Static balancing of link mechanism

Question

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?

Answer 1

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

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

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

4. 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.

5. d/dt(PE) = 0

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

6. 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.