I Archimedes' principle and the column of water

  • I
  • Thread starter Thread starter abrek
  • Start date Start date
  • Tags Tags
    Phisics
AI Thread Summary
The discussion centers on whether the volume of displaced water will equal the volume of the smaller container when it is lowered into a larger container of water. Key points include the impact of the height of the water column on resistance and the incompressibility of water, which may affect the volume released. The mechanics of the movable platform and the pressure exerted by the 2-meter water column are critical to understanding the system's behavior. Clarification is needed regarding the mass of the piston and its ability to maintain the column height. Overall, the outcome hinges on the interplay of pressure and displacement in the described setup.
abrek
Messages
14
Reaction score
1
TL;DR Summary
If go
If you assemble the structure shown in the picture into a large container of water, lower the smaller container onto a special movable platform (red). Will the volume of displaced water be equal to the volume of the smaller container (M) according to Archimedes’ principle, or will the fact that the release of liquid occurs at a height, the water column will create resistance and less water will come out in volume?
 

Attachments

  • Без названия1_20240505173630.png
    Без названия1_20240505173630.png
    5.8 KB · Views: 54
  • Без названия1_20240505174108.png
    Без названия1_20240505174108.png
    6.5 KB · Views: 60
  • Без названия1_20240505173630.png
    Без названия1_20240505173630.png
    5.8 KB · Views: 65
  • Без названия1_20240505174108.png
    Без названия1_20240505174108.png
    6.5 KB · Views: 49
Physics news on Phys.org
The question has nothing to do with the Archimedes principle. It has to do with the incompressibility of water.
 
abrek said:
.. will the fact that the release of liquid occurs at a height, the water column will create resistance and less water will come out in volume?
What do you think, and why?

Note that the "special movable platform (red)" shown in the initial condition must be physically restricted by its cylinder from moving up due to the static pressure created by the 2-meter column.
 
^^^^... as well as in the final condition.
 
I think the far left hand diagram needs some explanation. Does the red (lets call it a) piston have precisely the correct mass to keep the vertical column at 2m or is it restrained from rising?; this must be specified before going any further.

When M is lowered onto the piston, it will add a force downwards on the piston and water will be pushed out of the tube. The pressure at the bottom of the tube will always be 2m's worth whilst the tube is full.

I really don't know where this is going without more information.
 
Thread 'Gauss' law seems to imply instantaneous electric field'
Imagine a charged sphere at the origin connected through an open switch to a vertical grounded wire. We wish to find an expression for the horizontal component of the electric field at a distance ##\mathbf{r}## from the sphere as it discharges. By using the Lorenz gauge condition: $$\nabla \cdot \mathbf{A} + \frac{1}{c^2}\frac{\partial \phi}{\partial t}=0\tag{1}$$ we find the following retarded solutions to the Maxwell equations If we assume that...
Hello! Let's say I have a cavity resonant at 10 GHz with a Q factor of 1000. Given the Lorentzian shape of the cavity, I can also drive the cavity at, say 100 MHz. Of course the response will be very very weak, but non-zero given that the Loretzian shape never really reaches zero. I am trying to understand how are the magnetic and electric field distributions of the field at 100 MHz relative to the ones at 10 GHz? In particular, if inside the cavity I have some structure, such as 2 plates...
Back
Top