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On force in Young's modulus

  1. May 22, 2015 #1
    If a rod with cross section area A and length L is lying on a smooth table with one end being pulled by f newtons and other end by 2f newtons.
    What force I can use in calculating Young's Modulus and why?
    my doubt is whether it will be 3N or 1.5N and why?
    Stretched length is given.
  2. jcsd
  3. May 22, 2015 #2


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    Staff: Mentor

    It will be accelerated by a = F/m, not sitting still on the table, no?
  4. May 22, 2015 #3
    Thanks! I want to know value of force i.e., 1.5f or 3f.
  5. May 22, 2015 #4
  6. May 22, 2015 #5
    Then how can I calculate Young's modulus only under these given conditions.
  7. May 22, 2015 #6
    There is no physical condition with stable object by non zero external force. If I understand your problem.
  8. May 22, 2015 #7
    I need to find tensile stress on rod as I want to find elongation for calculating young's modulus
  9. May 22, 2015 #8


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    The tensile stress will vary throughout the length of the rod...density & some calculus is needed.
    Have you done any elongation or stress of a rotating rod type questions? This are similar.
  10. May 22, 2015 #9
    According to my understanding so far rod is moving here with accleration (2f-f) /m and it is also stretched having tensile stress f/2.
  11. May 22, 2015 #10
    Thanks billy! I have not done many problems as you asked.
  12. May 25, 2015 #11
    Stating same problem in a different wayhttps://mmi240.whatsapp.net/d/yywezFW6TcxsIWacjmME81VjKfY/AunnYXfkq2P8Blmy__wJQyRLQ4aL4z8kQXlZUs2wGVPB.jpg [Broken]
    Last edited by a moderator: May 7, 2017
  13. May 25, 2015 #12
    https://mmi130.whatsapp.net/d/l-CffzhTrS1R5PpPeDk30FVjLnU/Ard1QFFL7XB255nzwRl_0yw8Xxy2I-ChbVHnl1qLgWAY.jpg [Broken]
    Last edited by a moderator: May 7, 2017
  14. May 25, 2015 #13


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    No, as was stated earlier, you need to take into account that the tension varies throughout the rod.
  15. May 25, 2015 #14
    Then here in new case answer is F?
  16. May 26, 2015 #15


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    Imagine accelerating a long chain by pulling one end. Where is the chain most likely to break?
    The first link must support every other link. The last link supports only it's own self weight. Therefore, the stress & elongation in each link is not the same.
    Your problem looks at a chain with an opposing force on the last link.
    You need to analyse an infinitesimal section of your bar, set up an integral, find the integration constant from the boundary conditions, perform the integration over the length of the bar. (Or similar, It's been a while for me..)
    Every decent mechanics of materials textbook (I found Hibbeler good) will likely have multiple examples of the very similar rotating bar situation.
  17. May 26, 2015 #16
    Thanks!Why are you saying "very similar rotating bar". Specifically rotating.
    Also I want to know whether my question has a simple answer or the question is not complete or the question is irrelevant......(restated question)
  18. May 27, 2015 #17


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    They are very similar because they are analysed in a similar way.


    and exe 1.296 here

    There is enough information given to reach a meaningful answer. Whether it's simple or not would depend on the student.
  19. May 27, 2015 #18
    Thank you billy!
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