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Two identical springs in parallel are supporting a mass. One spring is twice as stiff. Which spring experiences more force?

I want to say that the stiffer spring experiences more force using Hooke's law but I am unsure.

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  • $\begingroup$ are the natural lengths of the two springs the same? $\endgroup$ Feb 17, 2022 at 19:04
  • $\begingroup$ yes they are since theyre identical $\endgroup$
    – hellinas8
    Feb 17, 2022 at 19:05
  • $\begingroup$ How are they "identical" if one is stiffer? The imprtant thing is to have the same deformation under given load (so you can apply the simplest case). They don't have to be identical. $\endgroup$
    – nasu
    Feb 18, 2022 at 4:23
  • $\begingroup$ I think as long as the springs are placed equally apart from the center of mass, it does not matter what the stiffness is, both springs will experience the same force. (if it's not then the mass will rotate until equilibrium is reached) $\endgroup$
    – Gary Ong
    Feb 18, 2022 at 6:32

3 Answers 3

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If this is an interview question then it is probably intentionally underspecified to see what questions you ask.

One of the pieces of information that is missing is how the attachment points between the springs and the mass are arranged. We don’t know whether the attachment points are symmetrically placed or not. But if we assume they are symmetrically placed then the forces exerted on the mass by the two springs must be the same (otherwise there will be a turning moment on the mass and it will not be in equilibrium). However, the extensions of the two springs will not be the same - the stiffer spring will extend by a smaller amount.

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Assume that a mass of $m$ kg is hung against gravity by these two springs (in parallel)

For first spring, spring constant = $k$

For second spring (stiffer), spring constant = $2k$

Now in equilibrium position net force on the mass is zero.

So,

$$F_{sp_1} + F_{sp_2} = mg$$

$$kx + 2kx = mg$$

Note that both the spring will elongate with same length from their natural length position

Now it is obvious from the second equation that the force exerted by the second spring (stiffer) is more.

Hope this helps.

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  • $\begingroup$ But if the springs are attached symmetrically to the mass and $F_{sp_1} \ne F_{sp_2}$, won't there be a turning moment on the mass, so it is not in equilibrium ? I think we have to assume $F_{sp_1} = F_{sp_2}$, in which case the extensions of the two springs are not the same. $\endgroup$
    – gandalf61
    Feb 17, 2022 at 19:25
  • $\begingroup$ @gandalf61 The OP has not specified this condition that's why I have explained the case in which there is no turning effect. Also these problems are usually asked ignoring the case you mentioned. $\endgroup$
    – Spencer
    Feb 17, 2022 at 19:32
  • $\begingroup$ It was an interview questions and it was exactly asked the way I wrote.. :/ $\endgroup$
    – hellinas8
    Feb 17, 2022 at 19:39
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Spring constant,
For parallel: $$k_{tot}=\sum_i k_i$$ For series $$k_{tot}=(\sum_i \frac{1}{k_i})^{-1}$$

Since you have two spring attached parallel. So spring constant for first one will be $k$ for second one $$k_{2nd}=k+k=2k$$ If spring constant for second one is different then you can write $k_{2nd}=k+k_2$

$$F_{2nd}=-k_{2nd}x$$

Which spring will experience more force?

That depends on value of $k$. First one will experience more force. $$F_1=-kx>-2kx=F_2$$ for the case $-1>-2$

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