Why a smaller spring will build up less force?

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Discussion Overview

The discussion revolves around the interpretation of the term "smaller spring" in the context of spring constants and their relationship to force. Participants explore the implications of this terminology in relation to analogies drawn between springs and capacitors, focusing on the definitions and physical characteristics that influence spring behavior.

Discussion Character

  • Debate/contested
  • Conceptual clarification
  • Exploratory

Main Points Raised

  • Some participants assert that a smaller spring refers to one with a smaller spring constant (K), which leads to less force being built up under the same conditions.
  • Others argue that cutting a spring in half typically doubles the spring constant, suggesting a misunderstanding of the term "smaller" as it relates to physical size versus spring constant.
  • There is a suggestion that "smaller" could refer to various physical attributes of the spring, such as coil spacing or wire thickness, which complicates the discussion.
  • Some participants emphasize the need for clarity in terminology, noting that the analogy between springs and capacitors may not hold in all contexts.
  • A few participants express that the analogy used in the article may lead to confusion and misinterpretation of the concepts being discussed.

Areas of Agreement / Disagreement

Participants do not reach a consensus on the interpretation of "smaller spring." There are multiple competing views regarding the meaning of the term and its implications for understanding spring behavior.

Contextual Notes

Limitations in the discussion include varying interpretations of "smaller," dependence on definitions of spring constants, and the potential for analogies to mislead rather than clarify the underlying physics.

harmyder
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While reading http://www.dartmouth.edu/~sullivan/22files/System_analogy_all.pdf article, i encountered the following sentence:

Thus, we see that K is analogous to 1/C , which makes sense, because with a given current for a given time, a smaller capacitor will build up more voltage, whereas with a given velocity for a given time, a smaller spring will build up less force.

I don't understand why smaller spring will build up less force? I know that F = K\int vdt and if v is the same so x is the same, and as K is the same, so force is the same.
 
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A smaller spring has a smaller K
 
Dale said:
A smaller spring has a smaller K
Why, usually if we cut spring in half, then the constant doubles.
 
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No, you misunderstand. In this case “smaller” is not referring to physical size. “Smaller” as it is used here directly means smaller K. Just like a “smaller” resistor means a lower resistance regardless of the fact that a lower resistance can be achieved by making it wider.
 
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"Smaller" could mean shorter, narrower coils or thinner wire. You would need to specify which meaning you are choosing before coming to a conclusion.
 
harmyder said:
Why, usually if we cut spring in half, then the constant doubles.

As Dale has stated, this is not the physical size! I can have two springs of identical size, and yet, they will have different spring constant k. This clearly indicates that k depends on many other factors.

Zz.
 
When talking about capacitors in a circuit it's common to use "large" and "small" to refer to the capacitance. You put small capacitors (nF or low uF) near integrated circuits for decoupling and large capacitors (high uF or mF) on the output of a power supply.
 
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Lord Crc said:
When talking about capacitors in a circuit it's common to use "large" and "small" to refer to the capacitance. You put small capacitors (nF or low uF) near integrated circuits for decoupling and large capacitors (high uF or mF) on the output of a power supply.
There is not a simple correspondence here. When we say a 'small Capacitor' we imply a 'small Capacitance'. Physical size is secondary until way down the line in circuit design. There are more design features in a spring that are of instant relevance.
Why not just bite the bullet and describe the spring in ways that allow a proper discussion of how it will perform in this launcher? Spring constant is a good place to start of course but length, compressibility (coil spacing) etc, are all very relevant here.
The electrical analogue is trivial in comparison with this project. How many of us have used spring calculations in a construction as often as calculations of circuits with capacitors in them?
 
Just goes to show sometimes analogies do more harm than good.
 
  • #10
I think "a smaller spring" in this article has no physical meaning except "a spring with a smaller k". It is just reading the equation, not trying to think about the material the spring is made of. It is comparing the role of k with the role of 1/C in the analogous equations.

The language is unfortunate.
 
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  • #11
Lord Crc said:
Just goes to show sometimes analogies do more harm than good.
Actually that goes to show that the analogy needs to be appropriate. There is a 'perfect' analogy where Force corresponds to PD and Charge corresponds to displacement. You can't just take two pairs of variables and expect them to exhibit the same rule.
 
  • #12
By smaller spring they mean a weaker spring. They hadn't considered that some people might interpret smaller spring to mean similar in construction but shorter.
 
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  • #13
Same old problem when a task is set without sufficient thought.
 
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