can't plug one into the other when they do not correlate.... different concepts.

I've seen a whole 700 chem class struggle with that concept last semester. which reminds me of the arguments here.

Don’t bring up chemistry, next we will be talking about moles…

Well then if we're discussing gravity we need to start talking about one of my favs the meter/second/second...weeeee! Oh oh how about Newtons kg-2 m2?

If in space, then sure. But only because the the masses are coming closer. But as soon as there is building mass on the side opposite the center... well, consider how the moon and sun's gravitational fields distorts the earth. https://www.thoughtco.com/land-tides-or-earth-tides-1435299

Or imagine a planet of 100% water, with the block of steel at the center. What would be the big gravitational force on that block of steel, and which way would this gravitational field accelerate that block? ;)

Smart ass! :D Of course saying that weight doesn't change leaves ambiguity about the weight measured. I mean, if you jump on a trampoline, are you not weightless in the air? There are no forces holding you in the air. You posses both mass and acceleration yet for a moment you are weightless - just as you would be in space.

Back to the ocean thing... If, as an engineering exercise, the questions arose of, how much force would be needed to suspend a 100_lb ingot of Aluminum in a freshwater lake?
would you answer 100_lbs ? Yeah, you probably would. -LOL

But would you show your work...
SG_Al = 2.6,
Suspension Force = 100-(100/2.6)= 61.54_lbf
Buoyancy Force = 100/2.6 = 38.46_lbf
Total Force = Buoyancy Force + Suspension Force = 100_lbf

Or would you leave those details of the answer hanging until your teacher marked "100_lbf" wrong? Not that I would ever do anything like that. :cool:

Pop quiz.

What’s the value of gravity of the center of the earth?

Are we assuming a static spherically symmetric perfect density?

And/or do you mean center of mass?

No, I mean the gravitational constant at the center of the earth.

Assume the earth has an even distribution of mass, or just don’t worry about the rounding error.

Gravitational constant is the point value, not to be confused with the small g of gravity at that point.

...point value of the total mass.

Used for planetary interaction calcs and such.

OK, smart Alec, give me the answer for both big G and little g.

Ok, the G is earth-sized massive whereas the g is zero.

If we had a tunnel thru the center you can imagine that you would fall to the center, with diminishing g as the center is approached and Coriolis kicking your ass the whole way down and back the other side a ways. ;)

Sorry, but my first thought was: "Ok, a thread started by someone who didn't pay attention in science class and missed the explanation of weight versus mass..."

I would argue with you, and tossing terms like "misunderstood" is a dangerous thing with physics.

If you accept the equivalency principle of ma=F=mg, then "weight" is the total sum of forces on an object to keep it un-accelerated in the local coordinate system. That would mean that the buoyancy force is absolutely part of it, and the weight of an object within water is less than the weight of that object outside of water...in fact, a floating object has zero weight.

The OP never mentioned physics. There was no stipulation, hence the long discussion.

I think you ASSUmed physics was the only way to weigh in on this question. ;)

clever bugger you are!

Really? What other branch of science might deal with this question?

How about Ocean science?

I've done some design work for deep (miles down) submersibles. Controlling neutral buoyancy is kind of a big deal.

Check out syntactic foams which resist this...

https://nautiluslive.org/sites/defau...?itok=U-Z8MCjc

Seems like someone couldn't find her G Spot.

Id say it only seems like things might weigh less because they fall to the bottom slowly. Both because there is still air in them and the resistence of the water

Thanks island and others!
What would be the result if you used seawater at SG of 1.022 for buoyancy force rather than freshwater at 1.00? If it makes no difference, that's the answer.

I don't think the depth makes any difference except for gravity and didn't mean to go down that black hole. I should have omitted the word "weight" from my original question and rephrased it as island did above.

If you want to really nerd out see https://www.physicsforums.com/threads/weight-of-objects-as-they-descend-into-the-earth.207148/

Essentially gravitational field becomes stronger in many places just below the earths crust. This being due to the crust being less dense enough to keep us a bit distance away from the really dense part of the planet. Fluffy crust? From there, the maximum, the field is thought to go approximately linearly to zero at the center.

Denser water will create a larger buoyant force than fresh water.

According to this link, the density of the ocean increases the deeper you get and is 1.07 at 10km.
https://www.britannica.com/science/seawater/Density-of-seawater-and-pressure

There are brine pools at the bottom of the ocean like an underwater lake. Like on Spongebobhttp://forums.pelicanparts.com/uploa...1657369602.jpg

I think you need to have another think about that.

I disagree that weight is the total sum of the forces, it’s one force, yes.

Pretty sure that if you do the math the point where the gravitational attraction is the strongest is somewhere partway to the core. It is increasing and you go deeper into the earth, up to a point, but it begins decreasing again.

I learned more than just the answer to my question. A layman's question to an engineer (or lawyer) needs to be unambiguous in terms but you can see the fault there. Also, there are some really sharp folks on here in various fields and interests...not just a few but an overwhelming number

Force (weight) is a vector quantity, meaning it has magnitude and direction. If gravity and buoyancy are acting on a mass, the net force (weight) is the vector addition of the two forces.

Yeah, I've seen some in nature documentary shows. It's cool and a bit crazy to see a "pool" underwater.

https://i.ytimg.com/vi/dgZD2STt_OU/maxresdefault.jpg

Exactly.

That leads to the question that is "Is weight the net force acting on the mass, or is it only the force due to gravity?"

If you look in Webster, it lists the first definition of weight as being something measured by weighing. And then if you look up the verb weigh, it says to measure something as by a scale. So, to me that says that weigh can be how something is measured on a scale which would make it a net force. But, I also absolutely agree that in a physics class at a basic level, weight would be considered the result of one force acting on the mass due to gravity, not the net force.

Using those two definitions, an item that sinks in the ocean is going to weight less in the ocean than in air due to bouyancy (first definition, net force). But also, an item will weigh the same whether it's in the water or not, with the only way to change that weight to determine the exact gravitational constant at the particular point where the item is in reference to earth (second definition, force due to gravity only).

A satellite in orbit is said to be weightless - where the forces (gravity and centripetal) are balanced.

Set that satellite on the moon the weight would measure quite a bit less than if weighed on earth.

So, I think it safe to conclude that the weight of an object is dependent on the environment.

The only way you can claim an item will weigh the same whether it's in the water or not is to imply that you are not concerned about anything other than the mass and/or gravitational field on the object changing.

BUT, that completely ignores the environment specified - and ignores the crux of the OP question, which was specifically the environmental affect of water on the weight of steel.

To avoid ambiguity the term mass is used to isolate, when needed, the physical bulk of an item. When asking about weight, we (engineers) typically consider that the person means mass. But when they throw the Q out with the environment variable (space, moon, ocean) we assume that they do mean to ask about weight (measurable) in that particular environment.

If a 1 kg wench goes down for a ride in a submarine, that 1kg wrench will still be a kg even though it is on a sub that can be considered to be weightless in the water.

This is because we typically consider weight to be the force measurement between solid objects. In this example, the force between the wrench and the sub bench that it rests, and the sub is weighed between the sub and the sea floor.