I Matter density on the bottom of a liquid container vs hanging in the middle of it

[roineust]

Why is the rough diamond in the following video held by a wire from above? Is it just so it will be easy to take the rock out of the water glass or is it to prevent the diamond from sinking to the bottom? If it is to prevent the diamond from sinking to the bottom, what is the reason for that ? Does matter push out the liquid exactly the same, if it is on the bottom of the glass or hanging from above? On the other hand, doesn't the wire make the rock less heavy, because it is hanging from it, while the wire is twisted from a holding rod on its other side? Doesn't this hanging of the rock from above, have any influence on the amount of water going up the water container? minute (1:43): Here it is

 

[Doc Al]

What is being measured is the weight of the displaced water. To do that properly, the only interaction you want between the diamond and the cup of water is the buoyant force. If the diamond is held suspended, then the added weight on the scale equals the weight of displaced water. If you let the diamond fall to the bottom, then the increase in weight measured would simply equal the weight of the diamond, not the weight of displaced water.

 

[roineust]

Doesn't the amount of displace water a result of the diamond weight? If the diamond is hanged from above, doesn't it take the weight down partially or totally, no mater what matter density it has? Does the water displacement has to do only with the diamond matter density and not at all with it's weight?

 

[anorlunda]

Suppose you pushed a balloon with the same volume as the diamond under the surface? But not touching the bottom. What would the scale read than?

 

[Doc Al]

Doesn't the amount of displace water a result of the diamond weight?

Most directly, the volume of water displaced will equal the volume of the diamond.

If the diamond is hangs from above, doesn't it take the weight down partially or totally, no mater what matter density it has?

We are not measuring the net force needed to support the diamond, but only the weight of the displaced water. (But yes, the wire does exert an upward force to make up the difference between buoyant force and the weight of the diamond. (The net force on the diamond, whether suspended or sitting on the bottom of the cup, must be zero.)

 

[roineust]

We are not measuring the net force needed to support the diamond, but only the weight of the displaced water. (But yes, the wire does exert an upward force to make up the difference between buoyant force and the weight of the diamond. (The net force on the diamond, whether suspended or sitting on the bottom of the cup, must be zero.)

I think i understand, but in this case it is a negative buoyant force and if it was a piece of wood it was positive buoyant force?

 

[Doc Al]

I think i understand, but in this case it is a negative buoyant force and if it was a piece of wood it was positive buoyant force?

The buoyant force on the object, whether diamond or wood, will always be upward. Since the diamond is more dense than water, you must suspend it as shown to measure the displaced water and thus the buoyant force. For a piece of wood that floats, you can just let it float.

 

[Doc Al]

Note: While the buoyant force on the object (the water on the object) is upward, the added force on the water (the object on the water) and thus on the scale is, of course, downward.

 

[roineust]

The buoyant force on the object, whether diamond or wood, will always be upward. Since the diamond is more dense than water, you must suspend it as shown to measure the displaced water and thus the buoyant force. For a piece of wood that floats, you can just let it float.

If you let a wooden cube float above the water, then although it is not all submerged inside the water, but just floats above the water, you still get from the amount of displaced water (knowing the matter density of both water and wood) the volume of that cube, is that saying correct? Or will amount of displaced water change as i push more and more of the cube inside the water?

 

[Doc Al]

If you let a wooden cube float above the water, then although it is not all submerged inside the water, but just floats above the water, you still get from the amount of displaced water (knowing the matter density of both water and wood) the volume of that cube, is that saying correct?

The weight of the displaced water will equal, per Archimedes' principle, the weight of the cube. Knowing the density of the cube allows you to calculate its volume.

Or will amount of displaced water change as i push more and more of the cube inside the water?

Absolutely. If you push down on the cube you will displace more water. The weight of the displaced water will equal the weight of the cube plus the additional force you exerted to hold it down. (If you just let it float, there is no additional force.)

 

[roineust]

Thanks, i wonder why i forget the buoyant force, each and every time i come to think of Archimedes' principle. What is there in buoyant force, that is so un-intuitive? I guess if we were human fish, it was more intuitive. (-:

 

[roineust]

I wonder, could there be found not even a minute difference for the amount of water displaced outside the water glass, for different shaped objects of the same kind of matter and volume, especially the kind of matter that you have to wholly submerge and hold hanging inside the water, so they won't touch the water glass bottom? For example, no difference between an upside coned object and a downside coned object of the same matter and volume, what so ever? Or between submerging an oblong shaped object horizontal and vertical?

 

[Doc Al]

For example, no difference between an upside coned object and a downside coned object of the same matter and volume, what so ever?

No difference whatsoever. It just depends on volume, not shape.