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Finding force that ground has on person

  1. Oct 1, 2016 #1
    1. The problem statement, all variables and given/known data
    A person jumps from the roof of a house 3.9-m high. When he strikes the ground below, he bends his knees so that his torso decelerates over an approximate distance of .70m. If the mass of his torso (excluding legs) is 42 kg, find a. his velocity just before his feet strike the ground, and b. the average force exerted on his torso by his legs during deceleration

    2. Relevant equations


    3. The attempt at a solution
    Had no trouble with a, got 8.7m/s as maximum velocity, however I seem to be wrong on b. Here is my work

    1/2at^2=0.70
    a=1.40/(t^2)
    a*t=8.7
    (1.40/t^2)*t=8.7
    (1.40/t)=8.7
    8.7t=1.40
    t=1.61
    8.7/.161=5.40 m/s^2
    42*54.0=m/s^2 upward acceleration
    42*54.0=2268 N
    My problem here is that everyone on the internet says that the force exerted on his torso by his legs IS 2268, but I feel like that should be only the net force, because gravity is still in play and the ground must have a higher force to counteract that. Why would the answer not be 2268+9.8*42=2679.6--->2700N (sig fig)??
     
  2. jcsd
  3. Oct 1, 2016 #2

    kuruman

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    Homework Helper
    Gold Member

    Hi Josepph_kijevski and welcome to PF.

    You are correct. The 2268 N is ma which is the same as the net force. The legs are providing the necessary force to support the weight of the torso and to provide the necessary acceleration to stop the torso from moving.

    Next time please include the relevant equations so that we know where you start from.
     
  4. Oct 1, 2016 #3
    Thank you!
     
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