Chapter 1.11, Problem 103RP

Determine the maximum amount of load, in kg, the balloon described in Prob. 1–101 can carry Answer: 900 kg

To determine

The maximum amount of load can carry by balloon.

Answer to Problem 103RP

The maximum amount of load can carry by balloon is $\underset{\_}{900\text{kg}}$.

Explanation of Solution

Show the free body diagram of the balloon.

Write the expression of volume of sphere of a balloon.

$V_{\text{balloon}}=4\pi r^3/3$ (I)

Write the expression of the buoyancy force acting on the balloon.

$F_B={\rho }_{\text{air}}g{V}_{\text{balloon}}$ (II)

Here, the density of air is ${\rho }_{{\text{H}}_{\text{2}}\text{O}}$, the volume of balloon is $V_{\text{balloon}}$, and the acceleration of gravity is $g$.

Write the expression of the mass of helium.

$m_{\text{He}}={\rho }_{\text{He}}{V}_{\text{balloon}}$ (III)

Here, the density of the helium is ${\rho }_{\text{He}}$ and volume of the balloon is $V_{\text{balloon}}$.

Write the expression of the total mass (buoyancy force and the weight force).

$\begin{array}{l}W=F_B\\ m_{\text{total}}g=F_B\\ m_{\text{total}}=\frac{F_B}{g}\end{array}$ (IV)

Here, the force acting on balloon is $F_B$ and the acceleration of gravity is $g$.

Write the expression of the maximum load can carried by balloon.

$m_{\text{people}}=m_{\text{total}}-m_{\text{He}}$ (V)

Here, the total mass of a balloon is $m_{\text{total}}$ and the mass of helium is $m_{\text{He}}$.

Conclusion:

Substitute $6\text{m}$ for $r$ in Equation (I).

$\begin{array}{c}{V}_{\text{balloon}}=4\pi {\left(6\text{m}\right)}^3/3\\ =904.8{\text{m}}^{\text{3}}\end{array}$.

Substitute $1.16\text{kg}/{\text{m}}^{\text{3}}$ for ${\rho }_{\text{air}}$, $9.81\text{m}/{\text{s}}^{\text{2}}$ for $g$, and $904.8{\text{m}}^{\text{3}}$ for $V_{\text{balloon}}$ in Equation (II).

$\begin{array}{c}{F}_B=\left(1.16\text{kg}/{\text{m}}^{\text{3}}\right)\left(9.81\text{m}/{\text{s}}^{\text{2}}\right)\left(904.8{\text{m}}^{\text{3}}\right)\\ =10296\text{kg}\cdot \text{m}/{\text{s}}^{\text{2}}\times \left(\frac{1\text{}\text{N}}{1\text{kg}\cdot \text{m}/{\text{s}}^{\text{2}}}\right)\\ =10296\text{N}\end{array}$

Substitute $1.16/7\text{kg}/{\text{m}}^{\text{3}}$ for ${\rho }_{\text{He}}$ and $904.8{\text{m}}^{\text{3}}$ for $V_{\text{balloon}}$ in Equation (III).

$\begin{array}{c}{m}_{\text{He}}=\left(\frac{1.16}{7}\text{kg}/{\text{m}}^{\text{3}}\right)\left(904.8{\text{m}}^{\text{3}}\right)\\ =149.9\text{kg}\end{array}$.

Substitute $10296\text{N}$ for $F_B$ and $9.81\text{m}/{\text{s}}^{\text{2}}$ for $g$ in Equation (IV).

$\begin{array}{c}{m}_{\text{total}}=\frac{10296\text{N}}{9.81\text{m}/{\text{s}}^{\text{2}}}\\ =1050\text{kg}\end{array}$

Substitute $1050\text{kg}$ for $m_{\text{total}}$ and $149.9\text{kg}$ for $m_{\text{He}}$ in Equation (V)

$\begin{array}{c}{m}_{\text{people}}=\left(1050\text{kg}\right)-\left(149.9\text{kg}\right)\\ =900\text{kg}\end{array}$

Thus, the maximum amount of load can carry by balloon is $\underset{\_}{900\text{kg}}$.