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When looking at Hooke's law for the entire spring the force is $kx$. But what happens when analyzing segments of the spring in order to look at the internal forces?

Imagine a spring of length $x$ that extends $\Delta{x}$ as shown below:

enter image description here

The internal forces need to be equal along the spring according to Newton's third law. The problem for me is if we analyze; for example the first part of the spring to a particular point, we see the force pulling this part to the right and left are both $k(\Delta{x})$, where $\Delta{x}$ is the extension of the entire spring. However when we write Hooke's law only for this small section of the spring we get the force $k(\Delta{x'})$, where $\Delta{x'}$ is the extension of this small part of the spring which is less than the entire extension of the spring. So analyzing only this part we get a contradiction? What is wrong here?

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2 Answers 2

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We normally write Hooke's law as:

$$ F = kx $$

but the more fundamental equation is:

$$ F = k' \frac{x}{\ell} $$

where $\ell$ is the original length of the spring. So the $k$ in Hooke's law is actually:

$$ k = \frac{k'}{\ell} $$

i.e. it depends on the length of the spring. The constant $k'$ is a fundamental property of the spring and is related to the thickness and stiffness of the wire and the geometry of the spring.

You can take a small part of the spring and find its extension, but if you do that you have to change the force constant $k$ to get the correct answer.

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I'm trying to add few details to @John Rennie's answer.

Let's assume that the spring is made of an elastic medium. The constitutive equation of a 1D elastic solid can be written as

$$N(x,t) = EA \, \varepsilon(x,t) \ ,$$

being $EA$ the axial stiffness (assuming homogeneous media, it's not a function of $x$), $N(x)$ the internal axial force at point $x$, and $\varepsilon(x)$ the local strain, that can be approximated as $u'(x)$ for small displacement and strain, i.e. the first derivative of the displacement (difference between the actual and the original position).

The governing equation of the axial dynamics of a 1-dimenisonal elastic medium with no distributed loads reads

$$0 = m \, \partial_{tt} u (x,t) - \partial_x N(x,t) \ .$$

For the elements with negligible inertia, the linear mass density goes to zero $m \rightarrow 0$ and the equation becomes

$$\partial_x N(x,t) = 0$$

implying an internal force independent from $x$, and equal to the external load (for boundary conditions of the problem) $$N(x,t) = F^{ext}(t) \ .$$

Thus:

  • internal force is independent from the $x$-coordinate, $N(x,t) = F^{ext}$

  • strain is independent from the $x$-coordinate, $u'(x,t) = \varepsilon(x,t) = \frac{F^{ext}}{EA}$

  • the change in length of the 1-d linear elements is obtained by means of integration of $u'(x,t)$ between the two ends of the element, i.e. from $x=0$ to $x = \ell$, and thus

    $$\Delta u(t) = u(\ell,t) - u(0,t) = \int_{x=0}^{\ell}\frac{F^{ext}(t)}{EA} = \frac{\ell}{EA} F^{ext} =: \frac{1}{K} F^{ext} \ ,$$

having defined the axial stiffness of the whole 1-d element as $K := \frac{EA}{\ell}$, so that $F = K \Delta u$

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  • $\begingroup$ Why do we take partial derivative of force with respect to x? $\endgroup$
    – LEXOR AI
    Commented Jun 23 at 18:23
  • $\begingroup$ Because the dynamical equations governing structure mechanics are PDE with displacement, strain, stress,... functions of the independent variable time $t$ and space $\mathbf{r}$. If you want the equation of axial dynamics in displacement formulation (~Navier-Cauchy) and not in mixed displacement-force formulation, it is $m \partial_{tt} u(x,t) - \partial_x (EA \partial_x u (x,t)) = f(x,t)$ $\endgroup$
    – basics
    Commented Jun 23 at 21:19

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