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Rotational and Translational Equilibrium Help Needed

  1. Jan 6, 2013 #1
    1. The problem statement, all variables and given/known data

    I am making a hanging mobile which needs to be done mathematically by calculating torque. The problem is, I can't seem to figure out how to solve for the distance.
    You see, all of the problems we did in class talked about finding mass, but to do this project, I already know the mass, I just need to know the distance between the objects that are hanging.

    Note: 0.0284 kg is the weight of the rod that is holding up the objects and the string holding the rod up is in the center of the rod, which is why the force up is the same distance away from the pivot point as the rod.

    2. Relevant equations

    Here is what my teacher gave me to work on this:

    1. Forceup = Forcedown
    2. τclockwise = τcounter clockwise
    3. Use an object as the pivot point if there is more than one object to solve for.

    3. The attempt at a solution

    Here is what I filled in those steps:

    1. Forceup = (2.3814 kg + .0452 kg + .0878 kg + 0.0284 kg)9.8 m/s2 = 3.96312 N.
    2. 2.3814 kg(9.8 m/s2)(0 m.) + 0.0284 kg(9.8 m/s2)(.125 m.) + 0.0878 kg(9.8 m/s2)(x m.) + 0.0452 kg(9.8 m/s2)(y m.) = 3.96312 N.(.125 m.)
    which is, if my math is correct:
    0.03479 Nm. + 0.86044 N (x m.) + 0.44296 N (y m.) = 0.49539 Nm.

    So my problem lies here: what do I do with the two different variables? I don't have a second equation nor do I know of a second equation that I can use to calculate the distance needed between them. What can I do to make this work?

    EDIT: Here is what the structure looks like: a string is holding up a rod from the middle of the rod so that the rod will be level when held from the string. There are three objects hanging from that rod, and the pivot point is the heaviest of the hanging objects.
    Last edited: Jan 6, 2013
  2. jcsd
  3. Jan 6, 2013 #2


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    Please describe the structure. It's hard to figure out from the numbers.
  4. Jan 6, 2013 #3
    Ok, just updated my question. Let me know if you need any more information
  5. Jan 6, 2013 #4


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    Are you putting one object at each end, and just want to know where to place the third? If so, don't put the heaviest at one end if it weighs more than the other two put together.
    The point where the string is attached is necessarily the 'pivot point', but I guess you mean the point you would take moments about. I don't see that it matters which point you take moments about. The point of string attachment looks the simplest, since then you don't care about the weight of the rod (which you don't, since it is attached in its middle, so it balances itself).
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