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Thanks for the help!

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

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...