# Archimedes principle again.

1. Jun 6, 2010

### pat666

1. The problem statement, all variables and given/known data

How much helium (density 0.18) is requires to lift a 340kg balloon. density of air is 1.28

2. Relevant equations

3. The attempt at a solution\
i dont know why i keep getting the wrong answer
buoyant force required is 340g which = rho g V
and i keep getting 1888.89m^3 which i can see is way to big, what i cant see is what is wrong with my procedure???

2. Jun 6, 2010

### Maybe_Memorie

In the question you have 340 kg, but in the answer you have 340 g. Which is it?

3. Jun 6, 2010

### pat666

g is gravity(force) sorry the balloon is 340Kg

4. Jun 6, 2010

### Maybe_Memorie

What is the correct answer supposed to be?

I'm getting an answer of $$V = 232.8767 m^3$$, but there is a high chance i'm wrong.

5. Jun 6, 2010

### rl.bhat

If V is the volume of the balloon, then the weight of the balloon with helium will be

W = (340 + V/0.18)g

Weight of teh displaced air = (V/0.18)*1.28*g

So 340 + V/0.18 = (V/0.18)(1.28)

Now solve for V.

6. Jun 6, 2010

### pat666

thanks rl.bhat, maybe memorie the correct answer is 309m^3 if you were interested

7. Jun 6, 2010

### pat666

rl.bhat when i solve that equation you gave me i get 218.6m^3 which is still significantly different to what my text book say..////??

8. Jun 6, 2010

### Staff: Mentor

The buoyant force equals the weight of the displaced air. Set that equal to the weight of the empty balloon plus the helium. Then you can solve for V.

9. Jun 6, 2010

### pat666

so F_b=w_air+w_helium
340g=1.28*9.81x+0.18*9.81x
that comes out to 232.877m^3 which is what someone previously said. maybe the answer in the text book is wrong???????

10. Jun 6, 2010

### Staff: Mentor

No, set the buoyant force equal to the weight of balloon + helium.
No, the text book is fine. For some reason you are setting the buoyant force equal to 340g. That's incorrect. (340g is the weight of the empty balloon.) Write an expression for the buoyant force in terms of the volume (what you call x).

11. Jun 6, 2010

### pat666

ok i still dont get this at all. w_air + w_helium = rho*g*v?????????? if thats right what is the fluid on the right side of the equation?

12. Jun 6, 2010

### Staff: Mentor

Why are you adding the weights of air and helium?

rhoair*g*V is the buoyant force (the weight of displaced air).

Set that equal to the weight of the balloon (which is given) plus the helium (express that in terms of V).

13. Jun 6, 2010

### pat666

THANKS Doc Al i finally have the correct answer although i dont understand why it works. I thought that the buoyant force the helium had to provide would be equal to the force due to gravity so that it could float up???

14. Jun 6, 2010

### Staff: Mentor

It does! But the buoyant force must lift both the helium and the empty balloon.

15. Jun 6, 2010

### pat666

ok i sort of get that, THANKS AGAIN ill consult my text book and see if i cant make some more sense of it.. thanks