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Vectors & Scalars

Vectors & Scalars. Vectors & Scalars. Vectors are measurements which have both magnitude (size) and a directional component. EXAMPLES OF VECTOR VALUES: Displacement Velocity Acceleration Force Direction counts in all of these measurements. Scalars are measurements which have only

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Vectors & Scalars

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  1. Vectors & Scalars

  2. Vectors & Scalars Vectors are measurements which have both magnitude (size) and a directional component. EXAMPLES OF VECTOR VALUES: Displacement Velocity Acceleration Force Direction counts in all of these measurements. Scalars are measurements which have only magnitude (size) and no directional component EXAMPLES OF SCALAR VALUES: Distance Speed Temperature

  3. LAKE TRANQUILITY A B Comparing Vector & Scalar Values Displacement (a vector) versus distance (a scalar) We want to get from point A to point B. If we follow the road around the lake our direction is always changing. There is no specific direction. The distance traveled on the road is a scalar quantity. A straight line between A and B is the displacement. It has a specific direction and is therefore a vector.

  4. 50 60 N 40 70 30 80 W E 20 90 S 10 SPEEDOMETER COMPASS Speed & Velocity Speed and velocity are not the same. Velocity requires a directional component and is therefore a vector quantity. Speed tells us how fast we are going but not which way. Speed is a scalar (direction doesn’t count!) VELOCITY +

  5. VECTOR Vectors are typically illustrated by drawing an ARROW F=10N • The direction of the physical quantity is given by the direction of the arrow. • The magnitude of the quantity is given by the length of the arrow and it is denoted by |F|. • It is a good habit to label your vectors in a diagram

  6. MAGNITUDE & DIRECTION • MAGNITUDE DISPLACEMENT - METERS, FEET, MILES ETC. FOR VELOCITY - METERS PER SECOND, FEET PER MINUTE, etc FORCE - NEWTONS, DYNES OR POUNDS. • DIRECTION - DEGREES - RADIANS - GEOGRAPHIC INDICATORS SUCH AS NORTH, EAST, SOUTH, NORTHEAST, ETC. - KNOWLEDGE OF COORDINATE GEOMETRY (X-Y PLANE)

  7. Physics Mathematics Left = - Up = + Down = - Right = + Rectangular Coordinates 90 o North y Quadrant II + Quadrant I - + 0 o East x West 180 o 360 o Quadrant III Quadrant IV - 270 o South

  8. 90O North +y 0O East West 180O +x - x 360O - y 270O South MEASURING THESAME DIRECTIONIN DIFFERENT WAYS 120O -240O 30O West of North 30O Left of +y 60O North of West 60O Above - x

  9. VECTOR NOTATIONS VECTOR NOTATION MAY TAKE SEVERAL DIFFERENT FORMS: • POLAR /CIRCULAR FORM INDICATES A MAGNITUDE VALUE AND A DIRECTIONAL VALUE. THE DIRECTION VALUE MAY BE IN DEGREES, RADIANS OR GEOGRAPHIC TERMS. • RECTANGULAR FORM IDENTIFIES THE X-Y COORDINATES OF THE VECTOR. THE VECTOR ITSELF EXTENDS FROM ORIGIN TO THE X-Y POINT. • EXAMPLES: (X = +10, Y = -10) or (10, -10) • THE MAGNITUDE OF THE VECTOR CAN BE FOUND USING THE PYTHAGOREAN THEOREM (102 + (-102))1/2 = 14.1 - THE DIRECTION CAN BE FOUND USING AN INVERSE TANGENT FUNCTION TAN-1 (10/10) = TAN-1 (1.0) = 45O SINCE X IS POSITIVE AND Y IS NEGATIVE THE ANGLE IS -45O AND IS IN QUADRANT IV OR 315O

  10. Circular Notation Circular notationdefines a vector by designating the vector’s magnitude |A| and angle θ relative to the +x axis. Using that notation the vector is written:

  11. 90O North +y West 180O +x - x 360O - y 270O South CIRCULAR NOTATIONS 315O -45O or 45O SOUTH OF EAST 0O East A=14 m • POLAR COORDINATES 14 METERS , θ=315O 14 METERS, θ= -45O 14 METERS, 45O SOUTH OF EAST

  12. Circular Notation Example: Force vector with magnitude 12 Newtons oriented at 210 degrees with the + x-axis. F = 12 N, θ = 210° F = 12 N, 30° South of West F = 12 N, θ = -150°

  13. Circular Notation F = 4 N, θ = 180° In this picture, we have a force vector of 4 Newtons oriented along the -x axis.

  14. Circular Notation C = 2, θ = 30° D = 4 , θ = -50° or D = 4, θ = 310° In the picture above we have 2 vectors C and D. How do we characterize the two vectors?

  15. Circular Notation Exercises: Draw the vectors graphically. 1. A= 6, 30° North of West 2. B= 3, θ = 330 ° 3. C= 20, 45° West of South 4. D= 5, θ = -240 ° 5. E= 8, 60° SE

  16. ADDITION/ SUBTRACTION OF VECTORS • TWO METHODS 1. GRAPHICAL METHOD (DRAWING) 2. COMPONENT METHOD (MATHEMATICAL). • GRAPHICAL ADDITION AND SUBTRACTION REQUIRES THAT EACH VECTOR BE REPRESENTED AS AN ARROW WITH A LENGTH PROPORTIONAL TO THE MAGNITUDE VALUE AND POINTED IN THE PROPER DIRECTION ASSIGNED TO THE VECTOR. (N ote: We need ruler and protractor)

  17. GRAPHIC REPRESENTATION OF VECTORS 30 METERS ,θ= 45O 30 METERS θ= 90O 50 METERS ,θ= 0O VECTOR ARROWS MAY BE DRAWN ANYWHERE ON THE PAGE AS LONG AS THE PROPER LENGTH AND DIRECTION ARE MAINTAINED = 10 METERS SCALE

  18. GRAPHICAL METHOD • VECTORS ARE ADDED GRAPHICALLY BY DRAWING EACH VECTOR TO SCALE AND ORIENTED IN THE PROPER DIRECTION. THE VECTOR ARROWS ARE PLACED HEAD TO TAIL. THE ORDER OF PLACEMENT DOES NOT AFFECT THE RESULT (A + B = B + A) • THE RESULT OF THE VECTOR ADDITION IS CALLED THE RESULTANT VECTOR. IT IS MEASURED FROM THE TAIL OF THE FIRST VECTOR ARROW TO THE HEAD OF THE LAST ADDED VECTOR ARROW. • THE LENGTH OF THE RESULTANT VECTOR ARROW CAN THEN BE MEASURED AND USING THE SCALE FACTOR CONVERTED TO THE CORRECT MAGNITUDE VALUE. THE DIRECTIONAL COMPONENT CAN BE MEASURED USING A PROTRACTOR.

  19. ALL VECTORS MUST BE DRAWN TO SCALE & POINTED IN THE PROPER DIRECTION Adding Vectors A D R B C B C A D + + + = A B C D R

  20. Vector A 30 METERS @ 45O C Vector B 50 METERS @ 0O B Vector C 30 METERS @ 90O A = 10 METERS SCALE Drawing Vectors to Scale To add the vectors Place them head to tail Angle is measured at 40o Resultant = 9 x 10 = 90 meters

  21. Vector A 30 METERS @ 45O Vector - A 30 METERS @ 225O WORKING WITH VECTORS GRAPHIC SUBTRACTION • IN ALGEBRA, A – B = A + (-B) OR IN OTHER WORDS, ADDING A NEGATIVE VALUE IS ACTUALLY SUBTRACTION. THIS IS ALSO TRUE IN VECTOR SUBTRACTION. IF WE ADD A NEGATIVE VECTOR B TO VECTOR A THIS IS REALLY SUBTRACTING VECTOR B FROM VECTOR A. • VECTOR VALUES CAN BE MADE NEGATIVE BY REVERSING THE VECTOR’S DIRECTION BY 180 DEGREES. IF VECTOR A IS 30 METERS DIRECTED AT 45 DEGREES (QUADRANT I), NEGATIVE VECTOR A IS 30 METERS AT 225 DEGREES (QUADRANT II).

  22. Subtracting Vectors A -C -D B B R A C = -C -D D = + -- = A B C D R + + ( - ) + ( - ) = A B C D R

  23. Graphical Method - Examples 1. A man walks 54.5 meters east, then another 30 meters east. Calculate his displacement relative to where he started? + 54.5 m, E 30 m, E Notice that the SIZE of the arrow conveys MAGNITUDE and the way it was drawn conveys DIRECTION. R= 84.5 m, E

  24. Graphical Method - Examples 2. You walk 50 meters 60 degrees north of west, then another 85 meters in opposite direction. Calculate your displacement relative to where you started? R=35, θ=300° N 60° W E 60° S

  25. COMPONENT METHOD • PYTHAGOREAN THEOREM • TRIGONOMETRIC FUNCTIONS • HORIZONTAL AND VERTICAL COMPONENTS OF THE VECTOR

  26. Fundamental Trigonometry Sin = A/C  C C C A A A Cos = B/ C  Tan  = A/B  B B B  = Arctan (A/B) A RIGHT TRIANGLE

  27. y + - + 0 radians  radians x 2 radians  - 3/2  radians Trig Function Signs Sin Cos Tan    Quadrant I + + + Quadrant II + -- /2 radians 90 o - - + Quadrant III - + - Quadrant IV 0 o 180 o    360 o 270 o

  28. Vector Components A Y COMPONENT X COMPONENT X COMPONENT Y COMPONENT C X COMPONENT Y COMPONENT B

  29. Vector Components X A Ay Y  Ax • is the angle between • the +x-axis and the vector. = COS  Ax A = SIN Ay A 

  30. VECTOR COMPONENTS • THE SIGNS OF THE X AND Y COMPONENTS DEPEND ON WHICH QUADRANT THE VECTOR LIES. • VECTORS IN QUADRANT I (0 TO 90 DEGREES) HAVE POSITIVE X ANDPOSITIVEY VALUES • VECTORS IN QUADRANT II (90 TO 180 DEGREES) HAVE NEGATIVE X VALUES AND POSITIVE Y VALUES. • VECTORS IN QUADRANT III (180 TO 270 DEGREES) HAVE NEGATIVE X VALUES AND NEGATIVE Y VALUES. • VECTORS IN QUADRANT IV (270 TO 360 DEGREES) HAVE POSITIVE X VALUES AND NEGATIVE Y VALUES.

  31. = COS  Ax A = SIN Ay A  VECTOR COMPONENTS WHAT ARE THE X AND Y COMPONENTS OF THE FOLLOWING VECTORS? • A =40 m,  = 60O ? 2. V= 60 m/s, = 2450 ?

  32. VECTOR COMPONENTS Exercises: Find the x and y components of the following vectors. 1. A= 100 N, 30° North of East 2. B= 30 km, θ = 330 ° 3. C= 20 m, 45° West of South 4. D= 50 m/s, θ = -240 ° 5. E= 80 N, 60° Southeast

  33. Analytical Method of Vector Addition 1. Find the x- and y-components of each vector. Ax = A cos q = Ay = A sin q = By = B sin q = Bx = B cos q = Cx = C cos q = Cy = C sin q = Rx= Ry = 2. Sum the x-components. This is the x-component of the resultant. 3. Sum the y-components. This is the y-component of the resultant. 4. Use the Pythagorean Theorem to find the magnitude of the resultant vector. Rx2 + Ry2 = R2

  34. Vector A 30 METERS @ 45O Vector B 50 METERS @ 0O = SIN Vy V  Vector C 30 METERS @ 90O = COS  Vx V ADDING & SUBTRACTING VECTORS USING COMPONENTS ADD THE FOLLOWING THREE VECTORS USING COMPONENTS • RESOLVE EACH INTO • X AND Y COMPONENTS

  35. COMPONENT METHOD • SOLVE EACH INTO X AND Y COMPONENTS • AX = 30 COS 450 = 21.2 AY = 30 SIN 450 = 21.2 • BX = 50 COS 00 = 50 BY = 50SIN 00 = 0 • CX = 30 COS 900 = 0 CY = 30SIN 900 = 30

  36. (2) ADD THE X COMPONENTS OF EACH VECTOR ADD THE Y COMPONENTS OF EACH VECTOR Rx =21.2 + 50 + 0 = +71.2 Ry= 21.2 + 0 + 30 = +51.2 (3) CONSTUCT A NEW RIGHT TRIANGLE USING THE Rx AS THE BASE AND RyAS THE OPPOSITE SIDE R Ry = +51.2 Rx = +71.2 THE HYPOTENUSE IS THE RESULTANT VECTOR

  37. Ry = +51.2 Rx = +71.2 (+71.2)2 + (+51.2)2 = 87.7 (4) USE THE PYTHAGOREAN THEOREM TO THE LENGTH (MAGNITUDE) OF THE RESULTANT VECTOR θ= TAN-1 (51.2/71.2) ANGLE = 35.7 O QUADRANT I (5) FIND THE ANGLE (DIRECTION) USING INVERSE TANGENT OF THE OPPOSITE SIDE OVER THE ADJACENT SIDE R = 87.7 m , θ = 35.7 O

  38. - + = A B C R Vector A 30 m @ 45O Vector B 50 m @ 0O + (- ) + = A B C R Vector C 30 m @ 90O SUBTRACTING VECTORS USING COMPONENTS Vector A 30 METERS @ 45O - B= 50 METERS @ 180O Vector C 30 METERS @ 90O

  39. 1. SOLVE THE X AND Y COMPONENTS • AX = 30COS 450 = 21.2 METERS • AY = 30SIN 450 = 21.2 METERS • BX = 50 COS 1800 = - 50 METERS • BY = 50SIN 1800 = 0 METERS • CX = 30 COS 900 = 0 METERS • CY = 30SIN 900 = 30 METERS

  40. (2) ADD THE X COMPONENTS OF EACH VECTOR ADD THE Y COMPONENTS OF EACH VECTOR Rx = SUM OF THE Xs = 21.2 + (-50) + 0 = -28.8 RY =SUM OF THE Ys = 21.2 + 0 + 30 = +51.2 (3) CONSTUCT A NEW RIGHT TRIANGLE USING THE Rx AS THE BASE AND RY AS THE OPPOSITE SIDE R RY = +51.2 Rx = -28.8 THE HYPOTENUSE IS THE RESULTANT VECTOR

  41. Ry = +51.2 Rx = -28.8 (-28.8)2 + (+51.2)2 = 58.7 (4) USE THE PYTHAGOREAN THEOREM TO THE FIND THE LENGTH (MAGNITUDE) OF THE RESULTANT VECTOR ANGLE TAN-1 (51.2/-28.8) ANGLE = -60.6 0 (1800 –60.60 ) = 119.40 QUADRANT II R = (5) FIND THE ANGLE (DIRECTION) USING INVERSE TANGENT OF THE OPPOSITE SIDE OVER THE ADJACENT SIDE RESULTANT = 58.7 METERS @ 119.4O

  42. Example A bear, searching for food wanders 35 meters east then 20 meters north. Frustrated, he wanders another 12 meters west then 6 meters south. Calculate the bear's displacement. 23 m, E - = 12 m, W - = 14 m, N 6 m, S 20 m, N 14 m, N 35 m, E R q 23 m, E The Final Answer:

  43. Example A boat moves with a velocity of 15 m/s, N in a river which flows with a velocity of 8.0 m/s, west. Calculate the boat's resultant velocity with respect to due north. 8.0 m/s, W 15 m/s, N Rv q The Final Answer :

  44. Ehra begins a three-day hiking trip by first walking 25 km, 30° south of east from her base camp. On the second day, she walks 40 km in a direction 60° north of east. On the third day, she walks 35 km, 50° south of west. What is her resultant displacement from base camp? • ICADEMY commuter jet departs NAIA and flies 175 km 30° East of North, turns and flies 150 km , 20° North of West, and finally flies 190 km west to its destination. What is its resultant displacement? • Find the resultant velocity vector for the following: • V1 = 89 km/h , 55° East of South • V2 = 46 km/h , 28° South of West • V3 = 37 km/h , South

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