
#1
Apr906, 11:35 AM

P: 41

Hi, I am having some trouble with the following question:
You (mass 70.1 kg) decide to take offliterallywith a helium balloon of mass 31.6 kg. The densities of air and helium are air = 1.321 kg/m3 and helium = 0.179 kg/m3. What volume of helium is needed to levitate you and the balloon? I think that you use the buoyant force equations, but I am not sure how. I tried doing it like this: F(buoy)=density(fluid)*V'*g or (101.7*9.81)=(1.321*V'*9.81) and solved for V' and got V'=76.987 m^3 but this is not correct. I cannot figure out how to do this problem, any help would be appreciated. Thanks!! 



#2
Apr906, 11:50 AM

Emeritus
Sci Advisor
PF Gold
P: 9,789

If we take the equation which you stated;
[tex]F_{b} = \rho V g[/tex] The 'thing' that is creating the bouyant force is the difference between the relative densities of the fluids hence in your case; [tex]F_{b} = ( \rho_{air}  \rho_{helium} )Vg[/tex] Hoot 



#3
Apr906, 11:53 AM

Sci Advisor
HW Helper
PF Gold
P: 12,016

First, you must state the total weight of your system:
[tex]W=W_{person}+W_{balloon}+W_{helium}[/tex] Note that your unknown volume appears in [itex]W_{helium}[/tex] Assuming that you and the balloon material contributes only negligibly to the total volume of the system, you can now find what the volume must be by equating this total weight with the buoyant force (which is the net pressure force needed to keep a volume V of air floating in..the air). 



#4
Apr906, 11:59 AM

HW Helper
P: 1,446

buoyant force problem
Try adding the weight of the helium in the balloon to the lefthand of your equation, the mass of the balloon probably do not include this additional weight
[tex]m_{helium}g=\rho_{helium} Vg[/tex] 



#5
Apr906, 04:43 PM

P: 41

so the volume must be 89.05 m^3, thanks everyone!



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