Question

A beam, the weight of which may be neglected, is supported by three identical springs. When a weight W is hung from the middle of the beam, the extension of each spring is x.

Click to see diagram: http://i164.photobucket.com/albums/u32/Codefusionlab/Springs.jpg

The middle spring and the weight are removed.
What is the extension when a weight of 2W is hung from the middle of the beam?

Answer 1

Each spring exerts a vertical force proportional to their extension on the beam, given by
F = k·x
Since the extension of all spring is equal the forces to.
Set up vertical force balance and find
F = W/3
=>
k = W/(3x)

In the new arrangement you have the double weight suspended by two springs at extension x' :
2F' = W' = 2W
<=>
k·x' = W
=>
x' = W/k = 3x