Mass and Acceleration of a Balloon in General Terms

[SniperArchety]

Homework Statement

Hot-air balloon of mass M is descending straight down with acceleration a. How much Mass must be discarded to make acceleration a the same magnitude, but directly up instead of directly down? Disregard wind resistance/lift.

Homework Equations

The only one my book discusses in this chapter is this one:
F = m * a

The Attempt at a Solution

Obviously I tried plugging into the equation given, F = ma. The initial force on the balloon with its initial mass must be F = M * a. Solving for M I got M = a / F. Now is where I get lost. I need to find a certain M such that a = -a? How exactly does gravity g factor into this equation? Is the initial acceleration (a - g), while the final acceleration (-a + g)?

I'm a calc II student so I understand the mathematics here, and I understand the notation of finding a general solution in terms of variables. I've just been stumped on this problem. In the grand scheme of things, this is the last problem of about a dozen for a bi-weekly homework assignment, so it shouldn't hurt my grade, but I would like to understand it better because sometimes we're randomly called to the board to show our work, and I don't want to feel like a hot air balloon.

 

[LowlyPion]

Welcome to PF.

No reason to get too far afield in calculus.

Consider that you have a constant Buoyant force from the air displacement.

That means then that you can write equations for before and after dropping.

Fb - Mg = -Ma

and after

Fb - mg = ma

That yields what the mass left over (m) that needs to remain to satisfy the conditions.

The mass jettisoned then is M - m.

 

[SniperArchety]

Sorry if I appear dense, but i don't believe we're supposed to consider buoyant forces (I don't even know what that is to be honest). I'm having a little difficulty following your steps here. You're saying initially that the Force of Buoyancy (Which I am just assuming is the force of the lift of the balloon from the fire burning on the inside) subtracted by M * g is equal to -Ma.. Could you please explain, if it's not too much trouble, what this means conceptually?

Also, we have an answer sheet (teacher gave it to us to make sure our work is good). The correct final answer for this problem is apparently:

2Ma / (a + g)

 

[LowlyPion]

Sorry if I appear dense, but i don't believe we're supposed to consider buoyant forces (I don't even know what that is to be honest). I'm having a little difficulty following your steps here. You're saying initially that the Force of Buoyancy (Which I am just assuming is the force of the lift of the balloon from the fire burning on the inside) subtracted by M * g is equal to -Ma.. Could you please explain, if it's not too much trouble, what this means conceptually?

Also, we have an answer sheet (teacher gave it to us to make sure our work is good). The correct final answer for this problem is apparently:

2Ma / (a + g)

Buoyant force is not needed to understand except that it is a lifting Force that remains constant (and you end up discarding) that acts against whatever mass there is. Whatever mass there is gets acted on by gravity g and is downward against whatever lifting force there is.

For instance take the 2 equations below and rearrange so Fb is on one side. Then set the two equations equal and Fb is no longer a consideration.

 

[SniperArchety]

Buoyant force is not needed to understand except that it is a lifting Force that remains constant (and you end up discarding) that acts against whatever mass there is. Whatever mass there is gets acted on by gravity g and is downward against whatever lifting force there is.

For instance take the 2 equations below and rearrange so Fb is on one side. Then set the two equations equal and Fb is no longer a consideration.

Doing that I got Fb = Mg - Ma AND Fb = mg + ma

Mg - Ma = mg + ma

Solving for m (mass discarded), I get

m = M(g - a) / (g - a)

Crossing out the term (g - a)

m = M

This doesn't quite make sense to me, and it's not inline with the answer the teacher gave us.
I'm going to try playing with the equations you gave me and see what I can come up with.

 

[LowlyPion]

This doesn't quite make sense to me, and it's not inline with the answer the teacher gave us.
I'm going to try playing with the equations you gave me and see what I can come up with.

I'm glad it doesn't make sense to you, because m is not the mass you threw overboard. m is the mass of the balloon AFTER the mass, call it X, is heaved.

ma + mg = Mg - Ma

That means that

m = M*(g - a)/(g + a)

X = M - m = M - M*(g - a)/(g + a) = M*(1 - (g - a)/(g + a))

Now simplify.

 

[SniperArchety]

I'm glad it doesn't make sense to you, because m is not the mass you threw overboard. m is the mass of the balloon AFTER the mass, call it X, is heaved.

ma + mg = Mg - Ma

That means that

m = M*(g - a)/(g + a)

X = M - m = M - M*(g - a)/(g + a) = M*(1 - (g - a)/(g + a))

Now simplify.

Awesome, I get it out. Thanks a lot I really appreciate it!