# Answer to Question #16341 in Mechanics | Relativity for WILLYN JEAN

Question #16341

WHEN A POLAR BEAR JUMPS ON AN ICEBERG, HE NOTICES THAT HIS 1870 NEWTON WEIGHT IS JUST SUFFICIENT TO SINK THE ICEBERG. WHAT IS THE WEIGHT OF THE ICEBERG?

Expert's answer

Let M be the weight and V be the volume of an iceberg. Then, according to the Archimedes' principle:

M*g + 1870[N] = p(water)*g*V, (1)

where p(water) is the density of water. As

V = M/p(ice),

where p(ice) is the density of ice, we can rewrite (1) as

M*g + 1870[N] = p(water)*g*M/p(ice).

Therefore

M = 1870[N]/(g*( p(water)/p(ice) - 1 )).

The density of ocean water p(water) = 1025 kg/m³ and the density of ice is p(ice) = 916.7 kg/m³, so

M = 1870[N]/(9.8[m/s²]*( 1025[kg/m³]/916.7[kg/m³] - 1 )) ≈ 1615[kg].

So, the weight of an iceberg is about 1615 kg.

M*g + 1870[N] = p(water)*g*V, (1)

where p(water) is the density of water. As

V = M/p(ice),

where p(ice) is the density of ice, we can rewrite (1) as

M*g + 1870[N] = p(water)*g*M/p(ice).

Therefore

M = 1870[N]/(g*( p(water)/p(ice) - 1 )).

The density of ocean water p(water) = 1025 kg/m³ and the density of ice is p(ice) = 916.7 kg/m³, so

M = 1870[N]/(9.8[m/s²]*( 1025[kg/m³]/916.7[kg/m³] - 1 )) ≈ 1615[kg].

So, the weight of an iceberg is about 1615 kg.

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