Solving a Fluid Volume Physics Problem Using Archimedes' Principle

[Rachel C]

I'm stuck on a practice problem for physics (and I have a test tomorrow afternoon!)

A small sculpture made of brass (ρ = 8470 kg/m3) is believed to have a secret central cavity. The weight of the sculpture in air is 15.76 N. When it is submerged in water, the weight is 13.86 N. What is the volume of the secret cavity?

I think this problem involves Archimedes’ Principle (Fbuoyant = Wdisplaced fluid ). And I know W = ρ*V*g. The density of water is 1000 kg/m3. I know the answer to the problem is 4 x 10-6 m3.

I just do not know how to get that answer because everything I come up with is wrong! I tried plugging in the weight of water, density of water, and gravity and solving for the volume displaced. Then I did the same thing with the weight of the sculpture in air, density of brass, and gravity and solving for the volume of brass. Then I subtracted that volume from the volume displaced by water… and came up with the wrong answer! I would appreciate any help! Thanks!

 

[dextercioby]

What's the formula for F_{buoyant}...?And what is this force equal to?

Daniel.

P.S.Assume equilibrium when the body is submerged.

 

[Andrew Mason]

A small sculpture made of brass (ρ = 8470 kg/m3) is believed to have a secret central cavity. The weight of the sculpture in air is 15.76 N. When it is submerged in water, the weight is 13.86 N. What is the volume of the secret cavity?

The volume of the brass (ie. just the brass, without the cavities) is the mass/density. The mass is the weight (mg) divided by g (15.76/9.8 = 1.61 kg). The volume of just the brass, therefore, is: 1.61/8470 = 1.9e-4 m^3

The buoyant force is equal to the weight of the water displaced. The volume of water displaced is equal to the volume of the brass and cavities (assuming the cavities are watertight).

The buoyant force of 15.76-13.86 = 1.9 N in water = 1.9/9.8 = .19 kg of water = 1.9e-4 m^3 This is the volume of water displaced.

So what does that tell you about the alleged cavity?

AM

 

[dextercioby]

Okay,what are you saying,really...?

Daniel.

 

[Andrew Mason]

Okay,what are you saying,really...?

I am saying that this cavity appears to have a hollow ring.

AM

 

[dextercioby]

That's a weird conclusion.Your calculations seem to show otherwise.

Judjing after your calculations,the statue is not hollow,isn't that right...?

Daniel.

 

[Andrew Mason]

That's a weird conclusion.Your calculations seem to show otherwise.

Right. The calculations show there is no cavity. Hence the suggested cavity has a 'hollow ring' - ie is false.

AM

 

[dextercioby]

I hope you're very much convinced that there is a cavity in the sculpture and that your calculations need "polishing"...

Daniel.

 

[Rachel C]

The volume of the brass (ie. just the brass, without the cavities) is the mass/density. The mass is the weight (mg) divided by g (15.76/9.8 = 1.61 kg). The volume of just the brass, therefore, is: 1.61/8470 = 1.9e-4 m^3

The buoyant force is equal to the weight of the water displaced. The volume of water displaced is equal to the volume of the brass and cavities (assuming the cavities are watertight).

The buoyant force of 15.76-13.86 = 1.9 N in water = 1.9/9.8 = .19 kg of water = 1.9e-4 m^3 This is the volume of water displaced.

So what does that tell you about the alleged cavity?

AM

Thanks Andrew! When I used 3 significant figures, I calculated that the volume of the cavity is 4 x 10-6 m3, which is the correct answer. Thanks again!