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Tension in the string

  1. Nov 30, 2015 #1
    1. The problem statement, all variables and given/known data

    ?temp_hash=935184fa0c5c76454084ea523756af8b.png

    Since the image is not clear ,the blocks on the left are 1 Kg(top) and 5kg(bottom) connected by a spring. On the right are 1Kg(top) and 2Kg(bottom) .The bottom right string is cut .

    2. Relevant equations


    3. The attempt at a solution


    Suppose T1 is the tension in string AB , T2 in string CD, T3 in EF and GH , T4(=kx) in the spring .

    Just after AB is cut T4 (i.e spring force) does not change . Tension in strings CD become T2' and in string EF and GH becomes T3' .

    Writing force balance equation for the two 1 Kg blocks and 2Kg block ,

    For the left 1Kg block , 6g - T3' = 1a

    For the right 1Kg block , T3' -T2' - 1g = 1a

    For the right 2Kg block , T2' - 2g = 2a

    Solving , I get T3' = 21g/4 , but this isn't an option .

    Please help me with the problem .

    Thanks
     

    Attached Files:

  2. jcsd
  3. Nov 30, 2015 #2

    haruspex

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    The form of the offered answer d strongly suggests a misprint. Why would they not specify it as 5g?
    I agree with your answer.
     
  4. Nov 30, 2015 #3
    Thanks . But the answer given is option c) .

    Option d) i.e 20g/4 is obtained under the assumption that the two blocks on the left (connected by the spring) have same acceleration just after the string is cut .

    Do you agree that just after the string is cut , the acceleration of 5Kg block on left is 0 whereas that of both the 1Kg blocks is 3g/4 ?
     
    Last edited: Nov 30, 2015
  5. Nov 30, 2015 #4

    haruspex

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    Yes, I agree with those accelerations.
    Doesn't that error give 4g?
     
  6. Nov 30, 2015 #5
    Yes , that gives 4g .

    Could you help me understand why tensions T2 and T3 should change when the string is cut ? Why doesn't T2 and T3 remain same just after the string is cut ?
     
  7. Nov 30, 2015 #6

    haruspex

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    Well, why would they? Masses are accelerating. If T3 stayed the same the top left mass would not move. T2 had to match the weight below it plus the tension in the string which has now been cut.
     
  8. Nov 30, 2015 #7

    PeroK

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    This problem to some extent exposes the assumptions you make when dealing with a "light, inextensible" string. If you think of the strings as slightly extensible (hence slightly stretched), then the behaviour of the string and spring should be the similar at the instant the system is released. Both should instantaneously retain their current tension until the system has had time to move.

    The assumption for the inextensible string, however, is that this change happens instantaneously (or, at least, in a very short time). Whereas, the assumption for the spring is that this takes time to happen (i.e. it's not instantaneous).
     
  9. Nov 30, 2015 #8
    Do you agree with the results in the OP and post#3 ?
     
  10. Nov 30, 2015 #9

    PeroK

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    @haruspex is never wrong!

    To add something, another way to do it is to calculate the acceleration of the 3 masses. The 5kg mass is instantaneously out of the equation and replaced by the 5g force in the spring. So:

    Net force on the 3-body system (down to the left): ##F = 5g + 1g - 3g = 3g##
    Total effective mass ##M = 4kg##.
    Acceleration of 3-body system ##a = 3g/4##

    Therefore:

    ##6g - T = 3g/4##
    ##T = 21g/4##
     
  11. Nov 30, 2015 #10
    I agree :smile:

    Nice !
     
  12. Nov 30, 2015 #11

    haruspex

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    Steady on there! Perhaps I do as well as the Captain of HMS Pinafore.
    What, never?
    No, never.
    What, never??
    Well, hardly ever!
     
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