Equivalent spring constant of a system

theBEAST · Oct 13, 2014

Homework Statement

and the attempt at a solution[/B]
I tried to find the equivalent sprint constant using two points of equivalence. The first is with respect to theta. The second is with respect to the datum x as shown in the images below. The two methods yield two different answers which is not what I expected. Did I do something wrong?

Handwritten notes on a worksheet titled Example 5 showing a pulley-spring system with linear springs and a torsional spring, with the solution deriving k_eq = k_t1 + k_t2 + k_3 l_2^2 + k_1 l_1^2 + k_2 l_2^2 in terms of theta.

Handwritten derivation of an equivalent spring constant keq, equating energy terms (1/2 keq x^2 = 1/2 k1 x^2 + 1/2 k2 x^2 + 1/2 k3 [(l2/l1) x]^2 + 1/2 k+1 (x/l1)^2 + 1/2 k+2 (x/l1)^2), with the result keq = k1 + k2 + k3 (l2/l1)^2 + k+1 (1/l1)^2 + k+2 (1/l1)^2 boxed, and a sketch of a triangular lever with y/l2 = x/l1.

Greg Bernhardt · Oct 19, 2014

Thanks for the post! Sorry you aren't generating responses at the moment. Do you have any further information, come to any new conclusions or is it possible to reword the post?

Equivalent spring constant of a system

Homework Statement

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