# Homework Help: Force and Newton's Laws

1. Sep 24, 2007

### 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 seperate quantity from the balloon.
I'm unsure of how to proceed from here,
any help is much appreciated

Thanks,

2. Sep 24, 2007

### learningphysics

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

3. Sep 25, 2007

### 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?

4. Sep 25, 2007

### learningphysics

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.

5. Sep 25, 2007

### 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

6. Sep 25, 2007

### learningphysics

You should use Mnew*g not Mg.

7. Sep 25, 2007

### 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

8. Sep 25, 2007

### learningphysics

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

9. Sep 25, 2007

### 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)

10. Sep 25, 2007

### learningphysics

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

11. Sep 25, 2007

### 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

12. Sep 25, 2007

### learningphysics

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

13. Sep 25, 2007

### Destrio

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

so leaving Mballast = M ?

14. Sep 25, 2007

### learningphysics

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

15. Sep 25, 2007

### Destrio

beautiful
that makes sense
thanks very much