# A mass is hung from the centre of an unstretched, horizontal wire. How do I work out the depression of the centre of the wire? [closed]

A wire of unstretched length $$l$$ is extended by a distance $$10^{-3}l$$ when a certain mass is hung from its bottom end. If this same wire is connected between two points that are a distance $$l$$ apart on the same horizontal level, and the same mass is hung from the midpoint, what is the depression $$y$$ of the midpoint?

I have been struggling with this question for too long, and I am totally stuck. My textbook states that the answer is $$l/20$$, but doesn't explain how or why. I would greatly appreciate any help you could give, and have described my own attempt below.

My attempt so far:

I have attempted to balance forces on the mass (balancing the vertical component of the tension in each string with the weight of the mass) and then to solve for tension, and equate tension with $$kx$$, with $$k$$ being given by $$1000mg/l$$ (as is easily deducible from the given information) and $$x$$ being given by $$2\sqrt{y^2+l^2/4}-l$$, as is easily obtainably by simply trigonometry. For tension I found, by geometry, that $$T=mg\sqrt{y^2+l^2/4}/2y$$. Equating all of this gives the below equation:

$$\frac{mg\sqrt{y^2+l^2/4}}{2y}=\frac{1000mg}{l}(2\sqrt{y^2+l^2/4}-l)$$

This expression is hideous and practically impossible to solve by hand, and I can't believe that this is the only way to do it.

• Maybe I'm missing something but it seems like a simple geometry problem and m and k are irrelevant. Trouble is when I do the trig functions, I get y=0.022 L and not the books y= 0.05 L. So I must be doing something wrong. – Bob D Mar 25 '19 at 19:26
• The solution to your equation is $y\approx l/20$ when you make the approximation $y\ll l$. – G. Smith Mar 25 '19 at 21:52
• The numerical solution is $y=0.050125\,l$ so the approximation is a good one. – G. Smith Mar 25 '19 at 22:05
• It seems like the textbook either made G. Smith's approximation or is erroneous. – Pancake_Senpai Mar 26 '19 at 11:13

My guess is that your textbook is wrong because I get the same result as you.

$$k=1000\frac{mg}{l}$$

With Newton:

$$\mathbf{T_1}+\mathbf{T_2}-\mathbf{mg}=0$$

Scalars:

$$2T\sin\theta =mg$$

$$\sin\theta=\frac{y}{H}$$

$$H=\sqrt{y^2+\frac{l^2}{4}}$$

$$T=\Big(H-\frac{l}{2}\Big)\times 1000\frac{mg}{l}=1000mg\Big(\sqrt{y^2+\frac{l^2}{4}}-\frac{l}{2}\Big)$$

$$2000\frac{1}{l}\Big(\sqrt{y^2+\frac{l^2}{4}}-\frac{l}{2}\Big)\times \frac{y}{\sqrt{y^2+\frac{l^2}{4}}}=1$$

• Your first vector equation needs subscripts to distinguish the two different $\mathbf{T}$ vectors. – G. Smith Mar 25 '19 at 22:08
• Their magnitudes are equal, but they have different directions so they aren't the same vector. – G. Smith Mar 25 '19 at 22:11