## Summersnow8 one year ago Find the resultant of each set of displacements: 60 km due south, then 90 km at 15 degrees north of west, and then 75 km at 45 degrees north of east. I know the answer is 38 km at 25 degrees north of west.

1. anonymous

So what do you need? How to get there?

2. Summersnow8

yeah, I need help with how to solve it.

3. anonymous

This seems like physics ._. @hba You good in Physics?

4. anonymous

Im new to physics lol

5. Summersnow8

6. princeharryyy

can you draw the figure

7. Summersnow8

a figure isn't given, but this is the figure i have drawn

8. princeharryyy

until you can't draw it you can't solve it.

9. Summersnow8

|dw:1436290559148:dw|

10. princeharryyy

try solving it now!

11. princeharryyy

ok! leave it! HOW DID YOU SOLVED IT!

12. Summersnow8

I don't know how to solve it, that's why I am asking for help

13. princeharryyy

oh!!!!

14. princeharryyy

ok

15. zepdrix

woops let fix your graph a sec :) 15 north of west means we start by facing west, and rotating up 15 degrees.

16. zepdrix

|dw:1436292130604:dw|

17. zepdrix

So what are we doing? Adding up these vectors? :O

18. zepdrix

That resultant shinanigans and all that? +_+

19. zepdrix

|dw:1436292221555:dw|

20. zepdrix

Let's figure out the components to each vector or force as I've written them :) For F_1, there is no x component, we've gone zero to the left or right. So our F_1 is simply $$\Large\rm \vec F_1=\left<0,-60\right>$$ Is this notation ok? Or do you prefer the i's and j's? :3

21. zepdrix

SummerPotatooooo, just chat in here XD don't need private message box lol

22. Summersnow8

what are i's and j's? I guess whatever is easiest

23. zepdrix

|dw:1436292562770:dw|Let's say we wanted to get to this point. We would go some distance to the right, and some distance upward. This is i that I've labeled, it's 1 unit long, in the x direction.

24. zepdrix

|dw:1436292657672:dw|So to get to that point, we'll need 3 i's. We moved 3 units horizontally.

25. zepdrix

And j is 1 unit vertically. So we need 4 j's.

26. Summersnow8

I've never learned that way....I know to use trigonometry

27. zepdrix

|dw:1436292736659:dw|So what I've done is, I've written this vector in component form.$\Large\rm v=3\hat i+4\hat j$ Our problem is kinda tricky. We're only given the length of the vector, so yes, you have to use trig to help you find each component of the vector.

28. zepdrix

Haven't learned that? :[ hmm

29. Summersnow8

I use x and y no i and j

30. Summersnow8

not*

31. zepdrix

ah ok :3

32. zepdrix

Let's look at the F_3 that I labeled on the graph. That one will be easiest so maybe we should start there.

33. Summersnow8

okay

34. zepdrix

|dw:1436292975469:dw|If we think of this as a triangle

35. zepdrix

We can use our trigonometry to get our x and y values.

36. Summersnow8

so x = 53 and y = 53

37. zepdrix

Mmm good good that sounds right! :) So here is a way we can write our F_3,$\Large\rm \vec F_3=53x+53y$

38. Summersnow8

This was my attempt at solving

39. zepdrix

Ok your process looks great! Let's look at displacement B a second though.

40. zepdrix

Notice that arrow is pointing to the left. That tells us that our x coordinate should come out to be negative, ya?

41. Summersnow8

okay

42. Summersnow8

so x= - 86.9 ?

43. zepdrix

mmm yah that looks correct!

44. Summersnow8

now A really confuses me

45. zepdrix

|dw:1436293376046:dw|Ah yes :) kind of hard to make a triangle with this one, huh?

46. Summersnow8

yeah... that's where i get stuck. one I solve that then I know to add up all the x's and y's and take the square root of x squared + y squared

47. zepdrix

If you can get used to this shortcut, then it might help when we get stuck. In the first triangle, you were able to find x by this formula: $$\large\rm x=|A|\cos(\theta)$$ Where |A| is the length of A, and theta is the angle formed. The reason your sign came out wrong was because you measured your angle from the negative branch of the x-axis.|dw:1436293621678:dw|If we had measured from the correct axis, using this large angle, then the negative sign would have come out correctly.

48. zepdrix

So for triangle A, we could have put $$\large\rm x=|A|\cos(\theta)$$ $$\large\rm x=90\cos(165^o)$$ to get that -86.9

49. zepdrix

AHHH that was all for letter B sorry sorry :( forgot to look back at the letters

50. zepdrix

So for A, if we measure from the positive side of the x-axis,|dw:1436293784227:dw|

51. zepdrix

Do you see how I got that angle? :O

52. Summersnow8

yes, I understand

53. zepdrix

$\large\rm x=|A|\cos(-90^o)$$\large\rm x=60\cos(-90^o)$

54. zepdrix

Similarly,$\large\rm y=|A|\sin(\theta)$$\large\rm y=60\sin(-90^o)$

55. zepdrix

This shortcut comes in handy when you're not able to draw a triangle :P that made things kinda tricky with it being on the axis hehe

56. Summersnow8

so x = 0 and y = -60

57. zepdrix

good good

58. zepdrix

And that makes sense if you look at the picture, right? we went 0 to the left or right, and we went 60 down.

59. Summersnow8

yeah that makes sense

60. Summersnow8

$\sqrt{33.9^{2}16^{2}} = 37.48$

61. Summersnow8

then -tan (16/33.9) = 25

62. Summersnow8

right?

63. zepdrix

The length of your resultant looks correct!$\sqrt{33.9^{2}\color{red}{+}16^{2}} = 37.48$ Angle um um um thinking :3

64. Summersnow8

I know the answer is 38 km at 25 degrees north of west. but how do i know the direction and i got 37 and not 38

65. zepdrix

You probably just didn't carry around enough decimals to get the correct rounded value :) no big deal.

66. zepdrix

For the angle,$\Large\rm \tan^{-1}\left(\frac{y}{x}\right)=\tan^{-1}\left(\frac{16}{-33.9}\right)=-25.69$Think about WHERE this angle is located, and then think about WHERE your x and y are located for your resultant.

67. zepdrix

|dw:1436294373840:dw|negative 25 degrees is somewhere around here

68. zepdrix

|dw:1436294434921:dw|But notice that our resultant vector should be somewhere in quadrant 2.

69. zepdrix

So tangent gave us the reference angle, we need the NEXT one though. So we have to ADD 180 degrees to the angle that we ended up with.

70. Summersnow8

so 155 degrees

71. zepdrix

|dw:1436294544290:dw|

72. zepdrix

|dw:1436294610322:dw|mm yessss

73. zepdrix

mmm what do you think? too confusing? :O

74. zepdrix

when dealing with inverse tangent things can get a little bit weird, you really have to pay attention to which quadrant your resultant is in

75. Summersnow8

yeah i get how we get the resultant but the answer has +25 degrees. So i really dont understand how we got NW

76. zepdrix

|dw:1436294829972:dw|So from facing West, to get to our resultant vector, we have to rotate north 25 degrees.

77. Summersnow8

but how do you go from -tan = -25 to the answer being 25? and knowing it will be NW, I don't understand. is there a different way of explaining

78. zepdrix

mmm thinking :)

79. Summersnow8

okay thanks

80. zepdrix

|dw:1436295041858:dw|I guess you can just remember that vertical angles are always congruent, so these two angles I labeled will always be the same. That goes back to geometry though :P

81. zepdrix

Keep in mind though, we didn't HAVE TO use NW for directional. Another acceptable answer would have been to use ... 65 degrees west of north.|dw:1436295194403:dw|

82. Summersnow8

so because the -tan is -25 and we cant have a - degree you go congruently to the + degree?

83. Summersnow8

so instead of -25 in the SE it is 25 in the NW

84. zepdrix

ya lemme try to clarify that a little bit :) the problem was NOT that the angle was negative. it was the fact that the angle was in quadrant 4, and our resultant vector was in quadrant 2. the problem was that the angle wasn't where it was supposed to be. But yes, that's a good little shortcut to remember maybe! :) This might come up in other problems as well... like... maybe inverse tangent spits out an angle in quadrant 1,|dw:1436295356582:dw|But your resultant x and y are clearly in the third quadrant.

85. Summersnow8

how did we know our resultant was in that quadrant though?

86. zepdrix

|dw:1436295443084:dw|It looks like your trick will work nicely for this type of problem. So instead of 19 degrees North of East, we would say our resultant is located 19 degrees South of West.

87. Summersnow8

because x= a - number? and y was +?

88. zepdrix

Because of the x and y that you end up with. The signs of those values tells you which quadrant you are in. Yes, good :)

89. Summersnow8

okay, I suppose I understand now, thank you :)

90. zepdrix

cool cool. i know i know, the math is kinda crazy \c:/ just keep up the hard work lady!