# Helium Balloon

1. Nov 17, 2007

### BuBbLeS01

1. The problem statement, all variables and given/known data
Under standard conditions, the density of air is 1.293 kg/m3 and that of helium is 0.178 kg/m3. A spherical helium balloon lifts a basket plus cargo of mass 269.0 kg. What must be the minimum diameter of the spherical balloon?

2. Relevant equations
p = F/A

3. The attempt at a solution
I don't know what equation to use to find F?

2. Nov 17, 2007

### rl.bhat

Body floats when the weight of the body is equal to the weight of the displaced fluid. In this question mass of the displace fluid is = volume x density of air.

Last edited: Nov 18, 2007
3. Nov 18, 2007

### chaoseverlasting

Rl Bhat has given you the equation of the buoyant force which counter acts against gravity. Gravitational force is mg[/tex]

4. Nov 18, 2007

### dynamicsolo

I believe the displaced fluid is air in this situation. (If you use helium in that role, the balloon will never float with that payload...)

5. Nov 18, 2007

### rl.bhat

Yes. It is air. The exact formula is
Volume of helium* denisity of helium*g + 269*g = Volume of air*denity of air*g. From this equation find the radius of the sphere.

6. Nov 18, 2007

### BuBbLeS01

Thank you for helping me! I am really confused about what formula to use for the buoyant force when the object floats, sinks, or stays still.

7. Nov 18, 2007

### dynamicsolo

It's always the same definition for the buoyant force: B equals the weight of the volume of displaced fluid. If the object is completely submerged, then the volume you use will be the complete volume of the object. If the object floats only partially submerged, then the volume you use will only be the volume of the object that is below the surface.

The question of how the object moves is in the balance between the buoyancy and the weight. The net force on the object will be B - mg = ma , with positive chosen as upward.

So if the object is denser than the fluid, B < mg and the object sinks (a < 0). If the object has an average density equal to fluid, then B = mg and a = 0 ; the object is said to be "neutrally buoyant".

If the average density is less than the fluid, B > mg , so the object will accelerate upward until it breaches the surface and "floats". It will rise above the surface until the volume of the object that is still submerged has a weight equal to the object's weight. This will lead to a lower buoyancy force B' , with B' = mg. As an example, ice has about 0.9 the density of water, so an ice cube (or an iceberg) floats so that only about 0.9 of its volume is below the water's surface. In that way, the weight of nine-tenths of its volume of water equals the weight of the ice cube.

Last edited: Nov 18, 2007
8. Nov 18, 2007

### BuBbLeS01

Ok I understand so far...now where I am confused is why the formula for this problem is...
Volume of helium* denisity of helium*g + 269*g = Volume of air*denity of air*g.
I thought we equate the buoyant force and the weight. And the buoyant force is just the volume of the fluid that is displaced. So why are multiplying it by all of that? And same for the right side.....I thought it was just the weight of the object (the balloon)?

9. Nov 18, 2007

### rl.bhat

Left hand side is the weight of the floating body and right hand side is the weight of displaced air. Weight of the empty balloon is not taken into account.

10. Nov 20, 2007

### BuBbLeS01

Volume of helium* denisity of helium*g + 269*g = Volume of air*denity of air*g

Vh * 0.178 kg/m3 * 9.8 m/s2 + 269kg * 9.8 m/s2 = Va * 1.293 kg/m3 * 9.8 m/s2

Vh = [(Va * 1.293 kg/m3 * 9.8 m/s2) - 269kg * 9.8 m/s2] / (0.178 kg/m3 * 9.8 m/s2)

So I use the volume helium to solve for the radius but how do I figure out the volume of air? Aren't they equal?

11. Nov 20, 2007

### rl.bhat

Yes.They are equal. Hence volume of the sphere = volume if air = volume of helium
= 269/(1,293 - 0.178) { Cancell 9.8 ms^-2 from both side}