Calculate the bouyancy of two spheres joined by a rope submerged in seawater

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Homework Help Overview

The discussion revolves around the buoyancy of two spheres connected by a rope, submerged in seawater. The spheres have different densities, with the lower sphere made of magnesium. The problem involves analyzing their behavior when submerged and the effects of cutting the rope.

Discussion Character

  • Exploratory, Conceptual clarification, Problem interpretation, Assumption checking

Approaches and Questions Raised

  • Participants explore the equations of motion (EOM) for both the sinking and rising spheres, questioning the assumptions about the densities and equilibrium of forces acting on the spheres. There is an emphasis on clarifying symbols and setting up equations for different scenarios.

Discussion Status

The discussion is ongoing, with participants seeking to clarify the problem's setup and the forces involved. Some guidance has been offered regarding the need for clear symbols and equations, but no consensus has been reached on the interpretation of the problem.

Contextual Notes

There is a lack of specific numerical values for radius, velocity, mass, or volume, which participants note may complicate the analysis. The original poster expresses uncertainty about the problem's validity, suggesting it may be more complex than initially perceived.

jnuz73hbn
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Homework Statement
Spheres in total 2, of equal volume are connected with a ship's rope (is not very long, nevertheless of unknown length, the rope is weightless) in sea water (Rho= 1200kg/m3). Both spheres sink down with constant speed. The lower sphere consists of magnesium (Rho= 1750 kg/m3). Underwater, the ship's rope is now cut. The upper sphere K1 moves with the same constant velocity (as it sank together with the lower sphere) v upwards again. We have not given the radius, the velocity, the mass or the value of the volume, how can we now find out, for example, what the density of the upper sphere is?
Relevant Equations
Fb= rho* g*h
m * g = mAl * g
V * ρ * g = VAl * ρAl * g
V * ρ * g = V * ρAl * g
ρ = ρAl
this does not work at all, because the upper ball must have a density smaller than that of seawater 1200kg/m3 or not?
 
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jnuz73hbn said:
Homework Statement: Spheres in total 2, of equal volume are connected with a ship's rope (is not very long, nevertheless of unknown length, the rope is weightless) in sea water (Rho= 1200kg/m3). Both spheres sink down with constant speed. The lower sphere consists of magnesium (Rho= 1750 kg/m3). Underwater, the ship's rope is now cut. The upper sphere K1 moves with the same constant velocity (as it sank together with the lower sphere) v upwards again. We have not given the radius, the velocity, the mass or the value of the volume, how can we now find out, for example, what the density of the upper sphere is?
Relevant Equations: Fb= rho* g*h

m * g = mAl * g
V * ρ * g = VAl * ρAl * g
V * ρ * g = V * ρAl * g
ρ = ρAl
this does not work at all, because the upper ball must have a density smaller than that of seawater 1200kg/m3 or not?
I think something is off with this question, but maybe I'm mistaken.

What do you get for the EOM of the two sinking spheres ( constant velocity ##v## ), and the EOM for the sphere that rising at ##v##?
 
erobz said:
I think something is off with this question, but maybe I'm mistaken.

What do you get for the EOM of the two sinking spheres ( constant velocity ##v## ), and the EOM for the sphere that rising at ##v##?
The balls are in equilibrium of forces
 
jnuz73hbn said:
The balls are in equilibrium of forces
Correct. So what is the EOM that describes the two balls sinking?
 
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Please explain your symbols. You have two situations:

1. balls connected and sinking
2. balls disconnected and one sinking, one rising

set up equations with clear symbols for each ball & case.Use ##\LaTeX##
1690327459942.png
 
jnuz73hbn said:
The balls are in equilibrium of forces
Right, but what are those forces?
 
erobz said:
I think something is off with this question, but maybe I'm mistaken.
@jnuz73hbn Just so you know the problem is ok, but it's a little sneakier than I originally suspected.
 

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