Cartesian Diver Watch me use my telekinesis power to push the diver down.

–

Reverse Cartesian My telekinesis power works both ways. Impressive sia?

–

Double Layer Golf Ball Float A golf ball is barely floating in brine. Cooking oil (which is less dense than brine) is going to be poured over the top. Will the golf ball sink deeper into the brine, float higher, or what?

–

Water See-Saw A tub of water is originally perched precariously on a wooden peg. When a copper bob is lowered into the tub, can the see-saw hold its balance?

–

Water Bridge A tub of water containing a boat is originally perched precariously on a wooden peg. When a boat embarks on a cruise around the tub, can the tub hold its balance?

When a foreign object is submerged in a fluid, the surrounding fluid will exert pressure forces perpendicularly into each point on the surface of the foreign object.

Because pressure increases with depth, the resultant of these pressure forces is an vertically upward force. This force is called the force of upthrust, U.

With some simple reasoning, we can deduce that the magnitude of the upthrust must be equal to the weight of the displaced fluid. Hence

To generate a larger upthrust, a body must displace more fluid.

Similar to weight, upthrust appears to act at a single point, the centre of gravity of the displaced fluid.

Compare the upthrust U

with the weight of the object W

When an object is fully submerged, V_{f} = V_{o}. This also represents the maximum upthrust the object can generate.

If ρ_{o }> ρ_{f }, then W > U_{max}.

The object will sink all the way down.

If ρ_{o }= ρ_{f}, then W = U_{max}.

The object will neither sink nor float.

It can hover at equilibrium at any depth in the fluid as long as it is fully submerged.

If ρ_{o }< ρ_{f}, then W < U_{max}.

The object will float partially submerged.

It will displace just enough fluid to generate the amount of upthrust exactly equal to its weight.

The lower the object’s density, the higher it will float.

Fact: Magnitude of upthrust is equal to the weight of fluid displaced.

–

When the cargo was in the boat:

Say the boat was on the right. Some people may think that the right side is heavier because of the additional weight of the boat and cargo. Other people may think that the right side is lighter because of the weight of missing (displaced) water.

Guess what? The missing water weighs exactly as much as the boat and cargo!!! Because the boat floats, the weight of the displaced water (which equals upthrust in magnitude) must be equal to the weight of boat and cargo. (Archimedes Principle).

So the missing water exactly compensates the weight of the boat and cargo. The weight is evenly distributed along the bridge.

–

When the cargo was by itself:

When the cargo was sunk on the right, the right side again has the additional weight of the cargo but less water.

But this time round, the missing water does not weigh as much as the cargo. The fact that the cargo sank meant that the weight of the displaced water (which equals upthrust in magnitude) is less than the weight of the cargo.

So there isn’t enough missing water to compensate the additional weight enough. So the right side is heavier, and the bridge toppled.

–

P.S. By the way, water bridges really exist. The Romans built them to transport water and boats. See aqueducts.

Should that side be heavier because of the bob’s “downthrust” (the downward action-reaction buddy of the upthrust force that the water exerts on the bob)?

Or should that side be lighter because of the missing (displaced) water’s weight?

Guess that, the “downthrust” is equal in magnitude to the missing (displaced) water’s weight!

Water has this amazing ability to spread the weight uniformly!

The balance is broken when the bob touches the bottom of the container. The additional downward normal contact force tips the balance.

When there is only the brine (see Figure 1), the weight of the brine displaced (labelled A) is equal to the weight of the golf ball.

Let’s imagine the golf ball floating at the same level as before after oil has been added on top (see Figure 2). Since the golf ball now displaces oil as well, it must receive an additional upthrust that is equal to the weight of the displaced oil (labelled B). This means that at this level, there is a net upward force acting on the the golf ball. So surely the golf ball will float higher until the weight of the displaced brine plus oil equals the weight of the golf ball again (see Figure 3).

–

Wait, are we certain that the Archimedes Principle is applicable even when an object is submerged in two different fluids? Let’s use the “water banana” trick again.

Imagine a golf ball that is made of oil above the fluid boundary, and brine below. (Basically, this imaginary golf ball is the displaced fluids.) Since such a golf ball will be at neutral buoyancy, it must be experiencing an upthrust that is equal to its weight, which is the weight of the displaced fluids. So Archimedes Principle works even there are two (or more) layers of fluids.

–

Still, how does the oil, which clearly exerts only downward pressure forces on the golf ball, results in additional upthrust? This is easily explained if we use a cube instead.

When there is only brine (see Figure 4), the cube receives only upward pressure forces^{1}. When oil is added (see Figure 5), besides resulting in downward pressure forces, it also results in an increase in the pressure in the brine. While the pressure at the top of the cube is increased by h_{1}ρg, the pressure at the bottom of the cube is increased by h_{2}ρg, resulting in a net increase in upthrust. Get it?

–

Let’s ignore atmospheric pressure since it will be cancelled out when we take the difference of upward and downward pressure forces.

With cylindrical bottles, the bottle becomes more pressurized when it is squeezed. With dettol bottles, whether the bottle becomes more or less pressurized depends which sides are being squeezed.

–

When the bottle is squeezed on the longer sides, the cross-sectional area becomes smaller (because we are making it more elongated).

This causes the water in the bottle to rise and press against the bottle cap, thus increasing the water pressure. The increased water pressure forces more water into the diver. Its weight of the diver overcomes the upthrust. And it sinks! This is the classic cartesian diver.

–

However, when the bottle is squeezed on the shorter sides, the cross-sectional area becomes larger (because we are making it more circular).

The water level in the bottle drops, causing the water pressure to decrease. The decreased water pressure draws water out of the diver (pushed out by the air pocket at the top of the diver). The weight of the diver drops below the upthrust, and it floats! This is the reverse cartesian diver.

I have built a variety of Cartesian divers over the years. Some divers have openings for water to go in and out, others are sealed. But the underlying principle is the same: The diver sinks when its weight is larger than the upthrust it receives from the surrounding water.

–

The diver in the above video has an opening at the bottom. At 0:11, you can see clearly water rising through the opening into the diver to fill up more of the air pocket on the top of the diver. This increases the weight of the diver, and it sinks.

The trigger for the water to rise up the diver was my hand squeezing the bottle. Squeezing the bottle caused the water level in the bottle to rise up and press against the bottle cap, thus increasing the water pressure in the bottle. The increased water pressure forces water into the diver, until the pressure of the air pocket in the diver matches the new water pressure.

–

In the above video, I used my fingers to directly compress the air pocket on top of the bottle. The increased air pressure means that the pressure in the water also increased. Water is forced into the diver, and the same thing happens.

–

The diver in the above video is a candy in an air-tight packaging. Needless to say, water cannot not go in and out of this diver. But it can be seen clearly at 0:19 that when the bottle is squeezed and unsqueezed, the candy shrinks and expands. As if the diver is breathing in and out! When the bottle is squeezed, the increased pressure forces on the candy squashes it. With a decreased volume, the candy displaces less water. The upthrust decreases, and it sinks.

When the bob is entering the water, it displaces more and more water. The water thus exerts a larger and larger upthrust on the bob. By Newton’s 3^{rd} Law, the bob exerts a larger and larger “downthrust” on the water (and beaker). Since the density of water is about 1 g cm^{-3}, every 1 cm^{3} displacement of water means an additional upthrust of 1 gram times 9.81 ms^{-2}, which is reflected by a 1 gram increase in the weighing balance reading.

When the bob is fully submerged, lowering it further does not lead to any increase in upthrust, since the amount of water displaced is the same (even though the pressure around the bob keeps increasing. Why?). This is reflected by the constant weighing balance reading.

When the bob touched the bottom of the measuring cylinder, it begins to exert a downward contact force on the cylinder. (And the tension in the string starts to weaken). This is reflected by the increase in the weighing balance reading.

When the bob has fully landed, the string is slack so tension has dropped completely to zero. The weight of the bob is now fully supported by upthrust and contact force, which are both reflected in the weighing balance reading.

The graph above shows how the forces of tension, upthrust and the normal contact force (that the beaker exerts on the bob) vary as the bob was lowered. (Note that the reading on the weighing balance corresponded to the upthrust and normal contact force.)