Faculty of Engineering
Engineering Physics I (BE121) Course
Fall 2012
Engineering physics I
Contents:
1.
2.
3.
4.
5.
6.
7.
8.
Units and dimensions.
Fluid mechanics.
Viscosity.
Heat transfer.
Thermodynamics.
Wave motion.
Sound waves.
Rotational motion.
Reference:
PHYSICS for Scientists and Engineers with Modern Physics Eighth Edition
Raymond A. Serway, John W. Jewett, Jr.
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Faculty of Engineering
Engineering Physics I (BE121) Course
Fall 2012
Chapter 1
Units and dimensions
Basic quantities and derived quantities:
We have six basic (fundamental) quantities which are mass, length,
time, temperature, current, and amount of substance. All other quantities can
be expressed in terms of these six quantities. These other quantities are
called derived quantities.
Systems of units:
There are three systems of units, shown in table below:
System
M.Kg.S (S.I.)
C.G.S (French)
F.P.S (English)
Units
Mass
kg
g
Pound
Length
m
cm
ft
Time
S
S
S
The S.I units uses also, the units of (A) and (K) for the electric current and
the temperature respectively.
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Faculty of Engineering
Engineering Physics I (BE121) Course
Fall 2012
Conversions between systems of units:
1 kg = 103 g.
1 m = 100 cm
1 foot = 12 inch.
Mass:
Length:
1 pound = 0.45 kg.
1 inch. = 2.54 cm
Dimensions:
The word dimension denotes the physical nature of the quantity. It is a
symbol given for the physical quantity that does not depend on the system of
units. Whether a distance is measured in units of feet, meter, or cm, it is still a
distance and we say its dimension is that of length.
We will use bracket [ ] to denote the dimension of a physical quantity.
Quantity
Length
Mass
Time
Dimension
L
M
T
Example:
What is the dimension of each one of the following physical
quantities: velocity, acceleration, force, work, power, pressure, and density?
Solution
length
time
[length] L
[v ] 
  LT 1
[time]
T
velocity 
Acceleration 
[a] 
change in velocity
time
[v] LT 1

 LT 2
[t ]
T
Force  mass  acceleration
[ F ]  [mass]  [a]  MLT 2
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Faculty of Engineering
Engineering Physics I (BE121) Course
Fall 2012
work  force  dis tan ce
[ w]  [ F ]  [dis tan ce]  ML2T 2
work
Time
[ work ] ML2T 2
[ power ] 

 ML2T 3
[time]
T
power 
Force
Area
[ force] MLT 2
[ pressure ] 

 ML1T 2
2
[area ]
L
mass
Density 
volume
[mass]
M
[density ] 
 3  ML3
[volume] L
pressure 
Dimensional analysis:
Dimension analysis is a powerful tool to:
1. Check whether a certain expression is correct or not.
2. Assist in derivation of simple physical expressions.
3. Convert from one system of units to another.
Notes on dimensional analysis:
1. Quantities can be added or subtracted only if they have the same
dimensions.
2. The L.H.S and the R.H.S of an equation must have the same
dimension.
Example:
Using dimensional analysis, check the validity of the following
relations:
i)
ii)
s  v0 t 
1 2
at ,
2
v f  v 0  as .
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Faculty of Engineering
Engineering Physics I (BE121) Course
Fall 2012
where (s) and vf are the displacement and final velocity of a particle after a
time (t) given that its initial velocity is (vo) and its acceleration is (a).
Solution
1
2
i) s  v0 t  at 2 ,
[L.H.S] = [s]=L
(1)
[R.H.S.] = [v0]×[t]+[a] × [t]2
= LT-1×T+LT-2 ×T2
=L+L
the expression is correct from the dimensional analysis point of view.
ii) v f  v0  as .
[L.H.S.] = [vf]=LT-1
[R.H.S.] = [v0] + [a] × [s]
= LT-1+ LT-2 ×L
= LT-1 + L2T-2
 The R.H.S is not correct, so the relation is not correct.
Example:
The acceleration (a) of a particle moving with uniform speed (v) in a
circle of radius (r) depends on both v and r. Using the dimensional analysis
get an expression for the acceleration. Given that the acceleration a is 1 m/s 2
when v is 3 m/s and r is 9m, find the dimensionless constant of
proportionality.
Solution
a= fn( v,r)
 a  v r
 a = k v r, where k is a dimensionless constant of proportionality,
[a]=[v] × [r]
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Faculty of Engineering
Engineering Physics I (BE121) Course
Fall 2012
LT-2= (LT-1)  × (L) 
LT-2= L+ T-
Comparing the power of T in both sides:
-=-2
 =2,
Comparing the power of L in both sides:
+ =1
 = -1,
 a = k v2 r-1
v2
 ak
r
To get the value of k,
v2
ak
r
2

3
1 k
9
k  1
Example:
The speed of a transverse pulse traveling on a taut spring depends on
the tension on spring (T) and its mass per unit length (). Using dimensional
analysis, find an expression for the speed (v) in terms of T and .
Solution
v= fn(T, )
v= k T
[v]= [T] × []
LT-1= (MLT-2) × (ML-1)
LT-1= M+ L- T-2
Comparing the power of T in both sides:
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-2=-1
 =0.5
Faculty of Engineering
Engineering Physics I (BE121) Course
Fall 2012
+= 0
Comparing the power of M in both sides:
vk
 =-0.5
T

Example:
The periodic time of complete oscillation (T) of a simple pendulum
depends on the string length (l) and the acceleration due to gravity (g), find
an expression for the oscillation time.
Solution
T= fn(l, g)
T= k lg
[T]= [l] × [g]
T= (L) × (LT-2)
T= L+ T-2
Comparing the power of T in both sides:
-2=1
 =-0.5
Comparing the power of L in both sides:
+= 0
 =0.5
 T k
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
g
Faculty of Engineering
Engineering Physics I (BE121) Course
Fall 2012
8
Faculty of Engineering
Engineering Physics I (BE121) Course
Fall 2012
Example:
Convert F= 15 N from S.I. to C.G.S and F.P.S.
Solution
F=15 N
[F]= MLT-2
In S.I the mass is measured in kg, the length is measured in m, and
time is in second, while in C.G.S the mass is measured in g, the length is
measured in cm, and time is in second, then:
2
 1000 g  100 cm  1s 

F  15 N  
 
 1kg  1m  1s 
 15  10 5
g.cm / s 2
In F.P.S the mass is measured in pound, the length is measured in
foot, and time is in second, then:
 1 pound  1 foot  1 s 

F  15 N  
 
0
.
45
kg
0
.
3048
m
 1 s 


 109 .36
2
pound. foot / s 2
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