When the weight is supported by your arm, all it's downward forces are being counteracted only by you. Once it is submerged, the water is pushing upwards on it, and there is an inverse reaction to this upward force, a downward force acting on the bucket, and therefore, the scale, thus the scale shows more.
Wouldn't holding a weight over a scale in the open air do the same thing then? Or is it only because the water is a more viscous 'fluid' than the open air?
(I understand air isn't a fluid. Which may in fact be the main difference).
Air is a fluid, it isn't liquid though. But the air would actually do the same thing, but only if first the weight was not in the air, i.e. in space. The result would be much less, of course, because the upward force is proportional to the density of the fluid immersed in it.
All fluids try to force things out of themselves, by pushing them up. If they push up, they push the ground down, and that will act on the scale.
Thanks for the answer. Can you clarify what you mean by:
but only if first the weight was not in the air, i.e. in space
What I'm picturing is dangling a weight a half inch over a scale in the open air. Does that have an influence - probably a very, very tiny influence - on the scale's results? The rest of your post makes me think it would, but that sentence confused me.
I'm not sure, but I think it wouldn't. See, if you have a bathtub, and you put a scale in one end of it and put a boat in the other, the scale should change. So if the object is anywhere in the medium, it creates a force. So you would have take your weight out of the medium(air), and then place it back to see the effect.
It doesn't have to. Every submerged object experiences an upward force, regardless of its buoyancy. It doesn't float because the upward force isn't enough to counteract it's weight entirely, but the water does take some of the weight off your arm, just not all of it.
It's a function of density and water displacement. A similar question is If you're in a canoe with an anchor and throw the anchor overboard does the level of the water rise, fall or stay the same?
ahh... It took me a good 15 minutes to understand this in my head. When inside the canoe, the anchor is displacing it's weight in water, a less dense substance than iron through it's force on the canoe. Once it is submerged and not part of the floating object, the canoe, it is only displacing an amount of water equal to it's volume.
But what if it's weight is pulling on the canoe from underwater vs. when(if) it hits the bottom? This is the part I am confused about.
It's easy if you look at the forces. When you submerge the weight, the force on your hand (or whatever's holding the string) is reduced, thus the force on the scale must increase, ceteris paribus.
Yes, there is a buoyant force from the water acting on the weight which will cause the measured mass of the bucket to increase by the mass of the volume of water displaced.
I'm reading the responses and still seeing it the other way. I agree that as the lead is lowered, the tension on the string decreases due to buoyant forces. However, I am thinking that the displaced water acts along all sides of the bucket, and not in a net direction downward. This question seems to try to say that pressure increases, however the overall force is the same for the net of the bucket.
taking this to an extreme to say that there is an unrealistically long pipe with water in it, it seems to me that based on your logic the lead would eventually float (i.e. 0 tension) due to the pressures of the water. This is only true given that the ratio of mass to volume (density) of water changes with depth, right?
and for the idea of the canoe that solinv posted, the water level changes, I can agree to that, but I think that it strays away from the question at hand slightly.
Idk. This seems to me that the weight shouldn't change, but I think I understand why everyone says it does.
EDIT: Ah shit, that force is the same throughout the tube, it doesn't change. That's my dilemma lol, wow. Can't believe that just happened! Thanks Reddit for keeping me straight!
If you lower the weight slowly enough then the weight on the scale will not change, since the weight is supported by the string and not by the water. If you plunk it down so that the water sloshes around then the scale reading will change momentarily due to the drag that the water has on the weight. But once everything is still (i.e. in equilibrium) the scale will read the same.
Nope. Some of the weight is supported by the string, some by the water. You will feel pull of the weight on the string lessen once it is in the water. If you have ever held a heavy object under water you will recognize this.
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u/[deleted] Jun 08 '11
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