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 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'.
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w=2aW/(a+g)
g is acceleration due to gravity.
 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.

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y2o2
 3 years ago
Best ResponseYou've already chosen the best response.0Well, i'm trying..but i really like it :D

y2o2
 3 years ago
Best ResponseYou've already chosen the best response.0Don't worry , it'll be solved

y2o2
 3 years ago
Best ResponseYou've already chosen the best response.0are you sure there is nothing about the mass ?!

arcticf0x
 3 years ago
Best ResponseYou've already chosen the best response.0yeah sure, mass isnt mentioned anywhere.

arcticf0x
 3 years ago
Best ResponseYou've already chosen the best response.0how about we eliminate mass after forming 2 equations?

arcticf0x
 3 years ago
Best ResponseYou've already chosen the best response.0nah, that gives W=2gw/a+g

JamesJ
 3 years ago
Best ResponseYou've already chosen the best response.1The 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 \( a_{initial} < 0 \).

JamesJ
 3 years ago
Best ResponseYou've already chosen the best response.1The 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.

JamesJ
 3 years ago
Best ResponseYou've already chosen the best response.1Equation one got chopped off. It is \[ F_{net}' = w + F_{net} \ \ \  (1) \]

JamesJ
 3 years ago
Best ResponseYou've already chosen the best response.1Actually, you'll need the more general form of equation (1) \[ F_{net}' = w  mg + F_b \] Anyway. Play around with those for a bit.

y2o2
 3 years ago
Best ResponseYou've already chosen the best response.0I solved it.............Finally

Ishaan94
 3 years ago
Best ResponseYou've already chosen the best response.2While going down, \[W  F_{air} = \frac{W}{g}a\] After removal of mass w. \[F_{air}(Ww) = \frac{Ww}{g}a\] Solve these now, eliminate \(F_{air}\).

Ishaan94
 3 years ago
Best ResponseYou've already chosen the best response.2After removal of weight* w.

arcticf0x
 3 years ago
Best ResponseYou've already chosen the best response.0Got it! Thanks y202, JamesJ, Ishan94 for you time and help. Much appreciated!!!

arcticf0x
 3 years ago
Best ResponseYou've already chosen the best response.0This was asked in the exam i am giving tomorrow, i hope they ask a similar one tomorrow :P
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