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Moment Calculations for a Machine

  1. Oct 28, 2008 #1
    Hi, this may seem like a silly question to ask, but it has to do with +/- signing when taking the moment about a point. I have images attached to this post to make things a little clearer.

    1. The problem statement, all variables and given/known data
    The compound-lever pruning shears shown can be adjusted by placing pin A at various ratchet positions on blade ACE. Knowing that 1.5-kN vertical forces are required to complete the pruning of a small branch, determine the magnitude P of the forces that must be applied to the handles when the shears are adjusted as shown.

    2. Relevant equations
    (Moment about C): [tex]\sum[/tex]MC = 0

    (Moment about D):[tex]\sum[/tex]MD = 0

    3. The attempt at a solution
    Using the FBD for the Blade ACE, I took the moment about C to find FAx and FAy, the horizontal and vertical components of FA, respectively:

    [tex]\sum[/tex]MC = 0 = (32 mm)x(1.5 kN) + (-10 mm)x(FAx) + (-28 mm)x(FAy)
    0 = 48 kN.mm - 10FAx - 28FAy

    [tex]\rightarrow[/tex] The dimensions provided (11 mm vertically and 13 mm horizontally) allow me to solve for the length of member AB through the Pythagorean theorem.

    c= [tex]\sqrt{a^2 + b^2}[/tex]
    c=[tex]\sqrt{(13)^2 + (11)^2}[/tex]
    c= 17.03 mm

    Therefore, instead of using FA sin[tex]\theta[/tex] and FA cos[tex]\theta[/tex] for FAy and FAx, respectively, I can just use their geometric equivalents:

    0 = 48 - 10([tex]\frac{13}{17.03}[/tex])FA - 28([tex]\frac{11}{17.03}[/tex])FA
    0 = 48 - 7.63FA - 18.09FA
    25.7FA = 48
    FA = 1.86 kN
    [tex]\rightarrow[/tex]Therefore, FAx = [tex]\frac{13}{17.03}[/tex](1.87 kN) = 1.43 kN
    [tex]\rightarrow[/tex]And FAy = [tex]\frac{11}{17.03}[/tex](1.87 kN) = 1.21 kN

    Now, based off the free body diagram for Handle ABD, in which member AB is a two-force member. With that in mind, I took the moment about D

    [tex]\sum[/tex]MD = 0 = (-70 mm)x(P) + (5 mm)x(1.43 kN) + (-15 mm)x(-1.21 kN)
    0 = -70P + 7.15 + 18.15
    70P = 25.3
    P = 0.361 kN

    However, when I checked my answer on the instructor's site, I saw that P actually equalled 0.157 kN, and that the Moment about D was calculated like this:

    [tex]\sum[/tex]MD = 0 = (-70 mm)x(P) - (5 mm)x(1.43 kN) + (15 mm)x(1.21 kN)
    0 = -70P - 7.15 + 18.15
    70P = 11
    P = 0.157 kN

    What I don't understand about this moment calculation is why we SUBTRACT the cross product of (5 mm)x(1.43 kN). Point B is located 5 mm above Point D, and up is positive in my frame of reference, and that the 1.43 kN force is moving toward the right, which is also positive in my my frame of reference. What reason is there to have a subtraction sign for that cross product, then?

    Sorry that my post is really long, but I really appreciate the time that anyone takes to read, think about, and answer this question that I have.

    Attached Files:

    Last edited: Oct 28, 2008
  2. jcsd
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