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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 ($V_{\text{displaced}}$) = $(4/3) \pi r^3$.

Weight of the displaced water = Buoyancy force ($B$) = $V_{\text{displaced}} \times$ density of water.

$$W_{\text{total}}= W_{\text{sphere}} + W_{\text{weight}}.$$

Then, the sphere will float when $B > W_{\text{total}}$, otherwise it will sink. Is this correct? 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
You've mostly got it. Note that the volume of the sphere is the maximum volume of water which can be displaced. Therefore, by setting $B=V_{\text{sphere}}\times\rho_{\text{water}}$, you're finding the maximum possible buoyancy force. As you say, if this does not exceed the sphere's total weight than the sphere sinks. –  DanielSank Oct 23 at 7:35

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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