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Angle between coupled forces

  1. Sep 30, 2015 #1
    1. The problem statement, all variables and given/known data
    The moment of the couple is 600k (N-m). What is the angle A?

    F = 100N located at (5,0)m and pointed in the positive x and positive y direction
    -F = 100N located at (0,4)m and pointed in the negative x and negative y direction

    2. Relevant equations
    M = rxF
    M = D


    3. The attempt at a solution

    M = 600k
    |F| = 100N
    r = <5,-4>

    rxF = a 3x3 determinant with
    <i , j , k>
    <5, -4, 0>
    <Fx , Fy, 0>
    =
    5Fy + 4Fx +0 = 600k
    but i've only got 1 equation and 2 unknowns now!
     
  2. jcsd
  3. Sep 30, 2015 #2

    haruspex

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    I'm guessing that your Fx and Fy are the x and y components of one of the two given forces. Your other equation, then, is the given magnitude of that force.
     
  4. Sep 30, 2015 #3
    I tried that.
    5Fy + 4Fx = 600
    25Fy^2 + 16Fx^2 = 10,000 <- (Magnitude formula squared)
    Simplifying:

    Fx = 150 - 5/4Fy
    25Fy^2 + 16(150- 4/5Fy)^2 = 10,000
    This yields a quadratic whose discriminant is
    6000^2 - 4*50*350000 < 0
     
  5. Sep 30, 2015 #4

    haruspex

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    why the 25 and the 16? If F has orthogonal components Fx and Fy, what is the magnitude of F?
     
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