Stasis with Unequal Rods

For me, this was not an easy one to solve. In fact, I think there is a lot more to this problem than meets the eye. I will first present the solution that worked for me.

Notice how the spring and the two rods form a triangle. More specifically, an isosceles triangle. This is because of the rods having an equal length. From here, two equations can be formed by calculating the torque about the origin.

The angle between either rod and the spring is:

$A = \frac{\theta_1 + \theta_2}{2}$

The above result follows from elementary geometry. Now the torque experienced by rod 1 about the origin is:

$T_1 = \frac{m_1 g L_1}{2}\cos{\theta_1} - F_s L_1 \sin{A}$

The torque experienced by rod 2 about the origin is:

$T_2 = \frac{m_2 g L_2}{2}\cos{\theta_2} - F_s L_2 \sin{A}$

Here,

$F_s = K\left(\sqrt{\left(\cos{\theta_1}+\cos{\theta_2}\right)^2 + \left(\sin{\theta_2}-\sin{\theta_1}\right)^2} - 1\right)$

At equilibrium, the net torques experienced by each rod must be zero. The angles at which this occur can be found by writing a simple code which sweeps through the angles within the specified range. I have not attached the code here now. From there, the answer comes out to be:

$\theta_1 = 66.0568^o$

Hence the final answer is $\boxed{6605}$ .

Now, arriving at this answer took me more than one try. Whenever I encounter an equilibrium problem which involves no dissipative forces, my first instinct is to minimise the potential energy of the system.

The total potential energy of the system is the sum of gravitational potential energies of the rods and the energy stored in the spring.

$V = \frac{m_1 g L_1}{2} \sin{\theta_1} + \frac{m_2 g L_2}{2} \sin{\theta_2} + \frac{1}{2} K \left(\sqrt{\left(\cos{\theta_1}+\cos{\theta_2}\right)^2 + \left(\sin{\theta_2}-\sin{\theta_1}\right)^2} - 1\right)^2$

Computing its gradient and equating it to zero (steps omitted here) would lead to a couple of nonlinear equations which would need to be solved numerically. I chose the Newton-Raphson method. The answer depends on the initial assumption. I was getting many weird equilibrium configurations but after some tries, found the required answer. Usually, the solution should be found with ease but that was not the case here. It involved quite a bit of work for me. And soon enough I realised why. The following 3D plot shows the variation of PE with the angles.

Given the odd shape of this surface, one can see that the required equilibrium point is definitely not stable. The odd shape also explains why the Newton Raphson iteration is so sensitive to the starting point.

I used energy approach mentioned by @Karan Chatrath which is very convenient in this case.

Physics says that equilibrium will occur once the potential energy of the system is minimal subject to the constraints. The rest of the problem is then of a mathematical kind. Here, it's obvious that there are two degrees of freedom, two independent coordinates - namely two angles. If we use $l$ to denote the length of the spring, by cosine theorem we have: $\begin{aligned}l^{2} &= L_{1}^{2} + L_{2}^{2} - 2L_{1}L_{2}\cos(\pi - \theta_{1} - \theta_{2}) \\ &= L_{1}^{2} + L_{2}^{2} + 2L_{1}L_{2}\cos(\theta_{1} + \theta_{2}) \\ 2l\frac{\partial l}{\partial \theta_{i}} &= -2L_{1}L_{2}\sin(\theta_{1} + \theta_{2}),\end{aligned}$ for $i = 0,1.$ Expression for potential energy is: $U = \frac{m_{1}gL_{1}\sin{\theta_{1}}}{2} + \frac{m_{2}gL_{2}\sin{\theta_{2}}}{2} + \frac{1}{2}k(l-L_{0})^{2}$ Equations are obtained by finding partial derivatives with respect to each coordinate and setting them to zero: $\begin{aligned} \frac{\partial U}{\partial \theta_{i}} &= \frac{m_{i}gL_{i}\cos{\theta_{i}}}{2} + k(l-L_{0})\frac{\partial l}{\partial \theta_{i}} \\ &= \frac{m_{i}gL_{i}\cos{\theta_{i}}}{2} - kL_{1}L_{2}\sin(\theta_{1} + \theta_{2})\left(1-\frac{L_{0}}{\sqrt{L_{1}^{2} + L_{2}^{2} + 2L_{1}L_{2}\cos(\theta_{1} + \theta_{2})}} \right ) = 0\end{aligned}$ I used GeoGebra to plot these equations in $\theta_{1}\theta_{2}$ -plane and find the intersection which satisfies given condition.

I used a hill-climbing algorithm which mutates the unknown angle parameters $\theta_1$ and $\theta_2$ until the net torque on each rod is zero. I have found the following solutions:

Solution 1
$\theta_1 \approx 66.054^\circ \\ \theta_2 \approx 35.733^\circ$

Solution 2
$\theta_1 \approx 89.587^\circ \\ \theta_2 \approx -89.175^\circ$

In the second solution, the two rods are nearly co-linear, almost lining up with the $y$ axis. Also, the second and third decimal points in solution 2 are not as consistent as they are for solution 1.

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import math
import random

# Constants

g = 10.0

k = 20.0
L0 = 1.0

m1 = 2.0
L1 = 1.0

m2 = 1.0
L2 = 1.0

#################################

delta = 0.001

minres = 99999999.0

theta1 = (math.pi/2.0)*random.random()
theta2 = math.pi - (math.pi/2.0)*random.random()

for j in range(0,4*10**6):

    if j > 2*10**6:
        delta = 0.0001

    theta1m = theta1 + delta * (-1.0 + 2.0 * random.random())  # mutated angles
    theta2m = theta2 + delta * (-1.0 + 2.0 * random.random())

    x1 = L1*math.cos(theta1m)  # rod end points
    y1 = L1*math.sin(theta1m)

    x2 = L2*math.cos(theta2m)
    y2 = L2*math.sin(theta2m)

    Dx = x2 - x1  # vector between rod end points
    Dy = y2 - y1

    D = math.hypot(Dx,Dy)

    ux = Dx/D   # unit vector between rod end points
    uy = Dy/D

    s = D - L0  # stretch
    Fs = k*s   # spring force magnitude

    Fs1x = Fs*ux  # spring force on rod 1
    Fs1y = Fs*uy

    Fs2x = -Fs1x  # spring force on rod 2
    Fs2y = -Fs1y

    r1x = x1  # lever arm for spring force on rod 1
    r1y = y1

    r2x = x2   # lever arm for spring force on rod 2
    r2y = y2

    Fg1x = 0.0     # gravity forces
    Fg1y = -m1*g

    Fg2x = 0.0
    Fg2y = -m2*g

    T1s = r1x*Fs1y - r1y*Fs1x   # spring torques
    T2s = r2x*Fs2y - r2y*Fs2x

    T1g = (r1x/2.0)*Fg1y - (r1y/2.0)*Fg1x  # gravity torques
    T2g = (r2x/2.0)*Fg2y - (r2y/2.0)*Fg2x

    T1 = T1s + T1g  # net torques
    T2 = T2s + T2g

    res = math.fabs(T1) + math.fabs(T2)   # residual is sum of torque magnitudes

    if res < minres:  # store mutated angles if residual becomes smaller

        minres = res

        theta1 = theta1m
        theta2 = theta2m

#################################

print minres
print ""

theta1_deg = 180.0*theta1/math.pi
theta2_deg = 180.0*theta2/math.pi

print theta1_deg
print math.floor(100.0*theta1_deg)
print ""
print theta2_deg
print (180.0 - theta2_deg)