QuestionApril 24, 2025

In the Pixar movie Up a man named Carl Fredricksen decides to sail away in his house by attaching thousands of helium balloons. Lets assume his house has a mass of 4.71cdot 10^4kg What minimum volume of helium would be needed to support a house of this weight? The density of helium is 0.164kg/m^3 and the density of air is 1.23kg/m^3

In the Pixar movie Up a man named Carl Fredricksen decides to sail away in his house by attaching thousands of helium balloons. Lets assume his house has a mass of 4.71cdot 10^4kg What minimum volume of helium would be needed to support a house of this weight? The density of helium is 0.164kg/m^3 and the density of air is 1.23kg/m^3
In the Pixar movie Up a man named Carl Fredricksen decides to sail away in his house by attaching thousands of helium balloons. Lets assume his house has a mass of 4.71cdot 10^4kg What minimum volume of helium would be needed to support a house of this weight? The density of helium is 0.164kg/m^3 and the density of air is 1.23kg/m^3

Solution
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Answer

4.22 \times 10^4 \, \text{m}^3 Explanation 1. Calculate the buoyant force needed The buoyant force must equal the weight of the house. Weight W = mg = 4.71 \times 10^4 \, \text{kg} \times 9.81 \, \text{m/s}^2 = 4.62 \times 10^5 \, \text{N}. 2. Use Archimedes' principle According to Archimedes' principle, the buoyant force is equal to the weight of the displaced air. **Buoyant Force = Volume \times \text{Density of Air} \times g**. 3. Solve for volume Set the buoyant force equal to the weight of the house: V \times 1.23 \, \text{kg/m}^3 \times 9.81 \, \text{m/s}^2 = 4.62 \times 10^5 \, \text{N}. Solve for V: V = \frac{4.62 \times 10^5}{1.23 \times 9.81} = 3.84 \times 10^4 \, \text{m}^3. 4. Adjust for helium density Since helium is less dense than air, we need more volume. **Net Lift = (\text{Density of Air} - \text{Density of Helium}) \times V \times g**. Solve for V: V = \frac{4.62 \times 10^5}{(1.23 - 0.164) \times 9.81} = 4.22 \times 10^4 \, \text{m}^3.

Explanation

1. Calculate the buoyant force needed<br /> The buoyant force must equal the weight of the house. Weight $W = mg = 4.71 \times 10^4 \, \text{kg} \times 9.81 \, \text{m/s}^2 = 4.62 \times 10^5 \, \text{N}$.<br />2. Use Archimedes' principle<br /> According to Archimedes' principle, the buoyant force is equal to the weight of the displaced air. **Buoyant Force = Volume \times \text{Density of Air} \times g**.<br />3. Solve for volume<br /> Set the buoyant force equal to the weight of the house: $V \times 1.23 \, \text{kg/m}^3 \times 9.81 \, \text{m/s}^2 = 4.62 \times 10^5 \, \text{N}$. Solve for $V$: $V = \frac{4.62 \times 10^5}{1.23 \times 9.81} = 3.84 \times 10^4 \, \text{m}^3$.<br />4. Adjust for helium density<br /> Since helium is less dense than air, we need more volume. **Net Lift = (\text{Density of Air} - \text{Density of Helium}) \times V \times g**. Solve for $V$: $V = \frac{4.62 \times 10^5}{(1.23 - 0.164) \times 9.81} = 4.22 \times 10^4 \, \text{m}^3$.
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