MODULE 1:
PHYSICAL QUANTITIES, UNITS AND VECTORS
MODULE 1: PHYSICAL QUANTITIES, UNITS AND VECTORS
1.1 Physics and its importance
The word physics comes from Greek, meaning “of nature” or “natural
philosophy”. Physics is concerned with the description of nature—that is, the
description and explanation of natural phenomena. In other words, physics is
concerned with how and why things work or behave the way they do.
Physics is an experimental science. Everything we know about the physical
world and about the principles that govern its behavior has been learned through
experiment, that is, through observations of the phenomena of nature. The ultimate
test of any physical theory is its agreement with experimental observations. These
observations usually involve measurements; thus physics is inherently a science of
experiment and measurements.
1.2 Physical Quantities
The study of Physics involves dealing with a lot of physical quantities. In
mechanics, we have the basic or fundamental quantities like mass, length and time.
All others are considered as derived quantities because they are obtained or
defined by simple relations between the fundamental quantities. The fundamental
quantities combined to form the derived quantity are sometimes called the
dimensions of the derived quantity.
Basic Quantities and Units
Unit
Symbol
Description
Definition
Meter
m
Measures
length
Distance light travel in 1/299792458 second.
Kilogram
kg
Measures
mass
Mass of a special platinum-iridium cylinder in
Paris.
Second
s
Measures
time
9192631770 oscillations of a special light
emitted by cesium-133 atoms.
Combinations of these basic quantities and units give various derived quantities
and units.
Example:
Force:1newton (N) = 1 kg-m/s2
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In the proper expression of physical quantities, there should at least be a
number (to indicate how large or how small the quantity is) and the unit (to indicate
the nature and type of the quantity). An expression that does not have one of these
two is meaningless.
1.3 Standards and Units
A standard is that quantity (usually in physical form i.e. an object) to which
other quantities being measured are compared. The measured quantity is then
expressed in terms of the standard, which now becomes the unit of the quantity.
Example: When we say the height of the building is 10 meters, it means
that the measured quantity (height of the building) is expressed in terms of
the length of an object (standard), which is considered to be one meter
long. Thus the "meter" is the unit for the height of a building.
1.4 Systems of Units:
There are two systems of units in common use: The English or British system
and the Metric system. A refinement of the old metric system was introduced in
1960 and is officially known as the International System of units or SI units. It is now
modern practice to use this system.
The English system is also known as the foot-pound-second (fps) system. The SI
system may be classified into the meter-kilogram-second (mks) system and the
Gaussian or centimeter-gram-second (cgs) system.
It is therefore necessary to look at the more salient aspects of the SI system:
1. In the SI system, the standard units for the different basic quantities are well
defined, clear and precise.
Example: For Length: 1 meter is defined as the distance traveled by
light in 1/299792458 sec
 The student is advised to look at the SI definitions for the other basic units.
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2. In the SI system of units, larger or smaller variations of these units are
obtained by attaching the proper prefix.
Prefixes in the Metric System or SI system
Prefix
Abbreviatio Power of 10
n
Prefix
Abbreviatio
n
Power of 10
exa
E
1018
deci
d
10-1
peta
P
1015
centi
c
10-2
tera
T
1012
milli
m
10-3
giga
G
109
micro

10-6
mega
M
106
nano
n
10-9
kilo
k
103
pico
p
10-12
hecto
h
102
femto
f
10-15
deka
da
101
atto
a
10-18
1.5 Unit Consistency and Unit Conversion
Unit consistency means that in a physical equation, each side of the
expression should have the same units otherwise the equation is an error.
Unit conversion is the process of changing the unit of a quantity to another
one within the same system or into another system. In physical computations, this is
usually done to attain unit consistency.
The process of unit conversion may be relatively easy but it has to be done in
an orderly manner to avoid errors.
One should also have considerable
knowledge of the needed conversion factors to be able to do it successfully.
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Steps in Performing Unit Conversion:
Example problems on Conversion of units:
1.
Convert 120 km/hr to mi/hr.
Solution:
120 𝑘𝑚
1𝑚𝑖
𝒎𝒊
×
= 𝟕𝟒. 𝟓𝟕
1 ℎ𝑟
1.6093 𝑘𝑚
𝒉𝒓
Note that the unit km cancels out due to division.
2.
As an astute observer walking around on continental crust (granite), you
might decide to test the hypothesis that the Earth is made entirely of granite.
You weigh a 1.00 ft3 piece of granite on your home scale and find that it
weighs 171 lbs. Thus you determine that the granite has a density of 171 lb/ft 3.
Convert your granite's density to g/cm 3.
Solution:
171 𝑙𝑏 1000𝑔
1𝑓𝑡
1𝑖𝑛
𝒈
𝑥
𝑥
𝑥
= 𝟐. 𝟕𝟒
𝑓𝑡
2.205𝑙𝑏 12𝑖𝑛
2.54𝑐𝑚
𝒄𝒎𝟑
Note that the units lb, ft3 and in3 cancels out due to division
Formative Problems: Practice conversion of units by solving the following problems.
1. The density of propane is 36.28 lb/ft3. Convert this to kg/m3. (Ans..581.67)
2. A box measures 3.12 ft in length, 0.0455 yd in width and 7.87 inches in height.
What is its volume in cubic centimeters? (Ans.. 7.91 x 103 cm3)
3. A block occupies 0.2587 ft3 . What is its volume in mm3 ? (Ans.. 7.326x106 mm3)
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1.6 Vectors and Vector Operations
Many physical quantities have magnitudes only but no direction. These are
called scalars. Examples are mass, time, density, temperature, etc. There are
however, many physical quantities such as force, velocity, displacement, etc.
which have directions as well as magnitude and these aspects always have to be
indicated when expressing these quantities. They are called vectors.
In physical computation and analyses, we have to be aware of the
difference between vectors and scalars because the mathematical treatments are
not the same. For example, we add scalars arithmetically but we cannot do the
same to vectors. Special methods are used.
1.6.1 BASIC ASPECTS ABOUT VECTORS
1. Vector Representation
a. Graphical representation - Vectors are represented by arrows.
b. Vector notations – Vectors are usually denoted with capital letters
written in boldface or with special markings. (A or A, B or B, etc.)
2. Indicating Directions of (coplanar) vectors:
METHOD 1: Using the angle θ that the vector makes with the “zero
reference line (usually the positive x-axis) ” measured going Counterclockwise.Illustration:
i.
Vector A = 3 units at 35o is a vector having a magnitude of 3 units,
and whose direction θA is 35o from the positive x-axis measured
going counter-clockwise.
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ii.
Vector A = 3 units at 1250 is a vector having a magnitude of 3 units,
and whose direction θA is 125o from the positive x-axis measured
going counter-clockwise
iii.
Vector A = 3 units at 2250 is a vector having a magnitude of 3 units,
and whose direction θA is 225o from the positive x-axis measured
going counter-clockwise.
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METHOD 2: Using Geographic Directions
Illustration: Let us assume that the figure below shows vector A = 3 units, θA = 25o
and vector B = 3 units, θB = 30o
The figure above means that:
i. Vector A = 3 units 25o EN (East of North)
ii. Vector A = 3 units 65o NE
iii. Vector B = 3 units 30o SW
iv. Vector B = 3 units 60o WS
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1.6.2 VECTOR OPERATIONS
1. VECTOR ADDITION AND SUBTRACTION:
Vector addition is the process of combining two or more vectors into
one. The combination is called the RESULTANT (R) of the vectors. Vector subtraction
is just like addition. In vector subtraction, the negative of one vector is added to the
other. For example, if two vectors A and B are to be added, the operation is
indicated as A + B. However, if vector B is to be subtracted from vector A, the
operation is indicated as A – B which is the same as A + (-B).

NOTE: The negative of a vector is another vector whose magnitude is the
same as the original vector but in the opposite direction.
METHODS OF VECTOR ADDITION
i.
The Algebraic method (for co-linear vectors only). Co-linear vectors are
vectors which lie along the same line.
Example: For the vectors shown in the diagram, determine a) their resultant; b) C A-D
E = 60 m
B = 20 m
C = 30 m
D = 25 m
A = 50 m
Solution:

For convenience we assign all vectors directed towards the right as
positive while all vectors directed towards the left are negative.
 Since vectors are co-linear simple arithmetic is applied
a) Resultant: R = A+B+C+D+E = 50m+(–20m)+(-30m)+ 25m+(-60m) = - 35 m, this
implies that the magnitude of the resultant vector has a magnitude of 35 m
and directed towards the left (negative sign)
b) C-A-D = C+(-A)+(-D) = -30m + (-50m ) + (-25m) = -101m, this implies that the
magnitude of C-A-D is 101m and directed towards the left (negative sign)
c) The Parallelogram Method
The procedure of "the parallelogram of vectors addition method" is
a.
b.
c.
d.
draw vector 1 using appropriate scale and in the direction of its action
from the tail of vector 1 draw vector 2 using the same scale in the direction of its action
complete the parallelogram by using vector 1 and 2 as sides of the parallelogram
the resulting vector R is represented in both magnitude and direction by the diagonal of the
parallelogram
A
A
A
R
B
B
B
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e.
Solve the resultant using sine law and cosine law
d) The Polygon method (Graphical method in determining the magnitude and
direction of the Resultant R)
Many vectors can be added together in this way by drawing the
successive vectors in a tip-to-tail fashion, as shown on the example below.
Scale: 1 cm = 1 unit
e) The Triangle method is similar to the Parallelogram Method but with the two
vectors connected from tip-to-tail.
Procedure:
a.
b.
c.
Construct the vector triangle by drawing the two vectors tip-to-tail. The vector that closes the
triangle is the resultant.
The resultant vector R of the two coplanar vectors can be calculated by trigonometry using
"the cosine law" for a non-right-angled triangle.
The angle between the vector and the resultant vector can be calculated using "the sine law"
for a non-right-angled triangle.
f) The Component Method
Ay
Many vector operations and analyses are carried
out using their
components. These are two or more vectors which when combined
or
A
added will give the original vector. For coplanar vectors (assumed to be on
the xy-plane) it is usually convenient to use two components which are
perpendicular to each other: one along the x-axis which is then called the xcomponent and the other one along the y-axis which is then called the ycomponent. These two components are collectively called the rectangular
components of the vector.
Determining the Components of a Vector
1. F1x is the magnitude of the x-component of vector F 1.
2. The sign of F1x is positive if it points in the positive x-direction,
negative if it points in the negative x-direction.
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3. F1y is the magnitude of the y-component of vector F1.
4. The sign of F1y is positive if it points in the positive y-direction,
negative if it points in the negative y-direction.
UNIT VECTORS (3 – dimensional vectors)
Let
Vectors having a magnitude of unity with no units. Its purpose is to
describe a direction in space.
𝚤 = unit vector pointing in the x – axis
𝚥̂ = unit vector pointing in the y – axis
𝑘 = unit vector pointing in the z – axis
y
𝚥̂
𝚤̂
x
𝑘
z
If A and B are in terms of their components:
A = Ax𝚤 + Ay𝚥 + Az𝑘
Addition:
and
B = Bx𝚤̂ + By𝚥̂ + Bz𝑘
A + B = (Ax + Bx)𝚤̂ +(Ay + By)𝚥 + (Az + Bz)𝑘
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PRODUCT OF VECTORS:
A) Scalar Product (Dot Product)
o Results to scalar quantity i.e. magnitude only, no direction.
A ∙ B = AB cos θ
B
Where A is the magnitude of vector A, and B is
the magnitude of vector B, and θ is the angle
between them.
θ
A

If θ = 90o, A∙B = AB cos 90o, cos 90o = 0, A∙B = 0; scalar product of 2
perpendicular vectors is always 0.

Using the unit vector computation:
A∙B = (Ax𝚤 + Ay𝚥 + Az𝑘 ) ∙ (Bx𝚤 + By𝚥 + Bz𝑘 )
A∙B = AxBx + AyBy + AzBz
B) Vector Product (Cross Product)
o Vector quantity with a direction perpendicular to the plane of the
vector and a magnitude given by:


A x B = AB sin θ
If A and B are parallel, θ = 0 or 180o then A x B = 0 since, sin 0 & sin 180o =
0.
There are always two directions perpendicular to a given plane. Use the
right hand rule.
USING VECTOR REPRESENTATION:
A x B = (Ax𝚤 + Ay𝚥 + Az𝑘 ) x (Bx𝚤 + By𝚥 + Bz𝑘 )
𝚤
Where:
𝚤 x𝚤=0
𝚤 x𝚥=𝑘
𝑘 x 𝚥 = -𝚤
𝚥x𝚥=0
𝚥 x𝚤 =-𝑘
𝚤 x 𝑘 = -𝚥
𝑘 x 𝑘 = 0𝚥 x 𝑘 = 𝚤
𝑘
𝑘 x𝚤 =𝚥
+
𝚥
vector product of two parallel vectors is always zero.
A x B = (AyBz – AzBy) ̂ + (AzBx – AxBz) ̂ + (AxBy = AyBx)𝒌
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SAMPLE PROBLEMS FOR VECTOR ADDITION:
1. Given are the following vector quantities:
A = 80 m due N
B = 40 m 300 NW
C = 60 m 150 NE
D = 50 m SE
Determine:
a. Magnitude and direction of the resultant of vectors A and B (using
Parallelogram Method)
b. Magnitude and direction of the resultant of vectors A and B (using
unit vectors)
c. Magnitude and direction of the Resultant of the four given vectors
using component method.
The given vectors drawn in the Cartesian-plane:
N
A=80m
B=40m
W
C=60m
300
150
45
E
0
D=50m
S
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a) Using PARALLELLOGRAM METHOD:
Let vector RAB represent the resultant of vectors A and B
N
RAB
A=80m
A=80m
1200
B=40m

300
W
E
S
By isolating the lower half of the parallelogram, our analysis in determining the
magnitude and direction of RAB can be determined by applying Sine and Cosine
Laws.
By cosine law:
RAB =
𝟖𝟎𝟐 + 𝟒𝟎𝟐 − 𝟐(𝟖𝟎)𝟒𝟎 𝐜𝐨𝐬 𝟏𝟐𝟎 = 𝟏𝟎𝟓. 𝟖𝟑𝒎
By sine law:
sin 𝛷 sin 120
=
105.83
80
⬚
Angle Φ = 40.89o
Direction of the resultant = 300 + 41.1436o = 71.1436o
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THEREFORE: RAB = 105.83m, 7.89 NW (or North of West)

Note that the Triangle method will have the same solution as the
parallelogram method. Connect the two vectors from tip-to-tail. The
Resultant is a vector drawn from the tail of the first vector to the tip of the
second vector.
B=40m
RAB
N
A=80m
W
E
S
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b) USING COMPONENT MET
N
A=80m
W
E
S
X-COMPONENT: AX =0
Y-COMPONENT: AY=80m
N
By
B=40m
300
W
E
Bx
X-COMPONENT: BX = 40 cos 30 = 34.64m
Y-COMPONENT: BY = 40 sin 30 = 20m
S
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By adding all x components of the vector: (vectors to the right “+”; vectors to the
left “-“), we have:
RABX = AX +BX = 0 +(-34.64) = - 34.64m
The negative sign means that the x-component of R AB is towards the left.
By adding all y-components of the vector: (vectors upward “+”; vectors
downward“-“), we have:
RABY = AY + BY = 80 + 20 = 100m
Since the value of the y-component of the resultant is positive, it means that it is
directed upwards.
The x and y components of the resultant can now be drawn as:
N
RAB
RABY = 100m
AB
W
E
RABX = 34.64m
S
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By Pythagorean Theorem:
RAB = √𝟑𝟒. 𝟔𝟒𝟐 + 𝟏𝟎𝟎𝟐 = 𝟏𝟎𝟓. 𝟖𝟑𝒎
From trigonometric functions of Right Triangles we have:
Tan θAB =
𝟏𝟎𝟎
𝟑𝟒.𝟔𝟒
θAB = 70.89O
THEREFORE: RAB = 105.83m, 7.89 NW (or North of West)
c). Magnitude and direction of the Resultant of the four given vectors
From Previous solution:
The Components of vector A are:
X-COMPONENT: AX =0
Y-COMPONENT: AY=80m
The Components of vector B are:
X-COMPONENT: BX = 40 cos 30 = 34.64m
Y-COMPONENT: BY = 40 sin 30 = 20m
For Vector C, the components are:
N
C=60m
Cy
150
W
Cx
E
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S
X-COMPONENT: CX = 60 cos 15 = 57.96m
Y-COMPONENT: CY = 60 sin 15 = 15.53m
For Vector D, the components are:
N
W
E
45
0
D=50m
Dy
Dx
S
X-COMPONENT: DX = 60 cos 45o = 42.43m
Y-COMPONENT: DY = 60 sin 15o = 42.43m
x-component of the Resultant:
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RX = AX + BX + CX + DX = 0 + (-34.64m) + 57.96m + 42.43m = 65.75m (to the right)
x-component of the Resultant:
RY = AY + BY + CY + DY = 80m + 20m + 15.53m + (-42.43m) = 73.1m (upward)
N
Ry
R
θ
W
Rx
E
S
RX = √𝟔𝟓. 𝟕𝟓𝟐 + 𝟕𝟑. 𝟏𝟐 = 𝟗𝟖. 𝟑𝟐𝒎
Tan θ =
𝟕𝟑.𝟏
𝟔𝟓.𝟕𝟓
θAB = 48.03O
THEREFORE: R = 98.32m, 48.03 NE (or North of EAST)
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Formative Problems:
1. A = 1km due south, B = 2km due west. Determine the resultant. (ans..2.24km,
63.4o W of S)
2. A = 72.4 m, 32.0° east of north, B = 57.3 m, 36.0° south of west, C = 17.8 m due
south. Determine the resultant. (ans..12.7m, 39o W of N)
SAMPLE PROBLEMS FOR DOT PRODUCT:
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SAMPLE PROBLEMS FOR CROSS PRODUCT:
EXERCISES:
1. Do the following conversions: (a) 15 m to ft, (b) 12 in to cm, (c) 30 days to sec
2. A football field is 300 ft long and 160 ft wide. What are the field’s dimensions
in meters and its area in square centimeters?
3. In the Bible, Noah is instructed to build an ark 300 cubits long, 50 cubits wide
and 30 cubits high. (A cubit was a unit of length based on the length of the
forearm and equal to half of a yard.) What would the dimensions of the ark
be in meters? What would its volume be in cubic meters? (Assume that the
ark was rectangular.)
4. Which is longer and by how many centimeters, a 100-m dash or a 100-yd
dash?
5. Vector is A =2.80 cm long and is above the x-axis in the first quadrant. Vector
is 1.90 cm long and is below the x-axis in the fourth quadrant.
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Find using triangle method for all the items:
a) A + B = C,
b) A – C
c) B + C
6. Three ropes pull on a large stone stuck in the ground, producing the vector
forces as shown in. Find the magnitude and direction of a fourth force on the
stone that will make the vector sum of the four forces zero.
DULE 2:
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