Calculation of the modulus of elasticity of a stretched string I posted concerning this question a little while ago, not asking for the answer but for an understanding of the setup. Well, I thought I would solve it but I seem to be unable to obtain the required answer. If anyone could help I'd be very grateful.
Here's the question again:

Here's my attempt:
We are told that for the weight attached to the unstretched string that AC = $\frac{4a}{3}$ and CB = $\frac{4a}{7}$ so that means that AB = $\frac{4a}{3}+\frac{4a}{7}=\frac{40a}{21}$ = natural length of string
Below is my diagram of the final situation:
My method will be by consideration of energy.
Once the string is stretched over the bowl, but before the weight falls to touch the inner surface, it will have elastic potential energy due to having been stretched. I'll calculate this.
Then I'll calculate the energy stored in the string due to both it's stretching over the diameter AND the weight having fallen.
The difference in these two energies will be equal to the loss in potential energy of the weight.
So, natural length = $\frac{40a}{21}$
Length once stretched over the diameter is $2a =\frac{42a}{21}$
and so extension due to this is $\frac{2a}{21}$ and so the elastic potential energy in the string due to only the stretching over the diameter is $\frac{\lambda(\frac{2}{21})^2a^2}{2*\frac{40}{21}a}=\frac{4}{441}\lambda a*\frac{21}{80}$
I'll leave it in this form for convenience later on.
Ok. from the diagram:

$x\,cos\,30+y\,cos\,60=2a$ therefore
$\frac{\sqrt{3}}{2}x+\frac{y}{2}=2a$ therefore
$\sqrt{3}x+y=4a$ call this equation 1
also
$h=x\,cos\,60=\frac{x}{2}$ and $h=y\,cos\,30=\frac{\sqrt{3}}{2}y$
and so $x=2h$ and $y=\frac{2h}{\sqrt{3}}$
And so, from equation 1, we have
$2\sqrt{3}h+\frac{2h}{\sqrt{3}}=4a$ therefore
h = $\frac{\sqrt{3}a}{2}$
and so
$x=\sqrt{3}a$ and $y=\frac{2}{\sqrt{3}}*\frac{\sqrt{3}}{2}a=a$
So the new length, due to stretching over diameter AND falling of weight = $a(1+\sqrt{3})$
So the extension is now:
$a(1+\sqrt{3})-\frac{40a}{21} = (\sqrt{3}-\frac{19}{21})a$
So energy now stored in string is:
$\frac{\lambda(\sqrt{3}-\frac{19}{21})^2a^2}{2\frac{40a}{21}}$ = $\frac{21}{80}\lambda(\sqrt{3}-\frac{19}{21})^2a$
So, the change in elastic potential energy in going from just stretched across the diameter to streched across the diameter AND having the weight fallen is:
$\frac{21}{80}\lambda a[(\sqrt{3}-\frac{19}{21})^2-\frac{4}{441}]$
and this can be equated to the loss in gravitational potential energy of the weight W giving:
$\frac{21}{80}\lambda a[(\sqrt{3}-\frac{19}{21})^2-\frac{4}{441}]=wh=\frac{\sqrt{3}}{2}wa$
So I have $\lambda$ in terms of w BUT I do not have $\lambda=w$
Is my physical reasoning incorrect?
If not, have I made a mathematical mistake(s)?
Thanks for any help,
Mitch.
 A: Assuming the Hooke's law
$$
F = \lambda\left(\frac{l-l_0}{l_0}\right)
$$
Calling
$$
|AC|_0 = a\frac 43\\
|CB|_0 = a\frac 47\\
|AC| = 2a\sin(\frac{\pi}{3})\\
|CB| = 2a\sin(\frac{\pi}{6})\\
F_{AC} = \lambda\left(\frac{|AC|-|AC|_0}{|AC|_0}\right)(-\cos(\frac{\pi}{6}),\sin(\frac{\pi}{6}))\\
F_{CB} = \lambda \left(\frac{|CB|-|CB|_0}{|CB|_0}\right)(\cos(\frac{\pi}{3}),\sin(\frac{\pi}{3}))\\
R = r(-\cos(\frac{\pi}{3}),\sin(\frac{\pi}{3}))\\
W = w(0,-1)
$$
we have in equilibrium
$$
F_{AC}+F_{CB}+R+W=0
$$
or
$$
\left\{
\begin{array}{c}
 -\frac{3 \sqrt{3} \left(\sqrt{3} a-\frac{4 a}{3}\right) \lambda }{8 a}+\frac{3 \lambda }{8}-\frac{r}{2}=0 \\
 \frac{3 \left(\sqrt{3} a-\frac{4 a}{3}\right) \lambda }{8 a}+\frac{3 \sqrt{3} \lambda }{8}+\frac{\sqrt{3} r}{2}-w=0 \\
\end{array}
\right.
$$
now solving for $\lambda, r$ we have
$$
\begin{array}{c}
 r=\left(\sqrt{3}-\frac{3}{2}\right) w \\
 \lambda =w \\
\end{array}
$$
NOTE
$R = $ normal surface reaction force
$\lambda = $ string elastic modulus.
$|\cdot| = $ stretched lenght
$|\cdot|_0 = $ unstrectched length.
A: I think the book is correct. Here's why.
The natural length of AC is $\frac{4}{3}a$ and in the final equilibrium position AC has been stretched to a length of $\sqrt{3}a$ and so the extension of AC is $(\sqrt{3}-\frac{4}{3})a$ and so the tension in AC is $T_1=\frac{\lambda(\sqrt{3}-\frac{4}{3})a}{\frac{4a}{3}}$=$\frac{3}{4}\lambda(\sqrt{3}-\frac{4}{3})$
Doing exactly the same for CB leads to the tension in CB is
$T_2=\frac{3}{4}\lambda$
Now, as we are in equilibrium we must have:
Let R = surface reaction force
and $\phi$=angle of reaction force to vertical
Resolving vertically:
$T_1\,cos\,60+T_2\,cos\,30+R\,cos\,\phi=w$
and Horizontally:
$T_1\,sin\,60-T_2\,sin\,30+R\,sin\,\phi=0$
And if we assume thhat $\lambda$=w, as we are assured it should, then this leads to the simultaneous equations:
$\frac{1}{2}\times\frac{3}{4}w(\sqrt{3}-\frac{4}{3})+\frac{\sqrt{3}}{2}\times\frac{3}{4}w+R\,cos\,\phi=w$
and
$\frac{\sqrt{3}}{2}\times\frac{3}{4}w(\sqrt{3}-\frac{4}{3})-\frac{1}{2}\times\frac{3}{4}w+R\,sin\,\phi=0$
And if we solve these we get:
$R=\frac{w}{2}(2\sqrt{3}-3)$
Which agrees with the book.
So it remains to show that $\lambda=w$ - if indeed the book is correct.
