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I took an empty clear sphere with radius r, put some weight inside this sphere, and put in into the water. I calculate the buoyancy force as:

Volume of the sphere = Volume of the water displaced (Vdisplaced) = (4/3)pi(r^3); Weight of the displaced water = Buoyancy force (B) = Vdisplaced * density of water; Wtotal= Wsphere + Wweight;

Then, the sphere will float when Buoyancy (B) > Wtotal, otherwise it will sink. I assume that I am right so far?? So the air inside the sphere will affect sphere's ability to float. My confusion is that I assume that I have already considered this effect during the aforementioned calculations (since density of the sphere itself is low). Is that right? Please note that the sphere is not flexible.

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Yup, you're right. Well, technically $B=\frac{4}{3}\pi r^3\rho$ only when the sphere sinks; if it floats, it'll be less than that, and will be whatever is required to have equilibrium. The air is included in the weight. –  DumpsterDoofus Apr 8 at 21:26

1 Answer 1

I don't think I follow you there; the force on the sphere is given by:

F = (weight of the water the sphere displaces) - (weight of the sphere)

The weight of the sphere there includes everything; the weight of the sphere itself, the air inside it, whatever weights you put in).

If F is positive it points upwards (the sphere floats to surface); if it it negative it points downward (the sphere sinks).

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