Tension force of a thread in a complex structure of six masless rods

[Kino Physics]

Homework Statement

On a weightless structure of articulated rods, a weight with mass 𝑚 is suspended. What is the tension force T of the red thread? The six rods form two identical diamonds.

Relevant Equations

Newton's Second Law; Law of Conservation of energy ( potentially)

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 is T2 in magnitude, which is equal to T1 due to the balance of forces in the x direction in the joint of the short and long rods , such that T1sin(a)=T2sin(a) and from there I get that T =mg.
However if I assume that the thread is extensible and solve the balance of forces with the thread acing as a spring by integration the work of the gravitational force I get that k(dl)^2/2= mg(2dl).
Thanks in advance!

 

[BvU]

Hello @Kino Physics , !

T =mg

is the wrong answer ...

You may assume the red thread is inextensible. And if you don't want that, then still the answer should be inependent of $k$.

k(dl)^2/2= mg(2dl)

$$ k (\Delta l)^2/2 = 2mg\Delta l\ \ \ ?$$is hard to read and unfinished: there is a relationship between $\Delta l$ and mg

 

[Kino Physics]

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!

 

[BvU]

No reason to apologise !
But you could do me a favour and explain the symbols ...

 

[Kino Physics]

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 the methods used to derive a solution involving an inextensive thread that would be very helpful! Thanks in advance!

 

[BvU]

My question was a disguised way to let you look again at $\Delta l$ for the spring and $\Delta l$ for the mass ... are they really equal ?

Goole pantograph

 

[Kino Physics]

For the mass the change in height is equal to 2$\Delta l$ . :)

 

[BvU]

Well done !

 

[Kino Physics]

Thanks for the help!

 

[BvU]

You are welcome. It's fun to help ..