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## Homework Statement

A polar bear partially 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 polar bears mass, assuming that the total volume of the ice is 10m[tex]^{3}[/tex] and the bears specific gravity is 1.0.

## Homework Equations

buoyant force- F[tex]_{B}[/tex]=[tex]\rho[/tex][tex]_{F}[/tex]V[tex]_{displaced}[/tex]g=[tex]\rho[/tex][tex]_{o}[/tex] V[tex]_{o}[/tex]g=F[tex]_{w}[/tex]

## The Attempt at a Solution

so first of i found the initial mount of water exposed before that damn polar bear climbs on using V[tex]_{disp}[/tex]=[tex]\frac{\rho_{o}V_{o}}{\rho_{F}}[/tex] where [tex]\rho[/tex][tex]_{o}[/tex] is 0.917x10[tex]^{3}[/tex] and [tex]\rho[/tex][tex]_{F}[/tex] is 1x10[tex]^{3}[/tex]. this turns out to be 9.17 so the amount exposed is 10 less that amount, therefore 0.83 m[tex]^{3}[/tex]. so since the amount exposed after the bear hops on is half the original amount it must be 0.415m[tex]^{3}[/tex] therefore the amount exposed is 9.585. so since the surface of the ice was orginally 0.83 above the water and afterwards it is 0.415 above the water that meens that a mass equivalent to 0.415 of volume must have been added to the ice. since m=[tex]\rho[/tex]V and [tex]\rho[/tex] for the polar bear is equal to that of water(its specific gravity is 1) then that mass is equal to (1x10[tex]^{3}[/tex])(0.415)=415kg. since only 70% of this beast is on the ice that must only be 70% of its total mass so didving 415 by 0.7 we get 592.9 which isnt the answer in the back of the book! (the actual answer is 790!)