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Rising balloons

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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 in Physics Problems by JD_PM (490 points)
edited Aug 18 by JD_PM

1 Answer

2 votes
 
Best answer

(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 by sammy gerbil (26,166 points)
edited Sep 7 by Einstein
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 $L=(N+1)l$.
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