Interesting Puzzle

I agree. GA

Archimedes' principle, basic science

Disagree, I can't resolve a mass (nut) suspended from a string as increasing the mass of the container & water.

Where/how does Buoyancy increase mass?

The string is restraining the 'gravity' of the nut from the 'gravity' of the container/water mass as I see it.

Or maybe I'm just in to deep ? 8-)

You won't weigh the nut directly, just the weight of the volume of water it displaces. Some weight of the nut will still be hanging by the string. If you had a scale on the string you would see the nut getting lighter as it enters the water, and the scale the cup is sitting on will show the same change only getting heavier.

Now that could be interesting.

What I'm considering though,

A) There is no addition or subtraction to the mass of either the container, the water, nor the nut.

B) There is a perception of an increase in mass registered by the scale.

Hmm.

A scale of the kind used for weighing things, like a spring scale, only measures weight (force). It can only determine mass by comparing the weight of an object to the weight of a known mass, weighed in the same location, so the strength of the gravitational field is constant. F=Mg.

Here is one way to look at it:

When the nut displaces water, the water depth gets greater. The pressure on the bottom is proportional to water depth. Force equals the bottom pressure times the bottom area. So with the same bottom area and increased depth, there will be additional force on the scale.

That is exactly the easiest way to describe it.

The easiest layman terms to use IMHO is that the head pressure increases with the increase in depth, and since the floor area stays the same, the scale shows more weight.

There is no more water than there was before... it has just moved.
If you have 10 cc of water in a wide glass it weights the same as 10 cc of water in a tall narrow glass.
That's a classic experiment you do with kids, they think the tall glass has more in it.
Del

As owner of a swimming pool, I can assure you that the 'weight' of an object decreases when it is immersed in the water. Therefore, the weight of the cup that the scale sees will increase.

I've been known to use the pool to cool various things after machining or welding.

Eww!

Homemade flotsom & jetsom.

Which brings up another interesting concept (as influenced by a dip in the Great Salt Lake as a lad.) , would this effect be affected by, say, the use of distilled water or ocean water (we will rule out heavy water due to difficulties w/procurement).

Which leads me to realize we are referring to an "affect" of a particular scenario & not a 'law of physics'.

tata for now

Jay

It's covered by earlier posts. Just replace "water" by "fluid". Archimedes still applies.

The salinity of the water will make a difference in the amount of water displaced by the nut,based on the specific gravity of the fluid,but the weight of displaced water will still be the same,Cargo ships have to know the salinity of foreign ports and oceans to adjust their ballast to maintain the ground clearance in that part of the ocean.

Your last statement is true, but you have amount (volume) and weight reversed.

Let me think about that. If You added say,10 pounds of weight,and displaced 10 pounds of water,the volume of water will be less as the Specific Gravity increases,but the weight of the water displaced will still be 10 pounds,even though it has less volume than plain water.

If amount means volume, the 1st sentence in #18 is wrong. The volume displaced must = the volume of body immersed. The weight of fluid displaced (hence upthrust on the body and apparent loss of weight) is proportional to the fluid density.

I agree.A poorly worded sentence on my part.

This nut is not buoyant. We can only talk about volume displacement. Whether it is water or honey the volume will be the same.

Having said that, it becomes interesting to work out exactly how much weight is added to the scale; water vs honey

Another interesting scenario to contemplate:

You are in a boat, with a solid carbon steel cylinder,6inch diameter, 1 foot long, and weighing 196 pounds,and you are in a pool or container of water .A very accurate level indicator is checking the water level.You drop the cylinder over the side and check the level of the water.Did the level rise or fall?

Clearly the level of water falls. The difference in volume is the difference between...

(Hey that's not fair you've edited the question.)

....between the volume of the cube cylinder and the volume of 200 196 pounds of water.

But here's another question: do you need to divide that volume by the area of the pond, or, the area of the pond minus the area of the boat?

My edit did not affect the result,only gave a more accurate weight,but only to 3 digits.

Do not need to know the actual level change,only asked if the level rises or falls.

But here's another question: do you need to divide that volume by the area of the pond, or, the area of the pond minus the area of the boat?

It gets worse than that. As the water level drops, the pond's area gets smaller. Without the weight, the boat rides higher and its "area" is smaller. So let's say the boat and the pond have vertical sides so the areas will be constant. It's a very tiny difference.

As far as the water level is concerned, moving the 196 lb weight from the boat is equivalent to subtracting 196 pounds of water from the pond. Dropping the weight into the pond is like adding a volume of water equal to the weight's volume.

If the boat is floating on the water, you would divide the change in volume by the area of the pond. As you subtract water, the pond's water level and boat will all drop equally.

In the OP,I only asked if the level rose or fell.All the other factors do not matter in order to answer the question.The shape of the object thrown overboard does not matter,as long it sinks and weighs more than water.The specific gravity of the object however will have an effect on how much the level will change.

Of course,the boat will change when the water level changes.It will not be suspended in the air!

You said; "The specific gravity of the object however will have an effect on how much the level will change."

Do you mean that <sg1 vs >sg1?

"The specific gravity of the object however will have an effect on how much the level will change."

I missed this at an earlier reading. NO! The specific gravity of the nut will only have an effect on the amount of water level change IF that specific gravity is less than the specific gravity of the water. There do exist wooden nuts that would float, so not all of their volume would submerge. If the nut floats, then not all of its volume is displacing water.

For any object with a density greater than that of water, the volume, and therefore the weight, of water displaced by lowering that object into the water will depend ONLY on the volume of the object. This of course is ignoring the volume of any air bubbles that get trapped in the object, as is likely to occur in the threads of a nut.

This sub thread has nothing to do with nuts (other than the participants): it's about HTRN's question in post #11.

Thanks for reminding me. I do forget easily...

I agree with your reply to #11. Originally, the boat was supporting the weight of the cylinder, so had to displace enough water to support that weight. when the cylinder is dropped overboard the boat will rise and the water level will drop. The volume of the cylinder is considerably less than the volume of water needed to support it in water.

Now on the other hand, where did the 196 pounds come from? I find the volume of a 6" diameter cylinder 12" long to be ≈339 cubic inches, and the density of carbon steel to be 0.284 pounds per cubic inch, which leads to a weight of 96.4 pounds.

Ha! being a pragmatist I did a quick experiment!
A small jar of water weighing 450grains, I dipped a length of steel in there being careful not to tough the bottom. The water level rose and so did the weight .
Quod erat demonstrandum.

Del