PHYSICS CHAPTER 1PHYSICS CHAPTER 1

CHAPTER 1:CHAPTER 1:Physical quantities and Physical quantities and

CHAPTER 1

PHYSICAL

QUANTITIES

AND

MEASUREMENTSPhysical quantities and Physical quantities and measurementsmeasurements

(3 Hours)(3 Hours)

1

MEASUREMENTS

UNIT FIZIKKOLEJ MATRIKULASI MELAKA

PHYSICS CHAPTER 1

Learning Outcome:

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

�� StateState basic quantities and their respective SI units: length basic quantities and their respective SI units: length (m), time (s), mass (kg), electrical current (A), temperature (m), time (s), mass (kg), electrical current (A), temperature (K), amount of substance (mol) and luminosity ((K), amount of substance (mol) and luminosity (cdcd).).

( Emphasis( Emphasis on units in calculation)on units in calculation)

1.1 Physical Quantities and Units (1 hours)

2

StateState derived quantities and their respective units and derived quantities and their respective units and symbols: velocity (m ssymbols: velocity (m s--11), acceleration (m s), acceleration (m s--22), work (J), ), work (J), force (N), pressure (Pa), energy (J), power (W) and force (N), pressure (Pa), energy (J), power (W) and frequency (Hz). frequency (Hz).

�� State and convert State and convert units with common SI prefixes.units with common SI prefixes.

2

PHYSICS CHAPTER 1

1.1 Physical Quantities and Units�� Physical quantityPhysical quantity is defined as a quantity which can be measured.quantity which can be measured.

� It can be categorized into 2 types

�� Basic (base) quantityBasic (base) quantity

�� Derived quantityDerived quantity

�� Basic quantityBasic quantity is defined as a quantity which cannot be derived quantity which cannot be derived from any physical quantities.from any physical quantities.

� Table 1.1 shows all the basic (base) quantities.� Table 1.1 shows all the basic (base) quantities.

3

Quantity Symbol SI Unit Symbol

Length l metre m

Mass m kilogram kg

Time t second s

Temperature T/θ kelvin K

Electric current I ampere A

Amount of substance N mole mol

Table 1.1Table 1.1

PHYSICS CHAPTER 1

�� Derived quantityDerived quantity is defined as a quantity which can be expressed quantity which can be expressed in term of base quantity.in term of base quantity.

� Table 1.2 shows some examples of derived quantity.

Derived quantity Symbol Formulae Unit

Velocity v s/t m s-1

Volume V l ×××× w ×××× t m 3

Acceleration a v/t m s-2

4

Acceleration a v/t m s

Density ρρρρ m/V kg m-3

Momentum p m ×××× v kg m s-1

Force F m ×××× a kg m s-2 @ N

Work W F ×××× s kg m2 s-2 @ J

Pressure P F/A N m-2 @ Pa

Frequency f 1/T s-1 @ Hz

Power P W/t kg m2 s-3 @ Js-1 @W

Table 1.2Table 1.2

PHYSICS CHAPTER 1

Prefix Multiple Symbol

tera × 1012 T

giga × 109 G

mega × 106 M

kilo × 103 k

� It is used for presenting larger and smaller values.for presenting larger and smaller values.

� Table 1.3 shows all the unit prefixes.

1.1.1 Unit Prefixes

kilo × 103 k

deci × 10−1 d

centi × 10−2 c

milli × 10−3 m

micro × 10−6 µ

nano × 10−9 n

pico × 10−12 p

5

� Examples:

� 5740000 m = 5740 km = 5.74 Mm

� 0.00000233 s = 2.33 × 10−6 s = 2.33 µs

Table 1.3Table 1.3

PHYSICS CHAPTER 1

Solve the following problems of unit conversion.

a. 15 mm2 = ? m2 b. 65 km h−1 = ? m s−1

c. 450 g cm−3 = ? kg m−3

Solution :Solution :

a. 15 mm2 = ? m2

Example 1.1 :

( ) ( )232m10mm 1 −=

×=−

h 1

m1065h km 65

31

6

b. 65 km h-1 = ? m s-1

11stst method :method :

×=−

s 3600

m1065h km 65

31

11 s m 81h km 65 −− =

( ) ( )m10mm 1 =262 m 10mm 1 −=

PHYSICS CHAPTER 1

=−

h 1

km 65h km 65 122ndnd method :method :

11 s m 18h km 65 −− =

=−

s 3600

h 1

km 1

m 1000

h 1

km 65h km 65 1

7

c. 450 g cm-3 = ? kg m-3

( )

=

−

−−

332

33

3

3

m 10

cm 1

g 1

kg 10

cm 1

g 450cm g 450

353 m kg 10.54cm g 450 −− ×=

PHYSICS CHAPTER 1

Follow Up Exercise

1. A hall bulletin board has an area of 250 cm2. What is this area in square meters ( m2 ) ?

2. The density of metal mercury is 13.6 g/cm3. What is this density as expressed in kg/m3

3. A sheet of paper has length 27.95 cm, width 8.5 cm and

8

3. A sheet of paper has length 27.95 cm, width 8.5 cm and thickness of 0.10 mm. What is the volume of a sheet of paper in m3 ?

4. Convert the following into its SI unit:(a) 80 km h–1 = ? m s–1

(b) 450 g cm–3 = ? kg m–3

(c) 15 dm3 = ? m3

(d) 450 K = ? ° C

PHYSICS CHAPTER 1

Learning Outcome:

At the end of this chapter, students should be able to:At the end of this chapter, students should be able to:a)a) DefineDefine scalar and vector quantities, scalar and vector quantities, b) Perform vector addition and subtraction operations b) Perform vector addition and subtraction operations

graphicallygraphically..

(Emphasise the meaning of positive and negative vectors)(Emphasise the meaning of positive and negative vectors)

c)c) Resolve vectorResolve vector into two perpendicular components (x and y into two perpendicular components (x and y axes)axes)

1.2 Scalars and Vectors (2 hours)

axes)axes)( Emphasise on resolving vector) ( Emphasise on resolving vector)

d) d) IllustrateIllustrate unit vectors ( ) in unit vectors ( ) in cartesiancartesian coordinate.coordinate.e) e) StateState the physical meaning of the physical meaning of dot (scalar) product dot (scalar) product

f) f) State State the physical meaning of cross ( vector) product:the physical meaning of cross ( vector) product:

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

9

ˆˆ ˆ, ,i j k

( ) ( )θABθBABA coscos ==•��

( ) ( )θABθBABA sinsin ==×��

PHYSICS CHAPTER 1

1.2 Scalars and Vectors

�� ScalarScalar quantity is defined as a quantity with magnitude quantity with magnitude only.

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

� Mathematics operational : ordinary algebra

�� Vector Vector quantity is defined as a quantity with both magnitude quantity with both magnitude �� Vector Vector quantity is defined as a quantity with both magnitude quantity with both magnitude & direction.& direction.

� e.g. displacement, velocity, acceleration, force, momentum, electric field, magnetic field and etc.

� Mathematics operational : vector algebra

10

PHYSICS CHAPTER 1

1.2.1 Vectors

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

Vector ALengthLength of an arrow– magnitudemagnitude of vector A

displacement velocity acceleration

DirectionDirection of arrow – directiondirection of vector A

s�

11

� Notation of magnitude of vectors.

v�

a�

s av

vv =�

aa =�

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

PHYSICS CHAPTER 1

P�

�� Two vectorsTwo vectors equal if both magnitude and directionmagnitude and direction are the same. same. (shown in figure 1.1)

� If vector A is multiplied by a scalar quantity k

Q�

QP��

=

Figure 1.1Figure 1.1

12

� If vector A is multiplied by a scalar quantity k

� Then, vector A is

� if kk = +ve= +ve, the vector is in the same directionsame direction as vector A.

� if k k = = --veve, the vector is in the opposite directionopposite direction of vector A.

Ak�

Ak�

A�

A�

−

PHYSICS CHAPTER 1

1.2.2 Direction of Vectors

� Can be represented by using:

a)a) Direction of compassDirection of compass, i.e east, west, north, south, north-east,

north-west, south-east and south-west

b)b) Angle with a reference lineAngle with a reference line

e.g. A boy throws a stone at a velocity of 20 m s-1, 50° above

horizontal.

y

13

50°

v�

x

y

0

PHYSICS CHAPTER 1

c)c) CartesianCartesian coordinates

� 2-Dimension (2-D)

m) 5 m, 1(),( == yxs�

s�

y/m

5

14

s

x/m10

PHYSICS CHAPTER 1� 3-Dimension (3-D)

3

m 2) 3, 4,(),,( == zyxs�

y/m

4 i + 3 j + 2 k s =�

15

s�

2

4

x/m

z/m

0

PHYSICS CHAPTER 1

Unit vectorsA unit vector is a vector that has a magnitude of 1 with no units.

Are use to specify a given direction in space.

i , j & k is used to represent unit vectors pointing in the positive x, y & z directions.

16

pointing in the positive x, y & z directions.

| | = | | = | | = 1i j k

PHYSICS CHAPTER 1

( )��

N,150 30=Fd)d) PolarPolar coordinates

F�

150°

17

e)e) DenotesDenotes with + or + or –– signssigns.+

+-

-

PHYSICS CHAPTER 1

1.2.3 Addition of Vectors� There are two methods involved in addition of vectors graphically i.e.

�� ParallelogramParallelogram

�� TriangleTriangle

� For example :

B�

A�

BA��

+

18

ParallelogramParallelogram TriangleTriangle

B

B�

A�

BA��

+

O

B�

A�

BA��

+

O

PHYSICS CHAPTER 1

� Triangle of vectors method:

a) Use a suitable scale to draw vector A.

b) From the head of vector A draw a line to represent the vector B.

c) Complete the triangle. Draw a line from the tail of vector A to the

head of vector B to represent the vector A + B.

ABBA����

+=+ Commutative RuleCommutative Rule

19

B�

A�

AB��

+

O

PHYSICS CHAPTER 1

RQP���

++� If there are more than 2 vectors therefore

� Use vector polygon and associative rule. E.g.

R�Q

�

P�

20

R�

Q�

P� ( )QP

��

+

( ) ( )RQPRQP������

++=++ Associative RuleAssociative Rule

( ) RQP���

++

PHYSICS CHAPTER 1

� Distributive Rule :

a.

b.

� For example :

Proof of case a:Proof of case a: let αααα = 2

( ) BABA����

ααα +=+

( ) AAA���

βαβα +=+number real are , βα

( ) ( )BABA����

+=+ 2α

21

( ) ( )BABA����

+=+ 2α

B�

A�

BA��

+

O ( )BA��

+2

PHYSICS CHAPTER 1

B�

2

BA��

22 +

BABA����

22 +=+αα

22

A�

2O

( ) BABA����

222 +=+ �

PHYSICS CHAPTER 1

Proof of case b:Proof of case b: let αααα = 2 and ββββ = 1

A�

( ) ( ) AAA���

312 =+=+ βα

A�

3

23

A�

3

AAAA����

12 +=+ βα

A�

2 A�+

A3

=

( ) AAA���

1212 +=+ �

PHYSICS CHAPTER 1

1.2.4 Subtraction of Vectors� For example :

D�C

�

DC��

−

D�

−

( )DCDC����

−+=−

24

ParallelogramParallelogram TriangleTriangle

O O

( )

C�

D�

−

DC��

−

C�

D�

−DC��

−

PHYSICS CHAPTER 1

� Vectors subtraction can be used

� to determine the velocity of one object relative to another object

i.e. to determine the relative velocity.

� to determine the change in velocity of a moving object.

1. Vector A has a magnitude of 8.00 units and 45° above the positive x

axis. Vector B also has a magnitude of 8.00 units and is directed along

the negative x axis. Using graphical methods and suitable scale to

Exercise 1 :

the negative x axis. Using graphical methods and suitable scale to

determine

a) b)

c) d)

(Hint : use 1 cm = 2.00 units)

25

BA��

+ BA��

−

B2A��

+ BA2��

−

PHYSICS CHAPTER 1

1.2.5 Resolving a Vector�� 11stst method method :

R�

yR�

θ

y

�� 22ndnd method method :

R�

yR� φ

y

φ

26

xR�

θ

0x

θR

Rx cos= θ R Rx cos=⇒

θR

Rysin= θ R Ry sin=⇒

xR�

0x

φsin=R

Rx φsin R Rx =⇒

φcos=R

Ryφcos R Ry =⇒

PHYSICS CHAPTER 1

� The magnitude of vector magnitude of vector R :

�� Direction of vectorDirection of vector R :

( ) ( )22or yx RRRR +=�

R R

� Vector R in terms of unit vectors written as

27

x

y

R

Rθ =tan

or

= −

x

y

R

Rθ

1tan

jRiRR yxˆˆ +=

�

PHYSICS CHAPTER 1

A car moves at a velocity of 50 m s-1 in a direction north 30° east.

Calculate the component of the velocity

a) due north. b) due east.

Solution :Solution :

Example 1.2 :

N

v�

30°

a)�

30vvN cos=

1s m 43.3 −=v

�

3050vN cos=or

�

60vvN sin=�

6050vN sin=

28

EW

S

Nv�

Ev�

v�30°

60°

b)

1s m 43.3 −=Nv

�

30vvE sin=

1s m 25 −=Ev

�

3050vE sin=or

�

60vvE cos=�

6050vE cos=

PHYSICS CHAPTER 1

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 :Solution :

Example 1.3 :

150°

F�

Sx

y Vector x-component y-component

29

Solution :Solution :

150°30°

F�

Sx

y

yF�

xF�

Vector x-component y-component

�30cosFFx −=

N 6.68−=xF

�30cos100−=xF

orF�

�150cosFFx =

N 6.68−=xF

�150cos100=xF

�30sinFFy =

N 05=yF

�30sin100=yF

or�150sinFFy =

N 05=yF

�150sin100=yF

PHYSICS CHAPTER 1

Example 1.4 : y

O

)N10(1F�

30o

x

30

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.

30o

O

)N30(2F�

30o

)N40(3F�

PHYSICS CHAPTER 1

30o

Solution :Solution :

O

y

x30o

xF2

�

1F�

2F�

60o

yF2

�

x3F�

31

3F�

y3F�

∑ ++== 321 FFFFF r

�����

∑ ∑+= yxr FFF���

xxxx FFFF 321

����

++=∑yyyy FFFF 321

����

++=∑

PHYSICS CHAPTER 1

Solution :Solution :

Vector x-component y-component

1F�

2F�

N 01 =xF 11 FF y =

N 011 =yF

�60cos302 −=xF

N 15−=F

�60sin302 =yF

N 62=F

32

3F�

2FN 152 −=xF N 622 =yF

�30cos403 −=xF

N 34.63 −=xF

�30sin403 −=yF

N 203 −=yF

Vector Vector sumsum

( ) ( )34.6510 −+−+=∑ xF

N 49.6−=∑ xF

( )20.02601 −++=∑ yF

N 16=∑ yF

PHYSICS CHAPTER 1

y

Solution :Solution :

The magnitude of the resultant force is

( ) ( )22

∑∑ += yxr FFF

N .125=rF

( ) ( )221649.6 +−=rF

33

xO

and

Its direction is 162162°°°°°°°° from positive xfrom positive x--axis OR 18axis OR 18°°°°°°°° above negative xabove negative x--axis.axis.

=

∑∑−

x

y

F

Fθ

1tan

�1849.6

16tan 1 −=

−= −θ

∑ yF�

∑ xF�

�162

rF�

18°

PHYSICS CHAPTER 1

A�

1. Vector has components Ax = 1.30 cm, Ay = 2.25 cm; vector

has components Bx = 4.10 cm, By = -3.75 cm. Determine

a) the components of the vector sum ,

b) the magnitude and direction of ,

c) the components of the vector ,

d) the magnitude and direction of . (Young & freedman,pg.35,no.1.42)

ANS. : 5.40 cm, ANS. : 5.40 cm, --1.50 cm; 5.60 cm, 3451.50 cm; 5.60 cm, 345°°°°°°°°; 2.80 cm, ; 2.80 cm, --6.00 cm; 6.00 cm;

Exercise 2 :

BA��

+BA��

+AB��

−AB��

−

B�

34

ANS. : 5.40 cm, ANS. : 5.40 cm, --1.50 cm; 5.60 cm, 3451.50 cm; 5.60 cm, 345°°°°°°°°; 2.80 cm, ; 2.80 cm, --6.00 cm; 6.00 cm;

6.62 cm, 2956.62 cm, 295°°°°°°°°

2. For the vectors and in Figure 1.2, use the method of vector

resolution to determine the magnitude and direction of

a) the vector sum ,

b) the vector sum ,

c) the vector difference ,

d) the vector difference .(Young & freedman,pg.35,no.1.39)

ANS. : 11.1 m sANS. : 11.1 m s--11, 77.6, 77.6°°°°°°°°; U think;; U think;

28.5 m s28.5 m s--11, 202, 202°°°°°°°°; 28.5 m s; 28.5 m s--11, 22.2, 22.2°°°°°°°°

A�

B�

BA��

+AB��

+BA��

−AB��

−

Figure 1.2Figure 1.2

y

x0

37.0°

( )-1s m 18.0B�

( )-1s m 12.0A�

PHYSICS CHAPTER 1

A�

3. Vector points in the negative x direction. Vector points at an

angle of 30° above the positive x axis. Vector has a magnitude of

15 m and points in a direction 40° below the positive x axis. Given

that , determine the magnitudes of and .(Walker,pg.78,no. 65)

ANS. : 28 m; 19 mANS. : 28 m; 19 m

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

Exercise 2 :

C�

B�

0=++ CBA���

A�

B�

35

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

Calculate the resultant vector of P, Q and R.

ANS. : 49.4 m sANS. : 49.4 m s−−−−−−−−22; 70.1; 70.1°°°°°°°° above + xabove + x--axisaxis

Figure 1.3Figure 1.3

y

x0

50°( )2s m 10 −R�

( )2s m 35 −P�

( )2s m 24 −Q�

PHYSICS CHAPTER 1

1.2.6 Unit Vectors

� notations –

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

of vector A.

A�

a

cba ˆ ,ˆ ,ˆ

1ˆ ==A

Aa �

�

� Unit vectors are dimensionless.

� Unit vector for 3 dimension axes :

36

aA

[ ] 1ˆ =a

)(@ˆ⇒- boldjjaxisy 1ˆˆˆ === kji

)(@ˆ⇒- boldiiaxisx

)(@ˆ⇒- boldkkaxisz

PHYSICS CHAPTER 1

x

z

y

k

j

i

� Vector can be written in term of unit vectors as :

� Magnitude of vector,

37

z

krjrirr zyxˆˆˆ ++=

�

( ) ( ) ( )2

z

2

y

2

x rrrr ++=

PHYSICS CHAPTER 1

� E.g. : ( )m ˆ2ˆ3ˆ4 kjis ++=�

( ) ( ) ( ) m 5.39234222

=++=s

j3

y/m

38

j3

x/m

z/m

0

s�

i4k2

PHYSICS CHAPTER 1

ab�

�

−

( )m ˆ6ˆ2ˆ kjia +−=�

Two vectors are given as:

Calculate

a) the vector and its magnitude,

b) the vector and its magnitude,

c) the vector and its magnitude.

Example 1.5 :

ba�

�

+

( )m ˆˆ3ˆ4 kjib +−=�

ba�

�

+2

39

c) the vector and its magnitude.

Solution :Solution :

a)

The magnitude,

( ) ibaba xxxˆ541 =+=+=+

��

( ) jbaba yyyˆ532 −=−−=+=+

��

( )m ˆ7ˆ5ˆ5 kjiba +−=+�

�

( ) kbaba zzzˆ716 =+=+=+

��

( ) ( ) ( ) m 9.95755222

=+−+=+ ba

ba +2

PHYSICS CHAPTER 1

b)

The magnitude,

( ) iabab xxxˆ314 =−=−=−

��

( ) ( ) jabab yyyˆ23 −=−−−=−=−

��

( )m ˆ5ˆˆ3 kjiab −−=−�

�

( ) kabab zzzˆ561 −=−=−=−

��

( ) ( ) ( ) m 5.92513222

=−+−+=− ab

40

The magnitude,

c)

The magnitude,

( ) ( ) ( ) m 5.92513 =−+−+=− ab

( ) ( ) ibaba xxxˆ641222 =+=+=+

��

( ) ( ) ( ) jbaba yyyˆ732222 −=−+−=+=+

��

( )m ˆ13ˆ7ˆ62 kjiba +−=+�

�

( ) ( ) kbaba zzzˆ1316222 =+=+=+

��

( ) ( ) ( ) m 15.913762222

=+−+=+ ba

PHYSICS CHAPTER 11.2.7 Multiplication of VectorsScalar (dot) productScalar (dot) product

� The physical meaning of the scalar productphysical meaning of the scalar product can be explained by

considering two vectors and as shown in Figure 1.4a.A�

B�

θ

A�

B�

Figure 1.4aFigure 1.4a

� Figure 1.4b shows the projection of vector onto the direction of

vector .

� Figure 1.4c shows the projection of vector onto the direction of

vector . 41

B�

A� B

�

θ

A�

B�

A�

B�

θBcos

Figure 1.4bFigure 1.4bθ

A�

B�

θAcosFigure 1.4cFigure 1.4c

( )ABABA����

toparallel ofcomponent =•

( )BABBA����

toparallel ofcomponent =•

PHYSICS CHAPTER 1

� From the Figure 1.4b, the scalar product can be defined as

meanwhile from the Figure 1.4c,

where

� The scalar product is a scalar quantityscalar quantity.

( )θBABA cos=•��

vectorsobetween tw angle :θ

( )θABAB cos=•��

� The scalar product is a scalar quantityscalar quantity.

� The angle θ ranges from 0° to 180 °.

� When

� The scalar product obeys the commutative law of multiplication commutative law of multiplication i.e.

42

��

90θ0 << scalar product is positivepositive��

180θ09 << scalar product is negativenegative�

90θ = scalar product is zerozero

ABBA����

•=•

PHYSICS CHAPTER 1

( ) ( ) 111 cosˆˆ 2===• o2

0iii

� Example of scalar product is work donework done by a constant force where the

expression is given by

� The scalar product of the unit vectors are shown below :

( ) ( )θFsθsFsFW coscos ==•=��

y

j

( ) ( ) 111 cosˆˆ 2===• o2

0jjj

( ) ( ) 111 cosˆˆ 2===• o2

0kkk

43

x

z

k

j

i 1ˆˆˆˆˆˆ =•=•=• kkjjii

( ) ( ) 111 cos ===• 0kkk

( )( ) 09 cosˆˆ ==• o011ji

0ˆˆˆˆˆˆ =•=•=• kikjji

( )( ) 09 cosˆˆ ==• o011ki

( )( ) 09 cosˆˆ ==• o011kj

PHYSICS CHAPTER 1

A�

Calculate the and the angle θ between vectors and for the

following problems.

a) b)

Solution :Solution :

a)

Example 1.6 :

BA��

• B�

( )( ) ( )( ) ( )( ) kkjjiiBA ˆˆ31ˆˆ21ˆˆ41 •−+•−−+•=•��

kjiA ˆˆˆ +−=�

kjiA ˆˆ3ˆ4 +−=�

kjiB ˆ3ˆ2ˆ4 −−=�

kjB ˆ3ˆ2 +=�

324 −+=• BA��

ANS.:ANS.:−−−−−−−−33; ; 9999..44°°°°°°°°

44

The magnitude of the vectors:

The angle θ ,

( ) ( ) ( ) 3111222

=+−+=A

324 −+=• BA

3=• BA��

( ) ( ) ( ) 29324222

=−+−+=B

θABBA cos=•��

=

•= −−

293

3coscos 11

AB

BAθ

��

�2.71=θ

PHYSICS CHAPTER 1

Referring to the vectors in Figure 1.5,

Example 1.7 :

Figure 1.5Figure 1.5

y

x0

( )m 1C�

( )m 2D�19°

25°

45

a) determine the scalar product between them.

b) express the resultant vector of C and D in unit vector.

Solution :Solution :

a) The angle between vectors C and D is

Therefore

2m 991.DC −=•��

( ) �1741925180 =+−=θ

θCDDC cos=•��

( )( ) �174cos21=

PHYSICS CHAPTER 1

b) Vectors C and D in unit vector are

and

jCiCC yxˆˆ +=

�

( ) ( ) ji ˆ25sin1ˆ25cos1 �� +−=

( )m ˆ42.0ˆ910 ji.C +−=�

( ) ( )jiD ˆ19sin2ˆ19cos2 ��

�

−+=

( )m ˆ65.0ˆ891 ji.D −=�

46

Hence ( ) ( ) jiDC ˆ65.042.0ˆ89.191.0 −++−=+��

( )m ˆ23.0ˆ98.0 ji −=

( )m ˆ65.0ˆ891 ji.D −=�

PHYSICS CHAPTER 1

Vector (cross) productVector (cross) product

� Consider two vectors :

� In general, the vector product is defined as

and its magnitudemagnitude is given by

krjqipB ˆˆˆ ++=�

kzjyixA ˆˆˆ ++=�

CBA���

=×

θABθBACBA sinsin ===×�����

where

� The angle θ ranges from 0° to 180 ° so the vector product always

positive positive value.

� Vector product is a vector quantityvector quantity.

� The direction of vector is determined by

47

θABθBACBA sinsin ===×�����

vectorsobetween tw angle :θ

RIGHTRIGHT--HAND RULEHAND RULE

C�

PHYSICS CHAPTER 1

� For example:

� How to use right hand rule :

� 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.

C�

B�

CBA���

=×

�� Direction of the vector product always perpendicular Direction of the vector product always perpendicular

to the plane containing the vectors andto the plane containing the vectors and .

48

A�

B� A

�

B

C�

CAB���

=×

ABBA����

×≠× but ( )ABBA����

×−=×

B�)(C

�

A�

PHYSICS CHAPTER 1

� The vector product of the unit vectors are shown below :

x

y

k

j

i

ijkkj ˆˆˆˆˆ =×−=×

kijji ˆˆˆˆˆ =×−=×

jkiik ˆˆˆˆˆ =×−=×

� Example of vector product is a magnetic force on the straight a magnetic force on the straight conductor carrying current places in magnetic fieldconductor carrying current places in magnetic field where the

expression is given by

49

z

i

0ˆˆˆˆˆˆ =×=×=× kkjjii

0in ˆˆ ==× o20siii

0in ˆˆ ==× o20sjjj

0in ˆˆ ==× o20skkk

( )BlIF���

×=

θIlBF sin=

PHYSICS CHAPTER 1

b)

c) The magnitude of vectors,

( ) ( )kjikjiBA ˆ5ˆ0ˆˆˆ2ˆ3 −+•−−−=•��

2=• BA��

( )( ) ( )( ) ( )( ) kkjjiiBA ˆˆ51ˆˆ02ˆˆ13 •−−+•−+•−=•��

503 ++−=• BA��

( ) ( ) ( ) 14123222

=−+−+−=A

50

Using the scalar (dot) productscalar (dot) product formula,

θABBA cos=•��

=

•= −−

2614

2coscos 11

AB

BAθ

��

�84=θ

( ) ( ) ( ) 14123222

=−+−+−=A

( ) ( ) ( ) 26501222

=−++=B

PHYSICS CHAPTER 1

kjickjibkjia ˆˆ2ˆ2 and ˆ2ˆ4ˆ; ˆ2ˆ3ˆ3 ++=+−−=−+=�

��

1. If vector and vector , determine

a) , b) , c) .

ANS. :ANS. :

2. Three vectors are given as follow :

Calculate

Exercise 3 :

46 ;26 ;ˆ2k

jia ˆ+ˆ= 53�

jib ˆ+ˆ= 42

�

ba�

�

× ba�

�

• ( ) bba��

�

•+

( )��

� ( )��

�

51

Calculate

a) , b) , c) .

ANS. :ANS. :

3. If vector and vector ,determine

a) the direction of

b) the angle between and .

ANS. : U think, 92.8ANS. : U think, 92.8°°°°°°°°

( )cba�

��

ו ( )cba�

��

+• ( )cba�

��

+×kji ˆ9ˆ11ˆ5 ;9 ;21 −−−−

kjiP ˆˆ2ˆ3 −+=�

kjiQ ˆ3ˆ4ˆ2 ++−=�

QP��

×P�

Q�

PHYSICS CHAPTER 1PHYSICS CHAPTER 1

THE END…

Next Chapter…

52

Next Chapter…CHAPTER 2 :

Kinematics of Linear Motion