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Math Help - Microlight Prob

  1. #1
    ADY
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    Microlight Prob

    Ok, so im having a slight problem with getting to the same resultant as suggested by my book. Perhaps you could help me...

    The microlights cruising speed in still air is 45 ms^-1, pointing in the direction of N 54◦ W but flies in a wind speed of 15.2 ms^-1 from a S 28W. We Take i to be 1 ms^-1 due east and j to be 1 ms^-1 due north.
    Vm - Velocity Microlight
    Vw - Velocity Wind

    So Vm has a magnitude of 45 and direction 28◦
    The wind comes from S 28◦ W & hence blows towards N 28◦ E for which the direction is -(90◦ -28◦) hence vector Vm has a magnitude 15.2 and direction -62

    The components are therefore

    Vm = 45 cos(28◦)i + 45sin(28◦)
    = 39.7326i + 21.1262J

    Vw = 15.2 cos(-62◦)i + 15.2cos (-62)J
    =-7.1360i + (-13.4208)J

    so i make the resultant

    v = Vm + Vw =

    39.7326 + (-7.1360)i + 21.1262 + (-13.4208)J

    =32.5966i & 7.7054

    The suggested resultant is -29.2698i + 39.8711J, so where am i going wrong, thanks guys!
    Last edited by mr fantastic; December 7th 2009 at 03:04 AM. Reason: Restored deleted question
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  2. #2
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    Hello, ADY!

    The microlight's cruising speed in still air is 45 m/s in the direction of N54W.
    but flies in a wind speed of 15.2 m/s from S28W.

    Find its resultant vector.

    The plane's vector has a magnitude of 45 and an angle of 144.

    . . v_p \:=\;\langle45\cos144^o,\;45\sin144^o\rangle


    The wind's vector has a magnitude of 15.2 and an angle of 62.

    . . v_w \;=\;\langle 15.2\cos62^o,\;15.2\sin62^o\rangle


    The resultant vector is:

    . . v_p + v_w \;=\;\langle45\cos144^o,\;45\sin144^o\rangle + \langle15.2\cos62^o,\;15.2\sin62^o\rangle

    . . . . . . . =\;\langle -29.26979699,\;39.87113976 \rangle

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  3. #3
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    Hello ADY
    Quote Originally Posted by ADY View Post
    Ok, so im having a slight problem with getting to the same resultant as suggested by my book. Perhaps you could help me...

    The microlights cruising speed in still air is 45 ms^-1, pointing in the direction of N 54◦ W but flies in a wind speed of 15.2 ms^-1 from a S 28W. We Take i to be 1 ms^-1 due east and j to be 1 ms^-1 due north.
    Vm - Velocity Microlight
    Vw - Velocity Wind

    So Vm has a magnitude of 45 and direction 28◦
    The wind comes from S 28◦ W & hence blows towards N 28◦ E for which the direction is -(90◦ -28◦) hence vector Vm has a magnitude 15.2 and direction -62

    The components are therefore

    Vm = 45 cos(28◦)i + 45sin(28◦)
    = 39.7326i + 21.1262J

    Vw = 15.2 cos(-62◦)i + 15.2cos (-62)J
    =-7.1360i + (-13.4208)J

    so i make the resultant

    v = Vm + Vw =

    39.7326 + (-7.1360)i + 21.1262 + (-13.4208)J

    =32.5966i & 7.7054

    The suggested resultant is -29.2698i + 39.8711J, so where am i going wrong, thanks guys!
    Thanks for showing us all your working, but I'm afraid I don't understand how you arrived at the angles you used in your calculations. v_m is in a direction N 54^o W, so its components are:
    -45\sin 54^o East and 45\cos54^o North
    So:
    v_m = -36.41\textbf{i}+26.45\textbf{j}
    And the wind (as you said) blows in a direction N 28^o E, so its components are:
    15.2\sin28^o East and 15.2\cos28^o North
    So:
    v_w = 7.14\textbf{i}+13.42\textbf{j}
    which gives a resultant velocity of:
    -29.27\textbf{i}+39.87\textbf{j}
    agreeing with the answer in the book.

    Grandad
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  4. #4
    ADY
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    Quote Originally Posted by Soroban View Post
    Hello, ADY!


    The plane's vector has a magnitude of 45 and an angle of 144.

    . . v_p \:=\;\langle45\cos144^o,\;45\sin144^o\rangle




    Thank you - so i was nearly there, all apart from the angle, so how is this derived?

    Thanks
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  5. #5
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    Hello, ADY!

    The angle of a vector is always measured from the positive x-axis.
    (Counter-clockwise: positive angle. .Clockwise: negative angle.)


    The plane's direction is N54W


    Code:
                    |
            *       |
              *     |
                *54|
                  * |
          - - - - - * - - - - -
                    |


    Measured from the positive x-axis, the angle is 144.
    Code:
            *
              *
                *   144
                  *
          - - - - - * - - - - -
                    |



    The wind's direction is from S28W.
    Code:
                    |
                    |
          - - - - - + - - - - -
                   /|
                  / |
                 /  |
                /28|
               /    |
                    |


    Measured from the positive x-axis, the angle is 62.

    Code:
                    |    /
                    |28/
                    |  /
                    | /
                    |/ 62
          - - - - - + - - - - -
                   /|
                  / |
                 /  |
                /28|
               /    |
                    |
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