CHAPTER 1: PHYSICAL QUANTITY AND MEASUREMENT

Chapter 01 Physics

CHAPTER 1:
Physical quantities and

measurements
(F2F:1.5 Hours)

1

Chapter 01 Physics

Overview:

Physical quantities
and measurements

Dimensions Scalar and Vectors
of physical vector multiplication
quantities
quantities Scalar and
vector
vector
resolution products

2

Chapter 01 Physics

Learning Outcome:

1.1 Dimensions of physical quantities (0.5 hour)
At the end of this chapter, students should be able to:

 Define dimension.
 Determine the dimensions of derived quantities.
 Verify the homogeneity of equations using dimensional

analysis.

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Chapter 01 Physics

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]

 Table 1.1 shows the dimension of basic quantities.

[Basic Quantity] Symbol Unit

[mass] or [m] M kg

[length] or [l] Lm

[time] or [t] Ts

[electric current] or [I] A @ I A

[temperature] or [T]  K

[amount of substance] N mole

Table 1.1 or [N] 4

Chapter 01 Physics

 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.

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Chapter 01 Physics

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

Example 1.1 :

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
or interval

v = s
t

v = L = LT−1

T

The S.I. unit of velocity is m s−1.

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b. a = v Chapter 01 Physics
t
c.p= m v
a = LT −1
p= (M)(LT −1)
T
p= MLT−1
a = LT−2
S.I. unit : kg m s−1.
Its unit is m s−2.
m e. F = m a
d. ρ = V 
F = (M)(LT −2 )
ρ = l m h
w F = MLT−2

ρ= M S.I. unit : kg m s−2.

LLL 7

ρ= ML−3

S.I. unit : kg m−3.

Chapter 01 Physics

Example 1.2 :

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

( )= M LT−2 L
T

A = L2 and T  = θ = ML2T−3 8

Chapter 01 Physics

= Pnett 9
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

Chapter 01 Physics

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

Example 1.3 :

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.

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Chapter 01 Physics

Solution :

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. 11

Chapter 01 Physics

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 L− 1 = T−1
1 LT −2 2 2

 f  =  1 g
 2π 

l 

Therefore the equation above is homogeneous or dimensionally
correct.

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Chapter 01 Physics

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.

ANS. : 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

dimension of k.

ANS. : M L-1 T-1 13

Chapter 01 Physics

3. Show that the equation below is dimensionally correct.

Q = πR4 (P1 − P2 )

8ηL

Where R is the inside radius of the tube, L is its length, P1-P2 is the

pressure difference between the ends,  is the coefficient of viscosity

(N s m-2) and Q is the volume rate of flow ( m3 s-1).

4. Acceleration is related to velocity and time by the following expression

a = vxt y

Determine the x and y values if the expression is dimensionally consistent.

ANS. : x= 1; y= −1

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Chapter 01 Physics

5. Bernoulli’s equation relating pressure P and velocity v of a fluid moving in a

horizontal plane is given as

P + 1 v2 = k

2
where  is the density of the fluid and k is a constant. Determine the

dimension of the constant k and its unit in terms of basic units.

ANS. : M L−1 T−2; kg m−1 s−2

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Chapter 01 Physics

Learning Outcome:

1.2 Scalars and Vectors (1 hour)
At the end of this chapter, students should be able to:
 Define scalar and vector quantities.
 Resolve vector into two perpendicular components (x and

y axes).
 Illustrate unit vectors (I, j, k) in Cartesian coordinate.

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Chapter 01 Physics

Learning Outcome:

1.2 Scalars and Vectors (1 hour)

 State the physical meaning of dot (scalar) product:

A • B = A (B cosθ ) = B (A cosθ )

 State the physicalmeaning of cross (vector) product:

A  B = A (B s in θ ) = B ( A s in θ )

Direction of cross product is determined by corkscrew
method or right hand rule.

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Chapter 01 Physics
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

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Chapter 01 Physics

Vectors

Vector A Length of an arrow– magnitude of vector A
Direction of arrow – direction of vector A

 Table 1.4 shows written form (notation) of vectors.

displacement velocity acceleration

  
s v a
v a
s
v (bold) a (bold)
Table 1.4 s (bold)

 Notation of magnitude of vectors.


v =v

 = a 19
a

Chapter 01 Physics

 Two vectors equal if both magnitude and direction are the same.

(shown in figure 1.1) 

 Q 
P P=Q

Figure 1.1

 If vector A is multiplied by a scalar quantity k

⚫ Then, vector A is kA 

 kA

A


−A

 if k = +ve, the vector is in the same direction as vector A.
 if k = -ve, the vector is in the opposite direction of vector A. 20

Chapter 01 Physics

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

y x
v

50

0

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Chapter 01 Physics

c) Cartesian coordinates

 2-Dimension (2-D)

 = (x, y) = (1 m, 5 m)
s

y/m

5
s

01 x/m

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Chapter 01 Physics

 3-Dimension (3-D) 
s = (x, y, z) = (4, 3, 2) m

y/m

3

0 4 x/m
2
23
z/m

Chapter 01 Physics

d) Polar coordinates ( )

F = 30 N,150


F

150

e) Denotes with + or – signs. + +
-
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Chapter 01 Physics

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

0  x  x
Rx Rx
0

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

Ry = sinθ  R y = R s in θ Rx = sin  Rx = R s in  Opposite
componen25t
R R

Chapter 01 Physics

 The magnitude of vector R :
( )

R or R =
(Rx )2 + Ry 2

 Direction of vector R :

tanθ = Ry or θ = tan −1  Ry 
Rx Rx

 Vector R in terms of unit vectors written as
R=
Rxiˆ + Ry ˆj

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Chapter 01 Physics

27

Chapter 01 Physics

Example 1.3 :

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 v sin 30 v cos 60
30 sin 30 30 cos 60
a) vN = or vN =
=  =

 60 vN vN = 15 m s −1
v30
W E

vW b)vW = v cos 30 or vW = v sin 60
= 30 cos 30 = 30 sin 60

S vW = 26 m s −1

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Chapter 01 Physics

Example 1.4 : 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

 210 Vector x-component y-component

Fx S x  Fx = −F cos 30 Fy = −F sin 30
F
30  = −100 cos 30 = −100 sin 30

 Fy F x = − 8 6 .6 N Fy = −50 N
F

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Chapter 01 Physics

Example 1.5 :
y

O 

 F3(40 N )

F 1 (1 0 N ) x

60o



F2 (3 0 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.

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Chapter 01 Physics

Solution :  y 
F2 F2y F3

60o O x

  31
F2 x F1

   
Fr = F = F1 +F2 + F3
 Fr =  Fx + Fy
 Fx = F1x + F2x + F3x
 Fy = F1y + F2 y + F3 y

Chapter 01 Physics

Solution :

Vector x-component (N) y-component (N)

 0 − F1 = − 1 0
F1

 − 30 cos60 3 0 s in 6 0 
F2 = −15 = 26

 0

F3 F3= 4 0  F y = (− 10 )+ 26 + 0

Vector Fx = 0 + (− 15 )+ 40 = 16

sum = 2 5

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Chapter 01 Physics

Solution :

The magnitude of the resultant force is

F r = ( )F x 2 + ( )F y 2

= (2 5 )2 + (1 6 )2

F r = 2 9 .7 N y

  andθ=tan−1  F y    3 2 .6 
 Fx  Fy Fr
x
θ = tan −1  1 6  = 3 2 .6  O Fx

 25 

Its direction is 32.6 from positive x-axis OR

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1.2.6 Unit Vectors Chapter 01 Physics

At the end of this lesson, students should be able to:
✓ Illustrate unit vectors (I, j, k) in Cartesian coordinate.

 notations – aˆ, bˆ, cˆ

 E.g. unit vector a – a vector with a magnitude of 1 unit in the

direction of vector A. 
A
aˆ = A = 1
A aˆ

 Unit vectors have no unit. iˆ = ˆj = kˆ = 1

 Unit vector for 3 dimension axes : 34

x - axis ⇒iˆ@i(bold)
y - axis ⇒ ˆj @ j(bold )
z - axis ⇒kˆ @ k(bold )

Chapter 01 Physics

y

ˆj x
kˆ iˆ

z

 Vector can be written inrte=rmroxfiˆu+nitrvyeˆcjt+orsraz ksˆ:

⚫ Magnitude of vector,

( )r = (rx )2 + ry 2 + (rz )2

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 Chapter 01 Physics
s=
( )⚫
E.g. : 4iˆ + 3 ˆj + 2kˆ m

s = (4)2 + (3)2 + (2)2 = 5.39 m

y/m

3 ˆj

2kˆ 0 4iˆ x/m

z/m

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Chapter 01 Physics

1.2.7 Multiplication of Vectors

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

✓ State the physical meaning of dot (scalar) product:

( ) ( )A • B = A B c o s θ = B A c o s θ

Scalar (dot) product A
 Physical meaning of the scalar product

A Figure 1.4a θ : angle between tw o vectors
BA

B cosθ OR 

 Figure 1.4c B
Figure 1.4b
B A cosθ

  
( )A • B= of ( )B • A = B component of A parallel to B

A• B=
A component B parallel to A

A(B cosθ) B • A = B ( Acos )

A • B = B • A = AB cos θ 37

Chapter 01 Physics

 The scalar product is resulting a scalar quantity.

 The angle  ranges from 0 to 180 .

⚫ When scalar product is positive

0  θ  90

90  θ  180 scalar product is negative

θ = 90 scalar product is zero

( ) ( )W = F • s = F s cos θ Example of scalar product is work done = s F cos θ

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Vector (cross) product Chapter 01 Physics

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

✓ State the physical m eaning of cross (vector) product:

A  B = A (B s in θ ) = B ( A s in θ )

Direction of cross product is determined by corkscrew method or right

hand rule.  

 The vector product is defined as A  B = C
   
and its magnitude : A B = C = A B sinθ = AB sinθ

θ : angle between tw o vectors

 Angle  from 0 to 180   vector product always positive.

 Vector product resulting a vector quantity.
 The direction of C is determined by

RIGHT-HAND GRIP 39
RULE

Chapter 01 Physics

 Physical meaning of the magnitude of vector product:

AA

 Figure 1.6a  B

B

A

B s in θ A s in θ

 Figure 1.6b   B
( )A B

A B
= A component of B perpendicular to A

= A(B sin θ)  Figure 1.6c

( )B  A = B component of ABperpAen=dBic(uAlasrintoθB)

 =   = A(B sin  ) = B(Asin ) 40
A B B A

Chapter 01 Physics

 How to determine direction of the cross product?

⚫ Method:

 Point the 4 fingers to the direction of the 1st vector.

 Swept the 4 fingers from the 1st vector towards the 2nd vector.

 The thumb shows the direction of the vector product. 

  
CA B = C

 B
A
 
B C  

A ( o u t of p a ge )    B A = C


( )A B  B  A but A B = − B  A
⚫ Direction of the vector product (C ) always perpendicular
to the plane containing the vectors Aand B.
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Chapter 01 Physics

 Example of vector product: a torque (moment of force) on a metre rule.

Vector form :  = r  F

Magnitude form :  = rF sin θ

42

Chapter 01 Physics

1. Given three vectors P, Q and R as shown in Figure 1.3.

( ) y ( )

Q 24 m s−2 P 35 m s−2

( ) 50 x
0
R 10 m s−2

Figure 1.3

Calculate the resultant vector of P, Q and R.

ANS. : 49.4 m s−2; 70.1 above + x-axis

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Chapter 01 Physics

THE END.

Next Chapter…

CHAPTER 2 :
Kinematics of Linear Motion

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