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The buoyant force does not get smaller as you sink

What is the buoyant force?

Imagine you’re hanging out with your friends on a Friday night, when your good friend Jacques texts you and asks you to join him on a trip to the bottom of the ocean. Jacques has a brand new submarine that he’s been itching to try out, and he wants you to come with him to check out some hydrothermal vents at the bottom of the Marianas Trench that he’s been talking about for weeks.
You don your itchiest wetsuit and climb aboard his submarine, which has suspiciously thick iron walls given how tiny and cramped it is inside. Jacques reminds you that the super-thick walls are necessary in order to survive the descent to the bottom of the Trench, since the external pressure down there is nearly 1000 times what you’re used to at sea level. The stiff walls hold the pressure inside constant even while the pressure outside increases. The walls themselves actually get compressed due to the gradually-increasing difference in pressure across them. The reason that they don’t collapse onto you and Jacques is that they compress (like springs), which counter balances the force arising from the pressure difference. So as the submarine dives deeper, the internal pressure (thankfully) stays the same… but the walls actually get thinner!
Figure of submarine walls compressing during descent
Recall that, at sea-level, you experience a pressure of about 101 kPa due to the weight of the Earth’s entire atmosphere sitting on top of you. Because air is not very dense this pressure only barely varies with elevation. For example, at the top of the Empire State Building the pressure is roughly 95 kPa. It’s only at colossal distances that the change in pressure becomes noticeable. At the top of Mount Everest the pressure is closer to 33 kPa. The decrease in pressure occurs simply because as you go higher there is less air above you pushing down on you.
Under water is a different situation. Because water is very dense, pressure rapidly increases with depth in the ocean. For every ten meters deeper you dive, the pressure of the surrounding water increases by an amount that’s equal to the total ambient pressure that you feel when you’re at the surface (101 kPa). So if you dive down to 10 m, the total pressure your body feels is now 202 kPa. If you dive even deeper to 20 m, you’ll feel 303 kPa! It’s important to remember that this pressure arises purely due to the combined weight of all the water that’s sitting above you. If you were diving on a planet with less gravity than Earth, the pressure you feel at 10 meters would be less than 202 kPa.
Jacque’s submarine is filled with air, which is much less dense than water. As you’d expect, the sub would float on the water’s surface for the same reason that boats and bubbles float. The force that allows the submarine to stay afloat is known as the buoyant force. In order to successfully descend, the submarine has to use a propeller that pushes against the buoyant force and drives the sub deeper into the ocean.
A strange property of the buoyant force is that it stays the same regardless of how deep you go; it is independent of the surrounding pressure. This means that, if you were watching Jacques’ submarine dive at a constant speed, it would appear that the propeller always spins at the same speed and that the engines consistently draw the same amount of fuel. Because water is incompressible, its density, stickiness, and other properties stay pretty much the same as you go deeper… and so the buoyant force stays the same as well.

What determines the size of the buoyant force?

Back on land, you decide to write down some equations to describe Jacque’s submarine. You start by making a free-body diagram describing the forces that push and pull on the submarine as it sinks. The first one is obviously gravity, which exerts a force Fg proportional to the mass of the sub (Msub)
Fg=Msubg
where g is the acceleration due to gravity, 9.8 m/s/s. But you know that gravity isn’t the only force that acts on the submarine. There must be a buoyant force that acts to counteract it. The submarine is filled with air (like a balloon), so you expect it would float on top of the water were it not for the action of its propeller: even though the steel walls of the submarine make it heavy, this added weight is offset by a huge amount of trapped air. This effect is similar to a balloon, which doesn’t float until it’s filled with air. The submarine thus requires a propeller that acts together with gravity to make the submarine sink, despite the buoyant force acting on it.
Another way of thinking about this is to realize that the submarine floats because it is less dense than the surrounding water. Even though the steel walls are definitely denser than the surrounding water, the enclosed air is much less dense, and so the overall density of the submarine is less than water and so it floats. So unlike a rock (denser than water) that immediately sinks when it’s dropped into the ocean, the submarine has to actively use energy (in the form of a spinning propeller) to push itself against the buoyant force and in the direction of gravity. The equation that gives you the size of the buoyant force is called Archimede’s principle, and it states:
Fb=dwaterVsubg
This says that the buoyant force Fb acting on the sub is determined by the density of water (dwater, which is roughly 1 g/cm3), the volume of the submarine (Vsub), and gravity (g). You might be surprised to see that “g” appears in the equation for buoyant force, which always points in the opposite direction to the weight of an object due to gravity. But there’s a good reason for this: the buoyant force on an object equals the total weight of water that it pushes out of the way. For bigger subs (larger Vsub), more water needs to get pushed out of the way, and so the buoyant force is larger.
We can combine this equation with the weight of the sub to get the net force acting on the submarine.
Ftotal=dwaterVsubgFpropellerMsubg
Fpropeller is the force that the propeller exerts during the dive. If Ftotal is positive, the submarine floats upwards, and if Ftotal is negative, the submarine sinks. Jacques can determine which one of these occurs by altering Fpropeller. The free-body diagram corresponding to this equation looks something like:
Figure of net force acting on the submarine
In the equation for Ftotal there is no term that is proportional to depth, which shows that the buoyant force does not get smaller as you sink. This is because water is incompressible. For this reason, a stiff object like a steel submersible (which has a roughly constant volume as it dives, since the compression of the walls is a tiny effect) displaces the same amount of water regardless of whether it’s just below the surface or right next to a sperm whale at the bottom of the Marianas trench. Because the volume is the same at any depth, and the density of water is the same at any depth, the total mass of displaced water (mass = volume x density) is the same at any depth—making the buoyant force constant.

Consider the following… the bends

The amount of dissolved gases in your bloodstream is related to the pressure that you’re sitting at. That means that when you scuba dive with a pressurized air tank, the amount of dissolved nitrogen in your bloodstream actually increases as you get deeper. If, at the end of a dive, you attempt to re-surface very quickly, this dissolved nitrogen will suddenly exit your bloodstream in the form of tiny bubbles that can disrupt or damage your blood vessels. This condition is known as decompression sickness or the bends, and it constitutes a life-threatening risk in deep-sea divers. The primary means by which divers can avoid developing the bends is by carefully controlling their rates of re-ascent during a very long dive. The reason that this condition is less of a risk for submarine-divers is because submarines have thick walls that hold out the external pressure, keeping the internal pressure pretty much constant.

Want to join the conversation?

  • blobby green style avatar for user Ian Kolata
    In the last paragraph it states, "That means that when you scuba dive with a pressurized air tank, the amount of dissolved nitrogen in your bloodstream actually decreases as you get deeper." but wouldn't the amount of gas dissolved in the blood INCREASE as the diver went deeper and the pressure increased?

    thanks,
    -IAN
    (37 votes)
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    • duskpin ultimate style avatar for user oskargonzalez
      I would think so because of Henry's law: At a constant temperature, the amount of a given gas that dissolves in a given type and volume of liquid is directly proportional to the partial pressure of that gas in equilibrium with that liquid. Higher pressures would force more nitrogen to dissolve and when you ascend the amount of gas is no longer soluble, releasing as tiny bubbles. Like opening a soda can.
      (23 votes)
  • blobby green style avatar for user josephine.gg.02
    im glad the sub isnt made of carbon fiber
    (9 votes)
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  • piceratops sapling style avatar for user azzopapj
    Do manufacturers of compressed air tanks for scuba diving intentionally decrease the partial pressure of nitrogen in their tanks to reduce the risk of getting the bends? If there's no nitrogen to 'compress' (dissolve with high pressure) into the blood, then I'm assuming there wouldn't be as much bubbling out of the blood as you ascend.
    (3 votes)
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    • male robot johnny style avatar for user Michael Hammer
      As a Nitrox Driver, I typically use either 32 percent to 36 percent Oxygen tanks. That is what NAUI and PADI consider Nitrox (Enriched Air) Diving. A regular Scuba Tank contains 21 percent O2. However, it should be noted that with a 32% to 36% mix, you will not be able to dive deeper with these percentages due to Oxygen Toxicity. Nitrox is used on dives above 100 feet. We usually use them for safety stops on ascent. This allows you too outgas more nitrogen. You guys are right, the deeper you go on Scuba the more saturated your tissues become, with Nitrogen. Enriched Air allows for safer ascents. Now days Dive computers use Reduced gradient bubble model (RGBM). Which track nitrogen bubbles that could start to come out of tissue as you ascent. Its pretty cool stuff. It was based off of old Navy Dive Tables, and gas fluid algorithms.
      (12 votes)
  • leaf blue style avatar for user Pavel Somesurname
    "the buoyant force does not get smaller as you sink" does that mean if i was originally more dense then water, that i would fall at constant velocity (even with drag) until i hit sea floor?
    (3 votes)
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    • male robot hal style avatar for user Mojo8427
      You should sink/fall with an acceleration until the drag you experienced through the water equaled the down forces. Just as you would experience an initial acceleration jumping out of a plane until you hit terminal velocity where there are no net forces acting on you.
      (3 votes)
  • blobby green style avatar for user Will Kenan
    Could someone explain practice question #8? How did you get .1m *.1m *.03m?
    (3 votes)
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  • blobby green style avatar for user Bailey Joan
    When the weight of an object equals the weight of the water displaced (or pushed away) by the object, the object is neutrally buoyant and easy to lift to the surface.
    A sailboat sank to the bottom of the ocean. The majority of the boat's weight is from the 0.25\text{ m}^30.25 m
    3
    0, point, 25, start text, space, m, end text, cubed of lead in its keel. The density of lead is approximately 11,000\text{ kg/m}^311,000 kg/m
    3
    11, comma, 000, start text, space, k, g, slash, m, end text, cubed. The density of seawater is approximately 1000\text{ kg/m}^31000 kg/m
    3
    1000, start text, space, k, g, slash, m, end text, cubed. Because gravity is pretty much the same everywhere on Earth, the weight of the sailboat can be assumed to be approximately equal to its mass.
    (2 votes)
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  • boggle yellow style avatar for user Anuja
    If an object was touching the bottom of a container, would there be a buoyant force acting on it?
    (2 votes)
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  • starky tree style avatar for user Gillian Chance Caudill
    So, in the last paragraph when you talk about deep sea diving with a pressurized tank, is it different from people diving without them? For example, do free divers have to slowly rise back up when they come from let's say, a 50-100 m dive?
    (2 votes)
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  • starky sapling style avatar for user Gavin Vasquez
    "F total = D water v sub ⋅g−F propeller −M sub ⋅g" I was confused about the purpose of the g - F propeller? Isn't the object floating?
    (1 vote)
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    • leaf green style avatar for user Nilay Gandhi
      The Dwater*Vsub*g is one term that represents the buoyancy of the submarine. The Fpropeller is the force of propeller that is exerted on the Sub, with the Msub*g being the mass of the sub times gravity. So from left to right you have the Buoyancy of the Sub - Propeller Force - Weight of the Sub = Total Force. So if we make the Fpropeller big (to go down into the water) the total force becomes negative, and if the total force is positive, you're taking the sub up.

      Yes the object is floating but recall that buoyancy doesn't change with how deep you go. Buoyancy is independent of the surrounding pressure.
      (2 votes)
  • blobby green style avatar for user billyyyyyy
    great - so I can't even save myself from drowning in case of an airplane crash and need to reach surface level - imagine surviving the crash but dying because of the tiny nitrogen bubbles ...
    (1 vote)
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