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Which equilibrium formula

  1. Feb 13, 2014 #1
    What formula (if there is one) would you use for a rod that is in equilibrium where there are n number of forces acting on the rod in either of the two perpedicular directions to the length of the rod, where each force is different, and where each force is acting at a different point along the length of the rod?

    And how would you derive it?
     
  2. jcsd
  3. Feb 13, 2014 #2

    SteamKing

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    It's not clear what you are asking here. Moments can be added together as long as they are about the same axis of rotation. If you can calculate the moment of one force, you can do the same for any number of forces.

    As far as the derivation, look to the definition of moment.
     
  4. Feb 13, 2014 #3

    collinsmark

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    The formula is
    [tex] \sum_n \left( \vec r_n \times \vec F_n \right) = 0 [/tex]
    where each number Fn represents a different, unique force.
    (There is the assumption here that the rod is at equilibrium and remains at rest (no acceleration, either linear or rotational.)

    Pick a location point -- any point -- in space. For each force acting on the rod, also note the location of where that force acts on the rod. Now find the displacement from your chosen location to the location of where that force Fn, acts on the rod. Call this displacement rn Now sum together all the cross products of each force and its corresponding displacement.

    Assuming the rod is in equilibrium (no accelerations), the above holds true no matter what you choose for your original location point. Your original, chosen location doesn't even have to be on the rod. It could be anywhere at all. All that's necessary is that once you choose a location you must stick with it for all displacements. Consistency is important.

    But some points are better choices than others. If you choose, as your chosen location, one of the locations where a force is acting on the rod, you can usually reduce the number of simultaneous equations immediately. While such a choice is not technically necessary, it can make the math easier.

    Simply start with Newton's second law of motion -- except use the "angular" version. Instead of saying, "the sum of all forces equal mass times acceleration," say, "the sum of all torques equal the moment of inertia times the angular acceleration." Now set the angular acceleration to zero since we know the system is in equilibrium (there is no angular acceleration).
     
    Last edited: Feb 13, 2014
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