# Why won't a block less dense than water fully submerge?

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?

• After a few read-throughs I was able to follow your question. Your first false statement is starts with "the object should be submerged totally in water". You know it's false so I don't see why you think it is so. I'm not sure how to answer your misconception if you don't provide some reasoning for why you think the object should be totally submerged. It seems like you think "the object will lose all of it's weight in water and as buoyant force is same as the weight of that object" is the reason. Is it? – Brandon Enright Nov 8 '14 at 7:30
• en.wikipedia.org/wiki/Archimedes%27_principle . The buoyant force is Upward, the force from the weight Downward, equal and opposite balancing the downward from the weight at .5. The object has the same weight, there is a force opposing gravity, the buoyant, that makes it float. – anna v Nov 8 '14 at 7:36
• Try thinking of it this way: to displace $1\: \mathrm{m^3}$ of water you need $1000\: \mathrm{kg}$ of downward force (ignore that kg isn't a force unit). The block only weighs $500\: \mathrm{kg}$ though so the $250\: \mathrm{kg}$ below the water and the $250\: \mathrm{kg}$ above the water combine to displace a total of $500\: \mathrm{kg}$ of water. If someone pushed down on the block with another $500\: \mathrm{kg}$ of force it would fully submerge and as soon as they let go a $500\: \mathrm{kg}$ upward force would push the block right back up to the surface. – Brandon Enright Nov 8 '14 at 7:39
• in our textbook it is said that ,,,, if buoyant force is W and weight of an object is w then,,, 1)object will float when W > w 2) object will be submerged and it wont float when W < w 3)object will float but it will be totally submerged when W = w. i hope you understand now where my misconception is.(is it a misconception ?) thanks for your response Brandon Enright – tahsin Nov 8 '14 at 8:43

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

• i think, if buoyant force = weight of that object then the object will be in equilibrium situation.... and equilibrium situation is "floating" in this case and how much of that object is submerged will be determined by density of that object.. so there is no such rule that if buoyant force=weight of object then the object will be totally submerged.. the thing is about equilibrium only, am i Right? [[my english is not so good , hope you will forgive me if i do any language mistake ]] – tahsin Nov 8 '14 at 18:02
• The equilibrium in the case of less density than water will leave part of the volume outside the water, yes. A denser than water body will fall to the bottom but with the acceleration modified by the buoyancy force – anna v Nov 8 '14 at 19:44

In this case you show that the net force acting on the object is zero in this situation, that is, half of the object is within the water and another half on the water. This position of the object is thus equilibrium position. If it would come totally on the water surface or within the water then the equilibrium will be lost, i.e. it comes to in-equilibrium position. Then object will try to return to its earlier equilibrium position.

That's why it will not sink.