# Intriguing physics compilation - Part 1

In this series of compilations of physics problems. I will simply put on some intriguing problems. This one is on mechanics.

## Two ropes with one mass point

### Question

Suppose there is one mass point $$M$$ whose mass is $$m$$. Two ropes which are both in length $$l$$ hook $$M$$ up with the ceiling. Each rope can support substance(s) whose mass up to $$T$$. The distance between two hook points which are used by two ropes is $$s$$. In order to hook the $$M$$ up, what is the minimum of $$l$$. ### Analysis

First, we need do force analysis on $$M$$. Obviously, we have:
\tag{1} \begin{cases}\begin{aligned} T_{1} &= \frac{F'}{2}+F_{3}\\ T_{2} &= \frac{F'}{2}+F_{4}\\ \left\vert\frac{F'}{2}\right\vert &= \frac{m|g|}{2} \end{aligned}\end{cases}

Because $$M$$ is at rest, it is easy to infer $$|F_{3}|=|F_{4}|$$ and thus we can even deduce $$|T_{1}|=|T_{2}|$$.
From the other side, $$|T_{1}|$$ and $$|T_{2}|$$ can be notated in another form: $|T_{1}|=|T_{2}|=\frac{m|g|}{2\cos{\frac{\theta}{2}}}\ ,\theta{}\in{}[0,\pi]$ As the given information from question, $$|T_{1}|\leq{T}$$ and $$|T_{2}|\leq{T}$$ should be fulfilled. And $$\cos{\frac{\theta}{2}}$$ can also be expressed as $$\frac{\sqrt{l^{2}-\frac{s^2}{4}}}{l}$$.
Thereby, we have: \tag{2} \begin{cases}\begin{aligned} \frac{m|g|}{2\cos{\frac{\theta}{2}}} &\leq |T|\\ \cos{\frac{\theta}{2}} &= \frac{\sqrt{l^{2}-\frac{s^2}{4}}}{l} \end{aligned}\end{cases} So let's check out the properties of (2).
\tag{3} \begin{aligned} &\because{} \cos{\frac{\theta}{2}} \propto{} \frac{1}{\theta{}} \\ Also &\because{} \frac{m|g|}{2\cos{\frac{\theta}{2}}} \propto{} \frac{1}{\cos{\frac{\theta}{2}}} \\ &\therefore{} \frac{m|g|}{2\cos{\frac{\theta}{2}}} \propto{} \theta \\ &\because{} \cos{\frac{\theta}{2}} = \frac{\sqrt{l^{2}-\frac{s^2}{4}}}{l} &\propto{} l \\ &\therefore |T| = \frac{m|g|}{2\cos{\frac{\theta}{2}}} \propto{} \frac{1}{l} \end{aligned}

Now, solve the equation set (2): \tag{4} \begin{aligned} \frac{m|g|}{\frac{2\sqrt{l^{2}-\frac{s^2}{4}}}{l}} &\leq |T| \\ m|g| &\leq \frac{2|T|\sqrt{l^{2}-\frac{s^2}{4}}}{l} \\ lm|g| &\leq 2|T|\sqrt{l^{2}-\frac{s^2}{4}} \\ (lm|g|)^{2} &\leq 4|T|^{2}(l^{2}-\frac{s^2}{4}) \\ (s|T|)^{2} &\leq (4|T|^{2}-m^{2}|g|^{2})l^{2} \\ l &\geq \frac{s|T|}{\sqrt{4|T|^{2}-m^{2}|g|^{2}}} \end{aligned} $\therefore{}l_{min} = \frac{s|T|}{\sqrt{4|T|^{2}-m^{2}|g|^{2}}}$

### Note

Many answers to this question fails to bear out the reason why $$|T|\propto{}\frac{1}{l}$$ in a formal way like (3), which is very crucial for one to comprehend this particular question thoroughly.

(To be continued)