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Vector question - North-east direction

  1. Aug 4, 2016 #1
    1. The problem statement, all variables and given/known data
    A particle undergoes three successive displacements given by ##S_1##=√2 m North- East, ##S_2##=2m due south and ##S_3##=4m 30 degree north of west , then find the magnitude of net displacement?

    2. Relevant equations
    ##S##= ##S_1## + ##S_2## + ##S_3##

    3. The attempt at a solution
    I don't understand how to represent these displacement in the form of vectors.
    I do have solution of this problem. According to the given solution
    ##S_1## = (√2 cos 45)i + (√2sin 45) j
    I can't comprehend how the simple North-east turned into this equation? And I think the question not at all mentions angle 45 . Is there any convention or thumb rule that I am missing while dealing with directions?
     
  2. jcsd
  3. Aug 4, 2016 #2

    A.T.

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    It mentions "North-East".
     
  4. Aug 4, 2016 #3
    Last edited: Aug 4, 2016
  5. Aug 4, 2016 #4
    direction.png
    Is this correct representation of the displacements in question?
     

    Attached Files:

  6. Aug 4, 2016 #5

    cnh1995

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    Yes.
     
  7. Aug 4, 2016 #6
    In solution ##S_3## is given to be (-4cos30)i +(4sin30)j
    I did not understand why is there negative sign in (-4 cos30)i
     
  8. Aug 4, 2016 #7

    cnh1995

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    Sign convention. Second quadrant has -ve x and +ve y. That way, S2 should be -2j.
     
  9. Aug 4, 2016 #8

    Ray Vickson

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    Look at your diagram in post #4. Is the x-component of ##S_3## positive, negative, or zero?
     
  10. Aug 4, 2016 #9

    David Lewis

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    It may be helpful to draw the vectors to scale, and represent coordinate axes with a thin line, and the vectors with a thick line. (That way you can tell them apart.) Also, if you arrange the vectors head to tail, you will be able to get an approximate answer quickly.
     
  11. Aug 5, 2016 #10
    Exactly, but in solution it's given to be +2j.
     
  12. Aug 5, 2016 #11

    cnh1995

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    It should be -2j. There might be a typo in the solution provided by the book.
     
  13. Aug 5, 2016 #12

    Nidum

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  14. Aug 15, 2016 #13
    I encountered one more problem
    Vector A points vertically upward and B points towards north.Vector product AXB is ?
    a)along west
    b)zero
    c)vertically downward
    d)along east
    I've been drawing north as an upward direction .
    NOR.png
    Going with that approach angle between A and B would be zero
     
  15. Aug 15, 2016 #14

    cnh1995

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    The wording implies north is not vertically upwards. Imagine you're standing on a ground. What does 'vertically upward' mean to you in this case?
     
  16. Aug 15, 2016 #15
    Towards the sky?
     
  17. Aug 15, 2016 #16

    cnh1995

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    Yep..
     
  18. Aug 15, 2016 #17
    So here shall I consider north in the usual way and vertically upward as coming out from page?
     
  19. Aug 15, 2016 #18

    cnh1995

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    Yes.
     
  20. Aug 15, 2016 #19
    That would be perpendicular to the plane containing A and B . It's either west or right. Using right hand screw rule I am getting along west as my answer .
     
  21. Aug 15, 2016 #20

    cnh1995

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    Are you sure?
    Edit: I (mis)read the problem as A being towards north. Your answer is right.
     
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