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.

• 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 '14 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 '14 at 7:35

Then, the sphere will float when B>Wtotal, otherwise it will sink. Is this correct? So the air inside the sphere will affect sphere's ability to float.

Yes but it will be a small effect. The density of air is nearly a thousand times less than that of water.

My confusion is that I assume that I have already considered this effect during the aforementioned calculations

The big question becomes how are you measuring your masses, both of the sphere, the objects you place in the sphere and the density of the water you are putting your sphere in?

Your scales are sitting in air, so they don't measure the actual weight of the object they measure the difference between the object's weight and the weight of the air displaced by the object.

So if you weigh everything, water, sphere and objects using normal scales in normal air you are implicitly taking account of the mass of the air by pretending everything (including the water) is slightly less dense than it really is.

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).

So the air inside the sphere will affect sphere's ability to float.

That's right. The sphere is a tiny bit heavier with the air in it than it would be if it were completely evacuated, so if you are doing a calculation that needs that level of accuracy you should include the weight of the air in $W_\text{total}$. The density of the sphere does not matter here; only its weight (including contents).

Of course, if the sphere were filled with water instead of air, you would have to include that weight instead. You will find that it almost completely cancels the buoyancy force (I say "almost" because the shell and the included mass are still displacing some water, so the weight of included water will be a bit less than the weight of displaced water).

I know what you're driving at but the best way to look at the situation is "the more complicated way" because it is the most complete way. You need to look at the free body diagram of the sphere. In your scenario, the weight force is acting down on the sphere and any buoyant force is acting up on the sphere. If the sphere's density, $\frac{W_{sphere}}{V_{sphere}}$, is greater than the density of the water, it will sink, otherwise, it will float.

Any voids inside the sphere, decrease the weight of the sphere, thereby decreasing its density. You could have a very thin-shelled empty, iron sphere of sufficient radius that would float on water because so much of its interior volume is empty space. On the other hand, you could fill a ping pong ball with iron and it would sink instead of floating like a normal ping pong ball does.

To fully analyze a hollow sphere with a variable weight inside of it and the sphere's relationship to floating on water would require quite a bit of math. The "simple" scenarios are much more complicated than one would initially believe. I mean let's face it, a beach ball floating on the water looks pretty simple right? Or is it? The answer is no it's not simple. The math involved in adequately describing a floating ball would be pretty darn intimidating to most people.

For an object submerged or partially submerged in a fluid, $$F_{net}= \rho gV-F_W$$ Where $\rho=$ the density of the fluid
$g=$ gravitational field strength (or acceleration due to gravity), $9.81\frac{m}{s^2}$ near the surface of the Earth
$V=$ the volume of the submerged portion of the object
$F_W=$ the weight of the object
As always, if you know all but one of the variables, you can find the other.
When $F_{net}=0$, the object floats, and, conversely, if the object floats $F_{net}=0.$
It's as simple as that.

If

$$\dfrac{W_{\text{total}}}{\frac43 \pi r^3}<1,$$

(water density),

then the sphere will float.The weight of air can be neglected.It comes into question if you are talking about lighter than air balloons and the like.