# Tension on string of submerged object.

1. May 16, 2013

### SherlockOhms

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. May 16, 2013

### rock.freak667

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.

3. May 16, 2013

### SherlockOhms

How would you go about calculating the density at a depth of 1000m?

4. May 16, 2013

### SteamKing

Staff Emeritus
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.

5. May 16, 2013

### haruspex

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.

6. May 16, 2013

### SherlockOhms

Brilliant. Thanks for that. A little confused as to why the 1000 metres was actually specified so.

7. May 16, 2013

### haruspex

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.

8. May 16, 2013

### SherlockOhms

That's true. Thanks for the explanations and help in general with that anyway.