Mass Hanging from Two Springs

The approach of my choice is the minimization of the potential energy of the mass.

Let the general coordinates of the mass be $(x,y$ ).

The potential energy of the mass is then:

$V = mgy + \frac{100}{2}\left(\sqrt{x^2 + y^2} - L\right)^2 + \frac{50}{2}\left(\sqrt{(6-x)^2 + y^2} - L\right)^2$

At stable equilibrium, the potential energy is globally minimized. The point at which this takes place is then used to compute the required answer. One can solve this by computing partial derivatives and using Newton-Raphson, however, I chose a brute force approach of parameter space sweeping.

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clear all
clc

% Given parameters:
L   = norm([6;0;0]-[3;-3.5;0]);
g   = 9.81;
M   = 30;
K1  = 100;
K2  = K1/2;

% Spatial resoloution:
dx  = 1e-2;
dy  = dx;

% Loop to compute the minimum potential energy:
V_min = 999;
for x = 0:dx:6
    for y = -8:dy:0

        % Sum of gravitational and spring PE:
        V = M*g*y + 0.5*K1*(sqrt(x^2 + y^2) - L)^2 + 0.5*K2*(sqrt((6 -x)^2 + y^2) - L)^2;

        % Checking for minimum:
        if V < V_min
            V_min = V;
            LC    = norm([6;0;0]-[x;y;0]); % distance from right support
            XC    = x;
            YC    = y;
        end

    end
end

ANSWER    =   LC;
% ANSWER = 7.2065;

I used a hill-climbing algorithm to solve. Let the unknown coordinates be $(x,y)$ . Randomly search the parameter space by introducing mutations. If the net force on the mass is smaller than the smallest seen so far, accept the mutations. If not, reject them. Eventually, the algorithm finds a point at which the net force is zero.

Interestingly, there are two equilibrium points, shown in the attached diagram. This problem asks for the stable equilibrium (shown in green). If the mass is perturbed starting at that point, it will eventually return. There is also an unstable equilibrium point, shown in red. Perturbing the mass in that state will result in the mass moving away and settling at the stable equilibrium point.

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import math
import random

m = 30.0
g = 9.81

k1 = 100.0
k2 = 50.0

L0 = math.hypot(3.0,3.5)

x1 = 0.0
y1 = 0.0

x2 = 6.0
y2 = 0.0

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

x = -20.0 + 40.0*random.random()
y = -20.0 + 40.0*random.random()

delta = 0.001

Fmin = 99999999.0

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

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

    if j > 3*10**6:
        delta = 0.00001

    xm = x + delta * (-1.0 + 2.0 * random.random())
    ym = y + delta * (-1.0 + 2.0 * random.random())

    L1x = x1 - xm
    L1y = y1 - ym

    L2x = x2 - xm
    L2y = y2 - ym

    L1 = math.hypot(L1x,L1y)
    L2 = math.hypot(L2x,L2y)

    u1x = L1x/L1
    u1y = L1y/L1

    u2x = L2x/L2
    u2y = L2y/L2

    s1 = L1 - L0
    s2 = L2 - L0

    F1 = k1*s1
    F2 = k2*s2

    F1x = F1*u1x
    F1y = F1*u1y

    F2x = F2*u2x
    F2y = F2*u2y

    Fx = F1x + F2x
    Fy = F1y + F2y - m*g

    F = math.hypot(Fx,Fy)

    if F < Fmin:

        Fmin = F

        x = xm
        y = ym

        L2_store = L2


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

if Fmin < 0.001:
    print Fmin
    print ""
    print x
    print y
    print ""
    print L2_store
else:
    print "you lose fella"


print ""
print "###########################"
print ""


# Results

#>>> ================================ RESTART ================================
#>>> 
#2.16952069902e-06

#-0.38322552068
#1.28713254333

#6.51170317444

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

#>>> ================================ RESTART ================================
#>>> 
#7.35506424333e-07

#2.26236643448
#-6.16051783713

#7.20568419317

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

#>>>