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What is the resultant vector

  1. Feb 13, 2005 #1
    what would the resultant vector be for two vectors both 66m in magnitude that look like this: /__ be?
     
  2. jcsd
  3. Feb 13, 2005 #2
    Is that the question? Just use the tip-to-tail method: Attach the tail of one vector to the tip of the other. I'm working on the assumption you're doing a qualitative and not quantitative answer because we would need to know the angle measurement then.
     
  4. Feb 13, 2005 #3

    actually its four vectors each 66m that form a parallerlogram and the angle is 62 degrees. it looks sumtin like this:

    ___
    /__/
     
  5. Feb 13, 2005 #4

    xanthym

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    IF {Angle Between Vectors}=(62 deg) THEN:
    {Magnitude of Resultant Vector} = 68 m


    ~~
     
  6. Feb 13, 2005 #5

    so to find the sum of all four would I just multiply 68 times 4?
     
  7. Feb 13, 2005 #6

    xanthym

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    {Sum All 4 Vectors} = (2)(66)Cos(31 deg) = 113 m
     
  8. Feb 13, 2005 #7

    thats not correct
     
  9. Feb 13, 2005 #8

    xanthym

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    In which direction is each vector pointing?


    ~~
     
  10. Feb 13, 2005 #9
    the diagram looks like this:

    ____
    /___/ with the vertical vectors pointing up and the horizontal ones pointing to the right
     
  11. Feb 13, 2005 #10

    xanthym

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    Magnitude{Sum All 4 Vectors} = (4)(66)Cos(31 deg) = 226 m


    ~~
     
    Last edited: Feb 13, 2005
  12. Feb 13, 2005 #11

    thats wrong too. do you even know what ur doing?
     
  13. Feb 13, 2005 #12
    /bites tongue about the above

    The direction of each vector in vector component form is:
    Let x = angle between the +x axis and the vector

    Magnitude*cos(x)i + Magnitude*sin(x)j

    If you sum each i component together and each j component together, you'll get the resultant vector. For the horizantal vectors, x = 0 and for the partially vertical ones x = 62. It's your job to do the arithmetic though.

    er I should note i and j are unit vectors in the +x and +y directions, respectively.
     
    Last edited by a moderator: Feb 13, 2005
  14. Feb 13, 2005 #13
    The parrelleogram rule is nice because once you move two vectors in head tail configuration, you can use the cos law to get the resultant of the two vectors.

    So you got an angle of 62 degrees, so the angle between the head and tail version of the vector is 28 degrees. Then, using the cosine law, we get
    [tex] 66^2+66^2-2(66)(66)(cos 118) = R^2[/tex]

    ... well, for the rest, consult your trig book.

    edit- whoops, subtracted from 90, not 180
     
    Last edited: Feb 14, 2005
  15. Feb 13, 2005 #14

    xanthym

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    Referencing the drawing below, the problem statement specifies 4 vectors given here which we are to sum:
    #1) A={P ---> Q}
    #2) B={Q ---> R}
    #3) C={P ---> S}
    #4) D={S ---> R}
    Code (Text):

                            Q         B
                             +---------------> R
                            /               /
                         A /               /
                          /               / D    |A|=|B|=|C|=|D| = 66 meters
                         /               /
                        / 62 deg        /
                     P +----------------> S
                               C
     
    Vector addition is geometrically performed by placing Heads to Tails of the vectors being added. Order does not matter. Thus:
    A + B = {P ---> Q} + {Q ---> R} = {P ---> R} = {Diagonal PR of PQRS}
    C + D = {P ---> S} + {S ---> R} = {P ---> R} = {Diagonal PR of PQRS}

    Thus:
    A + B + C + D = 2{P ---> R} = 2{Diagonal PR of PQRS}

    The length of Diagonal PR of the above Parallelogram is given by application of the Law of Cosines and using the fact that {Angle_P + Angle_Q = 180 deg}:
    {Length of PR} = sqrt{|A|^2 + |B|^2 - 2*|A|*|B|*cos(180 - 62)}
    = sqrt{(66)^2 + (66)^2 - 2*(66)*(66)*cos(118 deg)}
    = (113 meters)

    Using results from above, we multiply by 2 and get:
    Magnitude{A + B + C + D} = 2{113} = 226 meters


    ~~
     
  16. Feb 13, 2005 #15
    you are right, my mistake
     
  17. Feb 13, 2005 #16
    Another Vector Question

    I have a question. Vector A has x and y components of -20cm and 15 cm, respectively; vector B has x and y components of 10.9cm and -20 cm, respectively. If A - B + 3C = 0, what is the x component of C?

    I found out that C = -0.73 cm but thats not what the question is asking.
     
  18. Feb 13, 2005 #17

    xanthym

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    ---------- USE THE OTHER THREAD ----------
     
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