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?

Question

mrdoe · Answer

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

mrdoe · Answer

r is the displacement vector btw

anonymous · Answer

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

anonymous · Answer

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 · Answer

so why does the body move then again?

anonymous · Answer

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

anonymous · Answer

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

mrdoe · Answer

http://en.wikipedia.org/wiki/Torque

anonymous · Answer

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

anonymous · Answer

@vf321 is this 0 in here?

anonymous · Answer

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|