1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Tensile stress, radius and Young's modulus

  1. Apr 28, 2013 #1
    1. The problem statement, all variables and given/known data
    An Aluminum cable of length 3.5m has 15,000 N tensile force acting on it if the wire is only allowed to be stretched by 1mm before it breaks,
    What must be the radius of the wire if the Young's modulus of Al is 6.9 x 10^10 N/m^2?

    I am also supposed to find the tensile stress, but I am sure I could find that myself after finding out the radius


    2. Relevant equations
    F/A or F/pi*R^2


    3. The attempt at a solution
    I haven't tried attempting the solution because I am not sure about the equation to use. We didn't learn about tensile stress in class or Young's modulus. I tried looking in my book and those are the equations I found, but most of the equations have a diameter or a radius already given.
     
  2. jcsd
  3. Apr 28, 2013 #2
    Did you read the section regarding Young's modulus or modulii of elasticity in your textbook? That will help you solve this problem.
     
  4. Apr 28, 2013 #3
    I did look in my book and I my old notes. I just found my old notes because this is my second time taking Physics I. I thought there might be something in there. The book also shows that F/A = delta L/L. In my notes it mentions that delta L is the change in length over the original length. So would I say that delta L is the original length plus 1 mm (that I have to change to m) over the 3.5m? Then multiply that by the Young's modulus and that would be tensile stress? Although I still don't know how to find out the radius.
     
  5. Apr 28, 2013 #4

    SteamKing

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper

    Delta represents the change. Do you think that Delta L represents L + 1 mm?
     
  6. Apr 28, 2013 #5
    That's what I am thinking, but I don't know. I don't really have much else to go on. There might be more to go on and I just am not seeing it.
     
  7. Apr 28, 2013 #6

    SteamKing

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper

    There's less going on. Delta L represents the change in the length L.
     
  8. Apr 28, 2013 #7
    Have you come across the equation [itex]Y=\frac{FL}{AΔL}[/itex]. If not, I suggest looking for it in your textbook or some other reliable book. This helps you solve the problem. (Y is the young's modulus here).

    And as Steam King pointed out, ΔL is the change in length that the wire can tolerate here.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: Tensile stress, radius and Young's modulus
Loading...