Work calculation for lifting a Tetrahedron-shaped object from the water

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The discussion focuses on calculating the work done to lift a tetrahedron-shaped object from water using integrals. The problem is divided into three regions: fully submerged, partially submerged, and above the surface, with specific equations for work in each region. Participants emphasize the importance of using LaTeX for clarity and suggest that the buoyancy force should not be double-counted in the calculations. There are also suggestions to simplify the problem using known formulas for the tetrahedron's volume and center of gravity, which could eliminate the need for some integrals. Overall, the conversation aims to refine the calculations and clarify the integration process for accurate results.
  • #31
@Frabjous, I tried sending a PM (private message) to you (and @NODARman). But the system wouldn't let me send it to you. I know the problem requires a method using integration, but If you are interested in how to do it without any explicit integration, let me know.
 
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  • #32
Steve4Physics said:
@Frabjous, I tried sending a PM (private message) to you (and @NODARman). But the system wouldn't let me send it to you. I know the problem requires a method using integration, but If you are interested in how to do it without any explicit integration, let me know.
If you are talking about not having to integrate 1-(h/h₀)³, I’d be interested.
 
  • #33
Frabjous said:
If you are talking about not having to integrate 1-(h/h₀)³, I’d be interested.
I'll attempt to send it again.
 
  • #34
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.
 
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  • #35
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.
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.
 
  • #36
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.
The OP looks to me to still be doing an incorrect computation of the buoyant force.
 
  • #37
Charles Link said:
The OP looks to me to still be doing an incorrect computation of the buoyant force.
He’s calculating the work done by gravity and buoyancy of a fully submerged constant volume that is lifted height L-h.
 
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  • #38
Frabjous said:
He’s calculating the work done by gravity and buoyancy of a fully submerged constant volume that is lifted height L-h.
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. :)
 

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