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| 183_notes:youngs_modulus [2021/02/18 20:38] – [Hanging a mass from a platinum wire] stumptyl | 183_notes:youngs_modulus [2021/02/18 20:41] – [Determining the interatomic spring stiffness] stumptyl |
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| Earlier, you read how to [[183_notes:model_of_a_wire|add springs in parallel and in series]]. **In these notes, you will read about how the microscopic measurements of bond length and interatomic spring stiffness relate to macroscopic measures like [[http://en.wikipedia.org/wiki/Young's_modulus|Young's modulus]].** We will continue using Platinum wire as our example. | Earlier, you read how to [[183_notes:model_of_a_wire|add springs in parallel and in series]]. **In these notes, you will read about how the microscopic measurements of bond length and interatomic spring stiffness relate to macroscopic measures like [[http://en.wikipedia.org/wiki/Young's_modulus|Young's modulus]].** We will continue using Platinum wire as our example. |
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| ==== Hanging a mass from a platinum wire ==== | ===== Hanging a mass from a platinum wire ===== |
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| Consider a 2m long platinum wire ($L = 2m$) with a square cross section. That is, the wire is not "round" when viewed on end, but square. This wire is 1mm thick ($S = 1mm$); each side of the wire is 1mm. If you were to hang a 10kg weight, this 2m wire stretches by 1.166 mm ($s = 1.166mm$). | Consider a 2m long platinum wire ($L = 2m$) with a square cross section. That is, the wire is not "round" when viewed on end, but square. This wire is 1mm thick ($S = 1mm$); each side of the wire is 1mm. If you were to hang a 10kg weight, this 2m wire stretches by 1.166 mm ($s = 1.166mm$). |
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| === Determining the interatomic "spring stiffness" === | ==== Determining the interatomic "spring stiffness" ==== |
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| If you model the whole wire as a single spring, you can find the spring stiffness of the whole wire. From the [[183_notes:momentum_principle|momentum principle]] (momentum not changing), you can determine the this stiffness because the net force is zero. | If you model the whole wire as a single spring, you can find the spring stiffness of the whole wire. From the [[183_notes:momentum_principle|momentum principle]] (momentum not changing), you can determine the this stiffness because the net force is zero. |
| $$k_{s,wire} = \dfrac{mg}{s} = \dfrac{(10kg)(9.81 m/s^2)}{0.001166m} = 8.41\times10^4 N/m$$ | $$k_{s,wire} = \dfrac{mg}{s} = \dfrac{(10kg)(9.81 m/s^2)}{0.001166m} = 8.41\times10^4 N/m$$ |
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| This is very large spring constant because the wire (taken as a whole) is very stiff. | This is very large spring constant because the wire (taken as a whole) is very stiff. //Note: the units of N/m for k.// |
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| === Finding the number bonds in the wire === | ==== Finding the number bonds in the wire ==== |
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| [{{ 183_notes:mi3e_04-017.png?150|Using the model of a cubic lattice, you can determine the number of bonds in the wire and the number of bonds in a single chain.}}] | [{{ 183_notes:mi3e_04-017.png?150|Using the model of a cubic lattice, you can determine the number of bonds in the wire and the number of bonds in a single chain.}}] |