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What % volume of floating object is submerged?

  1. Dec 19, 2015 #1
    1. The problem statement, all variables and given/known data
    An object of 985 kg/cm^3 density is placed in water, which has a density of 1000 kg/m^3.
    What percentage of the object will be floating above the water?

    2. Relevant equations


    3. The attempt at a solution
    985/1000 = .985, or 98.5%. 100 - 98.5 = 1.5%. Therefore: 1.5% of this object will float

    Correct? Thoughts?
     
  2. jcsd
  3. Dec 19, 2015 #2
    Well, it's the right answer, but I don't see any analysis to convince me you did anything more than pull some numbers out of the air. I'd like to see some discussion of the physics, as applied to this problem. The objective is to learn how to apply the physics, and not just to guess the right answer to a specific problem.
     
  4. Dec 19, 2015 #3
    Did you mean to express the densities in two different units?
     
  5. Dec 19, 2015 #4

    SteamKing

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    Yeah, an object with a density of 985 kg/cc will sink like the proverbial stone. :rolleyes:
     
  6. Dec 20, 2015 #5
    OOPS!

    I meant: cubic meters, not cubic cm for the object's density

    At the time I originally posted, I had difficulty expressing my rationale behind my answer, but upon thinking about it later, I came up with the following, subject to approval...

    Obviously, an object which is less dense than another will float. But how much will submerge? When placed in the water, the floating object will continue to displace water, until the total weight of the water displaced is equal to the total weight of the object. Once this point is reached, the force of gravity is counterbalanced by the water's bouyancy, and it floats. This begs the question: "how much water will be displaced in terms of volume?" Well, since the water is more dense than the object, a smaller volume will be displaced. The displaced volume has an equal weight of the object. Thus: by dividing the less dense object by the denser fluid displaced, the percentage volume of the object that is submerged, which is equal to the volume of displaced water, is determined. Subtracting 100 from this give the percentage that floats.
     
  7. Dec 20, 2015 #6
    I'm not sure if your instructor will find that response acceptable. You don't explain quantitatively why the volume ratio should be exactly the same as the density ratio. There are ways to explain it using ratio reasoning, but the usual approach is to use equations. Start with the balance of forces, which you explained perfectly well, by the way, and then make substitutions for the quantities on both sides. Factors will cancel and you'll be left with an equation with a volume ratio on one side and a density ratio on the other.
     
  8. Dec 20, 2015 #7
    Would you be willing to give an example?
     
  9. Dec 20, 2015 #8
    Sure. An object has a density of 750 kg/m3 so 75% of it's volume is submerged when it floats on water.
     
  10. Dec 20, 2015 #9
    Let V be the volume of the object, and let f be the fraction of its volume below the surface. In terms of V and f, what is the volume below the surface? What is the displaced volume of water? What is the weight of the displaced volume of water? What is the buoyant force? What is the weight of the object? What is the equilibrium force balance on the object?

    Chet
     
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