The same 1.00 x 10^7 kg of steel is reshaped into a hull that can displace 1.00 x 10^5 m^3 of water. Taking water as 1.00 x 10^3 kg/m^3 and g as 9.80 m/s^2, what is the maximum buoyant force on the hull?
A1.00 x 10^8 N
B9.80 x 10^7 N
C1.28 x 10^9 N
D9.80 x 10^8 N
Answer & Solution
Correct answer: D. 9.80 x 10^8 N
1. The maximum buoyant force is the weight of the largest volume of water the hull can displace.
2. Mass of that water = density x volume = (1.00 x 10^3 kg/m^3)(1.00 x 10^5 m^3) = 1.00 x 10^8 kg.
3. Buoyant force = (1.00 x 10^8 kg)(9.80 m/s^2) = 9.80 x 10^8 N.
4. The steel's own weight is (1.00 x 10^7 kg)(9.80 m/s^2) = 9.80 x 10^7 N, so the hull can lift about ten times its own weight.
5. The value 9.80 x 10^7 N is exactly that weight of the steel, offered as the trap for confusing the two.
6. The value 1.00 x 10^8 N reports the displaced mass in kilograms as though it were a force.
7. The value 1.28 x 10^9 N carries over the submerged block's volume, which no longer applies once the steel is a hull.
_Source: OpenStax College Physics (CC BY 4.0), Ch 11 "Fluid Statics", section 11.7 Archimedes' Principle_
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