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!

What constant force does the floor exert

  1. Mar 9, 2015 #1
    1. The problem statement, all variables and given/known data
    A 74.4 kg person falls straight down from a 1.96 m height (measured from his feet) and lands with weight distributed equally on both feet. To soften the blow, he bends his knees so that it takes 0.906 s for him to stop once his feet touch the ground. What constant force does the floor exert on each foot while he's stopping? Give your answer in Newtons


    2. Relevant equations
    f=ma
    f=(m*v^2) / 2x
    F * t = m * Δv
    Δp=m(v-vo)
    Favg=Δp/Δt
    w=mg
    Xf=Xi+Vit+(at^2/2)
    Vf=Vi+at
    Vf^2=Vi^2+2a(Xf-Xi)

    3. The attempt at a solution
    The answer is 254.5 and I am trying to work backwards to see what I am doing incorrectly.

    Originally, I had lumped all variables into one equation ( f * t = m * Δv ) but now I think it's two parts? First system being A; person falling 1.96m and system B; ground receiving force and pushing back up as the person absorbs shock @ 0.0906s?

    I think the equation might be solved by calculating acceleration and solving for force with Newtons 2nd law, but I am unsure and that sounds completely wrong.

    Thanks in advance
     
  2. jcsd
  3. Mar 9, 2015 #2

    BvU

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    What was the speed when he reached the ground ?
     
  4. Mar 9, 2015 #3
    well he comes to rest so I would assume 0?
     
  5. Mar 9, 2015 #4

    SteamKing

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper

    His speed an instant before he comes to rest.
     
  6. Mar 9, 2015 #5
    Ah right, so 2.16? (1.96 / 0.906)
     
  7. Mar 9, 2015 #6

    SteamKing

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper

    Why do you say that?

    Think. The guy is falling. He falls 1.96 m. You know, gravity. Hint Hint
     
  8. Mar 9, 2015 #7
    Ok, Vf^2 = Vi^2 + 2ad ?

    That would be Vf^2 = 0 + 2(9.8)(1.96) = 38.416, is that right?
     
  9. Mar 9, 2015 #8

    haruspex

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    2016 Award

    Yes.
     
  10. Mar 9, 2015 #9
    Ok, I am stuck after this. I know initial velocity, final velocity, mass distance and time but i don't know where to go from there.
     
  11. Mar 9, 2015 #10

    SteamKing

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper

    You go back to the OP and solve for the unknown force, using the formula listed there.
     
  12. Mar 9, 2015 #11
    f=ma gives me 2858n when the answer should be 254.2?
     
  13. Mar 9, 2015 #12

    haruspex

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    2016 Award

    Please post your working.
    (I get an answer between the two. 254.2 is much too low. The force is greater than that when the person stands still.)
     
  14. Mar 9, 2015 #13
    f=ma as in; f=74.4 * 38.416

    Also 254.2 is the answer to the problem as generated via the program I am working this equation through - could this just be incorrect?
     
  15. Mar 9, 2015 #14

    haruspex

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    2016 Award

    Read your own equation:
    That's not an 'a' on the left.
    They've made a classic blunder. See if you can get the right answer, then guess what mistake they made.
     
  16. Mar 9, 2015 #15
    Oh whoops, a = (v - v0)/t

    a = (38.416 - 0)/0.906 = 42.201

    f=ma ---> f=74.4*42.201
    so f=3139n / 2 (split evenly between two feet) = 1569n?
     
  17. Mar 9, 2015 #16

    SteamKing

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper

    You're still guessing here. Go back and read Post #7 carefully. I don't understand why you are having so much trouble understanding your own work.
     
  18. Mar 9, 2015 #17
    I got it! I forgot to square root the final velocity.

    Vf = 6.1980

    a= 6.1980 / 0.906 = 6.8411

    6.8411 * 74.4 = 508.908

    508.908 /2 = 254.4
     
  19. Mar 9, 2015 #18

    BvU

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    And then after this 0.906 seconds (that's a pretty accurate time measurement they managed there ! (*) ) the force is zero :rolleyes:, right ?

    Reason I mention this accuracy is: the problem statement clearly says "falls straight down from a 1.96 m height (measured from his feet)" and it also says "bends his knees". Well, I'm too old to practice this, but I guestimate that his center of gravity after these 0.906 s is about 0.5 m lower than when standing up. Somehow that potential energy also has to be compensated. Ignoring that is an inaccuracy of around 25%. Makes those three-digit numbers look funny.



    (*) -- and I think it should be considerably shorter... anyone know a suitable Utube clip ? Maybe the last seconds of this one
     
  20. Mar 9, 2015 #19

    haruspex

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    2016 Award

    Indeed. The figures given require the athlete to have legs at least 3m long.
    If we take the given data on trust, the error is much greater.
    HJ^2, you can either use BvU's method of taking into account the extra PE lost in bending the legs (but be warned you'll get some crazy values), or you can think in terms of the forces acting on the athlete and the net acceleration. The two methods should produce the same result.
     
  21. Mar 10, 2015 #20

    BvU

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    Dear haru, we are criticizing the composer of the exercise. HJ is content with the 255 N -- even when standing still is already 365 N per foot...
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: What constant force does the floor exert
Loading...