Archimedes Principle: Pure gold or not?

[Phuf]

Homework Statement

Archimedes, prove your stuff. A royal crown (cola?) of pure gold is suspended in air by a thread and the tension in the thread is of magnitude T. When the crown is totally immersed in water, the magnitude of the tension in the thread is 0.872T. A) sketch a second law force diagram indicating the forces acting on the crown when suspended in the air. B) sketch a second law force diagram indicating the forces acting on the crown when suspended in water. C) Is the crown made of pure gold or of another metal with just gold plating? (the density of pure gold is 19.3x10^3 kg/m^3) Justify your answer.

Homework Equations

Fbuoy+Ftension-w=0

The Attempt at a Solution

I tried using Newtons second law which eneded up getting me that ROHobject(Vobject)+mg+ma=m and If you cancel the m's out you get Rohobject(Vobject)+g=0. I am confused on how to tell how this relates to whether the crown is made of pure gold or another metal.

 

[gneill]

Hi Phuf, Welcome to Physics Forums.

Please try to select thread titles that are more specific and descriptive of the actual problem or question being asked. Titles that are overly generic are frowned upon . I've modified your thread title this time.

Can you define the variables you've used in your attempt? What assumptions have you made (particularly involving the buoyancy of the object in air)?

 

[Phuf]

Sorry about the title! originally i tried to use ROHwater(vsubmerged)=ROHobject(Vobject) but I couldn't figure out Vsubmerged so I am giving this new equation a try. I use The density of gold for ROHobject and for Vobject I calculated it by setting it equal to 0.872g/ROHobject which gave me 4.43x10^-4. But after plugging that into the formula i derived I ended up getting 19309.81=0 which doesn't make sense at all to me.

 

[Phuf]

whoops i meant 18.36=0 not 19309.81, that number is from a differnet problem I've already completed.

 

[jbriggs444]

Sorry about the title! originally i tried to use ROHwater(vsubmerged)=ROHobject(Vobject)

As previously asked, what are the meanings of the variable names: "ROHwater", "vsubmerged", "ROHobject" and "Vobject".

What new equation are you using and what physical justification do you have for using it?

 

[Phuf]

Roh=density, So Id have the density of water multipleied by the submerged volume=the density of the object times the volume of the object. The justification i have for using it is that I was able to find all the variables for it since I knew the density of pure gold was 19.3x10^3, the density of water is 1000kg/m^3, and that the tension would be 0.872T.

 

[gneill]

Sorry about the title! originally i tried to use ROHwater(vsubmerged)=ROHobject(Vobject) but I couldn't figure out Vsubmerged...

Since the object is not floating you can't say anything about the mass of the water displaced versus the mass of the object. So that relationship would not have been helpful.

Since you are given a ratio of the tensions ($r = 0.872 = T_w/T_a$, where $T_w$ is the thread tension in water and $T_a$ is the tension in air) would it not make sense to write expressions for the two tensions and form a ratio from them?

By the way, if you click on the $\Sigma$ icon in the edit panel toolbar you will have access to characters from the Greek alphabet (along with other math symbols) to include in your text. This may help make the equations you write more clear. The $x_2$ and $x^2$ icons can be used to create subscripts and superscripts. LaTeX syntax for equations is also recognized, if you happen to be familiar with it.

 

[Phuf]

So T_air=weight=mg=ρVg and 0.872T_liquid = mg-ρVg so T_air-0.872T_liquid = ρ_waterVg

 

[Phuf]

So T_air=weight=mg=ρVg and 0.872T_liquid = mg-ρVg so T_air-0.872T_liquid = ρ_waterVg

it should be t air - t liquid

 

[gneill]

it should be t air - t liquid

Perhaps you should write out your equations again. Place one per line so they are easily distinguished and easy to quote.

 

[Phuf]

T_air=weight=mg=ρVg

and

0.872T_liquid = mg-ρVg

so T_air-0.872Tl_iquid = ρ_waterVg

 

[gneill]

T_air=weight=mg=ρVg

and

0.872T_liquid = mg-ρVg

I don't understand that last equation. Isn't ρVg equal to mg by the equation above it?

Ignore the ratio 0.872 for now. Just write an expression for the tension in liquid using V and the two densities (one for the water and an unknown one for the metal).