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Tension in a rope confusion

  1. Dec 31, 2015 #1
    • This homework problem was originally posted in another forum so there is no template
    Hi,

    I have a revision book that has an example in it for the tension on a rope
    if a 20 Newton force is hanging on the centre of a rope suspended between two points.

    The rope dips 25 Degrees either side. So effectively an upside down pair of right angled triangles is formed.
    The Hypoteneuse being 25 Degrees down.

    http://d:\Rope-example.png

    The Revision textbook Example shows how to find the Tension
    2TCos65(Degrees)=20N
    and shows T works out to be 24 Newtons.

    I can not see where the 2TCos65=20 comes from, at least its not intuitive to me.
    So I figured it out with a simultaneous equation. T1 = left T2 = right
    Where the forces are worked out, asuming everything is static.

    1) T1Cos25 + T2Cos65=0
    2) T1Sin65 + T2Sin25=20N

    From 1)
    3) T1=T2x0.466

    From 2)
    T1x0.966 + T2x0.422 = 20

    Substitute for T1 from 3)
    T2x0.466x0.906 + T2x0.422 = 20
    => T2 x 0.844 = 20
    Therfore T2 = 23.69N which is close enough to 24 Newtons in the example. I assume this is correct? (Maybe the book is wrong too?)

    I was pleased with myself, until I checked by substituting T2 back fot T1 to check T1 was the same. I assume T1 and T2 are equal tensions.
    But it is not, which I found very puzzling. I get about 11 Newtons. i.e put 23.69 in place of T2 in 3) above.

    Can anyone explain what I have done wrong? Signs perhaps?, unfortunately my math is not the greatest.

    Regards,
     

    Attached Files:

  2. jcsd
  3. Dec 31, 2015 #2
    Your Eqns. 1 and 2 are incorrect. Check the geometry more carefully. All the angles in both these equations should be 25. Also, in Eqn. 1, there should be a minus sign, not a plus sign.
     
  4. Dec 31, 2015 #3

    CWatters

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    Science Advisor
    Homework Helper

    It should be.

    Step by step...
    Its a static problem so the vertical forces sum to zero.
    I assume you know where the 65 degree angle is.
    The tension each side is T.
    The vertical component of each T is TCos65.
    There are two of them so the total upward force is 2TCos65.
    The downward force is 20N.
    Sum the forces to zero....
    2TCos65 - 20 = 0
    Rearrange to give
    2TCos65 = 20
     
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