Solving Flotation Problem: Estimate a Polar Bear's Mass

[WhackyWookie]

The problem is:

A polar bear partially supports herself by pulling part of her body out of the water onto a rectangular slab of ice. The ice (with specific gravity of 0.917) sinks down so that only half of what was once exposed now is exposed, and the bear has 70% of her volume and weight out of the water. Estimate the bear's mass, assuming that the total volume of ice is 10 m^3, and the bear's specific gravity is 1.0.

I have tried to solve it using the equivalence of the ratios between the volume of displaced water and volume of object and density of object and density of fluid, but have been consistently getting wrong answers. Any help is appreciated.

 

[AKG]

The problem is:

A polar bear partially supports herself by pulling part of her body out of the water onto a rectangular slab of ice. The ice (with specific gravity of 0.917) sinks down so that only half of what was once exposed now is exposed, and the bear has 70% of her volume and weight out of the water. Estimate the bear's mass, assuming that the total volume of ice is 10 m^3, and the bear's specific gravity is 1.0.

I have tried to solve it using the equivalence of the ratios between the volume of displaced water and volume of object and density of object and density of fluid, but have been consistently getting wrong answers. Any help is appreciated.

As a general approach, do it like this: You can find out what volume of ice is originally submerged (without the bear on it). Now, set up an equation where the net force is zero. The forces acting on the system would be the buoyant force on 30% of the bear, the buoyant force on the ice (and you should be able to figure out what volume of ice is in the water based on the previous calculation, and the fact that another 50% is submerged), the weight of the ice, and the weight of the bear. Solve for the weight of the bear, and thus determine the mass. Also, you can express the volume of the bear in terms of its specific gravity and its mass. So you'll have an "m" variable in two places, but still only one unknown and one equation, meaning you can still solve it.

 

[WhackyWookie]

thnx for ur explanation, but the answer I got using ur method is the same as I got before (592.9 kg), which is inconsistent with the 790 kg answer listed on the back of the book.

 

[AKG]

thnx for ur explanation, but the answer I got using ur method is the same as I got before (592.9 kg), which is inconsistent with the 790 kg answer listed on the back of the book.

Perhaps you should show your work, and make sure your units are correct.