Chapter 15, Problem 58P

(a)

To determine

The appropriate analysis model for the system when the balloon remains stationary.

Answer to Problem 58P

A particle in equilibrium model is the appropriate analysis model for the system when the balloon remains stationary.

Explanation of Solution

An analysis model is a simplified version of any physical system that strips away the unnecessary aspects of the situation. There are different kinds of analysis models such the particle under constant velocity, particle under constant acceleration and particle in equilibrium etc.

In the given situation, the balloon remains stationary which implies the balloon has zero acceleration. The net force on the balloon in any direction is zero. The balloon is in equilibrium. The particle in equilibrium is the appropriate analysis model for the given physical situation.

Conclusion:

Thus, a particle in equilibrium model is the appropriate analysis model for the system when the balloon remains stationary.

(b)

To determine

The force equation for the balloon.

Answer to Problem 58P

The force equation for the balloon is $B-F_b-F_{\text{He}}-F_s=0$.

Explanation of Solution

Write the equation for equilibrium.

  $\Sigma F_y=0$        (I)

Here, $\Sigma F_y$ is the net force acting on the balloon in the vertical direction.

Take the upward direction to be $+y$ direction.

Write the equation for $\Sigma F_y$ .

  $\Sigma F_y=B-F_b-F_{\text{He}}-F_s$

Here, $B$ is the buoyant force, $F_b$ is the weight of the balloon, $F_{\text{He}}$ is the weight of the helium and $F_s$ is the weight of the segment of the string above the ground.

Put the above equation in equation (I).

  $B-F_b-F_{\text{He}}-F_s=0$        (II)

Conclusion:

Therefore, the force equation for the balloon is $B-F_b-F_{\text{He}}-F_s=0$.

(c)

To determine

The expression for the mass of the segment of string.

Answer to Problem 58P

The expression for the mass of the segment of string is $m_s=\left({\rho }_{\text{air}}-{\rho }_{\text{He}}\right)\frac{4}{3}\pi r^3-m_b$.

Explanation of Solution

Write the equation for the buoyant force.

  $B={\rho }_{\text{air}}gV$        (III)

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

Write the equation for the weight of the helium.

  $F_{\text{He}}=m_{\text{He}}g$        (IV)

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

Write the equation for density of the helium.

  ${\rho }_{\text{He}}=\frac{m_{\text{He}}}{V}$

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

Rewrite the above equation for $m_{\text{He}}$.

  $m_{\text{He}}={\rho }_{\text{He}}V$

Put the above equation in equation (IV).

  $F_{\text{He}}={\rho }_{\text{He}}Vg$        (V)

Write the equation for the weight of the balloon.

  $F_b=m_{b}g$        (VI)

Here, $m_b$ is the mass of the balloon.

Write the equation for the weight of the segment of the string above the ground.

  $F_s=m_{s}g$        (VII)

Here, $m_s$ is the mass of the string.

Put equations (III), (V), (VI) and (VII) in equation (II) and rewrite it for $m_s$.

  $\begin{array}{c}{\rho }_{\text{air}}gV-m_{b}g-{\rho }_{\text{He}}Vg-m_{s}g=0\\ {\rho }_{\text{air}}V-m_b-{\rho }_{\text{He}}V=m_s\\ m_s=\left({\rho }_{\text{air}}-{\rho }_{\text{He}}\right)V-m_b\end{array}$        (VIII)

Write the equation for the volume of the balloon.

  $V=\frac{4}{3}\pi r^3$        (IX)

Here, $r$ is the radius of the balloon.

Conclusion:

Put equation (IX) in equation (VIII).

  $m_s=\left({\rho }_{\text{air}}-{\rho }_{\text{He}}\right)\frac{4}{3}\pi r^3-m_b$        (X)

Therefore, the expression for the mass of the segment of string is $m_s=\left({\rho }_{\text{air}}-{\rho }_{\text{He}}\right)\frac{4}{3}\pi r^3-m_b$.

(d)

To determine

The numerical value of the mass $m_s$.

Answer to Problem 58P

The numerical value of the mass $m_s$ is $0.0237\text{ kg}$.

Explanation of Solution

Equation (X) can be used to determine the numerical value of the mass $m_s$.

Conclusion:

The density of air is $1.20{\text{ kg/m}}^3$ and the density of helium is $0.179{\text{ kg/m}}^3$.

Substitute $1.20{\text{ kg/m}}^3$ for ${\rho }_{\text{air}}$ , $0.179{\text{ kg/m}}^3$ for ${\rho }_{\text{He}}$ , $0.400\text{ m}$ for $r$ and $0.250\text{ kg}$ for $m_b$ in equation (X) to find $m_s$.

  $\begin{array}{c}{m}_s=\left(1.20{\text{ kg/m}}^3-0.179{\text{ kg/m}}^3\right)\frac{4}{3}\pi {\left(0.400\text{ m}\right)}^3-0.250\text{ kg}\\ =0.0237\text{ kg}\end{array}$

Therefore, the numerical value of the mass $m_s$ is $0.0237\text{ kg}$.

(e)

To determine

The numerical value of the length $h$.

Answer to Problem 58P

The numerical value of the length $h$ is $0.948\text{ m}$.

Explanation of Solution

Write the equation for the mass $m_s$ in terms of the total mass of the string.

  $m_s=m\frac{h}{l}$

Here, $m$ is the total mass of the string, $h$ is the length of the string above the ground and $l$ is the total length of the string.

Rewrite the above equation for $h$ .

  $h=l\frac{m_s}{m}$        (XI)

Conclusion:

Substitute $2.00\text{ m}$ for $l$ , $0.0237\text{ kg}$ for $m_s$ and $0.050\text{ kg}$ for $m$ in equation (XI) to find $h$.

  $\begin{array}{c}h=\left(2.00\text{ m}\right)\frac{0.0237\text{ kg}}{0.050\text{ kg}}\\ =0.948\text{ m}\end{array}$

Therefore, the numerical value of the length $h$ is $0.948\text{ m}$.