2
$\begingroup$

I understand Hooke's Law to be 

F = kX

Where F is the tension applied to an object, X is the extension/change in length, and k is the spring constant, whose units are N/m.

I'm learning Mechanics 3 under the A Level Edexcel Maths specification (the M3 stuff is on page 70), and I am being told that Hooke's Law can also be defined as:

F = (λx)/l

Where x is the change in length, F is the tension and λ is the Modulus of Elasticity/ the Young's Modulus.

However, this would imply that the spring constant is equal to the Modulus of Elasticity, divided by the length. Instantly, if one already knows the units for the Modulus of Elasticity (Nm^-2), we can see that that would imply the spring constant is equal to Nm^-3, while the formula F = kX implies its units is N/metre.

I investigated further, and we know that modulus of elasticity is stress/strain, which is 

(Fl)/(Ax)

With a new variable being introduced as A, which is the cross-sectional area of the object.

We know that k = F/X, and do we can express the modulus of elasticity as

(kl)/A

Which would therefore mean that the spring constant k is (λA)/l

and not (λ/l)

We can check the units too, and (λA)/l gives N/metre.

So my question is: Why am I seeing two incompatible versions of Hooke's Law? The spring constant cannot be equal to F/x and λ/l, as the units aren't the same for one, and my above calculations also show that we need to include the cross-sectional area.

However, this is in the M3 syllabus and it's not anything new (although I haven't actually seen the formula anywhere else for k being λ/l) , so it's likely that there's something I'm not understanding.

Could anyone shed any light?

Improve this question
$\endgroup$

3 Answers 3

Reset to default
2
$\begingroup$

$F=kx$ is used in physics. $F=(\lambda / l)x$ is used in applied maths. Clearly $k=\lambda / l$.

$\lambda$ is called the modulus (of elasticity) but it is not the same as Young's Modulus.

To avoid confusion, decide which subject you are studying (physics or applied maths) then use the appropriate formula and ignore the other one.

Improve this answer
$\endgroup$
4
  • $\begingroup$ Could you explain the reasons behind this difference. The Wikipedia page on the modulus of elasticity equates it to the same thing as Young's Modulus. If it isn't, then what is it? $\endgroup$
    – 83457
    Oct 2, 2016 at 19:05
  • $\begingroup$ Different subjects use different concepts, just as different cultures use different languages, laws, currency, etc. Difficult to say why physicists prefer $k$ and mathematicians $\lambda$. $k$ is specific to a particular spring whereas $\lambda$ applies to all similar springs. ie If you double the length of the spring you must double $k$ but $\lambda$ stays the same. $\endgroup$
    – sammy gerbil
    Oct 2, 2016 at 19:21
  • $\begingroup$ The writer of this wiki article is referring to the 'modulus of elasticity' used in physics (ie Young's Modulus) and does not mention the modulus used in applied mathematics (no sign of $\lambda$ here), but this article is clearly talking about the modulus used in applied mathematics. $\endgroup$
    – sammy gerbil
    Oct 2, 2016 at 19:25
  • $\begingroup$ (Sorry, there is a $\lambda$ used in the last wiki article but it is not the same $\lambda$ as the one in applied mathematics, the one you are using.) $\endgroup$
    – sammy gerbil
    Oct 2, 2016 at 19:43
0
$\begingroup$

I'd say that the inclusion of the cross sectional area is a must. Perhaps the equation used a modulus which was already for a certain cross section? Otherwise it is easy to see that the cross sectional area must come into the expression: If you take a second spring, you double the cross sectional area. Then the force should also increase by the same factor. In the $F = kx$ version this is done by just setting $k \to 2k$. In the other example, the properties of the material does not change but there must be a factor 2 coming in as well.

Analysis with units must always succeed. If the units don't match up, something is wrong.

Improve this answer
$\endgroup$
4
  • $\begingroup$ that's what i thought, but this is not only in the m3 specification, but in their books as well. Whole exercises are devoted to using that formula, and it's in their exams. $\endgroup$
    – 83457
    Oct 2, 2016 at 14:22
  • $\begingroup$ Could it be that their definition of the modulus is different from the one that you assume? Are there examples or solved problems which have obviously wrong units? I'd assume that the book is consistent with itself but perhaps uses some other conventions than usual. $\endgroup$
    – Martin Ueding
    Oct 2, 2016 at 14:32
  • $\begingroup$ they refer to lamba specifically as the "modulus of elasticity" and in most of the questions you are given this modulus of elasticity, whose units is always in Newtons. So yeah it seems this isn't the same as the Young's modulus, but i've been looking around and i've seen no mention of another modulus of elasticity whose units is Newtons. $\endgroup$
    – 83457
    Oct 2, 2016 at 15:01
  • $\begingroup$ Their choice of unit defined how it is meant. And it seems like the cross sectional area is already factored in. This could make sense if you have certain predefined cross sections for say steel cables and compute with those. If you construct something with predefined parts then you don't really need to compute with “steel” but rather with a “10 mm steel cable” I guess. $\endgroup$
    – Martin Ueding
    Oct 2, 2016 at 15:10
0
$\begingroup$

Well. To focus on the original post, both constants can be used in maths or physics. If you are stuck to a particular scheme then they may be exclusively used but in general they are just alternative ways of looking at spring extensions.

The restoring Force $$ F=-kx \text{ or } F=-\lambda x/l $$

The second of these divides out the length and so any length of particular spring will have the same value whereas k will be proportional to the length.

Then the energy stored is integrated out as

$$ F=kx^2/2 \text{ or } F=\lambda x^2/2l $$

The units of $$\lambda \text{ are Newtons whereas the units of } k \text{ are Newtons/metre}$$

It perhaps inconvenient that these are both used from time to time but there you have it.

Improve this answer
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.