Chapter 20, Problem 63A

Referring to the previous problem, find the volume of the displaced air.

To determine

The displaced volume of air.

Answer to Problem 63A

The volume of displaced air is $1700{\text{m}}^3$ .

Explanation of Solution

Given:

The buoyant force acting on the balloon is equal to the weight of the displaced air.

The weight of the displaced air in the balloon is $20,000\text{N}$ .

Formula used:

The buoyant force acting on the balloon is given as follows:

  $F_{\text{B}}={\rho }_{\text{air}}Vg$

Here, ${\rho }_{\text{air}}$ is the density of air, $V$ is the volume of balloon or air, and $g$ is the acceleration due to gravity.

Rearrange the equation for $V$ .

  $V=\frac{F_{\text{B}}}{{\rho }_{air}g}$

Calculation:

The buoyant force acting on the balloon will be equal to the displaced air weight.

The balloon is neither accelerating upward nor downward, so the net force acting on it will be zero. The upward buoyant force will be equal to the weight of the balloon including the hot air. Therefore, the weight of the displaced air in the balloon is $20,000\text{N}$ .

Substitute $20,000\text{N}$ for $F_{\text{B}},1.2\text{kg}/{\text{m}}^3$ for ${\rho }_{\text{air}},$ and $9.8\text{m}/{\text{s}}^2$ for $g$ in the equation $V=\frac{F_{\text{B}}}{{\rho }_{\text{air}}g}$

  $\begin{array}{c}V=\frac{20,000\text{N}}{\left(1.2\text{kg}/{\text{m}}^3\right)\left(\text{9}\text{.8m}/{\text{s}}^2\right)}\\ \approx 1700\text{N}\end{array}$

Conclusion:

Therefore, the volume of displaced air is $1700{\text{m}}^3$ .