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Unbalanced beam / Torque (?)

  1. Mar 22, 2007 #1
    1. The problem statement, all variables and given/known data
    A beam of mass 25kg is suspended from a point 1.3m in from one side. Where must a 55kg mass be placed for the beam to be balanced?

    2. Relevant equations
    [tex]\tau = rF_ \bot [/tex]

    3. The attempt at a solution
    This problem has me totally confused. I no that the center of mass of the beam is at 1.35m (assuming that it is uniform). Since the beam is suspended off the center of mass the 55kg mass needs to placed predominantly on the 1.3m section of the beam so that the center of mass of the combined beam and mass is at 1.3m mark. However I am not sure on how to calculate this, I am assuming that a torque (?) calculation maybe required, however I'm not sure on how to apply.

    Many thanks for any suggestions and help

  2. jcsd
  3. Mar 22, 2007 #2
    I assume that 1.3m is the middle of the beam. I also assume that that is where the beam is suspended. One side is not suspended while the other is? If you put on the suspended side the beam falls while if you put it on the unsuspended side then the beam will remain still but not balanced. I believe that if the weight is put with the mass of the beam (in the middle) then it will still be balanced.
  4. Mar 22, 2007 #3


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    I think you missed out the part that the length of the beam is 2.7m. Unless youre doing a different question to the one im thinking.
  5. Mar 22, 2007 #4
    Leftward torque (or some call it moment) = Rightward moment

    for the beam to be balanced.

    To find torque, F = perpendicular distance * weight

    Assuming the beam is uniform,
    You know the torque on one side (the side that the weight of the beam is acting on), so use it to find the perpendicular distance of the weight on the other side
    Last edited: Mar 22, 2007
  6. Mar 24, 2007 #5
    thanks for the replies

    sorry i forgot to mention that the beam is 2.7 m, so it is not suspended from the center

    saplingg I'm an not too sure what you mean, sorry
  7. Mar 24, 2007 #6


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    Homework Helper

    Take where the beam is suspended as your pivot point. For the beam to balance, the sum of the torque around that point must be zero (this is what saplingg was getting at). There are two forces here, each providing torque. Use the equation you had in your original post.
  8. Mar 25, 2007 #7
    okay thanks for that hage567
    here is my working as i understand the problem:

    \sum \tau _{clockwise} = \sum \tau _{anticlockwise} \\
    \sum \tau = 0 \\
    \tau = rF_ \bot \\
    \tau _{righthandside} = 1.4\left( {\frac{{25}}{{2.7}} \times 1.4} \right)\left( {9.8} \right) = 177.8518\,N\,m \\
    \tau _{lefthandside} = 1.3\left( {\frac{{25}}{{2.7}} \times 1.3} \right)\left( {9.8} \right) = 153.3518 \\
    \sum \tau = \tau _{righthandside} - \tau _{lefthandside} - x \\
    x = \tau _{righthandside} - \tau _{lefthandside} = 24.5 \\
    \tau = rF_ \bot \\
    r = \frac{\tau }{{F_ \bot }} \\
    F_ \bot = 50\left( {9.8} \right) = \\
    r = \frac{\tau }{{F_ \bot }} = \frac{{24.5}}{{490}} = 0.05\,m \\

    from the above calculations the mass needs to be placed 0.05 m to the left hand side of the pivot point. is this correct?
  9. Mar 25, 2007 #8


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    Homework Helper

    I don't understand the (25/2.7 x 1.4) or the (25/2.7 x 1.3). Are you trying to take into account the mass of the beam at these points? I don't think you have to do that, if it is considered to be uniform.

    Think of it this way: you want to solve for r_m let's say. r_m is the distance from the pivot point to the 55 kg mass. Specify the centre of mass of the beam from the pivot point. That way, you can just set up (r_m)(F_m)=(r_b)(F_b). I think this is what you're basically doing, but you're making it too complicated.
  10. Mar 26, 2007 #9
    okay thanks hage567 ill look into it and see if i understand it
    thank you
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