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 \(l\).

nil

Analysis

First, we need do force analysis on \(M\).

nil

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)