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I was trying to do some elementary physics practice questions and came across this one that got me so confused...

like so

Here you see this spring system and I wanted to show that the the value of $k_1$ to ensure that a force $F$ displaces the spring system by a distance $x$ is given by $$k_1 = \frac{Fk_2}{2k_2x-2F}$$

I'm just so confused right now and don't know what to do.

I tried to get the $x$ distance equation in this case $$x = F(k_2 + 2k_1)/2k_1k_2$$

And now I'm just out of wits. Hopefully someone can shed some lights on this.

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closed as off-topic by John Rennie, user36790, ACuriousMind, heather, Jon Custer Oct 4 '16 at 12:35

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  • $\begingroup$ Let $\:n\:$ springs with constants $\:k_{1},k_{2},\cdots,k_{n}$. Try to prove that the equivalent constant $\:k_{\rm{total}}\:$ is \begin{align} k_{\rm{total}}^{\rm{parallel}} & =k_{1}+k_{2}+\cdots+k_{n} \quad \text{(parallel arrangement)} \tag{01}\\ &\\ \dfrac{1}{k_{\rm{total}}^{\rm{series}}} & =\dfrac{1}{k_{1}}+\dfrac{1}{k_{2}}+\cdots+\dfrac{1}{k_{n}}\quad \text{(series arrangement)} \tag{02} \end{align} $\endgroup$ – Frobenius Oct 4 '16 at 7:54
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just solve for $k_1$ from the above equation you got. you are there!

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  • $\begingroup$ no tricks here. its all very straightforward. the equivalent spring constant is $2k_2$||$k_1$. put F=$2k_2$||$k_1$x and then simply solve for $k_1$ $\endgroup$ – Prasad Mani Oct 4 '16 at 6:20
  • $\begingroup$ What does || mean in this case? $\endgroup$ – 2Xchampion Oct 4 '16 at 6:22
  • $\begingroup$ $2k_2$ and $k_1$ are in parallel combination. First the two springs of $k_1$ next to each other serve to add their spring constants. Its value is then $2k_1$($k_1$+$k_1$). Then the spring of $k_2$ which is alongside the two springs acts in such a way that the equivalent spring constant now becomes the parallel combination of $2k_2$ and $k_1$. That is, $k_{resultant}$=$\frac{(2k_1)(k_2)}{2k_1+k_2}$ $\endgroup$ – Prasad Mani Oct 4 '16 at 6:25
  • $\begingroup$ I might be wrong but it seems like it should be 2k1 and k2?... k1 has two springs tho... $\endgroup$ – 2Xchampion Oct 4 '16 at 6:27
  • $\begingroup$ oh yeah sorry. its $2k_1$. let me edit that $\endgroup$ – Prasad Mani Oct 4 '16 at 6:27
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You need to consider two extensions, the extensions of the top springs, and the extensions of the bottom spring. Call them $x_1$ and $x_2$ then $x=x_1 + x_2$. Then solve for the values of $x_1$ and $x_2$ given a force $F$. You also need to consider the forces on the central beam.

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  • $\begingroup$ The other guy said to solve for k1 from the X equation I already have... I'm even more confused now... Care to elaborate? $\endgroup$ – 2Xchampion Oct 4 '16 at 6:21
  • $\begingroup$ @CookieJar - I have no idea what that answer by the other individual means, looks like he didn't even read the question. I've given you some hints about how to solve the stated problem, can you try and take it using my methods, and see if that works? I don't want to just give you the answer. $\endgroup$ – Suzu Hirose Oct 4 '16 at 7:01

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