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PhysicsQ/ Adding vectors-equilateral triangle

  1. Sep 1, 2008 #1
    Please help me with this question, I am a complete physics newbie who is lost…
    The Question states:

    The drawing shows an equilateral triangle, each side of which has a length of 2.00 cm. Point charges are fixed at each corner, as shown. The 400 mC charge experiences a net force due to the charge q A and qB. This net force points vertically downward in the drawing and has a magnitude of 405 N. Determine the magnitude and algebraic sighs of the charges qA and qB.

    This is what I did:

    (1) I know that centimeters are supposed to be converted to meters
    (2) Two vectors A & B are added together to produce a third vector C = A + B .
    (3) When the triangle formed is not a right triangle:
    the magnitude of the vector sum can be found by using the Law of Cosines or the Law of Sines



    so now I am lost
  2. jcsd
  3. Sep 1, 2008 #2


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    Homework Helper

    Welcome to PF.

    First off: 1 cm = .01 m

    I can't see your picture. But you are right the basic idea for field values is vector addition.

    The rest of it ... I can see no drawing. I'm not exactly sure what you are doing math wise.
  4. Sep 1, 2008 #3


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    Staff Emeritus
    Science Advisor
    Gold Member

    Welcome to PF,

    Firstly, [itex]2cm = 2\times10^{-2}m[/itex] and not [itex]2\times10^{-6}m[/itex]. Secondly, have you tried splitting the vectors into horizontal and vertical components? Since the net force acts downwards, you know that the horizontal components must sum to zero.
  5. Sep 1, 2008 #4
    thank you
    alright this is where my problem is
    if: A B C
    ----= ----- = ---
    sine(θA) Sine (θB) Sine(θC)

    this is an equliateral triangle meaning:
    2*10^-2 2*10^-2 2*10^-2
    ---------= ----------- -----------
    1 1 1

    there is no answere
    and that is why I am stuck,.
  6. Sep 1, 2008 #5


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    Staff Emeritus
    Science Advisor
    Gold Member

    There is no need to resort to the laws of sines or cosines. As I said previously, try splitting each force vector into two components, one vertical and one horizontal.
  7. Sep 1, 2008 #6
    how would I do that?
    could you please elaborate..
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