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I know that the stifness of the following springs are different.

The two spring paradigms

But how to find their spring constant?

I think that

  • one has the tension force is fully transmitted along the solid (but is it both?)
  • but on which one the tension forces add their effects?

One will have $F=F_1=F_2$ for their tension force. I feel it's the first one but why? The other one should be $F=F_1+F_2$ but I don't understand why...

Then finding the spring constant should be easy. My problem is finding which paradigm to apply...

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  • $\begingroup$ please edit your question , so somebody can answer you , especially "one has the tension force is fully transmitted along the solid (but is it both?) " $\endgroup$ – Faiz Iqbal Feb 1 '16 at 12:32
  • $\begingroup$ You might want to consult Wikipedia for step-by-step derivations. But I would encourage you to try to derive it yourself based on the answers below. $\endgroup$ – Jonas Greitemann Feb 1 '16 at 12:58
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    $\begingroup$ Have you tried using Free Body Diagrams, or do you feel like you have advanced beyond the point where you need to use them? $\endgroup$ – Chet Miller Feb 1 '16 at 13:26
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As we know, for a mechanical system being in equilibrium means having total force equal to zero. Let's look now at the second spring on a picture b): to be in equilibrium, force from the first spring $F_{k1}$ (action of the first spring on the second) should equalise external force $F_{out}$. Thus, $F_{out}=-F'_{k1}$. This, in turn, after applying Newton's third law, will give that the force exerted on the first spring is equal to the external force. Thus, you should use condition $F_{k1}=F_{k1}$ for finding their spring constant.

For the b) case it is even simpler: you should notice that elongation of two springs should be equal. Thus, you have a different condition $\Delta x_{k1}=\Delta x_{k2}$.

Now it is straightforward to find both spring constants using $F=k\Delta x$ law.

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  • $\begingroup$ I understand that to be in equilibrium $F_{k1}$ should equalise external Force $F_{out}$ but why does it have $F_{out}=F'_{k1}$? Is it its derived form? Where does it comes from? $\endgroup$ – Revolucion for Monica Feb 3 '16 at 10:35
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Since the springs are connected in series, they will experience an equal amount of force as the tension is same inside both the springs in equilibrium. If the tension is non-zero, then some part of spring will accelerate and therefore equilibrium will be destroyed. In case of parallel springs, the force exerted by them will be the sum of the two forces as can be seen easily by making a free body diagram.

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The answer is the opposite of what you would have if the springs were resistors.

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  • $\begingroup$ technically this should have been a comment.. $\endgroup$ – Bruce Lee Feb 3 '16 at 11:07

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