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Vector problem

  1. Mar 20, 2009 #1
    given 3 forces working on one object with dimentions of 6x2

    F1=4500N (from the front at an anlge of 39)
    F2=9000N (from the front at an angle of -α )
    F3=2250N (grom the back at an angle of 180)

    see diagram below:
    http://picasaweb.google.com/devanlevin/DropBox?authkey=Gv1sRgCL_4l4PpvP_YsQE#5315263399457681474 [Broken]

    what must the angle α be in order for the total force applied to be parallel to the line AB, ?
    -------------------------------------------------------------------

    F=F1+F2+F3 (vectors)

    Fx=4500*cos(39) + 9000*cos(α) - 2250
    =2250(2*cos(39) + 4*cos(α) - 1)

    Fy=4500*sin(39) -9000*sin(α)
    =2250(2*sin(39)-4*sin(α))

    now in order for this vector to be parallel to AB, it must be at the same angle,-->

    2/6=[2250(2*sin(39)-4*sin(α))] / [2250(2*cos(39) + 4*cos(α) - 1)]

    (2*cos(39) + 4*cos(α) - 1) = (6*sin(39)-12*sin(α))
    cos(α)+3*sin(α) = 0.8054

    from here i am stuck, i need to find some trigonometrical identity to help me out, alternatively, another way to solve the problem
     
    Last edited by a moderator: May 4, 2017
  2. jcsd
  3. Mar 20, 2009 #2

    tiny-tim

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    Hi Dell! :smile:

    (i haven't checked how you got there, but …)

    Hint: to solve an equation like this, put it in the form cos(A)cos(α)+sin(A)sin(α) = … :wink:
     
  4. Mar 20, 2009 #3
    i dont understand, could you continue a bit
     
  5. Mar 20, 2009 #4

    tiny-tim

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    well, for example, if it was cos(α)+sin(α) = 0.8054, you could write that as

    cos(α)/√2 + sin(α)/√2 = 0.8054/√2

    ie cos(π/4)cos(α) + sin(π/4)sin(α) = 0.8054/√2

    so cos(α - π/4) = 0.8054/√2 :smile:
     
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