Chapter 10, Problem 25P

To determine

To calculate: The mass of the cargo that can be lifted by the spherical balloon.

Answer to Problem 25P

The mass of the cargo is $\text{920}\text{kg}$ .

Explanation of Solution

Given:

Density of air, ${\rho }_{air}=\text{1}\text{.29}{\text{kgm}}^{\text{-3}}$

Density of helium is, ${\rho }_{He}=0.79{\text{kgm}}^{\text{-3}}$

Radius, $r=7.35\text{m}$

Mass¸ $m=930\text{kg}$

Formula used:

According to Archimedes principle, there are two forces are acting on the object, one is its weight force and other is buoyancy force which is equal to the weight of the fluid that displaces its volume and is directed upward.

The buoyant force of the balloon equals to the sum of balloon’s weight, helium’s weight filled in the balloon and weight of the load. The buoyant force is equals to the weight of the air that is displaced by the balloon’s volume.

  $F_{\text{B}}=W_{\text{He}}+W_{\text{cargo}}+W_{\text{balloon}}$

Solution:

So, by substituting the values in the given formula is

  $\begin{array}{l}{F}_{\text{B}}=W_{\text{He}}+W_{\text{cargo}}+W_{\text{balloon}}\\ F_{\text{B}}=(m_{\text{He}}+m_{\text{cargo}}+m_{\text{balloon}})g\\ {\rho }_{\text{air}}Vg=\left(930+m_{\text{cargo}}+m_{\text{balloon}}\right)g\\ {\rho }_{\text{air}}V=\left({\rho }_{\text{He}}{V}_{\text{balloon}}+m_{\text{cargo}}+m_{\text{balloon}}\right)\\ \left({\rho }_{\text{air}}-{\rho }_{\text{He}}\right)V-930=m_{\text{cargo}}\\ \left(1.29-0.179\right)V-930=m_{\text{cargo}}\\ 1.111\times \frac{4}{3}\pi {\left(7.35\right)}^3-930=m_{\text{cargo}}\\ m_{\text{cargo}}=920\text{kg}\end{array}$

But the buoyant force is equals to the weight of the air that is displaced by the balloon’s volume, so the force is $920\text{kg}$ .

Conclusion:

The mass of the cargo is $920\text{kg}$ .