The discussion focuses on calculating the work done in lifting a tetrahedron-shaped object from water using integrals. Key parameters include the tetrahedron side length "a," water depth "L," and the density of iron "rho (Fe)." The work is divided into three regions: fully submerged, partially submerged, and above the water surface, with specific equations provided for each region. The total work is expressed as a combination of gravitational and buoyancy forces integrated over the respective heights.
PREREQUISITESStudents and professionals in physics, engineering, and mathematics who are involved in fluid mechanics, particularly those working on problems related to buoyancy and work calculations in lifting submerged objects.
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.