Tyrolean traverse - Forces Question

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The discussion centers on calculating the necessary sag of a rope for a Tyrolean traverse while ensuring it remains within safe tension limits. The rope must support a maximum tension of 2.7 kN, considering a safety factor of 10 from its breaking strength of 27 kN. Participants discuss the forces acting on the rope, including the weight of a 75 kg person and the need to resolve these forces into vertical and horizontal components. The equilibrium condition is established, leading to the equation involving tension and angle. Ultimately, the conversation concludes with the participant successfully determining the sag distance through trigonometric application.
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Homework Statement



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

Homework Equations



F = ma?

The Attempt at a Solution



tried but failed to understand, there's 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?
 
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What are you required to find? Also post the picture.
 
srry forgot about those just editted
 
copy and paste error srry

75kg
 
hanlon said:
srry forgot about those just editted

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


like that right

ehh Fg should be Force of the dude but w/e
 
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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?
 
thats one of the things I am having trouble with, I don't know how to calculate FT.

Is the vertical component the force of the dude hanging there, and if so what's the horizontal? what is the 27kN for? what's 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
 
hanlon said:
thats one of the things I am having trouble with, I don't know how to calculate FT.

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

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?
 
  • #10
vertical force would be (75kg)(9.8m/s^2) = 735N right?
 
  • #11
hanlon said:
vertical force would be (75kg)(9.8m/s^2) = 735N right?


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?
 
  • #12
equal to 0?
 
  • #13
hanlon said:
equal to 0?

Right so what is the summation of the forces vertically to get the angle?
 
  • #14
2ft = 735n?
 
  • #15
hanlon said:
2ft = 735n?

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

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?
 
  • #18
awesome got the answer, thanks for your help
 
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