183_notes:model_of_a_wire

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183_notes:model_of_a_wire [2015/09/20 12:16] – [Modeling the solid wire] caballero183_notes:model_of_a_wire [2021/02/18 20:35] – [Modeling a Solid Wire with Springs] stumptyl
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-===== Modeling a Solid Wire with Springs ======+Section 4.4 and 4.5 in Matter and Interactions (4th edition) 
  
-To understand how solids exert different forces, you must learn how the microscopic, ball and spring model relates to more macroscopic measures such as elongation/compression and force. To do this, we will need to model the interatomic bond between two atoms in a cubic lattice a spring. In these notes, you will read about the relationship between microscopic physical quantities and macroscopic ones as they relate to the extension or compression of solid materials.+===== Modeling a Solid Wire with Springs ======
  
 +To understand how solids exert different forces, you must learn how the microscopic, ball and spring model relates to more macroscopic measures such as elongation/compression and force. To do this, we will need to model the interatomic bond between two atoms in a cubic lattice a spring. **In these notes, you will read about the relationship between microscopic physical quantities and macroscopic ones as they relate to the extension or compression of solid materials.
 +**
 ==== Lecture Video ==== ==== Lecture Video ====
  
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 $$\rho = 21.45 \dfrac{g}{cm^3} \left(\dfrac{1kg}{10^3g}\right)\left(\dfrac{100 cm}{1m}\right)^3 = 21.45 \times 10^3 kg/m^3$$ $$\rho = 21.45 \dfrac{g}{cm^3} \left(\dfrac{1kg}{10^3g}\right)\left(\dfrac{100 cm}{1m}\right)^3 = 21.45 \times 10^3 kg/m^3$$
  
-Consider a cubic meter of Pt, which is a cube that is 1m long on each side((This amount of Pt would costs nearly $1,000,000.)). You can use this to find the number of Pt atoms in a cubic meter:+Consider a cubic meter of Pt, which is a cube that is 1m long on each side((This amount of Pt would costs nearly $700,000,000.)). You can use this to find the number of Pt atoms in a cubic meter:
  
 $$21.45\times10^3 \dfrac{kg}{m^3} \left(\dfrac{1 mol}{0.195 kg}\right) \left(\dfrac{6.02\times10^{23} atoms}{1 mol}\right) = 6.62\times10^{28}\:\mathrm{atoms}$$ $$21.45\times10^3 \dfrac{kg}{m^3} \left(\dfrac{1 mol}{0.195 kg}\right) \left(\dfrac{6.02\times10^{23} atoms}{1 mol}\right) = 6.62\times10^{28}\:\mathrm{atoms}$$
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 === Two springs connected end-to-end (series) === === Two springs connected end-to-end (series) ===
  
-[{{183_notes:series.png?150|Two springs connected end-to-end. }}]+[{{183_notes:series.png?300|Two springs connected end-to-end. }}]
  
  
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 Let's consider attaching a 100N ball to two 100N/m springs where each spring is connected to the ball and not to each other. In this case, both springs must stretch by the same amount. If the ball hangs motionless (no change in momentum), we can use the momentum principle to determine how much these springs stretch. Let's consider attaching a 100N ball to two 100N/m springs where each spring is connected to the ball and not to each other. In this case, both springs must stretch by the same amount. If the ball hangs motionless (no change in momentum), we can use the momentum principle to determine how much these springs stretch.
 +
 +[{{183_notes:parallel.png?300|Two springs connected side-by-side. }}]
 +
  
 $$\Delta \vec{p} = 0\:\:\:\mathrm{implies}\:\:\:\vec{F}_{net} = \vec{F}_{grav} + \vec{F}_{spring,1} + \vec{F}_{spring,2} = 0$$ $$\Delta \vec{p} = 0\:\:\:\mathrm{implies}\:\:\:\vec{F}_{net} = \vec{F}_{grav} + \vec{F}_{spring,1} + \vec{F}_{spring,2} = 0$$
  • 183_notes/model_of_a_wire.txt
  • Last modified: 2021/03/13 19:40
  • by stumptyl