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Pile Driver, 10000kg, 10m/s, strikes pile w/ what force?

  1. Sep 17, 2015 #1
    1. The problem statement, all variables and given/known data

    A pile driver with mass 10000kg strikes a pile with velocity 10m/s.

    a) What is the kinetic energy of the driver as it strikes the pile?
    b) If the pile is driven 20cm into the ground, what force is applied to the pile by the driver as it strikes the pile? Assume that all the kinetic energy of the driver is converted to work?


    2. Relevant equations

    1/2mv^2 = Fs

    3. The attempt at a solution


    a) .5(10,000)10^2 = 500,000 J

    b) ? Question: How can you know the Force simply by knowing how far the pile is driven into the ground? Doesn't it also depend on the material of the ground?

    If the ground is soft mud, it may travel 2 meters into the ground. If the ground is hard ice, it might travel .10 meters into the ground (just hypothetical distances since I don't actually know the distance a pile would travel in a given material...).

    The point is, the answer to the problem b) is:

    KE = F*s

    KE/s = F

    500,000J/.2m = 2,500,000N

    But if the ground was different, then the force of the driver applied to the pile would change.

    In soft mud, the force would be:

    500,000J/2 = 250,000N

    And in solid ice, the force would be:

    500,000J/.1 = 5,000,000N


    This looks like the further the driver pushes the pile, the less force the driver actually applied to the pile. Does that make sense? To me, the force should not depend on the material of the ground. If a 10,000kg driver drops on something, whether that thing is driven 0cm into the ground or 100m into the ground, the driver carries the same strength (or force). I mean, Mike Tyson doesn't throw a weaker punch depending on if he's punching a soft bag or a hard bag, right? The speed that his hand reaches at the point of impact and his intended follow through remains the same regardless of the material of punching bag, so why would his "force" change if he hasn't actually changed anything in his punch? Am I confusing applied force and net force some way?

    Thanks
     
  2. jcsd
  3. Sep 17, 2015 #2

    andrewkirk

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    The momentum is given up over a period of time. Since the change in momentum (impulse) is force x time, and the initial momentum is fixed, the longer the time it takes to stop, the less the force.

    That's why airbags and crash helmets work. The head takes longer to lose its momentum (by moving through polystyrene and high (but not too high) pressure inflated fabric respectively) and hence encounters less severe forces in doing so.

    By making some simple asusmptions such as constant deceleration ##a##, then using the equation ##v^2=u^2+2as##. We know everything except ##a## and can hence work out ##a##, and work out force by multiplying that by the mass of the thumper.
     
  4. Sep 17, 2015 #3

    rude man

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    You just have to understand that energy = force times distance.
    So for the same energy, the shorter the stopping distance, the larger the force has to be to maintain the product the same.
     
  5. Sep 18, 2015 #4

    haruspex

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    As others have replied already, that is correct. Perhaps it puzzles you because you conceptually confuse force with impulse. When the driver reaches the pile it doesn't 'have' any particular force; what it has is mass, velocity, momentum and KE. The force is determined by the resistance of the ground. The momentum and impulse, acting against this force, determine the time and distance over which the force will be exerted.
    (The question is badly worded. It should not ask for the force 'as the driver strikes the pile'. It should ask for the force while the driver is driving the pile.)

    One more thing: if you do simply divide the KE by the distance you will not get exactly the right answer. You need to bear in mind that gravity does not suddenly switch off when the driver contacts the pile. I have not checked whether that makes a significant difference to the answer here.

    Finally, the question assumes the pile is so light compared with the driver that its mass can be ignored. In many real examples, that is not true. You would have to solve the problem in two phases. The first phase is the collision of driver with pile, in which momentum is conserved but work is not.
     
  6. Sep 18, 2015 #5
    Andrewkirk, rude man, and haruspex,

    I have a tangent question that will lead back to this thread. I will ask it in a new post. I will return to this question as soon as I'm able to resolve it.

    Thank you.
     
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