Suppose we have an object of volume $1\: \mathrm{m^3}$. Mass of that object is $500\: \mathrm{kg}$, which means that the density of the object is $500\: \mathrm{kg/m^3}$.
If the object is in water it will float and half of it's volume ($0.5\: \mathrm{m^3}$) will be submerged in water (assuming that the density of water is $1000\: \mathrm{kg/m^3}$; as the object's density is half of water so half of it's will be submerged).
From the Archimedes principal we know that the object will displace the water of same mass as it. So the object will displace $500\: \mathrm{kg}$ water and $500\: \mathrm{kg}$ water = $0.5\: \mathrm{m^3}$ water.
We also know that the lost weight of an object = weight of water displaced by that object.
It means that the object will lose all of it's weight in water and as buoyant force is same as the weight of that object, the object should be submerged totally in water. But, that it is not possible, it will be submerged only half of it's volume. But how?
If the weight of displaced water is equal to weight of that object, shouldn't it be totally submerged?
Answer
From the Archimedes principal we know that the object will displace the water of same mass as it. So the object will displace 500kg water and 500kg water = 0.5m3 water.
We also know that the lost weight of an object = weight of water displaced by that object.
The object does not lose any weight. It is pushing down with its weight. The waters is pushing back up with an equal and opposite weight of volume .5 m3, displaced. Equilibrium. As the object is 1 m3 half of it is out of the water, since it did not displace it..
It means that the object will lose all of it's weight in water and as buoyant force is same as the weight of that object, the object should be submerged totally in water.
You are double counting. No weight/mass is lost. Just the forces acting on the body, gravity and buoyancy are in equal
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