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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!

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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!

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Doc Al

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Sounds good to me. Show what you did exactly.cscott said: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?

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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.

[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.

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Doc Al

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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:Doc Al said:

A polar bear paritally supports herself by pulling part of her body out of the water onto a rectangular slab of ice. The ice sinks down so that only half of what was once exposed now is exposed, and the bear has 70 percent of her volume (and weight) out of the water. Estimate the bear's mass, assuming that the total volume of the ice is 10 m^3, and the bear's specific gravity is 1.0.

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Doc Al

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Ah, ok. I guess the textbook answer is just wrong...?

Thanks for your help BTW.

Thanks for your help BTW.

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Doc Al

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Yep, seems wrong to me.cscott said:I guess the textbook answer is just wrong...?

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