% Homework07-F16 Madi -> Ahttps://www.webassign.net/web/Student/Assignment-Responses/submit?dep=140971 134Q11 Points Submissions Used 1/4 1/41/4 3/4 One mole of iron (6 x 1023 atoms) has a mass of 56 grams, and its density is 7.87 grams per cubic centimeter, so the center-to-center distance between atoms is 2.28 x 10-10 m. You have a long thin bar of iron, 2.9 m long, with a square cross section, 0.09 cm on a side You hang the rod vertically and attach a 84 kg mass to the bottom, and you observe that the bar becomes 1.47 cm longer. From these measurements, it is possible to determine the stiffness of one interatomic bond in iron. Top of wire Bottom of wire 1) What is the spring stiffness of the entire wire, considered as a single macroscopic (large scale), very stiff spring? ks = 156000 2) How many side-by-side atomic chains (long springs) are there in this wire? This is the same as the number of atoms on the bottom surface of the iron wire. Note that the cross-sectional area of one iron atom is (2.28 x 10-10)2 m2 Number of side-by-side long chains of atoms = 1211·558e13 N/m 3) How many interatomic bonds are there in one atomic chain running the length of the wire? Number of bonds in total length = 3 1272e10 4) What is the stiffness of a single interatomic \"spring\"? X N/m An interatomic bond in iron is stiffer than a slinky, but less stiff than a pogo stick. The stiffness of a single interatomic bond is very much smaller than the stiffness of the entire wire Additional Materials 12:15 PM 9/11/2016
The force applied to the bar is 84 kg = 84*9.81 = 824.04 N. The extension of the wire L = 0.0147 m, so the bar stiffness is F/L = 56057 N/m
The number of atoms in one layer of cross section is area of the bar divided by the area of one atom =0.0009²/(2.28*10^-10)² = 1.558*10^13.
No of bonds along the length is 2.9/(2.28*10^-10) = 1.272*10^10
The applied force is divided among all the bonds along the length, and then divided among all the atoms in a cross-section layer. The force applied to each atom in a layer is then 824.04/(1.558*10^13*1.272*10^10) = 4.16*10^-21 N
The strain (fractional length increase, L/L) on the bar is 0.0147/2.9 = 0.00507. This will also be the strain between the atom layers. The bond extension L is then 2.28*10^-10 * 0.00507 = 1.157*10^-12 m. The bond stiffness F/L = (4.16*10^-21)/(1.157*10^-12) = 3.59*10^-9 N/m
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