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Hammer and nail

  1. Oct 30, 2014 #1
    Hey everyone, this is a copy/paste from a another thread. I actually don't need help with finding an answer. Its a practice problem in the book so its all worked out. What I need is for someone to help explain to me why the normal force is the average...I'm having trouble understanding this part


    1. The problem statement, all variables and given/known data


    In a pile driver , a steel hammerhead with mass 200kg is lifted 3m above the top of a vertical I beam being driven into the ground. the hammer is then dropes driving the Ibeam7.4cm farther into the ground. The vertical rails that guide the hammerhead exert a constant 60N friction force on the hammerhead. Use the work energy theorem to find a) speed of the hammerhead just as it hits the I beam andb) the average force the hammerhead exerts on the I-Beam Ignore the air effects of the air.


    3. The attempt at a solution

    I managed to understand part a. Part b is where I am lost.

    The book shows the following work, and I need help understanding on of the steps.

    Wtotal = (w-f-n)s23

    Wtotal = (w-f-n)s23 = k3-k2

    *** n is the same as g right? and we solve for n? Since when is n the average force?***




    n = w-f - k3-k2/s23

    =1960N - 60N - 0J - 5700J / 0.074m

    =7900N

    the part underlined I understand.
    The part in bold is where I can't explain why.

    Read more: https://www.physicsforums.com
     
  2. jcsd
  3. Oct 30, 2014 #2

    haruspex

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    I need to know what all these variables represent.
    That link does not go to a specific thread.
     
  4. Oct 31, 2014 #3

    BvU

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    Good.

    Well, let me guess: 2 is when the hammer hits, 3 is when the 7.4 cm are done. s23 is then the 0.74 m.

    So what you have is some kind of energy equation:
    Ekin, 2+Epotential, 2 = Workbeam + Workfric +Epotential, 3

    with Workbeam the energy used to drive the I beam down
    and Workfric the energy lost in friction.

    Workbeam is s23 * average force, hence the average force.

    Your answer is a factor 10 off - or you made a typo.
    Probably forgot the brackets in

    1960N - 60N - ( 0J - 5700J ) / 0.074m

    AND a zero in the result.

    That makes it time-consuming to help you !
    Have pity with the poor helpers who desperately try to understand what you type: read what you typed as if you were one of them.
     
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