Discovering Lift Force in Case 2

AI Thread Summary
The discussion centers on understanding lift force in case 2, distinguishing it from buoyancy, which is defined by the volume of water displaced rather than the object's shape. It emphasizes that while lift force typically refers to dynamic situations like airplane wings, buoyancy pertains to static conditions. Calculating the resultant force in case 2 is possible, but the distribution of force across the submerged surface is complex, as each point experiences a normal force based on the pressure at that depth. The conversation clarifies that a force cannot equal pressure due to differing dimensions, and the correct relationship involves pressure multiplied by area. Overall, the thread highlights the nuances of lift and buoyancy in fluid dynamics.
yilbaris
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We all know the lift force in case 1, what about case 2 ?
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Note that there is also a force from the liquid pushing down on the additional wings you put in ...
 
Buoyant force depends only on the amount of water displaced, not the shape of the object displacing it.
 
yilbaris said:
We all know the lift force in case 1, what about case 2 ?View attachment 89160
If by "lift force", you actually mean "buoyancy", this can be calculated for case 2, although it is slightly more complicated than case 1.

Remember Archimedes' Principle:

https://en.wikipedia.org/wiki/Archimedes'_principle

Note: in technical use, "lift force" is usually reserved for a force which is created by a dynamic situation, like the lift generated by an airplane's wings when flying.
OTOH, "buoyancy" usually refers to a force created by a static situation, like the displacement of fluid by a floating object.
 
Thank you for answers,

? To calculate the resultant force is possible, but i wonder distribution of force. ?
 
Every point on the submerged surface of the object feels a normal force proportional to the pressure at that depth ## \rho g h ##
 
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MrAnchovy said:
Every point on the submerged surface of the object feels a normal force equal to the pressure at that depth ## \rho g h ##
A force can never equal a pressure, the dimensions are different. An area element is subjected to a normal force equal to the pressure multiplied by the area
 
Thank you all.

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Orodruin said:
A force can never equal a pressure, the dimensions are different. An area element is subjected to a normal force equal to the pressure multiplied by the area
I wanted to steer clear of elemental areas for the sake of clarity but I take your point; I have amended 'equal' to 'proportional'.
 
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