# Meteorological balloon

## 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

gneill
Mentor
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 ballon should be greater than the air density...but then ?

gneill
Mentor
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 ballon 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 ballon 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.

gneill
Mentor
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)?

gneill
Mentor
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.

The buoyant force = g(mass of the gas + cargo weight)?

gneill
Mentor
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.

i really need to understand buoyancy well..the formula that i have writen 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 ballon equal the volume of air displaced.
where all this will lead us?

(10.5+x)/20 = mair/1.18 ?

gneill
Mentor
i really need to understand buoyancy well..the formula that i have writen 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 ballon 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.

gneill
Mentor
First calculate the weight of air that is displaced by the balloon.

(10.5+x)/20 = mair/1.18 ?

I think my formula here is wrong

First calculate the weight of air that is displaced by the balloon.

how can we calculate that?

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

gneill
Mentor
Okay, make it simple. You have the volume. You have the density of air. What is the mass of air that is displaced?

and sorry for being slow to understand today

gneill
Mentor
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?

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?

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 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

gneill
Mentor
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.