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F =M(A) question for roofing contractor

  1. Aug 29, 2011 #1
    I know that F =M(A) and am looking for some helping figuring out the following scenario:
    A 200 lb. man wearing a full body harness is tied off to an anchor point on a roof. The roof 's angle is 22.5 degrees. If the man falls 25' (without falling off the roof),what would be the force exerted on the anchor point at the other end of the safety line?
     
  2. jcsd
  3. Aug 29, 2011 #2

    K^2

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    There is insufficient information to determine the answer. What you need is the Young's Modulus for the line.
     
  4. Aug 29, 2011 #3
    From what I read from the chart, the approx. Young's modulus for nylon is 2-4 GPa, or 290,000 - 580,000 lbf/in. squared
     
  5. Aug 29, 2011 #4

    K^2

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    It also depends on how the line is woven. I'm not familiar with ones used in construction, but the ones we use in climbing are made intentionally stretchy to minimize the force.

    But lets try 2GPa for an estimate, at least. Hopefully, the actual force will be lower.

    I forgot to ask what the diameter of the safety line is. And just to make sure I understand this correctly, the person falls 25' along the 22.5° incline, right?
     
  6. Aug 29, 2011 #5
    The line is 5/8" diameter, and is not as stretchy as what you likely use for climbing. And yes, the person falls along the 22.5 incline. (The safety equipment would keep him from actually free falling (gravity) off the edge of the roof.
     
  7. Aug 29, 2011 #6

    K^2

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    Ok. The simplest solution here is to consider energy balance, rather than forces directly. The line will absorb all of the kinetic energy of the falling worker. The maximum kinetic energy is the difference in potential energy before and after the fall:

    E=mgh = 90kg * 9.8m/s² * 7.6 meters * sin(22.5°) = 2,570J

    The safety line behaves as a spring. The spring coefficient is derived from Young's Modulus, ropes cross-section area, and length.

    k = Y*A/L = 2GPa * 0.00079 m² / 7.6 meters = 207,900 N/m

    The potential energy of the elastic is given by:

    U = (1/2)kx^2

    Where x is the maximum distance the line will stretch. Using E=U:

    x = sqrt(2*E/k) = sqrt(2*2,570 J / 207,900 (N/m)) = 0.15 meters.

    Finally, the force:

    F = kx = 31,000 N = 7,000 pounds of force.

    Keep in mind that this is the worst case scenario. It assumes that friction did not slow down the worker, that worker's own body did not flex, and that the nylon line was not designed to give.

    Realistically, the force would probably be smaller. But this is the only number I can give you that guarantees not to be an underestimate.
     
  8. Aug 29, 2011 #7
    0 lbf if the rope is unstreched.
     
  9. Aug 29, 2011 #8
    Thanks for the answer! (K2) We are dealing with new OSHA regulations, and this helps me understand some of their requirements.
     
    Last edited: Aug 29, 2011
  10. Aug 29, 2011 #9

    K^2

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    What do regs suggest for the anchoring strength?
     
  11. Aug 29, 2011 #10
    They're saying 5,000 lbs. of force, I initially thought that was high, but your equations show otherwise.
     
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