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A hanging wire made of an alloy of nickel with diameter 0.07 cm is initially 2.7 m long. When a 43 kg mass is hung from it, the wire stretches an amount 1.48 cm. A mole of nickel has a mass of 59 grams, and its density is 8.9 g/cm3. Based on these experimental measurements, what is Young's modulus for this alloy of nickel? Y = N/m2 As you've done before, from the mass of one mole and the density you can find the length of the interatomic bond (diameter of one atom). This is 2.22 10-10 m for nickel. As shown in the textbook, the micro quantity ks,i (the stiffness of one interatomic bond) can be related to the macro property Y: ks,i = N/m

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Answer:

Explanation:

  • Given that d = 0.07cm = 0.0007m, l = 2.7m, extension = 1.48cm = 0.0148m, mass = 43kg
  • force = 43 x 9.81 = 421.83N, Area = Pi x d^2/4 = 3.142 x 0.0007^2/4 = 3.849x10^-7m2
  • Calculate the stress = Force/Area = 421.83N / 3.849x10^-7m2

= 1095961236N/m^2

  • calculate the strain = extension/original length
  • = 0.0148 / 2.7 = 0.005481

  • Young modulus of elasticity = Stress/Strain = 1095961236N/m^2 / 0.005481 = 1.99 x 10^11Pa = Young's modulus for this alloy of nickel

  • Given the inter-atomic bond length = 2.22 10-10 m
  • To Calculate (the stiffness of one inter-atomic bond = Ks,i =Y x d
  • Ks,i = 1.99 x 10^11 x 2.22 10-10
  • Ks,i = 4.42 x 10 = 44.18N/m ; the stiffness of one inter-atomic bond
pristina

The Young's Modulus of the nickel alloy is obtained as [tex]1.9\times 10^{11}\,N/m^2[/tex].

The interatomic bond stiffness of nicket is obtained as 42.18 N/m.

Stress and Strain

Given that the diameter of the wire is, [tex]d = 0.07\,cm = 7\times 10^{-4}\,m[/tex].

So, the radius of the wire is, [tex]r=\frac{d}{2}= \frac{7\times 10^{-4}\,m}{2} =3.5\times 10^{-4}\,m[/tex].

The mass hung from the wire is, [tex]m=43\,kg[/tex].

The force exerted by the mass on the wire is the gravitational force on the mass. i.e.;

[tex]F = mg = 43\,kg \times 9.8\,m/s^2=421.4\,N[/tex]

The stress on the wire is given by,

[tex]\sigma =\frac{F}{A} = \frac{421.4\,N}{\pi \times (3.5 \times 10^{-4})^2 }=1.094\times 10^9\,N/m^2[/tex]

Initial length of the wire is, [tex]L =2.7\,m[/tex].

Extension of the wire is, [tex]\Delta L= 1.48\,cm =1.48\times 10^{-2}\,m[/tex]

Therefore the strain is given by;

[tex]\epsilon =\frac{\Delta L}{L}=\frac{1.48\times 10^{-2}\,m}{2.7\,m} =5.48\times 10^{-3}[/tex]

Now, the Young's Modulus can be given by;

[tex]Y=\frac{\sigma}{\epsilon} =\frac{1.09\times 10^9 \,N/m^2}{5.48\times 10^{-3}} =1.9\times 10^{11}\,N/m^2[/tex]

Given that the interatomic bond length is;

[tex]D= 2.22\times 10^{-10}\,m[/tex]

Therefore, the interatomic spring stiffness is given by;

[tex]k_{s,\, interatomic}=Y\times D=Y =(1.9\times 10^{11}\,N/m^2)\times (2.22\times 10^{-10}\,m)=42.18\,N/m[/tex]

Learn more about stress and strain:

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