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Principle of Mechanics
Chapter one
Measurements
1-1 Physical quantities
1-2 Basic physical quantities
1-3 Derived physical quantities
1-4 System of Units
1-5 Dimensional Theory
1-1: Physical Quantities
Physics is one of the most important branches of the basic science, because it provides basic principles and fundamental laws and formulas for other science branches in order to be used in their work. This make physics very important for most subjects such as chemistry, medicine, engineering, astronomy, earth science and other sciences. It also makes the study and understanding of what is happening around us easy.
The physical quantities measurement depends on the high precision and observations during the experimental work, while the student or the researcher is taking this measurement. This accurate measurement can be obtained by using the latest equipment.
Physical quantities are divided into two categories, which they known as basic quantities and derived quantities.
1-1-1: Basic physical quantities
The basic physical quantities are those quantities which do not need other physical quantities to define them; there are limited numbers of basic quantities. They can be represented by one unit and without having any particular direction. In mechanics, there are three basic physical quantities, mass, length and time. There are another quantities for other physics branches such as luminous intensity, thermodynamic temperature, amount of the substance, etc (see Table 1-1).
These physical quantities are defined as the amount of basic physical quantity, which is defined by only one unit, and they do not require another basic physical
quantity in their definition.
basic physical quantities Dimension Unit name Unit symbol
m المتر Length L الطول
s الثانية Time T الزمن
Kg كيلو جرام Mass M الكتلة
K الكلفن Temperature T درجة الحرارة
كمية المادةAmount
substance S المول mol
A األمبير Electric Current I شدة التيار
قوة اإلضاءةLuminous
Intensity I الشمعة Cd
Table (1-1): basic physical quantities
These physical quantities are defined as the amount of basic physical quantity, which is defined by only one unit, and they do not require another basic physical quantity in their definition.
1-1-2: Derived Physical Quantity
Derived physical quantities are those, which can be expressed by more than one basic physical quantity, this means that they need to be defined by more than one physical quantity. For example, speed is defined as the distance per time and its unit meter per second or kilometers per hour = x/t m/s
The units of the acceleration are known as the speed per time, and its unit is meter by second square meters or kilometers per hour square. Here we find ourselves needing two basic quantities to define units for the velocity and the acceleration. a= /t m/s2
Unit of the force is known as Newton, where the force is equal to the mass times acceleration and its unit is Newton. Newton is kilogram meter per second square, which is more than two basic physical quantities.
F = m a N (kg m/s2)
All these quantities are derived quantities, and they need more than one basic physical quantity to define their units. Sometimes a name of a scientist who discovered this phenomenon is used as the name of the unit as in the case of the units of the force as Newton for the purpose of simplicity. Newton is defined as kilograms meters per second square (N or kg ms-2). It is clear from the above that all physical quantities need to identify by a unit. These three examples illustrate the need to have a uniform and standard of measurement units as illustrated in Table 1-2.
الصيغة الرمز الكمية المشتقة
Area A المساحةالعرض× الطول
Velocity السرعةالزمن÷ المسافة
Accelerate a التسارعالزمن÷ السرعة
Density الكثافةالحجم÷ الكتلة
Force F القوةالتسارع× الكتلة
Table (1-2): Derived physical quantities
1-2: System of Units The units of measurement are a fundamental part of physical quantities and it required to define this quantity. The standard unit for a particular physical quantity is only valid for that quantity. Physical quantities are measured in terms of units; we
have already indicated that mechanics requires three fundamental quantities: length, mass, and time. The centimeter, meter and kilometer are examples of units of length, And the second, mint, hour and day are units of time, the kilogram and gram units of the mass.
To show the importance of the units we will take this example; if we say that a car move by speed of 120, this figure have no meaning it must follow by the appropriate unit which can be meter per second (m/s) or kilometers per hour (km/h) to determine the extent speed of this car. Another example if a mass of an object is 80; this not clear is it kg, gm, here lies the importance of the units. There are several conventional systems of units in which these quantities are expressed. The French system is centimeter-gram-second (cgs), the metric system is meter-kilogram-second (mks) and British system employs the foot, slug and pound, as their units.
Representatives of seventeen nations signed the Convention of the Metre (Convention
du Mètre) on 20th May 1875 in Paris. This diplomatic treaty provided the foundations
for the establishment of the Système International d'Unités (International System of
Units, international abbreviation SI) in 1960. Since then, national standards
laboratories have cooperated in the development of measurement standards that are
traceable to the SI.
Any organisation can achieve traceability to national standards through the correct use of
an appropriate traceable standard from NIS.
In 1971 at the General Conference of the measurements and weights in France, one system has been adopted which is known as System International Units or International System of Units, abbreviated SI units and popularly known as the metric system. The basic units of the system are the meter for measuring the length and kilograms as a measure of the mass, the second as a measure of the time (shown in the table (1-1).
basic physical quantities Dimension Unit name Unit symbol
m المتر Length L الطول
s الثانية Time T الزمن
Kg كيلو جرام Mass M الكتلة
K الكلفن Temperature T درجة الحرارة
mol المول Amount substance S كمية المادة
A األمبير Electric Current I شدة التيار
Cd الشمعة Luminous Intensity I قوة اإلضاءة
All other quantities in physics can be derived from these seven fundamental quantities. It is now has been applied in all parts of the world and one find that all textbooks using the SI units. The international system has become almost compulsory for all over the world. The four additional units in the table (1-1) there are symbols in parentheses are the ampere (A) for electric current, the Kelvin (K) for thermodynamic temperature and for temperature intervals, the candela (cd) for luminous intensity, and the mole (mol) for quantity of substance.
1-2-a: Standard length It has been agreed a long time to use the international standard of length which is meter. It has been defined as a bar made of alloy of platinum and Iridium materials; this was maintained by the Office of Weights and Measures International near Paris in France. In 1960, the scientists agreed to change this standard and to use the wavelength of the light emitted from Krypton 86 equal to 1,650,763.73. In 1983, the scientists agreed to change the standard meter to be the distance traveled by the light in a vacuum at a time of 1/ 299792458 of a second. The Scientists explained the recent change to be due to the constant value of the speed of light in vacuum, which are exactly equal 299792458 m/s.
1-2-b: Standard Mass The international unit for measuring the mass is the kilogram, which is a cylindrical alloy of platinum and Iridium maintained in the Office of Weights and Measures near Paris in France.
1-2-c: Standard time Prior to the Conference in Paris for the international unit for measuring the time, in 1967 the world was using the solar day as a measure of the time which equals to (1/86400) of the solar day. Later it was proved that this measure is not constant with time, because it changed by one second for each 30000 years. As a result, the scientists decided at the International Conference in 1967 to adopt the standard measure of time to be based on the cesium 133 radioactive isotope frequencies (133CS). It was agreed that one second is the measure unit of time and should equal to 9,192,631,770 oscillations emitted from the cesium-133 atom. Since the physical quantities have large and small amounts according to your need, thus in order to facilitate the expression of those amounts; the following table should be used
Symbol Multiple Unit Name
Da 10 Deka
H 102 Hecto
K 103 Kilo
M 106 Mega
G 109 Giga
T 1012 Tera
d 10-1 Deci
c 10-2 Centi
m 10-3 Milli
10-6 Micro
n 10-9 Nano
p 10-12 Pico
Table (1-3) Standard Prefixes
Example 1-1
The density of water is 1 gm/cm3. Find the density of water by the unit of Kilograms per cubic meter?
Solution
The density of water in Kilograms per cubic meter equal to
3
3
36
3
3
kg/m1000 m1
cm10
gm
kg
1000
1
cm
gm 1
gm/cm1 ρ
cm48.30in
cm54.2in)12(in12
cm02.120in
cm54.2in)25.47(in25.47
Example 1-2
If the length of the air condition (مكيف الحائط) is 47.25in and its width 12in. What are the air condition dimensions in centimeters and in meters? Solution (1) Note that 1 in = 2.54 cm,
Therefore the dimensions of the air condition by centimeter are:
The length is 120.02 cm and the width is 30.48 cm. (2)Note that the 1m = 100cm
Therefore the dimensions of the air condition by the meter are: length is 1.2m and the width is 3.1m.
m1.3m048.3100
m
cm
1cm)48.30(cm48.30
m2.1m2002.1100
m
cm
1cm)02.120(cm02.120
cm48.30in
cm54.2in)12(in12
cm02.120in
cm54.2in)25.47(in25.47
Therefore the dimensions of the air condition by the meter are: length is 1.2m and the width is 3.1m.
cm54.2 inch 1 inch , 12 ft 1 ft , 3 m 1
m56.8 = 44.91100xx21
m44.91=
m
cm 100
inch
cm54.2
ft
inch12
yard
ft 3 yards 100 =
2x
m100 = 1
x
Example 1-3
Two signs marked in a road (وضعت عالمتين على الطريق), one for distance 100 m and the other for 100 yard.
What is the difference between them in meters and feet?
Solution
The relationship between the international system (SI unit) and the British system is
Assume that the first distance is x1 and the second distance is x2 . 1) Finding the difference between the two distances in meters.
2) Finding the difference between the two distances in feet.
ft .. =
ftinch
inchcm .
mcm m . m = .xx
0828
1254210056856821
222cm45.6inch)1
inch
cm54.2(inch)1(
3633cm10)
m
cm100m1(m)1(
Example 1-4
Find the relationship between: A - Inch Square and centimeter square. B - Cubic meter and cubic centimeter. Solution A) Square inch is equal
B) Cubic meter
1-3: Dimensions Theory The students should be familiar with the physics laws, dimensions and measures of the physical quantities which they are studying, especially for those laws which students are using for the first time. Importantly, students should prove the correction of physics laws by using the dimensions theory. In this theory, each basic measureable physical quantity represented by a specific symbol written within square brackets is called a dimension. All other physical quantities can be derived as combination of the basic dimension. In this theory the symbol of the mass takes the symbol [M], the time [T] and length [L]. Using the square brackets [ ] to denote a physical quantity, and these symbols show only the measure of the quantity neither the unit nor the value.
Example 1-5
Verify the Newton’s second Law F = ma, note that “F” is the force which measured by Newton, “m” is the mass and measured by a kilogram and “a” is the acceleration measured by meters per second square?
Solution Left hand side:
Force “F” its unit is Newton and dimensions is [M LT-2]
Right hand side: “m” is the mass and its unit is kilogram (kg) and its dimension is “M”, The unit of the acceleration is distance divided by square of the time m/s2. Its dimension is [L/T2]. So the product of both (ma) gives the unit of the force which is Newton; its dimensions are [ML/T2].This means that the dimension of the right hand
side equals to the left hand side of the equation as
[ML/T2]= [ML/T2].
This should demonstrate the validity of Newton's Second Law.
Example 1-6
Verify the validity of Archimedes’s law F = V g. Note that: F is a symbol of the force with Newton “N” unit, V is a symbol of the volume and its unit is “m3 “, and the density has the unit of kilogram per cubic meter “kg/m3”. And g is the symbol of the gravitational acceleration with m/s2 unit. Solution Take each elements of the equation and find the dimensions of each one individually. Left hand side:
Force “F” its unit is Newton and its dimensions is [M LT-2] Right hand side:
The volume “V” with unit m3 and dimensions [L3] The density “” with unit kg/m3 and dimensions [M L-3] The gravitational acceleration “g” with dimensions [L T-2] By Equating the dimensions of the two sides of the equation F = V g, we find that:
22
2332
MLTMLT
LTMLLMLT
gvF
This proves that the corrections of Archimedes’s law.
Example 1-7
Verify the correction of the following wave frequency relationship
F/mL2
1f
by
using the theory of dimensions. Note that “f” is the frequency of the wire with unit “Hertz”, “F” is the force with unit “N”, “L” is the length of the wire with the unit “m” , and
m is the mass of the wire with the unit “kg”. Solution
Left hand side:
Since the frequency is the inverse of time, thus the dimensions of the frequency “f” is
1TT
1
Right hand side: Since the unit of the force “F” is “Newton” and “N” is equivalent to (kg m s-2 ), thus the dimensions of “F” is [M L T-2] , and the mass “m” has dimensions of [M], while the length “L” has dimensions of [L]. We can apply these dimensions to the right hand side of the equation and find that:
12
2
TTML
MLT
This means that the dimensions of the right hand equal to dimensions of the left hand This means that the relationship is correct.
Example 1-8
Prove that the numerical value of the surface tension (Ts) equals the numerical value of the energy surface (Es) using the theory of dimensions. Note that the surface energy is defined as the energy (E) divided by area (A) = Es
Surface tension force known as the force (F) divided by length (L) = Ts Solution Ts =F / L
Es =E / A
Left hand side: Now applying the theory of dimensions on Ts =F / L F has the dimensions [MLT-2] The dimensions of the length (L) = [L] ,
......(1) MTL
MLT
L
FT 2
2
s
Right hand side: “E” its unit is Joule which equals to kg m2 s-2. Thus the dimensions of “E” is [M L2T-2]
The dimensions of the area (A) is =[L2]
(2) ...... MTL
TML
A
EE 2
2
22
s
From equations (1,2), we find that both sides have equal dimensions, which means that the numerical value of surface tension Ts and surface energy Es are equal. This means that the relationship is true
Example 1-9
If the relationship between the distance (d) and acceleration (a)is given by the formula
YX ta2
1d
. Find the values of x and y which make this formula one of the laws of physics?
Solution By apply the dimensions theory we get: We must remember that any variable to the power zero is equal to one. For example T0=1, we can apply the dimension theory on this relation as: Left hand side: The dimension of “d” is [L]
Right hand side: The dimension of “a” is [L/T2], the dimension of “t” is [T]
By Equating the dimensions of the two sides of the equation
YXta
2
1d
we find that:
y2x-x0
y2x-x
yx2-0
T LL T
T T L
T LT L T
YX ta2
1d
2ta2
1d
From the last equation we find that the power of “L” in the left side equal to the power of L in the right side, thus x=1. In the same way the power of T (in the left side of the equation) equals to the power of T in the right side of the equation, so 0 = -2x + y. as a result of that we got: y=2.
Thus the relation
, becomes
Note that this relation, in the new form, is one of the known laws of physics.
Note that this relation, in the new form, is one of the known laws of physics.
