A block of solid steel of mass 1.00 x 10^7 kg is fully submerged in water. Steel has a density of 7.8 x 10^3 kg/m^3, water 1.00 x 10^3 kg/m^3, and g is 9.80 m/s^2. What buoyant force acts on the steel?
A9.8 x 10^7 N
B1.3 x 10^6 N
C1.3 x 10^7 N
D7.8 x 10^7 N
Answer & Solution
Correct answer: C. 1.3 x 10^7 N
1. The buoyant force is the weight of the water displaced, so first find how much water is displaced.
2. Since the steel is fully submerged, the displaced volume equals the volume of the steel.
3. Volume of steel = mass / density = (1.00 x 10^7 kg) / (7.8 x 10^3 kg/m^3) = 1.28 x 10^3 m^3.
4. Mass of water displaced = (1.00 x 10^3 kg/m^3)(1.28 x 10^3 m^3) = 1.28 x 10^6 kg.
5. Buoyant force = (1.28 x 10^6 kg)(9.80 m/s^2) = 1.3 x 10^7 N, rounded to two figures because the steel density has two.
6. The value 9.8 x 10^7 N is the weight of the steel itself, the trap for using the object's own mass instead of the displaced water's mass.
7. The value 1.3 x 10^6 N is the displaced mass in kilograms mistaken for a force, and 7.8 x 10^7 N misuses the density ratio.
_Source: OpenStax College Physics (CC BY 4.0), Ch 11 "Fluid Statics", section 11.7 Archimedes' Principle_
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