# Two different answers for an elastic strings question

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.

• 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? – Tom Ryan Jun 19 at 18:05
• I'm adding the tag physics – user10354138 Jun 19 at 18:08
• 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 – Shubham Johri Jun 19 at 18:10
• Are you sure it is $\angle APB=\theta$ not $\angle PAB=\theta$? You already have $\angle APB=90^\circ$ in the previous sentence. – user10354138 Jun 19 at 18:18
• @user10354138 oh yeah sorry I’ll fix that – Tom Ryan Jun 19 at 18:19

Both are correct, and the equality between the two identities will allow you to find $$\theta$$ and the other geometric parameters.
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 $$3\tan \theta =\frac{\cos\theta -3/4}{\sin\theta -1/4}.$$