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

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Homework Help Overview

The problem involves a lighter-than-air balloon that needs to rise a specific distance in a given time by dropping ballast. The context includes the balloon's dimensions and the density of air, with participants exploring the forces acting on the balloon, such as buoyant force and gravity.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the application of Bernoulli's equation and the forces acting on the balloon, including buoyant force and gravitational force. There are attempts to derive relationships between these forces and the balloon's acceleration. Some participants express confusion about the calculations and the assumptions regarding density and mass.

Discussion Status

The discussion is active, with various interpretations and approaches being explored. Some participants have offered guidance on identifying forces and drawing free body diagrams, while others are questioning the assumptions made about density and mass. There is no explicit consensus on the method to find the weight of the ballast needed.

Contextual Notes

Participants note that the balloon initially floats stationary, which implies a balance of forces. There is also mention of the need for specific values, such as density and mass, which are not fully established in the discussion.

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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?

Thanks!
 
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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)
 
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.
 
the Fnet is not zero... you know the balloon is accelatating upward, right?
 
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.
 
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...
 
AH HA! I did it, the answer comes out to be 1120 N. Hmm, well time well spent. Thanks Vinny.
 
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.
 
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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