Boyancy Problem

  • Thread starter Adam Y.
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  • #1
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An undersea research chamber for aquanauts is spherical with an external diameter of 6.0 m. The mass of the chamber, when occupied, is 75200 kg. It is anchored to the sea bottom by a cable.
(a) What is the buoyant force on the chamber?
N
(b) What is the tension in the cable?
N
What in am doing wrong? I assumed it's just the volume of the tank and then you find the weight of the water displaced. It's not working. I made sure that the diameter was changed into a radius for the volume of the sphere. Also as a sanity note, if you have a tank that sinks when it's empty it should still sink if it's full of air right?
 
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Answers and Replies

  • #2
dextercioby
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It's really simple.
a)[itex] F_ {APXIMN\Delta N\Sigma} = V_{sphere}\rho_{water} g [/tex]

That's the scalar.

For point b),apply Newton's second law for the sphere.How many forces act on the batiscaph?

Daniel.
 
  • #3
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It's really simple.
Yeah which is why I hate it when I get these problems wrong.:) I know that has to be the answer. I can't figure out why Im not getting the right number. Here is what I am getting. 1108353.888 Newtons.
 
  • #4
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dextercioby said:
[itex] F_ {APXIMN\Delta N\Sigma} = V_{sphere}\rho_{water} g [/tex]

Dex why are you writing hieroglyphs for very simple formulas :bugeye: .

Adam, [tex]F_B = \frac{4}{3}\pi r^3\rho_{w}g = F_B = \frac{4}{3}\pi (6m)^3(1000kg/m^3)(9.8m/s^2) = 8.87 * 10^6 N[/tex].
Right?
 
  • #5
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Right?
Yeah. Your equation is right. Though don't you need to divide the diamter by two to get the radius? I checked neither of those two possible solutions are correct.
 
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  • #6
The problem I did like this used

[tex]
\begin{multline*}
\Sigma F_{y}=F_{b}-F_{t}-F_{g}\\F_{g}=ma
\end{multline*}
[/tex]

with gravity and tension in the negative y direction.
 
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  • #7
dextercioby
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Ramollari,the radius is 3m...:wink:

Hey,hieroglyphs are from ancient Egypt.Those were capital Greek letters...Hey,you are Greek!! :rofl:

Daniel.
 

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