Mass on a stretch elastic string setup

Question

Could anyone dispel my ignorance please.

I must be missing something obvious.

I'm not looking for the answer here, I just can't visualize the setup.

It says that W is attached at point C of an unstretched string where $AC = \frac{4a}{3}$

But the unstretched string is of length a (its natural length) so how can one attach W at four thirds of a?

Thanks, Mitch.

Answer 1

$a$ is not the natural length; $a$ is the radius of your hemispherical bowl, so that $2a$ is the extended length. The point $C$ is $7/10$ of the way from $A$ to $B$, where the total distance from $A$ to $B$ is \begin{align} d &= \frac{4a}{3} + \frac{4a}{7}\\ & = 4a \left(\frac{1}{3} + \frac{1}{7} \right)\\ & = 4a \left(\frac{7}{21} + \frac{3}{21} \right)\\ & = 4a \left(\frac{10}{21}\right). \end{align}