Certain force to break surface tenson of Jelly

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SUMMARY

The discussion focuses on determining the maximum weight a structure can exert on jelly without breaking its surface tension. Participants confirm that the maximum weight is calculated by multiplying the surface tension (in N/m) by the length of the smallest side of the jelly's surface area. This method is applicable regardless of the jelly's shape, whether cubic or circular, as long as the appropriate dimensions are used. The conversation emphasizes the importance of understanding pressure and surface tension relationships in this context.

PREREQUISITES
  • Understanding of surface tension and its units (N/m)
  • Basic knowledge of pressure calculations (N/m²)
  • Familiarity with geometric properties of shapes (cubes, rectangles, circles)
  • Knowledge of fluid mechanics principles
NEXT STEPS
  • Research the relationship between surface tension and pressure in fluids
  • Explore calculations for maximum weight based on different shapes of jelly containers
  • Investigate the mechanical properties of jelly as a non-viscous fluid
  • Learn about experimental methods to measure surface tension accurately
USEFUL FOR

Students in engineering or physics, particularly those working on projects involving fluid mechanics and structural design, will benefit from this discussion.

Sam223344
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I am designing a structure that can rest upon jelly. I know how to determine the surface tension of the jelly in N/m. I can convert this to dynes/cm, J/m^2 etc but that is'nt the problem.
I need to figure out the maximum weight of the structure so that its force on the jelly does not break the surface tension. How do i do this?
 
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This is exactly like a similar question posted elsewhere here... is this some sort of homework question?
 
Multiply the surface tension by the length of one of the sides of the jello cube. If the surface is a jello rectangle, use the smaller of the two sides.
 
Can you explain why that gives the amount of weight the jelly can hold? It doesn't seem reasonable when I'm sure the same weight would break the surface of the jelly whether it was a 20cm cube or 100m cube.

I'm thinking it has something to do with the pressure that the structure exerts on the jelly surface which would depend on the weight and area of contact of the structure.
Ive done a lot of research into surface tension but I cannot find what I'm looking for. If anyone could provide some assistance it would be extremely helpful.
 
Jelly is not a viscous fluid. Have you found the mechanical properties?
 
I'm doing the same kind of thing as you and have been at dead ends every where to find my answers. How do you determine the surface tension to start with? And then how do you find the maximum weight that the jelly can actually hold? we are allowed a small footprint in our final structure.
 
Canterbury University students aye. I am also stuck, any help is welcome.
 
Is the jelly in a container or acted upon by any other forces?
 
Jelly is in a clear container, made up to a volume of 500ml. The constraints say that the jelly must have a depth, width and height of at LEAST one cm. So it is not acted on by other forces until the tower is added.
 
  • #10
Sam223344 said:
Can you explain why that gives the amount of weight the jelly can hold? It doesn't seem reasonable when I'm sure the same weight would break the surface of the jelly whether it was a 20cm cube or 100m cube.


I'm Guessing so that the units for pressure and surface tension are equal (i.e ST has units N/M and P has units N/M^2)..?
 
  • #11
isabelle said:
Multiply the surface tension by the length of one of the sides of the jello cube. If the surface is a jello rectangle, use the smaller of the two sides.

And what if the container is circular? Would you multiply it by its circumference?
 

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