Got Homework?
Connect with other students for help. It's a free community.
Here's the question you clicked on:
 0 viewing
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?
 one year ago
 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?
 one year ago
 one year ago

This Question is Closed

MrDoeBest ResponseYou've already chosen the best response.0
\[\tau=rF \sin \theta \] so plug it in, what does that tell you?
 one year ago

MrDoeBest ResponseYou've already chosen the best response.0
r is the displacement vector btw
 one year ago

srijitBest ResponseYou've already chosen the best response.0
dw:1357340856320:dw both the displaements appear to be the same
 one year ago

srijitBest ResponseYou've already chosen the best response.0
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
 one year ago

srijitBest ResponseYou've already chosen the best response.0
so why does the body move then again?
 one year ago

srijitBest ResponseYou've already chosen the best response.0
but the moment arm is that what matters right ? that is the perpendicular distance between the force and the axis of rotation..
 one year ago

srijitBest ResponseYou've already chosen the best response.0
but torque is just MOMENT ARM * force right so why would the angle theeta matter..??
 one year ago

MrDoeBest ResponseYou've already chosen the best response.0
http://en.wikipedia.org/wiki/Torque
 one year ago

vf321Best ResponseYou've already chosen the best response.1
Even if you just use the moment arm simplification, keeping track of your signs will still result in a consistent answer.
 one year ago

srijitBest ResponseYou've already chosen the best response.0
@vf321 is this 0 in here?
 one year ago

vf321Best ResponseYou've already chosen the best response.1
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
 one year ago

vf321Best ResponseYou've already chosen the best response.1
So long as the weights are equal, nothing will move.
 one year ago

srijitBest ResponseYou've already chosen the best response.0
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 ?
 one year ago

vf321Best ResponseYou've already chosen the best response.1
No, the blocks will not move. Why would they?
 one year ago

vf321Best ResponseYou've already chosen the best response.1
(equal mass assumption, of course)
 one year ago

vf321Best ResponseYou've already chosen the best response.1
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
 one year ago

srijitBest ResponseYou've already chosen the best response.0
@vf321 but it does from observation..
 one year ago

vf321Best ResponseYou've already chosen the best response.1
Of course! The reason we're not getting anywhere is b/c the problem is not welldefined.
 one year ago

vf321Best ResponseYou've already chosen the best response.1
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.
 one year ago

rajathsbhatBest ResponseYou've already chosen the best response.0
@vf321 is absolutely right! It has no reason to move. Try it out if you want.
 one year ago

srijitBest ResponseYou've already chosen the best response.0
but how can they remain tilted in there with 2 equal masses on either side ? 2 equal masses always stay on the same level..
 one year ago

rajathsbhatBest ResponseYou've already chosen the best response.0
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.
 one year ago

VincentLyon.FrBest ResponseYou've already chosen the best response.0
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
 one year ago
See more questions >>>
Your question is ready. Sign up for free to start getting answers.
spraguer
(Moderator)
5
→ View Detailed Profile
is replying to Can someone tell me what button the professor is hitting...
23
 Teamwork 19 Teammate
 Problem Solving 19 Hero
 Engagement 19 Mad Hatter
 You have blocked this person.
 ✔ You're a fan Checking fan status...
Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.