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manisar
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"Whenever you can, share. You never know who all will be able to see far away standing upon your shoulders!"

I write mainly on topics related to science and technology.

Sometimes, I create tools and animation.


How Do Objects Float?

May 16, 2020

Author - manisar

Multiple Explanations



Fun Fact

Saturn, the mighty Saturn, will float in water!

Saturn Globe

If one claims to have a scientific bent of mind and he has never wondered about these two at any time in his/her life, I'll doubt 🙂:

  1. How objects float in water
  2. How objects float and fly in air (like clouds, aeroplanes etc.)

For both of these, I'm not talking about the explanation in equations - yes equations help, and actually prove that something is possible. But here I'm interested in the basic physics part of the phenomenon - what's actually happening out there.

Just gonna talk about a bit on the first one here - how objects float in water (or any liquid). For a discussion on flying, check How Do Objects Fly?.

The main explanation that is taught in schools is based on the Archimedes' principle. But in order to understand this phenomenon more intuitively I found out that there are at least two other very legitimate explanations. Let me explain them here starting with the Archimedes principle.

1. Archimedes' Principle

It's a simple explanation (though not very intuitive) - the weight of an object immersed in a liquid is reduced by the weight of the amount of the liquid displaced by it. In order to apply this to the idea of flotation, let's imagine we have a cube with cross-sectional surface area equal to S. If it is immersed in water up to a depth d, the upward force on it will be given by the pressure at the bottom times the area of its bottom-facing surface (since the sideways forces cancel out). Remember that pressure at any depth d of any liquid is given by p=ρdg where ρ is the density of the liquid.

So, the moment this force becomes equal to mg - the weight of the cube, the object will start floating. Note that this may not happen at all in which case the object will sink.

Buyoncy Basic 1


This is a simple explanation based on a simple law. But let's see if there is something more intuitive.

2. Water Displaced Is Trying to Come Back!

This interpretation always made more sense to me. When the cube is immersed in water, it displaces some water. That water has to go somewhere.

Every vessel, howsoever big it is, has its boundaries.This means that in effect the water level has risen in the vessel.

So, if d denotes the submerged height of the cube, it means the volume of water displaced is S x d. This in turn means that this much volume of water is now floating above the original level of water.
Guess what, this amount of water also has some weight - just like our cube - this volume of water pushes the (rest of the) water in the same way as our cube does. The result - the downward force of this displaced water starts cancelling out the weight of the cube.

Buyoncy (by the idea of water displaced)


And finally, if and when the weight of this displaced water becomes equal to that of the cube, there is equilibrium (or the cube starts floating)!

$$\begin{align} \text{Weight of displaced water} & = \text{Mass of displaced water} \times g \\ & = V \times \rho \times g \\ & = (S \times d) \times \rho \times g \\ & = S \times \rho d g \end{align}$$ If we keep pushing the object down and down until: $$\begin{align} \text{Weight of displaced water} &= mg \text{ (the weight of the object)}\\ \text{we get: } mg &= S \times \rho d g \end{align}$$

We get the same equation as we got from Archimedes' principle!

So, to repeat, if the weight of this displaced water becomes equal to that of our cube, there has to be equilibrium - the two pieces of mass - our cube and the displaced water - both equal in weight - are now pushing the water downward at different places, and in a way, cancelling out each other!

(if you think about it, Pasca's law is fitting here in support of this interpretation)

3. The Density Interpretation

Between two objects of differing densities, if possible, the one with the greater density will assume the lower position. And, if the densities are same, there is no competition.

If the cube is already having a density lower than that of water, it's going to float. But what about a ship made of iron? Let's take that case.

Let's assume the density of iron is \( \rho' \). When some part of the ship sinks (i.e. goes below the surface of water), by its very design, inside its hollow part, there is air. For simplicity, let's assume the density of air to be zero and let's replace ship with a hollow iron cube with no top surface, with same cross-sectional area S.

Also assume that \(d\) units of height of the ship are in water when it's floating. Now, water doesn't know if the cube is hollow or not, nor does it know about the part of the cube above the water level. From its point of view there is something displacing part of it and that something has some volume \(V\). Let's try to find out the effective density of the cube from water's point of view.

$$ \begin{align} \text{Effective density of the cube} &= \text{mass of the cube} / \text{volume of submerged part of the cube}, or \\ \rho'' &= \text{mass of the cube} / \text{volume of water displaced} \\ &= \text{mass of the cube} / (\text{mass of water displaced}/\rho) \\ &= \rho \times (\text{mass of the cube}/\text{mass of water displaced}) \end{align}$$
On the right hand side of the equation above, mass of water displaced is not constant - it will keep on increasing as we keep submerging the cube, and guess what, the moment it becomes equal to the mass of the cube, \( \rho'' \) becomes equal to \( \rho \) ! That means, at that moment, the effective density of the cube becomes equal to that of water, and then, as we discussed, there is competition, no tug of war between the cube and the water for going lower. So, our ship floats!
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