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Find the volume of the object sticking out of the water

  1. Nov 17, 2007 #1
    1. The problem statement, all variables and given/known data
    One of the problems with debris in the ocean is that it is often difficult to see, because much of the object is under the surface of the water. An object with a density of 801 kg/m3 and a mass of 1039 kg is thrown into the ocean. Find the volume of the object sticking out of the water (use ρsea water = 1024 kg/m3).


    2. Relevant equations
    p = m/v

    3. The attempt at a solution
    I am not sure how to go about solving this problem?
     
    Last edited by a moderator: Jan 7, 2014
  2. jcsd
  3. Nov 17, 2007 #2

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    What is p=m/v here?

    Remember Archimedes' Principle? Use it. State it if you're not sure how to use it.
     
  4. Nov 17, 2007 #3
    Fb = Pf * Vf * g
    Not sure how to use it though.
    Pf = 1000kg/m^3
     
  5. Nov 17, 2007 #4

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    It means that the upward buoyant force on an object is equal to the weight of fluid it displaces. So, if d is the density of the fluid, V is the vol of the object that is under the fluid, then the upward force is d*V*g. Can you apply it now?
     
  6. Nov 17, 2007 #5
    Ok so I understand the concept but when I solve it do I use 0 for Fb and then solve for Vf? Or am I supposed to use the weight of the object there?
     
  7. Nov 17, 2007 #6

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    Let me illustrate by using the data given in this prob. Suppose V is the vol of the object that is submerged. Then Fb = weight of water with vol V = V*1024.

    This upward force balances the total weight of the body, since the body is floating. So,
    V*1024 = 1039*g.

    So, you know V. You also know that the total vol of the object = mass/density = 1039/801.

    Now, by subtracting, you can tell how much is sticking out.
     
  8. Nov 17, 2007 #7
    So is Fb always going to equal the density * V?
    okay I get an answer of 8.65 m^3
    it says incorrect I did 9.94 - 1.29
     
  9. Nov 17, 2007 #8

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    I didn't mean for you to just copy me blindly. By mistake I have written, "V*1024 = 1039*g", where it should be V*1024*g, because I'm equating weights. Try it now.
     
  10. Nov 17, 2007 #9
    Sorry I didn't know..I am still trying to figure this out...
    V * 1024 * g = 1039 * g
    V = (1039 * g) / (1024 * g) = 1.01

    1039/801 = 1.29

    So I subtract those volumes? I get 0.28...very small?
     
  11. Nov 17, 2007 #10

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    Right answer. What's the problem?

    The whole volume of the object is 1.29 cubicm. Sticking out portion is 1/5th of the whole. In an iceberg, only 1/12 of the whole sticks out.
     
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