## anonymous 2 years ago Stuck for a long time :(..........Please help me with vector addition! Illustration below. I need to find the direction and magnitude of the sum vector. The correct answer is 41.62, 42 degrees from x to -y

1. anonymous

|dw:1381109606282:dw| (Construct c = a + b)

2. anonymous

|dw:1381109717758:dw|

3. wolfe8

I believe you are supposed to add them like this: |dw:1381110194843:dw|

4. anonymous

?

5. anonymous

I was trying to use the component method (x and y components)

6. mathstudent55

What are the magnitudes of the two vectors?

7. anonymous

Oh! I forgot to write that! One second...

8. anonymous

|dw:1381111020150:dw|

9. mathstudent55

|dw:1381111017373:dw|

10. anonymous

255 or 75?

11. mathstudent55

$$R_x = A_x + B_x = 5 \cos20 + 5 \cos255$$ $$R_y = A_y + B_y = 5 \sin20 + 5 \sin255$$ $$R = \sqrt{R_x^2 + R_y^2}$$ $$\theta = \tan^{-1} \dfrac{R_y}{R_x}$$

12. mathstudent55

An angle of 75 deg down form the negative y-axis is an angle of 255 deg from the positive x-axis. The advantage of using the 255 deg angle is that the sin and cos will automatically be negative numbers since it's in the third quadrant.

13. anonymous

Oh!! Okay That makes sense!

14. anonymous

@mathstudent55 do you mind If I calculate that out to see if I get the correct answer? I've been half an hour, so please don't leave :)

15. mathstudent55

No problem.

16. anonymous

Hmmm I get 4.61 for magnitude?

17. anonymous

7 degrees?

18. mathstudent55

I got 4.62 magnitude, and -42.5 deg.

19. mathstudent55

What did you get for Rx and Ry?

20. anonymous

Rx: $$\ 5cos20+5cos255$$ $$\ =$$ $$\ -0.13...$$ Rx: $$\ 5sin20 + 5sin255$$ $$\ =$$ $$\ 0.416$$

21. anonymous

$$\ \huge \text{The latter should be } Ry.$$

22. mathstudent55

This is what I get for the components of the resultant. $$R_x = 5 \cos 20 + 5 \cos 255 = 4.6985 + (-1.29409) = 3.4044$$ $$R_y = 5 \sin 20 + 5 \sin 255 = 1.7101 + (-4.82963) = -3.11952$$

23. mathstudent55

$$R^2 = R_x^2 +R_y^2$$ $$R = \sqrt{(3.4044)^2 + (-3.11952)^2}$$ $$R = 4.6175$$ $$\theta = \tan^{-1}\dfrac{R_y}{R_x}$$ $$\theta = \tan^{-1} \dfrac{-3.11952}{3.4044}$$ $$\theta = 42.5^o$$

24. anonymous

Hmmm So perhaps the answer key is incorrect?

25. anonymous

I hope this not too much, but there is one other problem I am stuck on... this time with vector subtraction. Do you mind helping me with that?

26. anonymous

|dw:1381113129371:dw|

27. mathstudent55

a - b = a + (-b)

28. mathstudent55

The new vector is -b. Now find x and y components of vector a and vector -b. Add the components together and find the magnitude of the resultant. Then use the inverse tangent to find the angle of the resultant. |dw:1381113881005:dw|

29. mathstudent55

|dw:1381114004562:dw|

30. anonymous

Okay, and I would still add the components except add the OPPOSITE of the b components?

31. anonymous

would x still be cos and y still be sin in this case? I always forget when those are switched

32. mathstudent55

$$A_x = 8 \cos 135$$ $$A_y = 8 \sin 135$$ $$B_x = 4.5 \cos 240$$ $$B_y = 4.5 \sin 240$$ $$R_x = A_x + B_x = 8 \cos 135 + 4.5 \cos 240 = -7.90685$$ $$R_y = A_y + B_y = 8 \sin 135 + 4.5 \sin 240 = 1.75974$$

33. anonymous

Where do the 135 and 240 come from?

34. anonymous

It's subtraction, so shouldn't it be - 4.5cos240?

35. mathstudent55

Let me explain that again.

36. mathstudent55

|dw:1381114977171:dw|

37. mathstudent55

These are the original vectors A and B, ok?

38. anonymous

k

39. mathstudent55

|dw:1381115051999:dw|

40. mathstudent55

Instead of A + B, they want A - B, right?

41. anonymous

Yup

42. mathstudent55

Mathematically speaking, A + B is the same as A + (-B). For example, 4 - 3 = 4 + (-3)

43. anonymous

Yes

44. mathstudent55

So in order to perform the subtraction of vectors A - B, we can instead add vectors A and -B.

45. anonymous

Okay

46. mathstudent55

The first step now is to find what the vector -B is.

47. anonymous

k

48. mathstudent55

If vector $$B = B_x + B_y$$, then $$-B = -( B_x + B_y) = - B_x + (- B_y)$$

49. mathstudent55

If B is the vector in the figure below, |dw:1381115308130:dw|

50. anonymous

uhuh

51. mathstudent55

|dw:1381115353688:dw|

52. mathstudent55

Those are the components of B. Then this is -B: |dw:1381115399994:dw|

53. anonymous

ok

54. mathstudent55

Since vector B goes up to the right, vector -B goes down to the left.

55. mathstudent55

Now remember that we need to add vectors A and -B to subtract A - B.

56. mathstudent55

|dw:1381115515926:dw|

57. mathstudent55

There you have vector A and vector -B. Now we need to add them.

58. mathstudent55

First, let's translate the angles of the vectors to angles starting at the positive x axis.

59. mathstudent55

|dw:1381115630429:dw|

60. anonymous

okay

61. mathstudent55

Now we need to add a vector with magnitude 8 at 135 degrees and a vector of magnitude 4.5 at 240 degrees.

62. anonymous

k

63. mathstudent55

Now we get x and y components of both vectors.

64. anonymous

ok

65. mathstudent55

$$A_x = 8 \cos 135 = -5.65685$$ $$A_y = 8 \sin 135 = 5.65685$$ $$B_x = 4.5 \cos 240 = -2.25$$ $$B_y = 4.5 \sin 240 = -3.89711$$

66. anonymous

Ahhh! That makes more sense!

67. mathstudent55

Now to find the resultant we add the x components to find the x component of the resultant, and we do the same for the y components.

68. mathstudent55

$$R_x = A_x + B_x = -5.65685 + (-2.25) = -7.90685$$ $$R_y = A_y + B_y = 5.65685 + (-3.89711) = 1.75974$$

69. mathstudent55

Now that we have the x and y components of the resultant, we can use the Pythagorean theorem to find the magnitude of the resultant. $$R = \sqrt{R_x^2 + R_y^2}$$ $$R = 8.1$$ We find the angle by using the inverse tangent: $$\theta = \tan^{-1} \dfrac{R_y}{R_x}$$ $$\theta = \tan^{-1} \dfrac{1.75974}{-7.90685}$$ $$\theta = -12.5^o$$

70. mathstudent55

|dw:1381116568714:dw|

71. anonymous

Thank you so much @mathstudent55! You're help is invaluable! I was stuck on these for soooo long, thank you thank you!

72. mathstudent55

You're welcome. BTW, notice how I drew the direction in the last picture. The direction is 12.5 deg up from the negative x-axis, or 167.5 deg measured counterclockwise from the positive x-axis. The answer is 8.1 at 167.5 deg.