how dense fluid affect the buoyancy force?

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?

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I think you need to clarify what your question is. In the situation you're describing, the buoyancy force will be the same (because it is counteracting the same gravity force on the floating object as before). Since the density is larger, the displaced volume has to be smaller in order to give the same buoyancy force. –  jkej Apr 3 '13 at 23:51
Sorry for that. I just add my question. Yes, for the same object, in a denser liquid, the displaced volume is smaller but the density is larger, so how do we know which one dominates? or the change one both just cancel out so the buoyancy force stay unchanged? –  user1285419 Apr 4 '13 at 0:23
<<how do we know which one dominates?>> Mass dominates. The displaces mass is exactly the same, no matter which liquid, provided that the body is not completely sunk. Depending on the given density, the displaced volume adjusts automatically to match the required mass. But the displaced mass is fixed for a given, floating body. See my answer. –  Eduardo Guerras Valera Apr 4 '13 at 2:55
Hi Eduardo, I think you just point out the key to my question. That really what I am asking. Now come to the question, why the displaced mass is exactly the same for any liquid? It is pretty hard for me to understand this because we are dealing with two different liquid. I know you are right but just can see the straightforward reason –  user1285419 Apr 4 '13 at 3:15
It is exactly the same mass, no matter which liquid, as long as the object doesn't sink. The denser the liquid, the smaller the submerged part of the object. Think about it in this way: had the liquid an enormous density, you could nearly walk on top of it like Jesus. It is very far from our daily experience, because we never see or touch anything much different than water, but the example of the egg floating behaviour depending on the amount of solved salt is a very good clue. –  Eduardo Guerras Valera Apr 4 '13 at 3:46
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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)

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.

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Thanks. Yes, I understand that and I am thinking the same picture too. What really confusing me is is how to prove that in math? I know that for partially-submerged case, $P_{F1} = \Delta V_1 \rho_1 g$, here $\rho_1$ is the density of the liquid. If I select a liquid such that it's density $\rho_2 > \rho_1$ and $P_{F2} = \Delta V_2\rho_2 g$, how do I know that $P_{F2}>P_{F1}$ while $\Delta V_2<\Delta V_1$ but $\rho_2>\rho_1$? –  user1285419 Apr 4 '13 at 3:10
Draw a picture with a partially submerged vertical rectangle. Consider that it is only allowed to move vertically, and describe its position through a coordinate $y$. Then the $\Delta V_i$ is a function of $y$, and so is the displaced weight. Impose the equilibrium condition, that is, that the displaced weight must be equal to the total weight of the rectangle, $\Delta V_i$ and $P_{F_i}$ follow easily from that, and that you can compare what happens with different input density values. –  Eduardo Guerras Valera Apr 4 '13 at 3:31
I see now. Thanks a lot, it really helps :) –  user1285419 Apr 4 '13 at 3:56
You are welcome. To get a feeling for this simple questions is not unimportant. Physics that we may call "more advanced" rely deeply on these simple phenomena. For example, the Schwarzschild criterion for convection to happen inside stars like the Sun, is little more than this question, posed in a more sophisticated context. So it is very important to play with the simple phenomena and get intuition. –  Eduardo Guerras Valera Apr 4 '13 at 4:02

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.

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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.

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.

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Thanks for the detail proof. But ... I am afraid if I am not clarifying the question. What I am asking is: initially I have an object of density put into a liquid of density $\rho_1$, I observe that the object is partially submerge into the liquid. Then I place the same object into another denser liquid with density $\rho_2$ (i.e. $\rho_2 > \rho_1$). Why in the second case, the object is not submerge as deep as the first case? –  user1285419 Apr 4 '13 at 3:05
@joshphysics, Hallo Joshua! I see your reputation has increased nearly 2k since the last time I was here! –  Eduardo Guerras Valera Apr 4 '13 at 3:53
@user1285419 I probably should have bolded this statement that appears right before the horizontal line "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." –  joshphysics Apr 4 '13 at 19:31
@EduardoGuerrasValera Hello there. –  joshphysics Apr 4 '13 at 19:32