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Embarrassingly elementary physics question

  1. Jun 9, 2014 #1
    I know that if a beam is balanced on a central pivot point with equal weights on either end, the beam will be in equilibrium.

    If the pivot point were to to be moved to, say, a quarter of the distance from one end, then obviously more weight would be required on the "short" side of the beam to maintain equilibrium.

    What I am after is the simple formula for determining where on the beam should the pivot point be for any two given differing weights.

    I did say it was embarrassingly elementary ... :-(

    While I am at it, can I throw in a pendulum question as well ...

    Given two pendulums of identical construction (ie length and size of pendulum) , but the weight of the pendulum bob on pendulum A is ten times heavier than the weight of the pendulum bob on pendulum B, will pendulum B reach inertia before pendulum A (assuming that the initial start point is identical for both) or will both come to rest at the same time ?

    FWIW both pendulums are pivoting on identical ball races and the pendulum arm is rigid.
  2. jcsd
  3. Jun 9, 2014 #2


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    First one - you know sum of the arms equals the beam length, plus, you also know moments must be identical. This gives two equations describing the system. Solve, done. Generalize to any system with any number of moments/beams, and you are ready to solve (almost) any mechanical equilibrium problem.
  4. Jun 9, 2014 #3


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    Equilibrium will be attained when the torque from the one side is equal [and opposite] to the torque from the other side. "torque" is a term that roughly means "leverage". It is the product of force times perpendicular distance.

    If the masses are given by m1 and m2 and the horizontal distances from the pivot point to m1 and m2 are r1 and r2, if the beam is massless and if the masses are on opposite sides of the pivot point this means that:

    m1gr1 = m2gr2

    Together with information on the length of the beam (which creates an equation r1 + r2 = length), you have enough information to solve for r1 and r2.

    Even without the length, you have enough information to solve for the ratio between r1 and r2.
  5. Jun 9, 2014 #4
    Thanks jbriggs444 ... as you can tell I am no physicist, not even a student of physics ... I am a practical craftsman and this question is in connection with a kinetic sculpture which I have been commissioned to manufacture.

    So if I have this right, then assuming that the two weights are 10 kilos and 5 kilos, then the equation reads:

    10 x r1 = 5 x r2

    so ... 2 x r1 = r2

    so r1 = r2/2

    In practical terms that would mean that the pivot point for a beam with one weight twice as heavy as the other would be one third of the way along, am I right ?

    I do apologize for polluting the forum with such elementary enquiries , it won't happen again ...
  6. Jun 9, 2014 #5


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    It is a perfectly correct question for this section, nothing to worry about.
  7. Jun 9, 2014 #6
    Thanks Borek, but I know when I am out of my depth.

    I assume from the lack of response to the pendulum question that I have unwittingly transgressed the forum etiquette again ... maybe it's bad form to ask two questions in one post ? ...or maybe the answer is so obvious that no one would deign to reply.

    Like I say, I am not a physicist ... I'll try the same question on one or two of the (practical ) engineering forums I frequent, see what comes up. I could find out the answer experimentally, of course, but chances are somebody has already done this.
  8. Jun 9, 2014 #7


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    When you as more than one question in one thread you always risk getting only one asnwered.

    I would prefer some physicist to chime in. Do they start at the same level? Where is the energy lost? If it is not lost, pendulum won't ever stop. If it is lost, it can be lost either because of the air resistance, or because of the friction at the pivot point, and the answer depends on details of these.

    Assuming perfect pivot and both pendulums starting at the same height, the heavier one would work longer. Their initial speeds are identical, the air resistance is (identical size & speed) identical in both cases, so the one having more energy needs more time to lose it all.

    But I can be missing something.
  9. Jun 9, 2014 #8


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    In regards to question 1. The solution provided has assumed point masses on a mass-less beam. For practical applications, these two assumptions might not be (actually very probably not!) valid. Your masses might be distributed, and your beam might have mass (in which case, you HAVE TO take that into account!). I would NOT just carry out these equations blindly and expect a beam to balance at the end of the day.

    In regards to question 2. There's no way of giving a definitive answer to this question without knowledge of exactly how the energy is lost in each situation. If they both lose energy at an equal rate, then the heavier one should take longer to come to rest as Borek suggested.
  10. Jun 9, 2014 #9
    Again, thanks, Borek.

    I have a workshop in which I can mill, turn, drill, bore and surface grind any material to a level of accuracy which exceeds .0001" , so experimentally I can verify any postulate.

    I am envisaging a pendulum which consists of a 1" x 2" x 24" rectangular block of material with a bearing located at one end, the center of which would be exactly one inch in from the end and from each side ( the hole would be drilled / bored on the 2" face)

    So there are two pendulums, (I eschew the pedantic "pendula" ) in which pendulum A is constructed from a carefully dimensioned block of osmium, and the other is an equally carefully dimensioned block of balsa wood.

    The bearings in both pendulums are identical.

    The initial start is with both pendulums absolutely level, and then released.

    My gut feeling is that the osmium pendulum will oscillate for longer than the balsa wood pendulum .. but this is contrary to what classical physics would have us believe.

    I chose osmium as the total opposite to balsa wood btw ... I am well aware that I will never be able to machine a block of osmium ... so for practical purpose let's think brass or cast iron ... same difference.
  11. Jun 9, 2014 #10


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    Doesn't mean they will work exactly the same way under load.

    I assume by "ablsolutely level" you mean they both start from the same elevation, not from the bottom position :wink:

    Quite the opposite. I explained why I would expect the heavier one to swing for longer, and Matterwave post suggests I am not the only one to expect that.
  12. Jun 9, 2014 #11
    Thanks, Matterwave ... crossposted there ...
  13. Jun 9, 2014 #12
    And again, crossposted with Borek....
  14. Jun 9, 2014 #13
    My ( severely limited) understanding of classical physics is that the mass of the bob has no effect on the period of oscillation.

    My gut feeling (which will be put to the test in the course of this project) is that this is not in fact the case ... but of course the theoretical physics assumes a pendulum in which there is a massless *string* from which is suspended a known mass.

    In the real world, pendulums swing on bearings, and this may well be the crucial point.

    My gut feeling is that the mass of a osmium/brass/cast iron pendulum will overcome the frictional resistance of the bearing more efficiently than a balsa wood pendulum.
  15. Jun 9, 2014 #14
    Borek, starting off *absolutely level* means that the pendulum is released at exactly 90 degrees to the gravitational pull of the Earth ... if they were released at the * bottom position* they would start off *absolutely plumb* (and of course with no motion whatsoever) .. but again, I speak as a craftsman and not as a physicist ...:wink:
  16. Jun 9, 2014 #15


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    Period would be the same, but that is a separate thing from the time it takes to stop.

    Friction is - in a simple model - directly proportional to the force, so the energy lost to friction at the pivot point should be higher for a more heavy bob.

    And I am neither a physicist nor native English speaker, so I initially misunderstood what you wrote (now that you elaborated a little bit it is obvious what you meant).
  17. Jun 9, 2014 #16
    See .. this is where I get lost ... if you imagine the most efficient and most frictionless bearing from which a pendulum could swing, to my mind it would consist of a very accurately ground diamond upper block with a ( female) 90 degree channel machined into it, and then the end of the pendulum would have a ( male) V shaped end with maybe a 30 degree angle . These are totally theoretical ... I do realise that they could never be implemented in the real world.

    So ... what is the consensus ... in the real world, with commercial ball bearings at the pivot point, will the heavier pendulum take longer to reach an inertial state?
  18. Jun 9, 2014 #17


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    The decay time for a pendulum or any oscillator is easily described in terms of how many oscillations it takes for the amplitude to decay to half an original value. That's for an ideal case - where it decays exponentially and never actually gets to zero. This can also be looked at in terms of the fraction of energy lost per cycle. Relating this to your pendulums, there are losses that are proportional to the mass involved (bearing friction etc.) and there are losses due to air resistance etc. that don't depend on the mass.
    The balsa pendulum will not have much energy inherently, so it will decay faster because the air resistance (etc) losses per cycle, are greater as a proportion, compared with your iron pendulum.
  19. Jun 9, 2014 #18
    I just realised that I should have reversed the female /male thing ...

    Thanks for reply, sophiecentaur. .. I will never totally get my head around the physics involved, but at least i am now better equipped to make an experimental stab at the real thing.
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