anonymous
  • anonymous
when a body is weighed on an ordinary balance we demand that the arm should be horizontal if the weights on two pans are equal.suppose 2 equal weights are placed on either side, the arm is kept at an angle with the horizontal and released. is the torque of the 2 weights about the middle point(point of support) 0 ? is the net torque 0 ? if so why does the arm rotate and finally become horizontal?
Physics
chestercat
  • chestercat
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MrDoe
  • MrDoe
\[\tau=rF \sin \theta \] so plug it in, what does that tell you?
MrDoe
  • MrDoe
r is the displacement vector btw
anonymous
  • anonymous
|dw:1357340856320:dw| both the displaements appear to be the same

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anonymous
  • anonymous
as also they have the same force Mg acting on each of them one in clokwise and another in anticlockwise..so they cancel out and the net resultant torque is 0 about the centre
anonymous
  • anonymous
so why does the body move then again?
anonymous
  • anonymous
but the moment arm is that what matters right ? that is the perpendicular distance between the force and the axis of rotation..
anonymous
  • anonymous
but torque is just MOMENT ARM * force right so why would the angle theeta matter..??
MrDoe
  • MrDoe
http://en.wikipedia.org/wiki/Torque
anonymous
  • anonymous
Even if you just use the moment arm simplification, keeping track of your signs will still result in a consistent answer.
anonymous
  • anonymous
@vf321 is this 0 in here?
anonymous
  • anonymous
Yes, it is. In fact, if the weights are equal, the balance is ideal, etc., then the system will be static, just like an Atwood's machine where one of the weights is raised:|dw:1357342611368:dw|
anonymous
  • anonymous
So long as the weights are equal, nothing will move.
anonymous
  • anonymous
but in case of an atwood machine if one block is raised a bit above another and released then at that instant net torque is 0 still the block raised at a height comes down until both the blocks are at the same height right? why does it happen ?
anonymous
  • anonymous
No, the blocks will not move. Why would they?
anonymous
  • anonymous
(equal mass assumption, of course)
anonymous
  • anonymous
Yes, if you were to plug in the values for the two torques, they would equal zero. If the system was not in motion initially, it cannot suddenly start moving w/o violation of energy conservation
anonymous
  • anonymous
@vf321 but it does from observation..
anonymous
  • anonymous
|dw:1357343421063:dw|
anonymous
  • anonymous
Of course! The reason we're not getting anywhere is b/c the problem is not well-defined.
anonymous
  • anonymous
It depends on where your center of mass is. If your COM is perfectly in the middle of your axis of rotation, then what I described will happen.
anonymous
  • anonymous
@vf321 is absolutely right! It has no reason to move. Try it out if you want.
anonymous
  • anonymous
but how can they remain tilted in there with 2 equal masses on either side ? 2 equal masses always stay on the same level..
anonymous
  • anonymous
that's the beauty of it. No torque=no movement unless it's already moving. You should really experiment with this the next time you get your hands on a balance.
Vincent-Lyon.Fr
  • Vincent-Lyon.Fr
Probably because of the weight of the arm itself. Its centre of mass G is always below the fulcrum A, either by position of the fulcum such as: |dw:1357422838501:dw|. or by shape of the arm, such as: |dw:1357422965061:dw|

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