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Confusion regarding fundamental classical mechanics question!

  1. Sep 5, 2011 #1
    A block is resting on a frictionless surface as shown in the figure attached with this post. Calculate the minimum force F required so that the block will topple? The dimensions of the block, free body diagram and other details are there in the picture attached.


    Now, since the surface is frictionless, in the horizontal direction F=ma, where m=mass of the block and, a=the acceleration in the horizontal direction.
    In the vertical direction, N=mg, where N is normal reaction force by the surface. The third equation is the moment equation.


    The problem is very easy and fundamental, but I am confused as to whether I should balance the torques about the centre of mass or about the bottom right corner point? The difference between this and other toppling problem is that the block will accelerate as soon as the force is applied because the surface is frictionless, so it will kind of slide and topple instead of toppling at a fixed position. Please help me out with this!
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     

    Attached Files:

  2. jcsd
  3. Sep 5, 2011 #2

    tiny-tim

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    welcome to pf!

    hi rohitgupta! welcome to pf! :wink:
    the https://www.physicsforums.com/library.php?do=view_item&itemid=313" formula τ = Iα) works only when both L and ω are either about the centre of mass or about the centre of rotation

    in this case (unlike most cases), the point of contact is not the centre of rotation :redface:

    so use the centre of mass! :smile:
     
    Last edited by a moderator: Apr 26, 2017
  4. Sep 5, 2011 #3
    Hey thank you so much tiny tim! Yes it did occur to me that the moment equations need to be applied about the centre of mass, but I just can't find a strong explanation as to why the corner point in this case is not the centre of rotation because if I apply the moment equation about the centre of mass, then I am considering that the normal reaction acts at the corner point, which means that the body has rotated just a little bit about that point, isn't it? So can you please help me with this explanation!


    P.S: The moderator/ administrator please modify my first post, the three different paragraphs are the part of the three headings successively. I made a mistake by deleting them while posting, sorry about that!
     
  5. Sep 5, 2011 #4

    tiny-tim

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    hey rohitgupta! :smile:
    because the centre of rotation must be stationary, and the whole block will move (because the surface is frictionless) as soon as it's pushed, so there's no way the corner will be stationary :wink:

    (btw, the reason it's important is that the centre of rotation is the only point for which v = rω, which turns out to be an essential condition in proving L = Iω)

    It doesn't matter … the "template" is only there to make sure that everyone provides the right information … but you needn't provide it in the right place! :biggrin:
     
  6. Sep 6, 2011 #5
    Please don't get frustrated by my repeated questions, isn't the centre of mass also accelerating?
     
  7. Sep 6, 2011 #6

    tiny-tim

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    doesn't matter …

    you don't need v = rω to get L = Iω if you're using the centre of mass :smile:

    (you'd better study the derivation of L = Iω :wink:)
     
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