The discussion revolves around calculating the work done in lifting a regular tetrahedron-shaped object from water using integrals. The original poster presents parameters including the tetrahedron's side length, water depth, and material density, while expressing uncertainty about their approach and seeking feedback.
The discussion is ongoing, with participants providing guidance on using LaTeX for posting work and exploring different interpretations of the problem setup. There is no explicit consensus, but several productive directions have been suggested regarding the integration approach and the forces involved.
There are constraints regarding the format of submissions, as participants emphasize the importance of using LaTeX instead of images or PDFs. Additionally, the problem's requirement for integration is noted, though some participants suggest that standard results could be applied to simplify the process.
If you are talking about not having to integrate 1-(h/h₀)³, I’d be interested.Steve4Physics said:
I'll attempt to send it again.Frabjous said:If you are talking about not having to integrate 1-(h/h₀)³, I’d be interested.
It’s a work integral of a fully submerged volume. This is an incomplete thread because the discussion moved to a private conversation. I believe the OP is satisfied.Charles Link said:Maybe I misread the above posts, but IMO the buoyancy (from Archimedes principle) needs to be ## \int \rho g \,dV=\int \rho g A \, dh ##. and note: ## \rho ## is the density of the water, so it along with ## g ## can be in front of the integral.). post 26 is trying to do an integral of ## \int V \, dh ## , and that is incorrect.
The OP looks to me to still be doing an incorrect computation of the buoyant force.Frabjous said:It’s a work integral of a fully submerged volume. This is an incomplete thread because the discussion moved to a private conversation. I believe the OP is satisfied.
He’s calculating the work done by gravity and buoyancy of a fully submerged constant volume that is lifted height L-h.Charles Link said:The OP looks to me to still be doing an incorrect computation of the buoyant force.
Thank you=I just re-read the title of the post=the work to lift the object. I would expect to see ## W=\int F \cdot ds ##, but my mistake. :)Frabjous said:He’s calculating the work done by gravity and buoyancy of a fully submerged constant volume that is lifted height L-h.