2017 blocks

Treat the whole system as one big object. There is a vertical force of $2017 mg$ downward, and a horizontal force of $mg$ rightward.

The first string must supply $x$ and $y$ forces in this ratio to maintain equilibrium. Therefore, $\theta_1 = atan \left(\frac{x force}{y force}\right) = atan\left(\frac{1}{2017}\right) \approx 0.028$ degrees.

There is no need to evaluate $mg = (20.17)(9.80)$ Newtons. It eventually cancels out in the end. So we will just refer to it as $mg$ to avoid tedious work.

First let us consider the string connecting blocks $2017$ and $2016$ . Draw F.B.D 's and solve for tension in the string. We get, ( remember sin theta / cos theta = tan theta.)

T = $\sqrt 2$ mg. But T cos theta = T sin theta = mg.Thus $\theta = \tan^{-1}(1) = 45$ degrees. (between $0$ and $90$ ).

Now consider the string connecting masses $2016$ and $2015$ . Again solve for tension in the string.

T = $\sqrt 5$ mg. But T cos theta = 2 mg and T sin theta = mg. Thus $\theta$ = $\tan^{-1} (1/2)$ .

Let us consider a final case blocks $2014$ and $2015$ .

T = $\sqrt {17}$ mg. But T cos theta = 3 mg and T sin theta = mg. Thus $\theta = \tan^{-1} (1/3)$ .

Alright, you should have already seen this coming up.

When there is $1$ block below the string, angle = $\tan^{-1}(1)$ .

When there are $2$ blocks below the string, angle = $\tan^{-1}(1/2)$

When there are $3$ blocks below the string, angle = $\tan^{-1}(1/3)$ .

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When there are $2017$ blocks below the string, angle made by the string with the vertical = $\tan^{-1}( 1/2017)$ .

Amazing, isn't it? But that 's the sort of approach you might use if in a hurry. I will generalise it.

Okay, so again from our observations, we see that,

T sin theta remains constant = mg always.

But,

T cos theta = k mg (where k = integers 1, 2, 3 ...etc.)

This makes sense because as we go up a mass $mg$ keeps on getting added.

So, one component of the Tension remains constant whereas the other keeps on increasing.

So, for the $i^{th}$ block $T \sin \theta / T \cos \theta = mg / i mg$

Thus, $\tan \theta = 1 / i$ .

For block 1, $i = 2017$ ( because we are going backwards.).

So $\theta_1 = \tan^{-1} (1/2017) = 0.028$ degrees.

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Question was thought out by me.Hope you liked it.I tried to make it like those asked in Mathematical Olympiads.Just added a touch of Physics. .

Looking at the last block, we can call the tension in the last string $F_{T(2017)}$ set up the equations:

$F_{T(2017)} \sin \theta_{2017} \ - \ mg \ = \ 0 \ \Rightarrow \ F_{T(2017)} \sin \theta_{2017} \ = \ mg$ $F_{T(2017)} \cos \theta_{2017} \ - \ mg \ = \ 0 \ \Rightarrow \ F_{T(2017)} \cos \theta_{2017} \ = \ mg$

Now, for each successive block, in order for this system to be in equilibrium, we will have:

$F_{T(n+1)} \sin \theta_{n+1} \ - \ F_{T(n)} \sin \theta_{n} \ = \ 0$

We put a negative sign on the rope attached to the bottom of the block, as the force vector will be directed in the opposite direction as the force that it exerts on the block below it. Going back to $F_{T(2017)}$ , we can see the magnitude of the horizontal component of the tension within any of the strings will be $mg$ . Next, we have to realize that each time we successively travel one block farther up the string, we have:

$F_{T(n+1)} \cos \theta_{n+1} \ - \ F_{T(n)} \cos \theta_{n} \ - \ mg \ = \ 0 \ \Rightarrow \ F_{T(n+1)} \cos \theta_{n+1} \ = \ F_{T(n)} \cos \theta_{n} \ + \ mg$

As we have to account for the tension pulling the block upwards and downwards, as well as gravity pulling the block downwards. This is obviously a recursive formula from which we can draw a general form. Since $F_{T(2017)} \cos \theta_{2017} \ = \ mg$ , we get:

$|F_{T(n)}| \cos \theta_{n} \ = \ (2018 \ - \ n)mg$

So for the topmost string, we have $mg$ as the horizontal component of tension, and $2017mg$ as the vertical. We then take:

$\dfrac{F_{T(1)} \sin \theta_{1}}{F_{T(1)} \cos \theta_{1}} \ = \ \dfrac{1}{2017} \ \Rightarrow \ \tan \theta_{1} \ = \ \dfrac{1}{2017} \ \Rightarrow \ \theta_{1} \ = \ \arctan \Big( \dfrac{1}{2017} \Big) \ \approx \ 0.0284$