## arcticf0x Group Title 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. 2 years ago 2 years ago

1. arcticf0x Group Title

any luck y2o2?

2. y2o2 Group Title

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

3. y2o2 Group Title

Don't worry , it'll be solved

4. arcticf0x Group Title

ok! waiting!

5. y2o2 Group Title

are you sure there is nothing about the mass ?!

6. arcticf0x Group Title

yeah sure, mass isnt mentioned anywhere.

7. arcticf0x Group Title

mass

8. arcticf0x Group Title

how about we eliminate mass after forming 2 equations?

9. arcticf0x Group Title

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

10. JamesJ Group Title

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 Group Title

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 Group Title

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

13. JamesJ Group Title

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 Group Title

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

15. Ishaan94 Group Title

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 Group Title

After removal of weight* w.

17. y2o2 Group Title

18. arcticf0x Group Title

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

19. arcticf0x Group Title

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