# Derive the force from different spring configurations

1. Oct 28, 2005

### brentd49

Does anyone know a website that will derive the force from different spring configurations, i.e. two springs in series with different spring constants. I would just like to be able to understand and work with any combination of springs: two in series, connected to 3 in parallel connected to one in series...etc. I want to be able to understand from the general case. Halliday and Resnik does not go into detail at all, and I didn't have any luck googling. Thanks. -Brent

2. Oct 29, 2005

### brentd49

^bump

I'd be satisfied with a decent explanation of two springs in series with different spring constants.

3. Oct 29, 2005

### Integral

Staff Emeritus
For 2 springs in series you have a common force, each spring has a force of mg acting (neglecting the mass of the springs). So:

$$- x_1 k_1= mg$$
and
$$- x_2 K_2 = mg$$

For the pair of springs you can write:
$$-x_T K_T = mg$$

But we must have:
$$x_T = x_1 +x_2$$

For the total we now can write:
$$-(x_1 + x_2) K_T =mg$$

Observe that from our first relationships we have :

$$-x_n = \frac {mg} {k_n}$$

n= 1,2

so:
(now I can eliminate the pesky negative sign)
$$(\frac {mg} {k_1} + \frac {mg} {k_2})K_T = mg$$

Finally we get :

$$\frac 1 {K_t} = \frac 1 { k_1} + \frac 1 {k_2}$$

The key to the parallel case is that the displacement of the springs are equal. I'll let you do that one.

Last edited: Oct 29, 2005
4. Oct 29, 2005

### Q_Goest

The force on a spring equals the spring constant times the change in length.

(1) dF=ktotal dxtotal

and for a series of springs, dF is the same for each, so

(2) dF = k1 dx1 + k2 dx2 + k3 dx3 etc...
or dF = sum (kn dxn)

and the total distance x is
(3) dxtotal = sum (dF/kn)

Now substitute (3) into (1)

dF = ktotal dxtotal = ktotal sum(dF/kn)

Divide by dF

1 = ktotal sum (1/kn)

Put ktotal on the opposite side

ktotal = 1/ (sum (1/kn))

(sorry for the lousey presentation here.)

So the total spring constant for a bunch of springs in series is:

ktotal = 1 / ( sum (1/kn) )

Where
ktotal = total or equivalent spring constant of the springs in series
kn = one of the springs from 1 to n

Example:

For 3 springs with k's
1. 10 (lb/in)
2. 24 (lb/in)
3. 40 (lb/in)

The equivalent spring constant is

k = 1/ (1/10 + 1/24 + 1/40 )

k = 1/ (.1 + .0416667 + .025)

k = 6 (lb/in)

If you use the same basic logic, you should also be able to determine a more generic formula for springs in series and parallel.

Edit: I see Integral beat me to it! lol