New Balance Reading After Adding a Stone: X + Y - Z?

[songoku]

Homework Statement

A cup of water has weight X. A stone has weight Y in the air is immersed fully in the cup and it displaces water of weight Z. What will be the new reading on the balance after the stone is immersed inside the cup?

Homework Equations

The Attempt at a Solution

My guess that the new reading will be X + Y - Z, where X is the weight of the cup and Y - Z is additional weight by the stone (weight of stone in water). Am I correct?

Thanks

 

[Bandersnatch]

My guess that the new reading will be X + Y - Z, where X is the weight of the cup and Y - Z is additional weight by the stone (weight of stone in water)

So, if you were to use a stone that displaces all the water(Z=X), you'd make the water weightless?

 

[songoku]

So, if you were to use a stone that displaces all the water(Z=X), you'd make the water weightless?

errr...didn't cross my mind before

So, second guess, X + Z. The weight of water will be added by the weight displaced by the stone. Correct?

Thanks

 

[gneill]

I think you'd better define what the balance is weighing. What's on the balance?

Does the displaced water remain in the cup, or does it overflow and become lost?

 

[songoku]

I think you'd better define what the balance is weighing. What's on the balance?

Does the displaced water remain in the cup, or does it overflow and become lost?

The displaced water remains in the cup.

I think the balance weighs the normal force acting on it. My guess still X + Z

Thanks

 

[soothsayer]

I think the answer would be different based on whether the rock had just become fully emerged and was still sinking toward the bottom, or whether the rock had already come to rest at the bottom of the cup, would it not?

 

[gneill]

The displaced water remains in the cup.

I think the balance weighs the normal force acting on it.

So the balance is weighing the cup and all its contents, while the stone was weighed separately (yielding weight Y in air). That is: the cup is on the pan of the balance; the cup has water in it; the stone is added to the cup.

My guess still X + Z

That would depend upon whether the stone is allowed to sink to the bottom of the cup, or is being held in a fixed position under the water by some external force.

If you assume that the stone is allowed to sink freely and come to rest at the bottom of the cup (so there are no external forces acting), then what the balance reads should only depend upon the total mass and the total volume of air displaced by the "object" that is the cup+water+stone.

Does the total amount of air displaced by all the objects change when the stone is immersed?