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Statics of a particle, mechanics, F=ma

  1. Oct 13, 2016 #1

    Kajan thana

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    1. The problem statement, all variables and given/known data
    A smooth bead B is threaded on a light inextensible string. The ends of the string are attached to two fixed points A and C on the same horizontal level. The bead is held in equilibrium by horizontal force of magnitude 2N acting parallel to CA. The section of the string make angles of 60 and 30 with the horizontal.
    Find the tension in the string.

    2. Relevant equations
    Resolve the tension into X and Y component and then use F=Ma to solve it.

    3. The attempt at a solution
    Question 1) From the answer book, it says that the tension in the both string are same. Is it due to the word inextensible.
    Question 2) there is more than one way to approach this question, I tried to use Alternating angle as CA is parallel to 2N . Then Cos60=2/H and make the H the subject, I got 4 newtons, but this answer is wrong. Can someone please tell me if I had made any wrong assumption.

    Thank you in advance. Screen Shot 2016-10-13 at 23.17.35.png
     
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  3. Oct 13, 2016 #2

    BvU

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    Hi Kajan,
    well, a=0, so what did you find ? Show your work.
     
  4. Oct 13, 2016 #3

    Kajan thana

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    Thanks for the reply,

    When I do that way it gives me the right answer, but my question is why can I not use the alternating angle rule to find the tension? Have I made any wrong assumption?
     
  5. Oct 13, 2016 #4

    haruspex

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    It's more due to the word smooth. No string is cmpletely inextensible, so you should just think of it as having an arbitrarily high modulus. If the tensions were not equal then a tiny bit of string would slip through, barely changing the lengths, but equalising tension.
    Please explain how you get that.
     
  6. Oct 13, 2016 #5

    Kajan thana

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    Using alternating angle rule the angle of NBA is 60, then if we know the magnitude of adjacent is 2 and if we want to find out the hypotentuse then use H=A/cos theta same as 2/0.5=4
     
  7. Oct 13, 2016 #6

    haruspex

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    Why do you think that procedure will give the tension (which I presume is what H stands for)?
    H cosθ is the horizontal component of the tension in the left hand string. Why should that equal the applied force of 2N? They are acting in the same direction, and are opposed by the tension in the right hand portion of the string.
     
  8. Oct 14, 2016 #7

    Kajan thana

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    It makes sense now Haruspex.


    Thanks for your time.
     
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