Chapter 14, Problem 33AP

(a)

To determine

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

Answer to Problem 33AP

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

Explanation of Solution

Explanation:

Given information: 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

To determine: The force equation for the balloon for this model.

Answer to Problem 33AP

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

Explanation of Solution

Explanation:

Given information: 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 $1.20\text{kg}/{\text{m}}^{\text{3}}$.

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)

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

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

Answer to Problem 33AP

Answer: 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

Explanation:

Given information: 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 $1.20\text{kg}/{\text{m}}^{\text{3}}$.

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,

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

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

Formula to calculate the weight of the balloon is,

$F_{\text{b}}=m_{\text{b}}g$

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

Formula to calculate the weight of the helium gas is,

$F_{\text{He}}=m_{\text{He}}g$

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

Formula to calculate the weight of the string is,

$F_{\text{s}}=m_{\text{s}}g$

• $m_{\text{s}}$ 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,

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

• ${\rho }_{\text{He}}$ 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

To determine: The mass of the string.

Answer to Problem 33AP

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

Explanation of Solution

Explanation:

Given information: 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 $1.20\text{kg}/{\text{m}}^{\text{3}}$.

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

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

Answer to Problem 33AP

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

Explanation of Solution

Explanation:

Given information: 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 $1.20\text{kg}/{\text{m}}^{\text{3}}$.

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 $0.050\text{kg}$.