Aluminium covered with gold submerged into water

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Homework Help Overview

The discussion revolves around a physics problem involving a piece of aluminum covered with a gold shell, submerged in water. The problem requires participants to determine the weight of the gold covering based on the given weights and densities of the materials involved.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Problem interpretation, Assumption checking

Approaches and Questions Raised

  • Participants discuss the relationship between density and weight, explore the concept of buoyancy, and consider how to separate the effects of aluminum and gold in their calculations. There are attempts to formulate equations based on the principles of buoyancy and density.

Discussion Status

Participants are actively engaging with the problem, offering guidance and exploring various interpretations of the equations involved. Some have suggested focusing on the aluminum first to simplify the problem, while others are trying to incorporate the gold component into their equations. There is a recognition of the need to clarify the relationships between the variables involved.

Contextual Notes

Participants mention the importance of Archimedes' principle and the constraints of the problem, including the need to express the weight of the gold in terms of the given constants. There is an acknowledgment of potential errors in previous calculations and a desire to ensure the accuracy of their reasoning.

Artj0m
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Homework Statement



A piece of aluminum is completely covered with a gold shell to form an
object of weight W. When you suspend the object from a spring balance
and submerge it in water, the balance reads T. The density of water,
aluminum and gold is ρw, ρa, and ρg, respectively. The gravitational
acceleration is g.

What is the weight of the gold covering the object? Express
your answer in the given constants.

Homework Equations



m/V = Ro

The Attempt at a Solution


I don't know where to start
 
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Artj0m, Welcome to PF! What formulas do you know that relate density to weight?
 
Borg said:
Artj0m, Welcome to PF! What formulas do you know that relate density to weight?
m/v = ρ
 
Artj0m said:
m/v = ρ
OK. Let's ignore the gold for now to make the problem simplier. If you put the aluminum in the water, what happens to the volumn of the water? Do you see a possible relation that you can work with?
 
Borg said:
OK. Let's ignore the gold for now to make the problem simplier. If you put the aluminum in the water, what happens to the volumn of the water? Do you see a possible relation that you can work with?

Yes, the buoyancy. I understand the concept, but the problem is that I eventually have to separate the two metals.
 
Artj0m said:
Yes, the buoyancy. I understand the concept, but the problem is that I eventually have to separate the two metals.
In order to solve the final equation, yes. Let's ignore W (and the gold) for now though.
When the aluminum is put into the water, the water level will rise. What do you think the relationship is for the increased water level w.r.t. the aluminum?
 
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Borg said:
In order to solve the final equation, yes. Let's ignore W (and the gold) for now though.
When the aluminum is put into the water, the water level will rise. What do you think the relationship is for the increased water level w.r.t. the aluminum?

The amount of volume the water rises is equal to the volume of the object.
That I understand.
I'm just having troubles with writing a good equation.
 
So, what is the upward bouyancy force you can attribute to this volume, V, of displaced water? What is the formula?
 
Artj0m said:
The amount of volume the water rises is equal to the volume of the object.
That I understand.
I'm just having troubles with writing a good equation.
What is the relationship between the Vw increase and Va? Can you apply your formula from post 3 to gain more info?
 
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  • #10
Borg said:
What is the relationship between the Vw increase and Va? Can you apply your formula from post 3 to gain more info?

If I apply the formula, I know that Vw = mw / ρw = malu / ρw

Because of the Buoyancy, B = Walu = mw * g = malu * g
So mw = malu
 
  • #11
Artj0m said:
malu / ρw
Are you sure about this part? Typo?
 
  • #12
The piece of aluminum covered with a gold shell is in equilibrium so:

T=W-(ρw*V*g)
 
  • #13
  • #14
@Daan Janssen - The goal of the homework section isn't to give people the answers. It's to guide them to the answer so that they understand how to solve it for themselves.
 
  • #15
Borg said:
@Daan Janssen - The goal of the homework section isn't to give people the answers. It's to guide them to the answer so that they understand how to solve it for themselves.
Yes, sorry, I do understand that. I myself, just as Artj0m, don't know how to get to the answer. It is just there to insure we get the right answer out of it.

I've really been trying and I know I am very close to the right answer. I'm hoping you could guide us to it
 
  • #16
Daan Janssen said:
Yes, sorry, I do understand that. I myself, just as Artj0m, don't know how to get to the answer. It is just there to insure we get the right answer out of it.

I've really been trying and I know I am very close to the right answer. I'm hoping you could guide us to it
OK. However, I think that your answer is wrong.
 
  • #17
Borg said:
OK. However, I think that your answer is wrong.
The answer was taken out of a correction sheet from Eindhoven University. The Question was part of a final exam. I doubt that it would be wrong. It could still be ofcourse :p
 
  • #18
Daan Janssen said:
The answer was taken out of a correction sheet from Eindhoven University. The Question was part of a final exam. I doubt that it would be wrong. It could still be ofcourse :p
It wouldn't be the first time that I was wrong. I'm having someone check my work before I guide you both down the wrong path.
 
  • #19
Daan, although that is correct, that is not what the question is asking for. Artjom, I think you may have summed your forces incorrectly. Do you draw force diagrams? I find that drawing a diagram really helps me get the forces right.
 
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  • #20
Without the water, the spring balance will record the full weight of the metal. Dipping the metal into water reduces the balance reading. You should be able to come up with an equation involving expressions for these components.
 
  • #21
Borg said:
You might want to also review the Wiki article on Archimedes' principle.

I've read it and I have come to this.
Like you said, I only started with the aluminium

ρa/ρw = W/(W-T)

This leads to

W - [(ρa/ρw) * (W - T)] = 0

This comes pretty close to the answer in my book:

Wg = (ρg / (ρg - ρalu)) * (W - (ρa/ρw) * (W - T))
 
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  • #22
Good.

If you repeat the experiment with iron, the expression would be ρiron/ρw = W/(W-T) with different W and T of course.
If you repeat the experiment with an alloy (e.g. Au and Al !) , the expression would be ρalloy/ρw = W/(W-T) with again different W and T of course.

All that remains for this exercise is to find ρalloy for W-Wgold kilograms of Al and Wgold kilograms of Au, and then work around to Wgold.
 
  • #23
BvU said:
Good.

If you repeat the experiment with iron, the expression would be ρiron/ρw = W/(W-T) with different W and T of course.
If you repeat the experiment with an alloy (e.g. Au and Al !) , the expression would be ρalloy/ρw = W/(W-T) with again different W and T of course.

All that remains for this exercise is to find ρalloy for W-Wgold kilograms of Al and Wgold kilograms of Au, and then work around to Wgold.

I understand, I'm just struggling with the fact that I don't know where in the equation I have to place the Wgold.
Could you give a clue?
 
  • #24
Artj0m said:
I understand, I'm just struggling with the fact that I don't know where in the equation I have to place the Wgold.
Could you give a clue?
4ecc7fb6cd8f6295aa090bcec2010d8c.png

Now that you've solved this for Al, add in the gold component. You should get an equation that uses the initial given values that you can then use to solve for Wg.
 
  • #25
Artj0m said:
I understand, I'm just struggling with the fact that I don't know where in the equation I have to place the Wgold.
Could you give a clue?
Giving a clue boils down to to helping find a mixing rule for density, right ?
So if A kilograms of material with density ##\rho_A## and B kilograms of material with density ##\rho_B## are fused to A+B kilograms, what's the density ##\rho_{A+B}##?

(Big clue: you may add up the volumes and you can use your own relevant equation)
 
  • #26
BvU said:
Giving a clue boils down to to helping find a mixing rule for density, right ?
So if A kilograms of material with density ##\rho_A## and B kilograms of material with density ##\rho_B## are fused to A+B kilograms, what's the density ##\rho_{A+B}##?

(Big clue: you may add up the volumes and you can use your own relevant equation)

So ρa/ρw = W - Wg / W - T
I don't understand how I should add this equation with the Aluminium one.
 
  • #27
Try again: if A kilograms of material with density ##\rho_A## and B kilograms of material with density ##\rho_B## are fused to A+B kilograms, what's the volume ?
 
  • #28
VA+B = mA / ρA + mB / ρB = (mA*ρB + mB*ρA) / (ρA + ρB)
 
  • #29
Yep ! Next step: what is the density of mA+mB kilograms of stuff that has a volume VA+B ? :)
 
  • #30
BvU said:
Yep ! Next step: what is the density of mA+mB kilograms of stuff that has a volume VA+B ? :)

Just (ρA + ρB) right?
 

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