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Moments and couples

  1. Feb 6, 2012 #1
    Could someone take a look at what I am working on and tell me if I am on the right track?

    #1: Determine the moment of the 200 N force about point B. Please see attached picture for FBD.

    First I find the x and y components.

    x = 100N y= 173.2N, which are perpendicular and parallel to AB.

    Then I find the distance of the 15° angle.

    sin15(.5) = 0.13m
    cos15(.5) = 0.48

    So, now I add the forces to determine the moment about point B.

    Mβ = +(173.2)(.48) - (100)(.13) = 70.1 N.M c.c.w

    It would be of great appreciation is someone could help me.
    Thank you.

    Attached Files:

  2. jcsd
  3. Feb 6, 2012 #2
    2: Says, replace the two 5 KN forces with an equivalent system consisting of the 7 KN force applied at C and a second force applied at point D. Determine the magnitude and direction of the second force and the distance CD.

    Please see the picture for the FBD.

    How I did this.

    First I found the moment about point O for the two 5 KN forces.

    Mo = -(5KN)(3m) + (5KN)(1.5m) = -7.5 KN.M c.w

    Then I found the x and y components for the 7 KN force.

    7(cos30) = 6.06 KN 7(sin30) = 3.5KN

    Next I placed the 6.06 KN and 3.5KN force at point D and solve the moment about point C,( Cancelling those forces) and rearranging for the distance CD.

    Mc = (3.5)(0) - (6.06)(d) = -7.5 KN.M

    d = -7.5 - (3.5)(0) / -6.06 = 1.24 m

    So, Magnitude = 6.06KN c.w. and CD = 1.24m

    Attached Files:

    • 7kn.png
      File size:
      8.2 KB
    Last edited: Feb 6, 2012
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