Calculating extensions in elastic string

Question

Could anyone please help me with the following question?

Here's my failed line of reasoning so far:

My diagram:

You can see that I am assuming that P moves down to lie vertically under it's original position, is even this correct?

Otherwise:

I've tried a few ways, here's the most naive, thanks for any help:

Let extension in AP = $x_1$ and extension in PB be $x_2$ then:

$L_1=\frac{3}{20\,cos\,\theta}$

and so

$x_1=\frac{3}{20\,cos\,\theta}-\frac{3}{20}$

and

$L_2=\frac{1}{20\,sin\,\theta}$

and so

$x_2=\frac{1}{20\,sin\,\theta}-\frac{1}{20}$

Hence:

$\frac{x_1}{x_2}=\frac{\frac{60-60\,cos\,\theta}{400\,cos\,\theta}}{\frac{20-20\,sin\,\theta}{400\,sin\,\theta}}=\frac{60-60\,cos\,\theta}{\cos\theta}\times\frac{sin\,\theta}{20-20\,sin\,\theta}$

which can be simplified to:

$\frac{(3-3\,cos\,\theta)\,sin\,\theta}{(1-sin\,\theta)\,cos\,\theta}$

Which isn't the required answer. Thanks for any help.

Answer 1

We have

$$ |AP|_0 = 15\\ |BP|_0 = 5\\ r = \frac{|AP|_0+|BP|_0}{2}\\ |AP| = 2r\cos\theta\\ |BP| = 2r\sin\theta $$

then

$$ \frac{|AP|-|AP|_0}{|BP|-|BP|_0} = \frac{20 \cos (\theta )-15}{20 \sin (\theta )-5} = \frac{4 \cos (\theta )-3}{4 \sin (\theta )-1} $$

the solution follows as in Calculation of the modulus of elasticity of a stretched string

NOTE

with $\lambda$ the elastic modulus,

$$ \lambda\left(\frac{|AP|-|AP|_0}{|AP|_0}\right)\cos\theta = \lambda\left(\frac{|BP|-|BP|_0}{|BP|_0}\right)\sin\theta $$

(In equilibrium the horizontal projection for $F_{AP}$ and $F_{PB}$ are equal) then

$$ \frac{|AP|-|AP|_0}{|BP|-|BP|_0}=\frac{\sin\theta}{\cos\theta}\frac{|AP|_0}{|BP|_0} $$

or as expected

$$ \frac{4 \cos \theta-3}{4 \sin\theta-1} = 3\left(\frac{\sin\theta}{\cos\theta}\right) $$

Answer 2

Well, here's my other attempt which does not assume that P moves down vertically:

I'll be using Hooke's law $T=\frac{\lambda x}{a}$

Let h = the height of the triangle in the diagram, i.e. the distance through which P falls.

$h=L_2\,cos\,\theta=L_1\,sin\,\theta$

and so

$L_1=\frac{L_2\,cos\,\theta}{sin\,\theta}$ - call this "Equation A"

$L_2=\frac{L_1\,sin\,\theta}{cos\,\theta}$ - call this "Equation B"

Resolving forces horizontally:

$T_1\,cos\,\theta-T_2\,sin\,\theta=0$ - Call this "Equation 1"

and Hooke's law gives us:

$T_1=\frac{\lambda(L_1-\frac{3}{20})}{\frac{3}{20}}$

and

$T_2=\frac{\lambda(L_2-\frac{1}{20})}{\frac{1}{20}}$

And so Eqaution 1 gives us:

$\frac{20}{3}\lambda(L_1-\frac{3}{20})\,cos\,\theta-20\lambda(L_2-\frac{1}{20})\,sin\,\theta=0$

Substituting fot $L_1$ from Equation A gives:

$\lambda(\frac{20}{3}\times\frac{L_2\,cos\,\theta}{sin\,\theta}-1)\,cos\,\theta=\lambda(20L_2-1)\,sin\,\theta$

Which rearranges to:

$L_2=\frac{3\,sin\theta(cos\,\theta-sin\,\theta)}{20\,cos^2\theta-60\,sin^2\theta}$

Similarly substituting for $L_2$ from Equation B leads to:

$L_1=\frac{3\,cos\theta(cos\,\theta-sin\,\theta)}{20\,cos^2\theta-60\,sin^2\theta}$

Now we are asked to find $\frac{L_1 - \frac{3}{20}}{L_2-\frac{1}{20}}$

Which is equal to:

$\frac{60\,cos\,\theta(cos\,\theta-sin\,\theta)-3(20\,cos^2\theta-60\,sin^2\theta)}{20(cos^2\theta-60\,sin^2\theta)}/\frac{60\,sin\theta(cos\,\theta-sin\,\theta)-(20\,cos^2\theta-60\,sin^2\theta)}{20(cos^2\theta-60\,sin^2\theta)}$

Which simplifies to:

$\frac{9\,sin^2\theta-3\,sin\,\theta\,cos\,\theta}{3\,sin\,\theta\,cos\,\theta-cos^2\theta}$

Which is not the required answer.