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Rotational Equilibrium Problem

  1. Jan 14, 2014 #1
    Mass 1 is located at the far left end of a 90 cm beam. Mass 2 is located at the center of the beam, and mass 3 is located 30 cm from the center, on the right side. Mass 3 and mass 1 are the same. If the fulcrum is located 10 cm to the left of the center of the beam, what is the mass of mass 2 if the beam does not rotate?

    My attempt: 0 = (m)(g)(35) + (m)(g)(20) - (x)(g)(10)
    m corresponds to the two equal masses, and x corresponds to the mass we are looking for.

    The answer is that the beam must rotate. In other words, no matter what the mass is for Mass 2, the beam will always rotate.

    I just cannot see how this works. My attempt at an answer gave me that x is 5.5 times greater than that of m.
     
  2. jcsd
  3. Jan 14, 2014 #2

    collinsmark

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    Gold Member

    [red emphasis mine.]

    Yes, that's correct. No matter how much mass 2 is, an equilibrium cannot be reached.

    Take another look at the term I marked in red above.

    Is the term positive or negative? Where did the "20" come from (are you sure it shouldn't be some other number)?

    (It should help to draw out a diagram of the beam and masses.)
     
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