How much lighter is a stone in water?

[Yora]

Even when the bouyancy of an object is not high enough to make it float to the surface, it still should reduce the amount of energy required to lift it, right?

If we take an ordinary scale under water and put a stone on it to weigh it, how much would the weight change compared to weighing it in air?

[crayzz]

Equivalent to the weight of the displaced water. Objects with a higher density than water will sink, since their weight exceeds that of the displaced water, and vice versa.

[halfeye]

Even when the bouyancy of an object is not high enough to make it float to the surface, it still should reduce the amount of energy required to lift it, right?

If we take an ordinary scale under water and put a stone on it to weigh it, how much would the weight change compared to weighing it in air?

While you made it clear the body of the first post exactly what you mean, the answer to the title of the thread is different, the force of gravity on the stone (i.e. the effect of the Earth's gravity on the stone's mass) does not vary, the difference in weight is a local phenomenon due to the weight of the displaced water.

Interestingly, some pumice includes enough gas that it does float in water.

[Manga Shoggoth]

If we take an ordinary scale under water and put a stone on it to weigh it, how much would the weight change compared to weighing it in air?

Interestingly, if you do it the way you describe (having the scale under water), the stone should appear slightly lighter, depending on the mechanics of the scale.

The classic method is to weigh the stone with a spring balance. The stone will show its usual weight outside the water, and when the stone (but not the balance) is immersed will show a weight equal to the weight of the stone less the weight of its equivalent volume of water.

However... If you immerse the scale in water as well, the mechanism of the scale will also show a small upthrust, making the stone appear to be slightly lighter.

[Yora]

Oh yeah, the scale would also experience bouyancy. Hanging the stone into the water with a very low mass (or low volume?) string should mostly cancel out that effect.

So "generic rock" has a density of 2.7g/cm³, and water a density of 1. That means a 1000cm³ rock weighs about 2.7kg and displaces 1kg of water. An hypothetical perfect object with a density infinitesimally lower than water effectively "hovers" just under the water surface, so it's "weight" in water is 0. And this means the weight of any 1000cm³ object in water is reduced by 1kg, right?
This means a 2.7kg rock would appear to weigh only 1.7kg in water. Does that sound sound?

I had been wondering what happens when a diver is exploring loose piles of rocks on the sea floor and the rocks shift and fall on him. Because of fluid resistance, the rocks would fall noticeably slower than on land, but I was curious how hard they would be to lift. A 37% weight reduction is more than I expected, but freeing yourself from heavy objects would still be hard.
(Things get even weirder in space, where you can slowly push giant loads with just muscle power, but they then can easily crush you when you get in their way. But innertia is a whole different topic.)

Also, I think "light" is not a defined physical term.

[Radar]

Oh yeah, the scale would also experience bouyancy. Hanging the stone into the water with a very low mass (or low volume?) string should mostly cancel out that effect.

So "generic rock" has a density of 2.7g/cm³, and water a density of 1. That means a 1000cm³ rock weighs about 2.7kg and displaces 1kg of water. An hypothetical perfect object with a density infinitesimally lower than water effectively "hovers" just under the water surface, so it's "weight" in water is 0. And this means the weight of any 1000cm³ object in water is reduced by 1kg, right?
This means a 2.7kg rock would appear to weigh only 1.7kg in water. Does that sound sound?

I had been wondering what happens when a diver is exploring loose piles of rocks on the sea floor and the rocks shift and fall on him. Because of fluid resistance, the rocks would fall noticeably slower than on land, but I was curious how hard they would be to lift. A 37% weight reduction is more than I expected, but freeing yourself from heavy objects would still be hard.
(Things get even weirder in space, where you can slowly push giant loads with just muscle power, but they then can easily crush you when you get in their way. But innertia is a whole different topic.)

Also, I think "light" is not a defined physical term.

It's pretty much as you described. Pushing rocks underwater will most likely be more difficult as in order to put your strength to use, you need to have firm footing. Since a typical diver will be more or less buoyant, it will not be easy for him to exert force on the rocks.

As a side note, this change in perceived weight is used to measure volume of objects, since weight can be determined with far better precision than volume directly.

[NichG]

Oh yeah, the scale would also experience bouyancy. Hanging the stone into the water with a very low mass (or low volume?) string should mostly cancel out that effect.

So "generic rock" has a density of 2.7g/cm³, and water a density of 1. That means a 1000cm³ rock weighs about 2.7kg and displaces 1kg of water. An hypothetical perfect object with a density infinitesimally lower than water effectively "hovers" just under the water surface, so it's "weight" in water is 0. And this means the weight of any 1000cm³ object in water is reduced by 1kg, right?
This means a 2.7kg rock would appear to weigh only 1.7kg in water. Does that sound sound?

I had been wondering what happens when a diver is exploring loose piles of rocks on the sea floor and the rocks shift and fall on him. Because of fluid resistance, the rocks would fall noticeably slower than on land, but I was curious how hard they would be to lift. A 37% weight reduction is more than I expected, but freeing yourself from heavy objects would still be hard.
(Things get even weirder in space, where you can slowly push giant loads with just muscle power, but they then can easily crush you when you get in their way. But innertia is a whole different topic.)

Also, I think "light" is not a defined physical term.

I think the difficulty of freeing yourself would be more about leverage than weight that you need to shift. I feel like the fact that the diver is much closer to neutrally buoyant would make the kinds of motions that a person does to lift something behave weirdly. A straight up vertical pull part would be fine, but if you were pulling at an angle, you're going to have much less friction against the ground than you would on the surface because your own weight will be closer to 90% compensated by buoyancy, compared to only 40% for the rock. So you might find that you slide back and forth in a way that harms your ability to maintain a consistent grip.

[crayzz]

As a side note, this change in perceived weight is used to measure volume of objects, since weight can be determined with far better precision than volume directly.

The basic approach to this sort of thing is relatively straightforward: you use a special scale and do the math.

It's a major pain in the butt when e.g you're trying to find the volume of some geological rock samples and the samples are porous, which makes finding the true volume either obnoxious (because filling every cavity with water will take a long time) or effectively impossible (because you'll never be able to fill every cavity OR because you have no way of verifying that every cavity is filled and so can never trust your measurements).

Spent a good many Saturdays being mad at porous samples.

[Aeson]

While you made it clear the body of the first post exactly what you mean, the answer to the title of the thread is different, the force of gravity on the stone (i.e. the effect of the Earth's gravity on the stone's mass) does not vary, the difference in weight is a local phenomenon due to the weight of the displaced water.

While this is true enough for the sorts of experiments that you could do with e.g. your bathroom scale and a filled tub, it bears mentioning that Earth's gravitational field does have local variations in both direction and strength, and if you have sufficiently-sensitive equipment you might be able to detect those variations while conducting such an experiment. There are essentially two reasons for this:
1. Assuming Earth were a perfect sphere of radius R_E and uniformly-distributed mass M_E, an object of mass M at Earth's surface will experience a gravitational force G * M_E * M / R_E² whereas the same object at a depth of d below Earth's surface will experience a net gravitational force of G * M_e * M / R_e², where R_e = R_E - d and M_e = M_E * (R_e / R_E)³ - effectively, in moving beneath the planet's surface you have reduced the apparent mass of the planet, because the part of the shell of the planet further away from the center of mass than and "above" you exerts a gravitational pull which is exactly cancelled out by the part of the shell of the planet further away from the center of mass than and "below" you.

2. Because Earth is not a perfect sphere of uniform density, Earth's near-surface gravitational field doesn't look exactly like the gravitational field of a single point mass placed at the center of the planet; rather, it looks something more like the net gravitational field of several bodies in close proximity to one another - most simply, a big point mass at the nominal center of the planet and a bunch of smaller ones scattered around the nominal surface of the planet whose masses and locations reflect known surface defects (e.g. continents). Obviously, this isn't a perfect model, either - especially not if you're modeling entire continents as a single point mass and looking at the gravitational field close enough to that continent that you really shouldn't have treated it as a single point mass - but if used appropriately it can be a more accurate model of the net gravitational field than a more idealized model of the Earth could produce. In a particularly fine model of the planet, you could include your tub of water as one of the bodies contributing to the net gravitational field seen by your rock and so your model would predict some slight changes in the net gravitational force experienced by your rock as you move the rock around (or into) the tub.

This isn't really something that you're likely to need to concern yourself with if you're just immersing things in tubs of water unless maybe you have really accurate scales, but it is pertinent to some applications.

[Radar]

The basic approach to this sort of thing is relatively straightforward: you use a special scale and do the math.

It's a major pain in the butt when e.g you're trying to find the volume of some geological rock samples and the samples are porous, which makes finding the true volume either obnoxious (because filling every cavity with water will take a long time) or effectively impossible (because you'll never be able to fill every cavity OR because you have no way of verifying that every cavity is filled and so can never trust your measurements).

Spent a good many Saturdays being mad at porous samples.

One thing that comes to my mind is sulfur tetrafluoride: high enough density to make a significant difference (and to keep it inside some container with almost no mixing with air) and it will get into all the pores as long as you roll the sample a bit to get those stray air bubbles out. As scary as the stoichiometry looks like, that gas is rather inert. Fortunately never needed to do this kind of measurements myself, but I feel your pain.

[Mastikator]

You can measure the density of the stone with a scale, a jug filled with water and the stone.

Put the jug on the scale, then place the stone in the jug, the water is displaced and escapes the jug. Then take the stone out and calculate the difference in mass. The stone's volume is 1 liter per kg that was removed from the jug.

Then you weigh the stone, it's density = mass / volume.

The weight in water is it's (density -1kg/l) x volume.

Example, a 2.7kg stone displaces 1 liter water, it has the volume of 1 liter, it has the density of 2.7kg/l, and in water weighs 1.7kg.

[Lord Torath]

Oh yeah, the scale would also experience bouyancy. Hanging the stone into the water with a very low mass (or low volume?) string should mostly cancel out that effect.

So "generic rock" has a density of 2.7g/cm³, and water a density of 1. That means a 1000cm³ rock weighs about 2.7kg and displaces 1kg of water. An hypothetical perfect object with a density infinitesimally lower than water effectively "hovers" just under the water surface, so it's "weight" in water is 0. And this means the weight of any 1000cm³ object in water is reduced by 1kg, right?
This means a 2.7kg rock would appear to weigh only 1.7kg in water. Does that sound sound?

I had been wondering what happens when a diver is exploring loose piles of rocks on the sea floor and the rocks shift and fall on him. Because of fluid resistance, the rocks would fall noticeably slower than on land, but I was curious how hard they would be to lift. A 37% weight reduction is more than I expected, but freeing yourself from heavy objects would still be hard.
(Things get even weirder in space, where you can slowly push giant loads with just muscle power, but they then can easily crush you when you get in their way. But innertia is a whole different topic.)

Also, I think "light" is not a defined physical term.

Worth noting that kilograms are a unit of mass, not weight, and the mass of the submerged object will not change. Newtons are the units for force, and that is what will be reduced by the submergence. But yes, other than that, you're correct. Your 2.7 kg rock will weigh (very roughly) 27 newtons in air, and 17 newtons while submerged (assuming that the acceleration of gravity at the earth's surface - 9.807 m/s² is approximately 10m/s² for simpler math).

[Aeson]

Worth noting that kilograms are a unit of mass

Just as there's a pound-mass (lbm), there's a kilogram-force (kgf or kg_F, also called a kilopond (kp)), defined such that one kilogram-force is the force required to accelerate a one kilogram-mass object at one standard gravity (alternately, one kilogram-force is the weight of a one kilogram-mass object in one standard gravity), and just as it's pretty likely for someone using pound-mass conversationally to drop the -mass, it's pretty likely for someone using kilogram-force conversationally to drop the -force. It's not as common a unit now as it used to be since it wasn't made a standard unit when SI was introduced, but it is still in use in some applications and it's not that uncommon to see it in pre-SI metric stuff.

As long as you're working in normalized gravities and are concerned about a gravitational field whose strength is about a standard gravity, it's reasonable enough to say that a 2.7 kilogram(-mass) rock weighs 2.7 kilograms(-force).

[halfeye]

Worth noting that kilograms are a unit of mass, not weight, and the mass of the submerged object will not change. Newtons are the units for force, and that is what will be reduced by the submergence. But yes, other than that, you're correct. Your 2.7 kg rock will weigh (very roughly) 27 newtons in air, and 17 newtons while submerged (assuming that the acceleration of gravity at the earth's surface - 9.807 m/s² is approximately 10m/s² for simpler math).

Yeah, that's what I was apparently failing to say.

[Lord Torath]

Just as there's a pound-mass (lbm), there's a kilogram-force (kgf or kg_F,
*snip*
As long as you're working in normalized gravities and are concerned about a gravitational field whose strength is about a standard gravity, it's reasonable enough to say that a 2.7 kilogram(-mass) rock weighs 2.7 kilograms(-force).

Ok, fair, enough.