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Minimum force required to turn over box.

  1. Jan 1, 2014 #1
    Hi,
    1. The problem statement, all variables and given/known data
    I would like to determine the position, magnitude and direction of the minimum force F required to turn over the box shown in the attachment (width - B, height - H). The (uniform) box is placed on a horizontal plane. The coefficient of friction between the box and the plane is μ.


    2. Relevant equations



    3. The attempt at a solution
    Since τ=r x F, I believe the minimum force would have to be at either D or C (so that r is maximum). I picked C and placed my Fx and Fy there, where Fx=Fcosθ and Fy=Fsinθ and θ is the angle between F and the positive x axis. I was then trying to write the equations for the torques wrt to A. I figured that when the box is turned over, N equals zero and therefore there is no friction. I was then left with two effective forces, namely: F and mg.
    The equation I wrote down was:
    BFsinθ - FHcosθ-mgB/2=0
    Is that correct?
     

    Attached Files:

  2. jcsd
  3. Jan 2, 2014 #2

    tiny-tim

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    hi peripatein! happy new year! :smile:
    yes :smile:
    (how can N = 0 ?? :confused:)

    no, the criterion for any object to start to tip is that N goes through the edge of its base (in this case, A or B)

    do ∑moments = 0 to find when that happens :wink:

    (technically, this assumes that the box is not sliding

    if it does slide, you will also need to consider the acceleration)​
     
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