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Impact force equation

  1. Aug 10, 2015 #1
    Hello, I have been turning to this subject and tried to develop the right equation on my own.

    What I did was:
    mgh = ½kx2 (fall energy = spring energy)
    x2 = 2mgh/k (obviously devided by ½k)
    sqrt(2mgh/k) = x (taking out square root)
    F = mgh/x = mgh/sqrt(2mgh/k) = sqrt(½mghk) (energy /distance = force)

    But when I checked the internet for answers it was not the answer given.
    The correct equation is F = sqrt((mg)2 +2mghk)

    I definitely went wrong somewhere, could someone please help me understand how that equation was properly developed so that I can understand this. I am sort of embaressed to be mistaken.
    Thanks in advance.

    PS: I donno what perfix to put so I put intermediate.
  2. jcsd
  3. Aug 10, 2015 #2
    The " correct?" equation is not dimensionally correct.
  4. Aug 10, 2015 #3
    Which one?
    Also, can you explain in detail what is wrong about which?

    I got my new answer info from wikipedia, some inaccurations might occur.
  5. Aug 10, 2015 #4
    This is dimensionally incorrect. Force must have units of kg*m/sec2
  6. Aug 10, 2015 #5
    I see that you have made corrections in your original posting
  7. Aug 10, 2015 #6
    I see, I didn't account for that. I still think you are mistaken as sqrt((mg2)) is force squered and rooted and so is sqrt(2mghk) notice they are added to each other and not multiplied.
    m is mass (in MKS is kg)
    g is acceleration (in MKS is m/s^2)
    h is height (in MKS is m)
    k is spring constant (in MKS is kg/s^2)

    Edited: Might be.. it was a few hours ago so I don't remember. I will keep notes like so now onwards.
    Last edited: Aug 10, 2015
  8. Aug 10, 2015 #7


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    A bit more detail about the actual scenario could possibly be useful???
  9. Aug 10, 2015 #8
    I think you should approach the problem from a change in momentum point of view,

    Fp/Δt = (pf - pi)
  10. Aug 10, 2015 #9


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    Agreed - for most collision problems but there is always the problem that the force is not constant as a spring is compressed. It might be worth considering the time constant of the mass and the spring combination and treating it as a basic SHM problem.
    Is there supposed to be any damping or will the mass just bounce back?
  11. Aug 10, 2015 #10
    Basically. I will try not to be annoying with a big background story. Adam is a man attached to a rope, he weighs at m (kg). Adam fell from h (meters) and only has his trusty rope with spring current of k (kg/m^2) to block his fall. What force does Adam feels in Newtons.

    Normally I would try to solve it the same way using momentum and breaking time, but you do not know how much time it takes him to break or the strech of the rope, only its spring constant k.

    All in all it is an neutral equation to use, at what ever be the situation. for example if I had a rope with a defined k and height but I need to know what weight is too much. Or when I know the mass but need to calculate the wanted rope.
  12. Aug 10, 2015 #11
    assume the breaking force is constant
  13. Aug 10, 2015 #12
    I was thinking average force, but forget that.
  14. Aug 10, 2015 #13


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    mmm. I would not want to assume a constant force. It wouldn't be hard to get data about the elasticity of 'real' climbing ropes. Your scenario includes a spring constant k so I would think that's what you should include. (In fact, how else would you determine how long the rope tension would be operating for?)
    It wouldn't be hard to calculate the speed when the rope goes taught and then find the amplitude of the half cycle of mass / spring combination (SHM sums). That would produce the peak acceleration (= force).
  15. Aug 10, 2015 #14
    I think you can calculate the time to stop Adam at the end of the rope assuming spring-like behavior of the rope.
  16. Aug 10, 2015 #15


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    Exactly. It's called SHM. :wink:
    You don't need SHM to work out the distance stretched but the time / acceleration needs SHM approach.
  17. Aug 10, 2015 #16
    True, so I did what I needed to calculate the the distance strached, also with spring like motion calculations, and after that I devided the overall spring energy by the distance to get the force at the end of the stretch sequence. I was shocked to find it was to no avail as the correct answer by wikipedia is different.

    I still don't understand how they derived that equation from the Adams situation. If you'd like to view it https://en.wikipedia.org/wiki/Fall_factor it is the impact force there.
  18. Aug 10, 2015 #17
    Here us wear you went wrong 30f41230a373dbff09c3425bb5d59adf.png you forgot to note the energy stored in the stretch rope due to the mass of the climber when the climbers speed is zero . The length of the rope L is h - x.
  19. Aug 10, 2015 #18
    How does this help me? Getting from here to F = sqrt((mg)2 + 2mghk) is still vary uncleare and in fact was the main question of this form.
    I see how mghk is there, but the whole mg2 is a mystery for me. also I arrived at sqrt(½mghk).
  20. Aug 10, 2015 #19
    Note this is a quadratic equation in xmax,

    Just substitute Fmax/k for xmax and solve.

    I misspoke in post 17. The total fall distance is h + x so the potential energy from this fall distance is stored in the stretched spring.
  21. Aug 10, 2015 #20
    No, it is more annoying to post an incomplete question or not at all. :)
    And then people jump with answers before even knowing what is the question.

    Do you even know the length of the rope? Or you just assume that it has length h?
    This is very important. You cannot solve the problem without knowing the relationship between h and the length of the rope.
    Imagine that the height h is above the ground and the rope is also of length h. Or maybe even less. Then no matter how trusty is the rope, he may still hit the ground, doesn't he?
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