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Buoyancy: Salt Water vs. Fresh Water *February 26, 2009*

*Posted by Chris Sullivan in Training.*

Tags: Buoyancy, Dive Training, Divemaster, Diving, Diving Physics, Fresh Water, Salt Water, SCUBA, Scuba Calculations, Scuba Diving, Scuba Instructor, Scuba Training, Sea Water, Training

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Tags: Buoyancy, Dive Training, Divemaster, Diving, Diving Physics, Fresh Water, Salt Water, SCUBA, Scuba Calculations, Scuba Diving, Scuba Instructor, Scuba Training, Sea Water, Training

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Apparently this is a great source of confusion – or perhaps my previous writings about the subject have attracted many searches, and so biases my view of what divers are interested in. Anyway, I’m going to attempt to write something clear on the subject. Here are some of the searches on the topic that have found their way to my blog:

bouyancy of sea water

bouyancy salt

buoyancy salt water vs fresh scuba

buoyancy calculations salt water

buoyancy difference between salt water a

buoyancy fresh salt

buoyancy of salt & freshwater

buoyancy of sea water calculators

salt kg required for tank foot

salt water buoyancy calculation

salt water fresh water buoyancy

which is more buoyant fresh or salt water

why is bouancy more in salt water than fresh

why salt water is more buoyant than fresh

sea water density bar msw

why is salt water more buoyant

is salt or fresh water more buoyant

buoyancy salt water fresh water

Obviously people want to know. The answer is fundamentally simple – objects (including divers) are more buoyant in salt water than in fresh water because salt water is denser than fresh water. Why is salt water denser than fresh? Because salt is denser than water and so if you add it to water the resulting solution is denser than pure water. By how much? The salinity (saltiness) of the ocean varies, but the generally accepted average amount is 2.5%. So salt water weighs 2.5% more than the same volume (a gallon or litre, for example) of fresh water.

Buoyancy is an upward force equal to the weight of water displaced by the object. A cubic foot of fresh water weighs 62.4 pounds, so an object with a volume of 1 cubic foot would experience 62.4 pounds of upward force due to buoyancy when immersed in fresh water. Gravity will exert an opposite (downward) force equal to the object’s weight, so if the object weighs less than 62.4 pounds it will float. If the object weighs more than 62.4 pounds it will sink. If the object weighs exactly 62.4 pounds it will be *neutrally buoyant*, and will stay where it is unless pushed by something (current, turbulence, a diver, etc.).

In salt water, that same 1 cubic foot will displace 64 pounds, because that’s what a cubic foot of sea water (which you recall is heavier by 2.5 percent) weighs. So there is 1.6 pounds more buoyancy in salt water than in fresh. That means if an object with a volume of 1 cubic foot weighs 63 pounds it will float in salt water and sink in fresh water. So objects in salt water are more buoyant than objects in fresh water because salt water is denser than fresh.

Note that it is incorrect to say that salt water is more buoyant than fresh water. Objects in salt water are more buoyant than objects in fresh water. The buoyant force is exerted on an object, not the water itself.

Hope that clears things up!

P.S. The density of salt is 2.16 grams per cubic centimetre vs. the maximum density of fresh water at 1 gram per cubic centimetre. Sea water is about 3.5% salt by weight. A kilogram of sea water will have 35 grams of salt and 965 grams of fresh water (I’m ignoring the stuff that’s in sea water which isn’t water and salt). The volume of the water will be 965 cubic centimetres, while the volume of the salt will be 35 grams divided by the density of 2.16 which is 16.2 cc. The total volume of the components of this kilogram of sea water is 981.2 cc, versus a kilogram of fresh water at 1000 cc, so the combined density is almost 2% more with the salt than with fresh. This is a little less than the 2.5% real world difference – what’s going on?

My chemistry knowledge runs out at this point but my guess is that when salt dissolves in water, the salt molecules pack a little bit closer together with the water molecules than they do with each other, making the volume a bit smaller than the sum of the volumes of the salt and the water separately. If I find out the real answer I’ll post it here, unless someone beats me to it.

this is great!:P

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How much air (in cubic feet) is needed to lift 100 lbs under water. Ignoring the buoyancy of the weight material. How much does this differ in salt water?

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Hi Shawn,

To lift 100 lb you need to displace 100 lb of water, plus a little bit. For fresh water, which weighs 62.4 lb/cubic foot, you need 100 divided by 62.4 cubic feet of air, or just over 1.6 cubic feet. Remember though, that at depth, the volume of air from the tank that is required will increase with depth, so you need to multiply by the number of absolute atmospheres of pressure to get the volume used from the tank.

For instance, with our 100lb lift, if we were starting from 102′, which is 4 ata (absolute atmospheres), we then need 4×1.6=6.4 cubic feet of air from the tank. As the object ascends, the air will expand and either spill from the lift bag or increase the lift (and therefore the rate of ascent) if the bag can hold the extra volume. But a 100lb lift bag will hold 1.6 cubic feet of air and work at any depth, the difference is how much air you consume from your tank to fill it.

In salt water, which weighs 64lb/cubic foot, the volume is slightly less at 100/64 or 1.56 cubic feet. If you wanted to take the buoyancy of the object you are lifting into account, you would just subtract the volume of the object from that volume. So if the object’s volume was .46 cubic feet, you would only need 1.1 cubic feet of air.

Chris

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I have a camera witch is inside a glass container so it is submercable, am I to assume that the camera must wiegh at least 64lb to be nutrally bouyent in salt water?

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Only if it displaces a volume of 1 cubic foot or more. That would be one big camera!

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A small car ferry 6m wide, 12m long and 2m thick has a specific gravity of 0.80 how far, will it sink in fresh water then it is empty (unloaded)?

four cars, with a mass of 1500 kg each, are loaded on the ferry. how far will it sink in the water ??

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Sounds like a trick question! In fresh water the ferry displaces 6x12x2 cubic metres which is 144 metric tons (tonnes) of water. With a specific gravity of .8, the ferry weighs only 115.2 tonnes (80% of the displaced weight). Your 4 cars, weighing 6 tonnes, raises the total weight to 123.2, so it won’t sink (barring big waves and so on).

But that wasn’t the question. The question is how much further will it sink (or perhaps better put, how much lower will it lie in the water). Assuming the ferry is an oblong box, each tonne of additional weight will sink the ferry an additional 1/72 of a metre, because the ferry displaces 72 (6×12) tonnes for every metre it lies under the water. So 6 tonnes will sink it 1/12 of a metre which is 8 1/3 cm.

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what would be the answer of 100 cubic ft. of water = _______pounds= _______kilo= _______ounces

100cubic ft. salt water = ________pounds= _____kilo

100ton of water=______pounds= ______kilo

100ton of salt water= ______pounds=_______kilo

1000ton fresh water = _______cubic ft.=______cubic meter

1000ton salt water=______cubic water=______cubic ft.=______cubic meter

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Hm, not sure why you ask, but…

100 cu ft of fresh water is 6,242.8 pounds and 2,831.68 kg

100 cu ft of sea water is 6,398.87 pounds and 2,902.48 kg

100 tons of water (or 100 tons of anything) is 200,000 pounds or 90,718.4 kg

100 tons of salt water is 200,000 pounds or 90,718.4 kg

100 tons of fresh water is 3,203.69 cubic feet or 90.7184 cubic metres

100 tons of salt water is 3,125.55 cubic feet or 88.5058 cubic metres

This is all very easy to calculate using calchemy.

Chris

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Why does salt water sink?

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The salt which is dissolved in the water makes it denser than fresh water. Anything denser than the liquid it is in will sink eventually (ignoring surface tension, which is why some insects don’t sink when they walk on water, even though they are more dense).

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i want to know why it is possible to have a fresh water and salt water interface and they don’t mix as in the salt water doesn’t dissolve in the fresh water?

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Take a look at this Wikipedia article on Haloclines. I think the answer you’re looking for is that there has to be some force that mixes the fresh and salty water together. If there is no mixing, then the different densities of the water will cause them to remain separate. In the article, it even suggests an experiment in which you can create a halocline in a drinking glass.

–Chris

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that sure saved my butt for the buoyancy project

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I’m glad I could help. What was the buoyancy project?

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Im currently taking a scuba class, and on a recent test I got an answer wrong that I disagreed with; the instructor was able to tell me it was wrong, but didn’t really address my reason for disagreement… the question was:

“If an object is neutrally buoyant in salt water, it would mostly _______ in fresh water.”

a. Float (positively buoyant)

b. Sink (negatively buoyant)

c. Neither float nor sink (neutrally buoyant)

d. The question lacks enough information

-I chose d. based on the fact that the question didn’t address what the ‘object’ was… is it not at all possible for an object to be both neutrally buoyant in salt water AND neutrally buoyant in fresh water? Aren’t there several different levels of neutral buoyancy? Does the weight of the object in relation to the difference in weight/density of salt water not play a part?

Thanks

Thomas Houston

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Hi Thomas,

The correct answer is (b) sink. Buoyancy is equal to the weight of the water displaced by the object, so as the volume is the same in both cases, but that volume of fresh water weighs less than the same volume of salt water, the buoyancy is less and the formerly neutrally buoyant object will now sink. The object is question has exactly the same density as salt water, which is why it’s neutrally buoyant. This explains why we need less weight when diving in fresh water.

Chris

P.S. A physicist might argue the point. Some “objects” may have special properties, like individual atoms for instance, or stuff that dissolves.

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Can you get my sister to go in the water? She’s afraid that she’ll drown.

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Maybe. The owner of my local dive shop is really good at getting people to overcome their fears. I’ve been successful as well. Usually it’s a matter of being reassuring and patient, and starting off with little things in the shallow end of the pool.

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Will an object completely immersed sink

a) more in water

b) more in salt water

c) the same

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There’s a very similar question on the PADI open water exam. Goes something like “A neutrally buoyant object in salt water will ____ when in Fresh Water”

(a) sink (b) float (c) stay the same (d) impossible to tell.

Because salt water is denser, it provides more buoyancy for an object of a given volume. So in fresh water there will be less buoyancy, so the upward force is less. So the object will sink in fresh water. We know this intuitively I think because it is easier for a swimmer to float in the sea than in a fresh water lake, on in the Dead Sea, which has more salinity than regular sea water, people float really well, almost like they’re on a air mattress.

Sorry it took so long to respond. Had some issues to deal with.

Chris

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Why does a marble still sinks when put in salt water but not float?

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The marble won’t sink as fast as it will in fresh water, but marble (or glass, which “marbles” are made from) is more dense than salt water (a marble is heavier than its weight in water, salt or fresh).

The difference between salt and fresh water densities is only about 3%.

Hope that helps…

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Hi Chris,

I am wondering about a boat lift on fresh water. I am buying a boat that weighs 4300 dry weight, has a 90 gallon fuel tank, which full is basically another 720 lbs., and say 500 pounds more for other objects. Total together 5,520. How big of a boat lift would we need to lift the boat? Do we need to take into consideration the weight of the tank? I’m confused. We looked at a tank that was 9X6X30″- would this work and how big (lbs) is this. – THanks – Gala

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I don’t know a lot about boat lifts, but if you’re lifting it out of the water, it would at least have to be capable of lifting the full weight, as once it’s out of the water there’s no buoyancy available. I doubt the weight of the tank would be significant, and your calculation for the weight of the fuel is on the high side because you used the specific gravity of water (i.e. 1.0) instead of the specific gravity of gasoline which is .739 (unless it’s diesel, which is still <1 but higher than gasoline and covers a fairly wide range).

So the fuel, if gasoline, would actually weigh only about 532 lbs. The tank size of 9'x6'x2.5' is 270 cubic feet, and there's 0.133680556 US gallons in a cubic foot, so your proposed fuel tank would hold just over 2,000 gallons. For 90 gallons you need 12 cubic feet, say 2'x2'x3'.

Just for fun, let's think about the weight of the tank. Assume it is steel, which weighs .2836 lbs per cubic inch. The 2x2x3 tank from the previous paragraph has a surface area of 17 square feet, which is 2448 square inches. If the thickness of the walls of the tank is say 1/8th of an inch (that would be thicker than I'd expect), then the tank would contain 2448/8 or 306 cubic inches, so would weigh 306 x .2836 or just under 87 pounds. So you'd have about 620 lbs of tank and gasoline, all told.

Sounds like a nice boat. I hope you go diving on it.

Chris

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please sir i got a problem of a floating ship floats on surface of sea water of density 1030 transfers to river of density 1000 the immersed volume changes by 3 metres cube g equal 10 find immersed volume in sea , buoyant force in river and the weight of the ship plz help

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Hello Dee. I think I understand the problem.

This is very much like what happens to a diver moving from salt to fresh water, the diver will tend to sink more.

Let’s say the ship’s weight in tonnes (thousands of kg) is

x. The ship therefore displacesx/1.03 cubic metres of sea water, and x cubic metres of fresh water. We know that the displacement in fresh water is 3 cubic metres more than in salt water, sox–x/1.03 = 3. Therefore 1.03x-x = 3; .03x= 3,x=3/.03;x=100 tonnes (100,000 kg).I hope this helps

Chris

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can you explain floatation in sea water

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Sure. Things that float take up space. Take an iceberg for example. The iceberg takes up space so it has a certain volume. It also has weight. If the weight of iceberg is less than the weight of the same volume of water, it floats. We know that icebergs float, so we know that the same volume of water weighs more than the iceberg.

You may have heard that most of an iceberg is under water. This is where we get the saying “it’s only the tip of the iceberg” meaning there’s much more of the iceberg you can’t see than you can (unless, of course, you are diving near the iceberg, which can be very dangerous).

The part of the iceberg that is under water takes up space. If that space consisted of water, it would weigh exactly the same as the entire iceberg. That’s why we say the iceberg

displacesits weight in water. Not every object displaces its weight in water. A lead weight, for instance, displaces much less than its weight in water. That’s why if you release a dive weight in water it drop to the bottom (quickly enough that you don’t want to be under it, so be careful if you ever need to drop your weight belt).In summary the reason the iceberg (or anything else) floats, is that it is less dense than water, so it can displace

morethan its own weight in water. The extra volume, therefore, sits on top of the water instead of in it.LikeLike

How does salt effect water?

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When it comes to buoyancy, the addition of salt increases the density of water, making objects in it more buoyant. It also makes water more electrically conductive.

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Hey Chris,

I have a question about weighting in Scuba. Reading through your posts, sea water imparts 2.5% more bouyancy on objects (divers) than in fresh water. So if you’ve determined the correct amount of weight while in a fresh water pool you would need 2.5% more weight for sea water. If you needed 10 lbs of weight, would you need 10.25 lbs for sea water, or do you have to add 2.5% of weight to the entire weight of the diver? i.e. if the diver weights 250 lbs with all his or her gear and weights, then would you need 6.25lbs of extra weight in sea water? If the answer is the latter (which I believe is correct), then it seems like dive shops with pools should have scales to make this precise conversion between their pools and the ocean…

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You need to add 2.5 percent of the total weight. So if you’re 200lbs with all your gear on with your fresh water setup, then you need to add about 5lbs because that’s how much extra water you’ll be displacing (or 250lb total with 6.25 in your example)

Precision is not all that important. If you know what you weigh, add about 50lbs to that and go from there. A few ounces here and there isn’t going to make a huge difference. Remember that your lungs can vary your buoyancy by 5-10 lbs.

Thanks for the question,

Chris

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i have a question sir did buoyancy affect water animals ?if yes or no?how

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Buoyancy affects anything in the water, including water animals and even water itself. The principle of buoyancy applies to anything that takes up space in water. The object, like a water animal, displaces some weight of water, so there is an equivalent upward force.

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why does pen sink in kerosene but in oil it didn’t sink right to the bottom it stayed half way of the beaker??

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Take a look at this site. It shows that the specific gravity of Kerosene is .82, meaning that a given volume of Kerosene weighs 82% of the weight of the same volume of fresh water. Most oils seem to be around .89 to .94, meaning that they are heavier than Kerosene. So anything that has a density less than 82% of water will float in either liquid, anything more than 94% will sink in both, and anything between 82% and 89% will sink in Kerosene but float in oil, etc. etc.

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why more buoyant force is experianced by the object in salt solution

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Simply because salt water is approximately 2.5% denser than fresh water.

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How are you Chris !

Can you teach me about the weight difference betwen 80 tons weight of steel bar is now in the sea water and its weight above.

I would like to know the formula to calculate it.

Thank you in advance !

Arnold

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Arnold, You need to know the volume of 80 tons of steel. That is easy enough to calculate. A cubic foot of steel is .22 tons. So divide 80 by .22 and you will get the total number cubic feet of steel. Multiply that by 64 and this will be the number of pounds of upward buoyant force. Convert that from pounds to tons and subtract it from 80 to get the ‘weight’ of the steel in salt water (in fresh water, use 62.4 instead of 64 pounds per cubic foot). You see that it makes quite a difference, although you would still be unable to lift it yourself.

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Dear Chris,

Thank you for your reply ! I did not expect your kind answer.

But, what I am asking about current 80 tons is meaning that, It is already calculated by referring to volume.

Now I want to know the weight difference when 80 tons bar is gone down into the sea 1,000m (salt water).

It would be so much appreciated if you teach me some formula of calculation for this question.

Thank you in advance.

Arnold~~

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Hi Chris, if I had an approximate 80 foot motor yacht weighing approximately 160,000 pounds and it sits 5′ 8″ (draught) in salt water, is there any way to approximate how much deeper the vessel would sit in fresh water?

Thanks.

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Hi Mark,

I live near Lake Ontario so if you bring your yacht around so we can go for a spin we can answer that question empirically.

The tough part of the question is the shape of the hull, most importantly the area of the hull at the waterline. If it were a simple rectangular prism the answer would be fairly simple, but with a tapered hull it is much more difficult. However we can determine a maximum draught, with the actual draught being somewhat less, by assuming the area of the intersection of the hull and the waterline is constant, even though we know that it is probably tapered somewhat as you go deeper. A motor yacht is probably more rectangular than a sailing yacht so we’d have a decent approximation, I think.

What we know is the volume of the hull below the water line in both salt and fresh water. In salt water it is 160,000 pounds divided by 64 pounds per cubic foot yielding 2500 cubic feet. In fresh water the volume would be greater because the density of fresh water is only 62.4 pounds per cubic foot, at 2564. So with our rectangular shaped hull each cubic foot changes the draught by .0272 inches (i.e. 68″ divided by 2500 cubic feet is the number of inches that draught increases with every additional cubic foot of volume displaced. So as we’ve added 64 cubic feet of hull under the waterline it means the yacht is 64 times .0272 inches deeper, which is about 1 3/4″ additional draught.

That would be the maximum. A reasonable assumption is that surface area of the hull at the waterline increases the deeper the yacht lies so it will be something less than 1 3/4″ in practice.

Chris

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Arnold, the “weight” difference is 23,273 pounds. The depth doesn’t matter much as water doesn’t compress much, maybe a couple of percent at the deepest part of the ocean, so at most another 500 pounds. So it would take a little more than 68 tons of lift to raise 80 tons of steel in sea water.

Chris

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Hi Chris,

Thanks for the response and the explanation. I’m investigating bringing the yacht through the Erie canal so maybe we do need to get you on board to answer this empirically!

Many thanks again,

Mark

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I was just reading about the Erie Canal in a book on US History (“Flight of the Eagle” by Conrad Black). The Canal was built all the way to the Niagara instead of the Eastern end of Lake Ontario (which would have cost much less) because of fear of attack from Canada. This was around the time of the war of 1812. I’ll probably have to read the book a second time to take in all the detail but I’ve found it very interesting.

On a diving note, at the West End of Lake Ontario lie a couple of US ships that were sunk during the war. They are a technical depths (and cold temperatures) and are protected as heritage sites and war graves.

I’m glad we’re all friends now. Have a safe trip!

Chris

P.S. I remember an old song “The ear-eye-eeh was a rising, and the gin was getting low. And I scarcely think that we’ll get a drink ’til we get to Buff-a-lo-oh-oh, ’til will get to Buff-a-lo.” I hope you don’t have to endure such hardships on your own voyage.

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Good day Sir/Madam

Please assist me in providing solution on this concern;

I want tow an object on a water (salt water), the weight of the object is about 230 ton with a tug boat of about 40 ton.what is the force (pull) required to tow the object?

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Hi,

I’m afraid I can’t answer that question – as it is not a buoyancy question, although I think there are a lot of things that would depend on. A person can pull a very large boat with just a rope, so the answer would be ‘not much’, except if there are winds, waves and/or current. It would also depend on how fast you want to pull the boat. I expect that the size of the tugboat required would therefore depend very much on the conditions – an enclosed harbour on a windless day would require less power, and so on. If there were wind, then the surface area of both boats would be a factor, as would the direction and speed of the wind.

So this is a complex marine engineering problem.

Thanks for the question. Interesting to think about.

Chris

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Thank you very much for helping me clear things up, I had a project dealing with salt water, bouyancy, and density. This post helped me a lot with understanding these topics.

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A car ferry is rectangular with dimensions 15m wide, 90m long and has a mass of 2.7×10^6kg. When unloaded the centre of gravity is located 2m above the base of the ferry. While unloaded, will the ferry be stable?

If 60 cars with an average mass per car of 1500kg are loaded on the ferry, how much further will it sink into the water?

I have worked out that it will be stable but not sure how to calculate how far it has sunk into the water

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Hey Brodie,

OK it’s not all that hard. Based on those dimensions, the ferry (being 15 x 90m) presents a profile of 1,350 square metres to the water. The 60 cars weight 90,000 kg which is 90 metric tons. In fresh water, each cubic metre weighs 1 metric ton so 90 metric tons of will displace 90 cubic metres of water. If the 1,350 square metre bottom sank a full metre, that would displace 1,350 cubic metres. Clearly it will sink much less than that! In fact, in will sink a mere 90 / 1350 metres further into the water, which is a little less than 7cm.

The weight of ferry is 2700 metric tons. So without the cars it will lie 2 (2700/1350) metres below the surface.

Hope that helps!

Chris

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Why does sea water exert more pressure than fresh water

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Ok, so I have got this homework to weigh an aluminium block, pencil box, brass ornament and wooden block in air(mg), normal water and salt water so how am i supposed to do that. Then there is this question(if you can answer that then i would be more than grateful): ” if an object is submerged in water and salty water, how does this effect the results in the recorded data as it is experienced that to float in dead sea is more easier than in freshwater” to quote them directly!

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Because the salt dissolved in the sea water increases the density (by around 2.5%)

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A scale which allows you to hang an object from it would work, I think. You’d have to subtract the weight of the string in all cases, and that would be affected by immersion in water. How you handle this would depend on what level of precision and accuracy is required.

For a given object, the scale would read less when the object is in fresh water, and even less when in salt water, than in the air. The difference in weight in air vs. weight in water would be equal to the weight of the water displaced by the object. Salt water is more dense so the weight displaced would be greater.

Of course, if the object floats then the weight in water will be zero. Floating occurs when the object displaces a greater weight of water than it itself weighs.

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This is a really helpful blog! Thank you!!

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Good Afternoon, Mr. Sullivan,

Could you also answer the question i mentioned in my comment? It is kind of weird, actually! the English also seems quite off

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Hey there, thanks for the post.

The reason your math comes out with an error at the end of the post is because adding salt (or anything that dissolves) to water does not change the volume of the water. The atoms from the salt sit in the gaps between the water molecules, and the water molecules don’t change their mean separation.

Hence 1 litre of water plus 0.001 litres of salt results in 1.000 litres of salty water, not 1.001.

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Hi Jonathan,

Thanks for the comment. I thought something like that was going on. I suppose there is a limit to that phenomenon, as at some point the capacity of water to accommodate the salt has to end. Perhaps an experiment is in order, or at least finding the results of someone else’s experiment.

I’m glad you stopped by.

Chris

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Chris, Yes! You are right – there is a limit, when there is so much salt in that any extra will no longer dissolve. Then the surplus sits in suspension, making the water cloudy instead of clear, and this results in an increase in volume.

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Hai, thank you for the posting..

I have a question..what if an ball of solid cooper with diameter 1 meter has already sunk into the fresh water…How many volumes of hydrogen that we need to float the ball. Assuming we put hydrogen in buoyant..

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is it possible for a human being to walk on water by using this buoyancy force(upward force)????if it is possible,how??

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I think not. The force of buoyancy requires displacement of water. The amount of force depends on the displacement, so to support a 75kg person one would need to displace about 75l. A person’s density is about the same as water so most people are reasonably neutrally buoyant – i.e. that person displaces about 75l and the buoyant force is about the same as the gravitational force. So on earth, in fresh or salt water, with a normal human, there is no possibility of walking on water unaided. If one had footwear that was light but displaced sufficient volume, and could keep one’s balance, then sure, but that’s cheating.

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OK copper has a density of 8.96 g/cc, so a 100cm copper ball would weigh 4/3 * Pi * 50^3 * 8.96 grams, which is 4691.4 kg. That would require 4691.4 litres of volume to displace in fresh water, but of course the hydrogen weighs something. 4691.4 litres of hydrogen weighs 4691.4 * 2.016 grams or 9.46kg. So we meed to add that many litres to the volume to achieve neutral buoyancy (plus a bit more for the extra hydrogen), so we’re now at about 4701 litres. You would then need a little bit more to actually float, the amount depending on how high you want the flotation volume to stick out of the water. If you wanted it to be the same as an iceberg (1/7 out of the water), you’d need about 5,485 litres.

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Also “walk on water” implies only a small amount of the person’s feet are submerged, not their whole body, so the displaced volume would actually be tiny, just tens of cubic cm. The water would have to be much denser than mercury (or lead) for this tiny volume displacement to supply enough buoyancy to support their body weight. You cannot dissolve enough salt, or anything else, into water to make it so dense.

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Exactly

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Hi Chris loved your article. I have some questions regarding sea water and its current. If a 200kg object is dropped at a certain point in the ocean with a current of 1m/s uniform within a 50m water column how far would it be displaced from its original location horizontally?

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Hi Ryan, I can’t answer that one. My best guess at the answer is ‘not far’, but it would depend on a the size and shape of the object. 200kg of lead shaped like a teardrop pointing downward would drop very quickly. 200kg with a volume slightly less than 200 litres and shaped so it would drag in the water would travel a long way before settling to the bottom, and may well continue to travel along the bottom for that matter. Calculating this sort of thing with any accuracy is in the realm of fluid dynamics, which is an engineering discipline requiring advanced mathematics.

Thanks for your question

Chris

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