# Can you swim in a sea of Sulphur Hexafluoride?

In this demonstration, Sulphur Hexafluoride which is denser than air is poured onto a fish tank. Since it is denser than air, it accumulates in the bottom of the container and a buoyant force acts to keep the aluminum boat afloat in this sea of Sulphur Hexafluoride.

You can use the principles that you have learned in the last few sections to tackle this problem as well.

The only difference is that $\rho_f \rightarrow \text{Density of Sulphur Hexafluoride}$. And if $\rho_{avg} > \rho_f$, the boat would sink

Questions to ponder:

• People float on the dead sea because the density of sea water is greater than the average human density. If you consider a sea of Sulphur Hexafluoride, what are the conditions under which a human could possibly float on it?

Source for the Sulphur Hexafluoride demonstration – https://www.youtube.com/watch?v=1PJTq2xQiQ0

# Note on average density and why ships do not sink

Let’s ask a very generic question: I hand you an object and ask you to predict whether the object would float or sink. How would you go about doing that ? Well, you can measure the mass of the object and the volume of the object and can derive this quantity called Average Density ( $\rho_{avg}$) $\rho_{avg} = m_{object}/V_{object}$

It is the average density of the entire object as a whole. If this object is submerged in a fluid of density $\rho_f$, then we can draw the following force diagram:

If $\rho_{avg} > \rho_{f}$, we note that this generic object would sink and if $\rho_{avg} < \rho_{f}$ it would float!. Therefore in order to make any object float in water, you need to ensure its average density is less than the density of the fluid its submerged in!

Why does a ship stay afloat in sea?

A ship is full of air! Although it is made from iron which sinks in water but with all the air that it is full of, it’s average density ( $m_{ship}/V_{ship}$) drops down such that $\rho_{avg-ship} < \rho_{sea-water}$.

Fun Experiment:

If you drop some raisins in soda, you will notice that they raise up and fall down like so (Try it out!):

This is because air bubbles that form on the top of the raisin decrease its average density to the point that its able to make the raisin raise all the way from the bottom to the top. BUT once it reaches the top all the air bubbles escape into the atmosphere (its average density increases) and the raisin now falls down.

Questions to ponder:

• Why do people not sink in the dead sea ?
• How are submarines/divers able to move up and down the ocean ? How would you extend the average density argument in this case.
• Why do air bubbles in soda always want to raise up ?
• If the total load that needs to on a ship is 25 tons. What should be the total volume of the ship in order to remain afloat if the density of sea water is 1029 kg/m3,

# How do hot air balloons work?

One of the most surprising things about air that may not be intuitive is that it is a fluid and like any other fluids exerts a pressure on objects.

Standing on earth with layers of air above us, we are constantly being ‘weighed down’ by a pressure of ~1atm at all times.

All objects in air are also assisted by a buoyant force that is caused from the pressure difference between the top and bottom surfaces.

Let’s now consider the hot-air balloon in particular. A force diagram is probably the best way to start: Envelope is the the actual fabric which holds the air inside the balloon

Therefore we see that in order for the hot air balloon to float, we need to have the buoyant force compensate for the net weight from the balloon, the load on it and the air inside it: $F_B > w_{envelope} + w_{load} + w_{hot-air}$

Say you built a modest hot-air balloon that just barely managed to get off the ground. How would you make it go higher ?

You don’t want to play around with $w_{load}$ because you don’t want to throw out any of your passengers or your supplies. You can’t really play around with $w_{envelope}$ without re-making the balloon again.

BUT you can reduce the weight of the hot air inside the balloon ( $w_{hot-air} = m_{hot-air} g = (\rho_{hot-air} V_{hot-air}) g$) ! How ?

Recall the ideal gas law $P = \rho R T$. Assuming that the pressure and the volume of the hot-air balloon does not change we note that: $\rho_{hot-air} \propto 1/T_{hot-air}$

meaning that if we want to reduce the density of air, we just need to crank up the temperature.

And with a spectacular burner onboard our hot air balloon, we can easily increase the temperature of the air inside the balloon!

That’s pretty much how hot-air balloons work! You increase the temperature of the air inside the balloon to go up, decrease the temperature to go down and skillfully adjust the temperature to hover.

Questions to ponder:

• When you increase the temperature of air, its density decreases. What do you think happens to all the molecules that were previously inside ? Do they exit the balloon ?
• Different shaped hot-air balloons is a common sight. Do you think that the Buoyant force changes for each shape? (Review the formula for Buoyant force discussed in the previous post)
• How do you think Helium balloons work ?

# Buoyancy

Useful Preliminaries : $\rho = \frac{m}{V}$ $w = mg = \rho Vg$ $p = \frac{F}{A}$ $p = p_0 + \rho g h$

Pressure depends on depth:

In a tank filled with a fluid, the bottom most part of the tank experiences a higher pressure than the top most part of the tank.

This is simply because if you are at the bottom, there are more layers of the fluid above your head “weighing you down” than compared to the top where you have fewer layers.

Fact: Marina Trench is 1000 times the pressure at the sea level.

As a result of the pressure depending on depth, any object placed in a fluid experiences a pressure difference between its top and bottom surfaces.

The top is at a lower pressure and the bottom at a higher pressure. Therefore, there is an upward force called the ‘Buoyant Force’ that acts on the object when you submerge it in a fluid.

Anything that you submerge underwater will feel lighter than it actually is because the Buoyant force acts upward on the objects an help you counteract the effect of gravity. $F_{net} = mg - F_B$

(i) If $mg>F_B$, object sinks

(ii) If $mg, object floats

(i) If $mg=F_B$, object remains stationary. (It is neutrally buoyant)strived(i) If $mg=F_B$, object remains stationary. (It is neutrally buoyant)

Note on the Buoyant force:

• Air is also a fluid and also offers a Buoyant force on any object.
• The Buoyant force is a property of the fluid and has nothing to do with the nature of the object that you submerge. A 1 $m^3$ of Iron, Styrofoam, Lead, etc all would feel the same Buoyant force.

An equation to represent Buoyant force:

So far, all of this has been qualitative. Let’s try to obtain an expression for the Buoyant force that we can work with.

Consider an object submerged in a fluid with density $\rho_f$ as follows: $F_B = (P_{high} - P_{low}) A$ $F_B = \left[ p_0 + \rho_f g (h+d) - (p_0 + \rho_f g h ) \right] A$

where $\rho_f$ -> Density of the fluid. $F_B = \rho_f g (d A )$ $F_B = \rho_f g V_{object}$

Notice that the Buoyant force only depends on the volume occupied by the object and not it’s density and as a result all objects with the same volume irrespective of its density experience the same Buoyant force!

We can make this even better by realizing that $V_{object} = V_{fluid- displaced}$ because when an object submerges in water it pushes away all the fluid that was already there to occupy it for itself.

How much fluid does it need to displace ? For submerged objects it’s exactly how much volume it needs to accommodate the object in the fluid. Therefore we can rewrite the above formula like so: $F_B = (\rho_f V_{fluid-displaced}) g$ $F_B = m_f g$ (since $\rho = m/V )$

This is the Archimedes principle. It reads that the Buoyant force experienced by any object is equal to the weight of the fluid displaced.

Eureka!