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Sag in steel wire

  1. Feb 12, 2015 #1
    1. The problem statement, all variables and given/known data:A steel wire of radius 0.4 × 10–3 m and length 1 m is tightly clamped between points A and B which are separated by 1 m and in the same horizontal plane. A mass is hung from the middle point of the wire such that the middle point sags by 2 cm from the original position.Compute the mass of body.young modulus of steel:20 × 10^10 N/m2 )
    Relevant equations:=T= F =YAl/L
    Attempt to a solution::
    Elongation occurs as shown in figure. AD = BD = 0.50 m
    upload_2015-2-13_1-24-31.png
    In triangle ADC, Let AC = x
    That's why x =√(50^2 + 2^2=√(2500 + 4)
    = √(2504) = 50.03 cm
    = 50.03 * 10^-2m
    Elongation,l = x – 0.50
    = 50.03 – 50 = 0.03 cm
    Resolve T as shown in figure
    2T cos (90 – theta) = mg
    2T sin (theta) = mg
    2T *2*10^-2 divided by x=mg
    2T *2*10^-2 divided by 50.03 *10^-2=mg
    2*Y*A* l divided by 50*10^-2*2
    as we know T= F =YAl/L

    2 *Y*pi r^2*0.03*10^-2 divided by 50*10^-2*2*divided by 50.03=m *9.8
    That's why]m=2*20*10^10*3.14*(0.4*10^-3)^2*0.03*2 divided by 50*50.03*9.8
    m=0.492 kg

     
  2. jcsd
  3. Feb 12, 2015 #2
    I want to understand the diagram.
     
  4. Feb 12, 2015 #3

    Bystander

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    You've got the solution, which suggests you do understand the diagram. What questions have you got, or what bothers you?
     
  5. Feb 12, 2015 #4

    NascentOxygen

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    The diagram shape is not drawn to an accurate scale, it is exaggerated, and this is okay if it makes the diagram easier to understand.

    A and B are nails in the wall, and a steel wire is tightly stretched between A and B. When a weight is attached to that wire's midpoint, the wire stretches and sags into a V-shape. Steel wire is like a spring---it can be stretched by a force, and will return to its original length when that force is removed. The length that any column of steel will stretch can be calculated if you know its length, its cross-sectional area, and its Youngs Modulus. Every material has its own elasticity characteristic value known as its Young's Modulus. You can read up on it here: http://en.m.wikipedia.org/wiki/Young's_modulus
     
  6. Feb 12, 2015 #5
    Here A=Area Which area should I take?According to the formula above we should take area on which restoring/deforming force acts.But I don't know on which area restoring/deforming force is acting.
     
    Last edited: Feb 12, 2015
  7. Feb 12, 2015 #6

    Bystander

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    "A" is the cross-sectional area of the wire. You can use the area of the unstretched wire since it's going to be reduced very little by the stretching, and going back to the stress-strain problem, it's not all that clear which area you actually should use without knowing a little more about how "Y" was determined (you'll recall that there were two ways that was shown, 1) for a starting cross-sectional area, and 2) for the area when it had been reduced by stretching).
     
  8. Feb 13, 2015 #7
    So,we should take cross sectional area while calculating stress,not the area on which restoring/deforming force acts.
     
  9. Feb 13, 2015 #8

    Bystander

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    Both. They're not going to be enough different to be noticeable. Ignore me when I start throwing too much extra information. All you need on this problem is the cross-sectional area of the wire.
     
  10. Feb 13, 2015 #9
    OK.
     
  11. Feb 13, 2015 #10
    Means the cross sectional area and the area on which force acts both are one and the same thing.
     
  12. Feb 13, 2015 #11

    Bystander

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  13. Feb 13, 2015 #12
    But how?I mean how tension force or restoring force acts on cross sectional area of wire?
     
  14. Feb 13, 2015 #13

    Bystander

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    The wire is the only thing being stretched, or allowed to relax to the horizontal under the influence of the restoring force if the weight is removed.
     
  15. Feb 13, 2015 #14
    So,that means restoring force acts on the wire ,and we denote this force in terms of per unit area.That's why we will take stress is equal; to force divided by cross sectional area of wire .Right,sir?
     
  16. Feb 13, 2015 #15

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    I think that's the way to do it, yes.
     
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