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Center of Mass involving Dumbbell With Uneven Weigths

  1. Sep 23, 2012 #1
    1. The problem statement, all variables and given/known data
    You work in a plant that manufactures heavy
    dumbbells. Due to a manufacturing error, one side of a
    100 kg dumbbell was found to be 60 kg (M1) while the
    other was only 40 kg (M2). The mass of the bar itself is
    negligible. The factory has provided you and your
    colleague with special handles that permit one to lift one
    side of the dumbbell directly above the center of mass of
    that side. You and a colleague are asked to hand carry the
    dumbbell to the analysis lab on the other side of the
    compound. You have been assigned the heavy side (M1).
    You believe that if you lift your end of the dumbbell higher
    than your colleague, it will result in a fair, even distribution
    of weight. At what height above the center of mass of the
    opposite end of the dumbbell must you lift the center of
    mass of your end your end for this to be so? At what angle
    with respect to x does this occur?


    2. Relevant equations
    Cmass = [itex]\frac{M1*(L1)+ M2*(L1+L2)+M3*(L1+L2+L3)}{M1+M2+M3}[/itex]
    [itex]\tau[/itex] = F*d

    3. The attempt at a solution

    Setting up an equation where the two torques around the center of mass (which I got as .4 from M1 and .6 from M2) are equal to 0 in order to find the [itex]\theta[/itex] that would result in find the "y" value for how high we'd have to lift M1 in order to make it even.:

    0 = 60(.4)cos(180-[itex]\theta)[/itex] - 40(.6)cos([itex]\theta[/itex])
    0= 60(.4)sin(180-[itex]\theta)[/itex] - 40(.6)

    My friend assisted me with this problem, however, I don't fully understand why he arbitrarily chose to use cosine (I know that a right-triangle is involved, just not HOW) or why the first cosine is 180°-[itex]\theta[/itex].
     
  2. jcsd
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