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Tension in each rope

  1. Sep 16, 2016 #1
    1. The problem statement, all variables and given/known data
    A muscle exerciser consists of two steel ropes attached to the end of a strong spring contained in a telescopic tube. When the ropes are pulled sideways in opposite directions, the spring is compressed.

    The spring has an uncompressed length of 0.80 m. The force F (in N) required to compress the spring to a length x (in m) is calculated from the equation F = 500 (0.80 - x).

    The ropes are pulled with equal and opposite forces, P, so that the spring is compressed to a length of 0.60 m and the ropes make an angle of 30º with the length of the spring.

    0017ab74c2a6.jpg

    (a) Calculate: (i) the force, F, (ii) the work done in compressing the spring.
    (b) By considering the forces at A or B, calculate the tension in each rope.
    (c) By considering the forces at C or D, calculate the force, P.

    Answers: (a) (i) 100 N, (ii) 10 J, (b) 57.7 N, (c) 57.7 N

    2. The attempt at a solution
    (a) (i) F = 500 * (0.8 - 0.6) = 100 N
    (a) (ii) W = Fs = 100 * (0.8 - 0.6) = 20 J, which is wrong.

    I made a graph but I am not sure whether I noted the force correctly.
    9f9d5acc2f0d.jpg

    (b) I am looking for AC and I have AO (where O is the middle point) = 100 N and the angle is equal to 30 degrees. So AC = 100 / cos 30 = 115.5 N
    (c) Since PCA and PCA are same steel ropes, therefore they have the same forces which are equal to 115.5 N.

    What do I miss here?
     
  2. jcsd
  3. Sep 16, 2016 #2

    Doc Al

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    Staff: Mentor

    Good!

    Careful. The force is not constant. Instead: What's the spring potential energy stored in a compressed spring?

    Almost. Realize that two ropes pull down, so each one accounts for half of that total.

    The pulling force does not equal the tension in the ropes. What component of the tension force equals the force P? (There are two ropes there as well.)
     
  4. Sep 16, 2016 #3
    PE = 1/2 kx2? k = 100 N / 0.8 m = 125 N/m so PE = 0.5 * 125 * 0.22 = 2.5 J?

    So AC = 100 / cos 30 = 115.5 N. But it is the combined force of the two ropes, and the tension in one rope is 115.5 N / 2 = 57.75 N.

    Maybe CO and OD? CO = AC * cos 60 = 28.9 N. And since we also have CB therefore P = 28.9*2 (due to AC and CB) = 57.75 N. Is this the right logic?
     
  5. Sep 16, 2016 #4

    Doc Al

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    Staff: Mentor

    Right idea, but recalculate that value of k. The compression is only 0.2 m.

    Good!

    The force P is horizontal. Consider the horizontal forces acting at point C.
     
  6. Sep 16, 2016 #5
    k is equal to the tension required to produce unit of extension. So m in N / m is equal to the compressed part of the spring and not to it's normal length (as I wrongly calculated). As a result: PE = 1/2 * (100 / 0.2) * 0.22 = 10 J.

    Maybe CO and OD? CO = AC * cos 60 = 28.9 N. And since we also have CB therefore P = 28.9*2 (due to AC and CB) = 57.75 N. Is this the right logic?
     
  7. Sep 16, 2016 #6

    Doc Al

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    Staff: Mentor

    Good!

    Good!
     
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