Floating in Fresh & Salt Water: Buoyant Force & Wooden Boat Impact

[holly]

I'm really fired up about physics now. But I can't puzzle out the following:
Q. When you float in fresh water, the buoyant force that acts on you is equal to your weight. When you float higher in the high-density water of Salt Lake, the B.F. that acts on you is actually:
1) greater than your weight, 2) equal to your weight, or 3) less than your weight.
I am unsure about whether the buoyant force is affected by being something other than pure water. For example, if the buoyant force is the same no matter what liquid you are in, molasses or whatever, then I can say the answer is (2), equal. But, I have only examples of things floating in fresh water.

Q. A wooden boat is floating in a pond. It has a large chunk of iron on it. If you throw the chunk overboard, what will the water level of the pond do? Fall, Rise, or Stay the Same? I say FALL, but I would like a formula or a formalized way to understand this. I think of it as the boat's footprint sinking down in the water when the chunk of iron is in the boat, and then the footprint is less when the chunk goes overboard, and the chunk being denser than the pond water, it isn't displacing as much as when it was part of the FOOTPRINT, and thus the water falls. But that's giving me a headache.

Thanking you in advance for any help.

 

[Tom McCurdy]

Just a thought

I really dont' have any qualifications to judge your problem but I would say the answers are
1.) greater than your weight... just because it can allow for more weight and still be boyant, however the forces acting upon you are still equal.

2.) I would say rise... The force of the iron mg is probably pulling the boat down some, however its force is displaced along the whole boat allowing it to be pulled down only so far which thus increases the water level. however whent he weight is thrown overboard the entire weight is under water forcing its volume to be pushed up. i guess it would depend upon the weight and size of the block. I could see it also having the boat pulled to an extent in which the volume of the displacement is greater than the volume of the iron itself.

 

[cookiemonster]

Here's a question: If the buoyant force is greater than your weight, isn't there a net force acting upward on you? If there's a net force acting upward on you, wouldn't you magically fly into the air?

As for the second one, does a very low-mass object with the same "footprint" as a high-mass object displace the same amount of water?

cookiemonster

 

[holly]

(1) Tom, I think you are as lost as I am and we both ought to depend on Cookiemonster! Thank you for your answers, though. The more, the merrier...!
(2) Cook, by footprint I mean how FAR the foot is pressed into the water...the SHOE, so to speak, is staying the same size, but the chunk is pressing the whole LEG down. The toes, anyway. So, when the weight is thrown overboard and can't push the SHOE anymore, voila, the water level drops.

No?

Please, just give me the @#!$%^ answer, you KNOW you know it! I bet you have a formula for it, too, Mr. Sleeps In Class and Doesn't Take Notes. Ha!

Thanking you in advance...

 

[cookiemonster]

For #2, the first shoe gets smaller, but you just threw a second shoe into the water!

cookiemonster

 

[holly]

I'm telling you, I SEE the shoe, it is the same size it always was, but it's pushed in less! Isn't it? Isn't the piece I threw in denser than the water? Yes, I threw it in, but it's denser, right, it has a littler volume of displacement, doesn't it, than the amount it caused to be displaced by the big shoe? There are at least FIVE shoe-like questions involving ponds, puddles, boats, and kitchen sinks, and things are floating on sponges, boards, twigs and lilypads, so I MUST understand!

Whimpering sadly,
Holly

 

[cookiemonster]

Hehe, all right, all right.

When you throw the iron out of the boat, you just decreased the average density of the boat, so it's going to displace less water. But you also just threw something else into the water, which will increase the displacement of the water! In fact, it doesn't matter where the iron is! The iron + boat system doesn't change whether the iron is on the bottom of the ocean or the deck of the boat!

cookiemonster

 

[holly]

Maybe if I sleep on it I will understand. I just can't understand right now. The chunk of lead, once in the water, is displacing less volume of pond water than when it was pushing the boat in deeper. That, to me, means the water level falls, not stays the same. If it were a floaty type chunk of wood, I could see staying the same water level. But once in the water, honestly, I just see it as displacing less volume of water than it did before, because before it was on something floating. It doesn't seem a one-to-one sort of thing to me, since it sank. Float, yes. Sank, no. Sank means denser than pond water. Denser means displaced less pond water once in.

See, totally nonsense thinking. Monkey-think, as my ex used to call it. TOO TIRED TO THINK!

Thx for helping me, maybe it will "sink" in later. *sigh*

 

[holly]

BTW, you need an avatar, Cookiemonster!

 

[cookiemonster]

Why should the piece of lead be displacing more water when it's in the boat?

If you don't like the theory, try an experiment. Get a plastic bowl and marble, fill your sink (or tub) with water, measure the water level with the marble in the bowl(or just let a little water ring form, but that takes a bit longer), and then take the marble out of the bowl and throw it into the water and see what happens!

Avatar? Maybe. But I don't know what I'd use. =\

cookiemonster

 

[cookiemonster]

Heh, I knew there was some kind of principle for this.

Archimede's Principle: a body holly or partly immersed in a fluid is buoyed up with a force equal to the weight of the fluid displaced by the body.

The weight of the iron + boat system never changes, so neither does the fluid displaced!

cookiemonster

 

[NateTG]

The water level drops:

Consider the following - let's say that we have the iron attached to the bottom of the boat by rope instead. Now, when the rope is cut, the boat rises - displaces less water - if nothing else changes, then the water level would drop.

 

[cookiemonster]

Well, I managed to confuse myself half-way through again. Go me. Sorry about that, holly!

Still, Archimede's Principle is correct. The iron is no longer buoyed in the water, it sinks and is supported by the dirt.

cookiemonster

 

[holly]

Well, now it's as clear as the mud the chunk of metal is resting upon! But the answers, wrong or right, are much appreciated.

Okay, please help me clarify: Since the water level fell this time, it would also fall if a lump of gold fell from a swimming turtle's back, too, right? Ditto for an anvil falling from a raft?

BUT, what about an anvil initally floating anvil-side up tied to a piece of styrofoam? Is this the same as the anvil tied to the bottom of the boat NateTG mentioned? When the styrofoam flips and the iron is now in the water, still tied, not going downward, does the water level also fall? The thing is still floating...since it's either wholly or partially submerged, the same idea still applies?

Sorry, one last scenario, and then I think I'll have it: Someone throws something that floats from a large floating piece of wood. Water level will remain the same, because it isn't displacing anything since it is on top of the water's surface? I am shaky on that one, too. Floating is a special case, but I'm having trouble picturing it, I have looked at things floating, but it seems to me they are not really on TOP of the water not pushing at all...?

 

[Doc Al]

Originally posted by holly
Okay, please help me clarify: Since the water level fell this time, it would also fall if a lump of gold fell from a swimming turtle's back, too, right? Ditto for an anvil falling from a raft?

Yes. The question is always: Which situation displaces more water? When the system is "floating" it must displace a volume of water equal to the weight of the total system. Something that sinks, obviously is denser than water, so it displaces a volume of water less than its weight.

So if the system is "anvil + raft" and it's floating it displaces an amount of water that weighs equal to "anvil + raft". (I don't care where you attach the anvil.) Now, if you toss the anvil you have two separate systems. The raft still floats, so it displaces an amount of water equal to its weight. The anvil now sinks, so the amount of water it displaces is obviously less that its weight. Thus the net water displaced is less.

BUT, what about an anvil initally floating anvil-side up tied to a piece of styrofoam? Is this the same as the anvil tied to the bottom of the boat NateTG mentioned? When the styrofoam flips and the iron is now in the water, still tied, not going downward, does the water level also fall? The thing is still floating...since it's either wholly or partially submerged, the same idea still applies?

Read my answer above. It doesn't matter how the parts are attached (on top, underneath, tied to a string), as long as the system floats it will displace the same amount of water.

Sorry, one last scenario, and then I think I'll have it: Someone throws something that floats from a large floating piece of wood. Water level will remain the same, because it isn't displacing anything since it is on top of the water's surface? I am shaky on that one, too. Floating is a special case, but I'm having trouble picturing it, I have looked at things floating, but it seems to me they are not really on TOP of the water not pushing at all...?

Careful. If something is floating, it must displace water. The thing resting on top of the wood may not be touching the water, but its weight pushes the wood down further in the water.

The same analysis applies. Toss the thing on top in the water. The whole system (wood + thing) still floats, so the net displacement must still equal an amount of water equal to the weight of "wood + thing". No change in water level!

 

[holly]

Okay, thank you all. Doc Al, I printed out your answer and will put it with my questions & use it as a reference.

However, I am still struggling with the anvil tied to foam, as it seems to me that when it is flipped and is iron-side down, it will then not be floating any longer, but be partially submerged at this point? Since it will be partially sunk, we have a change in displacement? Or is this dependent upon the size of the piece of styrofoam it's tied to? I am trying to puzzle out exactly what the question, since it gives no details, is trying to hammer home...can't tell if it's about the special case or floating or if it's about Archimedes' principle. No eureka.

 

[Doc Al]

Originally posted by holly
However, I am still struggling with the anvil tied to foam, as it seems to me that when it is flipped and is iron-side down, it will then not be floating any longer, but be partially submerged at this point?

The "system" is either floating or not. As long as it doesn't sink to the bottom, it's floating.

Since it will be partially sunk, we have a change in displacement?

Nope. Archimedes's principle (and equilibrium) still holds. I don't care what side is up, the net displacement of water must be the same.

Or is this dependent upon the size of the piece of styrofoam it's tied to?

Nope. (Of course, if the styrofoam is too small, the system won't float.)

I am trying to puzzle out exactly what the question, since it gives no details, is trying to hammer home...can't tell if it's about the special case or floating or if it's about Archimedes' principle. No eureka.

Keep thinking! You'll get it.

Another way to think of things is treat the anvil and foam as two separate systems. But then you'd have to include the force that the anvil and foam exert on each other. You'll get the same answer though.

 

[holly]

Will this subject never be clear to me?

Okay, in reviewing what Tom McC said, I am now agreeing with him that the salt water is exerting greater buoyant effect than the plain water does. But how can the forces then be equal? I think they are not equal, and that's why things float higher in salt water? Tom McC, any more thoughts on it? Anyone? And then I'll go away for the weekend, promise.

As for Doc Al's assertion that the semi-floating styrofoam/iron thing is treated as floating, I will just have to take that on faith, because I just can't grasp it.

Thanks!

 

[Doc Al]

Originally posted by holly
Okay, in reviewing what Tom McC said, I am now agreeing with him that the salt water is exerting greater buoyant effect than the plain water does. But how can the forces then be equal? I think they are not equal, and that's why things float higher in salt water?

Salt water is denser than plain water, so you need to displace less of it to support your weight. That's why things float higher in salt water. But the bouyant force always equals the weight for any floating object.

As for Doc Al's assertion that the semi-floating styrofoam/iron thing is treated as floating, I will just have to take that on faith, because I just can't grasp it.

No need for faith. It's simple: If it doesn't sink, it floats. (That's what floating means, right?)

Maybe this will help. Let's say the iron block is glued onto the big piece of styrofoam. Let's look at two cases:

A) Let the thing float iron side up: the styrofoam will sink a certain amount in the water.

B) Flip it over so it floats iron side down: the styrofoam will still sink into the water, but not as much as in case A.

BUT the total amount of water displaced will be the same for both cases. In case A, the styrofoam does the displacing; in case B the iron displaces some and the styrofoam displaces some. But the amount of water displaced is ALWAYS equal to the weight of the floating system. (Which is the same no matter how you flip it.)

 

[holly]

Well, thank you. I continue to be fuzzy-minded, because it seems to me that there are two situations and the two situations are treated differently. Either something is floating, or it's wholly/partially submerged. I look at the drawings in the texts I pore over, and things are being drawn as suspended midway between the top of the water/pond/lake, as resting on the bottom, or as floating on the top, but the floating depictions vary from practically sitting on top to just barely with a nose up in the air. If the iron/styrofoam has sunk down enough, it's akin to the suspended-in-the-middle picture. So, I don't know how to place it, as one of the submerged (partially/wholly) items, or as a floating item.

Well, I'm thoroughly sick of the subject!

 

[Doc Al]

Originally posted by holly
I continue to be fuzzy-minded, because it seems to me that there are two situations and the two situations are treated differently. Either something is floating, or it's wholly/partially submerged.

Careful: I hope you don't think things can float on top of water without at least being partly submerged. (Ignoring special cases where surface tension, not bouyant force, supplies the upward force.)

I look at the drawings in the texts I pore over, and things are being drawn as suspended midway between the top of the water/pond/lake, as resting on the bottom, or as floating on the top, but the floating depictions vary from practically sitting on top to just barely with a nose up in the air. If the iron/styrofoam has sunk down enough, it's akin to the suspended-in-the-middle picture. So, I don't know how to place it, as one of the submerged (partially/wholly) items, or as a floating item.

To me, saying that something floats simply means that the bouyant force exactly equals the weight. How "high" it floats depends on the density of the object. If the density is less than water, the object will float on top of the water, partly submerged. If the (average) density is greater than water, there is no way it can float---it will sink to the bottom (totally submerged and sinking). Now if the density is equal to that of water, it will stay where you put it, totally submerged (totally submerged and floating). (I call that "floating", since it doesn't sink. But no matter if you call it floating or not, the key is that it is in equilibrium.)

Going back to our styrofoam/iron object: if you choose the right sizes of each and glue them together you can create an object whose average density is equal to water: that means it will "float" while totally submerged. Ignoring complications (compression, density variation by depth) you can push this thing two feet under water---and it will stay there.