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Tipping a Box

  1. Jan 15, 2009 #1
    1. The problem statement, all variables and given/known data
    A box is being dragged across the floor with a rope that is h distance above the floor. The coefficient of friction is 0.35. The box is 0.50m high and 0.25m wide. Find the force that just tips the box.


    2. Relevant equations
    Ff=uMg=3.43m/s2M
    Fg=Mg=9.80m/s2M
    Well,...that's about it.


    3. The attempt at a solution
    I don't understand the problem enough to draw a force/torque diagram! But I do know that h and M are going to be unknown variables.
     
    Last edited: Jan 15, 2009
  2. jcsd
  3. Jan 15, 2009 #2

    LowlyPion

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    There are two forces of interest.

    The weight acting through the center of mass.
    The force acting at height h.

    So when the force moment, F * h is greater than m*g*width/2 acting down, it should tip about the leading corner right?
     
  4. Jan 16, 2009 #3

    PhanthomJay

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    The problem should identify the mass of the box as M. Then M and h are given variables. F (the applied tipping force) is the unknown variable. I don't know why the problem left out the "M". But note that since the block is sliding and has overcome static friction without tipping, it must be accelerating. It can't be moving at constant speed at the 'tip over' point. The solution requires the use of simultaneous eqations involving sum of torques =0 , including the torque from the pseudo inertial force acting at the c.g (which is also not given) , and sum of forces in x direction, including the pseudo inertial force acting at the c.g , = 0. Poorly worded problem when you have to make your own assumptions.
     
    Last edited: Jan 16, 2009
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