# Volume buoyancy problem

1. Feb 7, 2010

### dymand68

1. The problem statement, all variables and given/known data
"friend with a mass of 79 kg ponders the idea of attaching a helium-filled balloon to himself to effectively reduce his weight by 21% when he climbs. He wonders what the approximate size of such a balloon would be"
solution is expressed as cubic meters
apology if this entry is miscategorized

2. Relevant equationsvolume =mass/density

3. The attempt at a solutionvolume= change in mass/ difference in density=16.59/(1.2-1.78)=16.27m^3
this amount is inaccurate, as is 13.27 m^3

Last edited: Feb 7, 2010
2. Feb 7, 2010

### mgb_phys

You have the right idea, you just need to be a bit more careful about working it out.

You want a bouyancy of 21% * 79kg = 16.59kg
bouyancy = volume * difference in density

Density of air = 1.27 kg/m^3 helium = 0.18 kg/m^2

3. Feb 7, 2010

### dymand68

presuming:
16.59 kg/1.09 kg/m^3=15.22 m^3 (volume)
neither 15 nor 15.22 m^3 is the correct solution- according to my online homework program

Last edited: Feb 7, 2010
4. Feb 7, 2010

### mgb_phys

Unless it gives you a value for the altitude or temperature, then assuming STP
Density 1.2754 - g/l (air) 0.1786 g/l (he) = 1.0968 g/l (or kg/m^3)
You need to lift 79*0.21 kg / 1.0968 kg/m^3 = 15.1 m^3

Last edited: Feb 7, 2010
5. Feb 7, 2010

### dymand68

these are the incorrect answers submitted:
13
13.2
13.27
13.28
16.95
16.23
16.26
1.38
16
18.08
18
15.22
15...14.9.....15.1

Last edited: Feb 7, 2010
6. Feb 7, 2010

### dymand68

exact question:
"A mountain-climber friend with a mass of 79 kg ponders the idea of attaching a helium-filled balloon to himself to effectively reduce his weight by 21% when he climbs. He wonders what the approximate size of such a balloon would be. Hearing of your physics skills, he asks you. What answer can you come up with, showing your calculations?"
-
solution is in m^3

-thank you for the explanation, i have no more available attempts...should find out the correct answer tomorrow, will update

Last edited: Feb 7, 2010
7. Feb 8, 2010

### dymand68

according to Paul G. Hewitt, via webassign.net;
the solution is:
volume = 79 kg*.21/1.2 kg/m^3= 13.825 m^3
no consideration of helium density and air density is considered 1.2