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183_notes:model_of_a_wire [2015/09/19 11:17] – [Modeling the interatomic bond as spring] caballero | 183_notes:model_of_a_wire [2021/03/13 19:40] (current) – [Modeling the interatomic bond as spring] 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, | + | ===== Modeling |
+ | To understand how solids exert different forces, you must learn how the microscopic, | ||
+ | ** | ||
==== Lecture Video ==== | ==== Lecture Video ==== | ||
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==== Modeling the interatomic bond as spring ==== | ==== Modeling the interatomic bond as spring ==== | ||
- | [{{ 183_notes:mi3e_04-011.png? | + | [{{ 183_notes:lattice_cube_spheres.png? |
To model the interatomic bond as a spring, we will need to first determine how " | To model the interatomic bond as a spring, we will need to first determine how " | ||
$$\rho = 21.45 \dfrac{g}{cm^3} \left(\dfrac{1kg}{10^3g}\right)\left(\dfrac{100 cm}{1m}\right)^3 = 21.45 \times 10^3 kg/m^3$$ | $$\rho = 21.45 \dfrac{g}{cm^3} \left(\dfrac{1kg}{10^3g}\right)\left(\dfrac{100 cm}{1m}\right)^3 = 21.45 \times 10^3 kg/m^3$$ | ||
- | 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, | + | 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, |
$$21.45\times10^3 \dfrac{kg}{m^3} \left(\dfrac{1 mol}{0.195 kg}\right) \left(\dfrac{6.02\times10^{23} atoms}{1 mol}\right) = 6.62\times10^{28}\: | $$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}\: | ||
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This is roughly the distance between atoms in a solid piece of Pt. We think of this as the " | This is roughly the distance between atoms in a solid piece of Pt. We think of this as the " | ||
- | ==== Modeling the solid wire ==== | + | ====== Modeling the solid wire ====== |
[{{ 183_notes: | [{{ 183_notes: | ||
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The simplest model we can use for a wire (beyond a single atomic chain), is to model it as many long parallel chains connected by springs. For you to understand this model, you will need to understand how to model two springs connected end-to-end (in series) and two springs connected side-by-side (in parallel). | The simplest model we can use for a wire (beyond a single atomic chain), is to model it as many long parallel chains connected by springs. For you to understand this model, you will need to understand how to model two springs connected end-to-end (in series) and two springs connected side-by-side (in parallel). | ||
- | === Two springs connected end-to-end (series) === | + | ===== Two springs connected end-to-end (series) ===== |
+ | |||
+ | [{{183_notes: | ||
Let's consider attaching a 100N ball to a single 100N/m spring. If we let the weight just hang motionless (no change in momentum), we know from the [[183_notes: | Let's consider attaching a 100N ball to a single 100N/m spring. If we let the weight just hang motionless (no change in momentum), we know from the [[183_notes: | ||
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$$s = \dfrac{mg}{k} = \dfrac{100N}{100N/ | $$s = \dfrac{mg}{k} = \dfrac{100N}{100N/ | ||
- | When we attach | + | Hence, |
- | == Modeling two end-to-end springs as one spring (effective spring constant) == | + | When we attach a second 100N/m spring to the end of the first and then attach the mass, both springs will stretch 1m. That is, the overall stretch of the spring-mass system is 2m. |
+ | |||
+ | //In series, each spring stretches as if the mass were attached to just that spring//, and the sum of all those stretches gives the overall stretch. | ||
+ | |||
+ | ==== Modeling two end-to-end springs as one spring (effective spring constant) | ||
You will often find it useful to consider a whole chain of springs as one spring. That is, you can model springs attached end-to-end as one hypothetical spring that stretches the same amount under the same load as the chain of springs. This is called using an effective spring constant. It's // | You will often find it useful to consider a whole chain of springs as one spring. That is, you can model springs attached end-to-end as one hypothetical spring that stretches the same amount under the same load as the chain of springs. This is called using an effective spring constant. It's // | ||
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$$k_{s,eff} = 50 N/m$$ | $$k_{s,eff} = 50 N/m$$ | ||
- | === Two springs side-by-side (parallel) === | + | ===== Two springs side-by-side (parallel) |
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: | ||
+ | |||
$$\Delta \vec{p} = 0\: | $$\Delta \vec{p} = 0\: | ||
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$$s = \dfrac{mg}{2k} = \dfrac{100N}{200N/ | $$s = \dfrac{mg}{2k} = \dfrac{100N}{200N/ | ||
- | When we attach a second 100N/m spring to the ball, the springs both stretch 0.5m. That is, the overall stretch of the spring-mass system is half of what it is with one spring. //In parallel, each spring stretches the same amount//. | + | When we attach a second 100N/m spring to the ball, the springs both stretch 0.5m. That is, the overall stretch of the spring-mass system is half of what it is with one spring. |
- | == Modeling two side-by-side springs as one spring (effective spring constant) == | + | //In parallel, each spring stretches the same amount//. |
+ | |||
+ | ==== Modeling two side-by-side springs as one spring (effective spring constant) == | ||
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$${k_{s, | $${k_{s, | ||
+ | |||
+ | This way of modeling end-to-end and side-by-side springs will be very useful for modeling [[183_notes: |