m is the mass of the weight
k is the spring constant
$$ \Delta l $$ is the diffrence between the unextended diagonal of the square that got stretched to a rhombus and the long diagonal of the rombus
g is the gravitational accelaration
This is the notation I used.
If you could give insight on...
Thak you for the swift reply! As you corretly guessed I meant to write
$$ k (\Delta l)^2/2 = 2mg\Delta l\ \ \ $$ And from there to derive that T=4mg
I am sorry for the inconvinience!
At first I tried solving the problemteh following way:
Due to symmetry let the rods connected to the green rod have tension forces in magnitde T1 => mg = 2T1cos(a), where a is half the angle formed by the two rods. From tere I got an expression from the longer rods in the force projected by them...