Calculating Buoyancy Force & Tension in Cable for 6m Spherical Chamber

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Homework Help Overview

The discussion revolves around calculating the buoyancy force and tension in a cable for a spherical undersea research chamber with a diameter of 6.0 m and a mass of 75200 kg. Participants are exploring the principles of buoyancy and forces acting on submerged objects.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • The original poster attempts to calculate the buoyant force using the volume of the sphere and the weight of the displaced water but expresses confusion over their results. Some participants suggest applying Newton's second law and question the assumptions made about the chamber's buoyancy when empty versus full.

Discussion Status

Participants are actively discussing the calculations and equations involved, with some providing formulas for buoyancy. There is acknowledgment of potential errors in the original poster's calculations, and a few participants are verifying the correctness of the equations used. Multiple interpretations of the problem are being explored, particularly regarding the radius and the forces acting on the chamber.

Contextual Notes

There is a mention of the chamber's behavior when empty versus full, raising questions about its buoyancy characteristics. Participants are also discussing the need for clarity on the radius calculation and the implications of the chamber's mass.

Adam Y.
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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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It's really simple.
a)F_ {APXIMN\Delta N\Sigma} = V_{sphere}\rho_{water} g [/tex]<br /> <br /> That&#039;s the scalar.<br /> <br /> For point b),apply Newton&#039;s second law for the sphere.How many forces act on the batiscaph?<br /> <br /> Daniel.
 
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 I am not getting the right number. Here is what I am getting. 1108353.888 Newtons.
 
dextercioby said:
F_ {APXIMN\Delta N\Sigma} = V_{sphere}\rho_{water} g [/tex]<br />
<br /> <br /> Dex why are you writing hieroglyphs for very simple formulas <img src="https://www.physicsforums.com/styles/physicsforums/xenforo/smilies/oldschool/bugeye.gif" class="smilie" loading="lazy" alt=":bugeye:" title="Bug Eye :bugeye:" data-shortname=":bugeye:" /> .<br /> <br /> Adam, 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. <br /> Right?
 
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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The problem I did like this used

<br /> \begin{multline*}<br /> \Sigma F_{y}=F_{b}-F_{t}-F_{g}\\F_{g}=ma<br /> \end{multline*}<br />

with gravity and tension in the negative y direction.
 
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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! :smile:

Daniel.
 

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