So I'm trying to get to the condition as shown in the picture below, where it says that if we have zero free length springs and k1r2=k2r1, then any configuration (pose) of the link is in equilibrium. So this means I need to construct the potential energy of this system and show that for the potential energy to be constant, I need to satisfy the condition below. So far no luck. Here the springs are assumed to have 0 natural length (as said in the text). Anyone could help me?



This is a typical VWM (Virtual Work Method) problem in Mechanics. But it seems there are some missing.

  1. Let's start with the energy equation:

d/dt(PE + KE) = Net Power

  1. There is no motion since it is in the static equilibrium. So, KE = 0.

  2. There is no external forces. So, Net power is also 0.

  3. PE = 0.5 * k_1 * s_1 - 0.5 * k_2 * s_2 + mgh

s_1: infinitesimal motion in the spring line 1 s_2: infinitesimal motion in the spring line 2

!Missing part 1: Either mass of the beam or plane should be clarified!

Let's suppose there is no gravity or the system is established in a horizontal plane.

  1. d/dt(PE) = 0

d/dt(PE) = k_1 * s_1 + k_2 * s_2 = 0

  1. So we need to relation with s_1, s_2 with r_1, r_2

At the end we convert your question into a geometry problem.

But it seems that there is no correlation between these parameters. Something an extra angle parameter is also missing.


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