1. ## Mechanics Problem

Question:

A Balloon of mass 490Kg is descending at a speed of 5 m/s with an acceleration of 0.2m/s^2. What mass must be thrown out to reduce the acell to zero.

I keep on getting an incorrect answer, but I'm not sure where I'm going wrong.

Here is how I am (trying) to work out the solution.

T.F = ma
mg - Upwards force = ma
- upwards force = ma - mg
Upwards force = (ma - mg) * -1

Then:

T.F = ma
mg - upwards force = ma = m*0 = 0
Mg = Upwards force
m = upwards force/g

Original M - new value for M = delta m = amount the mass must be reduced.

Can anyone enlighten me on where I'm going wrong please, perhaps the upwards force doesn't remain constant, or I'm missing a force in my modelling.

Any help is very much appreciated!!

2. Originally Posted by Ant
Question:

A Balloon of mass 490Kg is descending at a speed of 5 m/s with an acceleration of 0.2m/s^2. What mass must be thrown out to reduce the acell to zero.

I keep on getting an incorrect answer, but I'm not sure where I'm going wrong.

Here is how I am (trying) to work out the solution.

T.F = ma
mg - Upwards force = ma
- upwards force = ma - mg
Upwards force = (ma - mg) * -1

Then:

T.F = ma
mg - upwards force = ma = m*0 = 0
Mg = Upwards force
m = upwards force/g

Original M - new value for M = delta m = amount the mass must be reduced.

Can anyone enlighten me on where I'm going wrong please, perhaps the upwards force doesn't remain constant, or I'm missing a force in my modelling.

Any help is very much appreciated!!
Call the upward force B, for buoyant force. And call +y downward, as you have already done.

Then as the balloon is falling
$\sum F_y = mg - B = ma$

$B = mg - ma$

Now, throw out a mass $\Delta m$ and do it again. (The buoyant force will be constant as the volume of the balloon has not changed.)
$\sum F_y = (m - \Delta m)g - B = 0$

$mg - \Delta m g - (mg - ma) = 0$
(Where a is the acceleration from before.)

$\Delta m = \frac{ma}{g}$

-Dan

so, pluggin numbers in gives:

Change in Mass = (mg) / a
= (490 * 9.81) / 0.2
= 24010 Kg

isn't this impossible - going from 490kg to minus 23530 Kg?! The answer in my book says there must be a reduction in mass of 10kg. Perhaps there's an error in my book?

4. Originally Posted by Ant

so, pluggin numbers in gives:

Change in Mass = (mg) / a
= (490 * 9.81) / 0.2
= 24010 Kg

isn't this impossible - going from 490kg to minus 23530 Kg?! The answer in my book says there must be a reduction in mass of 10kg. Perhaps there's an error in my book?
Originally Posted by topsquark
$mg - \Delta m g - (mg - ma) = 0$
(Where a is the acceleration from before.)

$\Delta m = \frac{mg}{a}$
Tsk tsk! We both should have caught this. The solution to the equation above
$mg - \Delta m g - (mg - ma) = 0$
is
$\Delta m = \frac{ma}{g}$