Rock in water

This is more of a physics question then really anything relating to programming, but thought I'd ask anyway...

You have a rock tied to a rope, you submerge that rock in water, not too deep but it's completely covered. You see how much it weights; obviously it's going to be less then in the air. Now then, you take that same rock and put it much deeper in the water. The question is, will the weight be less, greater, or the same as when it was in the shallow water? The rock is not moving in either case, you're just holding it at the same depth.

Please let me know what you think and your reasons behind it.

I think it would be the same. The weight depends on the volum of water it displaces, since the stones volume is constant, the amount of water displaces is also constant. Hence I think the weight is the same.

On second thought, it would be ever so slightly greater, since it is closer to the centre of Earth, gravity would pull on it slightly stringer. But this effect would be miniscule.

If you're holding it there with a rope it has 0 weight :bigyello: .

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Originally posted by alkatran
If you're holding it there with a rope it has 0 weight :bigyello: .

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Wrong. It will still have a wieght, thought the force caused by it will be countered by the upwards force on the rope.

It also depends on how you measure the weight. If you're going by downward pressure on the rope, then it will be least in the air, next in the shallow water, and finally most heavy in the deep water. This is because of water (and air) pressures. If you measure the true weight, then they would all be the same.

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Originally posted by jemidiah
It also depends on how you measure the weight. If you're going by downward pressure on the rope, then it will be least in the air, next in the shallow water, and finally most heavy in the deep water. This is because of water (and air) pressures. If you measure the true weight, then they would all be the same.

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True weight? You mean mass?

Also we basically come to the same results; I myself said that the weight will not change while most of the others said the rock would get heavier in deep water. That I just don't get, water pressure would be acting on the rock from all sides, and so while it will increase with depth, it cancels itself out and does not cause the tension in the rope. As much as I tried to explain that people still said that it will get heavier so I'm kind of hoping that someone will do a better job of explaining one or the other then we were. As far as I see it, gravitational force does not change, the reason that rock is lighter when first submerged into water is due to buoyancy and the weight of the water displaced is subtracted from the rocks weight. Buoyancy force does not change with depth because the volume displaced is still the same as it was before, therefore, the only other force is tension and that should stay constant as long as the other two do.

Agree? Disagree?

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Originally posted by alkatran
If you're holding it there with a rope it has 0 weight :bigyello: .

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Damn, that must be only true in Canada :D j/k

totally agree. it will weigh the same before and after.

I thought weight was considered to be your "Force normale" (french, the force a static object gives to another.. like a table vs gravity) whereas a rope is a tension force...

wieght is:

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www.cogsci.princeton.edu/cgi-bin/webwn
the vertical force exerted by a mass as a result of gravity

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Mass is constant
gravity is near enough constant.
therefore wieght is constant.


problem with that method of explanation, is that according to that, it should also be the same in air as in water.

That's probably why I came to that conclusion.. the question just started me down the "normale" path.

the after having posted my previous post (i thought about the problem that it would always have the same weight. Then it clicked, of course it'll always have the same weight, but when it is in the water there will more upthrust so the stone will feel lighter.

weight is a constant basically as long as the gravitational field strength is constant and the mass is constant.

Contrary to popular beliefe water does compress a bit. For example water 2 miles down is denser than that at the surface.

In theory if the rock is not crushed by the water, it would sink and finally come to equilibrium at a point where the water is an equal density to the rock. The rock would be effectively weightless.

In reality though the rock would be pulverised before it ever sank to that depth. Unless it was perfectly spherical and contained no air cavities at all.

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Originally posted by wossname
Contrary to popular beliefe water does compress a bit. For example water 2 miles down is denser than that at the surface.

In theory if the rock is not crushed by the water, it would sink and finally come to equilibrium at a point where the water is an equal density to the rock. The rock would be effectively weightless.

In reality though the rock would be pulverised before it ever sank to that depth. Unless it was perfectly spherical and contained no air cavities at all.

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Denser than ice...!

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Originally posted by DiGiTaIErRoR
Denser than ice...!

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of course, otherwise ice would form on the bottom of the lake/pond/sea/body of water and the water-based life there would die.

At sea level, water is at its most dense when it is at 4 degrees C.

So ice is less dense than that.

Strange but true.

The object will weigh relatively the same regardless of the substance the object is in. It will "weigh" the exact same if it is in the water vs. if it is in the air.

The difference is how much downward force it would place on the item holding it. The water resistance pushing upwards makes the rock feel "lighter" in water, and makes it easier to move, vs virtually no air resistance supporting you.

You could encase the rock in cement and it will still weigh the same, it has nothing to do with the density of the water, etc.

There would be minor fluctuations based on the amount of water surrounding the rock, and the altitude of the rock.

If you are talking where someone holding onto the rope would feel the rock was the heaviest, you get into all sorts of variables, since as the person lowers the rope, they will have the weight of the rock, plus an ever increasing mass of rope. (1000m of rope gets heavy).

Assuming for purposes of arguement that the weight of the rope was either constant or ignored, the contribution of downward force the rock would have would be greatest at the air level, drastically reduced by placing it in the water, and somewhat reduced as it is lowered, as someone pointed out, the density of the water increases.

However, as I pointed out earlier, you'd have to lower it a long way for the density to become a factor, at which point you'd not only be holding up a rock, but a long length of rope as well.

Sorry this is a bit of a dead thread now my I think the following.

The mass of the rock remains constant.
The weight of the rock decreases the further down it goes, there is a slight increase in it's weight dur to it being nearer the center of gravity but I reckon the increasing density of water would conuteract this.

The whole 'why does ice float on water' thing i think is to do with the fact that the molecules actually align somehow when the water is frozen and take up more space than they do when they are all moving around happily and randomly in their water state

Ice floats because it is less dense than water.

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Originally posted by MixMaster
Sorry this is a bit of a dead thread now my I think the following.

The mass of the rock remains constant.
The weight of the rock decreases the further down it goes, there is a slight increase in it's weight dur to it being nearer the center of gravity but I reckon the increasing density of water would conuteract this.

The whole 'why does ice float on water' thing i think is to do with the fact that the molecules actually align somehow when the water is frozen and take up more space than they do when they are all moving around happily and randomly in their water state

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Acidic already said that weight is

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the vertical force exerted by a mass as a result of gravity

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That means that any other force is disregarded. If you're counting the upward push of water on the rock, why not count the upward push of solid ground on a rock? Or a scale? Everything is weightless except when it's accelerating downwards??

Couldn't weight be the amount of the net force (disregarding normal) in a given direction multiplied by its mass?

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Originally posted by Darkwraith
Couldn't weight be the amount of the net force (disregarding normal) in a given direction multiplied by its mass?

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My net force when standing up is 0. Thus my weight is 0. See the problem?

No, because if you disregard normal and consider that we are weighing your vertical force, your weight should be m * f = force of gravity.

Alright I've been hung, I'm being supported by rope tension. My net force is 0. My weight is 0.

(Goes off to lose some weight...) :afrog:

Isn't the upwards force of water considered normal force?

According to that, your vertical downward force (if we measure with a scale underneath you) is 0N meaning that you are weightless because all of your force is going through the rope. That rope has to be attached to something (lets say for our purposes, a scale). That scale (which if we face it to read the vertical upward force) should read -m * g, which would coororlate to a downward force of m*g.

This means that hanging yourself is not a valid weight-loss program. :)

In this problem, we have to find where the force of the water upward is equal to this net force.

Ah yes, but that means that what I'm being hung by weighs more due to me. The normal force is not linked to me directly, so I am assumed to be part of the contraption. When viewed seperately I am being affected by the rope which is being affected by the log, or whatever, which is being aff...

But honestly, this is an argument that neither side can win, because both sides are "correct".

My main point was the second one, the first was a joke.


Also, I hereby declare the earth to weigh whatever my weight is. The force of gravity I am exercing in it is the same as it is exercing on me. Everyone on the earth except me is considered part of it. (Congradulations, you just technicly lost 99.99% of your weight because you make up so little of the earth's mass!)

:lol:

We need a reference plane...

The original question has been answered, if you're arguing what wieght is, I posted a science dictionary defenition eariler. Use that one to see what mass, graviry, tension, and all the rest of the terms are if you need to.

Near the Earths surface you (and all objects) will always have a constant wieght. This is because gravity is constant and the mass is constant.

If a body is stationary the net force is always zero (ignoring rotationary forces).

In different mediums, the objects will feel lighter in more dense mediums as the upthrust (which will always be in the direction opposite to the weight) will partly counteract the wieght. In a sufficiently dense medium, the object will float, because it has a net force of zero. In fact it's weight is still constant, it is just totally counteracted by the upthrust. Obviously this balance is ruined as soon as you lift the body out of the liquid or push in into the liquid.

When something is being hung the object is still stationary hence it will have a net force of zero, but it will still weight the same. If you are being hung and you are taken to be part of the hanging contraption too, then the enitre contraption will have a net force of zero (as it is stationary (neither floating into the air nor sinking into the ground)), it's weight will have changed because your mass has been added onto it's mass thereby changing it's mass. It's net force remains zero because the ground beneath it will simply exert a larger counter force.

Why can't we consider weight to be force in a given direction disregarding normal or is there a special term for that?