How Hard Could Two Springs Be?

Classical Mechanics Level 4

In the x y xy -coordinate system, two massless ideal springs are anchored at their respective coordinates shown in the figure. Their other ends are attached to a massive point particle. The figure shows the unstretched spring lengths, spring constants, the particle mass, and the gravitational acceleration (all in standard SI units).

How far from ( 0 , 0 ) (0,0) does the particle reside when it is in equilibrium (to three decimal places)?


The answer is 1.871.

Steven Chase by Steven Chase , 37, USA

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2 solutions

Steven Chase
Mar 7, 2020

I used a hill-climbing algorithm to solve. The net force on the mass is zero when the mass is positioned at ( x , y ) ≈ ( 0.733 , − 1.721 ) (x,y) \approx (0.733, -1.721) . Simulation code is below, with comments 

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import math
import random

# Simulation params

x1 = 0.0
y1 = 0.0

x2 = 1.0
y2 = 0.0

L1 = 1.0
L2 = 1.0

k1 = 10.0
k2 = 30.0

m = 3.0
g = 10.0

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

# Randomly initialize coordinates of mass

x = 1.0*random.random()
y = -5.0*random.random()

# Mutation size parameter

delta = 0.001

Fmin = 99999999.0

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

for j in range(0,1*10**6):

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

    v1x = x1 - xm   # vectors from mass to spring anchor points
    v1y = y1 - ym

    v2x = x2 - xm
    v2y = y2 - ym

    v1 = math.hypot(v1x,v1y)   # stretched spring lengths
    v2 = math.hypot(v2x,v2y)

    u1x = v1x/v1                # unit vectors along springs
    u1y = v1y/v1

    u2x = v2x/v2
    u2y = v2y/v2

    s1 = v1 - L1               # spring stretches
    s2 = v2 - L2

    F1s = k1*s1                # spring force magnitudes
    F2s = k2*s2

    F1sx = F1s * u1x           # spring force vectors
    F1sy = F1s * u1y

    F2sx = F2s * u2x
    F2sy = F2s * u2y

    Fgx = 0.0                  # gravity force vector
    Fgy = -m*g

    Fx = F1sx + F2sx + Fgx     # net force vector
    Fy = F1sy + F2sy + Fgy

    F = math.hypot(Fx,Fy)      # net force magnitude

    if F < Fmin:                # accept mutations if net force decreases

        Fmin = F

        x = xm
        y = ym


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

print Fmin
print ""
print x
print y
print ""
print math.hypot(x,y)

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

#7.86113991781e-06

#0.732966784399
#-1.72114439906

#1.87071599914

0 Helpful 0 Interesting 0 Brilliant 0 Confused
Chan Tin Ping
Feb 3, 2018

Can someone post a solution without brute force. Pls~~~

0 Helpful 0 Interesting 0 Brilliant 0 Confused

I think this one requires either iteration or search. @Josh Silverman , what do you think?

Steven Chase - 3 years, 4 months ago

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Yes, when I solved this I had to write a minimization routine. I'll look up my notebook when I get to the office next week.

Josh Silverman Staff - 3 years, 4 months ago

@Steven Chase Thanks for posting your solution. I tried to find the discrepancy but was unable to resolve. I got x=1.154, y=-1.333 dist=1.764.

A Former Brilliant Member - 1 year, 3 months ago

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Both springs are pulling leftward on the mass in that case, so how do the horizontal forces balance?

Steven Chase - 1 year, 3 months ago

You're right. I was just checking that and saw that. Which is weird that a FEM program would yield an erroneous solution. Then I tried to minimize the potential energy and got the vertical to balance but not the horizontal again. I have too many variables and not enough equations. Maybe an iterative solution is the only way.

A Former Brilliant Member - 1 year, 3 months ago

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I'm intrigued. How does the FEM solution work?

Steven Chase - 1 year, 3 months ago

Did this by hand and have a difference of 0.15N in the horizontal which I guess is close enough and F1=10.58N, F2=23N. The FEM uses the stiffness method, it had the option to assign the spring constants and for each element. The nodes are just set as coordinates. Then you place a load on a node and fix the nodes at supports - I used a pin connection fixed in x and y. It works fine for trusses with stiffer members. Not sure why it didn't work.

A Former Brilliant Member - 1 year, 3 months ago

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In your new problem, are we supposed to assume that springs 1 and 2 both have natural length L1, and that springs 4 and 5 both have natural length L2?

Steven Chase - 1 year, 2 months ago

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No, spring (1 + 2)=L1 and spring (3+4)=L2.

A Former Brilliant Member - 1 year, 2 months ago

A few questions on the new problem posted today: 1) What is the gravity value? 2) Is there friction underneath the spring?

Steven Chase - 1 year, 2 months ago

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Updated per your questions. Thanks! No friction under spring. g=9.81

A Former Brilliant Member - 1 year, 2 months ago

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