Question

A spherical balloon has a radius of 7.15 m and is filled with helium. How large a cargo can it lift, assuming that the skin and structure of the balloon have a mass of 930 kg? Neglect the buoyant force on the cargo volume itself.

Short answer

The mass of the spherical balloon's cargo can lift is 672.31 kg.

Step-by-step solution

Understanding Newton's second law

According to Newton's Second Law of Motion, the weight of a body is equal to the product of the body's mass and acceleration due to gravity. It can be expressed in terms of Newtons.

Identification of the given data

The radius of the spherical balloon is r = 7.15m.

The mass of the skin and structure of the balloon is m_balloon = 930kg.

Determination of the mass of the cargo

Now, understand this question with the help of a diagram, as given below.

So, as you can see, four forces are acting on the spherical balloon.

Force in an upward direction tends to lift the balloon. It can be represented by,

F = ρ _air.V_{air .}g

Here, g is the acceleration due to gravity, ρ_air is the density of air which can be taken as 1.225 kg/m³ (standard value), V_air and is the volume of the air displaced by the balloon, and it is equal to the volume of the balloon, i.e.,

Substitute the value of r_balloon in the above formula.

$$\begin{array}{r}{V}_{\mathrm{b}\mathrm{a}\mathrm{l}\mathrm{l}\mathrm{o}\mathrm{o}\mathrm{n}}=\frac{4}{3}\pi {\left(7.15\right)}^3{\mathrm{m}}^3\\ V_{\mathrm{b}\mathrm{a}\mathrm{l}\mathrm{l}\mathrm{o}\mathrm{o}\mathrm{n}}=1531.11{\mathrm{m}}^3\end{array}$$

Substitute the values of ρ _air, V_{air ,}g in the formula for F.

Determination of the mass of the cargo

The balloon experiences force due to the weight of the skin and structure of the balloon. It can be represented by W_balloon.

W_balloon= m_balloon . g

Substitute the values in the above formula.

Force experienced by the balloon due to the weight of the helium gas filled in the balloon. It can be represented by W_He.

W_He = m_{He . g}

W_He = ρ_He.V_{He . g}

Here,ρ_He is the density of the Helium gas, which can be taken as 0.1785kg/m³ (standard value), and V_He is the volume of Helium gas present inside the spherical balloon, equal to the balloon's volume, i.e., 1531.1m³.

Substitute the values of ρ_He, V_{He, g} in the formula for W_He.

$$\begin{array}{r}{W}_{\mathrm{H}\mathrm{e}}=0.1785\mathrm{k}\mathrm{g}/\phantom{\mathrm{k}\mathrm{g}{\mathrm{m}}^3}\{{\mathrm{m}}^3\times 1531.11{\mathrm{m}}^3\times 9.81\mathrm{m}/\phantom{\mathrm{m}{\mathrm{s}}^2}\{{\mathrm{s}}^2\\ W_{\mathrm{H}\mathrm{e}}=2681.1\mathrm{k}\mathrm{g}\cdot \mathrm{m}/\phantom{\mathrm{k}\mathrm{g}\cdot \mathrm{m}{\mathrm{s}}^2}\{{\mathrm{s}}^2\\ W_{\mathrm{H}\mathrm{e}}=2681.1\mathrm{N}\end{array}$$

Force experienced by the balloon due to the weight of the cargo attached to the balloon. It can be represented by W_He.

W_cargo = m_cargo . g

Now, considering the spherical balloon in its equilibrium condition, we have,

$$\begin{array}{r}F-W_{\mathrm{b}\mathrm{a}\mathrm{l}\mathrm{l}\mathrm{o}\mathrm{o}\mathrm{n}}-W_{\mathrm{H}\mathrm{e}}-W_{\mathrm{c}\mathrm{a}\mathrm{r}\mathrm{g}\mathrm{o}}=0\\ m_{\mathrm{c}\mathrm{a}\mathrm{r}\mathrm{g}\mathrm{o}}\cdot g=F-W_{\mathrm{b}\mathrm{a}\mathrm{l}\mathrm{l}\mathrm{o}\mathrm{o}\mathrm{n}}-W_{\mathrm{H}\mathrm{e}}\end{array}$$

Substitute the values of all forces in the above formula.

$$\begin{array}{r}{m}_{cargo}\cdot g=18399.73-9123.3-2681.1\mathrm{N}\\ m_{cargo}=\frac{6595.33\mathrm{N}}{g}\\ m_{cargo}=\frac{6595.33\mathrm{k}\mathrm{g}\cdot \mathrm{m}/\phantom{\mathrm{k}\mathrm{g}\cdot \mathrm{m}{\mathrm{s}}^2}\{{\mathrm{s}}^2}{9.81\mathrm{m}/\phantom{\mathrm{m}{\mathrm{s}}^2}\{{\mathrm{s}}^2}\\ m_{cargo}=672.31\mathrm{k}\mathrm{g}\end{array}$$

Hence, the mass of the spherical balloon's cargo can lift is 672.31 Kg.