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Basic Buoyancy problem and analysis help

  1. Dec 9, 2012 #1
    1. The problem statement, all variables and given/known data
    A slab of ice floats on a freshwater lake. What minimum volume must the slab have for a 49 kg person to be able to stand on it without getting his or her feet wet?

    (use 920kg/m^3 as density of ice and 1000 kg/m^3 for the density of freshwater)

    2. Relevant equations
    Archimedes Principle:
    Fg = Fb
    mg = pVg

    3. The attempt at a solution
    Hi guys! First post. I have the solution to my problem, however I don't understand how I came to it.

    What I did: to set up the problem, I found the person's weight, and then set it equal to the weight of the ice to try to solve for volume. This didn't work, but I'm still not really sure what the quantity I found means.

    mg = rho_ice*V_ice*g
    49kg*9.8m/s^2 = 920 kg/m^3*V_ice*9.8m/s^2

    Solve for V_ice:
    V_ice = (49kg*9.8m/s^2)/(920kg/m^3*9.8m/s^2)
    V_ice = 0.053 m^3

    Since my physics class uses MasteringPhysics for homework input, I tried this answer but it was wrong, so I googled the question and found these forums, and this post: https://www.physicsforums.com/showthread.php?t=288259

    So although LunarJK set his initial problem up differently, I tried to use the second portion of his solution, as here:


    piceVice = pwaterVwater
    (920 kg/m^3)(0.053 m^3 + x) = (1000 kgm3)x
    0.053 m^3*920 kg/m^3+920 kg/m^3*x = 1000x
    48.76 = 1000x - 920x
    48.76 = x (1000-920)
    x = 0.6095 m^3 --> volume of ice under water (which I believe is the total volume of ice)

    The total volume of ice ended up being 0.61 m^3 according to MasteringPhysics, but I can't seem to wrap my mind around what my first equation found, and what the proper method of solving this should have been. I've been having a lot of trouble in this class with analysis and setting up problems properly. Any help explaining or pointers/links to help with analysis on future problems would be a huge life-saver. Thanks,

    Matthew
     
  2. jcsd
  3. Dec 9, 2012 #2

    gneill

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    Staff: Mentor

    Hi mwalter, Welcome to Physics Forums.

    The idea is to find a volume of ice which, when just fully submerged, will displace an amount of water whose weight is the same as that of the ice plus the person. That way the ice sinks until its top is level with the water while the person standing on it remains above the water level.

    What expression will give you the weight of the ice plus person? What expression will give you the weight of the water displaced?
     
  4. Dec 10, 2012 #3
    Hi gneill,

    Thinking more about this, and that the weight of the fluid displaced is equal to the buoyant force, I believe that:

    W_person + W_ice = W_water displaced, so
    m_person*g + p_ice*V_ice*g = p_water*V_water*g

    and that answers both of your questions. It probably answers mine as well, or you wouldn't have asked it. It looks like I didn't consider the entire system when constructing my original solution. I just tested that equation with the numbers given and it worked. It seems so ridiculously simple that way. argh.
     
  5. Dec 10, 2012 #4

    gneill

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    Staff: Mentor

    Okay, that looks good. Now, what would be a good choice for V_water? In other words, what part of the system do you want displacing the water?
     
  6. Dec 10, 2012 #5
    On this note, I recently helped someone at work understand this concept. He asked to what depth a ice cube sinks into water seeing that some of it stands out above the water.
    I also used the same theory of weight displacement. Gneill can you varify this or have I missed something in the concept?
    To make it easy I used a 1x1x1 m ice block, with the density of 917 kg/m^3.

    W_ice = W_water + W_air
    p_ice*V_ice*g = (p_water*V_water*g) + (p_air*V_air*g)
    917*1*9.8 = (1000*(1*1*d)*9.8) + (1.225*(1*1*(1-d))*9.8)
    with d the submerged distance.
    This gives d = 916.9 mm... So the ice cube will be 83 mm above the water level... Is this correct or have I explained it wrong?
     
  7. Dec 10, 2012 #6

    gneill

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    Staff: Mentor

    No, that's fine. Note that for typical small, dense objects we can usually ignore the buoyancy due to air displacement, particularly since we tend to weigh things in air to begin with!
     
  8. Dec 10, 2012 #7
    The person is standing on top of the ice without getting their feet wet, so the volume of the ice should be equal to the volume of the water displaced?

    The way I sort of solved it was:

    49.0kg * 9.8m/s^2 + 920kg/m^3*9.8m/s^2*V_ice = 1000 kg/m^3*9.8m/s^2*V_water
    480.2 N + 9016kg/m2s2*V_ice = 9800kg/m2s2*V_water
    480.2 N = 9800kg/m2s2*V_water - 9016kg/m2s2*V_ice

    From here, I did something that was mathematically incorrect and subtracted, and it worked, but I don't know how.

    EDIT: If V_ice = V_water then obviously that would explain how it worked. Is this a fluke?

    Matthew
     
  9. Dec 10, 2012 #8

    gneill

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    Staff: Mentor

    Yes, that's right. The ice alone is displacing the water.
    A suggested shortcut: Since every term was multiplied by g, cancel g before proceeding and save yourself a lot of calculator sweat.
    No fluke. It's the ice that displaces the water. Archimedes rules!
     
  10. Dec 10, 2012 #9
    Thanks gneill!
     
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