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Buoyancy question

  1. Nov 1, 2007 #1
    A spherical float has a volume of 0.17 m cubed and a mass of 54kg. It is held at the bottom of a pond by a rope. The pond is filled with water. The pond is 7.6m deep. The rope breaks.
    1. What is the radius of the float?
    I figure that I use r=the cubde root of (3V/4pi) But I am not sure how to do this even if its the right way.

    2. What density of fluid would allow the float to float on its surface with 25% of its volume above the liquid air interface?
    My teacher and I have no clue how to do this. Any help would be great.
     
  2. jcsd
  3. Nov 1, 2007 #2
    Post?

    Where has this post been moved to, or has it been removed? If so can anyone point me in a direction that can help with this question. As I stated before my teacher does not know how to this question either.
    Thanks
     
  4. Nov 1, 2007 #3

    Chi Meson

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    I'm so sorry for you and your teacher. This is basic stuff. What is your teacher doing "teaching" this? Well...

    1. yes you're right about the radius. How to do it? Use a calculator.

    2. Find the density of the sphere, since you know the mass and volume. And somewhere in the book, the teacher should have been able to find the part that explains that, for example, that the ratio of densities is equal to the submerged volume of a floating object.
     
  5. Nov 1, 2007 #4
    okay, so:

    the radius is the cubed root of (3 X 17.88 / (4 x 3.14). = 1.62, is that right?

    Also,
    The denisty is 54kg / 0.17m cubed= 317.65. Is that the denisity of the fluid that would allow the float to float? or would it be 25% of this density? i.e., 1270.6?

    My teacher is a graduate physics student, he teachers our physics lab. He said that this is a bonus question, since he does not know how to figure it out. Yes, I am shocked as well. Its sad that I actually pay tuition to get my answers answered online.
    Thanks for helping.
     
  6. Nov 2, 2007 #5

    Chi Meson

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    The floating object would have 75% of the density of the fluid.
     
  7. Nov 2, 2007 #6
    so thats 317.65 x 0.75=238.24 Is that the denisity of the fluid that would allow the float to float?

    and

    the radius is the cubed root of (3 X 17.88 / (4 x 3.14). = 1.62, is that right?

    Thanks for all your help. I could NOT have done this question without you!
     
  8. Nov 3, 2007 #7

    Chi Meson

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    The fluid must have greater density than the floating object! RIght? Does steel float? Do rocks float? (despite an earlier argument over whether or not ice is a rock, rocks sink!)

    So don't multiply by .75.
     
  9. Nov 3, 2007 #8
    Then what do I do with the density?
     
  10. Nov 3, 2007 #9

    Chi Meson

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    The density of the float (known) must be 75% of the density of the fluid (if 75% of the float is submerged). so:
    [tex] \rho_{float} = .75\rho_{fluid}[/tex]
     
  11. Nov 4, 2007 #10
    so:

    7.6m(54kg)=410.4(.75)=307.8

    is this right?
     
  12. Nov 4, 2007 #11

    learningphysics

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    No. Using Chi's equation:

    317.65 = 0.75*(density of fluid)

    solve for "density of fluid"
     
  13. Nov 4, 2007 #12
    so

    317.65/.75=423.5 is the density of the of the fluid thta would allow the float to float.

    Thank you VERY much for your help.
     
  14. Nov 4, 2007 #13

    learningphysics

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    yes, that's the answer.
     
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