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Moment of a force about a point

  1. Jan 28, 2009 #1
    1. The problem statement, all variables and given/known data
    This one's easy. A 2.9 lb force (P) is applied to a lever. Determine the value of alpha knowing that the moment of P about A is counterclockwise (+) and has a magnitude of 17 lb-in.


    2. Relevant equations

    Moment about a = (Force of x-component)(distance from a to Force point) + (F_y-comp)(dist. to foce)

    M_a = (F_x)(d_y) + (F_y)(d_x)

    Answer: alpha = 49.9 degrees or 59.4 degrees

    3. The attempt at a solution

    17 = (2.9sin(alpha))(4.8) + (2.9cos(alpha))(3.4)

    how do I solve for alpha?
     

    Attached Files:

  2. jcsd
  3. Jan 28, 2009 #2

    Doc Al

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    Staff: Mentor

    Use a trig identity to express sinα in terms of cosα (or vice versa). Then you can solve for cosα.
     
  4. Jan 28, 2009 #3
    can I use sin A = cos(pi/2 - A)?
     
  5. Jan 28, 2009 #4

    Doc Al

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    Staff: Mentor

    Use sin2α + cos2α = 1. (Rearrange it.)
     
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