Debating Object Buoyancy: Hollow vs Solid Spheres in Water

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SUMMARY

The discussion centers on the buoyancy of hollow versus solid spherical objects in water, both weighing ten pounds and having the same density and size. Participants conclude that both objects will either float or sink together, as buoyancy is determined solely by density relative to water. The hollow sphere's lower overall density allows it to float, while the solid sphere, despite having the same weight and density, will also float under the same conditions. Archimede's Principle governs this phenomenon, emphasizing that objects with equal densities to water will achieve equilibrium.

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  • Understanding of Archimede's Principle
  • Basic knowledge of density calculations (D = M/V)
  • Familiarity with buoyancy concepts
  • Concept of equilibrium in fluid mechanics
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I have been having a discussion with a friend for the past few days, and we are in sort of a stand still on a problem. We would greatly appreciate some input into our discussion.

The question is this: A hollow spherical object weighs ten pounds with a certain density, and size floats in water. Will a solid spherical object with the same weight, density, and size float in the same amount of water?

My argument is that no, the solid object will not float in the water. I believe this because the solid object has no air, and therefore has the lack of bouyancy needed to float in the water. Even though the displacement of water may be the same, the lack of air would keep the object from floating.

His argument is that yes, the solid object will float in the water. This is because the displacement of the water, and density of the water and object are the same and cancel out, causing it to float.

We would like to know who is right or wrong, and possibly an explanation why. Thanks for the help, we appreciate it :approve:
 
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What is important is the density of the object as a whole. The hollow sphere has a lower density because it has the same volume but less mass, so it can float when the solid sphere can't.
 
As long as the densities are equal, and the densities of both are less than that of water, both objects should float. The buoyant force that the object places down on the water is equal to the weight of the water the object displaces. The water then places an equal force in the opposite direction equal to the other force. If the water cannot sustain the balance, then the object would sink.

I would agree with your friend.
 
What? The problem said the masses were equal... and the volumes were equal, therefore they have equal densities.

D = \frac{M}{V}
 
I'm sorry, I thought he said they were made of the same material. Yes, they will either both float or both sink.
 
so the hollow part of the one object would not matter? i was always taught that no matter what the density of an object, if it was solid it would not float...i guess it makes sense that only density would matter though. I am still wondering something though, if i had a bowling ball weighing 10 lbs, it would float in water, but if i had another solid object of the same density it would shrink, why is that?
 
how could the two spheres have equal size & density? I was just confused, wouldn't the solid one have to be smaller because.. it's smaller?
 
Hey can someone help me with a math question right now? Please? Thanks.

Using a simple balance, could you find the one fake coin in a group of 27 coins. Here's the catch...you may only weigh the coins 3 times. You may not use anything else. How could this be accomplished?
 
The fake coin weighs different than the other coins.
 
  • #10
FlaMiNxSkullx05 said:
so the hollow part of the one object would not matter? i was always taught that no matter what the density of an object, if it was solid it would not float...i guess it makes sense that only density would matter though. I am still wondering something though, if i had a bowling ball weighing 10 lbs, it would float in water, but if i had another solid object of the same density it would shrink, why is that?
If you have any object of Mass "M" with solid (water-proof) boundaries, so that it has a defined Volume "V", then regardless of its shape or size or contents, the only factor determining whether it will float is whether its density given by (M/V) is less than water's density (1 gram/cm^3).

Two objects of the SAME DENSITY will either both float OR both sink in water even if their weights, shapes, contents, and sizes are different.

The physical phenomenon involved is called buoyancy, and its properties are described by Archimede's Principle. The following Web Sites provide further discussion of Archimede's Principle:
http://www.onr.navy.mil/focus/blowballast/sub/work2.htm
http://www.ac.wwu.edu/~vawter/PhysicsNet/Topics/Pressure/BouyantForce.html
http://www.plus2physics.com/hydrostatics/study_material.asp?chapter=3
http://www.bsharp.org/physics/stuff/icebergs.html



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  • #11
thank you all for helping me. I think i understand what is going on now, i believe i was confused earlier when i thought air was the determining factor when floating an object, i guess it is not. Another good point a teacher of mine brought up is that if the density of an object is the exact same as that of water's, what would the object do? would it sink, float, or kind of just sit in the water below the surface level?? because according to the original problem, the one density of the hollow object floated...therefore its density was less than that of waters, according to your equations, so if the density was equal to the waters it would technically come out at an equilibrium, making it a stable object..correct? i think i am starting to understand this now :bugeye:
 

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