## arcticf0x 3 years ago A balloon weighing 'W' newton descends with an acceleration of 'a'. If weight 'w' is removed from the balloon, the balloon has upward acceleration of 'a'. Show w=2aW/(a+g) g is acceleration due to gravity.

1. arcticf0x

any luck y2o2?

2. y2o2

Well, i'm trying..but i really like it :D

3. y2o2

Don't worry , it'll be solved

4. arcticf0x

ok! waiting!

5. y2o2

are you sure there is nothing about the mass ?!

6. arcticf0x

yeah sure, mass isnt mentioned anywhere.

7. arcticf0x

mass

8. arcticf0x

how about we eliminate mass after forming 2 equations?

9. arcticf0x

nah, that gives W=2gw/a+g

10. JamesJ

The balloon has two forces acting on it. Gravity and the upward force form the gas in the balloon. Call the second $$F_b$$. The first, $$F_g = -mg$$ where $$m$$ is the mass of the balloon. We're using the convention that upwards is positive and downward is negative. Hence $$F_b > 0$$ and $$F_g < 0$$. The net force acting on the balloon is $F_{net} = F_g(m_{initial}) + F_b$ We're told that that initially the balloon is descending. Hence $F_{net} = ma_{initial}$ where $$a_{initial} < 0$$.

11. JamesJ

The weight $$W$$ is the gravitational force, hence $$W = mg$$. We're told that a weight of $$w$$ is removed from the balloon. Hence the new weight of the balloon is $-F_g = W' = W - w = mg - w$ Hence the net force acting on the balloon now is $F_{net}' = F_g' + F_b = -(mg - w) + F_b = w + (-mg + F_b) = w + F_net \ \ \ ---(1)$ and we're told that now the balloon is now accelerating upward with the same magnitude of acceleration. Write $$-a = a_{initial}$$. Therefore $$a$$ is a positive quantity. Then we have $F_{net} = -ma \ \ \ ---(2)$ and $F_{net}' = ma \ \ \ --- (3)$ Now use the three equations--(1), (2) and (3)--to show the result you're looking for.

12. JamesJ

Equation one got chopped off. It is $F_{net}' = w + F_{net} \ \ \ -- (1)$

13. JamesJ

Actually, you'll need the more general form of equation (1) $F_{net}' = w - mg + F_b$ Anyway. Play around with those for a bit.

14. y2o2

I solved it.............Finally

15. Ishaan94

While going down, $W - F_{air} = \frac{W}{g}a$ After removal of mass w. $F_{air}-(W-w) = \frac{W-w}{g}a$ Solve these now, eliminate $$F_{air}$$.

16. Ishaan94

After removal of weight* w.

17. y2o2

18. arcticf0x

Got it! Thanks y202, JamesJ, Ishan94 for you time and help. Much appreciated!!!

19. arcticf0x

This was asked in the exam i am giving tomorrow, i hope they ask a similar one tomorrow :P