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Young's Modulus Formula / Steel Cable

  1. Jul 26, 2016 #1
    A steel cable 3.00 cm 2 in cross-sectional area has a mass of
    2.40 kg per meter of length. If 500 m of the cable is hung
    over a vertical cliff, how much does the cable stretch under
    its own weight? Take Y steel ϭ 2.00 ϫ 10 11 N/m 2 .


    Y=([L][/o]F)/(A*delta_L)

    My attempt of solution: http://docdro.id/Fft5plu [Broken]

    This problem is driving me crazy, the textbook says the correct answer is 0.0490m. Can anyone tell me where i went wrong?


    NOTE: This problem was taken from the textbook "Physics for Scientists and Engineers with modern physics. 7h ed, page 360, problem 56"
     
    Last edited by a moderator: May 8, 2017
  2. jcsd
  3. Jul 26, 2016 #2

    RUber

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    Homework Helper

    I can't see your work, the link is no good for me.
    Here is the process:
    You are given ## 2\times10^{11}\frac{N}{m^2} = \frac{500*F}{A*\Delta L }. ##
    (note that area must be changed to square meters)
    Rewriting to solve for Delta L:
    ## \Delta L = \frac{500*F}{A*(2\times10^{11}) }. ##
    Force acts linearly on the cable, so it can be averaged to get
    ##F= 500/2 * 2.4*9.8(N).##
    So
    ## \Delta L = \frac{500*(F)}{A*(2\times10^{11}) }. ##
     
  4. Jul 26, 2016 #3
    When i plug in the data into your eq. i get 0.016333333 m which is not the right answer. The link seems to works for me, can anyone else test the link i gave? Thank you Ruber for trying.
     
  5. Jul 26, 2016 #4

    RUber

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    Are you using .0003 for area or .0001?
     
  6. Jul 26, 2016 #5
    Wow! you were right, i plugged in a wrong value, thank you RUber! Could you tell me why the force must be averaged? What i did is use F=M_tot*g=(2.4Kg/m)(500m)(9.8m/s2), but this seems to be wrong.
     
  7. Jul 26, 2016 #6
    To get the average stress, you need to use half the weight. The local tension in the cable is $$T=\gamma gz_0$$ where z0 is the (unstretched) distance measured up from the bottom. The local stress in the cable is $$\sigma=\frac{\gamma g z_0}{A}$$. The local strain in the cable is $$\epsilon=\frac{\gamma g z_0}{YA}$$ The total stretched length of the cable is $$L=\int_0^{L_0}\left(1+\frac{\gamma g z_0}{YA}\right)dz_0$$
     
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