Physics question (helium balloon)?

Question

What volume V of helium is needed if a balloon is to just lift a load of 800 kg.? This number includes only the load and the weight of the empty balloon. Note the density of air is 1.29 kg/m cubed and the density of helium is 0.18 kg/m cubed.
Cool if you could include the math as to how you figured this out! Thanks

Answer 1

Archimedes principle for buoyancy is:
Fb = (mass of the surrounding fluid)*(gravity)
Fb = buoyant force
The mass of the surrounding fluid is air and can then be calculated as:
mass of surrounding fluid = (Density of Air)*(Volume of Balloon)
The buoyant force then can be written as
Fb = (Density of Air)*(Volume of Balloon)*(Gravity)

The opposing force to buoyancy is the gravity force and can be written for the helium balloon as:
Fg = (mass of helium + load of balloon)*(gravity)
Mass of Helium = (Density of Helium)*(Volume of Balloon)
So, then the equation looks like:
Fg = (Density of Helium)*(Volume of Balloon)*(gravity)
+ (load of balloon)*(gravity)

When a balloon is going up it has a larger buoyant force upward than the gravity force (Fg) downward. When they are equal then the net force is zero and can be written as:

Fnet = Fb - Fg = 0
or
Fb = Fg
The full equation then looks like:
(Density of Air)*(Volume of Balloon)*(Gravity) = (Density of Helium)*(Volume of Balloon)*(gravity)
+ (load of balloon)*(gravity)

Gravity cancels and the equation can then be algebraically arranged as:
(Volume of Balloon) = (load of balloon)/(Density of Air - Density of Helium)
or
V= m / (Air density - Helium density)

Plugging in the values that were stated in your problem we get:
800kg/(1.29kg/m³ - 0.18kg/m³) = 721m³

Answer 2

He has problems a million and a couple of astonishing, yet for 3 it rather is purely the quantity of the basket+the quantity of the balloon circumstances the density of the air, and then the stress of gravity. to that end: ((492.8+0.059)*a million.28)*9.eighty one regulate later whilst looking how many human beings could desire to extra healthful till now the balloon falls, employing that got here upon answer quite than the plenty smaller one with helium subtracted.

Answer 3

We will need to displace 800 kg of air to achieve equilibrium.

800 / 1.29 = 620.155 kg/m^3 of air = 800 k

Since helium is 0.18 kg/m^3 we need to displace an additional amount of air equal to
0.18 (620.155)+620.155 = 731.783 kg/m^3

Any amount higher than this, and the weight will be lifted.