5· 0-5 points M14 4.6.039 My Notes A hanging wire made of an alloy of iron with

Question

5· 0-5 points M14 4.6.039 My Notes A hanging wire made of an alloy of iron with diameter 0.06 cm is initially 2.9 m long. When a 29 kg mass is hung from it, the wire stretches an amount 1.46 cm. A mole of iron has a mass of 56 grams, and its density is 7.87 g/cm3. Based on these experimental measurements, what is Young's modulus for this alloy of iron? 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 2.28 x 10-10 m for iron. As shown in the textbook, the micro quantity ks, (the stiffness of one interatomic bond) can be related to the macro property Y: s,i N/m Additional Materials Section 4.6

Explanation / Answer

Y = / where is the applied stress and is the resulting strain.

= F/A = m*g/(*d²/4) = 29*9.81/(*0.0006²/4) N/m²

= Lw/Lw = 0.0146/2.9

So:

Y = / = 2*10^11 N/m²

The applied force (29*9.8 N) is distributed among all the bonds, which equals the number of atoms in the wire. The number of atoms in the wire Nb = moles*6.022*10^23; moles = wire mass/molecular mass

wire mass = density*Lw*Aw

Nb = 6.022*10^23*2.9*(*0.0006²/4)*7870 / 56 = 6.94 x 10^19

The force on each bond Fb = 29*9.81/Nb = 4.1x10^-18

The bond strain is the same as the wire strain; the bond extension is

Lb = *Lb (Lb = bond length) = (0.0146/2.9)x(2.28x10^-10) = 1.15 x 10^-12

The bond stiffness is ks,i = Fb/Lb = (4.1x10^-18)/( 1.15 x 10^-12)

                                                                        = 3.57*10^-6 N/m

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