-2016
Solution:
a) The instantaneous velocity is tangent to the path of
the car, and its magnitude is equal to the speed. Thus,
at point 1 the velocity is directed in the+�direction
and �
1=30��
−1
ො�. Similarly, at point 2 the velocity is
long the –�direction, and �
2=−30��
−1
ො�
(b) The average velocity is the displacement divided by
the elapsed time. The displacement is entirely along the
x direction, so∆Ԧ�=300�ො�.
Since ∆t = 40 s,
ҧ
Ԧ�=
∆Ԧ�
∆�
=
300�
40�
ො�=7.5��
−1
ො�
The average velocity during this time interval is directed
along the +�axis. Its magnitude is less than the speed
of 30��
−1
because the car does not travel in a
straight line.
Figure 2.11
Example 2.4 page 34:
A car travels halfway around an oval racetrack at a constant speed of 30��
−1
(Fig. 2.11).
(a) What are its instantaneous velocities at points 1and 2?
(b) It takes 40s to go from 1to 2, and these points are 300�apart. What is the average
velocity of the car during this time interval?

42. CHAPTER 2: MOTION IN TWO DIMENSIONS PREPARATORY YEAR 2015-2016 42
The acceleration in two dimensions
The average acceleration is defined by :
Ԧ�=
∆�
∆�
=
�
2−�
1
�
2−�
1
=�
�ො�+�
�ො�
Example 2.6 page 35:
InExercise2.4thevelocityofthecarchangedfrom�
1=30�s
−1
ො�
to�
2=−30�s
−1
ො�in40s.
Whatwastheaverageaccelerationofthecarinthattimeinterval?
Solution :
The average acceleration is defined as the velocity change divided by
elapsed time:
Ԧ�=
�2−�1
∆�
=
−30��
−1
ො�−30��
−1
ො�
40
=−1.5��
−2
ො�
Thus the average acceleration during the time the car goes from point 1
to point 2 is directed in the –�direction, or downward in Fig. (2.11b)

43. CHAPTER 2: MOTION IN TWO DIMENSIONS PREPARATORY YEAR 2015-2016 43
Thisexerciseillustratestwoimportantpoints:
1-Ifthevelocityisconstant,theaccelerationiszero,since�istherateofchangeofthe
velocity.
However,whenthespeedisconstant,theaccelerationmayormaynotbezero.Ifan
objectmovesataconstantspeedalongacurvedpath,itsvelocityischangingdirection,
anditisaccelerating.Wefeeltheeffectsofthisaccelerationwhenacarturnsquickly.
Theaccelerationiszeroonlywhenthespeedanddirectionofmotionarebothconstants.
2-Thedirectionsofthevelocityandaccelerationatanyinstantcanberelatedinmany
ways.
Themagnitudeanddirectionof�aredeterminedbyhow�ischanging.
Whenacarmovesalongastraightroad,theaccelerationisparalleltothevelocityif�is
increasingandoppositeif�isdecreasing.
Whenthemotionisalongacurvedpath,theaccelerationisatsomeangletothevelocity.

45. Chapter 3
NEWTON’S LAWS OF MOTION
COURSE TOPICS:
3.1 Force and weight
3.2-Density
3.3-Newton’s 1
st
law
3.4-Equilibrium
3.5-Newton’s 3
rd
law
3.6-Newton’s 2
nd
law
3.8-Some Examples of Newton’s Laws
3.12-Friction
Examples to be explained and solved:
3.1,3.2, 3.6,3.9 and 3.17
Homework Problems: 3.17, 3.49 and 3.54
CHAPTER 3: NEWTON’S LAWS OF MOTION PREPARATORY YEAR 2015-2016 45

46. Havinglearnedhowtodescribemotion,wecannowturntothemorefundamentalquestion
ofwhatcausesmotion.Inapreviouschaptersthemotionofanobjectwasdescribedwithout
consideringthecausesofmotion(Forces)andtheobjectinmotionwasconsideredlikea
pointparticlewithoutmass.Inthischapterweintroducetheconceptsofmassandforce.
Althoughtherearemanykindsofforcesinnature,theireffectsaredescribedaccuratelyby
threegenerallaws,firststatedbyIsaacNewton.
EventhoughtwentiethcenturyadvanceshaveshownthatNewton’slawsareinadequateat
theatomicscaleandatvelocitiescomparabletothespeedoflight(3x10
8
m/s).Butthese
lawsarefullyadequateformustapplicationsindifferentfieldssuchasastronomy,
biomechanics,geologyandengineering.
Introduction
CHAPTER 3: NEWTON’S LAWS OF MOTION PREPARATORY YEAR 2015-2016 46

47. 3.1 Force weight and gravitational mass
Aforceisavectorquantity,itrepresentstheabilitytoproducemotionortocauseanobject
tochangeitsstateofmotion.
Twokindsofforcescanbedistinguished:fieldforcesandcontactforces
Afieldforceisaforcethatactsthroughanemptyspace(gravitationalforce,electricforce,
magneticforce,…)
Acontactforceactsthroughacontactpointorsurface(pushingorpulling,frictionforce,tension
onastring,reactionforce,…)
If more than one force act on an object, the net forceor resultant is the sum of the individual
forces: Ԧ�
���=�
1+�
2+�
3+⋯.=σԦ�
Theweightisthemagnitudeofthegravitationalforceexertedbytheearthonanobjectof
mass�:�=��
Where�istheofgravityacceleration.
CHAPTER 3: NEWTON’S LAWS OF MOTION PREPARATORY YEAR 2015-2016 47

48. Themasscanbethoughtofasaquantityofmatter.Butfromamechanicalpointofviewit
canbeseenasresistancetothemotion.ThisresistanceisalsonamedINERTIAwhichisa
measureofhowisdifficulttomoveortochangethestateofmotionofanobject.
Butanobjectsubmittedtothegravitationalforcehasthegravitationalmasswhichisequalto
itsweightdividedbythegravityacceleration:�=�/�
ForExample,amanwhoweighs1000�ontheearthhasagravitationalmass:
�=�/�=1000�/9.80��
−2
=102��
Thesamemanonthemoonhastheweight:
�=��
����=102��×1.62Τ��
2
=165�
Ineverydayusage,massissometimesreferredtoas"weight",theunitsofwhichmaybepound
orkilograms
CHAPTER 3: NEWTON’S LAWS OF MOTION PREPARATORY YEAR 2015-2016 48

49. 3.2 Density
Thedensityofanobjectisanintrinsicpropertyofamaterial,microscopicallyrelatedtothe
atomicarrangement.Atthemacroscopicscale,thedensityisdefinedastheratioofmasstothe
volume:??????=
�
�
(itsunitin���
3
)
Twodifferentobjectsofthesamesize,madeupfromdifferentmaterials,havedifferentmasses
becausetheydon’thavethesamedensity.
Example 3.1 P 50:A cylindrical rod of aluminum has a
radius �=1.2��and a length L=2�.What is its
mass?
Solution: The density of aluminum is �
=2700���
3
.
The volume of a cylinder is �=��
2
�.
Then its mas is :
�=��=���
2
�=(2700���
3
)×�×(1.2×10
−2
�)
2
×2�=2.44��
CHAPTER 3: NEWTON’S LAWS OF MOTION PREPARATORY YEAR 2015-2016 49

50. CHAPTER 3: NEWTON’S LAWS OF MOTION PREPARATORY YEAR 2015-2016 50
3.3 Newton’s first law
Newton’s first law states that:
Every object continues in a state of rest, or of uniform motion in a straight line, unless it is
compelled to change that state by forces acting upon it.
An equivalent statement of the first law is that if there is no force on an object, or if there
is no net force when two or more forces act on the object, then:
(1) an object at rest remains at rest, and
(2) an object in motion continues to move with constant velocity.
Newton’s first law as stated does not hold true for someone who is accelerating.
�
���=�����ቐ
�=������������������������������
��
�=��������(�������������������������������)

51. CHAPTER 3: NEWTON’S LAWS OF MOTION PREPARATORY YEAR 2015-2016
3.4 Equilibrium
Newton’sfirstlawtellsusthatthestateofanobjectremainsunchangedwheneverthenet
forceontheobjectiszero(eventhoughtwoormoreforcesactuponit).Inthiscasetheobject
issaidtobeinequilibrium.
Thefirstlawappliestoobjectsinuniformmotionaswellastoobjectsatrest.
Therearethreetypesofequilibrium:unstable,stable,andneutral
1-Unstableequilibrium:asmalldisplacementleadstoan
unbalancedforcethatfurtherincreasethedisplacementfromthe
equilibriumlocation(ballinpositionA).
2-Stableequilibrium:asmalldisplacementleadstoan
unbalancedforcethattendstorestoretheobjecttothe
equilibriumlocation(ballinpositionB).
3-Neutralequilibrium:itisinequilibriumatanylocationnearC.

52. CHAPTER 3: NEWTON’S LAWS OF MOTION PREPARATORY YEAR 2015-2016 52
3.5 Newton’s third law
If one object exerts a force F on a second, then the second object exerts an equal but
opposite force -F on the first.
Forexample:supposeyouareatrestinaswimmingpool.Ifyoupushawallwithyourlegs,the
wallexertaforcethatpropelsyoufurtherintothepool.Thereactionforcethewallexertson
youisoppositeindirectiontotheactionforceyouexertonthepool.
Example 3. 2 P 52:
A woman has a mass of 60��. She is standing on a floor and remains at
rest. Find the normal force exerted on her by the floor.
Solution:
The woman is in equilibrium , then the net force on the woman must be
zero �+Ԧ�
??????=0. The normal force is the force exerted by the floor on
the woman. This force must have the same magnitude as her weight,
which acts downward. Symbolically:
�+Ԧ�
??????=0,�ℎ���=−Ԧ�
??????
Its magnitude is �
??????=��=60��×9,80
�
�
2
=588�
Note that the gravitational force and the normal force are not an action
and a reaction forces because they are applied on the same object.

53. 3.5 Newton’s second law
The second law states that:
Whenever there is a net force acting on an object , this object will undergoes an
accelerationin the same direction as the force.
The acceleration is proportional to the net force and inversely proportional to the mass.
Thus, we can relate the net force �and the acceleration a by Newton’s second law:
�=��
The proportionality constant mis the inertial mass of the object.
Note that if the net force is equal zero, then, the acceleration is also zero, which means that
the velocity is constant or zero. The second law is consistent with the first law
Example 3. 6 P55:A child pushes a sled across a frozen pond with a horizontal force of20�.
Assume friction is negligible.
a) If the sled accelerates at0.5��
−2
, what is its mass?
b) Another child with a mass of60��sits on the sled. What acceleration, the same force
produces now?
CHAPTER 3: NEWTON’S LAWS OF MOTION PREPARATORY YEAR 2015-2016 53

54. CHAPTER 3: NEWTON’S LAWS OF MOTION PREPARATORY YEAR 2015-2016 54
Solution of 3. 6 P55:
a)Thegravitational force and the normal force cancel each other. Thus the net force
exerted on the sled is a horizontal force with a magnitude of �=20�
According to the second law: �=
??????
�
=
20??????
0.5��
−2
=40��
b) The total mass of the sled becomes:
�=40��+60��=100��
Then the acceleration is:
�=
??????
�
=
20??????
100��
=0.2��
−2

55. 3.12 Friction
Frictionisaforcethatalwaysactstoresistthemotionofoneobjectonanother.
Frictionalforcesareveryimportant,sincetheymakeitpossibleforustowalk,use
wheeledvehicles,….
Frictionalforcescanexistbetweensolidsurfacesorbetweenfluids(viscousforces).
Viscousforcesarequitesmallcomparedtothefrictionbetweensolidsurfaces.Thusthe
useoflubricatingliquidsuchasoil,whichclingstothesurfaceofmetals,greatlyreduces
friction.
Whenwewalkorrun,wearenotconsciousof
anyfrictioninourkneesorotherlegjoints.These
joints,really,arelubricatedbysynovialfluid,
whichissqueezedthroughthecartilageliningthe
jointswhentheymove.Thislubricanttendstobe
absorbedwhenthejointisstationary,increasing
thefrictionandmakingiteasiertomaintaina
fixedposition.
CHAPTER 3: NEWTON’S LAWS OF MOTION PREPARATORY YEAR 2015-2016 55

56. Static friction
We distinguish two types of friction:
Static friction, between two surfaces at rest.
Kinetic friction, between two surfaces one moves against the other.
Static friction
Weconsiderablockatrestonahorizontalsurface(Fig.3.19).
SincetheblockisatrestFig.3.19a,thefirstlawrequiresthatthenetforceontheblockbezero.
Theverticalforcesaretheweight�andthenormalforce�,sowemusthave:�=�.
CHAPTER 3: NEWTON’S LAWS OF MOTION PREPARATORY YEAR 2015-2016 56

57. If there is no force applied in the horizontal direction and the block remains at rest (Fig 3.19 b),
the frictional force must be zero, according to the first law.
If we apply a small horizontal force �to the rights and if the block remains at rest (Fig. 3.19 c),
the friction force �
�can no longer be zero, since the first law requires that the net force on the
block be zero, then a frictional force opposite to the applied force must be appeared ( �
�=−�)
If �is gradually increased �
�increasesalso. Eventually when �become large enough, the block
begins to slide.
The static friction force attains a maximum value calledthe maximum static friction �
����
Experimentally it is found that �
����has the following properties:
1-�
����is independent of the contact area.
2-For a given pair of surfaces �
����is proportional to the normal force �:
CHAPTER 3: NEWTON’S LAWS OF MOTION PREPARATORY YEAR 2015-2016 57
�
����=�
��

58. ??????
�is the coefficient of static friction and �is the magnitude of the normal force.
The coefficient of static friction 
sdepends on:
The nature of the two surfaces in contact
Their cleanliness and smoothness
The amount of moisture present
For metal on metal: 
sis between 0.3and 1. When lubricating oils are used 
sis about 0.1. For
Teflon on metals, 
�≅0.04
CHAPTER 3: NEWTON’S LAWS OF MOTION PREPARATORY YEAR 2015-2016 58

59. The force necessary to keep an object sliding at constant velocity is smaller than that
required to start it moving.
Thus the sliding or kinetic friction�
�is less than �
����.
The kinetic friction is independent of the contact area, it satisfies:
Here 
kis the coefficient of kinetic frictionand is determined by nature of the two
surfaces.

kis nearly independent of the velocity (since f
k�
����then
k 
s)
Kinetic friction
CHAPTER 3: NEWTON’S LAWS OF MOTION PREPARATORY YEAR 2015-2016 59
�
�=�
��

60. Example3.17 p 149:
A 50−�block is on a flat, horizontal surface. (a) If a horizontal force �=20�is
applied and the block remains at rest ; what is the frictional force? (b) the block
starts to slide when �is increased to 40�. What is �
�? ( c) the block continues to
move at constant velocity if �is reduced to 32�. What is �
�?
Solution:
a)Since the block remains at rest �
�=�=20�
b)�
����=40�=�
��
??????then�
�=
�
??????���
????????????
=
40??????
50??????
=0.8
c)Whenthe block is moving at constant velocitywith�
�=�
��
??????
Then �
�=
�
�
??????
??????
=
32??????
50??????
=0.64
CHAPTER 3: NEWTON’S LAWS OF MOTION PREPARATORY YEAR 2015-2016 60

61. CHAPTER 4: STATICS PREPARATORY YEAR 2015-2016 61
Chapter 4
STATICS
4.1 Torque
4.2 Equilibrium of rigid bodies
4.3 The center of gravity
4.4 Stability and balance
4.5Levers and mechanical advantage
Examples to be explained and solved:
4.1, 4.3, 4.4, 4.6, 4.8, and 4.10
Homework Problems:
4.11, 4.18, 4.41, 4.45 and 4.55

62. Introduction
Staticsis the study of the forces acting on an object (rigid body ) that is in equilibrium and at rest.
The importance of Statics is to:
Find the forces acting on various parts of engineering structures, such as bridges or buildings, or
of biological structures, such as jaws, limbs, or backbones.
Understand the force multiplication or mechanical advantage obtained with simple machines,
such as the many levers found in the human body.
At the end of this chapter the student will be able to calculate the torque, the center of gravity of a
rigid body, evaluate the mechanical advantage of a mechanical system and to apply these concepts
to biological systems.
CHAPTER 4: STATICS PREPARATORY YEAR 2015-2016 62

63. 4.1 Torque
In figure 4.1 a child applies equal but opposite forces �
�and
�
�to opposite sides of a freely rotating seat. The seat will
begin to rotate, hence, it is not in equilibrium even though the
net force is zero, Ԧ�
1+Ԧ�
2=0. The rotation of the seat is due
to the Torque.
Torque is the ability of a force to cause rotation
CHAPTER 4: STATICS PREPARATORY YEAR 2015-2016 63
Supposeweneedtounscrewalargenutthatrustedintoplace.Tomaximizethetorque,weuse
thelongestwrenchavailableandexertaslargeaforceaspossibleandpullatrightangle.The
torqueisproportionaltothemagnitudeoftheforce,tothedistancefromtheaxisofrotationand
thedirectionoftheforceandtotheanglebetweenthem.

64. If a rigid body is able to rotate about a point P, and a force Ԧ�is applied on this rigid body, then
the magnitude of the torque about the point P is:
ris the distance from the pivot P to the point where the force acts on the objectand �is the
angle from the direction of �to the direction of Ԧ�.
The SI unit of torque is the Newton meter (N m)
Example 4.1:A mechanic holds a wrench 0.3�from the center of a nut. How large is the
torque applied to the nut if he pulls at right angles to the wrench with a force of 200�?
Solution:
Since he exerts the force at right angles to the wrench, the angle ���90°, and sin�=1,
Thus the torque is:
�=��sin�=0.3�×200�×1=60�.�
CHAPTER 4: STATICS PREPARATORY YEAR 2015-2016 64
??????=�����??????

65. Finding the magnitude of a torque using the lever arm:
1 Draw a line parallel to the force through the force’s point of application;
this line (dashed in the figure) is called the force’s line of action.
2 Draw a line from the rotation axis to the line of action. This line must be
perpendicular to both the axis and the line of action. The distance from the axis
to the line of action along this perpendicular line is the lever arm (r
⊥).
3 The magnitude of the torque is the magnitude of the force times the lever arm:
�=�
⊥.����=�.�
⊥
Where �
⊥=�sin�and�
⊥=�sin�
CHAPTER 4: STATICS PREPARATORY YEAR 2015-2016 65

66. The cross product
The vector productor cross product of two vectors Ԧ�and �is a
vector Ԧ�which is written as :
�=��
The magnitude of the vector Ԧ�is:�=��sin�
The direction of the vector Ԧ�is perpendicular to the plane
containing Ԧ�and �such that Ԧ�is specified by the right-hand rule;
By curling the fingers of the right hand from vector �to vector �,
the thumb points in the direction of �.
Example 4.2The vectors in the figure are all in the plane of the page. Find
the magnitude and direction of :
(a) ��
(b) ��
c) ��.
CHAPTER 4: STATICS PREPARATORY YEAR 2015-2016 66

67. Solution:
a) Ԧ�×Ԧ�=�.�.sin0°=0
The cross product of any parallel vectors is zero
(b) The magnitude of Ԧ�×�is:
Ԧ�×�=�.�.sin90°=4×5×1=20
We rotate our right palm from Ԧ�toward �. Our right thumb then points outof the page.
(c)The magnitude of Ԧ�×Ԧ�is:
Ԧ�×Ԧ�=�.�.sin30°=4×5×0.5=10
Now when we rotate our right palm through 30°from Ԧ�toward Ԧ�, our thumb points intothe
page.
CHAPTER 4: STATICS PREPARATORY YEAR 2015-2016 67

68. ACoupleisthetotaltorqueofapairofforceswithequal
magnitudesbutoppositedirectionsactingalongdifferentlinesof
action.
Couplesdonotexertanetforceonanobjecteventhoughthey
doexertanettorque.
Thenettorqueisindependentofthechoiceofthepointfrom
whichdistancesaremeasured.
Example4.3:Twoforceswithequalmagnitudesbut
oppositedirectionsactonanobjectwithdifferentlinesof
action(Fig4.8).Findthenettorqueontheobjectresulting
fromtheseforces.
Solution:Thetorqueresultingfromtheforceat�
1is
�
1=�
1�.ThetorqueaboutPresultingfromtheforceat
�
2is�
2=‐�
2�.(Theminussignindicatesaclockwise
torque).Thenettorqueis:�=�
1+�
2=�
1�‐�
2�
Thenτ=�
1‐�
2�=−��
Thenettorqueisindependentofthechoiceofthepoint
fromwhichdistancesaremeasured.
Couples
CHAPTER 4: STATICS PREPARATORY YEAR 2015-2016 68

69. 4.2 Equilibrium of rigid body
Two conditions for a rigid body to be in static equilibrium:
1. Translational equilibrium: The net force on the rigid body must be zero:σ�=�
2. Rotational equilibrium : The net torque on the rigid body about any point must be zero:
σ??????=�
Example 4.4Two children of weights �
1and �
2are balancedon a board pivoted about its
center.
(a) What is the ratio of their distances �
2/�
1from the pivot?
(b) If �
1=200�, �
2=400�and �
1=1�, what is �
2?
(For simplicity, we assume the board to be weightless; this will not affect the result.)
Solution:
a)The force �exerted by the support must balance the weights
of the two children so that the net force is zero:
�=�
1+�
2
The torques resulting from the weights (about P) are:
�
1=�
1�
1and �
2=−�
2�
2
Rotational equilibrium:�=�
1+�
2=0
�
1�
1−�
2�
2=0or
�
2
�
1
=
�
1
�
2
(b)�
2=�
1
�
1
�
2
=1�×
200??????
400??????
=0.5�
CHAPTER 4: STATICS PREPARATORY YEAR 2015-2016 69

70. Application to muscles and joints
Thetechniquesforcalculatingforcesandtorquesonbodiesinequilibriumcanbereadily
appliedtothehumanbody.Thisisofgreatuseinstudyingtheforcesonmuscles,bonesand
joints.
Generallyamuscleisattached,viatendons,to
twodifferentbones.Thepointsofattachment
arecalledinsertions.Thetwobonesareflexibly
connectedatajoint,suchasthoseattheelbow,
kneeandankle.Amuscleexertsapullwhenits
fibercontractunderstimulationbyanerve,but
itcannotexertapush.
CHAPTER 4: STATICS PREPARATORY YEAR 2015-2016 70

71. Example4.6Amodelfortheforearmintheposition
showninthefigureisapivotedbarsupportedbyacable.
Theweight�oftheforearmis12�andcanbetreated
asconcentratedatthepointshown.Findthetension
�exertedbythebicepsmuscleandtheforce�exerted
bytheelbowjoint.
Solution:
Applying the condition σ�=0,then �−�–�=0
Both �and �are unknown.
Calculating torques about the pivot P:
�produces no torque.
�produces a torque: �
�=−0.15×�
??????produces a torque: �
�=0.05�
At equilibrium the total torque is : −0.15�+0.05�=0
Then�=
0.15�
0.05
=
0.15×12
0.05
=36�
Replacing in the first equation:
�=�−�=36�−12�=24�
CHAPTER 4: STATICS PREPARATORY YEAR 2015-2016 71

72. 4.3 The centerof gravity
Theweightofabodycanbeconsideredasaforceactingthrougha
singlepointcalledthecenterofgravityorcenterofmass.
Thetorqueproducedbytheweightofarigidbodyisequaltothat
duetoaconcentratedobjectofthesameweightplacedatits
centerofgravity.
Locating the center of gravity:
For regular shapes: The C. G. of a uniformly dense symmetric object is at its geometric
center.
For less symmetric objects the C.G can be calculated mathematically or located
experimentally
CHAPTER 4: STATICS PREPARATORY YEAR 2015-2016 72

73. Calculating the position of the center of gravity:
Consider two weights �
1and �
2,their total torque about a
pivot �( arbitrarily chosen) is �=�
1�
1+�
2�
2.
This same torque is of the total weight as it is concentrated at
the center of gravity �: �=�(�
1+�
2).
By equating we obtain: �(�
1+�
2)=�
1�
1+�
2�
2. Then,
�=
�
1�
1+�
2�
2
�
1+�
2
If there are more than two weights, then the C. G. is:
The center of mass:
Replacing the weight by �=��, then �=
�
1�
1+�
2�
2+�
3�
3+⋯
�
1+�
3+�
3+⋯
is called center of mass.
If the C.G or the C.M has a component on the �−axis then, by the same way
�=
�
1�
1+�
2�
2+�
3�
3+⋯
�
1+�
3+�
3+⋯
CHAPTER 4: STATICS PREPARATORY YEAR 2015-2016 73
�=
�
1�
1+�
2�
2+�
3�
3+⋯
�
1+�
2+�
3+⋯

74. Example 4.8:A weightless plank 4�long has one concrete block at the left end, another at the
center, and two blocks at the right end. Find its center of gravity?
CHAPTER 4: STATICS PREPARATORY YEAR 2015-2016 74
Solution:
The total weight is�=�
1+�
2+�
3=4�
0
Then center of gravity is:
�=
�
1�
1+�
2�
2+�
3�
3
�
=
0+2��
0+4�×2�
0
4�
�
=
10��
0
4�
�
=2.5�
The center of gravity is between the center of the plank and the heavier end.

75. 4.4 Stability and balance
Balanceis a physical ability to control equilibrium and Stabilityis the degree of balance
Types of balance:dynamic and static
Dynamic balance, when a performer is
in motion.
Static balance: when an object remain over
a relatively fixed base.
Base of support:is the supporting area beneath the object that includes the points of contact
with the supporting surface and the area between them.
Human Base of Support is the area under
the feet (or shoes) including the area
between the feet. This area is traced from
toe to toe and from heel to heel.
CHAPTER 4: STATICS PREPARATORY YEAR 2015-2016 75

76. Stability criteria:
The major factors that affect the object’s stability and balance are:
1) The area of the support: The larger the base of support, the more stable the object
(Fig. 4. 18 a)
2) The mass of the object: The greater the mass the greater the stability
3) Position of the center of gravity: An object is in balance if its center of gravity is
above its base of support. Balance is less if the C.G. is near the edge of the Base of
support. ( Fig. 4. 18 b)
4)The height of the center of gravity: The higher the center of gravity, the more likely
that an object will be out of balance. ( Fig. 4.18 c)
CHAPTER 4: STATICS PREPARATORY YEAR 2015-2016 76
( Fig. 4. 18 a)
( Fig. 4. 18 b)
( Fig. 4. 18 c)

77. 4.4 Levers and Mechanical Advantage
Simplemachines,suchaslevers,pulleysaredesignedtoreducetheforceneededtoliftaheavy
load.Ineachcasethereisanappliedforce�
�andaloadforce�
�.
Themechanicaladvantageofamachineistheratioofthemagnitudesoftheloadforce�
�balanced
byanappliedforce�
�.
Class I: The fulcrum lies between �
�
and F
L.
ClassII:Thefulcrumisatone
end,�
�attheotherendand�
�
liesbetweenF
aandthefulcrum.
ClassIII:Thefulcrumisatoneend,
�
�attheotherendand�
�lies
between�
�andthefulcrum.
CHAPTER 4: STATICS PREPARATORY YEAR 2015-2016 77
Mechanical Advantage: �.�.=
�
�
��
Aleverin its simplest form is a rigid bar used with a fulcrum. The fulcrumis the point or support on
which a lever pivots. Three classes of levers can be distinguished:

78. Levers in human body
In human body, muscles, bones and joints act together to form levers.
Bones act as lever arms.
Joints act as pivots.
Muscles provide the applied forces to move loads.
Load forces are often the weights of the body parts that are moved or forces needed to lift,
push or pull things outside our bodies.
Iftheloadandappliedforcesareperpendiculartotheleverarminequilibrium,theirratiois
equaltotheratioof�
�and�
??????(distancesfromthefulcrum),then:
�.�=
�
??????
�
�
=
�
�
�
??????
WiththeforcesatrightangletothelevertheM.A.ofclassIIIleversisalwayslessthan1,
andofclassIIleversisalwaysgreaterthan1.
CHAPTER 4: STATICS PREPARATORY YEAR 2015-2016 78

79. CHAPTER 4: STATICS PREPARATORY YEAR 2015-2016 79
Examples of levers in the human body

80. Example 4.10:Suppose the load force �
�on a class one lever has a magnitude of 2000�. A
person exerts a force �
�=500�to balance the load.
(a)What is the ratio of the distances �
�and �
�?
(b)What isthe mechanical advantage of this lever?
Solution:
(a)The torque due to �
�is�
�=−�
��
�andthe torque due to�
�is�
�=�
��
??????.
Forbalance,thesemustsumtozero,so:
�
??????�
??????−�
��
�=0
and
�
�
�
??????
=
??????
??????
??????�
=
2000??????
500??????
=4
(b) The mechanical advantage of the lever is:
�.�.=�
??????/�
�=4
CHAPTER 4: STATICS PREPARATORY YEAR 2015-2016 80

81. CHAPTER 4: STATICS PREPARATORY YEAR 2015-2016 81
Theback
Whenthetrunkisbentforward(Fig.4.20a),thespinepivotsmainlyonthefifthlumbar
vertebra(Fig.4.20b).Toanalyzetheforcesinvolvedwhenthetrunkisbentat60°fromthe
verticalwiththearmshangingfreely,amodelisrepresentedin(Fig.4.20c).ThepivotpointA
isthefifthlumbarvertebra.TheleverarmABrepresentstheback.Theweightofthetrunkcan
berepresentedbyaweight�
1suspendedinthemiddle.Theheadandthearmsare
representedbyaweight�
2suspendedattheendoftheleverarm.Theerectorspinalismuscle
maintainsthepositionofthebackshownastheconnectionD-Cattachedatapointtwo-thirds
upthespine,Accordingtofigure4.21calculate(a)theforceF
mexertedbythespinalismuscle
and(b)thecompressionforceonthefifthlumbar

82. CHAPTER 4: STATICS PREPARATORY YEAR 2015-2016 82

83. CHAPTER 6: WORK AND ENERGY PREPARATORY YEAR 2015-2016 83
Chapter 6
WORK AND ENERGY
COURSE TOPICS
6.1 Work
6.2 Kinetic Energy
6.3 Potential Energy and Conservative
Forces
6.9 Power
Examples to be explained and solved:
6.1, 6.2, 6.3, 6.5 and 6.14
Homework Problems:6.2, 6.21, 6.31,
6.36 and 6.45
China's Deng Wei broke the world record as she won Olympic weightlifting gold in the
women's 63-kilogram category (Rio 2016)

84. 6.1 Work
TheworkdonebyaconstantforceԦ�isdefinedastheproductofthe
forcecomponentandthedisplacements:W=F
s.s
Where�isthedisplacementand�
�thecomponentoftheforceonthe
directionofthedisplacement.
Iftheforceandthedisplacementmakeanangle�,then�
�=�cos�,
andtheworkcanbewrittenas:
TheS.IunitofworkistheJoule(J).1J=1���
2
�
−2
=1N.m
Notethatnodisplacementisproducedbythecomponentoftheforceperpendiculartothe
directionofmotion,thenW
F⊥
=0
CHAPTER 6: WORK AND ENERGY PREPARATORY YEAR 2015-2016 84
�=�.�.cos�

85. Workcanbepositive,negative,orzeroasillustratedinthefigurebelow
CHAPTER 6: WORK AND ENERGY PREPARATORY YEAR 2015-2016 85
Example6.1:A600�forceisappliedbyamantoadresserthatmoves2�.Findthework
doneiftheforceanddisplacementare:
(a)parallel
(b)atrightangles
(c)oppositelydirected(Fig.3);wemayimaginethatthedresserisbeingslowedandbroughtto
rest.

86. Solution:
(a)�and�areparallel,so�=0°andcos(0)=1.Theworkis:
�=�.�.����=(600�)×(2�)×(1)=1200�
Themandoes1200�ofworkonthedresser.SinceFisparalleltos,F
s=F,andweobtain
thesameresultusingtheequationW=F
s×s
(b)�and�areperpendicular,so�=90°andcos(90
0
)=0
�=�.�.����600�×2�×0=0
No work is done when the force is at right angles to the displacement, since �
�=0
c)�and�areopposite,so�=180°andcos(180°)=−1.Theworkthenis:
�=�.�.����=600�×2�×−1=−1200�
Inthiscasetheworkdonebytheforceisnegative,sotheobjectisdoingworkontheman.
NotethathereFisoppositetos,so�
�=−�.
CHAPTER 6: WORK AND ENERGY PREPARATORY YEAR 2015-2016 86

87. Example6.2:Ahorsepullsabargealongacanalwitharopeinwhichthetensionis1000�
(Fig.4).Theropeisatanangleof10⁰withthetowpathandthedirectionofthebarge.
(a)Howmuchworkisdonebythehorseinpullingthebarge100�upstreamata
constantvelocity?
(b)Whatisthenetforceonthebarge?
Solution6.2:
(a)Theworkdonebytheconstantforce??????inmovingthebargeadistance�isgivenby:
�=�×�×cos�=(1000�)×(100�)×cos(10⁰)=9.85×10
4
�
(b)Sincethebargemovesataconstantvelocity,thesumofallforcesonitmustbezero.
TheremustbeanotherforceactingthatisnotshowninFig.4.
CHAPTER 6: WORK AND ENERGY PREPARATORY YEAR 2015-2016 87

88. 6.2 Kinetic Energy
The kinetic energyof an object of mass �and velocity vis defined by :
�=
1
2
��
2
Example:A car of mass �=1200��moving with a speed of 120Τ��ℎ, what is its kinetic energy?
The speed first should be converted into ��.Then �=
1
2
×1200��×33.3�.�
−12
=665��
Thekineticenergy of an object is a measure of the work an object can do by virtue of its motion.
The work-energy principle
The change in the kinetic energy of an object (final kinetic energy minus the initial kinetic energy) is
equal to the total work(W) done on it by all the acting forces.
CHAPTER 6: WORK AND ENERGY PREPARATORY YEAR 2015-2016 88
If the work is positive, the Kinetic energy increasesK>K
oifW>0
If the work is negative, the Kinetic energy decreases K<K
oifW<0
if the total work done by all the forces iszerothe kinetic energy remainsconstant
K=K
oifW=0
∆&#3627408446;=&#3627408458;

89. Example6.3:
Awomanpushesacar,initiallyatrest,towardachildbyexertingaconstanthorizontalforceԦ&#3627408441;of
magnitude5&#3627408449;throughadistanceof1&#3627408474;(figure7).(a)Howmuchworkisdoneonthecar?(b)
Whatisitsfinalkineticenergy?(c)Ifthecarhasamassof0.1&#3627408472;&#3627408468;whatisitsfinalspeed?
(Assumenoworkisdonebyfrictionalforces)
Solution 6.3:
(a) The force the woman exerts on the car is parallel to the displacement, so the work she does
on the car is:
&#3627408458;=&#3627408441;×&#3627408480;=5&#3627408449;×1&#3627408474;=5&#3627408445;
(b) The initial kinetic energy is &#3627408446;
&#3627408476;=0, so the final kinetic energy is :
&#3627408446;=&#3627408446;
&#3627408476;+&#3627408458;=0+5&#3627408445;=5&#3627408445;
(c) The final kinetic energy is &#3627408446;=
1
2
&#3627408474;&#3627408483;
2
,so
&#3627408483;=
2&#3627408446;
&#3627408474;
=
2×5&#3627408445;
0.1&#3627408472;&#3627408468;
=10&#3627408474;/&#3627408480;
CHAPTER 6: WORK AND ENERGY PREPARATORY YEAR 2015-2016 89

90. 6.3 Potential energy
Potential energyistheenergy associated with the position or the configuration of a mechanical
system.
In the figure below the work done by the gravitational force of an object of mass &#3627408474;raised from
an initial height ℎ
&#3627408476;to a height ℎis : &#3627408458;
&#3627408468;=&#3627408474;&#3627408468;ℎ
&#3627408476;−&#3627408474;&#3627408468;ℎ
The quantity &#3627408474;&#3627408468;ℎis defined as potential energy: &#3627408456;=&#3627408474;&#3627408468;ℎ+&#3627408456;
&#3627408479;&#3627408466;&#3627408467;
&#3627408456;
&#3627408479;&#3627408466;&#3627408467;is the potential energy of reference. For simplicity we can write:
CHAPTER 6: WORK AND ENERGY PREPARATORY YEAR 2015-2016 90
&#3627408456;=&#3627408474;&#3627408468;ℎ(
The change in potential energy when an object of
mass &#3627408474;is raised from an initial height ℎ
&#3627408476;to a height
ℎis opposite to the work done by the gravitational
force:
∆&#3627408456;=&#3627408456;−&#3627408456;
&#3627408476;=−&#3627408458;
&#3627408468;

91. CHAPTER 6: WORK AND ENERGY PREPARATORY YEAR 2015-2016
6.4 Total Energy
The total work can be equal to the sum of the work done by the applied forces &#3627408458;
&#3627408462;and thework
done by the gravitational force&#3627408458;
&#3627408468;
By the work-energy theorem:&#3627408446;=&#3627408446;
&#3627408476;+&#3627408458;
&#3627408455;&#3627408476;&#3627408481;&#3627408462;&#3627408473;=&#3627408446;
&#3627408476;+&#3627408458;
&#3627408462;+&#3627408458;
&#3627408468;
As &#3627408458;
&#3627408468;=−&#3627408456;−&#3627408456;
&#3627408476;
Then &#3627408446;=&#3627408446;
&#3627408476;−&#3627408456;−&#3627408456;
&#3627408476;+&#3627408458;
&#3627408462;
The last equationcanbewrittenas : &#3627408446;+&#3627408456;=&#3627408446;
&#3627408476;+&#3627408456;
&#3627408476;+&#3627408458;
&#3627408462;(a)
The sum of kinetic energy and the potential energy is the total mechanical energy: &#3627408440;=&#3627408446;+&#3627408456;
Equation (a) becomes : &#3627408440;=&#3627408440;
&#3627408476;+&#3627408458;
&#3627408462;or:
∆&#3627408440;=&#3627408440;−&#3627408440;
&#3627408476;=&#3627408458;
&#3627408462;
if the work done by the applied forces is zero (&#3627408458;
&#3627408462;=0)and only the gravitational force is doing
work, the mechanical energy is constant or conserved:
∆&#3627408440;=0or&#3627408446;
&#3627408476;+&#3627408456;
&#3627408476;=&#3627408446;+&#3627408456;

92. Example6.5:Awomanskisfromrestdownahill
20&#3627408474;height.Iffrictionisnegligible,whatisherspeedat
thebottomoftheslope?
Solution6.5:
Wecansolvethisproblembyapplyingthemechanicalenergytheorem∆&#3627408440;=&#3627408458;
&#3627408462;where&#3627408458;
&#3627408462;
istheworkdonebythenonconservativeforce.Theforcesactingontheskierarethe
normalreactionandtheweight.Thenormalreactionistheonlynonconservative,itswork
&#3627408458;
&#3627408462;=0asitisperpendiculartothedisplacement.
ThentheMechanicalenergyisconserved:&#3627408440;
&#3627408476;=&#3627408440;
&#3627408440;
&#3627408476;=&#3627408446;
&#3627408476;+&#3627408456;
&#3627408476;=&#3627408474;&#3627408468;ℎ(&#3627408446;
&#3627408476;=0,becausetheinitialspeediszero)
&#3627408440;=&#3627408446;+&#3627408456;=
1
2
&#3627408474;&#3627408483;
2
(wecantake&#3627408456;=0atℎ=0,asareferencepotential
energy)
Then&#3627408440;
&#3627408476;=&#3627408440;becomes&#3627408474;&#3627408468;ℎ+0=0+
1
2
&#3627408474;&#3627408483;
2
Finally&#3627408483;=2&#3627408468;ℎ=2×9.80&#3627408474;&#3627408480;
−2
×20&#3627408474;=19.8&#3627408474;&#3627408480;
−1
CHAPTER 6: WORK AND ENERGY PREPARATORY YEAR 2015-2016 92

93. CHAPTER 6: WORK AND ENERGY PREPARATORY YEAR 2015-2016
6.7 Power
When a system develops work Wduring a period of time∆&#3627408481;, the average poweris defined by :
The power is expressed in joule per second in the SI system, which is called Watt
Electrical energy is sold by kilowatt hour (&#3627408524;&#3627408510;&#3627408521;).
&#3627409359;&#3627408524;&#3627408510;&#3627408521;=&#3627409359;&#3627409358;&#3627409358;&#3627409358;&#3627408536;&#3627408514;&#3627408533;&#3627408533;×&#3627409361;&#3627409364;&#3627409358;&#3627409358;&#3627408532;=&#3627409361;.&#3627409364;×&#3627409359;&#3627409358;
&#3627409364;
??????
&#3627408503;
&#3627408514;&#3627408535;=
&#3627408510;
∆&#3627408533;
Example6.14(P152):A70−&#3627408472;&#3627408468;manrunsupaflightofstairs3&#3627408474;highin2&#3627408480;.(a)Howmuch
workdoeshedoagainstgravitationalforces?(b)whatishisaveragepoweroutput?
Solution:
(a)Themandevelopsaworkagainstthegravitationalforce,then&#3627408458;
&#3627408474;&#3627408462;&#3627408475;=−&#3627408458;
&#3627408468;
Since&#3627408458;
&#3627408468;=−∆&#3627408456;
Hence&#3627408458;
&#3627408474;&#3627408462;&#3627408475;=∆&#3627408456;=&#3627408474;&#3627408468;ℎ=(70&#3627408472;&#3627408468;)×(9.80&#3627408474;&#3627408480;
−2
)×(3&#3627408474;)=2060&#3627408445;
(b)&#3627408451;=
&#3627408458;&#3627408474;&#3627408462;&#3627408475;
∆&#3627408481;
=
2060??????
2&#3627408480;
=1030&#3627408484;&#3627408462;&#3627408481;&#3627408481;

95. CHAPTER 13: MECHANICS OF FLUIDS PREPARATORY YEAR 2015-2016 95
Chapter 13
MECHANICS OF NON VISCOUS
FLUIDS
COURSE TOPICS:
13.1-Archimedes’Principle
13.2-The equation of continuity, Streamline flow
13.3-Bernoulli’s Equation
13.4-Static consequence of Bernoulli’s equation
Examples to be explained and solved:
13.1, 13.2, 13.4, and 13.6
Homework Problems:
13.3, 13.12, 13.19 and 13.25

96. Introduction
Inthischapterwediscussfluidsatrestandnon-viscous(frictionless)fluidsmotion.Wefirst
developanunderstandingofwhyanobjectmayeithersinkorfloatinafluidatrest?
(Archimedes’principle).WethendeveloptheBernoulli’sEquation,whichputsworkand
energyconceptsintoaformsuitableforfluids.Then,wecanunderstandwhyfluidsin
connectedcontainerstendtohavethesamesurfacelevels?Andhowfluidsflowfromone
placetoanother?
Onthisdiscussion,theimportantconditionistheassumptionthatthefluidisincompressible:
agivenmassoffluidalwaysoccupiesthesamevolumethoughitsshapemaychange.
Influidmechanics,asagivenmassoffluiddoesnothaveafixedshape,thedensityand
pressurearecommonlyusedinsteadofmassandforce.
CHAPTER 13: MECHANICS OF FLUIDS PREPARATORY YEAR 2015-2016 96

97. 13.1 Archimedes’ principle
An object floating or submerged in a fluid experiences an upward or Buoyant force due to the
fluid.
Considerasolidofvolume&#3627408457;anddensity&#3627409164;completelysubmergedina
fluidofdensity&#3627409164;
&#3627408476;.Thefluiddisplacedbythesolidhasamass&#3627408474;
??????
=&#3627409164;
&#3627408476;&#3627408457;
??????,thenitweighs&#3627408484;
&#3627408476;=&#3627408474;
??????&#3627408468;=&#3627409164;
&#3627408476;&#3627408457;
??????&#3627408468;.
Thebuoyantforceisthen:
CHAPTER 13: MECHANICS OF FLUIDS PREPARATORY YEAR 2015-2016 97
The buoyant force is the resultant force exerted by fluid on the surface of a submerged solid.
Example:What is the magnitude of the buoyant force exerted on a piece of solid of volume &#3627408457;
=15&#3627408464;&#3627408474;
3
completely immersed in water.
Archimedes’ principle states that the buoyant force &#3627408437;on the object is equal to the
weight of the displaced fluid:
&#3627408437;=&#3627409164;
&#3627408476;&#3627408457;
??????g
Figure 13.1 : A solid
submerge in a fluid

98. Example13.1:Apieceofmetalofunknownvolume&#3627408457;issuspendedfromastring.Before
submersion,thetensioninthestringis10&#3627408449;.Whenthemetalissubmergedinwaterthe
tensionis8N.Thewaterdensityis&#3627409164;
&#3627408476;=10
3
&#3627408472;&#3627408468;/&#3627408474;
3
.(a)calculatethebuoyantforce(b)
calculatethevolumeofthepieceofmetal(c)Whatisthedensityofthemetal?
solution:
a) Before submersion (Fig. 13.2 a) the tension of the cord is equal to the weight of the piece of
metal : &#3627408455;
??????=&#3627408484;=&#3627408474;g=10&#3627408449;
After submersion (Fig. 13.2 b), since the object is in equilibrium: &#3627408455;
&#3627408467;+&#3627408437;−&#3627408484;=0
Then the buoyant force is &#3627408437;=&#3627408484;−&#3627408455;
&#3627408467;=&#3627408455;
??????−&#3627408455;
&#3627408467;=10&#3627408449;−8&#3627408449;=2&#3627408449;
b) &#3627408437;=&#3627409164;
&#3627408476;×&#3627408457;
??????×&#3627408468;. The piece of metal is completely submerged in the fluid, then the volume
of the displaced fluid is equal to the volume of the solid: &#3627408457;
??????=&#3627408457;. Hence, &#3627408457;=
&#3627408437;
&#3627409164;&#3627408476;&#3627408468;
=
&#3627408455;
??????−&#3627408455;
??????
&#3627409164;&#3627408476;&#3627408468;
c) The density of the piece of metal can be then calculated through &#3627409164;=
&#3627408474;
&#3627408457;
=
&#3627409164;
&#3627408476;&#3627408455;
??????
&#3627408455;
??????−&#3627408455;
??????
=5000&#3627408472;&#3627408468;/&#3627408474;
3
CHAPTER 13: MECHANICS OF FLUIDS PREPARATORY YEAR 2015-2016 98

99. Partiallysubmergedsolid
Ifanobjectofvolume&#3627408457;
&#3627408480;isnotcompletelyimmersedinafluid,the
displacedvolumeisequaltothesubmergedvolumeofthesolid&#3627408457;
&#3627408480;&#3627408482;&#3627408463;
(volumeofthepartofthesolidbelowthetopsurfaceofthefluid).
Thenaquantitywithoutunitcalledsubmergedfractionisdefinedby
theratioofthesubmergedvolumeandthetotalvolumeofthesolid
&#3627408457;????????????&#3627408463;
&#3627408457;??????
Byequatingthebuoyantforceandtheweightoftheobject&#3627408437;=&#3627408484;,
weget:&#3627409164;
&#3627408476;×&#3627408468;×&#3627408457;
&#3627408480;&#3627408482;&#3627408463;=&#3627409164;
&#3627408480;×&#3627408468;×&#3627408457;
&#3627408454;
Thesubmergedfractionisthenequaltotheratioofthedensityof
thesolidtothedensityofthefluid:
&#3627408509;
&#3627408532;&#3627408534;&#3627408515;
&#3627408509;&#3627408532;
=
??????
&#3627408532;
??????&#3627408528;
CHAPTER 13: MECHANICS OF FLUIDS PREPARATORY YEAR 2015-2016 99
Example 13.2:The density of ice is 920 &#3627408472;&#3627408468;/&#3627408474;
3
while that
of sea water is 1025&#3627408472;&#3627408468;/&#3627408474;
3
. What fraction of an iceberg
is submerged?
Solution:
&#3627408457;
????????????&#3627408463;
&#3627408457;
??????&#3627408476;&#3627408473;????????????
=
&#3627409164;
??????&#3627408476;&#3627408473;????????????
&#3627409164;
??????&#3627408473;??????????????????
=920/1025=0.89
About 90 % of an iceberg issubmergedin the sea.

100. 13.2 Equation of continuity
Theflowrate&#3627408452;isthevolumeofthefluidflowingpastapointinachannelperunittime:&#3627408452;=
∆&#3627408457;
∆&#3627408481;
TheS.Iunitoftheflowrateisthe&#3627408526;
&#3627409361;
/&#3627408532;.
Foranincompressiblefluid(&#3627409164;=&#3627408464;&#3627408476;&#3627408475;&#3627408480;&#3627408481;&#3627408462;&#3627408475;&#3627408481;)thevolumeoffluidthatpassesanysectionofthe
tubepersecondisunchanged.Thefluidthatentersoneendofthechannelsuchasapipeoran
arteryattheflowrate&#3627408452;
1,mustleavetheotherendatarate&#3627408452;
2whichisthesame.Thusthe
equationofcontinuitycanbewrittenas&#3627408452;1=&#3627408452;2.
Considerasectionofthetubewithcross-sectional
area&#3627408436;(Fig.13.4a)andsupposethatthefluidonthis
sectionhasthesamevelocity.Inthetime∆&#3627408481;thefluid
movesthedistance∆&#3627408485;=&#3627408483;∆&#3627408481;andthevolumeofthe
fluidcrossingthetubeis&#3627408457;=&#3627408436;&#3627408483;∆&#3627408481;.Theflowrateis
then&#3627408452;=&#3627408436;&#3627408483;
Theflowrateequalsthecross-sectionalareatimes
thevelocityofthefluid.
Forachannelwhosecrosssectionchangesfrom&#3627408436;
1
to&#3627408436;
2,thisresulttogetherwith&#3627408452;1=&#3627408452;2gives
anotherformofthecontinuityequation:
&#3627408436;
1&#3627408483;
1=&#3627408436;
2&#3627408483;
2
CHAPTER 13: MECHANICS OF FLUIDS PREPARATORY YEAR 2015-2016 100
Figure 13.4

101. Example 13.4;
A water pipe leading up to a hose a radius of 1&#3627408464;&#3627408474;.Water leaves the hose at a rate of 3 litres per
minute.
1.Find the velocity of the water in the pipe.
2.The hose has a radius of 0.5&#3627408464;&#3627408474;. What is the velocity of the water in the hose?
Solution:
1. The velocity (strictly speaking, the average velocity) can be found from the flow rate and the
area: &#3627408452;=&#3627408436;&#3627408483;
The flow rate is the same in the hose and in the pipe.
Using 1&#3627408473;&#3627408470;&#3627408481;&#3627408479;&#3627408466;=10
−3
&#3627408474;
3
and 1min=60&#3627408480;, the flow rate is then:
&#3627408452;=
∆&#3627408457;
∆&#3627408481;
=
3×10
−3
&#3627408474;
3
60&#3627408480;
=5×10
−5
&#3627408474;
3
/&#3627408480;
We will call the velocity and area in the pipe &#3627408483;
1and &#3627408436;
1, respectively.
Then, with &#3627408452;=&#3627408436;&#3627408483;, we have:
&#3627408483;
1=
&#3627408452;
&#3627408436;
1
=
&#3627408452;
&#3627409163;.&#3627408479;
1
2=
5×10
−5
&#3627408474;
3
/&#3627408480;
&#3627409163;0.01&#3627408474;
2=0.159&#3627408474;&#3627408480;
−1
2. The flow rate is constant, so &#3627408436;
1&#3627408483;
1=&#3627408436;
2&#3627408483;
2,and the velocity &#3627408483;
2in the hose is
&#3627408483;
2=&#3627408483;
1&#3627408436;
1/&#3627408436;
2=&#3627408483;
1(&#3627408479;
1
2
/&#3627408479;
2
2
)=0.637&#3627408474;/s
CHAPTER 13: MECHANICS OF FLUIDS PREPARATORY YEAR 2015-2016 101

102. 13.3 Bernoulli’s equation
Bernoulli’s equation can be used for the following conditions :
1-The fluid is incompressible, then its density remains constant.
2-The fluid is non-viscous (no mechanical energy is lost).
3-The flow is streamline, not turbulent.
4-The velocity of the fluid at any point does not change during the period of
observation. (This is called the steady-state assumption.)
If the above conditions are satisfied, then Bernoulli’s equation states that the pressure
plus the total mechanical energy per unit volume is constant everywhere in the fluid.
&#3627408451;+
1
2
&#3627409164;&#3627408483;
2
+&#3627409164;&#3627408468;&#3627408486;=&#3627408464;&#3627408476;&#3627408475;&#3627408480;&#3627408481;&#3627408462;&#3627408475;&#3627408481;
CHAPTER 13: MECHANICS OF FLUIDS PREPARATORY YEAR 2015-2016 102
Bernoulli’s equation can be written between two
sections as :
&#3627408503;
&#3627409359;+
&#3627409359;
&#3627409360;
??????&#3627408535;
&#3627409359;
&#3627409360;
+??????&#3627408520;&#3627408538;
&#3627409359;=&#3627408503;
&#3627409360;+
&#3627409359;
&#3627409360;
??????&#3627408535;
&#3627409360;
&#3627409360;
+??????&#3627408520;&#3627408538;
&#3627409360;

103. Case Schematicrepresentation Bernoulli’s Equation
Horizontaltube withnon-
uniforme size
&#3627408486;
1=&#3627408486;
2
&#3627408503;
&#3627409359;+
&#3627409359;
&#3627409360;
??????&#3627408535;
&#3627409359;
&#3627409360;
=&#3627408503;
&#3627409360;+
&#3627409359;
&#3627409360;
??????&#3627408535;
&#3627409360;
&#3627409360;
Non-Horizontal tube with
uniformsize.
&#3627408436;
1=&#3627408436;
2
&#3627408436;
1&#3627408483;
1=&#3627408436;
2&#3627408483;
2
Then&#3627408483;
1=&#3627408483;
2
&#3627408503;
&#3627409359;+??????&#3627408520;&#3627408538;
&#3627409359;=&#3627408503;
&#3627409360;+??????&#3627408520;&#3627408538;
&#3627409360;
Staticfluid(&#3627408535;=&#3627409358;)
&#3627408503;
&#3627409359;+??????&#3627408520;&#3627408538;
&#3627409359;=&#3627408503;
&#3627409360;+??????&#3627408520;&#3627408538;
&#3627409360;
HydrostaticEquation
CHAPTER 13: MECHANICS OF FLUIDS PREPARATORY YEAR 2015-2016 103
Specific forms of the Bernoulli’s equation

104. 13.4 Static consequences Bernoulli’s equation
When the fluid is at rest (&#3627408535;=&#3627409358;), Bernoulli’s equationis written as:
&#3627408451;+&#3627409164;&#3627408468;&#3627408486;=&#3627408464;&#3627408476;&#3627408475;&#3627408480;&#3627408481;&#3627408462;&#3627408475;&#3627408481;
Pressure in a fluid at rest:
the last form of the Bernoulli’s equationcan be used to calculate the pressure everywhere in the
fluid. For example, from the figure find the pressure at a point &#3627408437;in terms of the pressure at
surface and the depth.
Using Bernoulli’s equation we can write:
At the surface &#3627408454;
&#3627408436;, &#3627408451;
&#3627408436;+&#3627409164;&#3627408468;&#3627408486;
&#3627408436;=&#3627408464;&#3627408476;&#3627408475;&#3627408480;&#3627408481;&#3627408462;&#3627408475;&#3627408481;and at the surface &#3627408454;
&#3627408437;, &#3627408451;
&#3627408437;+&#3627409164;&#3627408468;&#3627408486;
&#3627408437;=&#3627408464;&#3627408476;&#3627408475;&#3627408480;&#3627408481;&#3627408462;&#3627408475;&#3627408481;.
Then &#3627408451;
&#3627408436;+&#3627409164;&#3627408468;&#3627408486;
&#3627408436;=&#3627408451;
&#3627408437;+&#3627409164;&#3627408468;&#3627408486;
&#3627408437;or &#3627408451;
&#3627408437;=&#3627408451;
&#3627408436;+&#3627409164;&#3627408468;&#3627408486;
&#3627408436;−&#3627408486;
&#3627408437;=&#3627408451;
&#3627408436;+&#3627409164;&#3627408468;&#3627408465;
Ifthe pressure at the surface &#3627408454;
&#3627408436;is equal to the atmospheric pressure so &#3627408451;
&#3627408436;=&#3627408451;
&#3627408462;&#3627408481;&#3627408474;
then : &#3627408451;
&#3627408437;=&#3627408451;
&#3627408462;&#3627408481;&#3627408474;+&#3627409164;&#3627408468;&#3627408465;
CHAPTER 13: MECHANICS OF FLUIDS PREPARATORY YEAR 2015-2016 104

105. This result shows that pressure at a depth &#3627408465;in a fluid at rest is equal to the surface pressure plus
the potential energy density change &#3627409164;&#3627408468;&#3627408465;corresponding to this depth.
Calculating &#3627408451;+&#3627409164;&#3627408468;&#3627408486;at points &#3627408437;and &#3627408439;gives: &#3627408451;
&#3627408437;+&#3627409164;&#3627408468;&#3627408486;
&#3627408437;=&#3627408451;
??????+&#3627409164;&#3627408468;&#3627408486;
??????,since&#3627408486;
&#3627408437;=&#3627408486;
??????then
&#3627408451;
&#3627408437;=&#3627408451;
??????
Thus the pressure at the same depth at two places in a fluid at rest is the same. The surfaces of
liquids at rest in connected containers of any shape must be at the same height if they are open to
the atmosphere.
CHAPTER 13: MECHANICS OF FLUIDS PREPARATORY YEAR 2015-2016 105
Example :The pressure at the floor is measured to be normal atmospheric pressure, its
value is &#3627408451;
&#3627408462;&#3627408481;&#3627408474;=1.013&#3627408463;&#3627408462;&#3627408479;.
1)How much is the pressure at a height of 1000&#3627408474;.
2)You are in scuba diving at a 10&#3627408474;depth, you feel pain in the ears. Explain why?
Solution:
Here,&#3627408465;=1000&#3627408474;.FromTable13.1(p.315)thedensityofairatatmosphericpressure
and0°&#3627408438;is&#3627409164;=1.29&#3627408472;&#3627408468;.&#3627408474;
−3
Thus:&#3627408451;
&#3627408437;=&#3627408451;
&#3627408462;&#3627408481;&#3627408474;+&#3627409164;&#3627408468;&#3627408465;=1.013×10
5
&#3627408451;&#3627408462;−1.29&#3627408472;&#3627408468;.&#3627408474;
−3
×9.8&#3627408474;.&#3627408480;
−2
×1000&#3627408474;
=88.7&#3627408472;&#3627408451;&#3627408462;

106. CHAPTER 13: MECHANICS OF FLUIDS PREPARATORY YEAR 2015-2016
Pressuremeasurement:
Themanometer:theopen-tubemanometerisaU-shapedtubeusedformeasuringgas
pressures(orliquidpressureifdoesn’tmixwiththemanometerfluid).Itcontainsaliquidthat
maybemercuryor,formeasurementsoflowpressures,wateroroil.
InFig.13.7thepressureofthegas(thepressuretomeasure)isequaltothepressureonthe
liquidattheleftarm&#3627408451;
&#3627408436;=&#3627408451;
&#3627408468;&#3627408462;&#3627408480;
Attherightarmthepressureofthemercuryis&#3627408451;
&#3627408437;=&#3627408451;
&#3627408462;&#3627408481;&#3627408474;+&#3627409164;&#3627408468;ℎ
As&#3627408451;
&#3627408436;=&#3627408451;
&#3627408437;(samelevel),then:
&#3627408451;
&#3627408468;&#3627408462;&#3627408480;=&#3627408451;
&#3627408462;&#3627408481;&#3627408474;+&#3627409164;&#3627408468;ℎ
Thus, a measurement of the height difference ℎof the
two columns determines the gas pressure &#3627408451;
&#3627408468;&#3627408462;&#3627408480;.
Thegaugepressure:isthedifferencebetweentheabsolutepressureandtheatmospheric
pressure.Intheaboveequation&#3627408451;
&#3627408468;&#3627408462;&#3627408480;istheabsolutepressure,thenthegaugepressureis&#3627408451;
&#3627408468;
=&#3627408451;
&#3627408468;&#3627408462;&#3627408480;−&#3627408451;
&#3627408462;&#3627408481;&#3627408474;
&#3627408451;
&#3627408468;=&#3627408451;
&#3627408468;&#3627408462;&#3627408480;−&#3627408451;
&#3627408462;&#3627408481;&#3627408474;=&#3627409164;&#3627408468;ℎ
Forexample,thebloodpressuregivenbyasphygmomanometeristhegaugepressure&#3627409164;&#3627408468;ℎ
Figure 13.7

107. CHAPTER 13: MECHANICS OF FLUIDS PREPARATORY YEAR 2015-2016
Cannulation:Inmanyexperimentswithanesthetizedanimals,thebloodpressureinan
arteryorveinismeasuredbythedirectinsertionintothevesselofacannula,whichisa
smallglassorplastictubecontainingsalinesolutionplusananticlottingagent.
Thesalinesolution,inturn,isincontactwiththefluidinamanometer.It’snecessaryto
havethesurfaceofcontactbetweenthesalinesolutionandthemanometerfluideitherat
thesamelevelastheinsertionpointofthecannulaortocorrectfortheheightdifference.
The pressure at the artery is:&#3627408451;
&#3627408437;=&#3627408451;
&#3627408462;&#3627408481;&#3627408474;+&#3627409164;&#3627408468;ℎ−&#3627409164;
&#3627408480;&#3627408468;ℎ′
Where &#3627409164;
&#3627408480;is the density of the saline solution and &#3627409164;is the density of the manometer fluid.
Figure 13.8: Measurement of blood
pressure by Cannulation.

108. CHAPTER 13: MECHANICS OF FLUIDS PREPARATORY YEAR 2015-2016
13.5 The role of gravity in circulation
Humanshaveadaptedtotheproblemsofmovingbloodupwardalargedistanceagainstthe
forceofgravity.Animalsthathavenot,suchas,snakes,eelsandevenrabbits,willdieifheld
headupwards;thebloodremainsinthelowerextremitiesandtheheartreceivesnoblood
fromthevenoussystem.Figure13.9showsthatintherecliningpositionthepressuresare
almostthesameeverywhere,thesmallpressuredropisduetotheviscousforces.Howeverthe
pressuresarequitedifferentinthestandingperson.

109. SincetheviscouseffectsaresmallwecanusetheBernoulli’sequation:
&#3627408451;+
1
2
&#3627409164;&#3627408483;
2
+&#3627409164;&#3627408468;ℎ=&#3627408464;&#3627408476;&#3627408475;&#3627408480;&#3627408481;&#3627408462;&#3627408475;&#3627408481;.Thevelocitiesatthethreearteriesareroughlyequal,sothe
term
1
2
&#3627409164;&#3627408483;
2
canbeignored.Hencethegaugepressuresattheheart&#3627408451;
??????,atthefoot&#3627408451;
??????andat
thebrain&#3627408451;
&#3627408437;arerelatedby:
&#3627408451;
??????=&#3627408451;
??????+&#3627409164;&#3627408468;ℎ
??????=&#3627408451;
&#3627408437;+&#3627409164;&#3627408468;ℎ
&#3627408437;
Typicalvaluesforadultscanbecalculatedforℎ
??????=1.3&#3627408474;,ℎ
&#3627408437;=1.7&#3627408474;andadensityofblood
&#3627409164;=1059.5&#3627408472;&#3627408468;/&#3627408474;
3
:
Wefindthat:
&#3627408451;
??????−&#3627408451;
??????=&#3627409164;&#3627408468;ℎ
??????=13.5&#3627408472;&#3627408451;&#3627408462;
&#3627408451;
??????−&#3627408451;
&#3627408437;=&#3627409164;&#3627408468;ℎ
&#3627408437;=1.7&#3627408472;&#3627408451;&#3627408462;
The pressures in the lower and upper parts of the body are very different when the person
is standing, although they are about equal when reclining.
Bloodreturnedtotheheart,atleastpartially,bythepumpingactionassociatedwith
breathingandbyflexingofskeletalmuscle,asinwalking.
Theimportanceoftheroleofgravityinthecirculationisillustratedbythefactthatsoldier
whoisrequiredtostandatstrictattentionmayfaintbecauseofinsufficientvenousreturn.
Toregainconsciousness,positionhastobealteredtohorizontal,thenpressureisequalized.
CHAPTER 13: MECHANICS OF FLUIDS PREPARATORY YEAR 2015-2016 109

110. EffectofAcceleration:
Whenapersonexperiencesanupwardacceleration&#3627408462;,hiseffectiveweightbecomes
&#3627408484;=&#3627408474;&#3627408468;+&#3627408462;.ApplyingBernoulli’sequationtothebrainandtheheartwith&#3627408468;replacedby
(&#3627408468;+&#3627408462;)weobtain:
&#3627408451;
??????+&#3627409164;(&#3627408468;+&#3627408462;)ℎ
??????=&#3627408451;
&#3627408437;+&#3627409164;(&#3627408468;+&#3627408462;)ℎ
&#3627408437;
Thenthepressureatthebrainwillbe:&#3627408451;
&#3627408437;=&#3627408451;
??????−&#3627409164;(&#3627408468;+&#3627408462;)(ℎ
&#3627408437;−ℎ
??????)
Thusthebloodpressureinthebrainwillbereducedevenfarther.Ithasbeenfoundthatifthe
accelerationistwoorthreetimes&#3627408468;(&#3627408462;=2&#3627408468;&#3627408476;&#3627408479;&#3627408462;=3&#3627408468;),ahumanwilllooseconsciousness
becauseofthecollapseofarteriesinthebrain.
Arelatedexperienceisthefellingoflight-headednessthatsometimesoccurswhenone
suddenlystandsup.
CHAPTER 13: MECHANICS OF FLUIDS PREPARATORY YEAR 2015-2016 110

111. CHAPTER 13: MECHANICS OF FLUIDS PREPARATORY YEAR 2015-2016
Duringacompleteheartpumpingcycle,thepressureintheheartandthecirculatorysystem
goesthroughbothamaximumSystolic(asthebloodispumpedfromtheheart)anda
minimumDiastolic(astheheartrelaxesandfillswithbloodreturnedfromtheveins)the
sphygmomanometerisusedtomeasuretheseextremepressures.
Thesphygmomanometer(Figure13.10)isusedintheupperhumanarmwhereitgivesvalues
nearlyclosetothepressureintheheart.Also,theupperarmcontainsasinglebonemakingthe
brachialarterylocatedthereeasytocompress.
Bloodpressuresareusuallypresentedas&#3627408454;&#3627408486;&#3627408480;&#3627408481;&#3627408476;&#3627408473;&#3627408470;&#3627408464;/&#3627408439;&#3627408470;&#3627408462;&#3627408480;&#3627408481;&#3627408476;&#3627408473;&#3627408470;&#3627408464;ratios.Typicalreadingsfor
restinghealthyadultareabout120/80intorr(mmHg),theborderlineforhighpressure
(hypertension)isusuallydefinedtobe140/90.Pressuresabovethatlevelneedsmedical
attention.
13.6 Blood Pressure-Sphygmomanometer

113. CHAPTER 14: VISCOUS FLUIDS PREPARATORY YEAR 2015-2016
113
Chapter 14
VISCOUS FLUIDS
COURSE TOPICS:
14.1 Viscosity
14.2Flow in circulatory system
14.3 Flow resistance
Examples to be explained and solved:
14.1, 14.5 and 14.6
Homework Problems:
14.3, 14.23 and 14.39

114. Introduction
Fluidsinmotionexhibitsomeeffectsoffrictionalorviscousforces.Whenevertheworkdone
againstthesedissipativeforcesiscomparabletothetotalworkdoneonthefluidorits
mechanicalenergy,Bernoulli’sequationcannotbeused.Nevertheless,itcanbeusedto
describeadequatelytheflowofbloodinthelargemainarteriesofamammal,butnotinthe
narrowerbloodvessels.
TheViscosityofafluidisameasureofitsresistancetoflowunderanappliedforce.Thegreater
theviscosity,thelargertheforcerequiredtomaintaintheflow,andthemoretheenergythatis
dissipated.Molasseshasahighviscosity,watersmallviscosityandairstillsmallerviscosity.
Webeginthischapterwithdefiningviscosity.Wethenexaminetheeffectsofviscousforcesin
theflowofthefluidinatube.Wethenexaminetheeffectofviscousforcesonheflowofafluid
inatube.
Viscosityisresponsibleforthedrag
forceexperiencedbyobjectmoving
throughafluid.
CHAPTER 13: MECHANICS OF FLUIDS PREPARATORY YEAR 2015-2016 114

115. 14.1 Viscosity
Viscosityisreadilydefinedbyconsideringasimpleexperiment.Thefigureshowtwoplates
separatedbyathinfluidlayer.Thelowerplateisheldfixed.Aforceisrequiredtomovetheupper
plateataconstantspeed.Thisforceisneededtoovercometheviscousforcesduetotheliquid
andisgreaterforahighlyviscousfluid.
Theforce&#3627408441;isobservedtobeproportionaltotheareaoftheplates&#3627408436;andtothevelocity
gradient
∆&#3627408483;
∆&#3627408486;
TheproportionalityfactoristhecoefficientofviscosityrepresentedbytheGreekletter
“eta”).
TheS.Iunitofviscosityisthe&#3627408472;&#3627408468;.&#3627408474;
−1
.&#3627408480;
−1
=&#3627408451;&#3627408462;.&#3627408480;
Thelargertheviscosity,thelargerforceneededtomovetheplateataconstantspeed.
CHAPTER 14: VISCOUS FLUIDS PREPARATORY YEAR 2015-2016
115
&#3627408441;=&#3627409154;&#3627408436;
Δ&#3627408483;
Δ&#3627408486;

116. The Relation between Viscosity and Temperature
•As the temperature increases,viscosity decreases for liquids.
•As the temperature increases, viscosity increases for gases.
•Because viscous forces are usually small, fluids are often used as lubricants to reduce friction.
Table 1: Typical values of viscosity in Pa.S
Temperature ℃ Castor Oil Water Air Normal BloodBlood Plasm
0 5.3 1.792×10
−3
1.71×10
−5
20 0.986 1.005×10
−3
1.81×10
−5
3.015×10
−3
1.810×10
−3
37 ------ 0.695×10
−3
1.87×10
−5
&#3627409360;.&#3627409358;&#3627409366;&#3627409362;×&#3627409359;&#3627409358;
−&#3627409361;
1.257×10
−3
40 0.231 0.656×10
−3
1.90×10
−5
60 0.080 0.469×10
−3
2.00×10
−5
80 0.030 0.357×10
−3
2.09×10
−5
100 0.017 0.287×10
−3
2.18×10
−5
CHAPTER 14: VISCOUS FLUIDS PREPARATORY YEAR 2015-2016
116

117. Example14.1page342
Anairtrackusedinphysicslecturedemonstrations,supportsacartthatridesonathincushionof
air1&#3627408474;&#3627408474;thickand0.04&#3627408474;
2
inarea.Iftheviscosityoftheairis1.8×10
−5
&#3627408451;&#3627408462;.&#3627408480;,findtheforce
requiredtomovethecartataconstantspeedof0.2&#3627408474;/&#3627408480;.
Solution:
Theforcerequiredis:&#3627408441;=&#3627409154;&#3627408436;
Δ&#3627408483;
Δ&#3627408486;
Then&#3627408441;=1.8×10
−5
&#3627408451;&#3627408462;.&#3627408480;0.04&#3627408474;
2
0.2&#3627408474;&#3627408480;
−1
10
−3
&#3627408474;
=1.44×10
−4
&#3627408449;
Thisisaverysmallforceandisconsistentwiththeobservationthatanairtrackis
frictionless.
CHAPTER 14: VISCOUS FLUIDS PREPARATORY YEAR 2015-2016
117

118. 14.2 The flow in the circulatory system
TheBlood
Bloodisconstantlyinmotion.Asleeporawakethebloodflowsinacirculationsystemat
almostthesamerate.Itbringsoxygenandnutritivesubstancestothecapillaries(smallest
bloodvessels)andremovesmetabolicwasteproductsandcarbondioxide,whicharethen
eliminatedfromthebodybytheexcretoryorgans.Thebloodcoordinatesactivitiesofvarious
organsbycarryingchemicalregulatorscalledhormones.Bloodregulatesbodytemperature
andprotectsthebodyagainstdisease.Itshasaliquidportionandasolidportion.Theliquid
portioniscalledplasma,whichisabout55%oftheblood'svolume.Thesolidportion(redcells,
whitecells,andothervitalfactors)makesuptheremaining45%.
Forourpurposes,itissufficienttotreatbloodasauniformfluidwithviscosity
&#3627409154;=2.084×10
−5
&#3627408451;&#3627408462;.&#3627408480;andadensity&#3627409164;=1059.5Τ&#3627408472;&#3627408468;&#3627408474;
3
atnormalbodytemperature.
Bloodaccountsfor7to9percentofthetotalbodyweight.Apersonweighing70kgwill
haveabout4to6litersofblood.
CHAPTER 14: VISCOUS FLUIDS PREPARATORY YEAR 2015-2016
118

119. Thecardiovascularsystem
Thecardiovascularsystemincludestheheart
(pump),andanextensivesystemofarteries,
vascularbedscontainingcapillaries,andveins
(vessels).Thearteriescarryblood(richin
oxygen)totheorgans,musclesandskin.Veins
transportthereturnedblood(poorinoxygen)
totherightatrium,andpassestotheright
ventricletobepumpedagainthroughthe
pulmonaryarterytothelungsforre-
oxygenation.Theleftatriumreceivesthenewly
oxygenatedbloodfromthelungsaswellasthe
pulmonaryveinwhichispassedtothestrong
leftventricletobedistributedthroughtheaorta
tothedifferentorgansofthebody.
CHAPTER 14: VISCOUS FLUIDS PREPARATORY YEAR 2015-2016
119

120. Table 2: Propertiesof the human cardiovascular system for typical adult. All pressures listed are
gauge pressures (1&#3627408462;&#3627408481;&#3627408474;=101.3&#3627408472;&#3627408451;&#3627408462;)
Meanpressure in large arteries 12.8&#3627408472;&#3627408451;&#3627408462;
Mean pressure in largeveins 1.07&#3627408472;&#3627408451;&#3627408462;
Volume blood (70-kg man) 5.2&#3627408473;&#3627408470;&#3627408481;&#3627408479;&#3627408466;&#3627408480;=5.2×10
−3
&#3627408474;
3
Time required for complete circulation (resting) 54 seconds
Heart flow rate 9.7×10
−5
&#3627408474;
3
&#3627408480;
−1
Viscosity of blood (37°C) 2.084×10
−3
&#3627408451;&#3627408462;.&#3627408480;
Density of blood(37 °C) 1059.5&#3627408472;&#3627408468;&#3627408474;
−3
CHAPTER 14: VISCOUS FLUIDS PREPARATORY YEAR 2015-2016
120

121. 14.3 The flow Resistance
TheFlowresistanceisdefinedingeneral,astheratioofthepressuredrop∆&#3627408451;totheflowrate&#3627408452;,
&#3627408453;
&#3627408467;=∆&#3627408451;/&#3627408452;,whethertheflowislaminarornot.
Forlaminarflow,theflowresistancecanbecalculatedusingthePoiseuill’slaw:
&#3627408453;
&#3627408467;=
8&#3627409154;&#3627408473;
&#3627409163;&#3627408453;
4
According to the Bernoulli’s theorem, for non viscous fluid flowing in a horizontal tube with
constant cross section the pressure is constant along the tube: &#3627408451;
1=&#3627408451;
2, then ∆&#3627408451;=0and &#3627408453;
&#3627408467;is
zero.
Nevertheless for aviscous fluid, a pressure drop is observed between the two cross sections,
this pressure drop is proportional to the flow rate &#3627408452;and defined by:
∆&#3627408451;=&#3627408451;
1−&#3627408451;
2=&#3627408453;
&#3627408467;&#3627408452;
CHAPTER 14: VISCOUS FLUIDS PREPARATORY YEAR 2015-2016
121

122. Example14.5page348:
Theaortaofanaverageadulthumanhasaradius1.3×10
−2
&#3627408474;.Whataretheresistanceand
thepressuredropovera0.2mdistance,assumingaflowrateof&#3627408436;=10
−4
&#3627408474;
3
&#3627408480;
−1
.
Solution:
&#3627408453;
&#3627408467;=
8&#3627409154;&#3627408473;
&#3627409163;&#3627408453;
4
=
8(2.084×10
−3
&#3627408451;&#3627408462;.&#3627408480;)(0.2&#3627408474;)
&#3627409163;(1.3×10
−2
&#3627408474;)
4
=37.2×10
4
&#3627408451;&#3627408462;.&#3627408480;.&#3627408474;
−3
The pressure drop over the distance 0.2 m is:
∆&#3627408451;=&#3627408453;
&#3627408467;&#3627408452;=(37.2×10
4
&#3627408451;&#3627408462;.&#3627408480;.&#3627408474;
−3
)(10
−4
&#3627408474;
3
&#3627408480;
−1
)=3.72Pa
Thisisverysmallvalueofthepressuredrop,comparedtothetotalpressuredropinthe
system,whichisabout13.3&#3627408472;&#3627408451;&#3627408462;.Mostoftheflowresistanceandpressuredropsoccurinthe
smallerarteriesandvascularbedsofthebody(Table14.4).
&#3627408473;&#3627408462;&#3627408475;&#3627408465;&#3627408453;arerespectivelythelengthandtheradiusofthetube,&#3627409154;istheviscosityofthe
fluid.Usuallytheflowresistanceinalargearteryissmall.Consequently,thepressuredrop
insucharteriesissmall.
Theunitofflowresistanceisthe&#3627408451;&#3627408462;.&#3627408480;&#3627408474;
−3
.
CHAPTER 14: VISCOUS FLUIDS PREPARATORY YEAR 2015-2016
122

123. Flowresistanceinbloodvessels
Theflowresistanceofacollectionofarteriescanbecalculatedbyconsideringeach
categoryofarteriesseparately(inseriesorinparallel).
Vesselsinparallel:
Supposetwovesselsinparallel(Fig.14.7)eachonecarriesitsequalshareofthe&#3627408453;
&#3627408467;
and&#3627408452;:∆&#3627408451;isthepressureacrossallofthearteries,then:
CHAPTER 14: VISCOUS FLUIDS PREPARATORY YEAR 2015-2016
123
&#3627408452;=&#3627408452;
1+&#3627408452;
2=
∆&#3627408451;
&#3627408453;
??????1
+
∆&#3627408451;
&#3627408453;
??????2
=∆&#3627408451;(
1
&#3627408453;
??????1
+
1
&#3627408453;
??????2
)=
∆&#3627408451;
&#3627408453;&#3627408477;
Anequivalentflowresistanceisobtainedby:
&#3627409359;
&#3627408505;
&#3627408529;
=
&#3627409359;
&#3627408505;
&#3627408519;&#3627409360;
+
&#3627409359;
&#3627408505;
&#3627408519;&#3627409360;
Onecaneasilyobservetheanalogywiththe
electricequivalentresistanceinelectriccircuitsin
parallel.
ForNvesselsinparallel:
&#3627409359;
&#3627408505;
&#3627408529;
=
&#3627409359;
&#3627408505;
&#3627408519;&#3627409360;
+
&#3627409359;
&#3627408505;
&#3627408519;&#3627409360;
+⋯….
&#3627409359;
&#3627408505;
&#3627408519;&#3627408501;
Figure 14.7 schematic representation
of two vessels in parallel
For &#3627408449;identical arteries, the equivalent flow resistance is:&#3627408453;
&#3627408477;=
&#3627408453;
??????
??????

124. Vesselsinseries:
Giventwovesselsconnectedinseries(Fig.14.8),thetotalpressuredropbetweentheendsof
thesevesselsis:
∆&#3627408451;=∆&#3627408451;
1+∆&#3627408451;
2=&#3627408452;×&#3627408453;
&#3627408467;1+&#3627408452;×&#3627408453;
&#3627408467;2=&#3627408452;×&#3627408453;
&#3627408467;1+&#3627408453;
&#3627408467;2=&#3627408452;×&#3627408453;
&#3627408480;
CHAPTER 14: VISCOUS FLUIDS PREPARATORY YEAR 2015-2016
124
Anequivalentflowresistanceisobtainedby:
&#3627408505;
&#3627408532;=&#3627408505;
&#3627408519;&#3627409359;+&#3627408505;
&#3627408519;&#3627409360;
Onecanalsoobservetheanalogywiththeelectricequivalentresistanceinelectriccircuitsin
series
ForNvesselsinseries:
&#3627408505;
&#3627408532;=&#3627408505;
&#3627408519;&#3627409359;+&#3627408505;
&#3627408519;&#3627409360;+⋯&#3627408505;
&#3627408519;&#3627408501;
Figure 14.8 Two vessels in series

125. Solution
Theresistanceofonecapillaryis:
Thenfor&#3627408449;=4.73×10
7
&#3627408464;&#3627408462;&#3627408477;&#3627408470;&#3627408473;&#3627408473;&#3627408462;&#3627408479;&#3627408470;&#3627408466;&#3627408480;inparallel:
&#3627408453;
&#3627408477;=
&#3627408453;
&#3627408467;
&#3627408449;
=
2.073×10
16
4.73×10
7
=4.38×10
5
&#3627408472;&#3627408451;&#3627408462;.&#3627408480;.&#3627408474;
−3
Example14.6page349:
FromTable14.2,theradiusofasinglecapillaryis4×10
−6
&#3627408474;anditslengthis10
−3
&#3627408474;.
Whatistheresistanceof4.73×10
7
capillariesinthemesentericvascularbedofadogif
theyareassumedtobeparallel?
CHAPTER 14: VISCOUS FLUIDS PREPARATORY YEAR 2015-2016
125

127. Chapter 24
MIRRORS, LENSES AND
IMAGING SYSTEMS
COURS TOPICS:
24.4 The Power of a Lens
24.7 The Human Eye
24.13 Optical Defects of the Eye
Examples to be explained and solved:
24.4, 24.6, 24.7, 24.12, and 24.13
Homework Problems:
24.63, 24.64, 24.67 and 24.68
CHAPTER 14: OPTICS PREPARATORY YEAR 2015-2016 127

128. Introduction
Cameras,microscopes,telescopesandthehumaneyeare
examplesofopticalsystemsthatemploylensesandinsome
casesmirrors.
Ordinarilylensesandmirrorsarelargecomparedtothe
wavelengthofthevisiblelight,sotheirprincipleeffectson
beamsoflightcanbediscussedwithoutincludinginterference
ordiffractionphenomena.Imageformationbysuch
instrumentscanbepreciselydeterminedby:
-Consideringthatthelightpropagatesinastraightline.
-Knowingthepropertiesofmirrorsandlenses.
Alongthispresentationwefocusonthehumaneyeandsome
ofitsdefects.
Optical microscope
Telescope
Human Eye
CHAPTER 14: OPTICS PREPARATORY YEAR 2015-2016 128

129. Lenses-definition and characteristics
Alensisapieceoftransparentmediumthatcanfocusatransmittedbeamoflightsoan
imageisformed.Thelensesinman-madeopticalinstrumentsareusuallymanufactured
fromglassorplastic,whilethelensinthehumaneyeisformedbyatransparent
membranefilledwithaclearfluid.Forourpurposeitssufficienttoconsiderthin,spherical
lenses.Thesehavetwosphericalsurfacesorasphericalsurfaceandaplanarsurface.The
thicknessissmallcomparedtotheradiiofthesurfaces.
CHAPTER 14: OPTICS PREPARATORY YEAR 2015-2016 129

130. Wecancategorizealllensesaseitherconvergingordiverging.Aconverginglensisthickeratits
centerthanattheedges,whiletheinverseistrueforadiverginglens.Aconverginglensbends
lightraystowarditsaxis,itrefractsparallelrayssothattheimageisformedatafocalpointF’
beyondthelens(figure24.2a),whileadiverginglenswillbendtheraysoutwardsothatthey
appeartocomefromafocalpointF’beforethelens(figure24.2b).Thedistancefromthecenter
ofthelenstothefocalpointiscalledfocallength.
The focal length is conventionally positive for converging lens and negative for a diverging lens.
CHAPTER 14: OPTICS PREPARATORY YEAR 2015-2016 130
Figure 24.2 converging and diverging lenses

131. Thefocallength
Thefocallength&#3627408467;dependsontheindexofrefraction&#3627408475;ofthelensandontheradiiofcurvature
(&#3627408453;
1&#3627408462;&#3627408475;&#3627408465;&#3627408453;
2)ofitssurfaces,itcanbecalculatedunsingThelensmarker’sformula:
1
&#3627408467;
=&#3627408475;−1
1
&#3627408453;1
+
1
&#3627408453;2
Thefollowingconventionsareusedtocharacterizelens
surfaces:
Aconvexsurfacehasapositiveradiusofcurvature.
Aconcavesurfacehasanegativeradiusofcurvature.
Aplanesurfacehasaninfiniteradiusofcurvature.
Thinlensformula
Theimageposition&#3627408480;’andtheobjectposition&#3627408480;
throughathinlensarerelatedbytheequation:
1
&#3627408454;
+
1
&#3627408454;′
=
1
&#3627408467;
CHAPTER 14: OPTICS PREPARATORY YEAR 2015-2016 131

132. 24.4 Power of lens, Aberration
In discussing lenses, it is often more convenient to deal with the reciprocal of the focal length f,
which is called the power of the lens (in Diopters: 1&#3627408439;&#3627408470;&#3627408476;&#3627408477;&#3627408481;&#3627408466;&#3627408479;=1&#3627408474;
−1
):
&#3627408451;=
1
&#3627408467;
For example,a lens of focal length &#3627408467;=−20&#3627408464;&#3627408474;has a power &#3627408451;=
1
−0.20&#3627408474;
=−2,5&#3627408439;&#3627408470;&#3627408476;&#3627408477;&#3627408481;&#3627408466;&#3627408479;&#3627408480;
A short focal-length lens, which bends light through large angle, has a large power.
CHAPTER 14: OPTICS PREPARATORY YEAR 2015-2016 132
Association of lenses Two thin lenses with focal lengths&#3627408467;
1and &#3627408467;
2placed next to
each other are equivalent to a single lens with a focal length &#3627408467;satisfying:
&#3627409359;
&#3627408519;
=
&#3627409359;
&#3627408519;&#3627409359;
+
&#3627409359;
&#3627408519;&#3627409360;
Alternatively, with &#3627408451;
1=
1
&#3627408467;1
and &#3627408451;
2=
1
&#3627408467;2
, the power of the pair of lenses is :
&#3627408503;=&#3627408503;
&#3627409359;+&#3627408503;
&#3627409360;
The powers of lenses in contact are simply added to find the net power.
For example, an ophthalmologist placing 3−&#3627408465;&#3627408470;&#3627408476;&#3627408477;&#3627408481;&#3627408466;&#3627408479;and 0.25−&#3627408465;&#3627408470;&#3627408476;&#3627408477;&#3627408481;&#3627408466;&#3627408479;lenses in front of a patient’s eye
immediately knows that the combination is equivalent to a single 3.25−&#3627408465;&#3627408470;&#3627408476;&#3627408477;&#3627408481;&#3627408466;&#3627408479;lens.

133. Aberration
Anylenssuffersfromvariouskindsofaberrations,whichlimitthesharpnessofitsimages.Inthis
subsection,wearegoingtoseetwotypesofaberrationsandhowtheycanbeminimizedor
eliminated.
Monochromaticaberrationsor(sphericalaberrations):occurforlightofsinglewavelength.
Lightraysneartheaxisofthelensarefocusedfartherthantheraysneartheedgeand
causetheimagetohaveasmalldiameter.
CHAPTER 14: OPTICS PREPARATORY YEAR 2015-2016 133

134. Chromaticaberration:Whenanobjectisilluminatedwithwhitelight,ifitsimageonascreenis
infocusforonecolourcomponent,itwillbeslightlyoutoffocusfortheothers.
CHAPTER 14: OPTICS PREPARATORY YEAR 2015-2016 134

135. Eliminatingaberration
Wecanillustratethesecancellationsbyconsideringadoublet,twolensesincontact.Lens
1hastwoconvexsidesandismadefromcrownglass.Lens2hasoneflatsideandone
concavesideandismadefromflintglass.Allofthecurvedsurfaceshaveradiiofcurvature
of10&#3627408464;&#3627408474;.Inbothtypesofglass,therefractiveindexvariesabout1percentoverthe
visiblespectrum.ThepowersP1andP2canbecalculatedusingthelensmaker’sequation:
&#3627408451;=
1
&#3627408467;
=&#3627408475;−1
1
&#3627408453;
1
+
1
&#3627408453;
2
The power of the two lenses when they are
placed next to each other is: &#3627408451;=&#3627408451;
1+&#3627408451;
2
CHAPTER 14: OPTICS PREPARATORY YEAR 2015-2016 135

136. Calculating the Power for each lens using the data available in the T able below
Crown lens:
Power of the red light : &#3627408451;
1&#3627408453;=1.517−1
1
0.1
+
1
0.1
=10.34&#3627408439;&#3627408470;&#3627408476;&#3627408477;&#3627408481;&#3627408466;&#3627408479;&#3627408480;
Power of the yellow light &#3627408451;
1??????=1.520−1
1
0.1
+
1
0.1
=10.40&#3627408439;&#3627408470;&#3627408476;&#3627408477;&#3627408481;&#3627408466;&#3627408479;&#3627408480;
Power of the Bleu light &#3627408451;
1&#3627408437;=1.527−1
1
0.1
+
1
0.1
=10.54&#3627408439;&#3627408470;&#3627408476;&#3627408477;&#3627408481;&#3627408466;&#3627408479;&#3627408480;
Flint lens:
Power of the red light &#3627408451;
2&#3627408453;=1.644−1−
1
0.1
+
1
∞
=−6.44&#3627408439;&#3627408470;&#3627408476;&#3627408477;&#3627408481;&#3627408466;&#3627408479;&#3627408480;
Power of the yellow light &#3627408451;
2??????=1.650−1−
1
0.1
+
1
∞
=−6.50&#3627408439;&#3627408470;&#3627408476;&#3627408477;&#3627408481;&#3627408466;&#3627408479;&#3627408480;
Power of the Bleu light &#3627408451;
2&#3627408437;=1.664−1−
1
0.1
+
1
∞
=−6.64&#3627408439;&#3627408470;&#3627408476;&#3627408477;&#3627408481;&#3627408466;&#3627408479;&#3627408480;
CHAPTER 14: OPTICS PREPARATORY YEAR 2015-2016 136
Refractiveindice Power (&#3627408491;&#3627408522;&#3627408528;&#3627408529;&#3627408533;&#3627408518;&#3627408531;&#3627408532;)
Wave length Crown lens L1 Flint lens L2 P1 ( crown) P2 (Flint) &#3627408451;=&#3627408451;1+&#3627408451;2
Red (656&#3627408475;&#3627408474;) 1.517 1.644 10.34 -6.44 3.90
Yellow (589&#3627408475;&#3627408474;) 1.520 1.650 10.40 -6.50 3.90
Blue (486nm) 1.527 1.664 10.54 -6.64 3.90

137. CHAPTER 14: OPTICS PREPARATORY YEAR 2015-2016 137
Eliminating chromatic
aberration with a doublet: two
lenses in contact

138. 24.7 The human eye
Theeyeballisapproximatelysphericalinshapewithadiameterabout2.3&#3627408464;&#3627408474;.Theshapeand
thefocallengthofthecrystallinelensarecontrolledbytheciliarymuscles.Iftheciliary
musclesarerelaxedthefrontsurfaceofthelensiskeptrelativelyflatandlightfromdistant
objectsisfocusedontheretina.Whentheciliarymusclescontractthelensassumemore
roundedshapeanditsfocallengthdecreasesbingingthenearbyintofocusontheretina.The
abilityofthelenstoadjustitsfocallengthiscalledaccommodation.
Medium Refractiveindex
Air 1
Water 1.333
Humors 1.336
Cornea -------
Crystalline 1.437
CHAPTER 14: OPTICS PREPARATORY YEAR 2015-2016 138
Table 1: Refractive index of some
substances

139. Power of Accommodation
The power of accommodationof the eye is the maximum variation of its power for focusing
on nearobjects &#3627408451;
&#3627408475;and distant(far) objects &#3627408451;
&#3627408467;.
CHAPTER 14: OPTICS PREPARATORY YEAR 2015-2016 139
At the far point (a person sees objects clearly at a distance &#3627408485;
&#3627408467;), the power P
fof the eye is:
&#3627408451;
&#3627408467;=
1
&#3627408485;
??????
+
1
??????
,&#3627408439;&#3627408470;&#3627408480;&#3627408481;ℎ&#3627408466;&#3627408465;&#3627408470;&#3627408462;&#3627408474;&#3627408466;&#3627408481;&#3627408466;&#3627408479;&#3627408476;&#3627408467;&#3627408481;ℎ&#3627408466;&#3627408466;&#3627408486;&#3627408466;&#3627408463;&#3627408462;&#3627408473;&#3627408473;:&#3627408439;≈2&#3627408464;&#3627408474;
For a person with normal vision the far point is at the infinity (&#3627408485;
&#3627408467;=), thenitspower is:
&#3627408451;
&#3627408467;=
1
∞
+
1
0.02
=50&#3627408465;&#3627408470;&#3627408476;&#3627408477;&#3627408481;&#3627408466;&#3627408479;&#3627408480;
When the eye adjusts its focal length so that it focuses on an object at the near point (the
object distance is &#3627408485;
&#3627408475;), the power of the eye is:
&#3627408451;
&#3627408475;=
1
&#3627408485;
&#3627408475;
+
1
&#3627408439;
, &#3627408485;
&#3627408475;:&#3627408470;&#3627408480;&#3627408481;ℎ&#3627408466;&#3627408475;&#3627408466;&#3627408462;&#3627408479;&#3627408477;&#3627408476;&#3627408470;&#3627408475;&#3627408481;
For a young adult with normal vision , &#3627408485;
&#3627408475;=25&#3627408464;&#3627408474;, then, &#3627408451;
&#3627408475;=
1
0.25&#3627408474;
+
1
0.02&#3627408474;
=54&#3627408465;&#3627408470;&#3627408476;&#3627408477;&#3627408481;&#3627408466;&#3627408479;&#3627408480;
For a young adult with normal vision &#3627408436;=54−50=4&#3627408439;&#3627408470;&#3627408476;&#3627408477;&#3627408481;&#3627408466;&#3627408479;&#3627408480;
Young children have a much greater power of accommodation, and often can read books held quite close to
their eyes. The accommodation decreases with aging, and most people find their near point gradually
recedes until they cannot read comfortably without corrective glasses.
&#3627408436;=&#3627408451;
&#3627408475;–&#3627408451;
&#3627408467;.

140. Thevisualacuity
Theacuity(clearnessofimages)ofatypicalpersonisabout5×10
−4
&#3627408479;&#3627408462;&#3627408465;.Objectswith
smallerangularseparationcannotbedistinguished(Fig24.11).Experimentsshowthatunder
optimalconditionsfewpeoplehaveanacuityoftwiceofthediffractionlimit(2×10
−4
&#3627408479;&#3627408462;&#3627408465;).
Nobodycanreachthediffractionlimit,thisfailureisduetothestructureoftheretina:
Visualacuitytestisoftenmeasuredaccordingto
thesizeoflettersviewedonaSnellenchart(Fig.
24.12)orthesizeofothersymbols,suchas
LandoltCsorTumblingE.
Figure 24.11
Figure 24.11 SlellenChart
CHAPTER 14: OPTICS PREPARATORY YEAR 2015-2016 140

141. TheEyesensitivity
Theminimumorthresholdintensityneededtoseeaflashoflightdependsonthe
wavelength.Thecorneaisopaquetowavelengthsshorterthan300&#3627408475;&#3627408474;,andthecrystalline
lenstowavelengthsbelow380&#3627408475;&#3627408474;,soultravioletlightdoesnotcontributetovision.The
sensitivityoftheeyegoestozerorapidlyabove700&#3627408475;&#3627408474;.Thephotosensitivemoleculesin
therodsanconesdonotrespondtothelongerwavelengths.Theconesareactiveonlyin
light-adaptedvision,whiletherodsarealwaysactive.
The sensitivity is greatest near 500&#3627408475;&#3627408474;
and 550&#3627408475;&#3627408474;for dark and light adapted
eyes , respectively. Both wavelengths
correspond to green light.
CHAPTER 14: OPTICS PREPARATORY YEAR 2015-2016 141

142. 24.13 Optical defects of the eye
Fourcommonopticaldefects(myopia,hypermetropia,presbyopiaandastigmatism)ofthe
eyecanbecorrectedbyuseofeyeglasses.Inthreeofthesedefectstheglassesareusedto
shifttheapparentpositionofanobject,sothatthedefectiveeyeisabletofocusproperly.In
thelast,astigmatismtheglassesareusedtocorrectthedistortionproducedbytheeye.
Myopia)ornearsightedness(:
Inthisdefectparallellightfromadistantobjectisfocusedbyrelaxedeyeatapointbefore
theretina(Fig.24.13).Anearsightedpersoncannotfocusclearlyonanobjectfartheraway
thanthefarpointlocatedatadistance&#3627408485;
&#3627408467;.Thisproblemarisesbecausethepoweroftheeye
istoogreat.Diverginglenseswithnegativepowerswillcompensateforthisdefect.
CHAPTER 14: OPTICS PREPARATORY YEAR 2015-2016 142
Figure 24.13: Schema of myopia

143. Example24.12page33:
Anearsightedmanhasafarpointatadistanceof0.2&#3627408474;.Hispowerofaccommodationis
4&#3627408439;&#3627408470;&#3627408476;&#3627408477;&#3627408481;&#3627408466;&#3627408479;&#3627408480;.(a)whatpowerlensesdoesheneedtoseedistantobjects?(b)Whatishisnear
pointwithoutglasses?(c)whatishisnearpointwiththeglasses?
Solution:
(a)Thefarpointisat&#3627408485;
&#3627408467;=0.2&#3627408474;thenthepoweroftheeyeis
&#3627408451;′
&#3627408467;=
1
&#3627408485;
??????
+
1
??????
=
1
0.2
+
1
0.02
=55&#3627408439;&#3627408470;&#3627408476;&#3627408477;&#3627408481;&#3627408466;&#3627408479;&#3627408480;(&#3627408439;isthediameteroftheeyeball).
Forafarpointattheinfinitythepoweroftheeyeis&#3627408451;
&#3627408467;=
1
∞
+
1
0.02
=50&#3627408439;&#3627408470;&#3627408476;&#3627408477;&#3627408481;&#3627408466;&#3627408479;&#3627408480;.
Acorrectivelensofpower&#3627408451;
&#3627408464;shouldbeused,where&#3627408451;
&#3627408464;+&#3627408451;′
&#3627408467;=&#3627408451;
&#3627408467;,
then&#3627408451;
&#3627408464;=&#3627408451;
&#3627408467;−&#3627408451;′
&#3627408467;=50−55=−5&#3627408465;&#3627408470;&#3627408476;&#3627408477;&#3627408481;&#3627408466;&#3627408479;&#3627408480;.
(b)Giventhepowerofaccommodation:&#3627408436;=&#3627408451;
&#3627408475;−&#3627408451;
&#3627408467;,then:
&#3627408451;
&#3627408475;=&#3627408436;+&#3627408451;
&#3627408467;=55+4=59&#3627408465;&#3627408470;&#3627408476;&#3627408477;&#3627408481;&#3627408466;&#3627408479;&#3627408480;.
Thenearpointsatisfying&#3627408451;′
&#3627408475;=
1
&#3627408485;
&#3627408475;
+
1
??????
=
1
&#3627408485;
&#3627408475;
+
1
0.02
=59&#3627408439;&#3627408470;&#3627408476;&#3627408477;&#3627408481;&#3627408466;&#3627408479;&#3627408480;is&#3627408485;′
&#3627408475;=0.11&#3627408474;
(c)Withtheeyeglassesthepoweroftheeyeis&#3627408451;
&#3627408475;=&#3627408436;+&#3627408451;
&#3627408467;=4+50=54&#3627408439;&#3627408470;&#3627408476;&#3627408477;&#3627408481;&#3627408466;&#3627408479;which
correspondsto&#3627408451;
&#3627408475;=
1
&#3627408485;′
&#3627408475;
+
1
0.02
=54dioptersand&#3627408485;
&#3627408475;=0,25&#3627408474;=25&#3627408464;&#3627408474;
CHAPTER 14: OPTICS PREPARATORY YEAR 2015-2016 143

144. Hypermetropia(orfarsightedness):
Hypermetropiastheoppositeofmyopia.Lightfromadistantobjectfocusestowardapoint
behindtheretina.WhentheLensisadjustedbytheciliarymusclestohaveitsmaximum
power,closerobjectsareblurred.Eyeglasseswithconverginglensessupplytheadditional
focusingpowerneeded.
Example 24.13 page 34:a woman has his near point 1&#3627408474;from her eyes. What power glasses
does she require to bring her near point to 25 cm from her eyes?
Solution:
&#3627408485;
&#3627408475;=1&#3627408474;&#3627408481;ℎ&#3627408466;&#3627408475;&#3627408451;
&#3627408475;=
1
&#3627408485;
&#3627408475;
+
1
??????
=
1
1&#3627408474;
+
1
0.02
=51&#3627408439;&#3627408470;&#3627408476;&#3627408477;&#3627408481;&#3627408466;&#3627408479;&#3627408480;
&#3627408485;
&#3627408475;=25&#3627408464;&#3627408474;&#3627408481;ℎ&#3627408466;&#3627408475;&#3627408451;′
&#3627408475;=
1
&#3627408485;
&#3627408475;
+
1
&#3627408439;
=
1
0.25
+
1
0.02
=54&#3627408439;&#3627408470;&#3627408476;&#3627408477;&#3627408481;&#3627408466;&#3627408479;&#3627408480;
Sheneedsaglassofpower+3Diopters(converginglens)
CHAPTER 14: OPTICS PREPARATORY YEAR 2015-2016 144

145. Presbyopia:
Accommodationisreducedwithage,whichistheresultofagradualweakeningofthe
ciliarymuscle.Thelensislessflexible.Thenearpointofapersonwhohasnormalvisionasa
youngadulteventuallyrecedesenoughsothatconverginglensesareneededforcloserwork
orreadingmuchlikepersonwithhypermetropia
Astigmatism:
Apersonwithastigmatismcannotfocuson
bothhorizontalandverticallines.Usually
thisisduetotheacorneathatisnot
perfectlyspherical.
CHAPTER 14: OPTICS PREPARATORY YEAR 2015-2016 145

147. CHAPTER 30: NUCLEAR PHYSICS PREPARATORY YEAR 2015-2016 147
Chapter 30
NUCLEAR PHYSICS
COURS TOPICS
30.1 Radioactivity
30.2 Half-life
Examples to be explained and solved:
30.1 and 30.2
Homework Problems:
30.1, 30.2, 30.3, 30.5 and 30.7

148. Introduction
Theatomicnucleusisaverysmalldenseobject.Itssizeis≈1fm(1fm=10
−15
&#3627408474;),10
−4
×thesizeofanatom.Thenucleusismadeupfromtwokindsofnucleons:protonsand
neutrons.Theprotonhasapositiveelectricalchargeequalinmagnitudetheelectron’scharge
andamassabout1840timesthemassoftheelectron.Theneutron(discoveredbyJames
Chadwickin1932)isaneutralparticleslightlyheavierthantheproton.
AnucleusisspecifiedbyitsatomicnumberZ(numberofprotons)anditsmassnumberA(
numberofneutronsandprotons&#3627408436;=&#3627408449;+&#3627408461;).ThenumberofneutronsisthenN=&#3627408436;−&#3627408461;
Example:
Theuranium238,&#3627408456;−238or
92
238
&#3627408456;has238nucleons,of
which92areprotonsand238−92=146neutrons.
Nuclideswhichhavethesameatomicnumberbutdifferent
numberofneutronsarecalledisotopessuchasDeuterium
1
2
&#3627408443;
andTritium
1
3
&#3627408443;,isotopesoftheHydrogennuclide
1
1
&#3627408443;
CHAPTER 30: NUCLEAR PHYSICS PREPARATORY YEAR 2015-2016 148

149. 30.1 Radioactivity
In1896HenriBecquerelnotedthaturaniumcompoundsproduceinvisibleradiationsthatcan
penetrateopaquecontainersandexposephotographicemulsion.Manyotherradionuclides
weresubsequentlyfound.
Theradioactivityorradioactivedecayisaspontaneousandrandomprocessinwhichthree
kindsofparticlescanbeemitted:
Gammadecay:aphotonofhighenergyisemitted.(greaterthanthoseofX-rays)
-Alphadecay:emissionoftheparticlealpha(
2
4
&#3627408443;
&#3627408466;)(processfollowedwithanuclear
transmutation)
-Betadecay:anelectron(&#3627409149;
−
)orapositron(&#3627409149;
+
)canbeemittedwithaneutrino.
CHAPTER 30: NUCLEAR PHYSICS PREPARATORY YEAR 2015-2016 149
Radioactivitydecay Nuclearreaction
Alpha
92
238
&#3627408456;→
90
234
&#3627408455;ℎ+
2
4
&#3627408443;
&#3627408466;A transmutationof th uranimu-
238 intothorium-234
Beta (&#3627409149;
+
)
Transmutationof a proton into
proton
9
18
&#3627408441;→
8
18
&#3627408450;+&#3627408466;
+
+&#3627409168;
&#3627408466;
Transmutationof th Thefluorine
-18 intooxygen.
Beta (&#3627409149;
−
)
Transmutationof a neutron into
proton
27
60
&#3627408438;&#3627408476;→
28
60
&#3627408449;&#3627408470;
∗
+&#3627408466;
−
+ҧ&#3627409168;
Transmutation of thecobalt -60
intoNi-60
Gamma
60
&#3627408449;&#3627408470;
∗
→
60
&#3627408449;&#3627408470;+2&#3627409150;
Unstable Nickel emits&#3627409150;−&#3627408479;&#3627408462;&#3627408486;,
the nucleusremains the same.

150. Theradioactivedecayofnucleiproducesseveraltypesof
ionizingradiationwithseveralmegaelectronvoltsperparticle.
Itwasfoundthatwhentheseradiationspassthroughthebody,
theycancausebiologicalhazardoussuchascelldamage,skin
burnsandcancer.
Radioactivity Caution
Ionizingradiationhavetheabilitytopenetratematter.
Alphaparticlesarestoppedbyasheetofpaper
BetaparticlescanbestoppedwithathinfoilofTinor
aluminum.
Gammaradiationisdampenedwhenitpenetrates
matter.Gammarayscanbestoppedfrom4metersof
lead.TungstenandtungstenalloyscanstopGamma
radiationwithmuchlessmassthanlead
CHAPTER 30: NUCLEAR PHYSICS PREPARATORY YEAR 2015-2016 150

151. 30.2 Half-life
Anucleardecayisarandomprocess,wecannotpredictinanywaywhenaspecific
nucleuswilldecay.Nevertheless,theprocesscanbecharacterizedbyaspecific
parameter:thehalf-life.
Ifattime&#3627408481;=0thereare&#3627408449;
&#3627408476;nuclei,thenonahalf-lifetimelater,&#3627408455;,anaverageof
??????0
2
willremain.
At&#3627408481;=2&#3627408455;,whentwohalf-liveshaveelapsed,halfofthese,orΤ&#3627408449;
&#3627408476;4nucleiwillbeleft;
At&#3627408481;=3&#3627408455;,Τ&#3627408449;
&#3627408476;8willbeleft,andsoone.
Dependingonthenuclide,thehalf-lifemayvaryfromasmallfractionofsecondto
billionsofyears(Table30.1).
Whentheelapsedtimeisnotanintegermultipleofthehalf-life,wecanfindthenumber
ofnucleiremainingasfollows:
Thechangeofthenumberofnuclei∆&#3627408449;inasmalltime∆&#3627408481;isproportionalto&#3627408449;and∆&#3627408481;:
∆&#3627408449;=−&#3627409158;&#3627408449;Δ&#3627408481;
CHAPTER 30: NUCLEAR PHYSICS PREPARATORY YEAR 2015-2016 151

152. Thelastequationimpliesthatifat&#3627408481;=0thereare
&#3627408449;
&#3627408476;nuclei,laterattime&#3627408481;thenumberofremaining
nucleiisgivenbytheequation:
CHAPTER 30: NUCLEAR PHYSICS PREPARATORY YEAR 2015-2016 152
Thisequationiscalledtheexponentialdecay
formula.
&#3627409158;istheconstantdecay.
Intermsoffractionofradioactivenuclei
remainingaftertime&#3627408481;weget:
??????
??????
&#3627408476;
=&#3627408466;
−&#3627409158;&#3627408481;
Agraphicalrepresentationof
??????
??????
&#3627408476;
isshowninFig.
30.3.
Wecaneasilyshowthatthedecayconstantis
relatedtothehalf-lifetimeby:
&#3627408449;=&#3627408449;
&#3627408476;&#3627408466;
−&#3627409158;&#3627408481;
Figure 30.3: Graphical representation of
the exponential decay function
&#3627409158;&#3627408455;=ln(2)

153. Formanyapplicationsitsmoreconvenienttoplotln&#3627408449;versus&#3627408481;becausetheresultinggraphisa
straightline.Thisresultfollowsfromtakingthenaturallogarithmof&#3627408449;=&#3627408449;
&#3627408476;&#3627408466;
−&#3627409158;&#3627408481;
toobtain:
ln&#3627408449;=ln&#3627408449;
&#3627408476;−&#3627409158;&#3627408481;
Theequationisoftheform:ln&#3627408449;=&#3627408462;+&#3627408463;&#3627408481;
Thiskindofgraph,plottedonasemi-logpaper,isusefulwhenwewishtodeterminethehalf-
lifeandthedecayconstantofaradioactivesample.
Example:Fromthegraphoffigure30.3find:(a)thehalf-life(b)thedecayconstant.
CHAPTER 30: NUCLEAR PHYSICS PREPARATORY YEAR 2015-2016 153

154. CHAPTER 30: NUCLEAR PHYSICS PREPARATORY YEAR 2015-2016 154
Theeffectivehalf-Life:
Theeffectivehalf-life&#3627408455;
&#3627408466;&#3627408467;&#3627408467;isacombinationbetweenthebiological&#3627408455;
&#3627408463;andthephysical&#3627408455;
&#3627408477;
half-lives:
Example30.1page474:Iodine131isusedin
thetreatmentofthyroiddisorder.Itshalf-life
timeis8.1days.Ifapatientingestsasmall
quantityof
131
&#3627408444;andnoneisexcretedfrom
thebody,whatfractionof
??????
??????&#3627408476;
remainsafter
8.1days,16.2days,60days?
Example30.2page475:
59
&#3627408441;isadministrated
toapatienttodiagnosebloodanomalies.
Finditseffectivehalf-life.
1
&#3627408455;
&#3627408466;&#3627408467;&#3627408467;
=
1
&#3627408455;
&#3627408477;
+
1
&#3627408455;
&#3627408463;

155. CHAPTER 31: IONIZING RADIATION PREPARATORY YEAR 2015-2016 155
Chapter 31
IONIZINGRADIATION
COURSE TOPICS:
31.1 The interaction of radiation with matter
31.2 Radiation Units
Examples to be explained and solved:
31.2, 31.3, 31.4 and 31.5
Homework Problems: 31.14, 31.16, 31.23 and
31.24

156. CHAPTER 31: IONIZING RADIATION PREPARATORY YEAR 2015-2016 156
Introduction
Theradioactivedecayofnucleiproducesseveralkindsofionizingradiationwithenergiesthat
aretypicallyseveralmillionelectronvolts(MeV)perparticle
Whenthisradiationpassesthroughmatter,itleavesatrailofionizedatomsalongitspath.
IonizingradiationincludesbothnuclearradiationandatomicX-rays.
Therearefourmajorcategoriesofradiationifinteresttous:
Positiveions,suchasalphaparticle
Electronandpositrons
Photons(gammaraysandX-rays)
Neutrons
31-1Theinteractionofradiationwithmatter

157. CHAPTER 31: IONIZING RADIATION PREPARATORY YEAR 2015-2016
Positive ions
Alpha particle, protons, and other positive ions have very short ranges in matter.
Roughly speaking, the average range of stopping distance varies inversely with the
density of the medium, so a 5-MeV alpha that can travel about 4 cm in air cannot
penetrate a sheet of paper or a layer of skin (Figure 31.1).
Electrons and Positrons
These produces a nuclear beta decays have ranges that are typically a hundred times
greater than those of alpha particles.
For example, a 1-Mev electron has a range in water or soft tissues of 0.4 cm (Figure
31.4).
Photons
Gamma rays and X-rays are both electromagnetic quanta or photons, but since the
gamma rays originate in nuclear rather than atomic processes, they typically have more
energy.
For example, a 1-MeV photon in water has a mean range of roughly 10 cm.
Neutrons
Neutrons are uncharged and produce ionization only indirectly. Since they
interact primarily with the small atomic nuclei rather than the atomic electrons, they
have a very long range in matter.
Neutrons with energies of a few million electron volts may travel a meter or so in water or
in animal tissues.

158. CHAPTER 31: IONIZING RADIATION PREPARATORY YEAR 2015-2016
31-2RadiationUnits
Fourtypesofradiationmeasurementsareusedinvariousapplications:
Sourceactivity
Exposure
Absorbeddose
Biologicallyequivalentdose
Sourceactivity
ThesourceactivityAisthedisintegrationrateofaradioactivematerialortherateofdecrease
inthenumberofradioactivenucleipresent.
Itismeasuredincuries(Ci),wherethecurieisdefinedbytherelationship:
&#3627409359;&#3627408412;&#3627408430;&#3627408427;??????&#3627408414;=&#3627409359;&#3627408386;??????=&#3627409361;.&#3627409365;×&#3627409359;&#3627409358;
&#3627409359;&#3627409358;
&#3627408413;??????&#3627408428;??????&#3627408423;&#3627408429;&#3627408414;??????&#3627408427;??????&#3627408429;??????&#3627408424;&#3627408423;&#3627408428;&#3627408425;&#3627408414;&#3627408427;&#3627408428;&#3627408414;&#3627408412;&#3627408424;&#3627408423;&#3627408413;

159. CHAPTER 31: IONIZING RADIATION PREPARATORY YEAR 2015-2016
Acurieisafairlylargeunit;a1-Ciradioactivesourcerequiresshieldingandcareful
handling.Radioactivesourcesusedinanintroductoryphysicslaboratoryexperiments
haveanactivitymeasuredinmicro-curies.
TheS.l.sourceactivityunitisthebecquerel(Bq),whichisonedisintegrationper
second.Itisexpectedtograduallyreplacethecurie.
&#3627409359;&#3627408386;??????=&#3627409361;.&#3627409365;×&#3627409359;&#3627409358;
&#3627409359;&#3627409358;
&#3627408385;&#3627408426;
Theactivityofasampleisrelatedtoitshalf-lifeT.
∆&#3627408449;=−&#3627409158;&#3627408449;Δ&#3627408481;and&#3627409158;=
0.693
&#3627408455;
,so&#3627408436;=
−Δ??????
Δ&#3627408481;
=&#3627409158;&#3627408449;
Theminussignisneededbecause∆&#3627408449;isthechangeinthenumberofnucleipresentandis
negative,whilethedisintegrationrateoractivityispositive

160. CHAPTER 31: IONIZING RADIATION PREPARATORY YEAR 2015-2016 160
Theactivityofasampleisrelatedtoitshalf-lifeT.
∆&#3627408449;=−&#3627409158;&#3627408449;Δ&#3627408481;and&#3627409158;=
0.693
&#3627408455;
,so&#3627408436;=
−Δ??????
Δ&#3627408481;
=&#3627409158;&#3627408449;
Theminussignisneededbecause∆&#3627408449;isthechangeinthenumberofnucleipresentandis
negative,whilethedisintegrationrateoractivityispositive.
Iftherearenmolesinthesample,thenthenumberofatomsis&#3627408449;=&#3627408475;&#3627408449;
&#3627408436;,whereAvogadro’s
number&#3627408449;
&#3627408436;=6.02×10
23
isthenumberofparticlesinamole.
Theactivityofnmolesofasampleisthen:&#3627408488;=
&#3627409358;.&#3627409364;&#3627409367;&#3627409361;
??????
&#3627408527;&#3627408501;
&#3627408488;
Example31.2
60-Cobetadecayswithahalf-lifeof5.27years=1.66x10
8
sinto60-Ni,whichthenpromptly
emitstwogammarays.Thesegammaraysarewidelyusedintreatingcancer.Whatisthemass
ofa1000-Cicobaltsource?

161. CHAPTE 31: IONIZING RADIATION PREPARATORY YEAR 2015-2016 161
ExposureandAbsorbedDose
Exposureindicatestheamountofradiationreachingamaterial.Itdepends
onthecharacteristicsofthebeamalone.
AbsorbedDoseindicatestheenergyabsorbedinthematerialfromthe
beam.Itdependsonthepropertiesofthematerialandthebeam.
ExposureisdefinedonlyforX-raysandgammarayswithenergiesupto3
MeV.
Itisdefinedastheamountofionizationproducedinaunitmassofdryairat
standardtemperatureandpressure(STP),1atmosphereand0
0
C.
The conventional unit is:
&#3627409359;&#3627408531;&#3627408528;&#3627408518;&#3627408527;&#3627408533;&#3627408520;&#3627408518;&#3627408527;=&#3627409359;??????=&#3627409360;.&#3627409363;&#3627409366;×&#3627409359;&#3627409358;
−&#3627409362;
&#3627408516;&#3627408528;&#3627408534;&#3627408525;&#3627408528;&#3627408526;&#3627408515;&#3627408529;&#3627408518;&#3627408531;&#3627408524;&#3627408522;&#3627408525;&#3627408528;&#3627408520;&#3627408531;&#3627408514;&#3627408526;

162. CHAPTER 31: IONIZING RADIATION PREPARATORY YEAR 2015-2016 162
TheAbsorbedDoseistheenergyimpartedbyionizingradiationtoaunitmassofabsorbing
tissue.Itismeasuredinrads.
&#3627409359;&#3627408531;&#3627408514;&#3627408517;=&#3627409358;.&#3627409358;&#3627409359;&#3627408523;&#3627408528;&#3627408534;&#3627408525;&#3627408518;&#3627408529;&#3627408518;&#3627408531;&#3627408524;&#3627408522;&#3627408525;&#3627408528;&#3627408520;&#3627408531;&#3627408514;&#3627408526;
theS.I.unitforabsorbeddoseistheGray(Gy).
&#3627409359;&#3627408494;&#3627408538;=&#3627409359;&#3627408523;&#3627408528;&#3627408534;&#3627408525;&#3627408518;&#3627408529;&#3627408518;&#3627408531;&#3627408524;&#3627408522;&#3627408525;&#3627408528;&#3627408520;&#3627408531;&#3627408514;&#3627408526;=&#3627409359;&#3627409358;&#3627409358;&#3627408531;&#3627408514;&#3627408517;
A1roentgenexposuretoX-raysorgammaraysproducesasofttissueabsorbeddoseof
approximately1rad
Example31.3
Livingtissuesexposedto10,000radsarecompletelydestroyed.Byhowmuchwillthis
absorbeddoseraisethetemperatureofthetissuesifnoneoftheheatislost?(Assumethat
thespecificheatofthetissueisthesameasthatofwater,c=4180J/kg.K).

163. CHAPTER 31: IONIZING RADIATION PREPARATORY YEAR 2015-2016 163
BiologicalQuantities
Theabsorbeddosereferstoaphysicaleffect:thetransferofenergytoamaterial.However,the
effectofradiationonbiologicalsystemsalsodependonthetypeofradiationanditsenergy.
Thequalityfactor(QF)ofaparticularradiationisdefinedbycomparingitseffectstothoseofa
standardkindofradiation,whichisusuallytakentobe200keVX-rays.
&#3627408504;&#3627408493;=
&#3627408491;&#3627408528;&#3627408532;&#3627408518;&#3627408528;&#3627408519;&#3627408511;&#3627408531;&#3627408514;&#3627408538;&#3627408532;(&#3627409360;&#3627409358;&#3627409358;&#3627408524;&#3627408518;&#3627408509;)
&#3627408491;&#3627408528;&#3627408532;&#3627408518;&#3627408528;&#3627408519;&#3627408514;&#3627408527;&#3627408538;&#3627408531;&#3627408514;&#3627408517;&#3627408522;&#3627408514;&#3627408533;&#3627408522;&#3627408528;&#3627408527;
Forexample,fastneutrons(withenergiesabove0.1MeV)haveQFofabout10forcausing
cataracts.Thismeansthatthedoseof200-keVX-raysneededtoproducecataractsis10timesthe
doserequiredforneutrons.

164. CHAPTER 31: IONIZING RADIATION PREPARATORY YEAR 2015-2016
TheQFvarieswiththeradiationtypeandenergy,withtheanimalspecies,andthe
biologicaleffectunderconsideration.
Seetable31.1
Theremandthemillirem=&#3627409359;&#3627409358;
−&#3627409361;
remaretheunitsusedindiscussionsofbiologicaleffects.
Forexample,inthecaseofcataractformation,1radof200-keVXraysand0.1radoffast
neutronseachproduce1remofdamage.
Biologicallyequivalentdose(inrems)isthephysicalabsorbeddose(inrads)timestheQF.
Biologically equivalentDose (rem) = absorbed Dose (rad) ×QF
InS.l.units,thebiologicallyequivalentdoseinsieverts(Sv)equalsthedoseingraystimesthe
QF.
Biologically equivalentDose (Sv) = absorbed Dose (Gy) ×QF
Seetable31.2

165. CHAPTER 31: IONIZING RADIATION PREPARATORY YEAR 2015-2016 165
Example31.4
Acancerisirradiatedwith1000radsof60-Cogammarays,whichhaveaQFof0.7.Findthe
exposureinroentgenandthebiologicallyequivalentdoseinrems.
Example31.5
Alaboratoryexperimentinaphysicsclassusesa10microcurie137-Cssource.Eachdecayemitsa
0.66MeVgammaray.(a)Howmanydecaysoccurperhour.(b)A60kgstudentstandingnearby
absorbs10percentofthegammarays.Whatisherabsorbeddoseinradsin1hour.(c)Thequality
factorQFis0.8.Findherbiologicallyequivalentdoseinrems.