Hooke's Law, three serial springs

[Lore]

Hello all, newbie here

Google led me to this thread:
https://www.physicsforums.com/showthread.php?t=70094"
which happens to be basically my question. I am doing an investigation on Hooke's Law for my VCE Physics class (Australian year 12, senior school).

Thanks to wikipedia's derivation and lots of blank staring, I finally understand how the 1/Keq = 1/k1+1/k2 formula shows up (excuse the cruddy layout).
I tried to find the relationship between three spring constants, but I failed my workbook and google failed me :(

Can anybody help me?

 

[Doc Al]

If you have the formula for two springs, use it for three. Take the first two, replace it by the effective spring constant. Now you're back to two "springs" in series--apply the formula again. See what happens.

Of course if you really understood the derivation for two springs, then you could apply the same logic directly to the three spring case.

 

[Lore]

Thanks :)

I did try what I think you mean (simply adding 1/k3 at the end of the previous equation) and got a similar result to my expected final value, but I wasn't sure if that mean't that it worked - using such small numbers.

Unfortunately my understanding doesn't cover logic, only the way the maths worked. When i tried doing the maths via the same method I found myself cancelling out k values that I needed :S

Thanks again :)

 

[Doc Al]

Since you are confident that the formula for two springs in series works, just use it twice. Replace k1 & k2 by their equivalent ke. Now you have ke and k3 in series and you can apply the formula again.

 

[Lore]

Just like electronics

 

[Kurret]

Hello!
Isnt the formula derivation on wikipedia way too advanced than it needs to be? Isnt it just possible to do like this:
The springs S1, S2,...,Sn, has spring coefficients K1, K2,...,Kn and Keq, and with length extensions X1, X2,...,Xn, and Xtot being to total length extension. Then:
$$X_{tot}=X_1+X_2+...+X_n$$
but $$F_m=X_mK_m\rightarrow X_m=\frac{F_m}{K_m}$$
thus$$\frac{F_{eq}}{K_{eq}}=\frac{F_1}{K_1}+...+\frac{F_n}{K_n}$$
Feq being the reaction force on the whole spring from the force making to total extension. but if you neglect the springs masses, you will after some force investigation realize that Feq=F1=F2=...=Fn (since at each shift between two springs, the upper spring must hold all the lower springs+weight, which is the same force as only the weight). Thus:
$$\frac{1}{K_{eq}}=\frac{1}{K_1}+...+\frac{1}{K_n}$$
hm I am not sure my motivation is totally correct, please correct me then

 

[Lore]

...
Little wonder it felt like I was beating my head against a wall..

You're absolutely correct (I think your last explanation was looking for the statement 'Fnet = 0'

The derivation on wikipedia was simply for two springs, so I think it was more a matter of proving it using the equations for that, but your version works much better :D

Also explains why it just keeps on going for many springs

Thankyou very much

 

[mysqlpress]

...
Little wonder it felt like I was beating my head against a wall..

You're absolutely correct (I think your last explanation was looking for the statement 'Fnet = 0'

The derivation on wikipedia was simply for two springs, so I think it was more a matter of proving it using the equations for that, but your version works much better :D

Also explains why it just keeps on going for many springs

Thankyou very much

in fact, it(the effective force constant) can be written as 1 over summation 1/k