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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.

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  1. undeadknight26
    • one year ago
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    So what do you need? How to get there?

  2. Summersnow8
    • one year ago
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    yeah, I need help with how to solve it.

  3. undeadknight26
    • one year ago
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    This seems like physics ._. @hba You good in Physics?

  4. undeadknight26
    • one year ago
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    Im new to physics lol

  5. Summersnow8
    • one year ago
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    ok, @undeadknight26 thanks for trying

  6. princeharryyy
    • one year ago
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    can you draw the figure

  7. Summersnow8
    • one year ago
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    a figure isn't given, but this is the figure i have drawn

  8. princeharryyy
    • one year ago
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    until you can't draw it you can't solve it.

  9. Summersnow8
    • one year ago
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    |dw:1436290559148:dw|

  10. princeharryyy
    • one year ago
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    try solving it now!

  11. princeharryyy
    • one year ago
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    ok! leave it! HOW DID YOU SOLVED IT!

  12. Summersnow8
    • one year ago
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    I don't know how to solve it, that's why I am asking for help

  13. princeharryyy
    • one year ago
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    oh!!!!

  14. princeharryyy
    • one year ago
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    ok

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

  16. zepdrix
    • one year ago
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    |dw:1436292130604:dw|

  17. zepdrix
    • one year ago
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    So what are we doing? Adding up these vectors? :O

  18. zepdrix
    • one year ago
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    That resultant shinanigans and all that? +_+

  19. zepdrix
    • one year ago
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    |dw:1436292221555:dw|

  20. zepdrix
    • one year ago
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    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
    • one year ago
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    SummerPotatooooo, just chat in here XD don't need private message box lol

  22. Summersnow8
    • one year ago
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    what are i's and j's? I guess whatever is easiest

  23. zepdrix
    • one year ago
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    |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
    • one year ago
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    |dw:1436292657672:dw|So to get to that point, we'll need 3 i's. We moved 3 units horizontally.

  25. zepdrix
    • one year ago
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    And j is 1 unit vertically. So we need 4 j's.

  26. Summersnow8
    • one year ago
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    I've never learned that way....I know to use trigonometry

  27. zepdrix
    • one year ago
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    |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
    • one year ago
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    Haven't learned that? :[ hmm

  29. Summersnow8
    • one year ago
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    I use x and y no i and j

  30. Summersnow8
    • one year ago
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    not*

  31. zepdrix
    • one year ago
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    ah ok :3

  32. zepdrix
    • one year ago
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    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
    • one year ago
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    okay

  34. zepdrix
    • one year ago
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    |dw:1436292975469:dw|If we think of this as a triangle

  35. zepdrix
    • one year ago
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    We can use our trigonometry to get our x and y values.

  36. Summersnow8
    • one year ago
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    so x = 53 and y = 53

  37. zepdrix
    • one year ago
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    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
    • one year ago
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    This was my attempt at solving

  39. zepdrix
    • one year ago
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    Ok your process looks great! Let's look at displacement B a second though.

  40. zepdrix
    • one year ago
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    Notice that arrow is pointing to the left. That tells us that our x coordinate should come out to be negative, ya?

  41. Summersnow8
    • one year ago
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    okay

  42. Summersnow8
    • one year ago
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    so x= - 86.9 ?

  43. zepdrix
    • one year ago
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    mmm yah that looks correct!

  44. Summersnow8
    • one year ago
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    now A really confuses me

  45. zepdrix
    • one year ago
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    |dw:1436293376046:dw|Ah yes :) kind of hard to make a triangle with this one, huh?

  46. Summersnow8
    • one year ago
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    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
    • one year ago
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    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
    • one year ago
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    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
    • one year ago
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    AHHH that was all for letter B sorry sorry :( forgot to look back at the letters

  50. zepdrix
    • one year ago
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    So for A, if we measure from the positive side of the x-axis,|dw:1436293784227:dw|

  51. zepdrix
    • one year ago
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    Do you see how I got that angle? :O

  52. Summersnow8
    • one year ago
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    yes, I understand

  53. zepdrix
    • one year ago
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    \[\large\rm x=|A|\cos(-90^o)\]\[\large\rm x=60\cos(-90^o)\]

  54. zepdrix
    • one year ago
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    Similarly,\[\large\rm y=|A|\sin(\theta)\]\[\large\rm y=60\sin(-90^o)\]

  55. zepdrix
    • one year ago
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    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
    • one year ago
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    so x = 0 and y = -60

  57. zepdrix
    • one year ago
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    good good

  58. zepdrix
    • one year ago
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    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
    • one year ago
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    yeah that makes sense

  60. Summersnow8
    • one year ago
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    \[\sqrt{33.9^{2}16^{2}} = 37.48\]

  61. Summersnow8
    • one year ago
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    then -tan (16/33.9) = 25

  62. Summersnow8
    • one year ago
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    right?

  63. zepdrix
    • one year ago
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    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
    • one year ago
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    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
    • one year ago
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    You probably just didn't carry around enough decimals to get the correct rounded value :) no big deal.

  66. zepdrix
    • one year ago
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    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
    • one year ago
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    |dw:1436294373840:dw|negative 25 degrees is somewhere around here

  68. zepdrix
    • one year ago
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    |dw:1436294434921:dw|But notice that our resultant vector should be somewhere in quadrant 2.

  69. zepdrix
    • one year ago
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    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
    • one year ago
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    so 155 degrees

  71. zepdrix
    • one year ago
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    |dw:1436294544290:dw|

  72. zepdrix
    • one year ago
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    |dw:1436294610322:dw|mm yessss

  73. zepdrix
    • one year ago
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    mmm what do you think? too confusing? :O

  74. zepdrix
    • one year ago
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    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
    • one year ago
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    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
    • one year ago
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    |dw:1436294829972:dw|So from facing West, to get to our resultant vector, we have to rotate north 25 degrees.

  77. Summersnow8
    • one year ago
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    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
    • one year ago
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    mmm thinking :)

  79. Summersnow8
    • one year ago
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    okay thanks

  80. zepdrix
    • one year ago
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    |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
    • one year ago
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    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
    • one year ago
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    so because the -tan is -25 and we cant have a - degree you go congruently to the + degree?

  83. Summersnow8
    • one year ago
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    so instead of -25 in the SE it is 25 in the NW

  84. zepdrix
    • one year ago
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    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
    • one year ago
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    how did we know our resultant was in that quadrant though?

  86. zepdrix
    • one year ago
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    |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
    • one year ago
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    because x= a - number? and y was +?

  88. zepdrix
    • one year ago
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    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
    • one year ago
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    okay, I suppose I understand now, thank you :)

  90. zepdrix
    • one year ago
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    cool cool. i know i know, the math is kinda crazy \c:/ just keep up the hard work lady!

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