Spring Force

Spring Force is the non-constant, elastic force exerted by a spring upon a system.

The Main Idea

The spring force models the force in a system due to the presence of a stretched or compressed spring. This force is based upon two factors of the spring: the spring's stiffness and the distance the spring has been stretched. The spring's stiffness is a constant that represents how much force is required to stretch or compress a spring over a certain distance. The distance the spring has been stretched is measured relative to the length of the spring at its relaxed length. This length is the length the spring exhibits when the net forces acting upon the spring are zero.

A Mathematical Model

L0 is the distance in the relaxed length, while L is the distance of the entire spring. s is the difference between these two variabels.

The magnitude of the spring force is represented by the equation [math]\displaystyle{ \vert \vec{F}_{spring} \vert = k_s \vert s \vert }[/math], where [math]\displaystyle{ \vert s \vert }[/math] is the absolute value of the stretch of the spring [math]\displaystyle{ s = L - L_0 }[/math].

[math]\displaystyle{ L_0 }[/math] is the relaxed length of the spring, when the spring is neither stretched nor compressed.

[math]\displaystyle{ L }[/math] is the length that the spring after it has been stretched or compressed.

[math]\displaystyle{ k_s }[/math] is the spring stiffness, which is a constant inherent to the property of the spring.

The spring force can also be modeled as a vector by the equation [math]\displaystyle{ \vec{F}_{spring} = -k_s s \hat{L} }[/math], where [math]\displaystyle{ \hat{L} }[/math] is the direction that the spring is stretched or compressed.

These models are linear models of the spring force and not the true force under every scenario. In a real system, the spring will deform when stretched and compressed, and thus change its spring constant; however, in a small enough stretches this deformation will be negligible and the linear representation will be reasonably accurate.

A Computational Model

A GlowScript model of a spring that prints out the spring force vector[1]

The spring force can be computed with a loop that updates the position of an object and an attached spring. Using a small enough time step, these quantities can be calculated to a reasonable degree of accuracy. The computational approach to spring force problems can be extremely useful, as the spring force is non-constant. For every change in position, which the spring force affects, the spring force changes. Because of this nature of the spring force, iterative calculations prove rather useful in its calculation. The code below demonstrates how knowing the value of the spring stiffness and relaxed length constants and keeping track of the spring position can be used to calculate and update the the spring force.

#GlowScript 1.1 VPython

L0 = 0.1 #the relaxed length of the spring
ks = 15 #spring constant

#plate holding spring end
plate= box(pos=vec(-.1,0,0), size=vec(.005,.1,.1))

# ball and spring objects
ball=sphere(pos=vec(-L0+.1,0,0), radius=0.02, color=color.red, make_trail=true)
spring=helix(pos=plate.pos, axis=ball.pos-plate.pos, radius=.02, coils=10)

ball.m = 0.1 #mass of the ball in kg
ball.p = ball.m * vec(0.5,0,0) #initial momentum

t = 0 #time
dt = 0.01 #size of the time step

#loops forever
while True: 
    rate(10) #100 calculations per second
    
    #length of the spring
    L = ball.pos - plate.pos
    
    #spring force
    Fs = -ks * (mag(L) - L0) * norm(L)
    
    #update the momentum of the ball
    ball.p = ball.p + Fs * dt
    
    #update the position of the ball
    ball.pos= ball.pos + ball.p * dt / ball.m
    
    #update the spring
    spring.axis = ball.pos - plate.pos
    

Examples

Simple

Suppose a spring has been calibrated to have a spring stiffness of 300 N/m. You pull on the spring at rest and observe that it stretches a length of 0.02 m. What is the magnitude of the force exerted by the spring?

[math]\displaystyle{ \vert \vec{F}_{spring} \vert = k_s \vert s \vert }[/math]

[math]\displaystyle{ \vert \vec{F}_{spring} \vert = (300 N/m)(0.02m) }[/math]

[math]\displaystyle{ \vert \vec{F}_{spring} \vert = 6N }[/math]

Middling

Suppose a spring sits at the origin with a spring stiffness of 15 N/m. The spring has a relaxed length of 0.32 m. The spring has a 5 kg mass on the end at position <0.06, -0.41, 0> m. What is the current force exerted by the spring?

[math]\displaystyle{ \vec{F}_{spring} = -k_s s \hat{L} }[/math]

[math]\displaystyle{ \vec{L} = \langle 0.06, -0.41, 0 \rangle m - \langle0,0,0 \rangle m }[/math]

[math]\displaystyle{ \vert \vec{L} \vert = \sqrt{(0.06m)^2 + (-0.41m)^2} = 0.1717m }[/math]

[math]\displaystyle{ \hat{L} = \frac{ \langle 0.06,-0.41, 0 \rangle m}{0.1717 m} = \langle 0.35, -2.39, 0 \rangle m }[/math]

[math]\displaystyle{ s = 0.1717m - 0.32m = -0.1483m }[/math]

[math]\displaystyle{ \vec{F}_{spring} = -(300N/m)(-0.1483m) \langle 0.35, -2.39, 0 \rangle m = \langle 15.57, -106.33, 0 \rangle N }[/math]

Difficult

A spring sitting vertically on a table. The spring has a relaxed length of 0.3 m and a spring stiffness of 4 N/m. A 0.8 kg block is stuck atop the spring. You push the block and spring down so the bottom of the block is 0.1 m from the surface of the table. What is the net force acting upon the block at this moment?

[math]\displaystyle{ \vec{F}_{net} = \vec{F}_{Spring} + \vec{F}_{g} }[/math]

[math]\displaystyle{ \vec{F}_{spring} = -k_s s \hat{L} }[/math]

[math]\displaystyle{ \vec{L} = \langle 0, 0.1, 0 \rangle m - \langle0,0,0 \rangle m }[/math]

[math]\displaystyle{ \vert \vec{L} \vert = \sqrt{(0.1m)^2} = 0.1m }[/math]

[math]\displaystyle{ \hat{L} = \frac{ \langle 0,0.1, 0 \rangle m}{0.1 m} = \langle 0, 1, 0 \rangle m }[/math]

[math]\displaystyle{ s = 0.1m - 0.3m = -0.2m }[/math]

[math]\displaystyle{ \vec{F}_{spring} = -(4N/m)(-0.2m) \langle 0, 1, 0 \rangle m = \langle 0, 0.8, 0 \rangle N }[/math]

[math]\displaystyle{ \vec{F}_{g} = \langle 0, -mg, 0 \rangle = \langle 0, -(0.8kg)(9.81m/s^2), 0\rangle = \langle 0, -7.848, 0\rangle }[/math]

[math]\displaystyle{ \vec{F}_{net} = \langle 0, 0.8, 0 \rangle N + \langle 0, -7.848, 0\rangle N }[/math]

[math]\displaystyle{ \vec{F}_{net} = \langle 0, -7.048, 0 \rangle N }[/math]

Connectedness

1. Springs harmonic oscillations can be entrancing, and the forces behind its motion can be just as compelling.
2. My major being CS, has many applications in modeling systems. The motion of springs is a very interesting system to model as its motion can both be harmonic and chaotic depending on the scenario.
3. Springs are often used in bungee jumping and car suspension systems.

History

The concept of spring force was first discovered in 1660 by Robert Hooke. Hooke's linear model of spring force expressed the force of the spring using the equation [math]\displaystyle{ F = kx }[/math], relating force, [math]\displaystyle{ F }[/math] and displacement, [math]\displaystyle{ x }[/math], by some constant [math]\displaystyle{ k }[/math]. This insight would lead Hooke to devise plans for a watch with a balance spring, that would make pocket watches much more accurate.

See also

Work Done By A Nonconstant Force
Spring Motion
Spring Potential Energy

External links

Hyper Physics[2]
Prof. Schatz Video Lecture[3]
Khan Academy[4]

References

Ruth W. Chabay, Bruce A. Sherwood (2011). Matter & Interactions. John Wiley & Sons, Inc. p. 64.
Hyper Physics[5]

--Arj (talk) 23:35, 5 December 2015 (EST)