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Determining mass distribution for team-lifting of arbitrary objects

  1. May 28, 2009 #1
    Hey all,

    New to the forum (obviously, since this is my first post).

    I've been scouring the internet trying to find some guidance on a problem I've been having, but have had no luck. Let me just say first off that my physics and math backgrounds are fairly elementary. I have minors in both areas, but it's been years since I've done anything in them.

    The problem I need to solve is the following: Let's say I have an arbitrary polygon. This polygon represents the top-down view of an object. We can assume also that it has no hidden surfaces (all faces of the object are either purely horizontal or purely vertical). I know the location of the vertices (in order), as well as the location of the centre of mass (calculated by triangulating the polygon, averaging the centres of the triangles weighted by their size).

    I have X number of people lifting this object, at X different points on its boundary. These points are also known. What I need to figure out is exactly what portion of the weight of the object each person is lifting, based on their positions, assuming a uniform density of the object.

    I feel as though, intuitively, this should be a relatively simple thing to calculate, but have had no luck in figuring it out or finding a solution....Could really use some guidance here...

    Edit: apologies if this is posted in the wrong place. Mods, feel free to move it if necessary (but it's NOT a homework question)
     
  2. jcsd
  3. May 28, 2009 #2

    tiny-tim

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    Welcome to PF!

    Hi Mikey-D! Welcome to PF! :smile:

    I don't think there is a solution, if more than 3 people are lifting.

    If two people lift a (one-dimensional) bar, you can find the forces, but if three people do, there are infinitely many solutions

    (just think … if you're the one in the middle, you could let go completely, and the other two would still support the bar :biggrin:)

    Similarly, with a polygon (instead of a bar), you can find the forces for 3 people, but for 4 or more, there are infinitely many solutions.
     
  4. May 28, 2009 #3
    Interesting...that makes sense. So any more than three people in two dimensions is (arguably) redundant. As long as the triangle formed by those three lifters constrain the CoM, any other lifters are just helping to lighten the load for anyone else on that side of the CoM. Make sense?

    ok. So after a little thought, I've come up with the following: obviously, the sum of vertical forces (down from the CoM and up from the three lifting points) must equal zero. Additionally, I should be able to look at the torques applied about any line passing through the CoM from the lifting points, and the torques on one side of the line should be equal to the torques on the other side of the line. By choosing lines passing through the CoM and any two of the lifting points, I should be able to obtain three different equations and solve for the forces applied at each lifting point.

    Make sense?

    From there, any additional lifters contributing will simply remove a proportional amount of necessary lifting force from lifters on the same side of the CoM as them. How much weight they can remove from others, however, is dependent on the new geometry they form with the original set of lifters. Any triangle which would exclude the CoM cannot bear the full weight, as it would cause the object to be dropped.
     
  5. May 29, 2009 #4

    tiny-tim

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    Yes … there are only three equations, and so they can't uniquely determine more than three unknowns! :biggrin:
    Not proportional … it can be anything.
    Sorry, I'm not following this geometrical argument … wherever he is, any force (up or down) by a new lifter can always be compensated for by the others.
     
  6. Jun 1, 2009 #5
    Sorry, but I appear to be a bit dense today...

    What are the three equations?
    1. Sum of vertical forces = 0
    2. Sum of torques about an arbitrary axis through the CoM = 0
    3. ? (I feel like I'm missing something really obvious here...)

    edit: seems my math was off. I can just pick a new arbitrary axis and re-evaluate torques about it, right?
     
  7. Jun 3, 2009 #6

    tiny-tim

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    Hi Mikey-D! :smile:

    (sorry to take so long :redface:)
    Replace #2 by Sum of torques about an arbitrary point

    torque is a vector (which in this case is always horizontal),

    so that gives you two equations! :smile:
    No, picking a new point (or a new parallel axis) will give you no new information …

    ie no information that you can't get more simply by using (linear) components of force. :wink:
     
  8. Jun 3, 2009 #7
    Sorry, Tim, but I'm missing something here...

    What are my two other equations then (other than sum of force = 0)?
     
  9. Jun 3, 2009 #8

    tiny-tim

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    r x F = 0

    since all the Fs are vertical and all the rs are horizontal, the r x Fs are horizontal, so that vector equation has two components … ie two equations. :smile:
     
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