How much ballast is needed to accelerate a descending research balloon upward?

[Destrio]

A research balloon of total mass M is descending vertically with downward acceleration a. how much ballast must be thrown from the car to give the balloon an upward acceleration a, assuming that the upward life of the air balloon does not change.

I figured that
let Fu = upward lift force
Fnet = Fg - Fu = mg - ma
and Fg must be greater than Fu since there is downward acceleration
I'm not sure if I want to use big M to represent the mass I throw out
or use little m and it be a separate quantity from the balloon.
I'm unsure of how to proceed from here,
any help is much appreciated

Thanks,

 

[learningphysics]

A research balloon of total mass M is descending vertically with downward acceleration a. how much ballast must be thrown from the car to give the balloon an upward acceleration a, assuming that the upward life of the air balloon does not change.

I figured that
let Fu = upward lift force
Fnet = Fg - Fu = mg - ma

Fnet = Ma

Fg - Fu = Ma

Mg - Fu = Ma etc...

Once you calculate Fu... you know it remains the same.

You want the upward acceleration = a

Fu - Fg = Mnew*a

 

[Destrio]

so if
upward acceleration = a
Fu - Fg = Mnew*a
a = (Fu - Fg) / Mnew

I would have to throw up M - Mnew ballast

how could I show this?

 

[learningphysics]

so if
upward acceleration = a
Fu - Fg = Mnew*a

substitute in the Fu calculated in the first part... substitute Fg = Mnew*g. Then solve for Mnew.

Then get the difference in masses between M and Mnew.

 

[Destrio]

the Fu = -Ma + Mg ?

Fu - Fg = Mnew*a
Fu - Mg = Mnew*a

-Ma + Mg - Mg = Mnew*a
-Ma = Mnew*a
Mnew = -Ma/a
Mnew = -M

Mnew - M= Mballast
-M - M = Mballast
-2M = Mballast

would this work?
Thanks

 

[learningphysics]

the Fu = -Ma + Mg ?

Fu - Fg = Mnew*a
Fu - Mg = Mnew*a

You should use Mnew*g not Mg.

 

[Destrio]

Fu = -Ma + Mnew*g
Fu - Fg = Mnew*a
Fu - Mnew*g = Mnew*a

will it change the rest?
If so, how can I cancel out the other terms

 

[learningphysics]

Fu = -Ma + Mnew*g
Fu - Fg = Mnew*a
Fu - Mnew*g = Mnew*a

will it change the rest?
If so, how can I cancel out the other terms

don't cancel anything... substitute in Fu from the first part (Fu does not change)... solve for Mnew.

 

[Destrio]

Fu = -Ma + Mnew*g
Fu - Fg = Mnew*a
Fu - Mnew*g = Mnew*a
-Ma + Mg - Mnew*g = Mnew*a
-Ma + Mg = Mnew*a + Mnew*g
M(-a+g) = Mnew(a+g)
Mnew = M(-a+g)/(a+g)

 

[learningphysics]

Fu = -Ma + Mnew*g
Fu - Fg = Mnew*a
Fu - Mnew*g = Mnew*a
-Ma + Mg - Mnew*g = Mnew*a
-Ma + Mg = Mnew*a + Mnew*g
M(-a+g) = Mnew(a+g)
Mnew = M(-a+g)/(a+g)

looks right. now get the difference... is M - Mnew.

 

[Destrio]

Mballast = M - Mnew
Mballast = M - M(-a+g)/(a+g)
Mballast = M[1 - (-a+g)/(a+g)]

Is there any more I can do, or is this the mass I have to toss off?

Thanks

 

[learningphysics]

Mballast = M - Mnew
Mballast = M - M(-a+g)/(a+g)
Mballast = M[1 - (-a+g)/(a+g)]

Is there any more I can do, or is this the mass I have to toss off?

Thanks

you can simplify a little... use a common denominator of a+g.

 

[Destrio]

how can I divide -a+g by a+g
would it be -1 + 1 = 0

so leaving Mballast = M ?

 

[learningphysics]

how can I divide -a+g by a+g
would it be -1 + 1 = 0

so leaving Mballast = M ?

M(1 - \frac{-a+g}{a+g}) = M(\frac{a+g + a - g}{a+g}) = M(\frac{2a}{a+g})

 

[Destrio]

beautiful
that makes sense
thanks very much