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Homework Help: Fluid Mechanics of swimming polar bear

  1. Apr 27, 2008 #1
    1. The problem statement, all variables and given/known data
    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.

    2. Relevant 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]

    3. 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!)
  2. jcsd
  3. Apr 27, 2008 #2

    Shooting Star

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    Homework Helper

    (The subscripts are self evident.)

    Use Archimedes' principle for all cases, and set up two equations.

    Initially, suppose x fraction of the ice-block was under the water. Then (1-x) fraction was above it.
    Afterward, x+(½)(1-x) = (½)(1+x) fraction of the ice is below water.

    (Omitting the 'g's in all equations.)

    For the case when only the ice was floating,
    weight of water displaced = weight of ice block =>

    xV_i \rho _w = V_i \rho _i

    When 70% of the bear’s body is on the ice, then (3/10) of the bear's volume is under the water =>

    (3/10)V_b \rho _w + (1/2)(1 + x)V_i \rho _w = V_b \rho _b + V_i \rho _i

    We have to assume values for the densities of water and ice, after x is eliminated from the two equations.
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