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

In summary, the conversation is about a person floating in a water pool with only their head exposed and trying to determine the average density of their body. The conversation includes equations, an attempt at solving the problem, and a discussion about the correct units and potential answers.
  • #1
aarunt1
5
0

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
 
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  • #2
Looks like you have selected wrong answer.

What is buoyancy definition?

Buoyancy equals body mass.

There is only one unknown, everything else cancels out.
 
  • #3
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.
 
  • #4
Hi aarunt1,

aarunt1 said:

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?

What are the units of the [itex]1.0\times 10^3[/itex] 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 [itex]10^2[/itex] instead of [itex]10^3[/itex]?
 

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

1. What is the "Stupid Archimedes Problem"?

The "Stupid Archimedes Problem" is a thought experiment that presents a paradoxical question about infinite geometric series. It asks what would happen if a person were to take one step halfway to a wall, and then continue halving the distance between themselves and the wall for an infinite number of steps.

2. What is the solution to the "Stupid Archimedes Problem"?

The solution to the "Stupid Archimedes Problem" is that the person would never reach the wall. This is because the number of steps required to reach the wall would also be infinite, as each step halves the remaining distance.

3. Who came up with the "Stupid Archimedes Problem"?

The "Stupid Archimedes Problem" was first described by the Greek mathematician Archimedes, who lived in the 3rd century BC. However, the specific version of the problem involving infinite geometric series was popularized by mathematician and philosopher Bertrand Russell in the 20th century.

4. What is the significance of the "Stupid Archimedes Problem"?

The "Stupid Archimedes Problem" is significant because it highlights the concept of infinite processes and the limitations of human understanding and perception. It also has practical applications in calculus and other mathematical fields.

5. Is the "Stupid Archimedes Problem" a real problem that can be solved?

No, the "Stupid Archimedes Problem" is a thought experiment and not a real problem that can be solved. It challenges our understanding of infinity and serves as a tool for exploring mathematical concepts and principles.

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