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arcticf0x

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

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  1. arcticf0x
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    any luck y2o2?

    • 2 years ago
  2. y2o2
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    Well, i'm trying..but i really like it :D

    • 2 years ago
  3. y2o2
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    Don't worry , it'll be solved

    • 2 years ago
  4. arcticf0x
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    ok! waiting!

    • 2 years ago
  5. y2o2
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    are you sure there is nothing about the mass ?!

    • 2 years ago
  6. arcticf0x
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    yeah sure, mass isnt mentioned anywhere.

    • 2 years ago
  7. arcticf0x
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    mass

    • 2 years ago
  8. arcticf0x
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    how about we eliminate mass after forming 2 equations?

    • 2 years ago
  9. arcticf0x
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    nah, that gives W=2gw/a+g

    • 2 years ago
  10. JamesJ
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    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 \( a_{initial} < 0 \).

    • 2 years ago
  11. JamesJ
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    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.

    • 2 years ago
  12. JamesJ
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    Equation one got chopped off. It is \[ F_{net}' = w + F_{net} \ \ \ -- (1) \]

    • 2 years ago
  13. JamesJ
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    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.

    • 2 years ago
  14. y2o2
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    I solved it.............Finally

    • 2 years ago
  15. Ishaan94
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    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 \(F_{air}\).

    • 2 years ago
  16. Ishaan94
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    After removal of weight* w.

    • 2 years ago
  17. y2o2
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    • 2 years ago
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  18. arcticf0x
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    Got it! Thanks y202, JamesJ, Ishan94 for you time and help. Much appreciated!!!

    • 2 years ago
  19. arcticf0x
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    This was asked in the exam i am giving tomorrow, i hope they ask a similar one tomorrow :P

    • 2 years ago
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