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Problem 13.33 - Floating gasoline

  1. Apr 18, 2015 #1
    1. The problem statement, all variables and given/known data
    k.png

    2. Relevant equations: Densities
    ##\rho = 680 ## gasoline
    ##\rho = 1000 ## freshwater
    ##\rho = 7800 ## steel

    3. 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 mg 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.
     
  2. jcsd
  3. Apr 18, 2015 #2

    Orodruin

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    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.
     
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