2. Richard J. Terwilliger by

3. Let’s look at some examples.

4. We’ll break this GREEN vector up into different RED vectors.

5. Here the GREEN vector is broken down into two RED vectors

6. The two RED vectors are called the of the GREEN vector.

7. Or, the of the two RED vectors is the GREEN vector.

8. The same GREEN vector can be broken up into 2 different RED vectors.

9. Or two different vectors!

10. There are possibilities!

11. A little later on we’ll show that two components at right angles are very helpful!

12. Here the same GREEN vector is broken up into 3 different component vectors

13. Again, the of the three RED vectors is the GREEN vector!

14. And, the GREEN vector can be broken into RED vectors

15. Let’s see another example.

16. This BLUE vector is broken down into several ORANGE vectors.

17. The tail of the BLUE vector starts at the same point as the tail of the first ORANGE vector. STARTING POINT

18. And they both end at the same point ENDING POINT STARTING POINT

19. The of the ORANGE vectors is the BLUE vector. E

20. E Or again, the of the BLUE vector are the ORANGE vectors..

21. E So what have we learned so far?

22. E A vector can be broken down into any number of components. The different arrangements of components are unlimited. Components are one of two or more vectors having a sum equal to a given vector. A resultant is the sum of the component vectors.

23. The process of finding the components of a vector given it’s magnitude and direction is called: E

24. E is an easy way to find the resultant of several vectors.

25. E To show how, let’s go back to two at right angles.. 90 o

26. E We’ll sketch in the x-axis so it goes through the TAIL of our original vector. X-axis TAIL

27. E X-axis Y-axis TAIL Then we’ll sketch the y-axis so it goes through the TAIL of our original vector, too.

28. Y-axis E In the diagram the red vector is called the horizontal or X- component. X-axis

29. Y-axis E The blue vector is called the vertical or Y- component. X-axis

30. Y-axis E To distinguish between the X and Y components, we label each vector with X and Y subscripts respectively. X-axis

31. Y-axis E To distinguish between the X and Y components, we label each vector with X and Y subscripts respectively. X-axis

32. Y-axis E To distinguish between the X and Y components, we label each vector with X and Y subscripts respectively. X-axis

33. E Knowing the original vector’s magnitude and direction we can solve for Vx and Vy using trigonometry. 90 o 0o0o 180 o 270 o

34. E Let’s label the original vector with it’s magnitude and direction. 0o0o 180 o 90 o 270 o

35. 90 o 270 o E Let’s label the original vector with it’s magnitude and direction. 0o0o 180 o

36. 90 o 270 o E To determine the value of the X-component (Vx) we need to use COSINE. 0o0o 180 o

37. 90 o 270 o E Remember COSINE? 0o0o 180 o

38. E 0o0o Cosine  = Adjacent Hypotenuse 90 o 270 o

39. 90 o 270 o We need to rearrange the equation: solving for the adjacent side.. Cosine  = Adjacent Hypotenuse E 0o0o 180 o

40. 90 o 270 o Cos  = Adj Hyp E 0o0o 180 o Adj = (Hyp) (Cos  ) therefore Vx = V cos 

41. 90 o 270 o E 0o0o 180 o Adj = (Hyp) (Cos  ) therefore Cos  = Adj Hyp Vx = V cos 

42. 90 o 270 o E 0o0o 180 o Vx = V cos  Vx = 36 m/s (cos 42 o ) Vx = 26.7 m/s

43. 90 o 270 o E 0o0o 180 o We can now solve for Vy using Sine.

44. 90 o 270 o E 0o0o 180 o Sine  = Opposite Hypotenuse

45. 90 o 270 o We are solving for the side opposite the 42 degree angle, Vy, therefore we’ll rearrange the equation solving for the opposite side. E 0o0o 180 o

46. 90 o 270 o therefore E 0o0o 180 o Vy = V sin  Opp = Hyp (Sin  ) Sine  = Opposite Hypotenuse

47. 90 o 270 o therefore E 0o0o 180 o Vy = V sin  Opp = Hyp (Sin  ) Sine  = Opposite Hypotenuse

48. 90 o 270 o E 0o0o 180 o Vy = V sin  Vy = 36 m/s (sin 42 o ) Vy = 24.1 m/s

49. 90 o 270 o E 0o0o 180 o We’ve solved for the horizontal vertical and components of our original vector!

50. 90 o 270 o E 0o0o 180 o Let’s REVIEW what we did.

51. V  REVIEW A vector can be broken down into two vectors at right angles A MATHEMATICAL METHOD 90 o

52. V  REVIEW A MATHEMATICAL METHOD The component that lies on the X-axis is called the horizontal or X - component The X-component is found using Vx = Vcos  Vx = Vcos 

53. V  REVIEW A MATHEMATICAL METHOD The component that lies on the Y-axis is called the vertical or Y- component The Y-component is found using Vy = Vsin  Vy = Vsin  Vx = Vcos 

54. Now that we know how to break vectors down into their X and Y components we can solve for the resultant of many vectors using:

55. Let’s look at a sample problem!

56. travels 4.2 m at an angle of 37 o north of east. The first putt 10 o east of north. The second putt travels 3.9 m at The third putt travels 2.6 m at an angle of 30 o south of west. The last putt heading 71 o east of south travels 1.8 m before dropping into the hole. What displacement was needed to sink the ball on the first putt? A golfer, putting on a green, requires four shots to “hole the ball”.

57. To solve the problem, using the component method, we’ll carry out the following sequence: 1. List the vectors 2. Solve for the X and Y components of each vector 3. Sum the X components and sum the Y components 5. Solve for the direction of the resultant using tangent 4. Solve for the magnitude of the resultant using the Pythagorean theorem

58. First, let’s list all of the vectors changing all of the directions to the number of degrees from east in a counterclockwise direction. 1. List the vectors

59. travels 4.2 m at an angle of 37 o north of east. The first putt 10 o east of north. The second putt travels 3.9 m at The third putt travels 2.6 m at an angle of 30 o south of west. The last putt heading 71 o east of south travels 1.8 m before dropping into the hole. What displacement was needed to sink the ball on the first putt? A golfer, putting on a green, requires four shots to “hole the ball”. 90 o 0o0o 180 o 270 o

60. travels 4.2 m at an angle of 37 o north of east. The first putt 10 o east of north. The second putt travels 3.9 m at The third putt travels 2.6 m at an angle of 30 o south of west. The last putt heading 71 o east of south travels 1.8 m before dropping into the hole. What displacement was needed to sink the ball on the first putt? A golfer, putting on a green, requires four shots to “hole the ball”. 90 o 0o0o 180 o 270 o d 1 = 4.2 m @ 37 o d 2 = 3.9 m @ 80 o d 3 = 2.6 m @ 210 o d 4 = 1.8 m @ 341 o 10 o east of north 30 o south of west 71 o east of south 37 o north of east. 4.2 m 3.9 m 2.6 m 1.8 m

61. Next, to make things easier to see, let’s set up a table that lists our vectors.. 1. List the vectors

62. First label all of the columns Vector (V) X- component Solve Y -component Solve

63. Remember the X component is cosine Vector (V) X- component Solve Y -component Solve Vector (V) X- component Solve Y -component Solve (Vcos  )

64. And the Y component is sine Vector (V) X- component Solve Y -component Solve Vector (V) X- component Solve Y -component Solve (Vcos  ) (Vsin  )

65. Vector (V) X- component Solve Y -component Solve (Vcos  ) (Vsin  ) Vector (V) X- component Solve Y -component Solve (Vcos  ) (Vsin  ) Next fill in the first column with the different vectors. 4.2 m @ 37 o 3.9 m @ 80 o 2.6 m @ 210 o 1.8 m @ 341 o

66. Vector (V) X- component Solve Y -component Solve (Vcos  ) (Vsin  ) Vector (V) X- component Solve Y -component Solve (Vcos  ) (Vsin  ) Substitute into the equations for X and then Y 4.2 m @ 37 o 3.9 m @ 80 o 2.6 m @ 210 o 1.8 m @ 341 o 4.2 m cos 37 o 3.9 m cos 80 o 2.6 m cos 210 o 1.8 m cos 341 o 4.2 m sin 37 o 3.9 m sin 80 o 2.6 m sin 210 o 1.8 m sin 341 o Remember to include units!

67. Now solve for the value of each X and Y component. Again, remember to include UNITS! Vector (V) X- component Solve Y -component Solve (Vcos  ) (Vsin  ) Vector (V) X- component Solve Y -component Solve (Vcos  ) (Vsin  ) 4.2 m @ 37 o 3.9 m @ 80 o 2.6 m @ 210 o 1.8 m @ 341 o 4.2 m cos 37 o 3.9 m cos 80 o 2.6 m cos 210 o 1.8 m cos 341 o 4.2 m sin 37 o 3.9 m sin 80 o 2.6 m sin 210 o 1.8 m sin 341 o 3.35 m 0.68 m -2.25 m 1.70 m 2.53 m 3.84 m -1.30 m -0.58 m

68. Sum the X components and then the Y components. Vector (V) X- component Solve Y -component Solve (Vcos  ) (Vsin  ) Vector (V) X- component Solve Y -component Solve (Vcos  ) (Vsin  ) 4.2 m @ 37 o 3.9 m @ 80 o 2.6 m @ 210 o 1.8 m @ 341 o 4.2 m cos 37 o 3.9 m cos 80 o 2.6 m cos 210 o 1.8 m cos 341 o 4.2 m sin 37 o 3.9 m sin 80 o 2.6 m sin 210 o 1.8 m sin 341 o 3.35 m 0.68 m -2.25 m 1.70 m 2.53 m 3.84 m -1.30 m -0.58 m Sum (  )  x = 3.48 m  Y = 4.49 m

69. Vector (V) X- component Solve Y -component Solve (Vcos  ) (Vsin  ) Vector (V) X- component Solve Y -component Solve (Vcos  ) (Vsin  ) 4.2 m @ 37 o 3.9 m @ 80 o 2.6 m @ 210 o 1.8 m @ 341 o 4.2 m cos 37 o 3.9 m cos 80 o 2.6 m cos 210 o 1.8 m cos 341 o 4.2 m sin 37 o 3.9 m sin 80 o 2.6 m sin 210 o 1.8 m sin 341 o 3.35 m 0.68 m -2.25 m 1.70 m 2.53 m 3.84 m -1.30 m -0.58 m Sum (  )  x = 3.48 m  Y = 4.49 m The four vectors are now broken down into two.

70. Vector (V) X- component Solve Y -component Solve (Vcos  ) (Vsin  ) Vector (V) X- component Solve Y -component Solve (Vcos  ) (Vsin  ) 4.2 m @ 37 o 3.9 m @ 80 o 2.6 m @ 210 o 1.8 m @ 341 o 4.2 m cos 37 o 3.9 m cos 80 o 2.6 m cos 210 o 1.8 m cos 341 o 4.2 m sin 37 o 3.9 m sin 80 o 2.6 m sin 210 o 1.8 m sin 341 o 3.35 m 0.68 m -2.25 m 1.70 m 2.53 m 3.84 m -1.30 m -0.58 m Sum (  )  x = 3.48 m  Y = 4.49 m One vector 3.48 m long lies on the X-axis.

71. Vector (V) X- component Solve Y -component Solve (Vcos  ) (Vsin  ) Vector (V) X- component Solve Y -component Solve (Vcos  ) (Vsin  ) 4.2 m @ 37 o 3.9 m @ 80 o 2.6 m @ 210 o 1.8 m @ 341 o 4.2 m cos 37 o 3.9 m cos 80 o 2.6 m cos 210 o 1.8 m cos 341 o 4.2 m sin 37 o 3.9 m sin 80 o 2.6 m sin 210 o 1.8 m sin 341 o 3.35 m 0.68 m -2.25 m 1.70 m 2.53 m 3.84 m -1.30 m -0.58 m Sum (  )  x = 3.48 m  Y = 4.49 m The other vector, 4.49 m long, lies on the Y-axis.

72. Vector (V) X- component Solve Y -component Solve (Vcos  ) (Vsin  ) Vector (V) X- component Solve Y -component Solve (Vcos  ) (Vsin  ) 4.2 m @ 37 o 3.9 m @ 80 o 2.6 m @ 210 o 1.8 m @ 341 o 4.2 m cos 37 o 3.9 m cos 80 o 2.6 m cos 210 o 1.8 m cos 341 o 4.2 m sin 37 o 3.9 m sin 80 o 2.6 m sin 210 o 1.8 m sin 341 o 3.35 m 0.68 m -2.25 m 1.70 m 2.53 m 3.84 m -1.30 m -0.58 m Sum (  )  x = 3.48 m  Y = 4.49 m We’ll plot these two vectors and determine their resultant.

73. Y  x = 3.48 m  Y = 4.49 m We’ll plot these two vectors and determine their resultant. X  x = 3.48 m  Y = 4.49 m

74. Y X  x = 3.48 m The resultant of these two vectors is the same as the resultant of our original four vectors!.  Y = 4.49 m

75. Y X  x = 3.48 m The Pythagorean Theorem will be used to determine the MAGNITUDE of the resultant. c 2 = a 2 +b 2 R 2 =  X 2 +  Y 2 R =  X 2 +  Y 2 R = (3.48m) 2 + (4.49m) 2 R = 5.68 m

76. Y X  Y = 4.49 m  x = 3.48 m We have now found the magnitude of our resultant to be 5.68 m long. c 2 = a 2 +b 2 R 2 =  X 2 +  Y 2 R =  X 2 +  Y 2 R = (3.48m) 2 + (4.49m) 2 R = 5.68 m

77. Y X  Y = 4.49 m  x = 3.48 m The last step is to find the DIRECTION our resultant is pointing.

78. Y X  Y = 4.49 m  Y = 3.48 m To find the angle we’ll use tangent. Tan  = adjacent opposite adj opp  = Tan -1 XX YY 3.48 m 4.49 m  = Tan -1  = = 52.2 o

79. Y X  Y = 4.49 m  Y = 3.48 m The RESULTANT is 5.68 m @ 52.2 o. Let’s go back to our original problem with the results.

80. travels 4.2 m at an angle of 37 o north of east. The first putt 10 o east of north. The second putt travels 3.9 m at The third putt travels 2.6 m at an angle of 30 o south of west. The last putt heading 71 o east of south travels 1.8 m before dropping into the hole. What displacement was needed to sink the ball on the first putt? A golfer, putting on a green, requires four shots to “hole the ball”.

81. Let’s SUMMARIZE the COMPONENT METHOD

82. List all the vectors. Break each vector down into an X and Y component using Cosine for X and Sine for Y. Sum all of the X components and then sum all of the Y components Using the Pythagorean Theorem solve for the MAGNITUDE of the resultant Using Tangent solve for the DIRECTION of the resultant.

83. Let’s go over that once again.

84. Vector (V) X-component Solve Y-component Solve xx YY 4.2 m @ 37 o 3.9 m @ 80 o 2.6 m @ 210 o 1.8 m @ 341 o List all the vectors.

85. Break each vector down into an X and Y component using Cosine for X and Sine for Y. Vector (V) X-component Solve Y-component Solve 4.2 m @ 37 o 3.9 m @ 80 o 2.6 m @ 210 o 1.8 m @ 341 o (Vcos  ) (Vsin  ) 4.2 m cos 37 o 3.9 m cos 80 o 2.6 m cos 210 o 1.8 m cos 341 o 4.2 m sin 37 o 3.9 m sin 80 o 2.6 m sin 210 o 1.8 m sin 341 o xx YY

86. Sum all of the X components and then sum all of the Y components Vector (V) X-component Solve Y-component Solve 4.2 m @ 37 o 3.9 m @ 80 o 2.6 m @ 210 o 1.8 m @ 341 o (Vcos  ) (Vsin  ) 4.2 m cos 37 o 3.9 m cos 80 o 2.6 m cos 210 o 1.8 m cos 341 o 4.2 m sin 37 o 3.9 m sin 80 o 2.6 m sin 210 o 1.8 m sin 341 o 3.35 m 0.68 m -2.25 m 1.70 m Sum (  )  x = 3.48 m 2.53 m 3.84 m -1.30 m -0.58 m  Y = 4.49 m xx YY  Y = 3.48 m  Y = 4.49 m

87. Vector (V) X-component Solve Y-component Solve 4.2 m @ 37 o 3.9 m @ 80 o 2.6 m @ 210 o 1.8 m @ 341 o (Vcos  ) (Vsin  ) 4.2 m cos 37 o 3.9 m cos 80 o 2.6 m cos 210 o 1.8 m cos 341 o 4.2 m sin 37 o 3.9 m sin 80 o 2.6 m sin 210 o 1.8 m sin 341 o 3.35 m 0.68 m -2.25 m 1.70 m Sum (  )  x = 3.48 m 2.53 m 3.84 m -1.30 m -0.58 m  Y = 4.49 m Using the Pythagorean Theorem solve for the MAGNITUDE of the resultant c 2 = a 2 +b 2 R 2 =  X 2 +  Y 2 R =  X 2 +  Y 2 R = (3.48m) 2 + (4.49m) 2 R = 5.68 m xx YY  Y = 3.48 m  Y = 4.49 m

88. Vector (V) X-component Solve Y-component Solve 4.2 m @ 37 o 3.9 m @ 80 o 2.6 m @ 210 o 1.8 m @ 341 o (Vcos  ) (Vsin  ) 4.2 m cos 37 o 3.9 m cos 80 o 2.6 m cos 210 o 1.8 m cos 341 o 4.2 m sin 37 o 3.9 m sin 80 o 2.6 m sin 210 o 1.8 m sin 341 o 3.35 m 0.68 m -2.25 m 1.70 m Sum (  )  x = 3.48 m 2.53 m 3.84 m -1.30 m -0.58 m  Y = 4.49 m c 2 = a 2 +b 2 R 2 =  X 2 +  Y 2 R =  X 2 +  Y 2 R = (3.48m) 2 + (4.49m) 2 R = 5.68 m xx YY  Y = 3.48 m  Y = 4.49 m Using Tangent solve for the DIRECTION of the resultant. Tan  = adjacent opposite adj opp  = Tan -1 xx YY 3.48 m 4.49 m  = Tan -1  = 52.2 o