Because the air is a fluid, we can apply the principle of Archimedes. To the flotation of the balloon; the air must exert on the balloon an upward buoyancy force, which must be at least equal and opposite to the sum of our body weight plus the weight of helium...
See
This Answer NowStart your subscription to unlock this answer and thousands more. Enjoy eNotes ad-free and cancel anytime.
Already a member? Log in here.
Because the air is a fluid, we can apply the principle of Archimedes. To the flotation of the balloon; the air must exert on the balloon an upward buoyancy force, which must be at least equal and opposite to the sum of our body weight plus the weight of helium in the balloon; in this case we are neglecting the weight of the balloon.
So we can write the following expression:
Fb = wp + wHe
Where:
Fb = ρair*Vair*g, is the buoyancy force exerted by air. ρair is the air density and Vair is the volume of air displaced by the balloon.
wp = mp*g, is the weight of the person.
wHe = mHe*g, is the weight of the helium inside the ballon.
Rewriting the equation:
ρair*Vair*g = (mp*g) + (mHe*g)
ρair*Vair = mp + mHe
We will express the mass of helium as the product of volume by the density; also keep in mind that the volume of air displaced is equal to the volume of helium in the balloon, then we have:
ρair*Vair = mp + (ρHe*VHe)
(ρair*Vair) – (ρHe*VHe) = mp
VHe = mp/(ρair - ρHe) = 89/(1.205 – 0.166)
VHe = 85.66 m^3
85.66 m^3, is the minimum volume from which the balloon begins to raise.