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Very EASY tension problem, problem.

  1. Sep 28, 2011 #1
    1. The problem statement, all variables and given/known data

    "Three situations involving a block with a weight of 20.0 N and a string are shown in the figure above. In case 1, there is only one string leading from the hook to the ceiling. In cases 2 and 3, the string passes through a hook on top of the block.

    [What the picture shows]: For case 2, an angle of 30 degrees is given and for case 3, an angle of 90 degrees is given. In each, the angle given is the angle "above the hook". That is, when one thinks of the triangle formed by the two ropes and the point-mass, we are given the angle made by the meeting of the two ropes, "pointing" down towards the ground.

    (a) Rank these situations based on the tension in the string, from largest to smallest (e.g., 3>1=2).

    (b) Calculate the tension in the string in case 3."

    2. Relevant equations

    Fnet = T - mg

    T = T1 + T2

    3. The attempt at a solution

    The most questionable assumption made is that the angles made with respect to the horizontal are equal. For instance, in case 2 the angle of 30 degrees is bisected so that we can deduce that the angle wrt the horizontal is 90 - 15 = 75. LIkewise with case 3 we are left with 45. It seems we have too little information to deduce the horizontal angles for an arbitrary (not necessarily equal) T1 and T2, so we've assumed that the bisection goes through as described.

    IF that's ok, then part (b) of the question should be simple. Considering a 45 degree angle wrt the horizontal (by bisecting 90 degrees), the equality of T1 and T2 implies that T = T1 + T2 = 2*T1 = 2*(10*sin(45))=20*sqrt(2) which is approximately 28.28427, which, to three significant digits should be 28.3.

    HOWEVER, the online submission system says this is wrong. As well as our answer to (a), for which we said that (case 1) < (case 2) < (case 3).

    So, MY QUESTION, is: Is the approach described the correct one? I've seen tension problems before, but usually have been given two angles to work with, not just the one center angle between the ropes.
    Last edited: Sep 28, 2011
  2. jcsd
  3. Sep 28, 2011 #2


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    If the angle between the 2 strings is 90 degrees and the 20 N weight hangs along the bisector of the angle, then the vertical component of the tension in each string is how much, from symmetry and Newton 1?? So then the tension in each string is ____?
  4. Sep 28, 2011 #3
    Wow. Thanks so much. I am an absolute fool.

    By symmetry deduce that resultant for each string is 10. Then since horizontal components cancel just add vertical components to get 2*10*sin(45) = 20/sqrt(2) apprx 14.1.

    Similar calculations show that 30 degrees gives 10, so (a) is 1>3>2.

    [These answers went through.]

    I don't know how I messed this up. Sad thing is I have a math degree (helping GF with physics hw). Ouch.

    Thanks again!
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