CHAPTER 1: PHYSICAL QUANTITIES & MEASUREMENTS

CHAPTER 1

PHYSICAL
QUANTITIES

&
MEASUREMENTS

KMM

CHAPTER 1 :
PHYSICAL QUANTITIES
AND MEASUREMENT

1

1.0 Physical Quantities and Measurements

1.1 1.2 1.3
Dimensions Scalars and Significant figures
of Physical and uncertainties
Quantities Vectors
analysis
(Laboratory

Works)

2

Learning Outcome:

1.1 Dimensions of Physical Quantities

At the end of this chapter, students should be able
to:

a) Define dimension.
b) Determine the dimensions of derived quantities.
c) Verify the homogeneity of equations using
dimensional analysis.

3

1.1 Dimension of Physical Quantities

At the end of this lesson, students should be able to:
✓ Define dimension.

 Dimension is defined as a technique or method which the
physical quantity can be expressed in terms of
combination of basic quantities.

 It can be written as [physical quantity or its symbol]
or

 Defined as a measure of physical quantity (without numerical
values ).

4

Table 1 shows the dimension of basic quantities

[Basic Quantity] Symbol Unit
kg
[mass] or [m] M m
[length] or [l] L s
[time] or [t] T A
[electric current] or [I] A@I K
[temperature] or [T]  mole
N
[amount of substance] or

[N]

Table 1

5

Derived Quantities 6

➢The dimensions of derived quantities may include
few or all dimensions in individual basic quantities.

➢In order to understand the technique to write
dimensions of a derived quantity, we consider the
case of force.

Force is defined as: F = m.a
[F] = [m] [a]

Table 2 shows the dimension of derived quantities.

[Quantity] Dimension
Distance
L

Area L2

Volume L3

Velocity LT-1

Acceleration LT-2

Energy ML2T-2

Table 2

7

Dimensional Analysis

➢Dimensional Analysis is a technique to check the
correctness of an equation or to assist in deriving an
equation.

➢Dimensions (length, mass, time, etc) can be treated
as algebraic quantities.

➢add, subtract, multiply, divide

➢Both sides of equation must have the same
dimensions.

8

 Dimension can be treated as algebraic quantities through the procedure
called dimensional analysis.

 The uses of dimensional analysis are
⚫ to determine the unit of the physical quantity.
⚫ to determine whether a physical equation is dimensionally correct
or not by using the principle of homogeneity.

Dimension on the L.H.S. = Dimension on the R.H.S

⚫ to derive/construct a physical equation.
 Note:

⚫ Dimension of dimensionless constant is 1,
e.g. [2] = 1, [refractive index] = 1

⚫ Dimensions cannot be added or subtracted.
⚫ The validity of an equation cannot determined by dimensional analysis.
⚫ The validity of an equation can only be determined by experiment.

9

Dimensional Homogeneity

The Principle of According to this principle, we can
Dimensional multiply physical quantities with same or
Homogeneity
different dimensional formulae at our
convenience, however no such rule
applies to addition and subtraction,
where only like physical quantities can

only be added or subtracted.

All additive terms in E.g. If P + Q P & Q both represent same
a physical equation
must have the same physical quantity.
dimensions.

When working with equations possessing units, an equation

is said to be dimension homogeneity when both sides of the

equation possess the same units. 10

At the end of this lesson, students should be able to:
✓ Determine the dimensions of derived quantities.

Example 1:

Determine a dimension and the S.I. unit for the following quantities:

a. Velocity b. Acceleration c. Linear momentum

d. Density e. Force

Solution :

a. Velocity  = change in displacement
time interval
or

v = s
t

v = L = LT−1

T

The S.I. unit of velocity is m s−1. 11

b. a = v c.p= mv
t
p= (M)(LT−1)
a = LT−1
p= MLT−1
T
S.I. unit : kg m s−1.
a = LT−2
e. F= ma
Its unit is m s−2.
m F = (M () LT−2 )
d. ρ = V 
F = MLT−2
ρ = l  m h
w S.I. unit : kg m s−2.

ρ= M

LLL

ρ= ML−3

S.I. unit : kg m−3. 12

Example 2:

Determine the unit of k in term of basic unit by using the equation below:

( )Pnett = kAT14 −T04

where Pnett is nett power radiated by a hot object, A is surface area , T1 and T0

are temperatures.

Solution :

Pnett  = W  = F s = mas
t  t t 

A = L2 and T  = θ ( )= M LT −2 L
T
= ML2T−3

13

= Pnett
A T14 − T04
( )k

 Pnett

T14 − T04
( )k=
A

Since T14 = T0 4 = T 4

thus k  = Pnett 
AT 4

= ( )ML2T−3
(L2 )(θ4 )

k  = MT −3θ−4

Therefore the unit of k is kg s−3 K−4 14

At the end of this lesson, students should be able to:
✓ Verify the homogeneity of equations using dimensional analysis.

Example 3:

Determine Whether the following expressions are dimensionally correct or not.

a. 2s = 2ut + at 2 where s, u, a and t represent the displacement, initial

velocity, acceleration and the time of an object respectively.

b. v2 = u 2 − 2gt where t, u, v and g represent the time, initial velocity, final

velocity and the gravitational acceleration respectively.

c. f = 1 g where f, l and g represent the frequency of a
2π l

simple pendulum , length of the simple pendulum and the gravitational

acceleration respectively.

15

Solution : 16

a. Dimension on the LHS : 2s = 2s = L

( )Dimension on the RHS : 2ut = 2ut = (1) LT−1 (T) = L

and

  ( )( )at2 = at2 = LT−2 T2 = L

Dimension on the LHS = dimension on the RHS
Hence the equation above is homogeneous or dimensionally correct.

  ( )b. Dimension on the LHS : v 2 = LT−1 2 = L2T−2
  ( )Dimension on the RHS : u 2 = LT−1 2 = L2T−2

and

( )2gt = 2gt= (1) LT−2 (T) = LT−1

   Thus v2 = u2  2gt

Therefore the equation above is not homogeneous or dimensionally
incorrect.

Solution : 1 
T 
 c. Dimension on the LHS :f = = T −1

Dimension on the RHS :  1 g  =  1 g 1 l− 1
 l   2π 2 2
 
2π

( )( ) ( )= 1 − 1
1 LT−2 2 L 2 = T−1

 f  =  1 g
 2π 
 l 

Therefore the equation above is homogeneous or dimensionally
correct.

17

Exercise 1.1:

1. Deduce the unit of (eta) in term of basic unit for the equation below:
Δv
F = η Δl
A
where F is the force, A is the area, v is the change in velocity and l is

the change in distance.

[kg m-1 s-1]

2. A sphere of radius r and density s falls in a liquid of density f. It

achieved a terminal velocity vT given by the following expression:

( )vT 2 r2g
= 9 k ρs − ρ f

where k is a constant and g is acceleration due to gravity. Determine the 18
dimension of k.

[M L-1 T-1]

Learning Outcome:

1.2 Scalars and Vectors

At the end of this chapter, students should be able to:
a) Define scalar and vector quantities.
b) Resolve vector into two perpendicular components
(x and y axes).
c) Determine resultant of vectors. (remarks: limit to
three vectors only).

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1.2 Scalars and Vectors

At the end of this lesson, students should be able to:
✓ Define scalar and vector quantities.
• Scalar quantity is defined as a quantity with magnitude only.

• e.g. mass, time, temperature, pressure, electric current, work, energy, power
and etc.

• Mathematics operational : ordinary algebra

• Vector quantity is defined as a quantity with both magnitude & direction.
• e.g. displacement, velocity, acceleration, force, momentum, electric field,
magnetic field and etc.
• Mathematics operational : vector algebra

20

21

1.2.1 Resolving a Vector

At the end of this lesson, students should be able to:
✓ Resolve vector into two perpendicular components (x and y axes).

• 1st method :  2nd method :

yy

     
Ry R Ry R

 x  x
Rx Rx
0 0

Rx = cosθ  Rx = R cosθ Ry = cos  R y = R cos Adjacent
component
R R

Ry = sin θ  R y = R s in θ Rx = sin   Rx = R s in  Opposite 22
R component
R

Resolving vector into 2 perpendicular components (x and y axes)

A vector may be expressed in terms of its components.

y

with the aid of trigonometry:

AA cos  = Ax  Ax = A cos

yθ A

Ax x sin  = Ay  Ay = Asin

A

23

Direction of Vectors

Can be represented by using:
a) Direction of compass, i.e East, West, North, South, North-East, North-West,
South-East and South-West etc.

N

300 eg.: 300 West of North
W
E

S

b) Angle with a reference line y 
e.g. A boy throws a stone at a v
velocity of 20 m s-1, 50 above
horizontal. 50

x

24

Example 4:

A car moves at a velocity of 30 m s-1 in a direction north 60 west. Calculate
the component of the velocity.

a) due north. b) due west.

Solution : N

→ a)vN = v sin 30∘ or vN = v cos 60∘
= 30 sin 30∘ = 30 cos 60∘
v v60
N vN = 15 m s−1

W →30 E
vW
b)vW = v cos 30∘ or vW = v sin 60∘
= 30 cos 30∘ = 30 sin 60∘

S vW = 26 m s−1

25

Example 4: 210
S
x


F

A particle S experienced a force of 100 N as shown in figure above.
Determine the x-component and the y-component of the force.

Solution : y

Vector x-component y-component

 210 Fx = −F cos30 Fy = −F sin 30
Fx
S x  = −100 cos30 = −100sin 30
30 F

Fy
 F x = − 8 6 .6 N Fy = −50 N
F

26

1.2 Scalars and Vectors

At the end of this lesson, students should be able to:
✓ Determine resultant vectors (remarks: limit to three vectors)

Steps of Adding Vectors using Components:

1) RESOLVE each vector into its x and y
components.

• Pay careful attention to signs: any component that
points along the negative x or y axis get a − sign.

2) ADD all the x components together to get the x
component of resultant.

• Do not add x components to y components.

Rx = Ax + Bx + any other Ry = Ay + By + any other

27

3) The MAGNITUDE and DIRECTION the RESULTANT
VECTOR is given by:

• Vector diagram drawn help to obtain the correct
position of the angle θ.

• θ is measured from x – axis.

| R |= Rx 2 + Ry 2 28

tan = Ry

Rx

• The magnitude of vector R :

( )

R or R =
(Rx )2 + Ry 2

• Direction of vector R : or θ = tan−1 Ry 
Rx
tanθ = Ry
Rx

29

Example 5: y

O F3(40 N)

F1(10 N) x

60o

F2 (30 N)

The figure above shows three forces F1, F2 and F3 acted on a particle O.
Calculate the magnitude and direction of the resultant force on particle O.

30

Solution : F2 F2y
:

60o O F3

F2 x F1 x

Fr = F = F1 + F2 + F3 31

 F r =
Fx + Fy

 Fx = F1x + F2 x + F3x
 F y = F1 y + F 2 y + F3 y

Solution : x-component (N) y-component (N)
Vector
1 = 0 1 = −10
F1
2 = 30 sin 60°
F2 2 = −30 cos 60° = 16
= -15
F3 1 =0
3 =40
Vector Σ = −10 + 26 + 0
sum Σ = 0 + −15 + 40 = 16
= 25

32

Solution :

The magnitude of the resultant force is

Fr = ( Fx )2 + ( Fy )2

= (25)2 + (16)2

Fr = 29.7 N y

θand=tan −1 Fy   Fy Fr 32.6∘
 Fx 
x
θ = tan −1 16  = 32.6∘ O
 25   Fx

Its direction is 32.6 above positive x-axis

33

Example 6:

The figure above shows three forces F1, F2 and F3
acted on a particle O. Determine the magnitude of the

x-component and y-component of each force.

34

Example 6 : Solution 
F2 y
 F2 y
F3 x 30o F1
 
F3 F2 x 60o

30o O x


F3 y

Vector x-component y-component

 F1x = 0 N F1 y = F1
F1
 F2x = −30 cos 60  F1y = 10 N
F2 F2x = −15 N
F3x = −40 cos 30  F2 y = 30 sin 60 
 F3x = −34.6 N F2 y = 26 N
F3 F3 y = −40 sin 30 

F3 y = −20 N

Example 7:

The magnitudes of the 3 displacement vectors shown in drawing. Determine
the resultant value when these vectors are added together.

Resolve the vector into x and y-components.

Ay = 10 cos 45֯
By = 5 sin 30֯

Bx = 5 cos 30֯ Ax=10 sin 45֯

Solution : Vector Component x Component y

A 10 sin 45 = 7.07 10 cos 45 = 7.07

B –5 cos 30 = –4.33 5 sin 30 = 2.50
C 0 –8
(A+B+C)
 = 2.74  = 1.57
x y

Magnitude of resultant vector y = 1.57 
→ R
x = 2.74
( A + B + C) = x2 + y2 = R

= (2.74) 2 + (1.57) 2
= 3.16 m

Solution :

Direction of resultant vector

Tan  =  y = 1.57 = 0.573
2.74
x

 = 29.81 above + x

Resultant vector , R = 3.16 m at 29.81° above positive x-axis

Exercise 1.2:

1. y 2. y ( )

( ) ( ) P 35 m s−2

B 18.0 m s-1 Q 24 m s−2 x

( ) 37.0 x ( ) 50
0
A 12.0 m s-1 R 10 m s−2

0

Figure 1 Figure 2

a) Calculate the x-component and y-component of each force.
b) Calculate the resultant magnitude and direction of the forces.

1. (a) ANS: x-component= 2.38 −1 , y-component= 10.83 −1
1. (b) ANS: 11.09 −1 , 77.09° above positive x-axis

2. (a) ANS: x-component= 16.81 −2, y-component= 46.50 −2 39
2. (b) ANS: 49.45 −2 , 70.13 ° above positive x-axis