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srijit

  • 2 years ago

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 ?

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  1. srijit
    • 2 years ago
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    @Vincent-Lyon.Fr

  2. Vincent-Lyon.Fr
    • 2 years ago
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    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?

  3. srijit
    • 2 years ago
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    3 forces have an angle 120' between them!

  4. srijit
    • 2 years ago
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    and passes through the centre of mass!

  5. Vincent-Lyon.Fr
    • 2 years ago
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    No, not necessarily. They do not need to have the same magnitude. They do not usually pass through the centre of mass.either.

  6. srijit
    • 2 years ago
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    no if the body is at rest then net force must be 0 ? so resultant force should be 0 ?

  7. Vincent-Lyon.Fr
    • 2 years ago
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    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?

  8. Vincent-Lyon.Fr
    • 2 years ago
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    *so that ladder is in equilibrium.

  9. srijit
    • 2 years ago
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    |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?

  10. srijit
    • 2 years ago
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    but what is even more confusing to me is to have an idea about whether the body stays in equilibrium or not..as in

  11. Vincent-Lyon.Fr
    • 2 years ago
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    You must draw only one force in A, not resolving it into N and f.

  12. srijit
    • 2 years ago
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    |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

  13. srijit
    • 2 years ago
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    ok but not resolving into components does give me an edge over resolving it? what is the advantage?

  14. Vincent-Lyon.Fr
    • 2 years ago
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    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?

  15. srijit
    • 2 years ago
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    ok sure..first i've to draw the free body diagram right?

  16. Vincent-Lyon.Fr
    • 2 years ago
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    Not necessarilyy. I'm not asking you to solve the problem. Only to draw reaction from ground in a plausible way.

  17. Vincent-Lyon.Fr
    • 2 years ago
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    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?"

  18. srijit
    • 2 years ago
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    |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?

  19. srijit
    • 2 years ago
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    |dw:1357930307800:dw|

  20. srijit
    • 2 years ago
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    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 ?

  21. Vincent-Lyon.Fr
    • 2 years ago
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    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|

  22. Vincent-Lyon.Fr
    • 2 years ago
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    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.

  23. srijit
    • 2 years ago
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    but in case the net force is not 0 about one point..torque would be different about different points then?

  24. Vincent-Lyon.Fr
    • 2 years ago
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    Definitely.

  25. srijit
    • 2 years ago
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    then how do we know whether it is in rotational equillibrium or not?

  26. Vincent-Lyon.Fr
    • 2 years ago
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    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.

  27. srijit
    • 2 years ago
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    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?

  28. Vincent-Lyon.Fr
    • 2 years ago
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    If it's in "translational equilibrium" it will rotate about an instant axis going through its centre of mass.

  29. srijit
    • 2 years ago
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    and if not in translational equilibrium also?

  30. srijit
    • 2 years ago
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    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!

  31. Vincent-Lyon.Fr
    • 2 years ago
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    "and if not in translational equilibrium also?" Well, it will simply be in (accelerated) motion.

  32. Vincent-Lyon.Fr
    • 2 years ago
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    " 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.

  33. srijit
    • 2 years ago
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    thank you! :)

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