FIKIRANKU SELALU INGATKAMU

3 kg + 4 kg = 7 kg

3 N + 4 N = ?

NxN 71

Vectors Quantities and Scalar QuantitiesVectors Quantities and Scalar Quantities

Vector Quantities are physical quantities Vector Quantities are physical quantities which have a magnitude or value and directionwhich have a magnitude or value and direction

Example : velocity, acceleration, Force, etcExample : velocity, acceleration, Force, etc Scalar quantities are physical quantities Scalar quantities are physical quantities

which have magnitude or value without which have magnitude or value without directiondirection

Example : Mass, Time, Temperature, volume , Example : Mass, Time, Temperature, volume , etcetc

Notation VectorsNotation Vectors The vectors quantities are written in bold type, The vectors quantities are written in bold type,

while italicization is used to represent the scalar while italicization is used to represent the scalar value/scalar quantities.value/scalar quantities.Exp: -The vector A is written as Exp: -The vector A is written as A A and the scalar and the scalar quantity is written quantity is written AA

The vector quantities can written with a The vector quantities can written with a distinguishing mark, such as an arrow.distinguishing mark, such as an arrow.Exp: - The vector A is written as A and the scalar Exp: - The vector A is written as A and the scalar

quantity is written A . quantity is written A .

A Vector can be expressed in diagram with a directed line segment.

Magnitude of vector

Direction of vector

A

Capture point

= direction of vector

NEGATIVE VECTORS

A

- A

Negative vectors is the vectors which have the same in magnitude but opposite in direction

To determine resultant vector with graphic method

1. Polygon method

2. Parallelogram method

BOXF1

F2

F1

F2

F1 + F2

BOXF1

F2

F3

F3

F1

F2

F1 + F2 + F3

TRIANGLE METHOD

POLYGON METHOD

F1

F2

FR = F1

+ F2

F2

F1

Parallelogram MethodParallelogram Method

BOX

Characteristic of Vectors addition

Resultant vector with analytical method

1.Cosine equation

2.Vector component method

F1

F2

FR = F1

+ F2

2F2

F1

1. The Magnitude and direction of vector resultant with cosines equation

1

cos2 2122

21 FFFFFR

sinsinsin 1

2

2

1 RFFF

The Magnitude of vector resultant :

The direction of vector resultant :

Two vectors form angle of 0Two vectors form angle of 000

21 FFFR

The magnitude resultant of vector

The direction of vector resultant a direction with both of vectors

F1

F2

F1 F2

FR = F1 + F2

Two vectors form angle of 180Two vectors form angle of 18000

(Two Vectors opposite each other)(Two Vectors opposite each other)

2121 FFifFFFR

1212 FFifFFFR

The magnitude resultant of vector

The direction of vector resultant a direction with the biggest vector

F1

F2

F1

F2FR = F1 - F2

Two vectors form angle of 90Two vectors form angle of 9000

(Two Vectors perpendicular (Two Vectors perpendicular each other) each other)

22

21 FFFR

1

2tanF

F

F1

F2

900

F R = F 1

+ F

2

The magnitude resultant of vector

The direction of vector resultant

Fx

FyF

2. 2. vectors resultant with component vectors resultant with component vectors methodvectors method

22yx FFF

22yx FFF

x

y

x

y

F

F

F

Ftan

X

y

Vectors components :

Fx = F cos Fy = F sin

The Magnitude of vectors resultant F:

For two or more vectors :

The Direction of vector resultant :

SAMPLE PROBLEMSAMPLE PROBLEM1.1. A vector of velocity (V) forms an angle 30A vector of velocity (V) forms an angle 300 0 with positive X with positive X

axis andaxis and the magnitude is 20 m/s. determine the the magnitude is 20 m/s. determine the magnitude of vector component!magnitude of vector component!

2.2. Two vector of velocity have base points which coincide, Two vector of velocity have base points which coincide,

those are vthose are v11= 3 m/s and v= 3 m/s and v2 2 = 4 m/s. if = 4 m/s. if = 60 = 6000.. find the find the

magnitude and direction of vector resultant. magnitude and direction of vector resultant.

3.3. Four velocity vector have magnitudes and directions as Four velocity vector have magnitudes and directions as follows :follows :

VV11 = 10 m/s , = 10 m/s , 11 = 0 = 000

VV22 = 12 m/s , = 12 m/s , 22 = 60 = 6000

VV33 = 10 m/s , = 10 m/s , 33 = 120 = 12000

VV44 = 6 m/s , = 6 m/s , 44 = 240 = 24000

Determine the magnitude and direction of vector resultant!Determine the magnitude and direction of vector resultant!

1.1.

sm

v

vv

sm

v

vv

y

y

x

x

/10

)2

1(20

30sin20

sin

/310

)32

1(20

30cos20

cos

0

0

SOLUTION

The components of vector

V

Vx

Vy

cos2 2122

21 vvvvv

022 60cos4.3.243

37

0

1

0

1

1

1,32

)5320,0(sin

5320,0sin

60sin37

3sin

sinsin

sinsin:

v

v

vvDirection

v1

V2

v

2.

3.3. vv1x1x = v = v11 cos cos 11 vv1y1y = v = v11 sin sin 11

= 10 cos 0= 10 cos 000 = 10 sin 0 = 10 sin 000

= 10 (1) = 10 (1) = 10 (0) = 10 (0) = 10= 10 = 0 = 0

vv2x2x = v = v22 cos cos 22 vv2y2y = v = v22 sin sin 22

= 12 cos 60= 12 cos 6000 = 12 sin 60 = 12 sin 6000

= 12 ( ) = 12 ( ) = 12 ( ) = 6 = 12 ( ) = 6 = 6= 6

vv3x3x = v = v33 cos cos 33 vv3y3y = v = v33 sin sin 33

= 10 cos 120= 10 cos 12000 = 10 sin 120 = 10 sin 12000

= 10 ( ) = 10 ( ) = 10 ( ) = 5 = 10 ( ) = 5 = - 5= - 5

vv4x4x = v = v44 cos cos 44 vv4y4y = v = v44 sin sin 44

= 6 cos 240= 6 cos 24000 = 6 sin 240 = 6 sin 24000

= 6 ( ) = 6 ( ) = 6 ( ) = -3 = 6 ( ) = -3 = -3= -3

2

1 32

13

2

1 3

2

13

32

1 3

2

1

No vectors direction Vx= V cos Vy = V sin

1

2

3

4

V1 = 10

V2 = 12

V3 = 10

V4 = 6

00

600

1200

2400

V1x = 10

V2x = 6

V3x = -5

V4x = -3

V1y = 0

V2y = 6

V3y = 5

V4y = -3

3

3

3

Table

8 xV 38 yV

The magnitude of result vector

16

256

19264

)38(8 22

22

yxR VVV

The direction of result vector

060

38

38tan

x

y

F

F

V1

V2

V3

V4

600600

600 V2x

V3x

V4x

V2y

V3y

V4yV2x = V2 cos 600

V3x = V3 cos 600

V4x = V4 cos 600 V4y = V4 sin 600

V2y = V2 sin 600

V3y = V3 sin 600

V1

V2

V3

V4

300300

300

V2x

V3x

V4x

V2y

V3y

V4yV2x = V2 sin 300

V3x = V3 sin 300

V4x = V4 sin 300 V4y = V4 cos 300

V2y = V2 cos 300

V3y = V3 cos 300

V1

V2

V3

V4

600

300

600 V2x

V3x

V4x

V2y

V3y

V4yV2x = V2 cos 600

V3x = V3 sin 600

V4x = V4 cos 600 V4y = V4 cos 600

V2y = V2 cos 600

V3y = V3 cos 600

UNIT VECTORUNIT VECTOR

Unit vector is a vector of which the Unit vector is a vector of which the magnitude equals to one and the direction magnitude equals to one and the direction is the same as the direction of vector is the same as the direction of vector component.component.

In three dimensional case there are 3 umit In three dimensional case there are 3 umit vector, that is vector, that is ii , , jj , , kk i i = unit vector in the same direction as x axis = unit vector in the same direction as x axis j j = unit vector in the same direction as y axis = unit vector in the same direction as y axis k k = unit vector in the same direction as z axis = unit vector in the same direction as z axis

Unit vector in three Unit vector in three dimensional casedimensional case

X

Y

Z

j

i

k

Vector A can be expressed by unit vector as Vector A can be expressed by unit vector as followsfollows

222zyx AAAA

AX i

Ay j

Az kA

A = AX i + Ay j + Az k

The magnitude of vector A can be expressed by

In one dimensional case, then Ay = Az = 0

In two dimensional case , then Az = 0

Y

Z

X

Vector MultiplicationVector Multiplication

Dot Product VectorDot Product Vector

Dot Product vector gives a scalar result, therefore the dot Dot Product vector gives a scalar result, therefore the dot product vector is also called scalar product vector.product vector is also called scalar product vector.

The dot product vector between A and B can be expressed The dot product vector between A and B can be expressed as follows :as follows :

AA .. B B = A B cos = A B cos

AA = vector A, = vector A, B B = vector B, A = the magnitude of vector A= vector B, A = the magnitude of vector A

B = the magnitude of vector B, B = the magnitude of vector B, = angle between = angle between AA and and BB

Dot product vector Characteristic a peer the unit vectorDot product vector Characteristic a peer the unit vector

i . ii . i = = j . jj . j = = k . kk . k = (1) (1) cos 0 = 1 = (1) (1) cos 0 = 1

i . ji . j = = i . ki . k = = j . kj . k = (1) (1) cos 90 = (1) (1) cos 9000 = 0 = 0

j . ij . i = = k . ik . i = = k . jk . j = (1) (1) cos 90 = (1) (1) cos 9000 = 0 = 0

If vector If vector AA and vector and vector BB written in unit vector notation : written in unit vector notation :

and and

So, dot product vector So, dot product vector A A and vector and vector BB is is

A . B A . B = (A= (AXX i i ++ AAyy j j ++ AAzz k k ) (B) (BXX i i ++ BByy j j ++ BBzz k k ))

= A= AXX i i BBXX i i + + AAXX i i BByy j j ++ AAXX i i BBz z k k + + AAyy j j BBXX i i + + AAyy j j BByy j j ++

AAyy j j BBzz k k ++ AAzz k k BBXX i i ++ AAzz k k BByy j j ++ AAzz k k BBzz k k

A . BA . B = A = AXX B BXX ++ AAyy B By y ++ AAzz B Bzz

A = AX i + Ay j + Az k B = BX i + By j + Bz k

Cross Product VectorCross Product Vector

Cross Product vector gives a new vector result, therefore Cross Product vector gives a new vector result, therefore the dot product vector is also called vector product.the dot product vector is also called vector product.

The Cross product vector between The Cross product vector between AA and and BB can be product can be product vector vector C, C, Which the magnitude is Which the magnitude is

C C = = AA XX B B = A B sin = A B sin

AA = vector A, = vector A, B B = vector B, A = the magnitude of vector A= vector B, A = the magnitude of vector A

B = the magnitude of vector B, B = the magnitude of vector B, = angle between = angle between AA and and BB

Cross product vector Characteristic a peer the unit vectorCross product vector Characteristic a peer the unit vector

i x ii x i = = j x jj x j = = k x kk x k = (1) (1) sin 0 = 0 = (1) (1) sin 0 = 0

i x ji x j = = kk j x i = -kj x i = -k

j x kj x k = = ii k x j = -ik x j = -i

k x i = jk x i = j i x k = -ji x k = -j

If vector If vector AA and vector and vector BB written in unit vector notation : written in unit vector notation :

and and

So, cross product vector So, cross product vector A A and vector and vector BB is is

A X B A X B = (A= (AXX i i ++ AAyy j j ++ AAzz k k ) (B) (BXX i i ++ BByy j j ++ BBzz k k ))

= A= AXX i i BBXX i i + + AAXX i i BByy j j ++ AAXX i i BBz z k k + + AAyy j j BBXX i i + + AAyy j j BByy j j ++

AAyy j j BBzz k k ++ AAzz k k BBXX i i ++ AAzz k k BByy j j ++ AAzz k k BBzz k k

A = AX i + Ay j + Az k B = BX i + By j + Bz k

= A= AXX i i BByy j j ++ AAXX i i BBz z k k + + AAyy j j BBXX i i ++ A Ayy j j BBzz k k ++ AAzz k k BBXX i i ++ AAzz k k BByy j j

= A= AXX B Byy kk ++ AAXX B Bz z (-(-jj)) + + AAyy B BXX (- (-kk) ) ++ A Ayy B Bzz ii ++ AAzz B BXX jj ++ AAzz B Byy(-(-ii))

= A= Ayy B Bzz ii ++ AAzz B Byy(-(-ii) ) ++ A Azz B BXX jj ++ A AXX B Bz z (-(-jj)) + + AAXX B Byy kk ++ A Ayy B BXX (- (-kk) )

= A= Ayy B Bzz ii -- AAzz B Byy((ii) ) ++ A Azz B BXX jj -- A AXX B Bz z ((jj)) + + AAXX B Byy kk -- A Ayy B BXX ( (kk) )

= A= Ayy B Bzz ii -- AAzz B Byy ii ++ A Azz B BXX jj -- A AXX B Bz z jj + + AAXX B Byy kk -- A Ayy B BXX kk

A X BA X B = = ((AAyy B Bzz -- AAzz B Byy)) ii ++ ((AAzz B BXX -- A AXX B Bzz)) jj + (+ (AAXX B Byy -- A Ayy B BXX)) kk

Cross product vector with determinant Cross product vector with determinant methodmethodA = AX i + Ay j + Az k

B = BX i + By j + Bz k

C = A x B

C =

i j k i j

Ax Ay Az Ax Ay

Bx By Bz Bx By

- negative

+ positive

C= A X B = Ay Bz i - Az By i + Az BX j - AX Bz j + AX By k - Ay BX k

C = A X B = (Ay Bz - Az By) i + (Az BX - AX Bz) j + (AX By - Ay BX) k

Cross product two Cross product two vectorsvectors

k

i

j

+

-

Positive

Negative

SAMPLE PROBLEMSAMPLE PROBLEMA = 2 i + 3 j + k A = AX i + Ay j + Az k

B = 4 i + 2 j - 2 k B = BX i + By j + Bz k

Determine : a A . B

b. A x B

1. a A . B = AX BX + Ay By + Az Bz

SOLUTION

= (2) (4) + (3) (2) + (1) (-2)

= 8 + 6 - 2

= 12

b.

C = A X B = (Ay Bz - Az By) i + (Az BX - AX Bz) j + (AX By - Ay BX) k

= ( (3) (-2) – (1) (2) ) i + ( (1)(4) – (2)(-2) ) j + ( (2) (2) – (3) (4) ) k

= - 8 i + 8 j - 8 k

222 )8()8()8( BxAC

Cross product vector with determinant Cross product vector with determinant methodmethod

222 )8()8()8( BxAC

A = 2i + 3 j + k

B = 4i + 2 j - 2 k

C = A x B

C =

i j k i j

2 3 1 2 3

4 2 -2 4 2

- negative

+ positive

C = A X B = -8 i + 8 j - 8 k

A = AX i + Ay j + Az k B = BX i + By j + Bz k

-6 i

4 j4 k

-12 k -2 i4 j

sin cos tan

00 0 1 0

300

450 1

600

900 1 0 -

1800 0 -1 0

2700 -1 0 -

3600 0 1 0

Sin, cos, tan table

22

12

2

1

2

13

2

13

3

1

32

12

13

NOTES

Kuadran II (1800 - )

Kuadran III ( 1800 + )

Kuadran IV ( 3600 - )

Example

Cos 1200 =……….

1200 = di kudran II ( hanya sin yang positive)

Cos 1200 = cos (180 - )

= cos (1800 – 600 )

= - cos 600 = 2

1

sin 2400 =……….

2400 = di kudran III ( hanya tan yang positive)

sin 2400 = sin (180 + )

= sin (1800 + 600 )

= - sin 600 = 32

1

Letak kuadran sudut sebuah vektor

Kuadran I II III IV

Fx

Fy

+

+

-

+

-

-

+

-