Certain force to break surface tenson of Jelly

[Sam223344]

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?

 

[Andy Resnick]

This is exactly like a similar question posted elsewhere here... is this some sort of homework question?

 

[isabelle]

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.

 

[Sam223344]

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.

 

[Andy Resnick]

Jelly is not a viscous fluid. Have you found the mechanical properties?

 

[abbyfitz80]

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.

 

[The Mailman]

Canterbury University students aye. I am also stuck, any help is welcome.

 

[WhoWee]

Is the jelly in a container or acted upon by any other forces?

 

[The Mailman]

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.

 

[The Mailman]

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)..?

 

[the-matrix]

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?