- #1
aarunt1
- 5
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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. Fb points up, Fg points down.
Then I said well its in static equilibrium so Fb = Fg.
Then I figured well Fg = rho*V so 60kg *(9.8 m/s2) = 588 N.
Then Fb = 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