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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.

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  • $\begingroup$ 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
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    $\begingroup$ 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

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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.

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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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