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srijit

  • one year ago

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?

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  1. MrDoe
    • one year ago
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    \[\tau=rF \sin \theta \] so plug it in, what does that tell you?

  2. MrDoe
    • one year ago
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    r is the displacement vector btw

  3. srijit
    • one year ago
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    |dw:1357340856320:dw| both the displaements appear to be the same

  4. srijit
    • one year ago
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    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

  5. srijit
    • one year ago
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    so why does the body move then again?

  6. srijit
    • one year ago
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    but the moment arm is that what matters right ? that is the perpendicular distance between the force and the axis of rotation..

  7. srijit
    • one year ago
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    but torque is just MOMENT ARM * force right so why would the angle theeta matter..??

  8. MrDoe
    • one year ago
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    http://en.wikipedia.org/wiki/Torque

  9. vf321
    • one year ago
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    Even if you just use the moment arm simplification, keeping track of your signs will still result in a consistent answer.

  10. srijit
    • one year ago
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    @vf321 is this 0 in here?

  11. vf321
    • one year ago
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    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|

  12. vf321
    • one year ago
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    So long as the weights are equal, nothing will move.

  13. srijit
    • one year ago
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    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 ?

  14. vf321
    • one year ago
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    No, the blocks will not move. Why would they?

  15. vf321
    • one year ago
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    (equal mass assumption, of course)

  16. vf321
    • one year ago
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    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

  17. srijit
    • one year ago
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    @vf321 but it does from observation..

  18. srijit
    • one year ago
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    |dw:1357343421063:dw|

  19. vf321
    • one year ago
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    Of course! The reason we're not getting anywhere is b/c the problem is not well-defined.

  20. vf321
    • one year ago
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    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.

  21. rajathsbhat
    • one year ago
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    @vf321 is absolutely right! It has no reason to move. Try it out if you want.

  22. srijit
    • one year ago
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    but how can they remain tilted in there with 2 equal masses on either side ? 2 equal masses always stay on the same level..

  23. rajathsbhat
    • one year ago
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    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.

  24. Vincent-Lyon.Fr
    • one year ago
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    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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