anonymous
  • anonymous
to test whether a body is in linear or rotational equilibrium or not what do we have to consider? in case of considering torques we have to consider it about what? if there is no particular axis of rotation known ?
MIT 8.01 Physics I Classical Mechanics, Fall 1999
  • Stacey Warren - Expert brainly.com
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chestercat
  • chestercat
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anonymous
  • anonymous
@Vincent-Lyon.Fr
Vincent-Lyon.Fr
  • Vincent-Lyon.Fr
Net force must be zero. Net torque about ANY POINT must be zero too, including centre of mass, instant point of rotation, origin of axes system, and so on and so forth. For instance, imagine a body at rest subject to 3 non parallel forces. What can you say about these 3 forces?
anonymous
  • anonymous
3 forces have an angle 120' between them!

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anonymous
  • anonymous
and passes through the centre of mass!
Vincent-Lyon.Fr
  • Vincent-Lyon.Fr
No, not necessarily. They do not need to have the same magnitude. They do not usually pass through the centre of mass.either.
anonymous
  • anonymous
no if the body is at rest then net force must be 0 ? so resultant force should be 0 ?
Vincent-Lyon.Fr
  • Vincent-Lyon.Fr
Here is an example. |dw:1357928491081:dw| The uniform ladder is resting on the wall. Contact with wall is friction-free. Can you find contact force with ground?
Vincent-Lyon.Fr
  • Vincent-Lyon.Fr
*so that ladder is in equilibrium.
anonymous
  • anonymous
|dw:1357928693270:dw| so here about point in contact with ground the gravity mg passing through centre of mass and the normal force provides torque baalancing each other...so we get the forces..and the normal force from wall is equal to the frictional force right?
anonymous
  • anonymous
but what is even more confusing to me is to have an idea about whether the body stays in equilibrium or not..as in
Vincent-Lyon.Fr
  • Vincent-Lyon.Fr
You must draw only one force in A, not resolving it into N and f.
anonymous
  • anonymous
|dw:1357928965333:dw| here the wall is smooth,the wege and floor is rough. now i am asked to discuss their rotational and linear equillibrim
anonymous
  • anonymous
ok but not resolving into components does give me an edge over resolving it? what is the advantage?
Vincent-Lyon.Fr
  • Vincent-Lyon.Fr
My example holds for the rod. But it will teach you something if you try to draw what I suggested in my example. Will you try?
anonymous
  • anonymous
ok sure..first i've to draw the free body diagram right?
Vincent-Lyon.Fr
  • Vincent-Lyon.Fr
Not necessarilyy. I'm not asking you to solve the problem. Only to draw reaction from ground in a plausible way.
Vincent-Lyon.Fr
  • Vincent-Lyon.Fr
It is related to what I wrote earlier: "Net torque about ANY POINT must be zero too, including centre of mass, instant point of rotation, origin of axes system, and so on and so forth. For instance, imagine a body at rest subject to 3 non parallel forces. What can you say about these 3 forces?"
anonymous
  • anonymous
|dw:1357929460737:dw| ok i am trying but just clarify me about one thing..that is in the free body diagram i drew the uniform rod may have no acceleration of centre of mass..they may happen to be stationary..but there is a net torque of the rod both about the centre of mass as also the point of contact..so it is not in rotational equilibrium?
anonymous
  • anonymous
|dw:1357930307800:dw|
anonymous
  • anonymous
and there's another doubt-if the body is found to have no net torque about a certain point does it necessarily mean that the body is in rotational equilibrium..or there must NOT exist a single point about which there IS a net torque to ensure the fact that it is in rotational equilibrium ?
Vincent-Lyon.Fr
  • Vincent-Lyon.Fr
It is quite complicated; The only thing I can say is that in order to have zero torque about point of action of left force (on the left), then normal force with wall my apply quite high up the side of the rod and be quite intense too. |dw:1357930546168:dw|
Vincent-Lyon.Fr
  • Vincent-Lyon.Fr
Equilibrium is when both net force and net torque are zero. If net force is zero, and net torque is zero about ONE point, then net torque will automatically be zero about any other point. A consequence is that if net torque is found to be non zero about a point, then equilibrium is not possible.
anonymous
  • anonymous
but in case the net force is not 0 about one point..torque would be different about different points then?
Vincent-Lyon.Fr
  • Vincent-Lyon.Fr
Definitely.
anonymous
  • anonymous
then how do we know whether it is in rotational equillibrium or not?
Vincent-Lyon.Fr
  • Vincent-Lyon.Fr
Then rotational equilibrium makes sense only in the frame of the centre of mass. In that case, you will need to work out net torque about centre of mass. Net force non zero AND net torque about CoM zero => so called rotational equilibrium. But "rotational equilibrium" is a pretty useless concept, I think.
anonymous
  • anonymous
ok! and another thing if let us suppose in this case that the body is not in rotational equilibrium then the body rotates about which point? how to get that?
Vincent-Lyon.Fr
  • Vincent-Lyon.Fr
If it's in "translational equilibrium" it will rotate about an instant axis going through its centre of mass.
anonymous
  • anonymous
and if not in translational equilibrium also?
anonymous
  • anonymous
another query- if the body here is in translational equilibrium(suppose)(that should acceleration of centre of mass 0 and hence it is stationary) but not in rotational equilibrium(say) then the rod here rotates around its centre of mass..but the centre of mass here remaining stationery and the rod is rotating about that..wouldn't that be way too awkward? i mean it then will rotate in mid-air!
Vincent-Lyon.Fr
  • Vincent-Lyon.Fr
"and if not in translational equilibrium also?" Well, it will simply be in (accelerated) motion.
Vincent-Lyon.Fr
  • Vincent-Lyon.Fr
" i mean it then will rotate in mid-air!" Not necessarily in mid air, but that's what happens to a pulley around its axis: net force is zero, because at every instant reaction by the axis balances the other forces. Then the motion will be a rotation about the axis. A compass needle will also be in translational equilibrium, but will oscillate in Earth's magnetic field.
anonymous
  • anonymous
thank you! :)

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