Example: Consider a piece of wood 10x20x20 cm with a density of 0.6 g/cm^3, in water that has a density of 1.0 g/cm^3.

The wood is less dense than water, so it will float. It weighs 10x20x20x0.6=2400 grams. That is the mass of water it will displace when it floats.

The volume of the wood is 4000 cm^3. That volume of water would weigh 4000 grams. How much additional weight must be added to the wood before the combination weighs 4000 g? (Answer: 1600 g). So a piece of, say, lead weighing 1599 g could be glued to the wood and it would still float---just barely.

Over the years since this was posted, I've learnt that the overall density needs to be above that of water. Basically, the overall density of the system determines whether it will sink or not. Calculate the weight over the volume and if it's below 1,000 kg/m³ (water), it'll float.

For an object partially, or totally, submerged in water, there is an upward buoyant force, equal to the volume of displaced water times the density of the water times g. Call it B = ρ_{water} * g * V_{d} This force acts upwards.

The weight of the object, W = m*g, pulls downwards.

The sum of these forces is ΣF = B -W = ρ_{water} * g * V_{d} - m*g

And in the case where object is floating on the water, ΣF = 0. Then solving for V_{d} gives you:

V_{d} = m/ρ_{water}

Which basically says that the object displaces a volume of with the same weight as itself.

By the way, this displaced volume is some number less than the total volume of the object. Call that v_{total}. So V_{d} kind of has to be somewhere in the range: 0 < V_{d} < V_{total}

Also it makes sense to think about what fraction of the object is submerged, i.e the ratio: V_{d} / V_{total} If this number is much smaller than one, then most of this object's volume is out of the water, and that's probably a good a thing if you're designing a boat. Also notice that for the case of an object floating on water, this ratio is equal to the ratio of the (average) density of the object to the density of water.

V_{d} / V_{total} = (m/V_{total})/ρ_{water} < 1

This suggests another trick. Supposing you know m, i.e. how much the whole thing is going to weigh. You just make V_{total} big enough that your average density is less than that of water m/V_{total} < ρ_{water}

Bio:I'm currently living in Mexico and I love building stuff. I speak English and Spanish but I prefer English and I think Instructables is great! https://sites.google.com/site/partontranslations/

active| newest | oldestThe wood is less dense than water, so it will float. It weighs 10x20x20x0.6=2400 grams. That is the

massof water it will displace when it floats.The volume of the wood is 4000 cm^3. That volume of

waterwould weigh 4000 grams. How much additional weight must be added to the wood before the combination weighs 4000 g? (Answer: 1600 g). So a piece of, say, lead weighing 1599 g could be glued to the wood and it would still float---just barely.Hope this clarifies.

I have two 20" X 20' made of aluminum. The complete structure weights 2000lbs.

How much weight can I put on it?

How much will it sink into the water with the 2000lbs?

Over the years since this was posted, I've learnt that the overall density needs to be above that of water. Basically, the overall density of the system determines whether it will sink or not. Calculate the weight over the volume and if it's below 1,000 kg/m³ (water), it'll float.

butts and booties all the way

In simplistic terms

An object displaces the same weight of water as it weighs

If you calculate in Kg 1 Kg of water is the same as 1 Ltr volume of water.

So an object of 100 ltr volume will when totally immersed displace 100 Kg of water.

When the amount of water displaced = the weight of the object it will be floating and will not go deeper into the water.

So a 50 KG object will displace 50 ltr of water if the volume of that object is 100 Kg it will only be half way into the water.

For mathematical calculations see other answers.

_{water}* g * V_{d}This force acts upwards.The weight of the object, W = m*g, pulls downwards.

The sum of these forces is ΣF = B -W = ρ

_{water}* g * V_{d}- m*gAnd in the case where object is floating on the water, ΣF = 0. Then solving for V

_{d}gives you:V

_{d}= m/ρ_{water}Which basically says that the object displaces a volume of with the same weight as itself.

By the way, this displaced volume is some number less than the total volume of the object. Call that v

_{total}. So V_{d}kind of has to be somewhere in the range: 0 < V_{d}< V_{total}Also it makes sense to think about what fraction of the object is submerged, i.e the ratio: V

_{d}/ V_{total}If this number is much smaller than one, then most of this object's volume is out of the water, and that's probably a good a thing if you're designing a boat. Also notice that for the case of an object floating on water, this ratio is equal to the ratio of the (average) density of the object to the density of water.V

_{d}/ V_{total}= (m/V_{total})/ρ_{water}< 1This suggests another trick. Supposing you know m, i.e. how much the whole thing is going to weigh. You just make V

_{total}big enough that your average density is less than that of water m/V_{total}< ρ_{water}Steve