Find the effective force constant ke

[Pupil]

This isn't really a homework problem, but I've been going over these Yale problem sets for physics 200 (http://oyc.yale.edu/physics/fundamentals-of-physics/content/resources/problem_set_3.pdf" ), and I'm having trouble with one of the questions.

Homework Statement

Two springs have the same unstretched length but different spring constants k1 and
k2 . Find the effective force constant ke if they are connected in series and in parallel.
(To ¯nd ke, imagine stretching the combination by an amount x and keeping track
of the force needed.)

Homework Equations

f = ma
fs = -kx
Pupil = awesome

The Attempt at a Solution

So I figured the parallel spring constant out fairly easily, but I'm having trouble with the series one. The way I imagined it is we tie spring with constant $$k_1$$ to the wall, and tie spring with constant $$k_2$$ to spring with constant $$k_1$$, and finally pull $$k_1$$ by some amount A. Spring with $$k_1$$ will stretch by some amount A1 and spring with $$k_2$$ will stretch by some amount $$A_2$$. Summing up, we have $$k_2A_2 + k_1A_1 = F$$. The problem is I don't know what $$A_1$$ and $$A_2$$ are. I know the springs won't stretch the same amount, so I'm stuck with this ugly expression. Help me get unstuck (don't tempt me to look at the answers)! Thanks!

EDIT: The title was supposed to be Spring Constant in Series...Don't know what happened.

 

[ideasrule]

Actually
$$k_2A_2 + k_1A_1 = F$$

isn't right. Think about the point where the two springs connect. The first spring applies k1A1 to the second, and the second applies k2A2 to the first, so F=k1A1=k2A2. The problem should be easy from there.

 

[Redbelly98]

https://www.physicsforums.com/latex_images/22/2291913-7.png

[/URL]

That's not quite right. Springs in series experience the same force, so

k₁A₁ = k₂A₂ = F​

However, the displacements add up, so we can also say

A_net = A₁ + A₂​

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EDIT: Hey ideasrule, great minds think alike

 

[rock.freak667]

x₁ is the extension produced by the force F in spring 1, similar for x₂

The force on the first spring=force on the second spring
so we have

$$F=k_1x_1=k_2x_2$$

Now the effective spring constant relates to F by this

F=k_ex (where x=x₁+x₂)

i.e. we have $F=k_e(x_1+x_2)$

looks a bit unsettling right? BUT we know that $F=k_1x_1=k_2x_2$

you can put either 'kx' equal (replace F) to the formula above, and replace the 'other' x using the relation in the last formula. You will understand what I mean when you do it.

For parallel:

Each spring will stretch by the same amount, so finding k_e is easier here.

EDIT: seems ideasrule and Redbelly98 can type faster than me

 

[Pupil]

Holy crap! The force of the second on the first is equal to the force of the first on the second. I forgot Newton's third law! *Face palm* Thanks guys. You rock!