How Much Weight Can a 20.0 m³ Meteorological Balloon Lift?

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Homework Help Overview

The problem involves a meteorological balloon with a volume of 20.0 m³ and a weight of 10.5 kg, exploring how much additional weight it can lift based on the principles of buoyancy and air density.

Discussion Character

  • Exploratory, Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • Participants discuss the relationship between the density of the gas inside the balloon and the surrounding air, questioning how buoyancy applies to the situation. Some express confusion about the principles of buoyancy and how to calculate the weight of displaced air.

Discussion Status

The discussion is active, with participants sharing insights about buoyancy forces and the calculations involved. Some have offered guidance on how to approach the problem, while others are still seeking clarity on the fundamental concepts.

Contextual Notes

Participants note the importance of understanding buoyancy in both gases and liquids, and there is a recognition of the need to calculate the weight of displaced air to determine lifting capacity. Confusion exists regarding the starting points for solving buoyancy-related problems.

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Homework Statement


A meteorological balloon has volume of 20.0 m3 and a weight of 10.5 kg (gas included).
The density of the air is 1.18 kg/m3 and acceleration of gravity 9.81 m/s2


Homework Equations


What amount of cargo in kg can the balloon lift?


The Attempt at a Solution


Can you please advise about the starting point, it has certainly something to do with the density of air.
I'm using the principle PV=nRT but it's not taking me any further
 
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Think buoyancy.
 
you're explaining me something difficult with something more difficult...some laws would be nice.
I think that to be able to lift some weight, the density inside the balloon should be greater than the air density...but then ?
 
chawki said:
you're explaining me something difficult with something more difficult...some laws would be nice.
I think that to be able to lift some weight, the density inside the balloon should be greater than the air density...but then ?

If the density inside the balloon were greater than that of air, then the balloon would be heavier than air and it would fall to the ground.

This is a buoyancy problem. I'm afraid there's no escaping it! You'll need to review your notes or text on the subject.

The balloon's volume is displacing a certain weight of air, replacing it with its own (less dense) weight of gas + balloon.
 
yes i was wrong about the density inside the ballon...it should be less than air of course..we heat air inside the balloon so that the density decrease.
you know, it would be nice to see the solution or half of it and i will try to understand it by my own..it have always worked for me.
but thank you for your help.
 
can someone tell me how can we use buoyancy here.
 
Buoyancy force acts upwards. Buoyant force magnitude is equal to the weight of the fluid (air in this case) displaced by the object. Whatever force is left over after you've subtracted out the weight of the object is the "lift" -- the lifting force available.
 
the lift force is like buyancy force (in air)?
 
chawki said:
the lift force is like buyancy force (in air)?

No, lift is what's left of the buoyancy force once the weight (due to gravity) of the object is subtracted.
 
  • #10
The buoyant force = g(mass of the gas + cargo weight)?
 
  • #11
chawki said:
The buoyant force = g(mass of the gas + cargo weight)?

Well, the "mass of the gas" should include the mass of the balloon's fabric and any other hardware associated with it, but yes that is the idea. When the forces balance as you have written, that is the maximum cargo weight, and the balloon and cargo will just hover.
 
  • #12
i really need to understand buoyancy well..the formula that i have written above..got it when googling about buoyancy..still don't understand it well.and if it is the same in all cases?
all i know so far is that if we contrast water with the case of air..i know that the volume of the balloon equal the volume of air displaced.
where all this will lead us?
 
  • #13
(10.5+x)/20 = mair/1.18 ?
 
  • #14
chawki said:
i really need to understand buoyancy well..the formula that i have written above..got it when googling about buoyancy..still don't understand it well.and if it is the same in all cases?
all i know so far is that if we contrast water with the case of air..i know that the volume of the balloon equal the volume of air displaced.
where all this will lead us?

Hopefully it will lead you to a class or textbook covering the topic! This is probably not the best place to learn a subject from scratch.

To answer your query, yes, buoyancy works in the same fashion for gases and liquids; any time you've got a displaced medium and gravity (or bulk acceleration) working.
 
  • #15
First calculate the weight of air that is displaced by the balloon.
 
  • #16
chawki said:
(10.5+x)/20 = mair/1.18 ?

I think my formula here is wrong
 
  • #17
gneill said:
First calculate the weight of air that is displaced by the balloon.

how can we calculate that?
 
  • #18
The weight of air that is displaced by the balloon is the same the weight of the ballon.
m*g=1.18*20*9.81=231.51N
 
  • #19
Okay, make it simple. You have the volume. You have the density of air. What is the mass of air that is displaced?
 
  • #20
please check post #18
and sorry for being slow to understand today
 
  • #21
chawki said:
please check post #18
and sorry for being slow to understand today

No problem. Some people think I'm like that all the time (of course, it depends on the topic!).

Okay, so you've determined that displaced air is producing a buoyancy force Fb = 231.51N.

Subtract from that the weight of the balloon and gas inside and you get

Fb - 10.5kg * 9.8m/s^2 = 231.51N - 103.0N = 128.5 N

That is then the net upward force available for lifting cargo.

How much mass can you lift with that force?
 
  • #22
why we don't go simply with the basic idea of The volume of balloon = the volume of air displaced
and then...20=(10.5+x)/1.18 and then we find x=13.1kg?
 
  • #23
gneill said:
No problem. Some people think I'm like that all the time (of course, it depends on the topic!).

Okay, so you've determined that displaced air is producing a buoyancy force Fb = 231.51N.

Subtract from that the weight of the balloon and gas inside and you get

Fb - 10.5kg * 9.8m/s^2 = 231.51N - 103.0N = 128.5 N

That is then the net upward force available for lifting cargo.

How much mass can you lift with that force?

OUPSOUPSOUPS... that's 128.5/9.81=13.1kg too :eek:
 
  • #24
Dear gneil
I go to sleep now..i have no idea how I'm thinking or how i think it in post 22
good night, i will continue this tomorrow
 
  • #25
I went the long way around so that you could see that there are forces involved; The displaced medium (air in this case) results in a buoyant force (in Newtons) when there is a gravitational field involved. You will come across problems involving buoyancy where it is essential that you understand where the forces involved are coming from, and in what directions they act.
 
  • #26
is buoyancy related to weight or volume?
in the case of a ship, what would be the condition for floating or sinking?
 
  • #27
chawki said:
is buoyancy related to weight or volume?
in the case of a ship, what would be the condition for floating or sinking?

A ship floats because it displaces its own weight in water before it's completely submerged -- it sinks until the weight of water displaced is equal to the weight of the ship. If it didn't displace its own weight in water before it is completely submerged, then its weight would be greater than the maximum possible buoyant force for the ship (the weight of water corresponding to the entire volume of the ship), and it would sink.
 
  • #28
Can you please write that mathemaically in equation
 
  • #29
Can we use Newton's second law to find Fbuoyancy?

m*g-Fbuoyancy=0
F=m*g
F=density of air*volume of balloon*g
F=1.18*20*9,81
F=231.5N
 
  • #30
See http://theory.uwinnipeg.ca/physics/fluids/node10.html" .
 
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