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\[\tau=rF \sin \theta \] so plug it in, what does that tell you?
r is the displacement vector btw
|dw:1357340856320:dw| both the displaements appear to be the same
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
so why does the body move then again?
but the moment arm is that what matters right ? that is the perpendicular distance between the force and the axis of rotation..
but torque is just MOMENT ARM * force right so why would the angle theeta matter..??
Even if you just use the moment arm simplification, keeping track of your signs will still result in a consistent answer.
@vf321 is this 0 in here?
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|
So long as the weights are equal, nothing will move.
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 ?
No, the blocks will not move. Why would they?
(equal mass assumption, of course)
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
@vf321 but it does from observation..
Of course! The reason we're not getting anywhere is b/c the problem is not well-defined.
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.
@vf321 is absolutely right! It has no reason to move. Try it out if you want.
but how can they remain tilted in there with 2 equal masses on either side ? 2 equal masses always stay on the same level..
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.
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|