# Balloons are often filled with helium gas because it weighs only about

Balloons are often filled with helium gas because it weighs only about one-seventh of what air weighs under identical conditions. The buoyancy force, which can be expressed as $$\displaystyle{F}_{{{b}}}=\rho_{{{a}{i}{r}}}{g}{V}$$ balloon will push the balloon upward. If the balloon has a diameter of 12 m and carries two people, 85 kg each, determine the acceleration of the balloon when it is first released. Assume the density of air is $$\displaystyle\rho={1.16}{k}\frac{{g}}{{m}^{{{3}}}}$$, and neglect the weight of the ropes and the cage. Answer: $$\displaystyle{22.4}\frac{{m}}{{s}^{{{2}}}}$$

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Jonathan Burroughs

Givens and Knowns:
$$\displaystyle\rho_{{{a}}}={1.16}{k}\frac{{g}}{{m}^{{{3}}}}$$
$$\displaystyle{D}={12}{m}$$
$$\displaystyle{m}_{{{p}}}={85}{k}{g}$$
Solution:
$$\displaystyle{a}={\frac{{{F}_{{\ne{t}}}}}{{{m}_{{\to{t}}}}}}$$
$$\displaystyle{F}_{{\ne{t}}}={F}_{{{B}}}-{W}$$
Where $$\displaystyle{F}_{{{B}}}$$ is the bouyancy force and W is the total weight.
$$\displaystyle{F}_{{{B}}}=\rho_{{{a}}}\times{V}{o}{l}\times{g}={1.16}\times{\frac{{\pi\times{4}\times{6}^{{{3}}}}}{{{3}}}}\times{9.81}={10296.02}$$
$$\displaystyle{m}_{{\to{t}}}={2}\times{m}_{{{p}}}+\rho_{{{h}{e}}}\times{V}{o}{l}={170}+{\frac{{{1.16}}}{{{7}}}}\times{904.78}={319.934}$$
$$\displaystyle{W}={g}\times{m}_{{\to{t}}}={9.81}\times{319.934}={3138.56}$$
(Note: $$\displaystyle\rho_{{{h}{e}}}={\frac{{\rho_{{{a}}}}}{{{7}}}}$$)
Now, we can calculate the net force exerted as follows,
$$\displaystyle{F}_{{\ne{t}}}={10296.02}-{3138.56}={7157.46}$$
$$\therefore a=\frac{7157.46}{319.934}=22.372$$

###### Not exactly what you’re looking for?
einfachmoipf

The mass of the helium is $$\displaystyle{\frac{{{1}}}{{{7}}}}$$ (weigth of air)
the radius of the balloon $$\displaystyle{r}={6}{m}$$
then the volume of the balloon is $$\displaystyle{V}={\left({\frac{{{4}}}{{{3}}}}\right)}\pi{r}^{{{3}}}$$
$$\displaystyle={\left({\frac{{{4}}}{{{3}}}}\right)}\pi{\left({6}\right)}^{{{3}}}$$
$$\displaystyle={904.32}{m}^{{{3}}}$$
the mass of each person $$\displaystyle{m}={85}{k}{g}$$
from Newton's law
$$\displaystyle{F}={F}{b}-{F}{g}$$
$$\displaystyle{m}{a}=\text{pair g Vballoon}-{\left[{2}\cdot{85}+{\left({\frac{{{1}}}{{{7}}}}\right)}{\left(\rho{a}{i}{r}\right)}\right]}{\left({9.8}\frac{{m}}{{s}^{{{2}}}}\right)}$$
$$\displaystyle={\left({1.16}\right)}{\left({9.8}\right)}{\left({904.32}\right)}-{\left[{2}\cdot{85}+{\left({\frac{{{1}}}{{{7}}}}\right)}{\left({904.32}\right)}{\left({1.16}\right)}\right]}{\left({9.8}\frac{{m}}{{s}^{{{2}}}}\right)}$$
$$\displaystyle={10280.3}-{3134.6}={7145.7}$$
$$\displaystyle{\left[{2}\cdot{85}+{\left({\frac{{{1}}}{{{7}}}}\right)}{\left({V}{b}{a}{l}{l}oo{n}\right)}{\left(\rho{a}{i}{r}\right)}\right]}{a}={7145.6}$$
therefore the acceleration is
$$\displaystyle{a}={\frac{{{7145.7}}}{{{319.85}}}}={22.34}\frac{{m}}{{s}^{{{2}}}}$$

nick1337

$$D=12m \\\rho_{he}=\rho_{air}/7 \rho_{air}=1.16kg/m^{3} \\m_{people}=85 \times 2=170kg \\F_{B}=\rho_{over} Vballon \\\omega=m_{total} \\m_{he}=\rho_{he} Vballon \\V_{balloon}^{2}=\frac{4}{3} \times R_{baloon} \\m_{he}=\frac{1}{7} \times 1.16 \times 904.78=149.93kg \\m_{total}=m_{he}+m_{people}=149.93+170=319.93kg \\\text{Applying Newton's 2nd law.}\ \sum f=m_{total}a \\\therefore F_{B}-\omega=m_{total}a \\\rho_{air} gVballon-m_{total} g=m_{total}a \\a=\frac{\rho_{air}V_{ball}-n_{total}}{m_{total}}g \\=\frac{(1.16)(904.78)-319.93}{319.93} \\=22.4m/s^{2}$$
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