Question

The density of ice is 917 kg/m^3, and the density of sea water is 1025 kg/m^3. A swimming polar bear climbs onto a piece of floating ice that has a volume of 6.78 m^3. What is the weight of the heaviest bear that the ice can support without sinking completely beneath the water?

Answer 1

Answer:

w_bear = 7175.95 N ,   m = 732.24 kg

Explanation:

Let's analyze the situation a bit, the ice block is in equilibrium with the thrust given by Archimedes' law that is directed towards the bottom, the weight of the ice and the weight of the bear.

           B -W_ice - w_bear = 0

           w_bear = B - W_ice

The thrust is given by

           B = ρ g V

           B = 1025 9.8 6.78

           B = 68 105.1 N

Note that we use the total volume of the block since the problem indicates that it is submerged.

The weight of the ice is

           W_ice = m g

     

the density is

          ρ_ice = m_ice / V

          m_ice = rho_ice V

we substitute

          W_ice = ρ_ice g V

           W_ice = 917 9.8 6.78

           W_ice = 60929.15 N

we substitute in the first equation

             w_bear = 68105.1 - 60929.15

              w_bear = 7175.95 N

the mass of this bear is

               w_bear = m g

               m = w_bear / g

               m = 7175.95 / 9.8

               m = 732.24 kg