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Archimedes principle

  1. Dec 11, 2008 #1
    1. The problem statement, all variables and given/known data
    Find: (a) volume of plank (b) bouyant force (c) density of plank
    mass plank = 877kg
    mass bricks loaded on plank = 2.3 * 74 = 170.2
    plank if floating level with surface of lake

    2. Relevant equations

    p = m/vol
    v = l w h
    Fb = pfluid x g x volume

    3. The attempt at a solution
    How does one calculate volume without known dimensions (only mass is know). It can not be derrived from density equation because density of plank is not given (and it varies a great deal)
  2. jcsd
  3. Dec 11, 2008 #2


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    Something floats when it's weight equals the weight of the displaced water
    So what is the total weight?
    How much buoyancy does it need - how much volume of water is displaced?
  4. Dec 11, 2008 #3
    total mass is mass of plank (877kg) plus mass of bricks loaded on top of plank (170.2kg)

    volume of plank and volume of water displaced is unknown. density of plank is unknown.

    I assumed that mass of water displaced was equal to total mass of loaded plank (877kg + 170.2 kg = 1047.2 kg) since the plank is floating even with surface of water.

    How do I find the density of the plank without knowing it's volume?

    How do I find the bouyant force?

    How do I vind the volume of the plank without knowing either density or dimensions?
  5. Dec 11, 2008 #4


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    Isn't the density of a material times the volume = weight?

    If you know the ratio of volume in terms of a known density, then isn't that going to tell you something about the ratio of their densities and hence the density of the unknown?

    Maybe the first 10 minutes of this lecture will help you better understand?

  6. Dec 11, 2008 #5


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    The volume of the plank underwater is equal to the volume of water displaced.
    You know the mass of water displaced and you know the density of water (since it says lake - assume fresh water)

    If it is floating then the weight down must equal the buoyant force up.
  7. Dec 11, 2008 #6
    yes, p = m/v. But in order to solve this equation, one must know two of the variables. In this case, neither the density of the object, nor the volume of the object is known.

    I have calculated the volume of water displaced by doing the following:
    Weight of fluid = (density of fluid)(volume of fluid)(g)
    if weight or mass of fluid displaced is equal to mass of object, then...
    1074.2 = mass of both fluid displaced and object, so...
    1047.2 = (1000cm^3)( v)(9.81m/s^2)
    v = .106.

    Then I did the following:
    Vobject = Vfluid(p solid / p fluid)
    .106 = .106 ( psolid/1000cm^3)
    psolid = 1.06m^3

    now I can use p = m/v
    1.06m^3 = 877kg/v
    827kgm^3 = v

    But still have to find bouyant force, if the above info is correct. I am not getting what the answer is supposed to be (final answer is given to us)
  8. Dec 11, 2008 #7


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    1000 kg/m³ is density of water.

    Your volume is 1.047 m³ isn't it?
  9. Dec 11, 2008 #8
    1.05 cubic meters is the density of the oak plank. But I have not been able to come up with this figure.

    I don't know if I am not converting something properly. My most recent attempt was off by a decimal place (.106 vs 1.06 when answer is 1.05 cubic meters)

    The volume quantity I have been able to surmise is 827 kg cubic meters (the given solution by the professor is 835 kg cubic meters.)

    The buoyant force I have calculated is no where near what the provided answer is (answer provided is 10300 N). I calculate 10398.6N The difference between my calcuation and the given answer is more than just rounding differences.

    Based on my previous post, can you tell where I went wrong with the applications of the formulas?
  10. Dec 11, 2008 #9


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    The volume is 1.047 m³. Not the density.

    Density is in units of kg/m³.

    Since you know that the volume of the plank and you know the mass is 877, then the density is 877/1.047 isn't it?

    Now the buoyant force is also virtually given because they loaded it down to until the water was just level. The total of those bricks and plank is your buoyant force isn't it?
  11. Dec 15, 2008 #10
    i got a similar problem with fluid dynamics , i got a test later in the week havent been prepped for it at all just been handed some sheets wondering if any one could help with an example i have got , as it will help me see what i need to do for future reference
    thanks in advance


    A cube of 0.25m length sides which has a weight of 50N is immersed in a tank of fluid .

    If the specific gravity of teh fluid is 0.8, calculate
    a) the density of the cube material
    b) the density of the fluid
    c) the resultant up-thrust of the cube


    50/0.8 = m
    m = 62.5 kg

    p=4000 kg/m3

    Thats all i can manage i havent really been taught this yet so any help would be appreciated
  12. Dec 15, 2008 #11


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    You've started on the wrong foot. The specific gravity is .8 and that expresses the ratio of the density of the liquid to that of water.

    The gravity you want to determine the mass with is the same old same old 9.8 m/s2.
  13. Dec 16, 2008 #12
    You have written mass of bricks = 170.2 I am assuming that this is in kg.
    Let M=mass of plank,
    V = volume of plank,
    p_plank = density of plank
    p_water = density of water

    Forces on the plan plank are
    1. Force of gravity = Mg downward
    2. Force by weight of bricks = 170.2*g downward
    3. Buoyant force V*p_water*g upward
    Plank is floating. So net force on it = 0
    Therefore M*g + 170.2*g - V*p_water*g = 0
    Divide by g,
    M + 170.2 - V*p_water = 0
    In the above, substitute M=877 kg and p_water=1000 kg/m^3. Then you can find V.
    That will give (a)volume of plank

    (b)Buoyant force=V*p_water*g
    In this, substitute V from (a), p_water=1000 kg/m^3, g = 9.8 m/s^2 and you will find buoyant force

    c) Mass is given. You have calculated volume. So you can calculate density.
  14. Dec 16, 2008 #13
    Ah yeah i see , is the specific gravity of the fluid the relative density of it then? so the density will be 800 i think
  15. Dec 16, 2008 #14


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    Yes specific gravity is the same as relative density (relative to water that is)
    It's not used much in metric countries because water has a density of 1g/cc so then density and relative density come out to the same thing (doesn't work quite as well in Kg/m^3)

    It's most important use is in beer - the specific gravity depends on the amount of sugar in the original brew and so the strength of the final beer.
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