1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: 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


    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


    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

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


    User Avatar
    Science Advisor
    Homework Helper
    2017 Award


    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.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted