# Need help with hot air balloon

1. Jun 4, 2005

### Swatch

Hi.
There is a hot air ballon with total mass M. There is an upward lift force of 2Mg/3 but the balloon is accelerating downward at a rate of g/3. The passenger needs to make the balloon lighter and starts throwing things of. What fraction of the total weight must he drop overboard so that the balloon accelerates upward at a rate of g/2.

I come up with the fact that the upward lift force must equal the weight plus the acceleration. But I get a wrong answear. The correct answear is 5/9. Can someone give me a hint to this problem please.

2. Jun 4, 2005

### lektor

dont worry trying a new idea

3. Jun 4, 2005

### Swatch

Have been trying all possible ways with the facts I have, have not been able to come up with the right answear.

4. Jun 4, 2005

### Staff: Mentor

Show your work and we can take a look.

Here are a few hints: The upward buoyant force (2Mg/3) will remain the same even after the passenger throws things out of the balloon. After the passenger tosses out some stuff, the new total mass of the balloon will be smaller. Call it "xM" (where x is some fraction). Now write Newton's 2nd law and solve for x.

5. Jun 4, 2005

### Swatch

OK. I tried solving for x and this is what I got.

2Mg/3=(M-xM)g + (M-xM)g/2 The upward force should be equal to the weight + acceleration.

So this should be:

2Mg/3 = (M-xM)3g/2
4Mg/3 = 3Mg - 3xMg
(4Mg/3) - 3Mg = -3xMg

or 5Mg = 9xMg or x=5/9

This is the right answear, so I guess this is one way of solving this.

In my former calculations I forgot to express the total mass as (M-xM) I just put xM. Guess that was the big mistake I did.

Thanks for the hint

6. Jun 4, 2005

### Staff: Mentor

Better to say it this way: The net force equals the mass x acceleration, and the net force equals the buoyant force minus the weight.

You used "xM" to represent the amount of mass tossed out. Good.