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A problem on elasticityhelp me !

  1. Mar 23, 2012 #1
    20p55oh.jpg
    A light rod is hung by two wires having same length of 2m and the same young's modulus of 8[itex]\times[/itex]1011Pa,but having different cross sectional areas of 1mm2 and 2mm2.
    The wire having the cross sectional area of 1mm2,gets an increment of 2mm in length because of a temperature increment.But the other wire remains same..
    A mass of 100kg is meant to be put on the rod to keep the rod horizontal..How much distance has it to be on the rod from point A...?

    PLEASE SOLVE THIS CITING REASONS...thanks !
     
  2. jcsd
  3. Mar 24, 2012 #2

    tiny-tim

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    welcome to pf!

    hi shalikadm! welcome to pf! :wink:

    show us what you've tried, and where you're stuck, and then we'll know how to help! :smile:
     
  4. Mar 24, 2012 #3
    Several things

    i want to know what the necessity that is to be fulfilled in order to keep it horizontally.
    whether it is to have same extension(e) for the two wires
    or to have the same total length(l+e) for both of the wires...
    or to fulfill e1mm2+2mm=e2mm2
    want to know whether to get the original length(l) of the 1mm2 wire as 2m or 2m+2mm

    Y=[itex]\frac{Fl}{Ae}[/itex]

    please help me...
     
  5. Mar 24, 2012 #4

    tiny-tim

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    after the temperature increment, one wire is longer

    apply the young's modulus equation to those lengths :smile:

    (of course, you need to extend both wires, to make them the same length)
     
  6. Mar 24, 2012 #5
    is it like this ?

    [itex]\downarrow[/itex]
    for the 1mm2wire,
    Y=[itex]\frac{T1*2.002}{1*10^-6*e1}[/itex][itex]\Rightarrow[/itex]1

    for the 2mm2wire,
    Y=[itex]\frac{T2*2}{2*10^-6*e2}[/itex][itex]\Rightarrow[/itex]2

    hence 1=2 ,
    [itex]\frac{T1*2.002}{1*10^-6*e1}[/itex]=[itex]\frac{T2*2}{2*10^-6*e2}[/itex][itex]\Rightarrow[/itex]A

    also,
    T1+T2=1000[itex]\Rightarrow[/itex]B

    and also ,
    taking moment around the center of 100kg's gravity[itex]\Rightarrow[/itex]C

    and then solving the A,B,C equations ?
    is that right ????
     
  7. Mar 24, 2012 #6

    tiny-tim

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    yes, but you'll also need an equation relating e1 and e2 :smile:

    (btw, please don't make the latex equations larger …

    it's not necessary …

    each reader can permanently adjust equation size by right-clicking on any equation and choosing "Scale All Math" :wink:)​
     
  8. Mar 24, 2012 #7
    how can i get a equation relating e1 and e2 ?
    sorry for larger Latex...:smile:
     
  9. Mar 24, 2012 #8

    tiny-tim

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    it'll be the equation saying the two wires are the same length :smile:
     
  10. Mar 24, 2012 #9
    Is it this ?(i have no idea)
    e1mm2+2mm=e2mm2
     
  11. Mar 24, 2012 #10

    tiny-tim

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    yes, so it's the same total length for both wires :smile:
     
  12. Mar 25, 2012 #11
    hope this help me...thanks a lot...!
     
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