Submit a question to our community and get an answer from real people.

When a polar bear on an iceberg notices that his 1870 N weight is just sufficient to sink the iceberg- WHAT IS THE WEIGHT OF THE ICEBERG?

Report as

Density of ice~.9. Weight of bear increases effective density to 1 so bear is 1/10 th the mass of the ice. Ice is 10x the weight of the bear.

Report as
Thank you for answering it :) but this is the only thing given by our teacher... and also the answer: 15,590 N except the last three numbers aren't the right ones according to her. Also i need a solution :/
Report as

F[up] = F[down] = 1870 N
to finish, you need the relation between buoyant force.and weight. Good luck!

Report as
Thank you for answering it :) but this is the only thing given by our teacher... and also the answer: 15,590 N except the last three numbers aren't the right ones according to her. Also i need a solution :/
Report as
This is a problem utilizing Archimedes' principle. Now the weight of the bear (1870N) is equal to weight of water that will be displaced when the ice sinks when he stands on the iceberg. Now ice is 0.9 as dense as water so the weight of the volume of ice to be submerged is 1683N. I.e.90% of the bear's weight. Since this represents 10% of the iceberg then the iceberg will weigh 16830N
Report as

bear 1870 N
water: 1000.0 kg/m^3 = 9806.65 N
917 kg/m^3 (assume 900 kg/m^3) = 8825.985 N
buoyant force = 9806.65 - 8825.985 = 980.665 N/m^3

volume = 1870 N / (980.665) = 1.906869 m^3
weight = 8825.985 N/m^3 * 1.960869 m^3 = 17306 N
Report as
Engineering details- it's the concept of how you get there that counts ,
pura vida
Report as