# FIKIRANKU SELALU INGAT KAMU 3 kg + 4 kg =7 kg 3 N + 4 N = ?

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FIKIRANKU SELALU INGAT KAMU

3 kg + 4 kg =7 kg 3 N + 4 N = ?

Vectors Quantities and Scalar Quantities Vector Quantities are physical quantities which have a magnitude or value and direction Vector Quantities are physical quantities which have a magnitude or value and direction Example : velocity, acceleration, Force, etc Scalar quantities are physical quantities which have magnitude or value without direction Scalar quantities are physical quantities which have magnitude or value without direction Example : Mass, Time, Temperature, volume, etc

Notation Vectors The vectors quantities are written in bold type, while italicization is used to represent the scalar value/scalar quantities. Exp: -The vector A is written as A and the scalar quantity is written A The vector quantities can written with a distinguishing mark, such as an arrow. Exp: - The vector A is written as A and the scalar 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

BOX F1F1 F2F2 F1F1 F2F2 F 1 + F 2 BOX F1F1 F2F2 F3F3 F3F3 F1F1 F2F2 F 1 + F 2 + F 3 TRIANGLE METHOD POLYGON METHOD

F1F1 F2F2 F R = F 1 + F 2 F2F2 F1F1 Parallelogram Method BOX

Resultant vector with analytical method 1.Cosine equation 2.Vector component method

F1F1 F2F2 F R = F 1 + F 2 2 F2F2 F1F1 1.The Magnitude and direction of vector resultant with cosines equation 1 The Magnitude of vector resultant : The direction of vector resultant :

Two vectors form angle of 0 0 The magnitude resultant of vector The direction of vector resultant a direction with both of vectors F1F1 F2F2 F1F1 F2F2 F R = F 1 + F 2

Two vectors form angle of 180 0 (Two Vectors opposite each other) The magnitude resultant of vector The direction of vector resultant a direction with the biggest vector F1F1 F2F2 F1F1 F2F2 F R = F 1 - F 2

Two vectors form angle of 90 0 (Two Vectors perpendicular each other) F1F1 F2F2 90 0 F R = F 1 + F 2 The magnitude resultant of vector The direction of vector resultant

Fx Fy F 2. vectors resultant with component vectors method 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 PROBLEM 1. A vector of velocity (V) forms an angle 30 0 with positive X axis and the magnitude is 20 m/s. determine the magnitude of vector component! 2. Two vector of velocity have base points which coincide, those are v 1 = 3 m/s and v 2 = 4 m/s. if = 60 0. find the magnitude and direction of vector resultant. 3. Four velocity vector have magnitudes and directions as follows : V 1 = 10 m/s, 1 = 0 0 V 2 = 12 m/s, 2 = 60 0 V 3 = 10 m/s, 3 = 120 0 V 4 = 6 m/s, 4 = 240 0 Determine the magnitude and direction of vector resultant!

1. SOLUTION The components of vector V VxVx VyVy

v1v1 V2V2 v 2.

3.v 1x = v 1 cos 1 v 1y = v 1 sin 1 = 10 cos 0 0 = 10 sin 0 0 = 10 cos 0 0 = 10 sin 0 0 = 10 (1) = 10 (0) = 10 (1) = 10 (0) = 10 = 0 = 10 = 0 v 2x = v 2 cos 2 v 2y = v 2 sin 2 = 12 cos 60 0 = 12 sin 60 0 = 12 cos 60 0 = 12 sin 60 0 = 12 ( ) = 12 ( ) = 6 = 12 ( ) = 12 ( ) = 6 = 6 = 6 v 3x = v 3 cos 3 v 3y = v 3 sin 3 = 10 cos 120 0 = 10 sin 120 0 = 10 cos 120 0 = 10 sin 120 0 = 10 ( ) = 10 ( ) = 5 = 10 ( ) = 10 ( ) = 5 = - 5 = - 5 v 4x = v 4 cos 4 v 4y = v 4 sin 4 v 4x = v 4 cos 4 v 4y = v 4 sin 4 = 6 cos 240 0 = 6 sin 240 0 = 6 cos 240 0 = 6 sin 240 0 = 6 ( ) = 6 ( ) = -3 = 6 ( ) = 6 ( ) = -3 = -3 = -3

Novectorsdirection V x = V cos V y = V sin 12341234 V 1 = 10 V 2 = 12 V 3 = 10 V 4 = 60 60 0 120 0 240 0 V 1x = 10 V 2x = 6 V 3x = -5 V 4x = -3 V 1y = 0 V 2y = 6 V 3y = 5 V 4y = -3 Table The magnitude of result vector The direction of result vector

V1V1 V2V2 V3V3 V4V4 60 0 V 2x V 3x V 4x V 2y V 3y V 4y V 2x = V 2 cos 60 0 V 3x = V 3 cos 60 0 V 4x = V 4 cos 60 0 V 4y = V 4 sin 60 0 V 2y = V 2 sin 60 0 V 3y = V 3 sin 60 0

V1V1 V2V2 V3V3 V4V4 30 0 V 2x V 3x V 4x V 2y V 3y V 4y V 2x = V 2 sin 30 0 V 3x = V 3 sin 30 0 V 4x = V 4 sin 30 0 V 4y = V 4 cos 30 0 V 2y = V 2 cos 30 0 V 3y = V 3 cos 30 0

V1V1 V2V2 V3V3 V4V4 60 0 30 0 60 0 V 2x V 3x V 4x V 2y V 3y V 4y V 2x = V 2 cos 60 0 V 3x = V 3 sin 60 0 V 4x = V 4 cos 60 0 V 4y = V 4 cos 60 0 V 2y = V 2 cos 60 0 V 3y = V 3 cos 60 0

UNIT VECTOR Unit vector is a vector of which the magnitude equals to one and the direction is the same as the direction of vector component. Unit vector is a vector of which the magnitude equals to one and the direction is the same as the direction of vector component. In three dimensional case there are 3 umit vector, that is i, j, k In three dimensional case there are 3 umit vector, that is i, j, k i = unit vector in the same direction as x axis i = unit vector in the same direction as x axis j = unit vector in the same direction as y axis j = unit vector in the same direction as y axis k = unit vector in the same direction as z axis k = unit vector in the same direction as z axis

Unit vector in three dimensional case X Y Z j i k

Vector A can be expressed by unit vector as follows A X i A y j A z k A A = A X i + Ay j + Az k The magnitude of vector A can be expressed by In one dimensional case, then A y = A z = 0 In two dimensional case, then A z = 0 Y Z X

Vector Multiplication Dot Product Vector Dot Product Vector Dot Product vector gives a scalar result, therefore the dot product vector is also called scalar product vector. The dot product vector between A and B can be expressed as follows : A. B = A B cos A. B = A B cos A = vector A, B = vector B, A = the magnitude of vector A B = the magnitude of vector B, = angle between A and B

Dot product vector Characteristic a peer the unit vector i. i = j. j = k. k = (1) (1) cos 0 = 1 i. j = i. k = j. k = (1) (1) cos 90 0 = 0 j. i = k. i = k. j = (1) (1) cos 90 0 = 0 If vector A and vector B written in unit vector notation : and So, dot product vector A and vector B is A. B = (A X i + A y j + A z k ) (B X i + B y j + B z k ) = A X i B X i + A X i B y j + A X i B z k + A y j B X i + A y j B y j + = A X i B X i + A X i B y j + A X i B z k + A y j B X i + A y j B y j + A y j B z k + A z k B X i + A z k B y j + A z k B z k A y j B z k + A z k B X i + A z k B y j + A z k B z k A. B = A X B X + A y B y + A z B z A. B = A X B X + A y B y + A z B z A = A X i + A y j + A z kB = B X i + B y j + B z k

Cross Product Vector Cross Product vector gives a new vector result, therefore the dot product vector is also called vector product. The Cross product vector between A and B can be product vector C, Which the magnitude is C = A X B = A B sin C = A X B = A B sin A = vector A, B = vector B, A = the magnitude of vector A B = the magnitude of vector B, = angle between A and B

Cross product vector Characteristic a peer the unit vector i x i = j x j = k x k = (1) (1) sin 0 = 0 i x j = k j x i = -k j x k = ik x j = -i k x i = ji x k = -j If vector A and vector B written in unit vector notation : and So, cross product vector A and vector B is A X B = (A X i + A y j + A z k ) (B X i + B y j + B z k ) = A X i B X i + A X i B y j + A X i B z k + A y j B X i + A y j B y j + = A X i B X i + A X i B y j + A X i B z k + A y j B X i + A y j B y j + A y j B z k + A z k B X i + A z k B y j + A z k B z k A y j B z k + A z k B X i + A z k B y j + A z k B z k A = A X i + A y j + A z kB = B X i + B y j + B z k

= A X i B y j + A X i B z k + A y j B X i + A y j B z k + A z k B X i + A z k B y j = A X B y k + A X B z (-j) + A y B X (-k) + A y B z i + A z B X j + A z B y (-i) = A y B z i + A z B y (-i) + A z B X j + A X B z (-j) + A X B y k + A y B X (-k) = A y B z i - A z B y (i) + A z B X j - A X B z (j) + A X B y k - A y B X (k) = A y B z i - A z B y i + A z B X j - A X B z j + A X B y k - A y B X k A X B = (A y B z - A z B y ) i + (A z B X - A X B z ) j + (A X B y - A y B X ) k A X B = (A y B z - A z B y ) i + (A z B X - A X B z ) j + (A X B y - A y B X ) k

Cross product vector with determinant method A = A X i + A y j + A z k B = B X i + B y j + B z k C = A x B C = i j k i j A x A y A z A x A y B x B y B z B x B y - negative + positive C= A X B = A y B z i - A z B y i + A z B X j - A X B z j + A X B y k - A y B X k C = A X B = (A y B z - A z B y ) i + (A z B X - A X B z ) j + (A X B y - A y B X ) k

Cross product two vectors k i j + - Positive Negative

SAMPLE PROBLEM A = 2 i + 3 j + kA = A X i + A y j + A z k B = 4 i + 2 j - 2 k B = B X i + B y j + B z k Determine : a A. B b. A x B 1. a A. B = A X B X + A y B y + A z B z SOLUTION = ( 2 ) ( 4 ) + ( 3 ) ( 2 ) + ( 1 ) ( -2 ) = 8 + 6 - 2 = 12

b.C = A X B = (A y B z - A z B y ) i + (A z B X - A X B z ) j + (A X B y - A y B X ) k = ( ( 3 ) ( -2) – ( 1 ) ( 2 ) ) i + ( ( 1 )( 4 ) – ( 2 )( -2 ) ) j + ( ( 2 ) ( 2 ) – ( 3 ) ( 4 ) ) k = - 8 i + 8 j - 8 k

Cross product vector with determinant method A = 2 i + 3 j + k B = 4 i + 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 = A X i + A y j + A z k B = B X i + B y j + B z k -6 i 4 j 4 k -12 k-2 i 4 j

sincostan 0 010 30 0 45 0 1 60 0 90 0 10- 180 0 00 270 0 0- 360 0 010 Sin, cos, tan table NOTES

Kuadran II (180 0 - ) Kuadran III ( 180 0 + ) Kuadran IV ( 360 0 - )

Example Cos 120 0 =………. 120 0 = di kudran II ( hanya sin yang positive) Cos 120 0 = cos (180 - ) = cos (180 0 – 60 0 ) = - cos 60 0 = sin 240 0 =………. 240 0 = di kudran III ( hanya tan yang positive) sin 240 0 = sin (180 + ) = sin (180 0 + 60 0 ) = - sin 60 0 =

Letak kuadran sudut sebuah vektor Kuadran IIIIIIIV FxFyFxFy ++++ -+-+ ---- +-+-

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