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FIKIRANKU SELALU INGAT KAMU

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

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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

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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.

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A Vector can be expressed in diagram with a directed line segment. Magnitude of vector Direction of vector A Capture point = direction of vector

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NEGATIVE VECTORS A - A Negative vectors is the vectors which have the same in magnitude but opposite in direction

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To determine resultant vector with graphic method 1. Polygon method 2. Parallelogram method

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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

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

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Characteristic of Vectors addition

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

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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 :

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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

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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

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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

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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 :

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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!

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1. SOLUTION The components of vector V VxVx VyVy

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v1v1 V2V2 v 2.

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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

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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

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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

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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

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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

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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

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

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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

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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

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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

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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

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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

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= 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

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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

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Cross product two vectors k i j + - Positive Negative

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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

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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

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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

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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

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

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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 =

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

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