I came across this question in a textbook and after a full day of going over it and consulting friends none of us can figure out why a particular approach to this question doesn't yield the correct answer.

The question

A light elastic string of natural length $0.2m$ has its ends attached to two fixed points $A$ and $B$ which are on the same horizontal level with $AB = 0.2m$. A particle of mass $5$kg is attached to the string at the point $P$ where $AP = 0.15m$. The system is released and P hangs in equilibrium below AB with $\angle{APB} = 90^{\circ}$. If $\angle{BAP} = {\theta}$, show that the ratio of the extension of $AP$ and $BP$ is $$\frac{4cos{\theta}-3}{4sin{\theta}-1}$$

The correct method

Let the extension of $AP$ be $x_1$ and the extension of $BP$ be $x_2$

$x_1 = 0.2cos{\theta} - 0.15$

$x_2 = 0.2sin{\theta} - 0.05$

$\frac{x_1}{x_2} = \frac{0.2cos{\theta} - 0.15}{0.2sin{\theta} - 0.05}$

$\therefore \frac{4cos{\theta}-3}{4sin{\theta}-1}$

Our method (seemingly incorrect)

With the same symbols for $AP$ and $BP$ and the tensions in $AP$ and $BP$ being $T_1$ and $T_2$ respectively.

Since the system is in equilibrium resolving horizontally gives:

$T_1cos{\theta} = T_2sin{\theta}$

$\therefore \frac{T_1}{T_2} = tan{\theta}$

Since for an elastic string $T = \frac{{\lambda}x}{l}$

$T_1 = \frac{{\lambda}x_1}{0.15}$

$T_2 = \frac{{\lambda}x_2}{0.05}$

$\therefore \frac{x_1}{x_2} = 3\frac{T_1}{T_2}$

$\therefore \frac{x_1}{x_2} = 3tan{\theta}$

The discrepancy here is not that I do not understand the first solution, but why the second method does not yield a correct result, the steps followed seem logical and correct to me.

  • $\begingroup$ I noticed that the classical-mechanics tag specifies it shouldn't be the only tag on a question, what other tags would be relevant here? $\endgroup$
    – Tom Ryan
    Jun 19, 2019 at 18:05
  • 1
    $\begingroup$ I'm adding the tag physics $\endgroup$ Jun 19, 2019 at 18:08
  • $\begingroup$ If I understand correctly, you used the relationship between stress, strain and Young's modulus. $T$ is not the stress -- $T/A$ is, and there is no reason to assume the two sections of the rope have the same cross-section area after stretching $\endgroup$ Jun 19, 2019 at 18:10
  • $\begingroup$ Are you sure it is $\angle APB=\theta$ not $\angle PAB=\theta$? You already have $\angle APB=90^\circ$ in the previous sentence. $\endgroup$ Jun 19, 2019 at 18:18
  • $\begingroup$ @user10354138 oh yeah sorry I’ll fix that $\endgroup$
    – Tom Ryan
    Jun 19, 2019 at 18:19

2 Answers 2


Both are correct, and the equality between the two identities will allow you to find $\theta$ and the other geometric parameters.
Note in fact that P will not move along the vertical from the original position, but will move somewhat towards B, since PB is more rigid than PA.


Your assumption that your second solution is incorrect is unwarranted. The "second solution" is an additional correct constraint as well, and serves to pin down $\theta$, which is, as a result, not arbitrary: As you increase the suspended weight from 0 to the critical value youo are considering, the angle $\angle{APB}$ decreases monotonically from $\pi$ to finally $\pi/2$.

Correspondingly, $\theta$ increases from 0 to crudely something like $\pi/7.9$, solving $$ 3\tan \theta =\frac{\cos\theta -3/4}{\sin\theta -1/4}. $$

There might be a pithy construction for all this, but I suspect you just wanted to discern you fallacy.


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