how dense fluid affect the buoyancy force?

Question

I read a story regarding the Archimedes' principle in a magazine of popular science and I am thinking of the following question: how does the density of the fluid change the buoyancy force for the same object? As we know, the Archimedes' principle tell that for any object in a fluid, the buoyancy force equals to the weight of the fluid displaced by the object. It is pretty straightforward. Now, if I have an object partially floating on the surface of a liquid, so we have

$$ F_b = \Delta V \rho g $$

where $\Delta V$ is the volume of the displaced liquid and $\rho$ is the density of the liquid. So what happen if we place the same object into a denser liquid? Physically or intuitively, since the liquid is denser, it is harder for the object to 'inject' into the liquid, so the buoyancy force should be bigger, so less part of the object submerge into the liquid. But if you look at the math, it seems not like this. Well, now $\rho$ is bigger, but the volume of displaced liquid will be smaller because it is harder to submerge the object into a denser liquid too. So how do we that for denser fluid, the same object will experience bigger buoyancy force instead of being the same?

So my question is from intuition, the same object in the denser liquid should submerse less than the case in less dense liquid. But from the math, it buoyancy force could stay the same or more. So how to prove from the math that our intuition is correct?

Answer 1

For completely submerged bodies the buoyance force, being simply equal to the weight of the displaced fluid, is stronger for a denser fluid.

But you know that the buoyancy force for a partially submerged body (like a sailing boat) must be equal to the weight of the body (unless the boat sinks or starts flying like a balloon).

Since the buoyant force is equal to the weight of the displaced fluid, a (non-sinking) boat displaces always the same mass, no matter which fluid, but more volume of a less dense fluid.

A classical example happens if you submerge an egg in water. It sinks to the bottom of the top. Then start adding salt, until eventually the egg will raise. See for example Tommy's webpage:

A quite different question is if a boat would happily float in a denser fluid like mercury, without turning upside down. The shape of the submerged part is very important for the stability. The buoyancy centre must be higher than the centre of mass, otherwise it will be unstable (that is why ballast is needed in many cases, to make a boat heavier in its underwater part... too much of the boat above water would result in a dangerous high centre of mass)

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EDIT: Ok, when the partially submerged body is in equilibrium, then $$W_{\text{displaced fluid}}=W_{\text{object}}$$ $$\rho g \Delta V = W_{object}$$

Since $g$ and the weight of the object $W_{\text{object}}$ are fixed, an increase in density means a decrease in the submerged volume, for the equation to hold.

Answer 2

For a floating object, the buoyancy force is equal to the gravity force on the object. Hence, the buoyancy force doesn't change with a denser fluid. Instead the displaced volume decreases to cancel out the effect of the increased fluid density.

For an object that doesn't float, the displaced volume needed for it to float is larger than the voulme of the object.

Answer 3

Let an object of volume $V_O$ and average density $\rho_O$ be given. Now, suppose we drop this object in a fluid of density $\rho_F$. Then we have the following claim that I prove below.

Claim. If (a) $\rho_O > \rho_F$, then the object will sink and thus become completely submerged, but if (b) $\rho_O < \rho_F$, then only part of the object will be submerged.

What does this mean? Well if the object is denser than the fluid, then it will sink and be completely submerged in which case the buoyant force it experiences will be $V_O\rho_F g$. As a result, if we increase the density of the fluid in such a way that the density of the object is still greater, then the buoyant force on the object will increase. In particular, in this case, the buoyant force on the object is always less than its weight, and it does change with changing fluid density.

On the other hand, if we increase the density of the fluid so much that it surpasses the density of the object, then the object will no longer be completely submerged. Instead, part of it will jut out beyond the surface of the fluid.

Now, suppose that the density of the object is less than the density of the fluid, and suppose that we change the density of the fluid, but in such a way that it remains greater than the density of the object. In this case, because the object is floating on the fluid, the buoyant force must always equal its weight. Thus in this case, if we change the density of the fluid, then the buoyant force will remain the same provided we don't change it so much that it becomes less than the density of the object. In these cases, the change in density of the fluid is precisely compensated for by the the change in submerged volume in order for the buoyant force to remain constant and equal to the weight of the object.

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Proof of claim. (a) Let $\rho_O > \rho_F$, and suppose, by way of contradiction, that the object is not completely submerged but instead is floating with some fraction $V_S$ of its volume submerged. Then the buoyant force $F_B$ on the object will, by Archimedes' Principle satisfy $$ F_B = \rho_F V_Sg $$ But the object is floating, so Newton's second law tells us that the buoyant force on the object must equal its weight $\rho_O V_O g$, so we have $$ \rho_F V_Sg = \rho_O V_O g $$ which implies that $\rho_F V_S = \rho_O V_O$, but this is a contradiction since $\rho_O > \rho_F$ and $V_O>V_S%$ by hypothesis. The proof of $(b)$ is similar.

Answer 4

The formula $Fb=ΔVρg$ hold whether you submerge an object or not. So for a completely submerged object the buoyant is stronger in dense liquids because $\Delta V$ is the same. If the buoyant force is bigger than the weight of the object, the object floats. It will move upwards until the forces are balanced:$$\Delta V\rho g=mg$$ $$\Delta V=\frac{m}{\rho}$$ Which means that if the object is floating, a higher fluid density means a lower submerged volume. So our intuition is correct.

Here's a picture of metal floating in mercury (a very dense liquid):

Answer 5

The buoyant force for a floating objects is always equal to its weight. When you increase the density of the liquid the submerged volume decreases.