Aluminium covered with gold submerged into water

In summary: 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.
  • #1
17
0

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
 
Last edited:
Physics news on Phys.org
  • #2
Artj0m, Welcome to PF! What formulas do you know that relate density to weight?
 
  • #3
Borg said:
Artj0m, Welcome to PF! What formulas do you know that relate density to weight?
m/v = ρ
 
  • #4
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?
 
  • #5
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.
 
  • #6
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?
 
Last edited:
  • #7
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.
 
  • #8
So, what is the upward bouyancy force you can attribute to this volume, V, of displaced water? What is the formula?
 
  • #9
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?
 
  • Like
Likes Artj0m
  • #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.
 
  • Like
Likes BvU
  • #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

ρaw = W/(W-T)

This leads to

W - [(ρaw) * (W - T)] = 0

This comes pretty close to the answer in my book:

Wg = (ρg / (ρg - ρalu)) * (W - (ρaw) * (W - T))
 
Last edited:
  • #22
Good.

If you repeat the experiment with iron, the expression would be ρironw = 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 ρalloyw = 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 ρironw = 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 ρalloyw = 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 ρaw = 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 = (mAB + mBA) / (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?
 
  • #31
Nope. (mA+mB) / VA+B. Check it out, do a numerical example and see that that isn't the same
 
  • #32
BvU said:
Nope. (mA+mB) / VA+B. Check it out, do a numerical example and see that that isn't the same

Oh I see! Because of ρ = M / V But how does this help me in the exercise?
 
  • #33
You already had (I removed the a, since it is valid for iron too and for any stuff):
Artj0m said:
ρ/ρw = W/(W-T)
So ρ = ρw * W/(W-T).

Since then you set up an expression for ρ (namely (mA+mB) / VA+B -- you still have to rewrite that to the given variables in the problem statement of this exercise).

And -- if we didn't make any mistakes -- equating this ρ to the earlier ρ leads to the book answer.
 
  • #34
BvU said:
You already had (I removed the a, since it is valid for iron too and for any stuff):
So ρ = ρw * W/(W-T).

Since then you set up an expression for ρ (namely (mA+mB) / VA+B -- you still have to rewrite that to the given variables in the problem statement of this exercise).

And -- if we didn't make any mistakes -- equating this ρ to the earlier ρ leads to the book answer.
For the ρ, I have:

ρg + a = [(WG + WA) / (WGA + WAG)] * (ρG + ρA)

So I will equate it with ρw * W/(W-T) and then I should get the correct answer right?
 
  • #35
You still have to get rid of WA .
 

Suggested for: Aluminium covered with gold submerged into water

Back
Top