1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Buoyancy (Archimede's Principle) Problem

  1. Jan 15, 2013 #1
    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.
     
  2. jcsd
  3. Jan 15, 2013 #2

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    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.
     
  4. Jan 15, 2013 #3
    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?
     
  5. Jan 15, 2013 #4

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    The person will sink until he/she displaces a volume of water that weighs 530N. Does that help?
     
  6. Jan 15, 2013 #5
    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?
     
  7. Jan 15, 2013 #6

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    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?
     
    Last edited: Jan 15, 2013
  8. Jan 15, 2013 #7
    Thank you very much!
     
  9. Jan 16, 2013 #8
    It seems that the specific gravities are intimately related to the density, making them useful in this problem.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook