Can anyone please help with this question?
A light elastic string AB has natural length l and modulus of elasticity 2mg. Another light elastic string CD has natural length l and modulus of elasticity 4mg. The strings are joined at their ends B and C and the end A is attached to a fixed point. A particle of mass m is hung from the end D and is at rest in equilibrium. Find the length AD.
this is what I did
Let Tension AB is T2 and tension CD be T1. since system is at rest
T2-T1 = mg (Equation 1)
AD = l +x +y +l (Where x and y are the extensions of AB and CD respectively)
So, AD = 2l +x+y
Using Hookes law
T2 = 2mgx/l
Similarly, T1 = 4mgy/l
substituting the above into equation 1
This is as far as I can go. I wasn't given the distance between the the fixed point and AD. Am I missing something? Kindly comment.