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Tension on string of submerged object.

  1. May 16, 2013 #1
    1. The problem statement, all variables and given/known data
    A hollow steel sphere of inner radius 0.9m and outer radius 1m is submeged 1000m below the surface of the sea. Take the density of water to be 1000 kg/m^3 and the density of steel to be 7.8 x 10^3. Calculate the tension in the wire to support the submerged sphere.


    2. Relevant equations
    B = ρ(f)V(f)g
    W = mg



    3. The attempt at a solution
    T + B = mg.
    Calculate T.
    Is this correct? The 1000m below sea level is what's making me doubt myself. Should this be factored into the calculation?
     
  2. jcsd
  3. May 16, 2013 #2

    rock.freak667

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    Your equation is correct however, I think the 1000 kg/m^3 for water they gave you would be at atmospheric pressure. You might need to get the density at 1000 m. However I don't think the value should vary by too much.
     
  4. May 16, 2013 #3
    How would you go about calculating the density at a depth of 1000m?
     
  5. May 16, 2013 #4

    SteamKing

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    Water is incompressible. The density of water at sea level is for all intents and purposes the same density at a depth of 1000 m.
     
  6. May 16, 2013 #5

    haruspex

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    According to http://en.wikipedia.org/wiki/Properties_of_water#Compressibility, density increase at 1km due to compression would be only about 0.5%. In practice, higher salinity would be more important. As against that, g would be a tiny bit less. I don't think you're expected to take any of that into account for this question, since it does not specify a salinity or pressure for the given density.
     
  7. May 16, 2013 #6
    Brilliant. Thanks for that. A little confused as to why the 1000 metres was actually specified so.
     
  8. May 16, 2013 #7

    haruspex

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    Providing irrelevant data in a question is a practice to be endorsed. Out in the real world, most available data are irrelevant, and recognising which are relevant is an important skill.
     
  9. May 16, 2013 #8
    That's true. Thanks for the explanations and help in general with that anyway.
     
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