Aluminium covered with gold submerged into water

[Artj0m]

You still have to get rid of W_A .

That's just W - W_G right?

 

[BvU]

Of course it is.

 

[BvU]

V_A+B = m_A / ρ_A + m_B / ρ_B = (m_A*ρ_B + m_B*ρ_A) / (ρ_A + ρ_B)

Before you go too far and get stuck anyway, I think there is something to be improved here. A sure sign of that is that the dimensions don't fit -- always a good check !

 

[Artj0m]

Before you go too far and get stuck anyway, I think there is something to be improved here. A sure sign of that is that the dimensions don't fit -- always a good check !

I'm stuck haha
What do you mean? I'm not sure I understand you correctly.

( I've changed the M into W, so it fits the equation ρ = ρw * W/(W-T))

 

[BvU]

What you found has a dimension mass. Should have been Volume

 

[Artj0m]

So the equation we made is not useful? Or did the equation had to be something like this?

m_A+B = ρ_A*V_A + ρ_B*V_B

 

[BvU]

V_A+B = m_A / ρ_A + m_B / ρ_B = (m_A*ρ_B + m_B*ρ_A) / (ρ_A + ρ_B)

Last + should have been a *

 

[mr166]

Here is how I would physically solve the problem.
1. Find the volume of the object.
2. Find the weight of the object.
3. Determine the density of the object.
4. Look up the density of that volume of AL and calculate the theoretical weight.
5. Subtract the theoretical weight from the true weight.
6. The difference is the weight of the gold.

Here is how I would do that.

1. Suspend the water tank from the spring scale.
2. Using a string of negligible volume and weight , suspend the object in the water without letting it touch the container. The change of weight in grams equals the volume of the object in CCs.
3. Let the object rest on the bottom of the container and measure the change in weight from reading in step one. This is the true weight of the object.

BTW buoyancy has nothing to do with the solution.

 

[BvU]

And with that we get the book answer ! Happily ever after (Have to admit it cost me a bit of effort too :) ) !

 

[Artj0m]

Thanks you guys! :)

 

[BvU]

Here is how I would physically solve the problem.
1. Find the volume of the object.
2. Find the weight of the object.
3. Determine the density of the object.
4. Look up the density of that volume of AL and calculate the theoretical weight.
5. Subtract the theoretical weight from the true weight.
6. The difference is the weight of the gold.

Here is how I would do that.

1. Suspend the water tank from the spring scale.
2. Using a string of negligible volume and weight , suspend the object in the water without letting it touch the container. The change of weight in grams equals the volume of the object in CCs.
3. Let the object rest on the bottom of the container and measure the change in weight. This is the true weight of the object.

BTW buoyancy has nothing to do with the solution.

Hello mr, welcome to PF :)

Nice to have you join in. So far I've been proceeeding on the hypothesis that post #1 is a homework exercise, not a lab practice (after all, gold is rather pricey :) !). And there isn't much physics in there, but a lot of juggling equations, something that can be pretty awkward (we needed over 40 posts already!).

I don't agree that buoyancy isn't involved. As far as I know buoyancy is our W-T !

But I do like your practical recipe. Do you agree that your 3. result minus your 1. result gives W, your 2. result minus your 1. result gives W-T. And with that we are back to the original exercise !

 

[mr166]

Hello mr, welcome to PF :)

Nice to have you join in. So far I've been proceeeding on the hypothesis that post #1 is a homework exercise, not a lab practice (after all, gold is rather pricey :) !). And there isn't much physics in there, but a lot of juggling equations, something that can be pretty awkward (we needed over 40 posts already!).

I don't agree that buoyancy isn't involved. As far as I know buoyancy is our W-T !

But I do like your practical recipe. Do you agree that your 3. result minus your 1. result gives W, your 2. result minus your 1. result gives W-T. And with that we are back to the original exercise !

After reviewing my reply, the the density of the object is not really relevant.

Only 1,2,4,5 and 6 are needed. Buoyancy is not a factor. Just the difference in weight between the empty, water filled, container and the container with the object resting on the bottom counts.

 

[Artj0m]

This has been a great help! Thanks! :)

 

[BvU]

After reviewing my reply, the the density of the object is not really relevant.

Only 1,2,4,5 and 6 are needed. Buoyancy is not a factor. Just the difference in weight between the empty, water filled, container and the container with the object resting on the bottom counts.

I really don't understand how you can say that. Buoyancy is W-T and it definitely is an actual factor in the answer. And of course the density of the object is relevant. We worked out this whole thing by equating the $\rho$ of the object as expressed in terms of W, W_Au, $\rho_{Al}$ and $\rho_{Au}$ on one hand and in terms of $\rho_{H_2 O}$, W, and W-T on the other.

 

[BvU]

This has been a great help! Thanks! :)

You're welcome. That's what PF is for.

 

[mr166]

I really don't understand how you can say that. Buoyancy is W-T and it definitely is an actual factor in the answer. And of course the density of the object is relevant. We worked out this whole thing by equating the $\rho$ of the object as expressed in terms of W, W_Au, $\rho_{Al}$ and $\rho_{Au}$ on one hand and in terms of $\rho_{H_2 O}$, W, and W-T on the other.

Yes you are right since Buoyancy = weight of displaced fluid and I was using the amount of displaced fluid to measure the volume of the object.