Buoyancy (Archimede's Principle) Problem

[Von Neumann]

Question:

Suppose a person weighing 530 Newtons is floating in a salt lake (concentration of 20% NaCl) with a specific gravity of 1.148. How much less of the person's body would be in the salt water as compared to if he were floating in ordinary water (w/ density 1.00g/cm^3)?

Comment:

One answer people keep giving me is that the weight "lost" is equal to the difference in specific gravities multiplied by the original weight. It doesn't seem obvious to me why someone would draw such a conclusion. I understand from Archimede's Principle that, as a result of the saltwater being more dense than ordinary water, less water must be displaced in order to balance the constant downward force mg of the person. However, I am having trouble expressing this mathematically.

 

[Dick]

Well, what volume of water is displaced to support 530N? Now the same question for salt water? The difference is what you are looking for.

 

[Von Neumann]

I really appreciate your help, but I'm really not sure how to go about finding the displaced water. The buoyant force must equal the weight of the displaced water, as well as the weight of the person because the system is in equilibrium, right?

 

[Dick]

I really appreciate your help, but I'm really not sure how to go about finding the displaced water. The buoyant force must equal the weight of the displaced water, as well as the weight of the person because the system is in equilibrium, right?

The person will sink until he/she displaces a volume of water that weighs 530N. Does that help?

 

[Von Neumann]

I think I understand!

F=mg
F=ρVg
=>V=F/(ρg)

So in freshwater:

V=530N/(1000kg/m^3*9.8m/s^2)=0.054m^3

And in salt water:

V=530N/(1148kg/m^3*9.8m/s^2)=0.047m^3

Is this correct?

 

[Dick]

I think I understand!

F=mg
F=ρVg
=>V=F/(ρg)

So in freshwater:

V=530N/(1000kg/m^3*9.8m/s^2)=0.054m^3

And in salt water:

V=530N/(1148kg/m^3*9.8m/s^2)=0.047m^3

Is this correct?

Yes. The difference in the volumes are what they are looking for. Or possibly the ratio. You can see how that would be connected with the ratio of specific gravities, right?

 

[Von Neumann]

Thank you very much!

 

[Von Neumann]

It seems that the specific gravities are intimately related to the density, making them useful in this problem.