Question

An airship contains 6*10^3 m^3 of helium. Helium has a density of 0.180 kg m^-3. Given that the air has a density of 1.20 kg m^-3 at the height at which the airship is floating in equilibrium, what load is th airship carrying?

Answer 1

An airship contains $6*10^3 \mathrm{~m}^3$ of helium. Helium has a density of $0.180 \mathrm{~kg} \mathrm{~m}^{\wedge} - 3$. Given that the air has a density of $1.20 \mathrm{~kg} \mathrm{~m}^{\wedge} - 3$ at the height at which the airship is floating in equilibrium, what load is the airship carrying?

Solution

$W_{L}$ - the weight of the load that the airship can carry in equilibrium at an altitude where the density of air is $1.20\mathrm{kg / m}^3$

$$W _ {H e} + W _ {L} = F _ {B} \rightarrow W _ {L} = F _ {B} - W _ {H e}$$

So

$$W _ {L} = \rho_ {\text {air}} V _ {\text {ship}} g - \rho_ {\text {He}} V _ {\text {ship}} g = (\rho_ {\text {air}} - \rho_ {\text {He}}) V _ {\text {ship}} g$$ $$W _ {L} = (1.2 - 0.18) * 6 * 10 ^ {3} * 9.8 = 5.997 * 10 ^ {4} N \approx 6 * 10 ^ {4} N$$

Answer: $6 * 10^{4} N$.