Question

A spherical balloon has a radius of 6.95m and is filled with helium. The density of helium is 0.179 kg/m3, and the density of air is 1.29 kg/m3. The skin and structure of the balloon has a mass of 960kg. Neglect the buoyant force on the cargo volume itself. Determine the largest mass of cargo the balloon can lift.

Answer 1

Concepts and reason

The concepts used to solve this question is the density, force due to gravity and volume. At first, determine the density of the helium inside the baloon. Later calculate the mass of baloon and at last determine the mass of cargo that the baloon can lift.

Fundamentals

Volume is defined as the three dimensional space occupied by the object. The volume of the sphere is defined as, \(V=\frac{4 \pi}{3} r^{3}\)

Here, \(V\) is the volume of the sphere and \(r\) is the radius of the sphere. The mass per unit volume is defined as the density. It is measured in kilogram per meter cube. The density of the object is defined as, \(\rho=\frac{m}{V}\)

Here, \(\rho\) is the density of the object and \(m\) is the mass of the object. The force with which the earth is pulling is called as force of gravity. It is defined as, \(F=m g\)

Here, \(g\) is the acceleration due to gravity.

 

The force of gravity acting on the balloon is, \(F_{\mathrm{B}}=\rho_{\mathrm{B}} V g\)

Here, \(\rho_{\mathrm{B}}\) is the density of helium in balloon. Substitute \(0.179 \mathrm{~kg} / \mathrm{m}^{3}\) for \(\rho_{\mathrm{B}}\)

\(F_{\mathrm{B}}=\left(0.179 \mathrm{~kg} / \mathrm{m}^{3}\right) V g\)

The thrust force \(\left(F_{\mathrm{T}}\right)\) acting on the balloon is, \(F_{\mathrm{T}}=\rho_{\mathrm{air}} V g\)

Here, \(\rho_{\text {air }}\) is the density of air. Substitute \(1.29 \mathrm{~kg} / \mathrm{m}^{3}\) for \(\rho_{\text {air }}\)

\(F_{\mathrm{T}}=\left(1.29 \mathrm{~kg} / \mathrm{m}^{3}\right) V g\)

The force of gravity acts on the balloon in downward direction and the thrust force acts on the balloon upwards.

 

The force balance on the balloon is,

$$ F_{\mathrm{B}}+m_{\mathrm{C}} g+m_{\mathrm{B}} g=F_{\mathrm{T}} $$

Here, \(m_{\mathrm{C}}\) is the mass of cargo and \(m_{\mathrm{B}}\) is the mass of balloon. Substitute \(\left(0.179 \mathrm{~kg} / \mathrm{m}^{3}\right) V g\) for \(F_{\mathrm{B}}\) and \(\left(1.29 \mathrm{~kg} / \mathrm{m}^{3}\right) V g\) for \(F_{\mathrm{T}}\)

$$ \begin{array}{c} F_{\mathrm{B}}+m_{\mathrm{C}} g+m_{\mathrm{B}} g=F_{\mathrm{T}} \\ \left(0.179 \mathrm{~kg} / \mathrm{m}^{3}\right) V g+m_{\mathrm{C}} g+m_{\mathrm{B}} g=\left(1.29 \mathrm{~kg} / \mathrm{m}^{3}\right) V g \\ m_{\mathrm{C}}+m_{\mathrm{B}}=V\left(\left(1.29 \mathrm{~kg} / \mathrm{m}^{3}\right)-\left(0.179 \mathrm{~kg} / \mathrm{m}^{3}\right)\right) \end{array} $$

The above equation is modified as,

\(m_{\mathrm{C}}=V\left(\left(1.29 \mathrm{~kg} / \mathrm{m}^{3}\right)-\left(0.179 \mathrm{~kg} / \mathrm{m}^{3}\right)\right)-m_{\mathrm{B}} \ldots \ldots(1)\)

The volume of the helium inside the balloon is, \(V=\frac{4 \pi}{3} r^{3}\)

Substitute \(6.95 \mathrm{~m}\) for \(r\)

\(V=\frac{4 \pi}{3}(6.95 \mathrm{~m})^{3}\)

\(=1405.47 \mathrm{~m}^{3}\)

Substitute \(1405.47 \mathrm{~m}^{3}\) for \(V\) and \(960 \mathrm{~kg}\) for \(m_{\mathrm{B}}\) in the equation (1)

$$ m_{\mathrm{C}}=V\left(\left(1.29 \mathrm{~kg} / \mathrm{m}^{3}\right)-\left(0.179 \mathrm{~kg} / \mathrm{m}^{3}\right)\right)-m_{\mathrm{B}} $$

\(=\left(1405.47 \mathrm{~m}^{3}\right)\left(\left(1.29 \mathrm{~kg} / \mathrm{m}^{3}\right)-\left(0.179 \mathrm{~kg} / \mathrm{m}^{3}\right)\right)-960 \mathrm{~kg}\)

$$ =601.48 \mathrm{~kg} $$

The mass of the cargo that the balloon can lift is \(601.48 \mathrm{~kg}\).

The only upward force acting on the balloon is the thrust force and by balancing the upwards and the downward forces, the mass of the cargo that the balloon can lift has been determined.

The mass of the cargo that the balloon can lift is \(601.48 \mathrm{~kg}\).