# Pulling a spring from both ends with different forces [closed]

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?

• How are you going to apply different forces to the ends of the spring without the sping accelerating away? Commented Nov 11, 2020 at 13:20
• but there would still be an extension in its length right? Commented Nov 11, 2020 at 13:35
• Yes, there will be extension.
– Gert
Commented Nov 11, 2020 at 13:38
• @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. Commented Nov 11, 2020 at 13:58
• @BioPhysicist How does the wall-attached sping have unequal forces acting on it? Commented Nov 11, 2020 at 14:01

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 $$F_L$$.
• @HaydenSoares Yes, that is correct for the equilibrium length $x=x_R-x_L-\ell$ Commented Nov 12, 2020 at 13:44