Floating objects and measuring their weight

1. Nov 1, 2012

K41

So I came across a question on the internet and its confused me a lot.

We have 200g of water inside a cup. We neglect the mass of the cup completely and assume no water is lost. We also assume that the gravitational field is uniform within our experiment room.

So we place the cup on some scales and it weighs 1.962N. (0.2 * 9.81).

Now we have a 10g mass object which comes in two different volumes. One volume is such that the object sinks. The other volume is such that the object floats.

So the question was asked whether the scales would record different weights for the case where the object sank and the case where the object floats.

Now the reply for the question given was that the weight (the one measured by the scales) would stay the same, regardless. But I've seen other replies on the internet that start talking about "apparent" weight etc etc and getting confused..

2. Nov 1, 2012

someGorilla

Why don't you do the experiment?

3. Nov 1, 2012

K41

I don't have the equipment lol.?

4. Nov 1, 2012

AJ Bentley

OK - this is a bit of confusion between the simple matter of conservation of mass and Archimedes' Principle.

In the latter, the apparent weight is how much something seems to weigh basically if you weigh it under-water.
If you try to weigh yourself in a swimming pool using a pair of bathroom scales, you apparently weigh almost nothing or even a negative weight.

Meanwhile, the whole swimming pool increased in total weight by the amount of your own body when you got in.

5. Nov 1, 2012

K41

Ah okay, I understand apparent weight now. Another misunderstanding I have is the following:

The Buoyancy force acts upwards in this case and so it opposes the weight. Firstly, how do we know what direction it acts in and secondly, if it opposes the weight, what mechanisms still cause the scale to register the additional weight of the object?

6. Nov 1, 2012

Staff: Mentor

In the direction of the (negative) pressure gradient of the medium - which is simply "up" in a gravitational field.

Buoyancy of water (and most other everyday objects) in air is often negligible. In addition, water is nearly incompressible - it keeps its buoyancy all the time in the atmosphere, so it is sufficient to look at its apparent weight (weight in vacuum minus buoyancy).

7. Nov 1, 2012

AJ Bentley

The idea of apparent weight is a short-cut.

The true situation is that the water level rises slightly by an amount proportional to the volume of the object. That mass of water presses down into the water, increasing the pressure. The net result of that is a force on the underside of the object to balance the extra weight of water. It's kind of difficult to put into words but think about it. It's just like putting the water on a seesaw with the object - a balancing thing

Anyway, the net effect is that the object appears to lose weight by an amount equal to it's own volume of water - but it's only an 'illusion'. The object hasn't really lost weight.

8. Nov 1, 2012

K41

See, the way I keep thinking it, is I have a free body force diagram drawn such that on the object, we have weight going down, buoyancy going upwards. Then on the water we have a weight going down. So the forces going down would be just the weight of the water and so the scales would read 200g. Now I know that is wrong because we've added mass into the system but i can't picture the newton third law forces.

9. Nov 1, 2012

HallsofIvy

What everyone has been trying to tell you is that "bouyancy" is purely "internal". What ever "bouyancy" force the water applies upward will add to the weight registered. If the cup and water weight 200 g and you add a 10 g mass, whether it floats or sinks to the bottom of the cup, the scale will register the same as if you just put a 210 g mass on it to begin with.

10. Nov 1, 2012

sophiecentaur

If that were not true, I am sure someone would invent a perpetual motion machine based on the weight difference. So it's just gotta be true.

11. Nov 1, 2012

JustinRyan

If we have 2 identical containers with the same amount of water, add to each an object with identical volume but slightly differing densities. Such that they both float, fully submerged without touching the bottom. They are suspended at differing heights in the water.

Is the weight of the container given purely by the pressure on the bottom of the container? As identical amounts of water are displaced, would this not be equal?

12. Nov 1, 2012

nasu

You cannot have all these conditions simultaneously. If they have same volume but different densities they cannot both float when fully submerged.
They will have different weights but same buoyant force when fully submerged.
If one floats when fully submerged, the other one will either sink or will rise above the water level.

13. Nov 2, 2012

Staff: Mentor

If the densities are very similar and you take the compression of water into account, it is possible. But then the water level is different in both containers (the same volume corresponds to different amounts of water), and the scales will give different values.

14. Nov 2, 2012

nasu

Right, if we consider the compressibility of water.
I am sorry, I did not see that this was not one of the OP' posts.
I would not think that the OP questions would benefit from considering this effect.

15. Nov 3, 2012

Jupiter6

I think you're mixing the frames of reference which is what the intended purpose of the question was. They're trying to confuse you.

If you were just to consider the water and the object alone, you'd have the object weight and buoyancy force. The water mass would not matter.

If you consider the water, object AND scale...the scale only cares about the weight upon it. In other words, while the buoyancy force and weight of the object may seem to cancel out in the water-object frame, it won't here.

Think of it this way. While the water is providing a buoyancy force, that only decides whether the object floats or sinks in the water. The mass is still there as far as the scale is concerned. As far as the scale is concerned, the weight of the object is added to the mass of the water.You could freeze the water and drop the ice cube and the object on the scale, same reading.

To grasp this, think about what keeps an airplane up. Ultimately the ground does.

Frame of reference is everything.

16. Nov 3, 2012

JustinRyan

apologies for the divergence from the original question. Thank you for your answers.

17. Nov 3, 2012

K41

Yup, I see it more clearly now with frame of references.

Thanks everyone!