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Tension in a Rope

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  1. Dec 5, 2016 #1
    1. The problem statement, all variables and given/known data
    1k8rj.jpg
    So it's given the pipe has a inside diameter of 60cm and outside diameter of 70cm. the two ropes AC and AB are separated by a spreader bar. Wants us to find tension in the ropes. Also give is the density of concrete which is 2320kg.

    2. Relevant equations
    pi(r)^2*L=Area

    3. The attempt at a solution

    pi(.7-.6m)^2*2.5*2320=23215Newtons

    Can we just then divide that answer by sqrt of 2 because the ropes are equal? If no that is where I get lost is how to do the other part of the math.

    Thanks!
     
  2. jcsd
  3. Dec 5, 2016 #2
    Hi Noreturn, congrats on the first post !!!.
    :welcome:

    ##\pi r^2l## =volume

    Do:
    1) : - Make a FBD.

    2) :- Take net force upwards and equate it to net force downwards. For this you have to take the component of all the tension forces.

    If I interpreted the diagram correctly then there are four tension forces.

    3) :- find the tension. YOU ARE DONE !!!!
     
  4. Dec 5, 2016 #3
    So I guess I found the volume which I can use to find the density of the concrete pipe. Then can I dive that by 4 since all four ropes have the same angle and should disperse the force evenly?

    So 23215/4=5803.75 or 5804N in each rope
     
  5. Dec 5, 2016 #4
    I think you mean mass of the pipe.
    Yes they will disperse evenly but only dividing by 4 does not work as weight and the tension force are not along the same line. Tension makes and angle with the weight. you need to account for it also.
     
  6. Dec 5, 2016 #5
    So we dont know the length of AC or BC. So it would be something along the lines of

    Sin(45)+sin(45)+mass=0 right? Or what is next? Or teacher isn't very good at explaining these, he didn't even teach us these. Basically a engineering class to show us what we will be doing in engineering but didn't explain how to do any of the problems.
     
  7. Dec 5, 2016 #6
    Ok answer this question first :-

    asxasc.png
    What is the missing side ?
    Hint :- use trignometry.
     
  8. Dec 5, 2016 #7
    cos(45)=?/T

    Using Soh Cah Toa
     
  9. Dec 5, 2016 #8
    So, ##? = T cos(45)##.
    Now back to your question
    1k8rj.jpg

    What is x here in terms of T ? (Sorry for the bad editing, the angle is 45)
     
  10. Dec 5, 2016 #9
    Oh, then the T will be what I said earlier. So 5804=Tsin45

    So t=8208.1N or 8208N
     
  11. Dec 5, 2016 #10

    CWatters

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    Perhaps go back a step..

    How do you get the area of one end to be π(D-d)2?
     
  12. Dec 5, 2016 #11
    Yes correct, assuming your calculations are correct.
     
  13. Dec 5, 2016 #12
    Well it would be outer diameter-inner diameter * length. Or in other words the thickness of the pipe is 10cm or .1m
     
  14. Dec 5, 2016 #13
    Is that how you would get the mass of the pipe? the divide that by 4 then plug that in for ?
     
  15. Dec 5, 2016 #14

    CWatters

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    The area of an annulus is the area of the outer circle minus the area of the inner circle...eg...


    Pi*352 - Pi*302

    Then multiply that by the length to get the volume.
     
  16. Dec 5, 2016 #15

    CWatters

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    No, once you have the right mass calculate the weight. Then follow what Buffy has been saying about a FBD.

    Edit: Actually yes, dividing by 4 gets you the tension in the 4 short vertical ropes.
     
    Last edited: Dec 6, 2016
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