Buoyancy and Archimedes' Principle of steel

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SUMMARY

The discussion centers on calculating the total volume of steel that can be used in constructing a 260 L steel drum filled with gasoline, ensuring it floats in fresh water. Utilizing Archimedes' principle, the equation P = ρ * V * g is applied, where the densities of water (1000 kg/m³), gasoline (680 kg/m³), and steel (7800 kg/m³) are critical. The initial calculation yielded an incorrect volume of steel (1.1 * 10^-2 m³), prompting a reevaluation that suggests considering the combined displacement of both the gasoline and the steel. The corrected volume of steel is proposed to be approximately 10.66 L.

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The_Fritz
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Homework Statement


Because gasoline is less dense than water, drums containing gasoline will float in water. Suppose a 260 L steel drum is completely full of gasoline.
What total volume of steel can be used in making the drum if the gasoline-filled drum is to float in fresh water?


Homework Equations


P=\rho*V*g
Archimedes' principle - the buoyant force on an object immersed in a fluid is equal to the weight of the fluid displaced by that object.


The Attempt at a Solution


\rho_steel*V_steel*g+\rho_gas*V*g=\rho_water*V*g

Solve for V_steel: *gravity cancels*
V_steel= (\rho_water*V)-(\rho_gas*V)/(\rho_steel)

plug in values:
density water = 1000 kg/m^3
density gas = 680 kg/m^3
density steel = 7800 kg/m^3
Volume of drum= 260 L = .26 m^3

Answer: 1.1 * 10^-2 (which was not the correct answer)
 
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The_Fritz said:
\rho_steel*V_steel*g+\rho_gas*V*g=\rho_water*V*g

Steel displaces water as well.
 
Last edited:
One assumption that you make which may be throwing you off is that the volume of steel is negligible. In other words there are two volumes, one of gas and one of steel which displace a total volume of water. See if that helps. Otherwise me answer of 10.66L agrees with yours.
EDIT: Simultaneous post there.
 

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