Tension on string of submerged object.

In summary, the problem involves calculating the tension in a wire supporting a submerged hollow steel sphere at a depth of 1000m below the sea surface. Using the equations B = ρ(f)V(f)g and W = mg, the density of water and steel, and the fact that water is incompressible, the density at a depth of 1000m can be assumed to be the same as at sea level. Therefore, there is no need to factor in the depth in the calculation. However, the given density of water may need to be adjusted for salinity, but this is not specified in the problem.
  • #1
SherlockOhms
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Homework Statement


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.


Homework Equations


B = ρ(f)V(f)g
W = mg



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?
 
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  • #2
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
How would you go about calculating the density at a depth of 1000m?
 
  • #4
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
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
Brilliant. Thanks for that. A little confused as to why the 1000 metres was actually specified so.
 
  • #7
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
That's true. Thanks for the explanations and help in general with that anyway.
 

FAQ: Tension on string of submerged object.

1. What is tension in the context of a submerged object?

Tension is the force acting on a string or a cable that is attached to a submerged object. It is the force that resists the movement of the object.

2. How does the depth of the submerged object affect the tension on the string?

The depth of the submerged object does not directly affect the tension on the string. However, as the depth increases, the weight of the object and the surrounding water will increase, causing an increase in the tension on the string.

3. What factors affect the tension on the string of a submerged object?

The tension on the string of a submerged object is affected by the weight of the object, the depth of the water, the buoyancy force, and any external forces acting on the object.

4. How can tension on the string of a submerged object be calculated?

Tension on the string of a submerged object can be calculated using the equation T = W + B + F, where T is the tension, W is the weight of the object, B is the buoyancy force, and F is any external force acting on the object.

5. Does the shape of the submerged object have an impact on the tension of the string?

Yes, the shape of the submerged object can have an impact on the tension of the string. Objects with a larger surface area will experience more resistance from the surrounding water, resulting in a higher tension on the string compared to objects with a smaller surface area.

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