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 forces1. 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 h1ρg, the pressure at the bottom of the cube is increased by h2ρg, 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.
Static and kinetic coefficient of friction
If we think of the water as one body, then normal contact force (of bucket on water) must be larger when the bucket is accelerating upward, and smaller when the bucket is accelerating downward.
Likewise, if we look within the water itself, then the water pressure must also be larger when the bucket is accelerating upward, and smaller when the bucket is accelerating downward.
By thinking about the pressure forces required to accelerate a column of water of height h, we can deduce that the water pressure at depth h must be hρ(g+a).
Since upthrust is the net resultant force from all the pressure forces, the upthrust that the water exerts on the ball must also be affected the same way. Thus the upthrust is now ρV(g+a). We can actually think of (g+a) as the apparent g of the water.
So when the bucket is stationary, the ball experiences zero net force because ρVg – mg = 0.
When the bucket is undergoing acceleration a, the ball experiences a net force of ma because ρV(g+a) – mg = ma. This implies the ball must also be accelerating at a.
Amazingly, when the water becomes more or less pressurized in order to accelerate itself at a, it also pushes exerts a larger or smaller upthrust on the ball that ensures the ball also accelerates at a, for the same amount of water displaced. That’s why the ball does not become more or less submerged.
Actually, this conclusion is obvious if we remember that the floating ball and the displaced water have the same mass. If the displaced water accelerates up and down at the same rate as the rest of the water (which it obviously must), then the floating ball must also accelerate up and down at the same rate as the water.
When stationary, the upthrust must balance the tension in the rubber band plus the weight of the ping pong ball.
ρVg = T + mg
When the bucket is accelerated upward at a, the apparent g of the water becomes g+a. The water thus exerts a larger upthrust on the ball than before. Applying N2L on the ball,
ρV(g+a) – T – mg = ma’
ρVa = ma’
Since water is more dense than ping pong ball, ρV > m. Hence, while the bucket accelerates at a, the ping pong ball accelerates at >a, and the rubber becomes more stretched.
On the other hand, when the bucket has a downward acceleration of a, the apparent g of the water becomes g-a. The water exert a smaller upthrust on the ball than before. Applying N2L on the ball,
T + mg – ρV(g–a) – = ma’
ρVa = ma’
Since water is more dense than ping pong ball, ρV > m. Hence, while the bucket has a downward acceleration of a, the ping pong ball has a downward acceleration >a, and the rubber becomes less stretched.
The simplest case to consider is if the bucket is released from rest. As the bucket free falls, the upthrust acting on the ball disappears. While the bucket and the water free fall at g, the ball accelerates at larger than g, because it would be pulled down by gravitational force plus the tension in the rubber band.
The terminal velocity of a falling body is attained when its weight is balanced by the drag force.
Let’s assume that the drag force is proportional to the velocity of the falling body. So we write the drag force as kv, where k is a constant that is dependent on how streamline the falling body is.
We can then solve for the terminal velocity vt for the single sheet of paper dropped horizontally:
kvt = mg
vt = mg/k
When a stack of papers are dropped (instead of a single sheet), k is unchanged (since it is the same area pushing into air), but m is of course multiplied by the number of sheets of paper in the sheet. For a stack of 20 sheets, the terminal velocity would be increased by 20 times.
For the single sheet dropped vertically, m is unchanged, but k is reduced dramatically (since now only a tiny area is pushing against air). Assuming the k is directly proportional to the cross-sectional paper, k would be reduced by more than 1000 times (because the cross-sectional area is reduced from 216 mm x 297 mm to about 0.05 mm x 297 mm). This implies that the terminal velocity would be increased by more than 1000 times.
This video highlights the fact that the friction between two surfaces increases if the two surfaces are pressed more strongly into each other. (f=μN)
For example, the grip that the stick exerts on the beans (f) is large only if stick is wedged tightly among the beans (leading to large N).
Same thing for the paper. Only if the paper is pressed hard horizontally against the stick will the vertical friction be large enough.
When we “pin” papers on the fridge using fridge magnets, it is not the magnetic force that holds the papers in position. In fact, the magnetic forces do not even act on the papers. What the magnets do is to ensure a large normal contact force , which then allows for a large friction.
Lastly, note that in all the examples in this video, friction was the force that resulted in motion. It is a common misconception that force opposes motion. This is wrong. Friction opposes relative motion between two surfaces, not motion per se.
The water exerts an upward pressure force on the card (F=P.A). If the pressure force is larger than the brass weight, then the card is pressed upward against the tube, and the pig is safe.
Since hydro-static pressure is proportional to the depth of the fluid (P = hρg), the pressure force decreases as the tube is raised slowly towards the water surface. There will come a point where the pressure force is less than the brass weight, the card drops away from the tube, and the pig is doomed.
Let’s do a rough calculation. The mass of the brass weights is 300 g. The inner radius of the tube is 8.6 cm.
P.A = mg
hρg.πr2 = mg
h(1000).(3.14)(0.0432) = 0.300
h = 5.17 cm
By equating the weight to the upward pressure force, we estimate the minimum depth to support the brass weights to be about 5 cm. Which looks about right in the video.
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.
The friction between two pages is quite weak. So how did friction grow to be so incredibly strong in this demonstration? 2 reasons:
When we interleave N pages on one side with N+1 pages on the other side, each side now experiences a total of 2N frictions.
Remember also that friction depends on how hard the two surfaces are pressed perpendicularly into each other. Consider the bottom page: it is now weighed down by 2N pages instead of just one. The next bottom-most page is weighed down by 2N-1 pages. And so on until the top page where the normal contact force is back to the weight of one single page.
So the total friction is multiplied by (1+2+…2N). The sum of this arithmetic progression is about 2N2.
It takes just over 30 pages on each side to amplify the friction by about 2000 times!