- #1
samy4408
- 62
- 9
I found 2 formulas about surface tension -- which one is correct?
- Context: Undergrad
- Thread starter samy4408
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SUMMARY
The discussion focuses on two formulas for calculating surface tension: ##\gamma_{\text{disk}}=\dfrac{W}{C}## for a solid disk and ##\gamma_{\text{ring}}=\dfrac{W}{2C}## for a thin ring. The choice of formula depends on the specific scenario being analyzed. The formulas derive from the concept of surface tension as a force resulting from a uniform distribution of tiny parallel springs around the interface of an object and a fluid, with the spring constant represented by ##\gamma##. The ring's longer interface necessitates a lower spring constant to support the same weight as the disk.
PREREQUISITES- Understanding of basic physics concepts, particularly fluid mechanics.
- Familiarity with the mathematical representation of physical formulas.
- Knowledge of surface tension and its implications in fluid dynamics.
- Ability to interpret and manipulate variables in equations.
- Research the derivation of surface tension formulas in fluid mechanics.
- Explore the role of surface tension in different fluid interfaces.
- Study the applications of surface tension in engineering and material science.
- Learn about the effects of temperature and impurities on surface tension.
Students and professionals in physics, engineers working with fluid dynamics, and researchers interested in material properties and surface interactions will benefit from this discussion.
- #2
- 15,877
- 9,047
Can you provide the context or references of how these formulas were derived?
- #3
samy4408
- 62
- 9
I found the first one by typing surface tension formula on google , and the second :kuruman said:
- #4
- 15,877
- 9,047
Which you use depends on what question you wish to answer. Say you have a solid disk and a very thin ring both of circumference ##C## and weight ##W## floating in a fluid. In the case of the disk, ##\gamma_{\text{disk}}=\dfrac{W}{C}##; in the case of the ring, ##\gamma_{\text{ring}}=\dfrac{W}{2C}.##
You can imagine the surface tension force as the resultant of a uniform distribution of tiny parallel springs around the length of the interface of the object and the fluid with ##\gamma## playing the role of the spring constant. The ring has twice as long an interface (on the inside and outside) as the disk and therefore twice as many springs. Thus, half the spring constant is required to support the same weight.
You can imagine the surface tension force as the resultant of a uniform distribution of tiny parallel springs around the length of the interface of the object and the fluid with ##\gamma## playing the role of the spring constant. The ring has twice as long an interface (on the inside and outside) as the disk and therefore twice as many springs. Thus, half the spring constant is required to support the same weight.