4
$\begingroup$

Across StackExchange, you see two explanations for why an object floats:

  1. The buoyant force, equal to weight of displaced water, is in equilibrium with the object's weight.
  2. The density of the object is less than the density of water.

Explanation #1 is great, but #2 strikes me as incomplete/imprecise for non-convex objects? My question: Is there a sensible, concise definition of an object's volume and mass such that an object floats if and only if its density (i.e. mass/volume) is less than the density of water? If so, are explanation #1 and #2 of equal quality?

Example where density explanation looks problematic:

(i) identical objects except orientation. (ii) red object floats, blue object sinks

enter image description here

  • The red and blue objects are identical except the blue object is rotated approx. 95 degrees.
    • The mass of the red and blue objects are identical.
    • I would think a definition of volume would be invariant to rotation? Hence, red and blue objects have identical volume, hence identical density?
  • There exists a mass for the identical objects such that the red object floats while the blue object sinks if placed on top of a body of water.

The orientation of the red object leads to greater water displacement than the blue object when placed on water. What's going on is an obvious application of Archimedes Principle. On the other hand, an explanation involving density seems rather more complicated? For some density explanation to work, an object's volume must include all air in the interior of the object below waterline (which will depend on the orientation of the object). Fine definition of volume and density? Should explanation #1 be preferred or #2? Or entirely equivalent?

$\endgroup$
2
$\begingroup$

Buoyancy is equal to the weight of the displaced liquid. If you insist on a density formulation, you have to define "instantaneous density" as total mass divided by the submerged volume. When an object is first in contact with the water this number will be very high and the object initially "sinks". As it gets more submerged, the "volume" increases although the mass remains constant. Thus it may reach equilibrium depending on shape and orientation.

That simple definition covers all your cases, and even explains things like capsizing.

$\endgroup$
  • $\begingroup$ Perfect. Clear and concise! $\endgroup$ – Matthew Gunn Dec 9 '15 at 16:56
2
$\begingroup$
  1. The buoyant force from displaced water is in equilibrium with the downward force on an object due to gravity.

  2. The density of the object is less than the density of water.

The apparent "incompleteness/impreciseness" of 2. only arises from a particular interpretation of that statement.

If by density of the object we understand the average density of the object over the entire volume of the object, then 2. is always true, whether the object is a solid piece of expanded polystyrene (density approx. $25\:\mathrm{kg/m^3}$) or a suitably hollowed out shell of made of uranium (density approx. $19100\:\mathrm{kg/m^3}$). In both cases the objects will float provided their densities are lower than the fluid in which we suspend it.

If by density however we mean density of the material of a hollow and open object is made of, things change significantly. A ship made of uranium may float well in water as long as the true density of the part that is submerged below the water line is sufficiently smaller than that of water.

But when our uranium ship starts taking water on board (for whatever reason), the true density of the part that is submerged below the water line increases and the ship will be lower in the water. More water taken on will eventually sink it because its true density now well exceeds that of water.

Similarly if we lower the uranium ship into the water stern or bow first, the true density will always well exceeds that of water and the ship will never float and always sink.

$\endgroup$
  • $\begingroup$ Let $\rho$ denote density. Am I correct in saying an object floats iff $\rho_{fluid} \geq \rho_{object} \cdot \frac{V_{object}}{V_{immersed}}$? If an object is fully immersed, $V_{object} = V_{immersed}$ and flotation depends only on an object's density. If an object is partially immersed, it also depends on the ratio of $V_{object}$ to $V_{immersed}$ or alternatively, we could define "true density" as $\frac{M_{object}}{V_{immersed}}$ and flotation depends on that. $\endgroup$ – Matthew Gunn Dec 8 '15 at 1:36
  • 2
    $\begingroup$ You're close but we just don't really do it that way. For a closed object the true density is of course simple its total mass divided by its total volume. For a concave object that is much harder to define because it depends how the object lays in the water due to its mass distribution. That's why in that case we simply use Archimedes: the concave object will float if the weight of the displaced water exceeds the weight of the whole object. For irregular shapes that can be really difficult to model and requires knowing the centre of mass of the object. $\endgroup$ – Gert Dec 8 '15 at 2:52
2
$\begingroup$

The buoyant force arises from the pressure difference between any two given points on the body when submerged in a liquid. Writing force equations on the body gives us the buoyant force is equal to $\rho Vg$ where $\rho$ is the density of the liquid. So it "seems" that the buoyant force is due to the weight of the displaced water but yes, that is an easy way of looking at it. Now, lets say a body of density $\rho_{0}$ is completely immersed in water. It's downward force (the weight) will be $\rho_{0}Vg$ where V is volume of the object. So the net force on the object will be $(\rho-\rho_{0})Vg$ (taking upwards positive). If $\rho >\rho_{0}$ the object will float as there is a net force in the upward direction. Keep in mine that here for $\rho_{0}Vg$ to be weight of the object, $\rho_{0}$ should be the effective density, that is total mass divided by total volume. Introducing air cavities increases total volume without increasing total mass and so the value of $\rho_{0}Vg$ becomes lesser than $\rho Vg$. This holds true for all objects. You are correct in saying the red object's volume must include all air below the water line thus bringing down effective density. Remember, it is the volume of water displaced and not volume of the object.

However, if you had the blue object rotated by 95 degrees, there would be a torque about center of mass due to the buoyant force as buoyant force appears to act at the center of mass of the displaced liquid. So the object would tip over and (probably) float even if it sinks at first. It would sink at first as the air enclosed would escape out and not remain within the boundary of the object so it wouldn't account of displaced volume, quite unlike the first case. However, if you somehow managed to fully immerse the object inside the liquid and it's density is greater than that of the liquid i.e the bo=buoyant force from the displaced liquid is less than the weight of the object, it will sink.

So both your explanations for why an object floats is the same. When something floats, it floats such that $\rho_{0}V g= \rho V_{immersed}g$. Since $V_{immersed}<V$, $\rho_{0}<\rho$. So the more precise explanation here is the first one, buoyant force from displaced water less than weight of water. The effective density being less than that of water is a consequence of this inequality. When fully submerged, both are the same.

$\endgroup$
2
$\begingroup$

Both explanations hold, but they are tricking you at the surface. But we must use them both to explain the surface-float behavior.

Let's consider the object you have shown - a shell.

When submerged in water, the object displaces a water volume exactly equal to its own volume. The density explanation holds. And so does the volume explanation.

When placed at the surface to float, it displaces some water, which is more than the actual volume submerged at the moment. This is simply because of the geometry. The water cannot reach the inside of the shell. In here we still have air, which is much lighter than the object. So, much more water is displaced than what equals the submerged part of the object. The displaced water volume equals the volume of the submerged part of the object PLUS the volume of the air, which is now also below the water surface since the water cannot reach it.

The average density of everything submerged may be equal to or smaller than the water density at some moment. When that is the case, the object will not sink anymore.

Therefore the buoyancy force from the water is much larger now than what it would be if the object was pushed under the water to let the water fill all space around the object.

That is why a metal ship can "trick" the water into applying a much larger buoyancy force on it than what it would do if the geometry was different. If this buoyancy force exceeds the weight of everything submerged, the object floats. And this is achieved, because much of what is submerged is air.

$\endgroup$
  • $\begingroup$ sed 's/dispersed/displaced/g' answer.txt $\endgroup$ – Floris Dec 9 '15 at 19:27

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.