Medical physics
The medical physics overlaps the two very
large fields of medicine and physics and study :- The application of physics to the function of
the human body , it is called the physics of
physiology .
- The physics of the instruments are used in
medicine .
- The medical application of : The laser light
 UV light
 IR light
 Nuclear radiation
 radiological physics
Forces on and in the Body
The fundamental origins of forces :1. The gravitational force :- From Newton law :- There is a force of
attraction between any two objects .
F = mg
g = acceleration of gravity
m = The mass
f = The force
- Ex: our weight is due to the attraction between
the earth and our bodies .
- The medical effects of gravitational force :a- The formation of varicose veins in the legs as
the venous blood on its way to the heart .
b- The medical effect of gravity on the skeleton
( on the bones ) , in some way contributes to
healthy bones , if a person becomes wightless ,
such as in an orbiting satellite , he may lose
some bone minerals .
2- The electrical force :
This force is more complicated than gravity
since it involves attractive and repulsive forces
between static electrical charges as well as
magnetic produced by moving electrical charges
(electric currents).
Ex: The electrical force between an electron (`e )
and a proton (P + ) in hydrogen atom is about 1039
times greater than the gravitational force between
them .
(e+P)
atom
ele. force in H atom
 1039 > ( e + p )
att. force in H
Ex:- our bodies are basically electrical machines :
- The forces produced by the muscles are
electrical charges attracting or repelling other
electrical charges .
- The control of the muscles .
3- The nuclear force :It acts as the force to hold the nucleus together
against the repulsive forces produced by the
protons on each other .
The types of problems involving forces on body :- statics , where the body in equilibium .
- dynamics , when the body is accelerated .
- friction , is involved in both statics and
dynamics .
Statics
Objects are stationary ( static ) they are in a state
of equilibrium when:1. The sum of forces in any direction is zero
 Fi = 0 , f1 + f2 + f3 + f4 = 0 ( First condition
of equilibrium )
2. The sum of the torgues about any axis is zero :
  = 0 ( Second condition of equilibrium )
=F.l
where  = The torque
f = the force
l = The vertical distance from the pivot
( fulcrum point ) to the line action of the force .
cw = ccw
sum of clock wise torque = sum of counter clock
wise torque .
F2
Fi=0
F1
F3
F4
( Fulcrum point )
A

L
F
The kind of levers in the body :Many of the muscle and bone systems of the body
acts as levers .
1-
First class lever :
The fulcrum point ( P ) between the resistance
force ( weight ) W and the effort force the muscle
force ( M ) .
W
M
P
Ex: in the head of human :
W = The weight
M = Muscle force
F = the force at the fulcrum point ( P )
2- Second Mclass
level :
W
( Feet )
P
f
3- Third class lever ( The hand )
M
f
W
Ex: The lever system in the body is the case of the
biceps muscle and the radius bone acting to
support a weight ( W ) in the hand .
M
R
W
)The Forces and dimensions )
4Cm
30Cm
W
Where R = the reaction force of the humerus on
the ulna .
M = the muscle force supplies by the biceps .
W = the weight in the hand .
M
R
W
P
4Cm
14Cm H
30Cm
)The Forces and dimensions )
W
The force and dimensions where the weight of the
tissue and bones of the hand and arm ( H ) at their
center of gravity .
From this example , find the value of M ?
Solution :Two torques :
1due to the weight W
Torque about point P
Frictional Forces :
A force which resist the motion between two
surfaces ( in contact) depend on the nature of the
surface and independent on the area of the surface
Ffricitional forces :1- static friction ( Fs ) :
The effective force between surfaces that are at
rest with respect to one another .
Fs = s N
Where :s = coefficient of static friction ( used to find the
maximum resistance force on an object can exert
before it starts to move ).
N = normal force .
2- Kinetic friction ( Fk ) :
The effective force between surfaces that are in
relative motion .
Fk = k N  Fk  N
Where : Fk = coefficient of kinetic friction .
s > k ( Force to start the motion is a greater than
needed to keep it moving )
Friction in the body :The walking :
As the heel of the foot touches the ground a
force is transmitted from the foot to the ground Fig
( 1 ) , we can resolve this force into horizontal and
vertical components .
Fh
N
Fig. (1-a)
Friction between the heel and surface prevents the
foot from slipping forward .
Fn = the horizontal reaction component supplied
by frictional forces .
N = Normal force , supplied by the surface .
The maximum force of friction F is F =  N
Fh
N
Fig (1-b)
When the foot leaves the ground , Fh prevents the
toe from slipping backward .
Measurment ( Fh ) :The value of the horizontal force component (
Fh ) of the heel as it strikes the ground when a
person is walking .
Fh  0.15W , W = mg ( person's weight )
When  less than 0.15 , his foot slips .
- The oefficient of friction in the bone joints is
usually much lower than in engineering type
materials . If a disease of the joint exists the
friction may become large , the synovial fluid
in the joint is involved in the lubrication .
- Some of organs of human body are lurbricated
by a slippery mucus covering to minimize the
friction .
- The saliva used when we chew food acts as a
lubricant .
Ex: The mass of ( 10 Kg ) is pulled along a
horizontal surface at constant velocity by aforce
50N and which make an angle of 25o with
horizontal what is the coefficient of k between
block ( mass ) and the plane ?
Solution :N
F sin 25
25o
Fk
F
F cos 25
W ( 10 x 9.8 )
Fk = k N
 k 
Fk
N
Fk = F cos 25 = 50 N x 0.92 = 46.2 N
N= ( 10 x 9.8 ) N – F sin 25 = 98N – ( 50 x 0.38 )
Normal force  N = 79N  Newton
 k 
46.2 N
 0.58
79 N
Dynamics :
The forces on the body uder the constant
acceleration or deceleration of one dimensional
motion .
The Newton's second law , forces equal mass
times acceleration , can be written without vector
notation as :F = ma
where : F = The force
m = the mass
a = acceleration
m
the unit , F { N , dyne } , m { g , Kg } , a { cm
,
}
s
s
also F = The change of momentum over a short
interval of time .
2
2
m
F
t
m = change of momentum =  ( mv )
m = mass , v = velocity of this mass
t = interval of time
Ex: A 60 Kg person walking at 1 m sec-1 bumps
into a wall and stops in a distance of 2.5 cm in
about 0.05 sec what is the force developed on
impact ?
Solution: (mv ) = momentum before impact = momentum
after impact
60 Kg x 1 ms-1 = 60 Kg x 0
 (mv ) = 60 Kg x 1 m s -1 – 0 = 60 Kg m s-1
(mv) 60kgms1
F 

 1200 Kgm / s 2  1200 N
t
0.05s
 2 times of weight , N = Kg
m s-2
The application of a dynamic force in the body :- The work done by the heart beats ( systole ) 60
g of blood is given a velocity of 1 m/sec upward
in about 0.1 sec . The upward momentum is
0.06 Kg x 1 ms-1 and due to third Newton's law
producing downward reaction force on the rest
of the body is 0.06 Kgs-1 /0.1 sec or 0.6N.
- The deceleration of the body take place
through compression of the padding of the feet
when the person jumps from a height of 1m ,
under these condition , the body is traveling at
4.5 ms-1 just prior to hitting , and if the
padding collapses by 1cm the body stops in
0.005 sec .
F ( foot ) = 100 of body's weight (W), force in
the (Leg) = 100 W.
- The large velocity of modern cars the riders
have a larger momentum than when walking ,
in accident the car stops in a short time ,
producing very large forces the results of these
forces on the passengers can be broken bones ,
injuries , …death , ….
- Consider the case of "Whiplash" A person
sitting in an auto that is struck from behind
will often suffer a whiplash injury of the neck (
cervical region of the spine ) when the car is
struck , forces act through the seat forcing the
trunk of the body ahead while the inertia of the
head causes it to stay in place , leading to
severe stretching of the neck .
The method reduce the injuries from accidents :-
seatbelts in automobiles .
Airbags and shoulder belts .
Safety devices .
The energy – absorbing steering column
reduces the deceleration forces during a
collision by increasing the time the trunk of the
body takes to come to stop .
- If there is no safety devices , there is many
injuries , when the auto traveling at 15 m/sec
that stops in 0.5 m due to collision , the
passenger's head and body are thrown against
the dash and stopped , if the dash is padded ,
the decelerative effect may be minimized , if
the dash is not padded or if the head strikes a
metal surface , forces far beyond that of human
tolerance occur and severe head injuries or
death can be expected .
The effects of the acceleration :1increase or decrease in the body's weight .
2changes in internal hydrostatic pressure .
3distortion of the elastic tissues of the body .
4the tendency of soilds with different
densities suspended in a liquid to separate .
5at a large acceleration the body loss control
because it does not have adequate muscle force
to work against the large acceleration forces
and the blood may pool in various region of the
body . if g 3g the blood do not flow to the
brain causes black out and unconsciousness .
The centrifuge way :Is away to increase apparent weight , it is
especially useful for separating a suspension in a
liquid , it speed up the sedimentation that occurs at
a slow rate under the force of gravity .
Stoke law:Let us consider sedimentation of small
spherical obejects of density P in a solution of
density Po in a gravitaional field ( g ) . We know
that falling objects reach a maximum ( terminal )
velocity due to viscosity effects . Stoke has shown
that for a spherical object of radius ( a ) , the
retarding force Fd and terminal velocity V are
related by :Fd = 6aV …………….stoke law
Where :
Fd = retarding force
V = Terminal force
 = viscosity of liquid
When the particle is moving at a constant
speed , the Fd is an equilibrium with the difference
between the downward Fg and the upward buoyant
force :
Fd = Fg - F B
Fg = force of gravity = 43 a3g
Fb = the bouyant force = 43 o g
Fd = 6aV – retarding force
 the terminal or sedimentation velocity :
2a 2
V 
g (   o )
9
The Medical use of terminal velocity :- From some disease such as rheumatic heart
disease , the red blood cells clump together and
the effective radius increase and also increase
sedimentation velocity .
- From the disease such as :
Hemolytic jaundice and sickle cell .
Anemia , the red blood cells change shape or
breack, the radius decrease , then the rate of
terminal of these cells is slower than normal .
- Determination of the hematocrit :
V  gravitational acceleration ( g )
g increase by means of a centrifuge .
geff effective acceleration = 4 2f2r
f= the rotation rate in revolutions per second .
r= the position on the radius of the centrifuge
where the solution is located .
hematocrit depends on :
- radius of centrifuge
- speed and duration of centrifugation .
- Ultracetrifuge used for molecular weight or large
macromolecules, for protein research use
40.000 – 100.000 rpm.
W = W x 30 cm
= 30 W acting clockwise ………….( 1 )
2due to M ( Muscle force )
M= 4M …………..( 2 ) counter clockwise
hence the arm in equilibrium :
W = M 4M – 30 W = 0
 M = 7.5 W
If W = 100 N  M = 7.5 x 100 = 750 N
If neglected the weight of the forearme and hand
(H).
If we H = 15 N : W = 5 N
In equilibrium  = 0
5N x 30 x 10-2 M + 15N x 14 x 10-2 M = M x 4 x
10-2 M
clockwise = counter clockwise
360 N . M = 4 M . N
M = 90 N
The effect of the angle arm on the muscle force :1- If the forearm at an angle  to the horizontal
as
shown
below
Fig ( 1 ) :
R
M

H
W
If take the torque about the joint , M remains
constant as  changes .
The change of M with  , the length of biceps
muscle changes of M with  , each muscle has a
min . length to which it can be contracted and a
max length to which it can be streched and still
function at these two extremes the M can exerts is
essentially zero . At some point in between the
muscle can produce it max force as shown fig. ( 2 )
.
If the biceps pulls vertically  does not affect
the M required but it does affect the length of the
biceps muscle ( L ) . Fig ( 2 )
The maximum force of muscle at its optimum
length ( L ) is 3.1 x 107 N/M2
2- The arm can be raised and held out horizontally
. Fig ( 3 ) the arm raised and held horizontally
from the shoulder by the deltoid muscle .
T * 18 cm
R

36 cm
W1
W2
72 cm
Fig ( 3 ) raising the arm
Where : T = is tension in the deltoid muscle fixed
at the angle  .
R = is the reaction force on the shoulder
joint .
W1= the weight of the arm at center of
gravity .
W2= weight in the hand .
What is the force needed to held up the arm under
the state of raising the arm ?
T ( tension ) at angle  resolved into two
componenty
.
T
sin

,
T cos  and taking the sum of the torques about
the shoulder joint .
T sin 
T
T cos 

cw =ccw
36 W1 + 72 W2 = 18 T sin 
 the tension
T
2W  4W21
sin 
if  = 16o , W1= 68N , W2=45N
 T = 1145 N the force needed to hold up the arm
is large .
The tension in the Lumbar ( Lower back ) region :
- The force at the fifth Lumber vertebra ( LS )
with the body tipped forward at 60o to the
vertical and with a weight of 255 N in the
hands .
- The lifting heavy objects from in correct
position is thought to be a primary cause of low
back paint due to large the forces are in the
Lumbar region of the back , and yield a large
pressure in the discs separating the vertebra
back .
P
F
F
P
, PF
2
A
r
If F is heigh  P is large on the disc and give the
pain in the back in fig ( 4 ) the pressure in the
third Lumbar disc for an adult in different
positions .
A= picking up 20 Kg in correctry without
bending the Knees .
B = picking up 20 Kg correctly by bending the
knees .
Ex: Lifting a weight of 225 N , calculate the force
at the ( L5 ) with the body tipped forward at 60o to
the vertical .
Where :
W1= weight of body trunk , W2= weight of arms ,
head and 225 N
A=rigid fulcrum at LS , AB = body trunk ( boom
).
AD 
2
1
AB, AE  AB ,
3
2
T = tension in muscle .
R= reaction force at L5 .
W1=320N
,
W2=382N
,
T=3380N
Fx=3225N,Fy=1748N ,
R  Fx2  Fy2 , R = 3803N.
,
The physics of skeleton
The function of the bone :
1. Support
2. locomotion .
3. protection of organs .
4. storage of chemicals .
5. nourishment .
6. transmission of sound ( middle ear ) .
- The spinal column in human body .
- The kind of vertebrae .
- The bone kind – the shape – the structure of
the bone .
- The mechanical properties of bone :- Bone is composed of small hard bone mineral
crystal attached t a soft flexible collagen matrix
.
- Some standard physics measurements on a
piece of compact bone.
- The density of compact bone is 1.9 g/cm3 and
change with the age of the body .
Young's modulus of elasticity :How much forces is needed to break the bone
by compression , tension and twisting .
- When the bone placed under tension or
compression there is change in its length from
the stress – strain curve in fig ( 1 ) .
Stress = FA =N/mm2
Strain = LL
Stress = 120 N/m2
L
 0.015 at fracture
L
The strain LL increase linearly at first with the
stress FA ( hooks's law )
If F increases the L increase more rapidly and the
bone breaks at stress of 120 N mm-2 .
The ratio of stress to strain in the initial linear
portion is called young's modulus Y
Y=
LF
AL
, Ybone = 1.8 x 1010 N/m2 .
Ex: Man with mass of 100 Kg standing on the one
leg has a 1 M shaft of bone with average crosssectional area of 3 cm2 find :1. The pressure in Pa .
2. The amount of shortening in this bone .
P=
F
A
. F = Mg = 100 x 10 = 103 N
P = 103N/ 3 x 10-4m2
= 13 x 107 Pa
= 3 x 10 6 Pa
L =
Y=
LF
AY
=
LF
AL
1x103
3 x10  4 x1.8 x1010
10-4 m
tension elongate in L due to
F
A
stress
LF
AY
L =
compression
Shorting in the length of the bone of its length ( L )
Measurment of bone mineral :
1. X – ray technique :
In this technique use x – ray beam with
different energyies in fig ( 1 ) and the absorption of
radiation by calcium varies rapidly with range of
the energy . x – ray beam contains much scattered
radiation when it reaches the film , the film is not a
reproducible detector , then this technique is not
useful for measurment of bone mineral due to
beam is in homogenous [ hetergeneous beam ].
2. Photon absorpotiometry :The problems with x – ray techniques were
climinated by using:1. monoenergetic x – ray or gamma ray source,
monochromatic light.
2. anarrow beam to reduce the scatter .
3. scintillation detector that detects all photons .
The strength of bone depends on the mass of bone
mineral
present
[ decreases , 1 to 2% per year ] in this technique
the bone to be measured is immersing in auniform
thickness of soft tissue .
The absorption :
I* o is the inteasity of beam before enters the bone .
I is the inteasity of beam after enters the beam [
transmition beam ]
 the bone mineral mass ( MB ) at any point in the
beam is proportional the Log (I*o/I)
MB  Log
MB = K Log
 I o* 
 
 I 
 I o* 
 
 I 
g/cm2
Where K = constant .
The pressure
The pressure ( P ) :
The force per unit area ( gas , liquid , solid )
P 
F
A
F= force , A = area
The unit of the pressure :- metric system :
dynes / cm2 , N/m2
- SI unit Pa ( pascal ) = Newton/m2
N=newton , m2= square meter
- The most common method of indicating
pressure in medicine is by the height of a
column of mercury (Hg).
The pressure in fluid :If we have a column of liquid ( blood ) , ( water
) of cross section A ( U tube manometer for
measuring the pressure ( P ) .
U tube manometer
Po
h2
P
h1
h2h1=h
Weight of the column of liquid of its height ( h )
= force of column on the bottom
mass ( m ) of a height ( h ) of the liquid
= Ah times its mass per unit volume
= Ah 
Where A = Area , h=height ,  = density of liquid

m
v
, m = mass , v = volume
W = f …………………( 1 )
W =mg ……………..( 2 ) , g = accerleration of
gravity
m = Ah  ……………...( 3 )
W = Ah g  This weight ( W ) is the force
downward of the column on the area ( A ) at the
bottom .

W
 gh
A
P = gh
If this column is open to atomsphere ( Po)
the pressure on the bottom of liquid column is :
P = Po + gh
Po = 1.01 x 105 dyn/cm2
h ?? air column
PHg = gh = ( 13.6 x 980 cm/sec2 ) h
1.01 x 105 dyn/cm2 = 13.6 g/cm3 x 980 x h
h =76 cm
The types of recording pressures :1. total ( absolute ) pressure :
P = Po + gh
2. gauge pressure P - Po = gh
- if the pressure below Po is called negative
pressure .
ex: pressure inside the Lung below Po ( inspire ).
- If the pressure above Po is called positive
pressure .
Measurement of pressure :
1. The pressure in the bladder can be measured
by passing a catheter with a pressure sensor
into the bladder through the urinary passage (
urethra ) .
2. direct cystometry a needle is passed through
the wall of the abdomen directly into the
bladder .
In this method ( cystometric method ) give
information about the bladder and the pressure
more than the catheter technique .
Example :
What height of water will produce the same
pressure as 120 mm Hg ?
P ( 120 mm Hg ) = gh = 13.6 g/cm3 x 980 cm/sec2 x
12 cm
= 1.6 x 105 dynes/cm2
for water
1.6 x 105 dynes / cm2 = 1.0 g/cm3 x 980 cm/sec2 x h
cm H2O
h  163 cm H2O
 The h of H2O can be obtained by multiplying the
h of Hg by 13.6
- Eye pressure
- Pressure in the digestive system
- Pressure in the skeleton
- Pressure in the urinary bladder
Ex: Positive pressure is used in blood transfusions ,
contaner is placed 1 m above a vein with a venous
pressure of 2 mm Hg , what is the net pressure
acting to transfer the blood into the vein .
P of d blood =
blood
x103 mm
 Hg
= 76.5 mm Hg
 of blood = 1.04 g/cm3
 of Hg = 13.6 g/cm3
 of H2O= 1.0 g/cm3
Net pressure = 76.5 – 2 = 74.5 mm Hg
H. W:
P of air ??
 of air = 1.3 x 10 -3 g /cm3
Negative pressure ( suction ) is to drain body
cavities . In the drainage arrangement . the
negative pressure supplied to the collection bottle
is 100 mm Hg and the top end of the tube is 37 cm
above the end negative pressure at the lower end of
the tube .
Pressure of 37 cm of water =
H O
x370mm
 Hg
2
Net pressure = 100 mm Hg – 370
 1.0 


 13.6 
= 73 mm Hg
1 mm Hg = 1.3 x 103 dyn g / cm2
1 mm Hg =
N/m2
1 Pa = 9.869 x 10-6 atm
1 atm = 1 atmosphere = 1.013 x 105 N/M2
= 1.013 x 105 Pa
1 cm Hg = 1.333 x 103 N/M2
= 1.316 x 10-2 atm
1 Pa = 7.501 x 10-4 cm Hg
1 atm = 76 cm Hg = 760 mm Hg
Physics of the cardiovascular system
CVS :The blood , blood vessels and the heart .
The blood :
7% of the body mass or 4.5 Kg ( ~ 4.5 Liters )
in a 64 Kg person .
The component of the blood :1red blood cell ( erythrocytes )
- Flat disk ( 7 m in diameter ( dia. )
- 45% of the volume of the blood
- 5 x 106 cells/mm3 of the blood
2The blood plasma
Clear fluid 55% of the blood
3White blood cell ( Leukocytes )
- ( 9 – 15 m dia. )
- 8000 cell's/mm3
- in some infection in the body , the number
of this cell's increases like , cancer ,
Leukemia .
4Platelets
- 1~ 4 m in dia.
- 3 x 105 platelets /mm3 of blood.
The blood vessels :
1- arteries 2-veins
3-capillary bed
The heart :
- double pump
pulmonary 20% of (
blood )
- circulatory systems
systemic (general )
80% of blood
- The heavier and stronger muscular walls on
the left side of the heart and the circular shape
of the left ventricle produces the high pressure
to circulate the blood .
The valves of the heart :
- natural valves
flow of blood only in the correct direction .
- artificial ( mechanical ) valves
Work done by the heart :
- The work of the heart from the contraction of
the heart muscles forces .
- 80 ML of blood through the Lungs from the
right ventricle and a similar volume to the
systemic circulation from the left ventricle .
- The work done by the heart ( W )
W = PV
Where P = constant pressure
V = volume pumped
- The pressures in two pumps of the heart are
not the same , in the pulmonary system the
pressure is very low due to the low resistance of
the blood vessels in the Lungs .
- Physical ( W ) is ( average ( P ) x volume of
blood is pumped )
The pressures of blood throughout the circulatory
system :
Let us start with the blood in the left side of
the heart and follow
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LV ( 120 – 125 ) mm Hg  arteries
120 mm
Hg
( systolic pressure )
80
 arterioles ( 80  25 ) mmHg  cappillaries ( 25
– 7) mmHg  veins ( 7 – 5 ) mmHg RV (25
mmHg) Lung (25) LA ( 7 – 8 mmHg) 
LV(120 mmHg)
W = Pdv
The P of blood at systolic is 120 mmHg
The pressure at the resting phase diastole is 80
mmHg.
Example :
If the average pressure is 100 mmHg (100
mmHg=1.4x 105 dynes /cm2) and 80 mL of blood is
pumped each second ( pulse rate of 60/min ) what
is the ( W ) of the heart .
W per second = P V = 80 x 1.4 x 105
= 1.1 x 107 ergs
= 1.1 J/sec = 1.1 W .
Blood pressures and its measurment :The measurment of blood pressure by two
method:1. direct method
2. indirect method
1. Direct method :
hollow needle blood vessel  pressure of
blood transmits through cather ( hollow plastic
tube ) to the pressure transducer .
2. Indirect method :
The instrument is used to measure the pressure
is called sphygmomanometer .
- Sphygmomanometer : ( manometer )
pressurecuff – gauge on the upper arm –
stethoscope .
- The measurement :
- stethoscope placed over the brachal artery at
the elbow .
- The pressure cuff is inflated rapidly to a
pressure used to stop the How of blood ~ 130
MMHg
- The air in the cuff is gradually released .
- The pressure in the cuff drops below the
systolic pressure The turbulent flow of boold
through the artery causes sound vibrations 
Sound heard by stethoscope  K – sound ( Korotk
off sound )  K sound  systolic pressure level .
- If the pressure in the cuff reduce , K sounds
become Louder and then begin to fade  K
sound die out  diastolic pressure .
Blood pressure 
Systolic (contraction)
Diastolic (relaxation)
120
80
12
8
Normal blood pressure =
mm Hg
Arterial blood pressure  cm Hg
Venous blood pressure ( 8 – 16 ) mm H2O ( blood )
Mean of blood pressure = dias. P + 13 [syst – dias ]
Pulse of blood pressure =[ syst. – dia. ]
The effect of accerleration :- The pressure in the arteries in the body varies
from one point to another due to he
gravitational forces .
- Shows direct measurments of blood pressure
made on a standing person open glass tube
manometers are shown connected to arteries in
:
1. foot { P1 = 200 mm Hg , h = 2.6 m }
2. upper arm { P2 = 100 mm Hg , h = 1.3 m }
3. hed { P3 = 60 mm Hg , h = 0.8 m }
The blood rises to the same level , P1 at ( 1 ) > P2 >
P3 due to g  gravitational force .
P1 = gh [ by column of blood ( of height h )
between heart and foot ] + pressure of the heart .
  density of the blood ( 1.04 g/cm3 )
h of blood = 13 times of hHg
- If P of blood 

120
80
1560
1040
mm Hg
mm of blood
- If g on earth = 3g  blood rise only 43 cm
above the heart .
- P is the same on all point in the body when the
body is horizontal ( not standing ).
Laplace Law :
How the tension in the wall of a tube related to
the radius of the tube and the pressure inside the
tube .
Consider a long tube of radius ( r ) carrying
blood at pressure P Fig ( 1 ) uniform section of a
tube of length ( l ) wall thickness ( t ) and radius ( r
).
L
t
r
P1
Pe
The pressure across the blood vessel ( tube ) wall is
called transmural pressure (Ptm)
Ptm = Pi – Pe
(mean ) = Patiral
Pi = internal blood pressure = P
= Pvenous = P
If Pe is very small
Ptm = Pi = P
Pe = external tissue pressure
The tension in the wall :
The pressure inside the tube is uniform on the
wall and can mathematically divide the tube in half
as shown in Fig ( 2 ) .
The force trying to seprate the upper X lower
halves is the Ptm times an area ( A ) .
F = PtmA
A by integration , that area against
the pressure acts is the diameter ( R ) of the tube
times ( l ) .
F = Ptm ( Rl )
 R=2r
F =2rl Ptm = 2rl P
Ptm = P = P
In oreder , the tube in equilibrium
pushing force = tension force ( T ) ( holding two
halves )
2rl P = 2Tl
T = tension force per unit length
T = rP
Laplace law
If r is small  T is small and there is no break in
this wall .
Table ( 1 )
Aorta ( r = 1.2 cm )
Typical artery ( r =
0.5 cm )
Small capillary ( r =
6 x 10-4 cm )
Small vein ( r = 2 x
10-2 )
Vena cava ( r = 1.5
cm )
Typical pressures and T
P
=
100
mm
Hg
T=156,000 dynes/cm
= 1.3 x 105 dynes/cm2
P = 90 mm Hg  T = 60,000
dynes/cm
P = 30 mm Hg  T = 24
dynes/cm
P = 15 mm Hg  T = 400
dynes/cm
P = 10 mm Hg  T = 20,000
dynes/cm
Bernoullis Principle :Based on the low of conservation of energy :
The velocity of blood ( fluid ) increase at the
narrow section of the part at r2
PMin
VMax
The work done on the any system W
W = KE + PE
W = P V = P M
KE = Kinetic energy = 12 MV2
PE = Potential energy = Mgh
P M =
1
2
MV2
+
Mgh
P = 12  V2 + gh
( P 1 – P2 ) =
P1 + gh1 +
1
2
1
2
( V -  V ) + g ( h2 – h1 )
2
2
2
1
 V = P2 +gh2 +
2
1
1
2
V
2
2
If there is no friction with the walls of a tube then
the pressure decrease where the velocity of fluid
flow increase : P  V1
If the system is heart  the work ( W )
Used for – raise the pressure in the arteries
- impart KE to the blood
- work against friction
- increase the velocity of blood
Example :From the left dise of normal human heart , the
difference in height from the ventricle to the aorta
is ( 15 cm ) , the velocity of the blood at this point
as 40 cm/sec . Find the work done for each 1 cm3 of
1 g of blood ?
Where is :blood = 1.04 g/cm3  1.0 g/cm3
P l artery = 80 mm Hg ( dias )
( aorta ) = 120 mm Hg ( syst )
initial velocity before systole = 0
venous drain pressure = 0
left arterial pressure prior to contraction = 0
V1, P1 , h1  velocity , pressure and height of the
blood entering the heart .
V2, P2 , h2  velocity , pressure and height of the
blood leaving the heart .
The work done by heart is W :
W = P V
According to Bernouollis principle
W = KE +  PE + P


KE =
velocity change
( work of velocity change ) = 12 x 1g ( 40
cm/sec)2
= 8 x 102 dyne cm
1
2
2
m
V

V
2
1
= 2
PE =
height change
= mg ( h2 – h1 )
( work of height change ) = 1g x 980 cm/sec2
x 15 cm
= 0.15 x 105 dyne cm
P =
pressure change
= m ( P2 – P1 )
( work of pressure change ) =
1g
1g / cm3
x 120 mm
Hg
= 1 cm3 x 12 cm Hg
= 12 x 1.3 x 104 dyne/cm2
= 1.6 x 105 dyne cm
P >> KE
P. E  10% of P
The work per 1 cm3 of blood pumped = 1.7 x 105
dyne cm
The work per 60 cm3 of blood pumped = 107 dyne
cm
( the work of LV )
The work by RV = 16 work by LV
Total work by the leavt = ( 1 + 16 ) 107 = 1.16 x 107
dyne cm
Note : the pressure in pulmonary artery is 120 mm
Hg
The medical important of venturi effect :
If any artery is narrowed by internal plagues
or by external pressure resulting from a tumor 
then the blood pressure in the constricted region of
the artery will fall very dramatically
Vmax
Pmin
A1 > A2
V1 < V2  P1 > P2

V1A1 = V2 A2 equation of continuity
the venturi tube method has been adapted to
measure the flow velocity of blood in arteries .
Example :
A liquid of density 950 Kg m-3 flows in a
horizontal pipe of radius 4.5 cm in a section of the
tube of restricted radius 3.2 cm , the liquid
pressure is 1.5 x 103 Nm-2 less than in the main pipe
calculate the velocity of the liquid in the pipe .
Solution :
Let the pressure in the main tube is P1
Let the pressure in the restricted tube is P2
P1 > P2  V1 < V2  r1 > r2
P1 = P2 + 1.5 x 103 N m-2
From Bernoulis principle
P + 12 V2 + gh = constant
the pipe is horizontal there is No P. E
P1 + 12  V = P2 + 12  V

2
1
2
2
P1 – P2 = 12 ( V - V )
2
2
2
1
P2 + 1.5 x 10+3 – P2 = 12 ( V - V )
2
2
2
1
1.5 x 10+3 = 12 ( V - V )
-2
( V - V ) = 3x10 = 3950
=
31.6
x
10
………….( 1 )
x10
2
2
2
2
2
1
3
2
1
3
from equation of continuity :
A1V1 = A2V2
 r V1 =  r V2
r V1 = r V2
2
1
2
2
2
1
2
2
V1 =
 r2 
 
 r1 
2
V2 …………….( 2 )
from eq. ( 1 ) and eq. ( 2 ) we get :
-2
V -  r  V = 31.6 x 10
r 
2
2
1-
4
2
2
2

1

 r2

 r1



4
= 31.6Vx 10
2
2
-2
1-
 3 .2 


 4 .5 
4
= 31.6Vx 10
-2
2
2
31.6 x 10-2
V22
1- 0.3 =
 0.7 V = 31.6 x 10-2
2
2
V2 =
31.6 x 10-2
0.7
m
The flow of blood :
The volume rate of flow through the tube of a
liquid of viscosity  is given by :
Q=
r 4 P
8L
[ Poiseuille's equation ]
Where :
Q is volume rate of flow
 cm   cm 
ML

 
 ,
Q = V = Volume
S
time
min
sec
3

3


Q = Vt = ALt = A  
A = cross – section area of tube
 = velocity of liquid ( blood )
The physical factors effect the [ Q ] :1 Q = A 
Q  
the blood velocity ( ) is related in an inverse way
to the total cross – sectional area of the vessels
carries the blood
 ( average velocity ) =
Q
A
( flow rate ) Q =  x A
 in aorte  30 cm/sec
 in capillary = 1 mm/sec
The relation between the velocity of blood ( l ) and
the cross – sectional area ( A ) in the circulatory
system .
2- The pressure difference ( P ) from one end to
the other end of the blood vessel :
P1
P
P2
l
Q ~ ( P1 – P2 ) = P
3- the length of the vessel Q ~ L1
4- The pressure gradiant Q ~ LP ,
mmHg
cm
5- The radius of the blood vessel Q  r4
Q =
a 4 P
8 l
[ applied to rigid tubes ]
but not applied exatly to blood vessel due to change
of radius through the heart rate .
The viscosity of blood :Q  1 if  is high Q is low
In the real fluid ( blood , water ) is
characterized by beingcompressible and by there
being internal forces acting in it , there is frictional
forces when the fluid is moving and such forces
lead to a loss of mechanical energy .
The property of a blood that determines the
magnitude of these disspitive forces is known as the
viscosity (  ) .
The unit of  :Cgs unit is poise ( P )
SI unit is pascal second ( PaS )
PaS = 10 P  1 Kgm-1S-1 = 10 P
1 P = 0.1 PaS
1P = dyne . cm-1 . sec-1
The dependence of  :
1. percentage of red blood cell in the blood (
hematocrit )
hematocrit increase  is increase
If  increase when hematocrit
polycrthemia ) deacrease in Q.
>
75%
 depend on the temperature ( T )
If T is large ,  is small and Q is high
If T is small ,  is high and Q is small
T of human from normal 37oC to 0oC
Increase the  of blood by a factor of 2.5 .
The flow rate through a tube depends on the :
1. pressure difference :
P = pressure , f = Q = flow
F2 = 2F1
2. The length of the tube :
F2= 2F1 at the same pressure
3. The viscosity of the ( fluid ) ( blood ) , water
F2= 2F1
4. The radius of the tube
(
F2 = 16 F1  Q = f = r4
The radius has the largest effect on flow rate of
the liquid .
The kind of blood flow :
1. Laminar flow :It is a slow , smooth and quietly flow of blood
in most blood vessels .
The properties of Laminar flow :
- the layers of blood in contact with the walls of
the blood vessels is essentially stationary .
- the next to the outside layer is moving slowly
and the layer in center of the vessel is moving
more rapidly .
- the effect of Laminar flow on the distribution
of red blood cells in the circulatory system is
not uniform .
( parobla shape )
There are more in the center than at the edges
And this produces two effects :
1. red blood cells in small vessel from the side of a
main vessel will be slightly due to the skimming
effect .
2. hematocrit in the extremities is high .
2. Turbulent flow :It is rapid and noisy flow of blood , for example
, where the blood is flowing rapidly past the heart
valves . The heart sounds heard with a stethoscope
are caused by turbulent flow , measurement of
blood pressure , the constriction produced by the
pressure cuff on the arm produces turbulent flow
and the resulting vibrations can be detected by
stethoscope on the brachial artery .
3. Critical flow :
The velocity of blood increase by reducing the
radius of the tube , it wiill reach a critical velocity (
Vc ) when Laminar flow changes into turbulent
flow :
L . f  T . f at V = Vc
Vc is critical velocity of blood
Vc  r ,  = viscosity of blood ,  = density of blood
, r = radius of vessel
Vc = K r , K = constant of proportional
K = Reynold number
K = 1000 (fluid, blood , ….) flowing (in long tube of
constant diameter)

But if there is bend or obstruction , K < 103  Vc is
lower
Example : Find the Vc of blood in the aorta of
radius 1 cm ( adult )

Vc = K r at normal state
blood
37
o
C
= 4 x 10-3 pas , of
= 1.04 g cm-3 103 Kg/m3 , r=1cm , K=103
 Vc =
103 x4 x103 pas
103 Kgm3 x10 2 m
Vc = 0.4 m/sec , the range of velocity of blood in
Aorta ( 0 – 0.5 m s-1 ) and thus the flow is turbulent
during part of the systole .
The efficiency of flowing :
From poiseuilles equation :

r 4P 
Q  8L 


Q P  R =
8L
r 4
the volume flow depends upon the pressure drop
(P ) per unit length of the tube :
L . f  T . f at V = Vc  P = Pc Q = Qc
At critical point at P = Pc , the flow becomes
turbulent and a great deal of energy is converted to
the KE of eddies .
From the curve
( slope )Lf >> ( slope )Tf
 Q 
 Q 

  

 P  Lf
 P Tf
increase in pressure  increase in Laminar flow
rate than in the turbulent flow rate .
the laminar flow is more efficilent than turbulent
flow .
flow through an artery with an obstruction :at normal artery ( 1 )
VA = QA at P1
VA = QA at P2 for artery ( 2 )
P2 > P give the same Q
If Q = VB at P1 for ( 1 )
at P2 for ( 2 )  need much pressure to give the
same Q
Heart sound :-The opening and closing of the heart valves
contribute to the heart sounds and the turbulent
flow occurs at these times and the vibrations
produced are in the audible rang .
- fig below show the sound heard with a
stethoscope from normal heart .
Cardiovascular diseases :
1- hypertension ( high blood pressure )
W = P dv  work done by the heart  tension of
heart muscle times how long it acts .
High blood  causes the muscle tension 
increase the work load of heart .
2- Fast heart rate ( tachy cardia )
increase the work load since the time of the heart
muscle spend contracting increasing .
3- heart attack
4- congestive heart failure :
causes by an enlargement of the heart from
laplace law :
T = rp , if r of the heart is large , the tension ( T ) of
the heart muscle must be large .
If r is double  T is double  at the same
pressure how the muscle of the heart is stretched ,
it may be not be able to produce sufficient force to
maintain normal circulation .
5. heart valve defect :
1. stenosis ( not open wide )
not close enough
( using artificial valve )
2. in sufficiency
6.blood vessels disease :
– aneurysm
is a weakening in the wall of an artery resulting
in an increase in it diameter ( R ) .
if R is increase  increase the tension in the wall of
the vessel .
- formation of sclerotic plagues can causes
turbulent flow and produce a murmur .
 artery  bagues  narrow region  r is
reduce  velocity of blood increase and decrease
in wall pressure due to Bernolli effect
 if this blaque may dislodge and travel with the
blood until it lodges in smaller artery and will shut
off the blood supply to the ( brain ) , it will produce
a stroke another type of cerebrovascular accident .
Question for CVS
1. A liquid of density 950 Kg m-3 flows in a
horizontal pipe of radius 4.5 cm in a section of
the tube of restricted radius 3.2 cm the liquid
pressure is 1.5 x 103 Nm-2 less than in the main
tube find the velocity of the liquid in the pipe ?
2. The level of blood in the bottle using for blood
transfusion is 1.3 meter above the needle which
has an internal diameter of 0.36 mm and 3 cm
in length . In one nunute 4.5 cm3 of liquid pass
through the needle . Find the viscosity of blood
at 37o C .
3. The radius of the aorta in human is 1 cm and
the cardiac output is 5 x 10-3 m2 /minute . what
is the average velocity of flow in the aorta ?
4. An artery with a 3 mm radius is partially
blocled with to plaque ; in the constricted
region the effective radius is 2mm and the
average blood velocity is 50 cm sec-1 find :
a. What is the average velocity of the blood in the
unconstricted region .
b. For the blood in the constricted region , find
the equivelent pressure due to the KE of the lood .