How much weight must be dropped to make a balloon rise 105m in 15s?

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In summary, the conversation is about determining the weight that needs to be dropped from a lighter-than-air balloon in order for it to rise 105m in 15.0s. The participants discuss using Bernoulli's equation and solving for the weight of the ballast, as well as considering the acceleration and forces acting on the balloon. They ultimately determine that the weight needed to be dropped is 1120 N, with the clarification that the balloon should have the same density as air to stay afloat.
  • #1
Winner
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Yay, hopefully I can gather some pointers for this question.

A lighter-than-air balloon and its load of passengers and ballast are floating stationary above the earth. Ballast is weight (of negligible volume) that can be dropped overboard to make the balloon rise. The radius of this balloon is 6.25m. Assuming a constant value of 1.29 kg/m^3 for the density of air, determine how much weight must be dropped overboard to make the balloons rise 105 m in 15.0s.

I thinking this will involve Bernoulli's equation:

P1+1/2pv^2 + pgy=P2+1/2pv^2 +pgy.

Ok so P1 doesn't equal P2. Density should be same they cancel.

P1+1/2v^2 +gy=P2 +1/2V^2 +gy.

Now I'm stuck, P1 doesn't equal P2. y1 doesn't equal y2. Velocity is same? 105m/15s= 7m/s.

Am I on the right track? :grumpy:

Thanks!
 
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  • #2
first, find the acceleration of the balloon... I assume you can handle this part
second, find the force act on the balloon... easy?
draw the free body diagram and figure out what equation you need for the up force...(not your original one)
 
  • #3
Ok, so there are two forces I see acting on the balloon. Gravity acting downard and buoyant force acting up. So Fnet=FB-Fg so, pVg=mg. pV=g? Something's wrong, I'm not using acceleration, which I found to be 0.93m/s^2.
 
  • #4
the Fnet is not zero... you know the balloon is accelatating upward, right?
 
  • #5
I thought about that. But even if I went Fnet=FB-Fg and ma=FB-Fg and ma=pVg-mg and I solve for m = pVg/ (a+g), I dont' see how this will give me the weight of the ballasts vs weight of the balloon.
 
  • #6
and at the begining, the balloon is floating stationary. Fnet = 0 at that case... and you can easily find what is the mass of the balloon before dropping the weight...
 
  • #7
AH HA! I did it, the answer comes out to be 1120 N. Hmm, well time well spent. Thanks Vinny.
 
  • #8
I still don't get how you guys came up with that. you've got density x volume = mass...but you don't know the density OR the mass. Shine some light on me! its bugging me feeling so stupid.
 
  • #9
I still don't get how you guys came up with that. you've got density x volume = mass...but you don't know the density OR the mass. Shine some light on me! its bugging me feeling so stupid.
Wrong, you know the density, after you have the density, the mass is just a piece of cake!
the balloon it stays in air without going up/down... that's mean the balloon should have same density as air... Do you know why?
imagine what will happen if the balloon has higher/lower density than air...
 
  • #10
^yeah...apply Archimedes' principle, in other words.
 

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