(adsbygoogle = window.adsbygoogle || []).push({}); 1. The problem statement, all variables and given/known data

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?

2. Relevant equations

Archimedes: F = rho*V*g

mass = rho*V

3. 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 thats 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/s^{2}) = 588 N.

Then F_{b}= 588 N right?

So I said 588 N = 1.0x10^{3}*V*(9.8 m/s^{2})

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 10^{3}kg/m^{3}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:

9.2x10^{3}kg/m^{3}i figured it wasn't this close to water's density

9.610^{3}kg/m^{3}i figured it wasn't this close to water's density

1.0x10^{3}kg/m^{3}obviously not this one

1.04x10^{3}kg/m^{3}obviously not this one either

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# Stupid Archimedes Problem

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