Chapter 1.11, Problem 102RP

Balloons are often filled with helium gas because it weighs only about one-seventh of what air weighs under identical conditions. The buoyancy force which can be expressed as F_b = ρ_airgV_balloon, will push the balloon upward. If the balloon has a diameter of 12 m and carries two people 85 kg each determine the acceleration of the balloon when it is first released. Assume the density of air is ρ = 1 16 kg/m³, and neglect the weight of the ropes and the cage. Answer: 22.4 m/s²

To determine

The acceleration of the balloon when it is first released.

Answer to Problem 102RP

The acceleration of the balloon when it is first released is $\underset{\_}{22.4\text{m}/{\text{s}}^{\text{2}}}$.

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 can carried by balloon.

$m_{\text{total}}=m_{\text{He}}+m_{\text{people}}$ (IV)

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

Write the expression of the total weight.

$W=m_{\text{total}}g$ (V)

Write the expression of net force acting on the balloon.

$F_{\text{net}}=F_B-W$ (VI)

Write the expression of acceleration.

$a=\frac{F_{\text{net}}}{m_{\text{total}}}$ (VII)

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 $2\times 85\text{kg}$ for $m_{\text{people}}$ and $149.9\text{kg}$ for $m_{\text{He}}$ in Equation (V)

$\begin{array}{c}{m}_{\text{total}}=\left(149.9\text{kg}\right)+\left(2\times 85\text{kg}\right)\\ =319.9\text{kg}\end{array}$

Substitute $319.9\text{kg}$ for $m_{\text{total}}$ and $9.81\text{m}/{\text{s}}^{\text{2}}$ for $g$ in Equation (VI).

$\begin{array}{c}W=\left(319.9\text{kg}\right)\left(9.81\text{m}/{\text{s}}^{\text{2}}\right)\\ =3138\text{kg}\cdot \text{m}/{\text{s}}^{\text{2}}\times \left(\frac{1\text{N}}{1\text{kg}\cdot \text{m}/{\text{s}}^{\text{2}}}\right)\\ =3138\text{N}\end{array}$

Substitute $10296\text{N}$ for $F_B$ and $3138\text{N}$ for $W$ in Equation (VII).

$\begin{array}{c}{F}_{\text{net}}=10296\text{N}-3138\text{N}\\ =\text{7157}\text{N}\end{array}$.

Substitute $7157\text{N}$ for $F_{\text{net}}$ and $319.9\text{kg}$ for $m_{\text{total}}$ in Equation (VIII).

$\begin{array}{c}a=\frac{\left(7157\text{N}\right)}{\left(319.9\text{kg}\right)}\\ =\frac{\left(7157\text{N}\right)}{\left(319.9\text{kg}\right)}\times \left(\frac{1\text{kg}\cdot \text{m}/{\text{s}}^{\text{2}}}{1\text{N}}\right)\\ =22.4\text{m}/{\text{s}}^{\text{2}}\end{array}$

Thus, the acceleration of the balloon when it is first released is $\underset{\_}{22.4\text{m}/{\text{s}}^{\text{2}}}$.