91Ó°ÊÓ

First, the mass of the steel drum and the mass of the fluids are converted to weight. This can be done using the formula \(\text{Weight} = \text{mass} \times \text{gravity}\). The gravity of Earth can be taken as \(9.81 \, \text{m/s}^2\). So the weight of the drum is \(16 \, \text{kg} \times 9.81 \, \text{m/s}^2 = 157.86 \, \text{N}\).

The volume of water and gasoline that would be displaced equals the volume of the drum, namely \(0.23 \, \text{m}^3\). The mass of fluid that would be displaced can be found using the relation \(\text{mass} = \text{density} \times \text{volume}\). For water, which has a density of \(1000 \, \text{kg/m}^3\), this gives: \(\text{mass of water} = 1000 \, \text{kg/m}^3 \times 0.23 \, \text{m}^3 = 230 \, \text{kg}\). For gasoline with a density of \(860 \, \text{kg/m}^3\), this gives: \(\text{mass of gasoline} = 860 \, \text{kg/m}^3 \times 0.23 \, \text{m}^3 = 198 \, \text{kg}\). These masses can now be converted to weights as done in step 1.

In this last step, compare the weight of the drum filled with water or gasoline to the weight of the displaced water or gasoline. The drum will float on the fluid if the weight of the fluid displaced is greater than or equal to the weight of the drum with fluid inside. The weight of water displaced is \(230 \, \text{kg} \times 9.81 \, \text{m/s}^2 = 2255.3 \, \text{N}\). Thus, when filled with water, the drum will not float because \(2255.3 \, \text{N} < 2255.3 \, \text{N} + 157.86 \, \text{N}\). The weight of gasoline displaced is \(198 \, \text{kg} \times 9.81 \, \text{m/s}^2 = 1941.8 \, \text{N}\). So when filled with gasoline, the drum will float because \(1941.8 \, \text{N} > 1941.8 \, \text{N} + 157.86 \, \text{N}\).