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2 people pulling a spring with equal forces from opposite ends is identical to pulling it from a rigid wall, but how to calculate its extension if its pulled from both ends with different forces? Should the mean of the forces be taken?

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  • $\begingroup$ How are you going to apply different forces to the ends of the spring without the sping accelerating away? $\endgroup$
    – mike stone
    Commented Nov 11, 2020 at 13:20
  • $\begingroup$ but there would still be an extension in its length right? $\endgroup$ Commented Nov 11, 2020 at 13:35
  • $\begingroup$ Yes, there will be extension. $\endgroup$
    – Gert
    Commented Nov 11, 2020 at 13:38
  • $\begingroup$ @mikestone I don't think the OP said anything about it not accelerating away. Plus just the center of mass will accelerate, you can still pull on the ends while this happens. $\endgroup$ Commented Nov 11, 2020 at 13:58
  • $\begingroup$ @BioPhysicist How does the wall-attached sping have unequal forces acting on it? $\endgroup$
    – mike stone
    Commented Nov 11, 2020 at 14:01

1 Answer 1

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If we assume Hooke's law holds, then for a spring constant $k$ with resting length $\ell$, the spring force is given by $F=\pm k(x_R-x_L-\ell)$, where $x_R$ and $x_L$ are the positions of the right and left ends of the spring respectively (the $\pm$ sign is to take care of which side of the spring you are looking at).

Now, if you assume identical masses $m$ are attached to each end of the spring, and that a force $F_R$ acts to the right on the right side and a force of $F_L$ acts to the left on the left side, you should be able to use Newton's second law to determine the equations of motion for $x_R$ and $x_L$, and more importantly the equation of motion for the separation $x_R-x_L$ to find where the equilibrium is obtained given $F_R$ and $F_L$.

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  • $\begingroup$ I understood the equation 'F=±k(xR−xL−ℓ')' , but I not sure about how to determine the equations of motion for xR and xL.. $\endgroup$ Commented Nov 12, 2020 at 13:28
  • $\begingroup$ I tried using the 'assuming identical masses idea', F_r - kx = ma, kx - F_l = ma (right and left masses respectively), therefore F_r - kx =kx - F_l => kx = (F_r + F_l) / 2 $\endgroup$ Commented Nov 12, 2020 at 13:32
  • $\begingroup$ Is what I did above logically right? I assumed that the spring applies equal force on each mass and that the entire system is accelerating uniformly , i.e the extended length of the spring is constant $\endgroup$ Commented Nov 12, 2020 at 13:34
  • $\begingroup$ @HaydenSoares Yes, that is correct for the equilibrium length $x=x_R-x_L-\ell$ $\endgroup$ Commented Nov 12, 2020 at 13:44
  • $\begingroup$ but does assuming that there are 2 masses at each end not prove the above relation for the general case in which the spring is unconnected to anything? $\endgroup$ Commented Nov 12, 2020 at 13:48

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