1_PHYSICAL Quantities to define the laws of physics

Physical quantities
 SI Units
Scalars and vectors
CHAPTER 1: Physical Quantities &
Units
1

Objectives
2
1. Show an understanding that all physical quantities
consist of a numerical magnitude and a unit
2. Express derived units as products or quotients of
the base units and use the named units listed in
this syllabus as appropriate
3. Use base units to check the homogeneity of
physical equations
4. Recall the following SI base quantities and their
units: mass (kg), length (m), time (s), current
(A), temperature (K), amount of substance
(mol)

Physical Quantity
 Is a property of something
which can be measured
 Example : length,
temperature, time.
 Every physical quantity
has a numerical value & a
unit
 Can be divided into
1. Base quantities
2. Derived quantities
 For standardization, unit of
physical quantities should
be given in SI unit.
3
length of 8 m
Numerical
value
unit
Physical Quantity

SI Base Quantities
4
Quantity SI Unit
Abbr.
 Length metre m
 Mass kilogram kg
 Time second s
 Electric current ampere A
 Thermodynamic temp Kelvin K
 Amount of a substance mole mol

SI Derived Quantities
5
Is combination of base quantities whether by
multiplication or division, but never by addition
and/or subtraction
Quantities Derivation unit
Area Wide (m) x length (m) = m2 squared meter
Volume Wide (m) x length (m) x
height (m) x
= m3 cubic meter
Density mass (kg) / volume (m3) = kg/m3 kilogram per cubic meter
Speed Distance (m) / time (s) = m/s meter per second
Acceleration Velocity (m/s)/ time (s) = m/s2 meter per second squared

Derived Quantities (cont)
6
 Force : N (Newton) = kgms-2
 Frequency : Hz (Hertz) = s-1
 Pressure : Pa (Pascal) =___________
 Energy : J (Joule) = kgm2s-2
 Power : W (Watt) =J/s =_____________
 Charge : C (Coulomb) = A s

Uses of base quantities
7
1. To find units of unknown quantities in an equation
 Example :
 What is the unit of h in the equation E= hf (E = energy of
a photon of light & f is its frequency)?
 E = hf
 h = E/f
 Unit of h = kg m2 s-2 / s-1 = kg m2 s-1
2. Can be used to check the homogeneity of a physical
equation
 Homogeneous/ balanced in an equation means each
term in the equation has the same base units
 Example (see slide after)

Uses of base quantities (cont)
8
 Example: Use base units to show that the following equation is
homogeneous
 Work done = gain in kinetic energy + gain in gravitational potential energy
 Terms here are : work, kinetic energy (gain) and gravitational potential
energy
 Work done = force x distance move (in direction of force)
= kgms-2 x m = kgm2s-2
 K.E = (½) mass x speed2 , so base unit is
= kg x m2s-2 ; as pure number (1/2) has no unit.
= kgm2s-2
 G.P.E = mass x gravitational field strength g x distance
= kg x ms-2 x m
= kgm2s-2
 So, as all terms have the same base units, so, the equation is
homogeneous

Exercise
9
Given : Energy = mass x (speed of light)2
1. The thermal energy Q is needed to melt a solid of mass m
without any change of temperature is given by the equation
Q = mL , where L is a constant
Find the base units of L
(Ans: m2s-2)

Learning outcome
10
 use the following prefixes and their symbols to
indicate decimal submultiples or multiples of both
base and derived units: pico (p), nano (n), micro
(μ), milli (m), centi (c), deci (d), kilo (k), mega
(M), giga (G), tera (T)

SI Prefixes
11
 A unit is often expressed with a prefix
 For example, the meter is may be written with the
prefix ‘kilo’ as the kilometer.
 The prefix represents a power of ten.
 In this case, the power of ten is 103

SI Prefixes
13
 Example
 Calculate the number of micrograms in 1.0 milligram
 1.0 mg = 1.0 x 10-3 g
 1.0 mg = 1.0 x 10-6 g
 So, 1 mg = (1.0 x 10-3 g/ 1.0 x 10-6 g) x 1.0 mg
 = 1.0 x 10-3 mg

Exercises
14
1. Calculate the area in cm2 of the top of a table
with sides of 1.2 m and 0.9 m (1.08 x 104 cm2)
2. Determine the number of cubic metres in one
cubic kilometre (1.0 x 109m3)

Exercises
15
1) Write down, using scientific notation, the values of the
following quantities
a) 6.8 pF
b) 32 mC
c) 60 GW
2) How many electric fires, each rated at 2.5 kW, can be
powered from a generator providing 2.0 MW of electric
power
3) An atom of gold has a diameter of 0.26 nm and the
diameter of its nucleus is 5.6 x 10-3 pm. Calculate the ratio
of the diameter of the atom to that of the nucleus

Mgh
l
T 
2

Exercises (cont)
16
4. Determine the base units of the following quantities:
a) Energy (= force x distance)
b) Specific heat capacity ( thermal energy change =
mass x specific heat capacity x temperature
change)
5. The period T of a pendulum of mass M is given by the
expression
 Where g is the gravitational field strength & h is a
length
 Determine the base units of the constant l

Learning Outcome
17
 show an understanding of and use the
conventions for labeling graph axes and table
columns as set out in the ASE publication Signs,
Symbols and Systematics (The ASE Companion to
16–19 Science, 2000)
 make reasonable estimates of physical quantities
included within the syllabus

Convention for symbol & units
18
When labeling graph & table :
1. Write symbol for a physical quantity first
2. Then, write a forward slash
3. Finally, write the unit
 Example
 Acceleration and time (physical quantity)
 Symbol for acceleration = a, time = t
 Unit for acceleration = ms-2, unit for time, s
a/ms-2 t/s

Order of magnitude of quantities
19
1. It is very useful if we able to estimate the size of
physical quantities.
 Valuable when planning & carrying out experiments
 As such, when a student plotting a graph of acceleration
of a free fall object, it will much easier for him/her to know
that acceleration of free fall at the earth’s surface is
9.81ms-2. If a value of 980 ms-2 is calculated, then this is
obviously wrong

Some values of distance quantities
20
Distance type Distance/m
Diameter of a nucleus 6 x 10-15
Diameter of an atom 3 x 10-10
Distance from earth to
sun
1.5 x 1011
Diameter of a hair 5 x 10-4
Diameter of a galaxy 1.2 x 1021

Exercises
21
Make estimates of the following quantities
1. the speed of sound in air
2. the density of air at room temperature and
pressure
3. the mass of a protractor
4. the volume, of a head of an adult person

Learning outcome
22
 Distinguish between scalar and vector quantities
and give examples of each
 Add and subtract coplanar vectors
 Represent a vector as two perpendicular
components.

Scalars & vectors quantity
23
 Scalar quantity = a quantity which can fully described by
magnitude only
 Vector quantity = a quantity which can fully described by
magnitude & direction
 (But, remember, all physical quantities have unit)
 Examples
 Mass = scalar
 Weight = vector
 Electric current = vector
 Temperature = scalar

Some example of scalar and vector
quantities
24
Quantity Scalar Vector
Mass √
Weight √
Speed √
Velocity √
Force √
Electric
current
√
Temperature √

Coplanar vector
25
 When two vectors are on the same plane then
they are known as coplanar vector

Vector representation
26
 One way to represent a vector is by means of an
arrow
 The direction of an arrow is the direction of the vector
quantity
 The length of the arrow, drawn to scale, represents
its magnitude
N
S
Scale : 1 unit represents velocity of
5ms-1
a) Velocity of 15 ms-1, due east
b) Velocity _____, due ____

Vector representation
27
 To describe a vector A , we will use:
 The bold font: A
 Or an arrow above the vector:
 In the pictures, we will always show vectors as
arrows
 Arrows point the direction of vector
 To describe the magnitude of a vector we will use
absolute value sign: or just A
 Magnitude is always positive,
 the magnitude of a vector is equal to the length of a
vector (if vector is drawn on scale)
A
tail
head
A

28
Moving Vectors
 Moving a vector does
not change it. A vector
is only defined by its
magnitude and
direction, not starting
location

Understanding Vector Directions
To accurately draw a given vector, start at the second direction
and move the given degrees to the first direction.
N
S
E
W
30° N of E
Start on the East
origin and turn 30° to
the North

Exercise: Graphical
Representation
 5.0 m/s East
(suggested scale: 1 cm = 1 m/s)
 300 Newtons 60° South of East
(suggested scale: 1 cm = 100 N)
 0.40 m 25° East of North
(suggested scale: 5 cm = 0.1 m)

Adding Vectors
31
 When adding vectors (only same type of vectors
can be added together) , their directions must be
taken into account
 Units must be the same
 We have 3 method to add vectors
1. Graphical Methods
 Use scale drawings
2. Geometry Methods
3. Component Methods

1. Graphical Addition of Vectors
1. Tip-To-Tail Method
“tip to tail” meaning the arrow-point end of one
vector is placed at the non arrow-point end of the
other vector
1. Pick appropriate scale, write it down.
2. Use a ruler & protractor, draw 1st vector to scale in
appropriate direction, label.
3. Start at tip of 1st vector, draw 2nd vector to scale,
label.
4. Connect the vectors starting at the tail end of the
1st and ending with the tip of the last vector. This =
sum of the original vectors, its called the resultant
vector.

Graphical Addition of Vectors (cont.)
Tip-To-Tail Method
5. Measure the magnitude of R.V. with a ruler. Use
your scale and convert this length to its actual
amt. and record with units.
6. Measure the direction of R.V. with a protractor and
add this value along with the direction after the
magnitude.
Example: 3.0 m/s 20° S of E

Example
34
 A ship is travelling due North with a speed of 12kmh-1
relative to the water. There is a current in the water flowing at
4.0 kmh-1in an easterly direction relative to the shore.
Determine the velocity of the ship relative to the shore by
scale drawing
 Determine Scale: let say 1cm = 2kmh-1
 Draw vector according to scale &
 Connect them as tip to tail
 Draw the resultant vector
 Measure the length of resultant vector as magnitude
& using protractor, find its angle/direction
N
12kmh-1
6 cm
4kmh-1
2 cm
 12.6 kmh-1
12.6 kmh-1 in a
direction 18o East of
North

Graphical Addition of Vectors
(cont.)
35
 You are given vectors A & B (in scale drawing). Find the
resultant vectors of A & B
Arrange the
vectors in a tip
to tail fashion.
The resultant is drawn
from the tail of the first
to the tip of the last
vector.
Vector triangle

(cont.)
Add the Given Vectors
A + B + C + D
 When you have many
vectors, just keep repeating
the process until all are
included
 The resultant is still drawn
from the origin of the first
vector to the end of the last
vector
 Then, the resultant value
can be obtained using scale
diagram
B
A



B
A



C
B
A





B
A




(cont.)
37
2. Parallelogram Method
A parallelogram is a four sided figure with
opposite sides parallel and equal in length.
 Draw the vectors required to sum up using arrows to
indicate the direction.
 Form a parallelogram with the vector.
 The resultant vector (R.V) is the line joining from the point
with 2 tails to the point with 2 tips.

Parallelogram Method
38
60o 60o
B
B
A
A
 Measure the resultant vector (length as the
magnitude and its angle,  as the direction)


Graphical Subtraction of Vectors
 Vectors in the same direction can be simply added
 Subtraction of vectors is actually the addition of vectors in an
opposite direction or addition of a negative vector
 The negative of a same vector has the same magnitude, but
in the 180° opposite direction.
 Example: 20 N – 30 N
 = 20 N + (-30N)
=
20 N - 30 N
+ -10 N

Graphical Subtraction of Vectors
(cont.)
 Subtracting Vectors act in direction between 0o to 180o
Graphically
Flip one
vector.
Then
proceed to
add the
vectors
The resultant is
drawn
from the tail of the
first
to the head of the
last
C

Adding Vectors Using Geometry
Method
Right Triangle
a
c
b
A
B
C
c is the hypotenuse
c2 = a2 + b2
sin = o/h cos = a/h tan = o/a
A + B + C = 180°
tan A = a/b
tan B = b/a
B = 180° – (A + 90°)
Quick Review

For right triangles:
 If the two vectors are at 90o use Pythagoras’
Theorem
 Right angles triangle is a triangle which result from
two vectors which are perpendicular to each other
1. Draw a tip to tail sketch first.
2. To determine the magnitude of the resultant
 Use the Pythagorean theorem.
3. To determine the direction
1. Use the tangent function.

These Laws Work for Any
Triangle.
a
c
b
C
B A
A + B + C = 180°
Law of sines:
a = b = c
sin A sin B sin C
Law of cosines:
c2 = a2 + b2 –2ab Cos C
b2 = a2 + c2 –2ac Cos B
a2 = b2 + c2 –2bc Cos A

Use the Law of:
 Sines when you know:
 2 angles and an opposite side
 2 sides and an opposite angle
 Cosines when you
know:
 2 sides and the angle
between them

Example using trigonometry
 A fielder in a cricket match
throws the ball to the wicket
keeper. At one moment of
time, the ball has a
horizontal velocity of 16 ms-1
and a velocity in the
vertically upward direction of
8.9 ms-1. Determine for the
ball :
a) its resultant speed
b) The direction in which it
is travelling relative to
the horizontal
 Right triangles
45
16 ms-1
8.9 ms-1
 R2 = 8.92 + 162
 = 18.3 ms-1
 tan  = 8.9 ms-1/16 ms-1
= 29o to horizontal
R

R
8.9 ms-1
16 ms-1

Eg.using trigonometry (Non- Right
Triangle)
 Two forces act at a point P as
shown below. Using Trigonometry
function, determine the resultant
force
Find magnitude of R
 c2 = a2 + b2 –2ab Cos C
 C = 140o, a =50 N, b = 80 N, c = R
 R2 = 502 + 802 – 2(50)(80) Cos
140o
 R = 123 N
2. Find the direction of R
b/sin B = c/sin C
(80/ sin ) = (123/ sin 140o)
sin  = (80 x sin 140o / 123)
 = 25o to the 50 N force in
anticlockwise direction
50N
80N
40o
We know, two sides
and
angles between
them,
so, find magnitude
using cosine rule
80N
R
50N
40o
50N
P
P


Resolution of vectors
47
 Any vector can be described as having both x
and y components in a coordinate system
 The process of breaking a single vector into
its x and y components is called vector
resolution.
 Vectors are said to be in equilibrium if their
sum is equal to zero.
E + R = 0 E = - R
E = 5 N R = 5 N
at 180 ° at 0°

Solving Vector Problems by Component
Method
 Each vector is replaced by 2 perpendicular vectors
called components.
 Vector component diagram always yields a right
triangle x and y component
 Add the x-components and the y-components to find
the x- and y-components of the resultant.
 Use the Pythagorean theorem and the tangent
function to find the magnitude and direction of the
resultant.

Vector Resolution
49
h
y = h sin 
x = h cos 

x
y

Components of Force:
50
 The pulling force by the buy can be resolve into X
and Y component
x
y

Example: Adding vectors by
component method
51
5 N at 30°
6 N at 135°
x y
5 cos 30° = +4.33 5 sin 30° = +2.5
6 cos 45 ° = - 4.24 6 sin 45 ° = + 4.24
+ 0.09 + 6.74
R = (0.09)2 + (6.74)2 = 6.74 N
 = arctan 6.74/0.09 = 89.2°

10 km at 30
6 km at 120
Solve the following problem using the
component method.

Equilibrium for 2 Coplanar
Forces
53
 We can understand that if two coplanar forces are
unbalanced, there will be a resultant force
 But, if forces are in equilibrium, it means that they are
balanced
 Two balanced forces are equal in magnitude but opposite in
direction to the other.
The free body
diagram shows
that the resultant
force is zero.
The free body
diagram shows
that there is a
resultant force.

Equilibrium for 3 Coplanar Forces
1. The forces form into a
closed triangle.
2. The directions of the
forces go round the
triangle.
3. That does not mean
that there are no
forces; the forces
balance each other
out
54
 3 coplanar forces act at a point are in equilibrium if they
form a closed triangle vectors.

Equilibrium for Coplanar Forces
55
You will have done an
experiment like this to show
the idea:
Each mass exerts a force
equivalent to its weight. We can
put the all force vectors tip -to-
tail:

1_PHYSICAL Quantities to define the laws of physics

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