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?
[Physics] Can tapwater have more buoyancy than saltwater
buoyancy
Related Solutions
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.
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.
Best Answer
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.