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Homework Help: Buoyant force help

  1. Apr 27, 2006 #1
    70% of a mass is supported by a slab of ice and the ice sinks down so that only half of what was previously exposed now is exposed. What is the mass assuming that the ice has a volume of 10 m^3 and the mass has a specific gravity of 1.0?

    Why can't I use the buoyant force of the ice before and after the extra weight is added and subtract to get the weight of the object itself? I get 539.5 kg while the textbook says 790 kg. Only hints please!
     
  2. jcsd
  3. Apr 27, 2006 #2

    Doc Al

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    Staff: Mentor

    Sounds good to me. Show what you did exactly.
     
  4. Apr 27, 2006 #3
    Alright, well since [itex]F_B = w[/itex] when an object is floating, the buoyant force is [itex]F_B = \rho_{ice} Vg = (0.917 \times 10^3)(10)(9.8) = 9.0 \times 10^4N[/itex].

    [tex]\frac{0.917 \times 10^3}{1.00 \times 10^3} + \frac{1}{2}(1 - 0.917) = 0.9585
    [/tex]
    for the fraction of the ice submerged once the unkown mass is put on.

    With this, the buoyant force is [itex]F_b = (1.00 \times 10^3)(0.9585 \cdot 10)(9.8) = 9.4 \times 10^4N[/itex]

    [tex]W_{object} = 9.4 \times 10^4 - 9.0 \times 10^4 = 4.0 \times 10^3N[/tex]

    [tex]\frac{4.0 \times 10^4 \cdot 1.3}{9.8} = 530 kg[/tex]

    I rounded the numbers this time.
     
    Last edited: Apr 27, 2006
  5. Apr 27, 2006 #4

    Doc Al

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    Your method looks OK to me (assuming we are interpreting the problem correctly--why do they specify the specific gravity of the mass?) but I would divide by 0.7 instead of multiply by 1.3.
     
  6. Apr 27, 2006 #5
    The fact that they gave the specific gravity had me starting to work with volume but I could never get any sensible answers. If it makes any difference, this is how the question is worded in the book:

     
  7. Apr 27, 2006 #6

    Doc Al

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    Since the specific gravity is one, it makes no difference--the buoyant force on the bear equals the weight of her submerged portion. (Were it something else, it would matter.)
     
  8. Apr 27, 2006 #7
    Ah, ok. I guess the textbook answer is just wrong...?

    Thanks for your help BTW.
     
  9. Apr 27, 2006 #8

    Doc Al

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    Yep, seems wrong to me.
     
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