Maintaining Submerged Objects: Understanding Density and Height

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AirForceOne
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How do you keep a simple object submerged underwater at a certain height from the surface of the water? What would its density have to be? I'm guessing it's very close to the density of water.

Sorry if that was a very basic/dumb question...
 
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If the object's density is equal to the water's density then it will have neutral buoyancy, which means it won't float or sink, rather it will stay where you put it. If it's any less it will float and any more and it will sink. With water of a constant uniform density you won't be able to make an object that will sink to a certain depth then stay there.

In practice though, actual bodies of water will have slightly varying densities, both with time and depth. In addition, there will be upward and downward currents that will move an object that even if it is equal in density to the surrounding water. This will make it difficult to build anything that is constantly neutrally buoyant. A more practical solution may be to simply make the object slightly buoyant and then anchor it to something on the bottom to keep it the certain distance from the surface.
 
DaleSwanson said:
If the object's density is equal to the water's density then it will have neutral buoyancy, which means it won't float or sink, rather it will stay where you put it. If it's any less it will float and any more and it will sink. With water of a constant uniform density you won't be able to make an object that will sink to a certain depth then stay there.

In practice though, actual bodies of water will have slightly varying densities, both with time and depth. In addition, there will be upward and downward currents that will move an object that even if it is equal in density to the surrounding water. This will make it difficult to build anything that is constantly neutrally buoyant. A more practical solution may be to simply make the object slightly buoyant and then anchor it to something on the bottom to keep it the certain distance from the surface.

This answered my question perfectly. Thank you.