# Rising balloons

47 views

There are N balloons full of helium separated by equal intervals attached to a weightless rope of length L. Each balloon has a rising force F.  a) The second figure shows force diagram of $i$ balloon. Prove horizontal component of $T_i$ (named $T_H$) is equal in all segments.

b)Deduce $F$ equation.

c) Prove $tg\theta_o = -tg\theta_{N+1} = \frac{NF}{2T_H}$

d) Based on $tg\theta_o = -tg\theta_{N+1} = \frac{NF}{2T_H}$ and both diagrams prove the following expressions:

$$tg\theta_i = \frac{(N-2i)F}{2T_H}$$

$$y_i = \frac{L}{N + 1}\sum_{j=0}^{i-1} cos\theta_j$$

$$y_i = \frac{L}{N + 1}\sum_{j=0}^{i-1} sin\theta_j$$

What I tried:

a) We know the balloons are in equilibrium:

$\sum F_y = 0$ and $\sum F_x = 0$

The exercise tells us that:

$$T_{i-1}cos\theta_{i-1} = T_o cos\theta_o = T_H$$

But I am not able to find a general trigonometric equation so as to prove it. Any clue?

b) Based on second diagram ($i$ balloon):

$$\sum F_y = 0$$

$$F = T_{i-1}sin\theta_{i-1} - T_i sin\theta_i$$

c) By extending segment 0 (red line pointing down with $\theta_o$ angle) and applying trigonometry:

$$tg\theta_o = \frac{NF}{2T_H}$$

But what I do not get is why:

$$tg\theta_o = -tg\theta_{N+1} = \frac{NF}{2T_H}$$

I know this issue is related to symmetry but I do not see it.

d) In order to proof the following expression:

$$tg\theta_i = \frac{(N-2i)F}{2T_H}$$

We have to assume:

$$tg\theta_{i-1} = tg\theta_o$$

Once we do it and apply $\frac{F}{T_H}$ we get it.

However my issue here is: We know horizontal component is the same in both cases. What I do not know is why the vertical component is $\frac{NF}{2}$ in balloon $i$ case. How could we proof $tg\theta_{i-1} = tg\theta_o$?

In order to prove the last two we have to start with defining the length between two balloons which I saw is:

$$l = \frac{L}{N +1}$$

The reason for N+1 is that whatever balloon we pick there is another to account for in order to have the segment? I am not sure why it is divided by N + 1.

Once here I do not how to carry on in order to get:

$$y_i = \frac{L}{N + 1}\sum_{j=0}^{i-1} cos\theta_j$$

$$y_i = \frac{L}{N + 1}\sum_{j=0}^{i-1} sin\theta_j$$

Thank you

asked Aug 18, 2018
edited Aug 18, 2018

(a) Each segment is in static equilibrium under the forces acting on it. The only forces with a horizontal component are the tensions in the segments of rope. So the horizontal tension forces pulling on the left and right sides of each segment are equal. For example $T_1\cos\theta_1=T_3\cos\theta_3$ and $T_2\cos\theta_2=T_4\cos\theta_4$ etc. This argument applies for all segments, so it means that the horizontal component of tension in each segment is the same : $T_i \cos\theta_i=T_H=\text{ a constant}$.

(c) $\tan \theta_{N+1}= -\tan \theta_0$ because $\theta_{N+1}= -\theta_0$.

This occurs because the slope of the rope segments changes from +ve to -ve as you move from left to right.

$\theta$ is +ve when measured anticlockwise from the $+x$ direction. After reaching the middle of the rope $\theta$ becomes -ve because the angle is then measured clockwise from the $+x$ direction.

(d) I do not see where you are asked to prove or assume that $\tan \theta_{i-1}=\tan \theta_0$. This is not correct. It implies that $\theta_i=\theta_0$ which is obviously wrong.

$$\sin \theta=\tan \theta \cos\theta$$ therefore $$F=T_0 \sin\theta_0 -T_1\sin\theta_1=T_0\cos\theta_0 \tan \theta_0 -T_1\cos\theta_1 \tan\theta_1=T_H(\tan \theta_0 -\tan \theta_1)$$ $$F=T_H(\tan \theta_1-\tan \theta_2)$$ $$F=T_H(\tan\theta_2-\tan\theta_3)$$ and so on. In general $$F=T_H (\tan \theta_{i-1} -\tan \theta_i )$$

Adding these i equations together we get $$iF=T_H( \tan_0-\tan \theta_i)$$ $$\tan \theta_i=\tan \theta_0 -\frac{iF}{T_H}=\frac{NF}{2T_H} -\frac{2iF}{2T_H}=\frac{(N-2i)F}{2T_H}$$

The equations for $x_i, y_i$ are derived from : $$y_i =l\sin \theta_0 +l\sin\theta_1 +l\sin\theta_2 +\text{...}+l \sin \theta_{i-1}$$ $$=l\sum_0^{i-1} \sin\theta_j$$ and similar for $x_i$.

answered Aug 18, 2018 by (27,556 points)
edited Mar 4
Sorry. Text editor is not working properly.
No problem at all, it can be understood. All my doubts were clarified except for one thing. Why length between two balloons is l = $\frac{L}{N +1}$?
There are $N$ balloons and $N+1$ segments of rope attached to them, because the rope ends are not attached to balloons. The total length of rope is $\text{ no. of rods x length of each rod }=(N+1)l$. But this is also equal to $L$. Therefore $L=(N+1)l$.