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How to find distance that the rope must sag

  1. Feb 5, 2016 #1
    • Moved from technical physics section, so missing the homework template
    I'm stuck on a homework question. It states, "Bob traverses a chasm by stringing a rope between a tree on one side of the chasm and a tree on the opposite side, 25 m away. Assume the rope can provide a tension force of up to 29 kN before breaking, and use a "safety factor" of 10 (that is, the rope should only be required to undergo a tension force of 2.9 kN). Bob's mass is 52.0 kg . Determine the distance x that the rope must sag at a point halfway across if it is to be within its recommended safety range."

    The given information:
    T=2900N
    m=52.0Kg
    and the equation I am using is ma=∑F, which is ma=T-mg

    I'm not sure how to post my free body diagram, but I know that I am trying to find the sin of θ when i connect the two tension forces. This is where i'm stuck. I don't know how to get the value of the angle from only knowing the length of one side.

    Any help is appreciated,
    thank you
     
  2. jcsd
  3. Feb 5, 2016 #2

    CWatters

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    See the "upload" button to post a picture.

    You have to work out sin θ from the max allowable tension not the length of sides, that comes later. Post your free body diagram and equations for the vertical equilibrium.
     
  4. Feb 5, 2016 #3
    I was only using ma=∑F so far. Since it it is not accelerating i worked the equation into: 0=2Tsinθ-mg.
    Plugging in the information given it should be (52.0Kg)(9.8m/s^2)=2(2900N)sinθ.
    Which would make sinθ=.087862069, but I don't think this is correct since it doesn't have any units
     

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  5. Feb 5, 2016 #4

    SteamKing

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    The sine of an angle is not supposed to have any units.

    You do have to take the sine of the angle and figure the tension in the rope with it.
     
  6. Feb 5, 2016 #5
    So would the tension in this rope be found by multiplying 2900Nsinθ since that should give me the y value for the rope?
     
  7. Feb 5, 2016 #6

    SteamKing

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    I misspoke earlier. 2900 N is the allowable tension in this rope, once the factor of safety is accounted for.

    What you need to find is the allowable sag in this rope such that the allowable tension of 2900 N is not exceeded. Use the sine of the angle to do that.
     
  8. Feb 5, 2016 #7
    So the tension should be (.5)(52.0kg)(9.8m/s^2)sinθ, which equals 22.38725518. I'm unsure about the units since it would still be in Newtons and not meters. Am I missing a step to change it to meters?
     
  9. Feb 5, 2016 #8

    CWatters

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    No that's not correct (wrong track. You already know the tension is 2900N).

    Earlier you complained you couldn't calculate the dimensions of the triangle with just one side. Now you have a side and an angle.
     
  10. Feb 5, 2016 #9
    Ok so the length of the known side is 12.5m. To find the other side I use 12.5msinθ which gives me 1.098275863m. Is this correct?
     
  11. Feb 5, 2016 #10
    I found the answer. Thanks for all your help!:smile:
     
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