Consider a man whose weight is w 249 lb Find the minimum ra

Consider a man whose weight is w = 249 lb. Find the minimum radius (m) of a helium balloon that will lift him off the ground. The density of helium gas is 0.178 kg/m3 , and the density of air is 1.29 kg/m3 . Neglect the weight of the deflated balloon itself. Assume that the shape of the balloon is a sphere. Neglect the buoyant force of the air on the man. (b) If we replace the helium with hydrogen gas, how much more weight (lb) can the balloon lift? The density of hydrogen gas is 0.089 kg/m3 . (Find how much weight in excess of w the same balloon inflated to the same volume can lift.) Caution: the answer is not w anymore. The buoyant force is the same. Only the weight of the gas has changed.

Solution

part1

Volume of baloon , V = [4/3]r^3.

Mass of air displaced V*1.29 kg/m^3

Mass of helium = V*0.178

Mass difference = V* [1.112]


Total mass = 153 lb + V*0.178 = 69.4kg + V*0.178 kg
V* 1.112 = 69.4+ V*0.178
0.934 V = 69.4
V = 74.3 m^3.

[4/3]r^3 = . 74.3

r = 2.61 m .
part 2.
Density difference between helium and hydrogen is
0.089 kg /m^3.
The excess mass it can lift is V *0.089
= 74.3*0.089 = 6.61 kg
= 14.57 lb.