# Can tapwater have more buoyancy than saltwater?

An object can float in saltwater while sinking in tap water since the density of tap water is less than that of saltwater. Then the weight of tap water displaced is larger than that of saltwater, but buoyant force is larger in saltwater, which contradicts Archimedes principle, which states buoyancy is equal to the weight of liquid displaced. How is that possible?

• The buoyant force is proportional to the density of the liquid times the volume of liquid displaced by the solid object. Thus, if you submerge the same object in two different fluids, the volume displaced will be the same. The buoyant force will be larger in the fluid with the larger density. Think of honey vs. water if you need a better contrast. – honeste_vivere Aug 31 '17 at 15:33

Then weight of tapwater displaced is more than that of saltwater but buoyant force is more in saltwater [...]

No, why would the weight of the displaced tap water be more? The object only sinks if its total density is higher than that of the surrounding water. In all other cases it will float. Therefore the sinking object displaces water that is lighter than itself.

In the salt water, however, in order to float, the same object will displace water of weight equal to its own weight, which is more than in the above case.

An object can float in saltwater while sinking in tap water since the density of tap water is less than that of saltwater.

That's correct.

weight of tap water displaced is larger than that of saltwater

That's incorrect.

In short: Buoyancy is the weight of the displaced water, so stating that "weight of tap water displaced is larger than that of saltwater" is equivalent to saying that the buoyancy of tap water is greater than that of tap water, directly contradicting the first statement, that it "can float in saltwater while sinking in tap water".

In other words, the volume of displaced tap water is larger than that of saltwater, but by hypothesis the displaced tap water weight is lower than that of the object (since it sinks). While the weight of the displaced salt water equals the objects weight, and is therefore larger than that of tap water.

Densities

What you can obtain is relations between the involved densities:

Imagine yourself slowly lowering an object of density $\rho_o$ and volume $V_o$ into salt water. It starts displacing an increasing volume of salt water $V_s$ and being subjected to buoyancy $B_s=\rho_sV_sg$.

Since the object floats, $B_s$ equals the object's weight, $W=\rho_oV_og$, before the object is fully submerged, i.e., for a volume $V_s<V_o$. That is,

$$B_s = W \;\;\Longrightarrow\;\; \rho_sV_sg=\rho_oV_og \;\;\Longrightarrow\;\; \frac{\rho_s}{\rho_o} = \frac{V_o}{V_s} >1\;\;\Longrightarrow\;\; \rho_s>\rho_o.$$

Now, the same experiment with fresh water, in which by hypothesis the object sinks, reveals that even at full submersion, the object experiences a buoyancy smaller than its weight:

$$B_f<W \;\;\Longrightarrow \;\;\rho_fV_og < \rho_oV_og \;\;\Longrightarrow \;\; \rho_f<\rho_o.$$