Question

A balloon of weight $w$ newton descends with acceleration $fm/s^2$ the weight that must be thrown out in order to give equal upward acceleration to the balloon will be

Answer 1

Step 1: Given Data

Weight of the balloon $=w$

Acceleration $=fm/s^2$

Let the new weight of the balloon be $w-w\text{'}$.

Tension $T$ will be acting upwards in both balloons.

Let the acceleration due to gravity be $g$.

Let $m$ be the mass of the balloon.

Step 2: EOM for First Case

In the first case, the mass of the balloon can be given as

$m=\frac{w}{g}$

Since the balloon is going down, therefore, the equation of motion can be written as,

$w-T=\frac{w}{g}f$ $\left(1\right)$

Step 3: EOM for Second Case

In the second case, the mass of the balloon can be given as

$m=\frac{w-w\text{'}}{g}$

Since the balloon is going up, therefore, the equation of motion can be written as,

$T-\left(w-w\text{'}\right)=\frac{w-w\text{'}}{g}f$ $\left(2\right)$

Step 4: Calculate the Required Weight

Upon adding equation $\left(1\right)$ and $\left(2\right)$ we get,

$w-T+T-w+w\text{'}=\frac{w}{g}f+\frac{w-w\text{'}}{g}f$

$\Rightarrow w\text{'}=\frac{w}{g}f+\frac{w}{g}f-\frac{w\text{'}}{g}f$

$\Rightarrow w\text{'}+\frac{w\text{'}}{g}f=\frac{2w}{g}f$

$\Rightarrow w\text{'}\left(1+\frac{f}{g}\right)=\frac{2w}{g}f$

$\Rightarrow w\text{'}\left(\frac{g+f}{g}\right)=\frac{2w}{g}f$

$\Rightarrow w\text{'}=\frac{2wf}{g+f}$

Hence, if we throw out the weight of $\frac{2wf}{g+f}$the balloon will move upward with an equal acceleration.