Question

A large weather balloon whose mass is$\mathbf{226}\mathbf{\text{\hspace{0.17em}kg}}$is filled with helium gas until its volume is$\mathbf{325}{\mathbf{\text{\hspace{0.17em}m}}}^{\mathbf{\text{3}}}$. Assume the density of air isrole="math" localid="1663758896876" $\mathbf{1.2}\raisebox{1ex}{${\displaystyle \mathbf{\text{kg}}}$}\!\left/ \!\raisebox{-1ex}{${\mathbf{\text{m}}}^{\mathbf{\text{3}}}$}\right.$and the density of helium isrole="math" localid="1663758915409" $\mathbf{0.179}\raisebox{1ex}{${\displaystyle \mathbf{\text{kg}}}$}\!\left/ \!\raisebox{-1ex}{${\mathbf{\text{m}}}^{\mathbf{\text{3}}}$}\right.$.

(a) Calculate the buoyant force acting on the balloon.

(b) Find the net force on the balloon and determine whether the balloon will rise or fall after it is released.

(c) What additional mass can the balloon support in equilibrium?

Short answer

(a) The buoyant force acting on the balloon is $\text{3822\hspace{0.17em}N}$.

(b) The net force on the balloon is $1037.08\text{\hspace{0.17em}N}$. The balloon rises when released.

(c) The balloon can support an additional $105.82\text{\hspace{0.17em}kg}$in equilibrium.

Step-by-step solution

Identification of the given data:

The given data can be listed below.

• Mass of the balloon, $m=226\text{\hspace{0.17em}kg}$
• Volume of the balloon, $V=325{\text{\hspace{0.17em}m}}^{\text{3}}$
• Density of air, $d_a=1.2\text{\hspace{0.17em}kg}/{\text{m}}^{\text{3}}$
• Density of helium,$d_h=0.179\text{\hspace{0.17em}kg}/{\text{m}}^{\text{3}}$

Buoyancy and gravitational force:

The buoyant force from a liquid of density $\mathit{d}$when volume of the liquid displaced $\mathit{v}$ is,

$\mathit{B}\mathbf{=}\mathit{d}\mathit{g}\mathit{v}$ ..... (1)

Here, $\mathit{g}$is the acceleration due to gravity having value

$\mathit{g}\mathbf{=}\mathbf{9}\mathbf{.8}\mathbf{\text{\hspace{0.17em}m}}\mathbf{/}{\mathbf{\text{s}}}^{\mathbf{\text{2}}}$

The gravitational force on a body of mass $\mathit{m}$is,

$\mathit{F}\mathbf{=}\mathit{m}\mathit{g}$ ..... (2)

(a) Determining the buoyancy force on the balloon:

Volume of displaced air is equal to the volume of the balloon. Thus from equation (1),

$B=d_{a}gV$

Substitute all the value in the above equation,

$\begin{array}{rcl}B& =& 1.2\text{\hspace{0.17em}kg}/{\text{m}}^{\text{3}}\times 9.8\text{\hspace{0.17em}m}/{\text{s}}^{\text{2}}\times 325{\text{\hspace{0.17em}m}}^{\text{3}}\\ & =& 3822\cdot \left(1\text{\hspace{0.17em}kg}\cdot \text{m}/{\text{s}}^{\text{2}}\times \frac{{\displaystyle 1\text{\hspace{0.17em}N}}}{{\displaystyle 1\text{\hspace{0.17em}kg}\cdot \text{m}/{\text{s}}^{\text{2}}}}\right)\\ & =& 3822\text{N}\end{array}$

Thus, the required buoyancy is $3822\text{N}$.

(b) Determining the net force on the balloon:

The buoyancy obtained in the previous step acts upward. The net downward force is the weight of the balloon. From equation (II) the net weight of the filled balloon is,

$W=(m+d_{h}V)g$

Substitute all the value in the above equation,

$\begin{array}{rcl}W& =& (226\text{\hspace{0.17em}kg}+0.179\text{\hspace{0.17em}kg}/{\text{m}}^{\text{3}}\times 325{\text{\hspace{0.17em}m}}^{\text{3}})\times 9.8\text{\hspace{0.17em}m}/{\text{s}}^{\text{2}}\\ & =& 2784.92\cdot \left(1\text{\hspace{0.17em}kg}\cdot \text{m}/{\text{s}}^{\text{2}}\times \frac{{\displaystyle 1\text{\hspace{0.17em}N}}}{{\displaystyle 1\text{\hspace{0.17em}kg}\cdot \text{m}/{\text{s}}^{\text{2}}}}\right)\\ & =& 2784.92\text{N}\end{array}$

Thus, the net force on the block is,

$\begin{array}{rcl}B-W& =& 3822\text{\hspace{0.17em}N}-2784.92\text{\hspace{0.17em}N}\\ & =& 1037.08\text{\hspace{0.17em}N}\end{array}$

This is directed upward and hence the balloon moves up.

(c) Determining the mass the balloon can carry at equilibrium:

At equilibrium the net force on the balloon is zero. This can be obtained by increasing the mass on the balloon which increases the downward gravitational force which balances the net upward force obtained in the previous step. Let the extra mass be $m_e$.

Thus from equation (2),

$m_{e}g=1037.08\text{\hspace{0.17em}N}$

$\begin{array}{rcl}{m}_e& =& \frac{1037.08\text{\hspace{0.17em}N}}{9.8\text{\hspace{0.17em}m}/{\text{s}}^{\text{2}}}\\ & =& 105.82\cdot \left(1\text{\hspace{0.17em}N}\times \frac{{\displaystyle 1\text{\hspace{0.17em}kg}\cdot \text{m}/{\text{s}}^{\text{2}}}}{{\displaystyle 1\text{\hspace{0.17em}N}}}\right)\cdot \left(\frac{{\displaystyle 1}}{{\displaystyle 1\text{\hspace{0.17em}m}/{\text{s}}^{\text{2}}}}\right)\\ & =& 105.82\text{\hspace{0.17em}kg}\end{array}$

Hence, the required mass is $\begin{array}{rcl}& & 105.82\text{\hspace{0.17em}kg}\end{array}$.