Chapter 14, Problem 14.60AP

(a)

To determine

The appropriate model to describe the system when balloon is stationary.

Answer to Problem 14.60AP

The appropriate model to describe the system is particle in equilibrium.

Explanation of Solution

The mass of the balloon is $0.250\text{kg}$ tied to a uniform length $2.00\text{m}$ and mass $0.050\text{kg}$. The balloon is spherical with radius $0.4\text{m}$ and the density of the air is $1.20\text{kg}/{\text{m}}^{\text{3}}$.

If a system remains stationary, the sum of all forces acted on a system in all direction vertical as well as horizontal is equal to zero. This condition is also called is equilibrium condition.

Conclusion:

Therefore, appropriate model to describe the system is particle in equilibrium.

(b)

To determine

The force equation for the balloon for this model.

Answer to Problem 14.60AP

The force equation for the balloon for this model is $B-F_{\text{b}}-F_{\text{He}}-F_{\text{s}}=0$.

Explanation of Solution

In equilibrium condition, sum of all forces in vertical direction is equal to zero.

    $\begin{array}{c}{\displaystyle \sum F_{\text{y}}}=0\\ {\displaystyle \sum F_{\text{y}}}=B-F_{\text{b}}-F_{\text{He}}-F_{\text{s}}=0\end{array}$        (I)

Here, $B$ is buoyant force, $F_{\text{b}}$ is the weight of the balloon (envelope), $F_{\text{He}}$ is the weight of the helium gas and $F_{\text{s}}$ is the weight of the string.

Conclusion:

Therefore, the force equation for the balloon for this model is ${\displaystyle \sum F_{\text{y}}}=B-F_{\text{b}}-F_{\text{He}}-F_{\text{s}}=0$.

(c)

To determine

The mass of the string in terms of $m_{\text{b}}$, $r$, ${\rho }_{\text{He}}$ and ${\rho }_{\text{air}}$.

Answer to Problem 14.60AP

The mass of the string in the terms of $m_{\text{b}}$, $r$, ${\rho }_{\text{He}}$ and ${\rho }_{\text{air}}$ is $\frac{4}{3}\pi r^3\left({\rho }_{\text{air}}-{\rho }_{\text{He}}\right)-m_{\text{b}}$.

Explanation of Solution

From equation (I),

    $B-F_{\text{b}}-F_{\text{He}}-F_{\text{s}}=0$

The buoyant force act on the balloon is equal to the displaced volume of the air by the balloon.

Formula to calculate the buoyant force acting on the balloon is,

    $B={\rho }_{\text{air}}g\times \frac{4}{3}\pi r^3$

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

Formula to calculate the weight of the balloon is,

    $F_{\text{b}}=m_{\text{b}}g$

Here, $m_{\text{b}}$ is the mass of the balloon.

Formula to calculate the weight of the helium gas is,

    $F_{\text{He}}=m_{\text{He}}g$

Here, $m_{\text{He}}$ is the mass of the helium gas.

Formula to calculate the weight of the string is,

    $F_{\text{s}}=m_{\text{s}}g$

Here, $m_{\text{s}}$ is the mass of the string.

Substitute ${\rho }_{\text{air}}g\times \frac{4}{3}\pi r^3$ for $B$, $m_{\text{He}}g$ for $F_{\text{He}}$, $m_{\text{b}}g$ for $F_{\text{b}}$ and $m_{\text{s}}g$ for $F_{\text{s}}$ in equation (I).

    ${\rho }_{\text{air}}g\times \frac{4}{3}\pi r^3-m_{\text{b}}g-m_{\text{He}}g-m_{\text{s}}g=0$

Formula to calculate the mass of the helium gas is,

    $m_{\text{He}}={\rho }_{\text{He}}\times \frac{4}{3}\pi r^3$

Here, ${\rho }_{\text{He}}$ is the density of the helium gas.

Substitute ${\rho }_{\text{He}}\times \frac{4}{3}\pi r^3$ for $m_{\text{He}}$ in above expression.

    ${\rho }_{\text{air}}g\times \frac{4}{3}\pi r^3-m_{\text{b}}g-{\rho }_{\text{He}}\times \frac{4}{3}\pi r^{3}g-m_{\text{s}}g=0$

Rearrange the above expression for $m_{\text{s}}$

    $\begin{array}{c}{m}_{\text{s}}={\rho }_{\text{air}}\times \frac{4}{3}\pi r^3-{\rho }_{\text{He}}\times \frac{4}{3}\pi r^3-m_{\text{b}}\\ m_{\text{s}}=\frac{4}{3}\pi r^3\left({\rho }_{\text{air}}-{\rho }_{\text{He}}\right)-m_{\text{b}}\end{array}$

Conclusion:

Therefore, the mass of the string in terms of $m_{\text{b}}$, $r$, ${\rho }_{\text{He}}$ and ${\rho }_{\text{air}}$ is $\frac{4}{3}\pi r^3\left({\rho }_{\text{air}}-{\rho }_{\text{He}}\right)-m_{\text{b}}$.

(d)

To determine

The mass of the string.

Answer to Problem 14.60AP

The mass of the string is $0.0237\text{kg}$.

Explanation of Solution

From equation (II),

    $m_{\text{s}}=\frac{4}{3}\pi r^3\left({\rho }_{\text{air}}-{\rho }_{\text{He}}\right)-m_{\text{b}}$

Substitute $1.20\text{kg}/{\text{m}}^{\text{3}}$ for ${\rho }_{\text{air}}$, $0.250\text{kg}$ for $m_{\text{b}}$, $0.4\text{m}$ for $r$ and $0.179\text{kg}/{\text{m}}^{\text{3}}$ for ${\rho }_{\text{He}}$ to find $m_{\text{s}}$.

    $\begin{array}{c}{m}_{\text{s}}=\frac{4}{3}\pi {\left(0.4\text{m}\right)}^3\left(1.20\text{kg}/{\text{m}}^{\text{3}}-0.179\text{kg}/{\text{m}}^{\text{3}}\right)-0.250\text{kg}\\ =0.0237\text{kg}\end{array}$

Conclusion:

Therefore, the mass of the string is $0.0237\text{kg}$.

(e)

To determine

The length $h$ of the string if mass of the string is $0.050\text{kg}$.

Answer to Problem 14.60AP

The length $h$ of the string is $0.948\text{m}$ if mass of the string is $0.050\text{kg}$.

Explanation of Solution

From equation (II),

    $m_{\text{s}}=\frac{4}{3}\pi r^3\left({\rho }_{\text{air}}-{\rho }_{\text{He}}\right)-m_{\text{b}}$

The mass of the string of height $h$ is equal to the $m_{\text{s}}\times \frac{h}{l}$.

Substitute $m_{\text{s}}\times \frac{h}{l}$ for $m_{\text{s}}$ in above expression.

    $m_{\text{s}}\times \frac{h}{l}=\frac{4}{3}\pi r^3\left({\rho }_{\text{air}}-{\rho }_{\text{He}}\right)-m_{\text{b}}$

Substitute $1.20\text{kg}/{\text{m}}^{\text{3}}$ for ${\rho }_{\text{air}}$, $0.250\text{kg}$ for $m_{\text{b}}$, $0.4\text{m}$ for $r$, $0.050\text{kg}$ for $m_{\text{s}}$$2.0\text{m}$ for $l$ and $0.179\text{kg}/{\text{m}}^{\text{3}}$ for ${\rho }_{\text{He}}$ to find $h$.

    $\begin{array}{c}0.050\text{kg}\times \frac{h}{2.0\text{m}}=\frac{4}{3}\pi {\left(0.4\text{m}\right)}^3\left(1.20\text{kg}/{\text{m}}^{\text{3}}-0.179\text{kg}/{\text{m}}^{\text{3}}\right)-0.250\text{kg}\\ h=\frac{0.0237\text{kg}\times 2.0\text{m}}{0.050\text{kg}}\\ =0.948\text{m}\end{array}$

Conclusion:

Therefore, the length $h$ of the string is $0.948\text{m}$ if mass of the string is $0.050\text{kg}$.