How Much Weight Can a 20.0 m³ Meteorological Balloon Lift?

[gneill]

so the weight of the balloon is simply 10.5*9.81=103.005 N ?

Yes, that is the weight of the balloon and the gas inside it if the buoyancy is ignored.

 

[chawki]

Yes, that is the weight of the balloon and the gas inside it if the buoyancy is ignored.

While the weight of the air displaced is 1.18*20*9.81=231.51 N ?

 

[gneill]

Yes.

You know, I think we went through all this before...

 

[chawki]

Yes.

You know, I think we went through all this before...

yes i know. just let me get it right and step by step because I'm confused about it...

we have the weight of the displaced air, and the weight of the ballon.
and by law,
The buoyant force= weight of the displaced air-weight of the balloon ?
The buoyant force= 231.51-103.005 = 128.505 N (that's the buoyant force for a balloon of 10.5kg with volume of 20m³ and under the density of 1.18kg/m³)

 

[gneill]

The buoyant force is the equal to the weight of the displaced air. Period.

The lift force is equal to the buoyant force minus the weight of the materials making up the balloon -- that is, it is the force available for lifting cargo (you don't count the balloon as cargo).

 

[chawki]

The buoyant force is the equal to the weight of the displaced air. Period.

The lift force is equal to the buoyant force minus the weight of the materials making up the balloon -- that is, it is the force available for lifting cargo (you don't count the balloon as cargo).

ok. so Fb=weight of the materials making up the balloon + weight of cargo ?
but what is the weight of the materials making up the balloon here

 

[gneill]

ok. so Fb=weight of the materials making up the balloon + weight of cargo ?

If the balloon is hovering and not accelerating either up or down, yes.

but what is the weight of the materials making up the balloon here

Read the original problem statement.

 

[chawki]

this is seriously word complications..that's when i get lost
in post#23 you said the buoyant force is 231.51 N
isn't 128.505 N ?

 

[gneill]

The buoyant force is the equal to the weight of the displaced air. Nothing else. It's the upward force due to the displacement of the air. It does not include any other forces (like weights, loads, other external forces).

 

[chawki]

the buoyant force is the equal to the weight of the displaced air. Nothing else. It's the upward force due to the displacement of the air. It does not include any other forces (like weights, loads, other external forces).

... ... ...

 

[chawki]

... ... ...

i think we are always supposed to calculate the buoyant force, and if:
the buoyant force = weight of displaced fluid-weight of the objcet
and if:
the buoyant force is = the weight of the displaced air

HOW CAN THESE EQUATIONS BE SAME?

 

[gneill]

Forces of interest:

1. Weight of balloon
2. Buoyant force
3. Weight of cargo

The lift is what remains of the buoyant force when the weight of the balloon is subtracted... it's the force available to lift the cargo. In this case it's equal to the weight of the cargo, since the problem specifies the maximum cargo.

Note that the balloon will not be able to rise with its cargo because there's no unbalanced lift force left over to provide for acceleration. This is a condition referred to as "neutral buoyancy", where an object will neither sink nor rise.

 

[gneill]

i think we are always supposed to calculate the buoyant force, and if:
the buoyant force = weight of displaced fluid-weight of the objcet
and if:
the buoyant force is = the weight of the displaced air

HOW CAN THESE EQUATIONS BE SAME?

Why do you want to say that the buoyant force is equal to the weight of the displaced fluid minus the weight of the object? It is not. You are confusing lift with buoyant force.

Now, read carefully: The buoyant force is equal to the weight of the displaced air.
That is it. That is all. Nothing else. No other interpretation. No other factors. End of story.

To find out how much cargo can be lifted, you take the buoyant force and subtract the weight of the balloon. Whatever force remains is available to lift cargo.

 

[chawki]

Forces of interest:

1. Weight of balloon
2. Buoyant force
3. Weight of cargo

The lift is what remains of the buoyant force when the weight of the balloon is subtracted... it's the force available to lift the cargo. In this case it's equal to the weight of the cargo, since the problem specifies the maximum cargo.

Note that the balloon will not be able to rise with its cargo because there's no unbalanced lift force left over to provide for acceleration. This is a condition referred to as "neutral buoyancy", where an object will neither sink nor rise.

Buoyancy force=weight of ballon+the weight of cargo(lift force) ?

 

[gneill]

Yes. That is the equation that describes the situation put forth in the problem statement. So you should be able to solve for the weight of the cargo, right?

 

[chawki]

Yes. That is the equation that describes the situation put forth in the problem statement. So you should be able to solve for the weight of the cargo, right?

i found that by applying second law of Newton at a statioanary state...
but you didn't answer what i asked in post#43

 

[gneill]

i found that by applying second law of Newton at a statioanary state...
but you didn't answer what i asked in post#43

What did you ask? You said:

"in post#23 you said the buoyant force is 231.51 N
isn't 128.505 N ?"

That question is answered! The buoyant force is 231.51 N, the weight of the displaced air.
It is not 128.505 N, the weight of the displaced air minus the weight of the balloon; that is the lift.

I cannot see how to make it any clearer. The only forces of interest are the buoyant force, the weight of the balloon, the lift (buoyancy - weight of balloon), and the weight of the cargo (which is set equal in magnitude to the lift for this problem).

 

[chawki]

What did you ask? You said:

"in post#23 you said the buoyant force is 231.51 N
isn't 128.505 N ?"

That question is answered! The buoyant force is 231.51 N, the weight of the displaced air.
It is not 128.505 N, the weight of the displaced air minus the weight of the balloon; that is the lift.

I cannot see how to make it any clearer. The only forces of interest are the buoyant force, the weight of the balloon, the lift (buoyancy - weight of balloon), and the weight of the cargo (which is set equal in magnitude to the lift for this problem).

NOW IT'S CLEAR...because i have been working with the idea of:
buoyancy force=weight of displaced air - weight of ballon
which is WRONG