## anonymous 3 years 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?

1. MrDoe

$\tau=rF \sin \theta$ so plug it in, what does that tell you?

2. MrDoe

r is the displacement vector btw

3. anonymous

|dw:1357340856320:dw| both the displaements appear to be the same

4. 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

5. anonymous

so why does the body move then again?

6. anonymous

but the moment arm is that what matters right ? that is the perpendicular distance between the force and the axis of rotation..

7. anonymous

but torque is just MOMENT ARM * force right so why would the angle theeta matter..??

8. MrDoe
9. anonymous

Even if you just use the moment arm simplification, keeping track of your signs will still result in a consistent answer.

10. anonymous

@vf321 is this 0 in here?

11. 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|

12. anonymous

So long as the weights are equal, nothing will move.

13. 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 ?

14. anonymous

No, the blocks will not move. Why would they?

15. anonymous

(equal mass assumption, of course)

16. 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

17. anonymous

@vf321 but it does from observation..

18. anonymous

|dw:1357343421063:dw|

19. anonymous

Of course! The reason we're not getting anywhere is b/c the problem is not well-defined.

20. 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.

21. anonymous

@vf321 is absolutely right! It has no reason to move. Try it out if you want.

22. 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..

23. 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.

24. anonymous

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|