Answer to Question #248891 in Physics for Zee

Question

A 5450 m3 blimp circles Busch Stadium during the World Series, suspended in the earth’s 1.21

kg/m3 atmosphere. The density of the helium in the blimp is 0.178 kg/m3. a) What is the

buoyant force that suspends the blimp in the air? b) How does this buoyant force compare to

the blimp’s weight? c) How much weight, in addition to the helium, can the blimp carry and

still continue to maintain a constant altitude?

Answer 1

a) The buoyancy force is

$$F_b=\rho_\text{Air}gV=1.21·9.8·5450=64630\text{ N}.$$

b) The weight is

$$F=\rho_\text{He}gV=0.178·9.8·5450=9507\text{ N}.$$

The buoyant force is 55000 N greater.

c) This weight equals the difference between the buoyant force and weight of the blimp:

$$W=F_b-F=55119\text{ N}.$$