How Does Submarine Buoyancy Affect Its Movement and Stability?

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SUMMARY

The discussion focuses on the principles of buoyancy as they relate to a 3000m³ submarine filled with air and submerged in seawater. Key equations include the buoyancy equation: Buoyancy = density × Volume × Gravity, and the weight equation: weight = buoyancy. Participants calculated the mass of the submarine as 4,203,870 kg and the apparent weight as 41,197,926 kg. The discussion also addresses the effects of filling the submarine with water, reducing the air volume by 75%, and the resulting acceleration and travel speed at 1000 feet.

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  • Understanding of buoyancy principles and Archimedes' principle
  • Familiarity with basic physics equations involving mass, weight, and acceleration
  • Knowledge of fluid density, specifically seawater and air densities
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Bouyancy

Question:
A 3000m^3 submarine of mass, m, is filled with air(density air =1.29kg/m^3) and is floating, submergerd, in seawater(density = 1400kg/m^3)
a) draw diagram.

b) Write an equation that that equates weight of the sub to the buoyant force( include weight of air)?


c) Use the equation to calculate the mass of the sub?


d) what's the apparent weight of the sub?

e)Sub fills with water to reduce the volume of air by 75%, write the equations and find the acceleration? is it B-mg=ma? or am i missing something?


f) how fast is it traveling at 1000 feet?

***thanks in advance!

Relevant Equations:
Bouyancy = density X Volume X Gravity

Attempts at Question:
A)-----i did that-------
B)W=B so i got mg=(bouyancy of sub in water) X (bouyancy of air on top of sub)
C)Then i figured m=V(Pw + Pa) where Pw is density of water and Pa is density of air.
so, m=4203870 kg?
D) i suppose that's just mass x gravity
so, weight= 41197926 kg?
E)This is the part that mostly confuses me, i said Fnet= Bouyancy-Weight =ma
but i feel like somethings missing and I am not sure what the bouyancy would be since there is air still in it.
F)Once i get here i suppose i just use Vf^2=Vi^2 + a(Xf-Xi) ?
 
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The question is not very clear.
It doesn't say whether the mass of the sub includes the air in it (probably not because of the "include weight of air" in (b).
The air is inside the sub, adding to its mass if that assumption is correct, so (b) would be
weight = buoyancy
(m + Pa*V)g = Pw*V*g
However, this assumes that the entire volume of the sub is air, which is quite impossible. A large part of the volume will be steel, not air.

For (e) I would write mg - buoyancy = ma (taking down to be positive)
and I would calculate m as the answer to (c) minus the mass of .75*V*Pa + .75*V*Pw to account for 75% of the air being replaced by water.

Good luck with this!
 
Ya, several people i talked to have a problem with this question because it seems so vague. But what i think the question is asking for is basics of this problem, not accounting for everything, which doesn't make much sense in the end.
 

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