How Do You Calculate Average Body Density Using Archimedes' Principle?

[aarunt1]

Homework Statement

A 60kg person floats vertically in a water pool with just her head, of volume 2.5L, exposed. What is the average density of her body?

Homework Equations

Archimedes: F = rho*V*g
mass = rho*V

The Attempt at a Solution

I got this problem on my midterm and I'm pissed off that I couldn't figure out the answer. I've had similar problems to this but none that had this many missing variables and I can't even find anything in the book to help me out after I just took the test. This was supposed to be an "easy" question so that's why I'm pissed and felt like punching my teacher in the face after the test.

I first drew a free body diagram where the person is in the water with only the head out of it. F_b points up, F_g points down.

Then I said well its in static equilibrium so F_b = F_g.

Then I figured well F_g = rho*V so 60kg *(9.8 m/s²) = 588 N.

Then F_b = 588 N right?

So I said 588 N = 1.0x10³*V*(9.8 m/s²)
so V = 0.06 L ?WTF?

The person is obviously not floating in 60 mL of water so I was pretty much stuck here and didn't know what the hell to do.

But since it happened to be a multiple choice question I went with an average density of 8.4 x 10³ kg/m³ since I already know the person is less dense than water and I remembered an iceberg example where the iceberg was 89% under water because the density of the fresh water was 89% of the density of the salt water. I only had common sense to justify my answer so I figured since the whole head was sticking out of the water that was about 84% of the body submerged.

The other answers were:

9.2x10³ kg/m³ i figured it wasn't this close to water's density
9.610³ kg/m³ i figured it wasn't this close to water's density
1.0x10³ kg/m³ obviously not this one
1.04x10³ kg/m³ obviously not this one either

 

[Borek]

Looks like you have selected wrong answer.

What is buoyancy definition?

Buoyancy equals body mass.

There is only one unknown, everything else cancels out.

 

[aarunt1]

Are you serious? Dammit! What the hell is the buoyancy definition then? All we learned about buoyancy was Archimedes Principle. As I stated: F = rho*V*g

Please explain how you got to your conclusion because I can't as I've said. There are too many unknowns or I'm just blind and am not seeing something obvious.

 

[alphysicist]

Hi aarunt1,

Homework Statement

A 60kg person floats vertically in a water pool with just her head, of volume 2.5L, exposed. What is the average density of her body?

Homework Equations

Archimedes: F = rho*V*g
mass = rho*V

The Attempt at a Solution

I got this problem on my midterm and I'm pissed off that I couldn't figure out the answer. I've had similar problems to this but none that had this many missing variables and I can't even find anything in the book to help me out after I just took the test. This was supposed to be an "easy" question so that's why I'm pissed and felt like punching my teacher in the face after the test.

I first drew a free body diagram where the person is in the water with only the head out of it. F_b points up, F_g points down.

Then I said well its in static equilibrium so F_b = F_g.

Then I figured well F_g = rho*V so 60kg *(9.8 m/s²) = 588 N.

Then F_b = 588 N right?

So I said 588 N = 1.0x10³*V*(9.8 m/s²)
so V = 0.06 L ?WTF?

What are the units of the $1.0\times 10^3$ factor? You said the force is 588 N; what is a Newton in terms of meters, seconds, etc? Based on those units, what are the units of the volume 0.06?

The person is obviously not floating in 60 mL of water so I was pretty much stuck here and didn't know what the hell to do.

But since it happened to be a multiple choice question I went with an average density of 8.4 x 10³ kg/m³ since I already know the person is less dense than water and I remembered an iceberg example where the iceberg was 89% under water because the density of the fresh water was 89% of the density of the salt water. I only had common sense to justify my answer so I figured since the whole head was sticking out of the water that was about 84% of the body submerged.

The other answers were:

9.2x10³ kg/m³ i figured it wasn't this close to water's density
9.610³ kg/m³ i figured it wasn't this close to water's density
1.0x10³ kg/m³ obviously not this one
1.04x10³ kg/m³ obviously not this one either

Are the answers beginning with 8.4, 9.2, and 9.6 supposed to be multiplied by $10^2$ instead of $10^3$?