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Buoyancy Problem: Slab of Ice on Water

  1. Aug 26, 2012 #1
    1. A slab of ice floats on a freshwater lake. What minimum volume must the slab have for a 45.0 kg woman to be able to stand on it without getting her feet wet?



    2. Archimedes Principal as well as Newton's Second Law will be applied



    3. The attempt at a solution

    I know that the volume of water that is being displaced will equal the combined weight of the woman and the slab that is pushing down on the water.

    This gives the equation:
    ρVg = Weightwoman + Weightice

    ***ρVg is the same as [itex]\frac{m}{V}[/itex]***

    Applying this gives:

    ρVg = g(45.0-kg + mice)
    ρV = (45.0-kg + mice)

    I know that the density for the freshwater lake will simply be 1.00 x 103 kgm-3 and the density for ice will be 0.92 x 103 kgm-3.

    Now, this is where I am stuck:

    V = [itex]\frac{(45.0-kg + mice)}{ρwater}[/itex]

    I am leaning towards saying that V should be the volume that has been displaced by the combined weight of the woman and the ice. And if I can figure out how to find that V, then I can plug it into the entire equation to give me the mass of the ice. Using the new found mass and the density of ice, I can find out the volume of the slab of ice.
     
  2. jcsd
  3. Aug 26, 2012 #2

    ehild

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    Do you know what V means? The volume of what? What displaces the water? The feet of the women can not be wet, so she is not immersed, she does not displace any water.

    ehild
     
  4. Aug 26, 2012 #3
    I apologize. The V, in this case, is referring to the volume of the water that is being displaced by the woman's weight on the slab of ice.
     
  5. Aug 26, 2012 #4

    ehild

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    The water is displaced by the immersed volume of a floating body. The volume of he displaced water is the same as the volume of the immersed part of the body. See picture. It shows the slab of ice. It is partly immersed in the water (the green part) but totally if the woman stands of it. But the women is outside of the water!
    If the women stands on the ice the total volume of ice is immersed. How do you get that volume of ice from its mass and density?

    The volume of the immersed part of a body displaces water of equal volume. The weight of a floating body is equal to the weight of the water displaced, the volume of the immersed part multiplied by the density of water and g.

    Here the weight of the ice + woman must be equal to the weight of the water displaced. The volume of the displaced water is the same as the volume of ice immersed, it is the total volume of ice now.

    m(ice)+M(women)=V(ice)ρ(water)

    But you know that m(ice)=ρ(ice)V(ice):

    ρ(ice)V(ice)+M(women)=V(ice)ρ(water)

    isolate V(ice).


    ehild
     

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  6. Aug 27, 2012 #5
    The volume of ice can be obtained by multiplying together it's density and it's mass. But the mass of the ice is something that we do not have. We do have the mass of the woman. If the ice is completely submerged within the water, as shown by your diagram, then shouldn't that just be equal to the woman's weight? We can find the volume that is being displaced using the mass of the woman and the density of the water? Which should be 0.045m-3. I still feel slightly hesitant as to why we divided the woman's mass by a 1000. I've been under the impression that it had to be the mass of the woman AND the ice.



    Carrying out this calculation gave me 441 N.



    Here I am slightly confused. I thought that it should be ρ(ice)V(ice) = ρ(water)V(water)

    (920 kg/m3)(45kg + X) = (1000 kg/m3)x

    Solving for x should give 0.5625 m3.

    I am still slightly confused by this problem.
     
  7. Aug 27, 2012 #6
    So I've been thinking a bit more about this problem. And I realized the following.

    1) The volume of water being displaced should equal the volume of the ice because the slab of ice is completely submerged within the water. The weight of the woman acts as an outside factor, keeping the ice submerged.

    2) The Buoyant Force exerted by the water will be acting upward and will thus be supporting both the woman and the slab of ice.

    We already know that the Buoyant Force exerted by the water should equal to the ρwaterViceg.

    We are tasked with finding what Vice is. Picture a free body diagram for this question. There will be a buoyant force acting upwards as well as a force, weight, acting downwards. This force is equal to (mwoman + mice)g.

    And we can set them equal to one another, giving:

    ρwaterViceg = (mwoman + mice)g.
    ρwaterVice = (mwoman + mice)
    mice) = ρiceVice
    ∴ ρwaterVice - ρiceVice = mwoman

    Vice = mwomanwaterVice - ρiceVice

    And plugging in the values gives the correct answer.


    This did make me come up with a few more questions.

    Is this situation applicable to similar situations? i.e. a piece of wood floating on some oil.

    Will the V of the liquid being displaced always be equal to the volume of the object that is immersed in the fluid?
     
  8. Aug 27, 2012 #7

    ehild

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    The volume of ice can be obtained by dividing its mass by the density.

    The weight(mass) of ice is not the same as the weight (mass) of the women. You need the volume of the ice slab. It is unknown, denote it with something, say V.

    The ice with the woman on it will float if the buoyant force is equal to the combined weight of ice and woman.

    If V is the volume of ice, the mass of ice is ρ(ice)V=920 V.
    You get the buoyant force if multiply the the displaced mass of water by g. If you immerse an object of volume V into water it displaces V volume of water.
    So the weight of ice+woman= buoyant force →ρ(ice)V+45=ρ(water)V

    ehild
     
  9. Aug 27, 2012 #8
    I understand :)! Would you mind looking at the post I made before your post? I would appreciate it quite a bit.
     
  10. Aug 27, 2012 #9

    ehild

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    If you you meant
    Vice = mwoman/(ρwater - ρice)
    then it is correct. :smile:

    The volume of the liquid displaced is equal to the volume of that part of the floating object which is immersed into the fluid.

    ehild
     
  11. Aug 27, 2012 #10
    I did indeed mean Vice = mwoman/(ρwater - ρice).

    What do you mean by THAT part of the floating object. As in, for the question we did above, the ice was the part that was immersed, therefore it's volume was that of the volume of liquid being displaced? I think I understand now.
     
  12. Aug 27, 2012 #11

    ehild

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    As an example, the ice would not be immersed into the water totally without the woman.
    You can find the immersed ratio by calculating the buoyant force as V(immersed)ρ(water)g. That is equal to the weight of the ice: ρ(ice)V(ice)g.
    So V(immersed)/V(ice)=ρ(ice)(water)=0.92. 92% of an iceberg is immersed into water so only a small part is seen, that is why the Titanic sank...(in case of seawater a bit less is immersed, as the density of the salty water is higher)

    ehild
     
  13. Aug 27, 2012 #12
    That is actually quite interesting :) And because the woman is standing on the ice, it is submerged in at 100% correct? Without her standing on it, it would only be at 92%. Oh, and Vimmersed is simply referring to the slab of ice that is in the water, it's not referring to something specific, therefore we don't have an exact specific value for it, correct? I am sorry for the questions, I'm just trying to thoroughly understand this!

    And the equation you used is simply setting the volume of the Buoyant Force being exerted by water to the weight of the ice?
     
  14. Aug 27, 2012 #13

    ehild

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    Yes, you are going to fully understand floating:smile:

    ehild
     
  15. Aug 27, 2012 #14
    And just one more question (for now!), the equation you have used is the Buoyant Force (the product of the density of water, gravity and the volume of ice? that has been immersed) that is being exerted by the water = the weight of the slab of ice, which can be found by the density of the ice, the Volume of the ice, and the gravity?
     
  16. Aug 27, 2012 #15

    ehild

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    The immersed volume of ice which is the same as the volume of water displaced.

    ehild
     
  17. Aug 27, 2012 #16
    I cannot thank you enough. Most people wouldn't have sat and answered all my questions :) I have learned quite a bit and I hope that this helps me out in my next homework problem :) Thanks a ton! May I ask you for more help should I need it? My discussion professor isn't quite good at answering our questions and we end up far more confused than learning anything.
     
  18. Aug 27, 2012 #17

    ehild

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    You are welcome. I will help gladly if I see your next problems, and I can solve them myself.

    ehild
     
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