1. Limited time only! Sign up for a free 30min personal 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!

Young's Modulus; Extension of a wire

  1. May 28, 2013 #1
    1. The problem statement, all variables and given/known data

    This is from A.P. French, Vibrations and Waves, Problem 3-7

    A wire of unstretched length l0 is extended by a distance of 10-3l0 when a certain mass is hung from its bottom end. If this same wire is [turned to be horizontal] and the same mass is hung from the midpoint of the wire [...], what is the depression y of the midpoint, and what is the tension in the wire?

    2. Relevant equations

    [itex]F = -\frac{AY}{l_0}*x[/itex]

    [itex]F = k*x[/itex]

    [itex]k = -\frac{AY}{l_0}[/itex]

    3. The attempt at a solution

    The correct answer is y = l0 / 20
    and T = 5 * mg

    I have not recreated these solutions.

    I have two triangles to consider:

    one of lengths, with base l0/2, height y and hypotenuse l0 + x

    one of Forces, with base Ti, height mg/2 and hypotenuse T = -\frac{AY}{l_0}*x

    I've tried a lot of things. I'm missing something fundamental.
    If you can get the correct answer, I would appreciate a little direction.
     
  2. jcsd
  3. May 29, 2013 #2
    Wire extension

    Does this make sense?
    At this stage my answer seem to be out by a factor 2.
     

    Attached Files:

    Last edited: May 29, 2013
  4. May 29, 2013 #3
    For forces,

    how did you get kl0/2000 for the horizontal aspect?
     
  5. May 29, 2013 #4
    From the conditions on the vertical orientation of the wire, you get that

    YA = 1000 mg

    Let y represent the distance that the center of the wire moves down. Then, from the pythagorean theorem,
    [tex]\left( \frac{l}{2}\right) ^2=\left( \frac{l_0}{2}\right) ^2+y^2[/tex]
    Solving for l/2:
    [tex]\left( \frac{l}{2}\right) =\sqrt{\left( \frac{l_0}{2}\right) ^2+y^2}≈\left( \frac{l_0}{2}\right)+\frac{y^2}{l_0}[/tex]
    From this, [itex]strain=\frac{l/2-l_0/2}{l_0/2}=2\left(\frac{y}{l_0}\right)^2[/itex]

    Let T represent the tension in the wire. Force balance on hanging weight:
    [tex]2T\frac{y}{l_0/2}=mg[/tex]

    How is the tension related to Y, A, and the strain in the wire?
     
  6. May 30, 2013 #5
    1/2 mg = k lo /1000
     
  7. May 30, 2013 #6
    This is incorrect. We already showed before that YA=kl0=1000 mg, based on the weight being in equilibrium when the weight is hung from the wire vertically. The question I was asking was, what is the general relation between the tension and strain, to wit:

    T = YA (strain) = kl0 (strain)

    In the case of the horizontal wire, this gives

    T = YA (2 (y/l0)^2)=1000mg (2 (y/l0)^2)

    See what you get when you substitute this into the force balance.
     
  8. May 30, 2013 #7
    Typo, I wanted to say

    1/2 mg = k lo /2000

    This I concluded from the information given in the initial problem statement.
    The 1/2 factor comes from the fact that only half of the magnitude of the weight is used for the base of the inverted right angled triangle in the force diagram. Do you agree with that?
     
  9. May 30, 2013 #8
    No. If this is the force balance for the weight hanging from the horizontal wire, then it is incorrect. From the free body diagram, if T is the tension in the wire, you have to determine the component of T in the vertical direction (see the force balance equation that I gave in my earlier post). Also, the strain and tension in the horizontal wire are not the same as in the vertical wire.

    Chet
     
  10. May 30, 2013 #9
    I used a vector diagram for the forces turned through 90o
    where W is the weight and T the tensions in the wire.
    One can then use the fact that the two triangles are congruent to compare the ratio of the sides.
     

    Attached Files:

    Last edited: May 30, 2013
  11. May 30, 2013 #10
    This diagram is not what is described in the problem statement (for which the book solution and my solution apply). Think of a long horizontal clothesline from which you hang a small weight at the center.
     
  12. May 30, 2013 #11
    Taylor expansions need to be more popular in my repertoire.
    And using sinx = tanx was also wise.
    These were stumbling points - certainly.

    Using your approximations, I can easily calculate the Tension if I assume y = l0/20
    However, I'm still having problems calculating y.

    May I have a less thorough suggestion?
     
  13. May 30, 2013 #12
    You are basically dealing with 3 key types of equations in this analysis:

    1. The Strain-Displacement Equation
    2. The Force Equilibrium Equation
    3. The Stress Strain Equation

    In the end, you will combine these three equations to get the solution to your problem.

    Strain-Displacement Equation:
    The Strain-Displacement Equation is a geometric equation which, in this problem, expresses the tensile strain ε in the wire as a function of the downward displacement at the center of the wire y. We already derived this relationship, and it is given by:
    [tex]\epsilon=2\left(\frac{y}{l_0}\right)^2[/tex]
    Force-Equilibrium Equation:
    This is the force balance on the weight, and expresses the tension in the wire T as a function of the displacement y. We already derived this relationship, and it is given by:
    [tex]2T\left(\frac{y}{l_0/2}\right)=mg[/tex]
    Stress Strain Equation:
    This is basically Hooke's law, and expresses the tension in the wire as a function of the tensile strain in the wire:
    [tex]T=YA\epsilon[/tex]
    But, in Part I of the problem, we showed that, [itex]YA=1000mg[/itex]
    Therefore, for this problem, the Hooke's law relationship becomes:
    [tex]T=1000mg\epsilon[/tex]

    Now, all you need to do is combine these three equations:
    [tex]\epsilon=2\left(\frac{y}{l_0}\right)^2[/tex]
    [tex]2T\left(\frac{y}{l_0/2}\right)=mg[/tex]
    [tex]T=1000mg\epsilon[/tex]

    You do this by eliminating the tension T and the strain ε from the equations. See what you get when you do this.
     
  14. May 30, 2013 #13
    Damn.

    I appreciate all the help. I've arrived at the right answer.
    But, damn.

    I did not do well on this question.

    Thanks mate
     
  15. May 30, 2013 #14
    Hey AJ. Don't feel bad. I've had lots of practical experience with solid deformational mechanics. In my judgement, this was not an Introductory Physics problem. Just hang in there.

    Chet
     
  16. Aug 9, 2013 #15
    Thank you so much for this explanation. Something that comes to mind:
    How did you know to make this approximation?
    [tex]\sqrt{\left( \frac{l_0}{2}\right) ^2+y^2}≈\left( \frac{l_0}{2}\right)+\frac{y^2}{l_0}[/tex]
     
  17. Aug 10, 2013 #16
    Hi MBIGRAS. Welcome to Physics Forums.

    Answer to your question: From experience with clotheslines, I just knew that y was going to be small compared to l0/2. So, I just used the binomial expansion (a+b)n with n = 1/2 and a = (l0/2)2.

    Chet
     
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: Young's Modulus; Extension of a wire
Loading...