Question

In many locations, such as Lake Washington in Seattle, floating bridges are preferable to conventional bridges. Such a bridge can be constructed out of concrete pontoons, which are essentially concrete boxes filled with air, Styrofoam, or another extremely low-density material. Suppose a floating bridge pontoon is constructed out of concrete and Styrofoam, which have densities of \(2200 \mathrm{~kg} / \mathrm{m}^{3}\) and \(50.0 \mathrm{~kg} / \mathrm{m}^{3}\). What must the volume ratio of concrete to Styrofoam be if the pontoon is to float with \(35.0 \%\) of its overall volume above water?

Short answer

Answer: The volume ratio of concrete to Styrofoam is approximately 0.0606.

Step-by-step solution

Understanding buoyancy

According to Archimedes' principle, the buoyancy force acting on an object submerged in a fluid (in this case, water) is equal to the weight of the fluid displaced by the object's volume. The equation for this principle is: \(F_{buoyancy} = \rho_{fluid} \cdot V_{displaced} \cdot g\) where \(\rho_{fluid}\) is the density of the fluid, \(V_{displaced}\) is the volume of fluid displaced, and \(g\) is the acceleration due to gravity. For the pontoon to float with 35% of its overall volume above water, the buoyancy force must be equal to the weight of the whole pontoon.

Setting up the problem using variables

Let's denote the volume ratio of concrete to Styrofoam as \(R\). The total volume of the pontoon (\(V_{total}\)) can be represented as: \(V_{total} = V_{concrete} + V_{styrofoam}\) or in terms of the ratio, \(R\): \(V_{total} = R \cdot V_{styrofoam} + V_{styrofoam}\) Now, the mass of each material present in the pontoon can be found using the equation: \(mass = density \cdot volume\) So, \(m_{concrete} = 2200 \cdot R \cdot V_{styrofoam}\), and \(m_{styrofoam} = 50 \cdot V_{styrofoam}\). The weight of the pontoon is the sum of the weights of concrete and Styrofoam: \(W_{total} = m_{concrete} \cdot g + m_{styrofoam} \cdot g\)

Setting up the buoyancy condition

Since 35% of the pontoon's overall volume stays above water, it means that 65% of the volume is submerged, displacing the same volume of water. Therefore, \(V_{displaced} = 0.65 \cdot V_{total}\) Now, according to Archimedes' principle, \(F_{buoyancy} = W_{total}\). Using this condition, we can set up the following equation: \(\rho_{fluid} \cdot (0.65 \cdot V_{total}) \cdot g = (2200 \cdot R \cdot V_{styrofoam} + 50 \cdot V_{styrofoam}) \cdot g\)

Solving for the volume ratio

Now, we have an equation relating the volume ratio, \(R\), to the buoyancy force. We can solve this equation for \(R\). First, rearrange the equation by cancelling out some terms: \(0.65 \cdot (\rho_{fluid} \cdot V_{total}) = (2200 \cdot R + 50) \cdot V_{styrofoam}\) Next, expand the equation: \(0.65 \cdot (\rho_{fluid} \cdot (R \cdot V_{styrofoam} + V_{styrofoam})) = (2200 \cdot R + 50) \cdot V_{styrofoam}\) Now, we want to solve for R. Divide both sides by \(V_{styrofoam}\): \(0.65 \cdot (\rho_{fluid} \cdot (R + 1)) = 2200 \cdot R + 50\) Substitute the given density of water, which is \(1000 \mathrm{~kg} / \mathrm{m}^{3}\): \(0.65 \cdot (1000 \cdot (R + 1)) = 2200 \cdot R + 50\) Solve for R: \(R \approx 0.0606\)

Conclusion

The volume ratio of concrete to Styrofoam in the floating bridge pontoon must be approximately 0.0606 for it to float with 35% of its overall volume above water.