Archimedes Principle - A Rock and Water Magic Trick

[James William Hall]

I understand how to measure specific gravity of a rock. One weighs the rock then tares a beaker of water and then suspends the rock on a string to observe the “weight” gain of displaced water. My brain however could never get how, without adding a drop more water or adding the mass of the rock to the bottom of the beaker the weight goes up. No additional mass added. Yes, I know Archimedes’ principle and can do the math but I can’t help thinking it is a magic trick that everyone pretends to understand.

 

[kuruman]

If you understand Archimedes's principle, then you understand that the water exerts a buoyant force that is equal to the weight of the displaced water and is directed up.

If you also understand Newton's Third law, then you should understand that the rock exerts a reaction force on the water that is equal in magnitude to the buoyant force but has opposite direction. In other words, the immersed rock exerts on the water a force that is down and has magnitude equal to the weight of the water it displaces.

The scale dutifully reports the normal force on the surface of its top plate, which in this case is the weight of the water plus the reaction force equal to the weight that the rock displaces. That's because the water is at rest and has two down forces acting on it, one from the rock and one from the Earth.

If you have a kitchen scale, do this simple experiment. Fill a drinking cup with water, put it on the scale and tare the scale. Then stick your finger in the water and watch what the scale reads. The deeper you go, the more it reads. Magic? Absolutely not if you understand two basic physical principles.

 

[sophiecentaur]

One weighs the rock then tares a beaker of water and then suspends the rock on a string to observe the “weight” gain of displaced water.

As above, weigh the beaker plus its water on scales.
The reduction of the measured weight of the rock is due to the fact that the water is pushing it up. The reduction in that measured weight will be exactly the same as the gain in weight of the beaker. So 'total weight' is the same before and after.
Note: You have to avoid any overflow of water when you add the rock. In the story, Archimedes would have had to mop up all the spilled water and include that in any calculations.

 

[jbriggs444]

Yes, I know Archimedes’ principle and can do the math

So your concern is that you can grok it piecemeal but cannot grok it in fullness. Yes, at some point we all have to "shut up and calculate".

 

[jack action]

First, the scale doesn't measure masses, it measures forces.

Second, the force $F$ at the bottom of the beaker is the water pressure $p$ at the bottom times the area of the beaker's bottom $A$:
$$F= pA$$
And the pressure at the bottom is $p=\rho g h$.

Therefore if the water height increases and everything else stays the same, the force at the bottom must also increase. (more info)

Before you ask why a tall glass weighs the same as a short glass of equal volume, remember that the short glass will have a smaller area.

without adding a drop more water or adding the mass of the rock to the bottom

But the rock does "rest" on the water. It might be easier to imagine a floating boat: It doesn't sink but its full weight is added to the water. With a dense rock that wants to sink, part of its weight is added to the water, the other part being supported by the string attached to it.

 

[Ibix]

Might be interesting to imagine that the rock is suspended from a string attached to the ceiling by a spring, which you can use to measure the weight of the rock. What happens to the spring reading of weight if you extend the string so that the rock submerges?

 

[Mister T]

My brain however could never get how, without adding a drop more water or adding the mass of the rock to the bottom of the beaker the weight goes up. No additional mass added.

You probably think that the rock is held up entirely by the string. But that is not the case. The surrounding fluid pushes up on the rock, too. Thus the string exerts an upward force on the rock that is weaker than the weight of the rock.

Another way to look at it is to not have any fluid. Just let the table top exert a force on the rock thus lowering the amount of force the string exerts on the rock. No magic involved.

 

[kuruman]

Another way to look at it is to not have any fluid. Just let the table top exert a force on the rock thus lowering the amount of force the string exerts on the rock. No magic involved.

Did you mean "spring" instead of "string"? I think that pushing on a string would be a bit difficult.

 

[Mister T]

Did you mean "spring" instead of "string"? I think that pushing on a string would be a bit difficult.

The string pulls (upward) on the rock.

 

[James William Hall]

"So your concern is that you can grok it piecemeal but cannot grok it in fullness. Yes, at some point we all have to "shut up and calculate".

Yes, well said. If we are honest we must all end up at that point in our education though for most here it will be a bit further along than than the Archimedes Principle. I would never admit that I have the same feeling when I see a paper clip leap across space to attach itself to a magnet. But no worries, when we do the calculation and get the right answer we still get an "A" on the test in school or at work--even if we know in our heart that something is not. quite. right.

 

[A.T.]

My brain however could never get how, without adding a drop more water...

The water level raises, and a higher water column needs to be supported by more force from the bottom of the beaker.

 

[Ibix]

But no worries, when we do the calculation and get the right answer we still get an "A" on the test in school or at work--even if we know in our heart that something is not. quite. right.

More precisely, when we do the calculations we accurately predict what will happen if we actually do the experiment. Which model should we trust? Our intuitive model that sees the weight reading change in reality and goes "that's not right" or the formal mathematical model that predicted exactly what happened and explained why it happened?

 

[James William Hall]

Quite right, Ibix. I'm afraid I've expressed myself poorly. I, and we all, trust the mathematical models accepted by the community. My now elusive point was, and is, that I have an admittedly primitive emotional feeling about it over which I have no control to feel--but total control to ignore, which I do. Maybe no one else has feelings that physics can sometimes look spooky no matter the math, so I think I had better stop digging the hole. Many thanks everyone for your contributions.

 

[Ibix]

Maybe no one else has feelings that physics can sometimes look spooky no matter the math,

I understand this. The first time I watched a helium baloon drift towards the front of an accelerating vehicle was a strange thing for me, even though I knew why it happened. Personally, I take it as an opportunity to reflect on why I have that "I don't believe it" reaction. In the case of the helium balloon all I needed to do was add the air to my mental picture and see the balloon floating in it like a tethered submerged bouy in water, and then it made sense.

I'd suggest thinking carefully about exactly what it is that you aren't seeing in your mind's eye. A better model of the world is always better.

 

[Dullard]

There are a lot of things that can seem counter-intuitive, depending on the quality of your intuition. Intuition can be re-trained. When you look at this situation as you presented it, I understand what you're saying. As an individual of limited intellect (but breath-taking stubbornness) I have developed strategies for coping with that situation - I examine the boundaries of the problem:

I know that a rock hanging from a string (no water, ignore atmosphere)) would produce a tension in the string equal to the weight of the rock.

I don't have any problem understanding why the scale reading would increase by the weight of the 'rock' if the rock was floating (a 'cork, in this case) in the water (no string). If I add a small bit of weight to the cork, it will still float (new equilibrium) - with larger displacement of water. The added weight would show up on the scale. The buoyant force must be larger with the weight added.

If I change the string to a (skinny) stick and force the cork completely underwater, It makes sense that the scale reading would 'go up' by the force that I am applying via the stick. I know that water isn't (detectably) compressible, so depth doesn't change anything once the cork is fully submerged. It also makes simple 'intuitive' sense that the magnitude of that force is exactly the buoyant force less the weight of the cork.

looking at those explicit cases, it is apparent (for a submerged rock) that:

There is a buoyant (upward) force exerted on the rock. This force exists whether (or not) it exceeds the weight of the rock. The string tension is reduced by an amount equal to the buoyant force; The stick force is increased by the buoyant force on the cork. Rock/Cork are examples of the same problem.

 

[A.T.]

My brain however could never get how, without adding a drop more water ...

Maybe you should consider using a "rock" with the same density as water. It will float neutrally when fully submerged, thus the force in the string will go to zero. So this is just like adding more water.

 

[Charles Link]

I think the thing that may be puzzling the OP here is something that has come up before on this topic=
Archimedes principle is better stated as the buoyant force is the weight of the "effective" volume of water displaced. e.g. You can float a small object weighing one pound or more (that is shaped like the beaker) with only an ounce of water in a beaker=the "effective" volume of water displaced is the water that would be there below the waterline of the object.

 

[Meir Achuz]

You have to go back to a 'free body diagram', if they still teach that.
Draw a circle around the bucket of water and the rock.
Then you don't even need Archimedes.

 

[A.T.]

You can float a small object weighing one pound or more (that is shaped like the beaker) with only an ounce of water in a beaker=the "effective" volume of water displaced is the water that would be there below the waterline of the object.

I don't think that this is the issue of the OP. It rather seems to be the fallacy in thinking that, if a rock hangs from a string which is taut, then the string always supports the full weight of the rock.

 

[Charles Link]

It may be of interest to the OP though, and part of what may seem like magic that you can get a substantial buoyant force with even a small amount of water, as in my example above, (post 17).

 

[DaveC426913]

It may be of interest to the OP though, and part of what may seem like magic that you can get a substantial buoyant force with even a small amount of water, as in my example above, (post 17).

I am still trying to find a practical way to prove (Mythbusters-style) that you can float an ocean liner in a bucket of water.

 

[jbriggs444]

I am still trying to find a practical way to prove (Mythbusters-style) that you can float an ocean liner in a bucket of water.

Ocean liner: $700 M (purchase price).
Ocean liner rental, 1 day @ 10% per annum: $70M per annum / 365 days = $200 K (give or take)
Bucket of water: a buck fifty

Not practical on what they pay Science Advisors around here.

Wait a minute... buy low, sell high. Where did I put my tide tables!

*Whips out calculator*... 200 million kg times 9.8 m/s² times 2 meters is around 4 gigajoules. One kwh is about 4 megajoules and sells for maybe $0.10. So that's $100 gained from selling high! [less the cost of the generator]

 

[tech99]

The water level raises, and a higher water column needs to be supported by more force from the bottom of the beaker.

But the mass of water and hence its gravitational attraction is unchanged. A tall cylinder weighs the same as a squat cylinder.

 

[DaveC426913]

I meant to say you can float an ocean liner in a buckets-worth of water. You cannot float an ocean liner in a bucket of water (unless the bucket is several stories tall.)

 

[A.T.]

But the mass of water and hence its gravitational attraction is unchanged.

See my suggestion in post #16:

Maybe you should consider using a "rock" with the same density as water. It will float neutrally when fully submerged, thus the force in the string will go to zero. So this is just like adding more water.

 

[jack action]

But the mass of water and hence its gravitational attraction is unchanged. A tall cylinder weighs the same as a squat cylinder.

Yes, but the area it rests on is smaller, thus a higher pressure. Higher pressure equals more stress in the material which can then break.

Imagine yourself standing on the ground and not sinking. Then stand on top of a pointy spike and the spike sinks into the ground. Same weight, different pressure.

 

[James William Hall]

It occurs to me that AI would be unable to think something that is easily explained mathematically could still seem spooky and deserving of more thought--something maybe not quite right. I don't know if this is a weakness in humans or in AI.

 

[jack action]

It occurs to me that AI would be unable to think something that is easily explained mathematically could still seem spooky and deserving of more thought--something maybe not quite right. I don't know if this is a weakness in humans or in AI.

It's a weakness in AI.

What you are describing has nothing to do with intelligence, just a machine that repeats steps that were defined for it to do. Intelligence sees beyond the already known steps and can imagine something else. Sometimes what is imagined does not correspond to reality but it makes wonderful stories that encourage humans to imagine even more scenarios.

 

[nasu]

But the mass of water and hence its gravitational attraction is unchanged. A tall cylinder weighs the same as a squat cylinder.

But if you put a weight on top of either one they push harder on whatever they sit on. The scale does not measure weight but the force that pushes on it. The fact that the weight (as in gravitational attraction) of the water does not change is a red herring.

 

[Ken G]

Although equivalent things have been said above already, I would say the goal is to give your intuition better training, moreso than substituting your intuition with math (which is not necessarily the goal of physics understanding, it is more like giving up on the effort). Of course sometimes we have little choice when the math gets impenetrable, but this is not a situation of impenetrable math, so we should be able to find a way to train the intuition instead.

One way to train the intuition here would be to hold the string from which the rock is dangling, so you would have a visceral connection to the external force. Now you would notice that the force you apply is reduced when the rock goes in the water, and by exactly the amount the scale increases. So your intuition starts to understand that the scale is ultimately "taking over" responsibility for the missing force you feel. If you feel it, your intuition will accept it, without mathematical calculation of any kind.

Then to follow up, push a rubber ducky of the same volume into the water, and note the scale rises the same amount as before. Again you feel this force, which now is downward, so it's even clearer that the difference in force you are experiencing as the object enters the water should show up on the scale.