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Homework Help: Tyrolean traverse - Forces Question

  1. May 29, 2010 #1
    1. The problem statement, all variables and given/known data

    Christian is making a Tyrolean traverse as shown in the figure. That is, he traverses a chasm by stringing a rope between a tree on one side of the chasm and a tree on the opposite side, 22 away. The rope must sag sufficiently so it won't break. Assume the rope can provide a tension force of up to 27 before breaking, and use a "safety factor" of 10 (that is, the rope should only be required to undergo a tension force of 2.7 ) at the center of the Tyrolean traverse.

    Determine the distance that the rope must sag if it is to be within its recommended safety range and Christian's mass is 75.0kg .
    Express your answer using two significant figures.

    GIANCOLI.ch04.p31.jpg


    2. Relevant equations

    F = ma?

    3. The attempt at a solution

    tried but failed to understand, theres nothing about this type of question in my textbook or if there is I don't understand the connection between them. Can someone give me some hints?
     
    Last edited: May 29, 2010
  2. jcsd
  3. May 29, 2010 #2

    rock.freak667

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    What are you required to find? Also post the picture.
     
  4. May 29, 2010 #3
    srry forgot about those just editted
     
  5. May 29, 2010 #4
    copy and paste error srry

    75kg
     
  6. May 29, 2010 #5

    rock.freak667

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    Ok, well since the man hangs at the middle, then can you draw a free body diagram with the forces acting on the rope?
     
  7. May 29, 2010 #6
    [PLAIN]http://img175.imageshack.us/img175/7074/9999k.png [Broken]


    like that right

    ehh Fg should be Force of the dude but w/e
     
    Last edited by a moderator: May 4, 2017
  8. May 29, 2010 #7

    rock.freak667

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    Yes like that, but draw if you draw in a dotted line to connect the FT forces and call the the angle made θ. What are the vertical and horizontal components of the forces FT?
     
  9. May 29, 2010 #8
    thats one of the things Im having trouble with, I dont know how to calculate FT.

    Is the vertical component the force of the dude hanging there, and if so whats the horizontal? what is the 27kN for? whats the factor of 10 for?

    srry for asking so many questions but I can't seem to wrap my head around how to place the values
     
  10. May 29, 2010 #9

    rock.freak667

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    The factor of safety basically changes the value of the tension. Normally you would find for 27kN. But since a factor of safety is used, you design for 2.7kN/10 = 2.7 kN

    So the value of tension you want is 2.7kN. This is your FT

    In your free body diagram, FT acts at the angle θ right? So what are the vertical forces of the force FT?
     
  11. May 29, 2010 #10
    vertical force would be (75kg)(9.8m/s^2) = 735N right?
     
  12. May 29, 2010 #11

    rock.freak667

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    Yes that is one force. But remember, the forces FT act at an angle and hence can be split into vertical and horizontal components.

    When you do that, then you know the man hanging is in equilibrium, so what should the sum of the forces vertically be?
     
  13. May 29, 2010 #12
    equal to 0?
     
  14. May 29, 2010 #13

    rock.freak667

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    Right so what is the summation of the forces vertically to get the angle?
     
  15. May 29, 2010 #14
    2ft = 735n?
     
  16. May 29, 2010 #15

    rock.freak667

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    You are leaving out the angle. Draw the angle like I suggested. Do you know how to split a force into components?
     
  17. May 29, 2010 #16
    ok I get it now, F = 2Tsin(theta) -mg
     
  18. May 29, 2010 #17

    rock.freak667

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    Right yes!. So since you know F=0 for equilibrium and you want T=2.7 kN, what is the angle θ?

    When you get that, it becomes a simple application of trigonometry to find the distance x.

    Since he is in the middle, what is the distance from him to either side?
     
  19. May 29, 2010 #18
    awesome got the answer, thx for your help
     
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