Dynamics calculations

Discussion in 'Engineers' Talk' started by BiancoTheGiraffe, Mar 4, 2021.

  1. BiancoTheGiraffe

    BiancoTheGiraffe Screwfix Select

    Was installing a safety wire today and was curious about the method used to calculate the force it needed to take, but unfortunately I last did those sort of calculations nearly 20 years ago during my engineering degree!

    It's particularly annoying that I can't remember what to do, seeing as engineering maths was about the only part of the course that I was any good at!

    I've calculated the energy that a 100kg person would be carrying after they've fallen 2 metres so I've got the answer in Joules, but I can't remember how to convert that into the equivalent static mass.

    Anyone here able to help?

    This is for study only, I'm not planning on designing anything safety critical based on this!
     
  2. 92Charlie92

    92Charlie92 New Member

    Ignoring air resistance, after falling 2 metres from standstill, a person would be falling at 6.2m/s (v^2 = u^2 + 2as). The energy is 0.5mv^2 = 1960J. Assuming all this energy is transferred to an extension of the wire, the energy stored in a safety wire to bring said person to a rest is the integral of kx, so 0.5*k*x^2 (=1960J). Use this to find out x for your given safety wire stiffness, K. x = sqrt(3920/k)

    static mass leading to an extension of the wire is mg = kx, so m = kx/g. Plugging in x from above will give you M = (k/g)*sqrt(3920/k). That's the theory behind such a calculation, but in practice you'd want to use a big safety factor.
     
    BiancoTheGiraffe likes this.
  3. spannerw

    spannerw Screwfix Select

    Yeah I thought that too:p
     

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