Problem 13.33 - Floating gasoline

[Calpalned]

Homework Statement

Homework Equations

: Densities[/B]
$\rho = 680$ gasoline
$\rho = 1000$ freshwater
$\rho = 7800$ steel

The Attempt at a Solution

I first converted 210 liters to volume in m³ by dividing 210 by 1000. Then I used this volume and the given density to find the mass of the gasoline. $\rho = \frac{m}{v} = \frac{m}{0.21} = 680$ Therefore, $m_g = 142.8$kg. The buoyant force can be determined by multiplying m_g with 9.8. $F_b = 1399.44 N$. That value is equal to the weight of the gasoline. We know that because the drum floats, the buoyant force is not at its maximum. I will find the maximum buoyant force that water can exert on an object with a volume of 0.21 m³, and then use that to determine the maximum volume of steel. $m_{max}=1000(0.21) = 210 N$ Thus, $F_{buo-max}=2058 N$. 2058 - 1399.44 = 658.56 N. The amount of steel used must be equal to or less than 658.56 N. Dividing 658.56, we see that the mass of steel is 67.2 kg. Using $\rho = \frac{m}{v} = \frac{67.2}{V} = 7800$ my answer is $0.008615$ which is wrong. The correct answer, as seen in the picture above, it 0.0099. Where's my error? Thank you all.

 

[Orodruin]

The steel also has a volume, leading to additional buoyancy. The buoyancy is neutral when the total mass divided by total volume of the gasoline+steel combination is equal to the water density.