Determining the ratio of volumes

[1729]

Homework Statement

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) $\frac{V_{\mathrm{gold}}}{V_{\mathrm{silver}}}=\frac{56}{44}$

Homework Equations

(i) $V=\frac{m}{\rho}$

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?

 

[haruspex]

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?

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?

 

[SteamKing]

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?

 

[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.

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

$m_{total}-x$

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?

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.
So, the answer is: "no"?

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?

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.

 

[Chestermiller]

The easiest thing is to work with the volumes of gold and silver. Let V_g be the volume of gold in the crown and let V_s be the volume of silver in the crown. In terms of V_g, what is the weight of gold in the sample? In terms of V_s, what is the weight of silver in the sample? These have to add up to the total weight. In terms of V_g and V_s, 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 V_g and V_s). You should now have two equations in the two unknowns V_g and V_s. What are these two equations, and what are their solutions?

Chet

 

[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?

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

 

[Chestermiller]

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?

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

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

Chet

 

[1729]

Great, thank you!