Hot air balloon buoyancy problem

AI Thread Summary
The discussion revolves around a hot air balloon buoyancy problem where the initial calculations for the weight to be dropped were deemed incorrect. The user calculated a need to drop 114.6 kg based on buoyancy and net force equations but later realized the problem required the weight in Newtons, which should be 1124 N. Participants noted that the problem's parameters seemed unrealistic, questioning the feasibility of a balloon with a 12-meter diameter rising under such a force while neglecting air resistance. The conversation highlights the importance of clarity in distinguishing between mass and weight in physics problems. Overall, the thread emphasizes the complexities and potential oversights in buoyancy calculations.
Jpyhsics
Messages
84
Reaction score
2
Homework Statement
A hot-air balloon is initially floating (stationary) a few metres above the ground. The radius of the spherical balloon is 6.25 m, and the density of surrounding air is 1.29 kg m-3. "Ballast" is weight of negligible volume that can be dropped to make the balloon rise. How much weight would you have to drop overboard to make the balloon rise 105 m in 15.0 s? You may assume that the density of surrounding air remains constant.
Relevant Equations
Archimedes principle: Fb=rho*gV
Vsphere= 4/3*pi*r^3
Initially the hot air balloon is stationary so...
FB=Fg
ρgV=mg
m=1319.2...kg
FB=Fg=12941N

In the air...
a=2d/t^2=0.933...m/s^2
Fnet=FB-Fg
(1319-x)(0.93...)=12941-(1319-x)g
x=114.6 kg

but apparently this is wrong?...
 
Physics news on Phys.org
Jpyhsics said:
Problem Statement: A hot-air balloon is initially floating (stationary) a few metres above the ground. The radius of the spherical balloon is 6.25 m, and the density of surrounding air is 1.29 kg m-3. "Ballast" is weight of negligible volume that can be dropped to make the balloon rise. How much weight would you have to drop overboard to make the balloon rise 105 m in 15.0 s? You may assume that the density of surrounding air remains constant.
Relevant Equations: Archimedes principle: Fb=rho*gV
Vsphere= 4/3*pi*r^3

Initially the hot air balloon is stationary so...
FB=Fg
ρgV=mg
m=1319.2...kg
FB=Fg=12941N

In the air...
a=2d/t^2=0.933...m/s^2
Fnet=FB-Fg
(1319-x)(0.93...)=12941-(1319-x)g
x=114.6 kg

but apparently this is wrong?...
Why do you think it is wrong?
 
I didn't calculate using the given values, but equations you have used seems to be correct.
Hint: your solution would be more readable if you work only with symbolic names for the quantities throughout your calculation, and put in the given values in the end.
 
Jpyhsics said:
but apparently this is wrong?...
Do you know what the answer is claimed to be?
 
ehild said:
Why do you think it is wrong?
I was marked wrong by the system.
 
haruspex said:
Do you know what the answer is claimed to be?
no...sadly
 
Jpyhsics said:
I was marked wrong by the system.
Try to enter the result with three digits .
 
ehild said:
Try to enter the result with three digits .
So I figured out that they were asking for weight and not mass, but thanks everyone.
 
The answer was supposed to be 1124 N.
 
  • #10
Jpyhsics said:
So I figured out that they were asking for weight and not mass, but thanks everyone.
Yes, it was weight... It is easy to overlook.
 
  • #11
The stated problem seems cringe-worthy. We are expected to have a 12 meter diameter balloon rising at an eventual rate of 14 meters per second under a force of a mere 1124 N while neglecting air resistance?

And neglecting the fact that for the balloon to deflect upwards, some volume of the surrounding air must deflect downward?

12 meter diameter is similar to that of a deployed parachute and 1124N is similar to the weight of a skydiver. The computed result might be good to one significant digit. Four is ludicrous.
 
Back
Top