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Toppling of a box

  1. Apr 18, 2014 #1

    adjacent

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    I don't know if "toppling" is even a word.:uhh:

    This is not a homework,just a question.
    https://www.physicsforums.com/attachment.php?attachmentid=68762&stc=1&d=1397811697
    If I apply a force at A,the box will topple and fall.(Probably because of friction)
    If I apply a force at B,the box will move(Probably because the force overcame the friction)

    What's causing it? Torque?
    As the distance increase,the effect of force on friction decreases but why?

    P.S I am a Grade 10 student.
     
  2. jcsd
  3. Apr 18, 2014 #2

    SteamKing

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    'toppling' is the present participle of the verb 'to topple', just like 'running' is the present participle of the verb 'to run'.

    What can you say about the location of A or B w.r.t. the center of gravity of the box?
     
  4. Apr 18, 2014 #3
    There is a problem with the attachment, I cannot see the picture. Generally, it is safe to say that "toppling" involves torque.
     
  5. Apr 18, 2014 #4

    adjacent

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    attachment.php?attachmentid=68767&stc=1&d=1397814593.png
    Here is the image.

    I don't think this problem involves Center of Gravity.When I push B,the box does not topple at all.
     

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  6. Apr 18, 2014 #5
    If the force at B is great enough, the box will topple. Counter-clockwise.
     
  7. Apr 18, 2014 #6

    Radarithm

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    When you apply the torque at the bottom, there is friction to oppose it (I think).
    When you apply it to the top (A), there is no friction to oppose it except air resistance.
    Are the forces equal?

    EDIT: the reason the friction decreases as the distance increases is because of kinetic friction being less effective than static friction.
     
  8. Apr 18, 2014 #7

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    No.We are applying the same force on both A and B. But it only topples when we apply the force on A.
    It only moves horizontally to right when we apply a force on B.
    So there should be some relationship between the distance the force is applied and whether it should topple or not.
     
  9. Apr 18, 2014 #8

    adjacent

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    I don't think so.There is still the friction at the bottom.If there is not friction,it would only move without rotating(It will rotate about the center of gravity.But here it is rotating about a point at the bottom)
    Yes.
     
  10. Apr 18, 2014 #9
    This really depends on the magnitude of the force. A force great enough, applied either at A or B, will have the box topple, in different directions.

    The minimal forces required for that at A or B may be different, however. Radarithm explained one reason for that.
     
  11. Apr 18, 2014 #10

    adjacent

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    What do you mean by "less effective"?
     
  12. Apr 18, 2014 #11

    Radarithm

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    [itex]\mu_s N\ > \mu_k N[/itex]

    Where mu sub s is the coefficient of static friction, the other one is kinetic friction.
     
  13. Apr 18, 2014 #12

    Radarithm

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  14. Apr 18, 2014 #13
    For what I know it is torque, initially the box is at rest , when the box topples it is because you disturbed its equilibrium and there is a net torque due to its weight.

    In both cases,
    You apply a force F at the point A or B,
    Consider its bottom right point as the axis of rotation.
    It is clearly visible that the moment arm of A is greater than B.
    Note : The box toppled because the angular inertia of this box is less and it was easier to overcome the torque, otherwise its angular inertia will be more and the linear equilibrium would have been easier to disturb.
    Note 2 : As someone stated earlier if you apply a big enough force at B , the box would topple the other way , this is also because it was easier to turn it with the center of box as axis , also note that it is totally dependent on the static friction,
    Since while toppling the axis of rotation is stationary w.r.t the floor.
     
  15. Apr 18, 2014 #14

    adjacent

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    uhh. What's angular inertia and linear equilibrium? :eek:
     
  16. Apr 18, 2014 #15

    SteamKing

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    This is where you start studying physics to find the answers.
     
  17. Apr 18, 2014 #16

    adjacent

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    As I said before,I'm just a Gr.10 student.
    OK,I'll try to solve this after A-levels and see if I can really solve it. :smile:
     
  18. Apr 18, 2014 #17

    Radarithm

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    Just like how a Torque is the angular counterpart of a Force, Angular inertia is the angular counterpart of linear inertia, which is an object's "resistance" to forces.
     
  19. Apr 18, 2014 #18

    Just like in linear (straight , no rotation) motion , we have mass/inertia defined as the force required to produce a unit acceleration ,
    By Newton's law,
    m = F/a,
    So the analogous angular variables,
    Angular Inertia (I) = Torque ([itex]\tau[/itex]) / Angular acceleration([itex]\alpha[/itex])

    When I meant linear equilibrium , I meant the one involving linear variables , while the other type of equilibrium is angular.

    Hope that clear things up.
     
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