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