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Average force?

  1. Dec 2, 2007 #1
    Average force is change in momentum, divided by change in time, correct?

    So when I land on the ground after jumping, how can I calculate the average force I exert on the ground, (while I decelerate from velocity at moment I touch ground to zero velocity).

    I calculated my velocity just when I touch ground. Clearly my final is zero, so my change in momentum is just my initial velocity times my mass. Divide this by the time taken, found from initial velocity and acceleration, and I should get average force, right?

    But it doesnt work out to the value I should get. Am I going wrong in my idea here? Should I be doing it a different way???
  2. jcsd
  3. Dec 2, 2007 #2


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    Your formula is correct, but:

    To find the average force you exert on the ground you want to find the change in momentum you experience, while interacting with the ground. So, the change in momentum while interacting with the ground would be:

    [tex]\Delta p= mv_f - m(0)[/tex] where v_f is the speed right before you hit the ground, and, obviously, zero is the speed right after you hit the ground.

    Then, the time in the equation for average force would be the time you are in contact with the ground before you stop, not how long it took you to fall. Does this make sense?
  4. Dec 2, 2007 #3
    I think you misunderstood me.

    The time I am using is the time I was in contact with the ground.
    I have the distance fallen, so I used the kinematic equations of constant acceleration to work out my velocity when I just touch the ground.

    I then worked out my acceleration, (from a distance: bend in the knees: I would be accelerating over this distance) to zero velocity.

    Using these I worked out the time.
  5. Dec 2, 2007 #4


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    OK then those would be the correct values. It could be that your method has some source of error in it.
  6. Dec 2, 2007 #5
    MMmmm yea, thanks for the input anyway. I'll get back to this thread after i have a closer look!!
  7. Dec 3, 2007 #6
    Nope, it's still not right.

    I have velocity at moment I touch ground. I have acceleration (constant). I have the distance I accelerate over. I have my mass.

    But my answer is not right. I don't know where I can possibly be going wrong!!!!
  8. Dec 3, 2007 #7
    I solved it. I had to ADD acceleration due to gravity to my deceleration.

    I m not actually sure why though Anyone explain?
  9. Dec 3, 2007 #8


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    F_net(average) = m(dv/dt). The net force includes both your weight and the force from the ground.
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