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Puzzle with Vectors-almost there!

  1. Apr 22, 2008 #1
    [SOLVED] Puzzle with Vectors--almost there!

    1. The problem statement, all variables and given/known data
    The sum of three vectors is zero. The magnitude of the first vector is twice the magnitude of the second. The first and second vectors are perpendicular. The direction of the third vector is along the negative x-axis. What are the directions of the other two vectors? There are two possible sets of answers. For both sets of answers, give the directions in standard polar form



    2. Relevant equations



    3. The attempt at a solution

    To make two perpendicular vectors with one twice the length of the other:

    Set up a right triangle with side lengths 1 and 2 and therefore hypontenuse=root of 5

    Because the third vector(hypontenuse) is along the negative x direction, the triangle is placed on a coordinate system with the hypotenuse like that. However, so the sum of all three is 0, vector 1 must be projected over the y-axis, and vector 2 over the y and x-axis. So they all cancel out, because vector 3 can be any negative value, root 5 was just used to determine the angles of the other vectors

    Therefore, the angle of the first vector is 180-arcsin(1/root5)=26.6 degrees and the second is 360-(180-90-(180-26.6)=296.6 degrees

    Assuming this is all correct, I still have no clue what the second possible set of answers would be. Any insights?
     

    Attached Files:

  2. jcsd
  3. Apr 22, 2008 #2

    alphysicist

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    Hi mistymoon38,

    You actually have both solutions present on your attachment. The first diagram corresponds to one set of directions, and the second corresponds to the other.

    (Note that in the first diagram, vector 1 is pointing to the right and upwards, and in the second it is pointing to the right and downwards.)

    What do you get for the second set of directions?
     
  4. Apr 22, 2008 #3
    I got 153.4 and 63.4 degrees. But the sin of these (y component) is the same, not opposite of each other so they dont cancel out. Plus, the x components of these angles are negative, so how can they cancel out with the 3rd vectors x component, since it is negative (whatever we want) ?
     
  5. Apr 22, 2008 #4
    What if I moved the vector L=2 to the fourth quadrant and vector L=1 to the first quadrant?
     
  6. Apr 22, 2008 #5

    alphysicist

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    That's what you have in your first diagram; the actual position of the vectors doesn't matter for displacements. You can move them whereever you like to add them together, etc. What matters is the direction. The fourth quadrant is for vectors to the right and downwards, and that is the direction of vector 2 in the lefthand diagram and vector 1 in the righthand diagram.

    You can also see the two options directly by looking back at your first diagram. There you had the two vectors above the third vector. Now just flip the triangle so that it lies directly below the third vector.
     
  7. Apr 23, 2008 #6
    So I flipped the triangl of diagram one and got angles of 206.5 and 63.4. Their y components sums to zero which is good but the only problem is their x components add to a negative number. And since vector 3 is negative, can they ever sum 0?
     
  8. Apr 23, 2008 #7
    OK I finally got it.... 333.4 and 63.4 are the angles, just flip the diagram and put it on the positive side...I was just having trouble visualizing what you were saying and what I myself was trying to think....Thanks!
     
  9. Apr 23, 2008 #8

    alphysicist

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    How did you get 206.5 degrees? (I did get about 63 degrees for the other one.)
     
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