Solving Weight Equations: Jar & Stone

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Homework Help Overview

The problem involves a jar containing water on a balance, with a stone that has different weights when measured in air and when submerged in water. Participants are exploring the implications of buoyancy and how it affects the readings on the balance when the stone is immersed.

Discussion Character

  • Exploratory, Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • Participants discuss the apparent weight of the stone in water and question why it is not simply added to the weight of the jar. There is consideration of buoyancy and the forces involved when the stone is submerged.

Discussion Status

There are various interpretations of the problem, with some participants suggesting that the stone's weight should be considered differently when submerged. Guidance on the effects of buoyancy and equilibrium is present, but no consensus has been reached regarding the final weight calculation.

Contextual Notes

Some participants note the ambiguity in the wording of the problem, which may lead to different assumptions about the stone's position in the water during measurement.

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Homework Statement



A jar containing water is kept on the pan of a compression type balance which reads 500 gf. A piece of stone is suspended from the balance and reads 400 gf. The stone when lowered in water weighs 300 gf. What will be the weight of the jar and stone (weighed together) when the stone is immersed in water?

Homework Equations



Weight= Weight of jar+Vdg, ie.e. the upthrust because of Newton's third law of motion of equal reaction (opposite.)

The Attempt at a Solution



So the weight=500+(400-300)=600 gf. This answer matches with the answer given at the back of the book. But please can you explain the reason why the apparent weight of the stone is not added to the reading?
I mean to say that it should be equal to 500+100+300. Do you agree??
 
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The problem is worded a bit awkwardly. I am assuming that the stone weighs 400 gm when weighed conventionally and 300 grams when suspended in water. Why might it weigh less in the latter case? (hint: consider buoyancy)

But that should have no bearing on the total weight of rock plus water when the two are taken together.
 
It takes a lil bit of imagination to acknowledge this:

I'm supposing that you know why the net weight of the stone is reduced; its because of the upthrust which equals the weight of water displaced. Effectively this means that the water is exerting a force on the stone via upthrust and since the system is in equilibrium and from Newton's third law the stone should exert an equal and opposite force on the liquid. This force adds to its weight. Thus reading should become 500 + 100 = 600 gf!
 
Sorry, I thought it was a trick question, and believed the rock was allowed to drop to the bottom for the final weigh-in. My bad for not having read more carefully. Or am I still missing something?
 
600 gf is ok provided the stone is sinking through the water. When it hits the bottom total weight goes back to 900 gf.
 
Dick said:
600 gf is ok provided the stone is sinking through the water. When it hits the bottom total weight goes back to 900 gf.

True. However I assumed from the question that the stone was somehow suspended in the liquid.
 
Yea, poor wording IMO.
 
Yeah, so did I at first. But the question doesn't actually SAY that, does it? A complete answer should probably state both weights.
 
Yes, but the latter question is much more simple isn't it?! So we might assume the worse!
 

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