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!

Ok, I got part of this question right (Not HW)

  1. Apr 15, 2009 #1
    There is a diver of mass 72Kg who stands 2.4m away from the pivot of the diving board, 3.2m above the water.

    "The water brings the diver to rest when his centre of mass is 1.6 m below the surface of the
    water. Calculate the average total upward force acting on the diver which brings his vertical
    velocity to zero."

    I calculate the speed as he enters the water to be 7.92 which is correct.

    u=7.92
    v=0
    a=?
    s=1.6

    a comes out to be -19.6m/s^2

    F=ma

    F= 72x |19.6| = 1411N

    Yet, in the answers, it says that you have to plus 706 to this to get the total upward force. I understand that 706 comes from the mass(72) x g(9.8).

    It tells me that 1411 is the decelarating force, and that 2117N is the toal upward force. Can I just ask, why is the extra 706 needed? Is it to do with Newtons first law, stating that every force has an equal and opposite force? But even so, doesnt the F=Ma in my calculation take care of that force?

    Thanks
     
  2. jcsd
  3. Apr 15, 2009 #2
    Big hint:
    |Fnet|=|Fupwards - Fgravitation|= m |a|

    :)
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook