Determining the ratio of volumes

1. Sep 20, 2015

1729

1. The problem statement, all variables and given/known data
A crown is made out of silver and gold. Its weight is 58.8 N in air, and 54.8 N when submerged in water.
What is the ratio of $V_{\mathrm{gold}}$ to $V_{\mathrm{silver}}$ in the crown?

$\rho_{\mathrm{gold}}=19300\frac{\mathrm{kg}}{\mathrm{m^3}}$
$\rho_{\mathrm{silver}}=10100\frac{\mathrm{kg}}{\mathrm{m^3}}$
$\rho_{\mathrm{water}}=1000\frac{\mathrm{kg}}{\mathrm{m^3}}$

(a) $\frac{V_{\mathrm{gold}}}{V_{\mathrm{silver}}}=\frac{50}{50}$
(b) $\frac{V_{\mathrm{gold}}}{V_{\mathrm{silver}}}=\frac{44}{56}$
(c) $\frac{V_{\mathrm{gold}}}{V_{\mathrm{silver}}}=\frac{40}{60}$
(d) $\frac{V_{\mathrm{gold}}}{V_{\mathrm{silver}}}=\frac{56}{44}$

2. Relevant equations
(i) $V=\frac{m}{\rho}$

3. The attempt at a solution
The volume of water displaced by submerging the crown is equal to the volume of the crown itself.
The crown weighs 58.8 N in air, which implies the mass is 6.0 kg. It weighs 54.8 N when submerged, which implies the mass in 5.6 kg in water. Since the difference in mass is equal to 0.4 kg, this means that by formula (i), the volume of the crown is 0.0004 cubic metres.

Since this is the total volume of the crown, we can put this in an equation.
$V_{crown}=\frac{m_{gold}}{\rho_{gold}}+\frac{m_{silver}}{\rho_{silver}}$

How do I continue from this point? Would it be legal to rewrite the masses as $m=\frac{F}{g}$ and thus conclude (a) is the correct answer?

2. Sep 20, 2015

haruspex

Didn't understand your question at the end.
You have one equation and two unknowns. You just need one more equation. Can you think of one involving the total mass you worked out?

3. Sep 20, 2015

SteamKing

Staff Emeritus
Let's say that x = mass of gold in the crown. What is the mass of silver?

If you apply the density of each material to the respective masses, do you also obtain the total volume of the crown, as determined from submerging it in water?

4. Sep 20, 2015

1729

Thanks for the input guys. I feel like this question is way easier than I make it out to be, but I just can't connect the dots.
$m_{total}-x$
I don't know. I plugged these numbers in the equation and solved for $x$:
$0.0004 \ \mathrm{m^3} = \frac{x}{19300 \ \mathrm{\frac{\ kg}{m^3}}}+\frac{6.0 \mathrm{kg}-x}{10100 \ \mathrm{\frac{kg}{m^3}}}$
$\Leftrightarrow x=4.1 \ \mathrm{kg}$
If you were to calculate the volumes using $V=\frac{m}{\rho}$, and were to check the ratio it produces, you would notice it satisfies none of the four possible answers given. My intuition would give you an affirmative answer to your question, but I can't ignore this calculation that contradicts that assertion.
Just ignore the last question, I've worked it out to be false.

I don't see what other equation I would need. I've tried rewriting the masses in relation to each other -- see the calculation above -- but I suppose that's a dead end.

5. Sep 20, 2015

Staff: Mentor

The easiest thing is to work with the volumes of gold and silver. Let Vg be the volume of gold in the crown and let Vs be the volume of silver in the crown. In terms of Vg, what is the weight of gold in the sample? In terms of Vs, what is the weight of silver in the sample? These have to add up to the total weight. In terms of Vg and Vs, what is the total volume of the crown? What is the upward force exerted by the water on the crown when the crown is submerged (in terms of Vg and Vs). You should now have two equations in the two unknowns Vg and Vs. What are these two equations, and what are their solutions?

Chet

6. Sep 20, 2015

1729

Hi Chet,

Thank you for taking the time to reply.
After spending a multitude of hours on this problem today, it seems that I've formulated a linear system whose solutions appear to indicate (a) is the correct answer. Could you please have a look at it and confirm whether or not my reasoning is incorrect?

$4 \ \mathrm{N}=\rho_{water}g(V_g+V_s)$
$58.8 \ \mathrm{N}=\rho_gV_gg+\rho_sV_sg$
$\Leftrightarrow V_g=V_s$ which implies $\frac{V_g}{V_s}=\frac{50}{50}$

7. Sep 20, 2015

Staff: Mentor

Yes. Your formulation is exactly what I envisioned. I haven't checked your "arithmetic."

Chet

8. Sep 20, 2015

1729

Great, thank you!