Chapter 15, Problem 77PQ

(a)

To determine

The buoyant force acting on the Strato-lab balloon.

Answer to Problem 77PQ

The buoyant force acting on the Strato-lab balloon is $\underset{\_}{3.6\times {10}^5\text{N}}$.

Explanation of Solution

The volume displaced is equal to the volume of balloon. Consider volume is spherical in shape, and volume displaced will be the volume of sphere.

Write the expression for volume displaced.

  $V_{\text{disp}}=\frac{4}{3}\pi r^3$                                                                                                               (I)

Here, $r$ is the radius of ball.

Write the expression for buoyant force.

  $F_B={\rho }_f{V}_{\text{disp}}g$                                                                                                          (II)

Here, ${\rho }_f$ is the density of fluid, $g$ is the acceleration due to gravity.

Substitute, equation (I) in (II).

  $F_B=\frac{4}{3}\pi r^{3}g{\rho }_f$                                                                                                      (III)

Conclusion:

Substitute, $19.5\text{m}$ for $r$, $1.20{\text{kg/m}}^3$ for ${\rho }_f$, and $9.81{\text{m/s}}^2$ for $g$ in equation (III).

  $\begin{array}{c}{F}_B=\frac{4}{3}\pi {\left(19.5\text{m}\right)}^3\left(9.81{\text{m/s}}^2\right)\left(1.20{\text{kg/m}}^3\right)\\ =3.66\times {10}^5\text{N}\end{array}$

Therefore, the buoyant force acting on the Strato-lab balloon is $\underset{\_}{3.6\times {10}^5\text{N}}$.

(b)

To determine

The net force acting on lab.

Answer to Problem 77PQ

The net force acting on lab is $\underset{\_}{3.05\times {10}^5\text{N}}$.

Explanation of Solution

Write the expression for total force acting on lab.

  ${\displaystyle \sum F_y}=F_{\text{B}}-F_{\text{g,bal}}-F_{\text{g,He}}$                                                                          (IV)

Here, $F_{\text{g,bal}}$ is the force due to the weight of balloon, and $F_{\text{g,He}}$ is the force due to the weight of Helium.

Substitute, $m_{\text{balg}}$ for $F_{\text{g,bal}}$, and ${\rho }_{\text{He}}g{V}_{\text{b}}$ for $F_{\text{g,He}}$ in equation (V).

  $\begin{array}{c}{\displaystyle \sum F_y}=F_{\text{B}}-m_{\text{bal}}g-{\rho }_{\text{He}}g{V}_{\text{b}}\\ =F_{\text{B}}-\left(m_{\text{bal}}+{\rho }_{\text{He}}{V}_b\right)g\end{array}$                                                                                  (V)

Here, $m_{\text{bal}}$ is the mass of the balloon, ${\rho }_{\text{He }}$ is the density of helium gas, $V_b$ is the volume of balloon.

Conclusion:

Substitute, $3.65\times {10}^5\text{N}$ for $F_B$, $595\text{kg}$ for $m_{\text{bal}}$, $0.179{\text{kg/m}}^3$ for ${\rho }_{\text{He }}$, $3.11\times {10}^4{\text{m}}^3$ for $V_b$, and $9.8{\text{m/s}}^2$ for $g$ in equation (V).

  $\begin{array}{c}{\displaystyle \sum F_y}=3.65\times {10}^5\text{N}-\left(595\text{kg}+\left(0.179{\text{kg/m}}^3\right)\left(3.11\times {10}^4{\text{m}}^3\right)\right)9.8{\text{m/s}}^2\\ =3.05\times {10}^5\text{N}\end{array}$

Since, ${\displaystyle \sum F_y=m{a}_y>0}$, is positive balloon will rise.

Therefore, the net force acting on lab is $\underset{\_}{3.05\times {10}^5\text{N}}$.

(c)

To determine

The mass carried by Strato-lab.

Answer to Problem 77PQ

The mass carried by Strato-lab is $\underset{\_}{3.11\times {10}^4\text{kg}}$.

Explanation of Solution

Modify equation (V) by including gravitational force due to the extra payload.

  $\begin{array}{l}{\displaystyle \sum F_y}=F_{\text{B}}-F_{\text{g,bal}}-F_{\text{g,He}}-F_{\text{g,load}}=0\\ F_{\text{g,load}}=\left(F_{\text{B}}-F_{\text{g,bal}}-F_{\text{g,He}}\right)=3.05\times {10}^5\text{N}\end{array}$                                                             (VI)

Write the expression for the mass of the extra payload.

  $m_{\text{load}}=\frac{F_{\text{g, load}}}{g}$                                                                                                        (VII)

Conclusion:

Compare equation (VI) and (VII), and substitute, $9.8{\text{m/s}}^2$ for $g$, and $3.05\times {10}^5\text{N}$ for $F_{\text{g,load}}$ in equation (VII).

  $\begin{array}{c}{m}_{\text{load}}=\frac{3.05\times {10}^5\text{N}}{9.8{\text{m/s}}^2}\\ =3.11\times {10}^4\text{kg}\end{array}$

Therefore, the mass carried by Strato-lab is $\underset{\_}{3.11\times {10}^4\text{kg}}$.