May 16, 2020

Author - manisar

Saturn, the mighty Saturn, will float in water!

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

- How objects float in water
- How objects float in air (like clouds, aeroplanes etc.)

Just gonna talk about a bit on the first one here - how objects float in water (or any liquid).

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.

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

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.

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)!

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.

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.

Ad

Copyright © RandomPearls.com 2020