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Physics on a stake in the ground

  1. Mar 3, 2007 #1
    i'm trying to figure out how much force it would take to remove a three inch piton driven into rock or concrete. For example, assuming the concrete in a 50 square foot parking lot was more than three inches deep, and you drove a 3 inch piton into the center of it, how many pounds of force would it take to pull it out? The only factor i havent been able to find is how much force the rock, which you forced apart while driving in the piton, is pressing against the sides of the piton or stake. if anyone can tell me or lead me to anywhere or formula that would be helpful, thanks in advance. I've found that the coefficient of friction of steel on concrete to be .5 (on one surface of the piton) so probably closer to .8 figuring the other sides.... i'm really just looking for a rough answer:tongue2:
     
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  3. Mar 3, 2007 #2

    Meir Achuz

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    Why not just do it experimentally hanging weights on a piton in a vertical position (just not at great altitude).
     
  4. Mar 3, 2007 #3
    If you multiply the force pressing on the sides by the friction coefficient 0.5 then you will get a rough but decent approximation to the force needed to pull out the spike. I know it sounds too simplistic, but not even the area of the surface of the spike will matter.

    To get any better than this approximation you will have to go experimental. Also, you might be able to calibrate your coefficient of static friction.
     
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