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?
Asked
Active
Viewed 724 times
0
-
How are you going to apply different forces to the ends of the spring without the sping accelerating away? – mike stone Nov 11 '20 at 13:20
-
but there would still be an extension in its length right? – Hayden Soares Nov 11 '20 at 13:35
-
Yes, there will be extension. – Gert Nov 11 '20 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. – BioPhysicist Nov 11 '20 at 13:58
-
@BioPhysicist How does the wall-attached sping have unequal forces acting on it? – mike stone Nov 11 '20 at 14:01
-
@mikestone How else would the center of mass move? – BioPhysicist Nov 11 '20 at 14:25
-
OP, have you tried writing out the equations of motion? – BioPhysicist Nov 11 '20 at 14:41
-
@VincentThacker That's not the same thing. This question is asking about pulling on each end with a different force. – BioPhysicist Nov 12 '20 at 14:04
-
@BioPhysicist It is the same thing. In most problems the spring is assumed to be massless, and a massless spring cannot have different forces on each end. There are numerous such questions all over this site, for example this one which has itself been marked a duplicate of another. – Vincent Thacker Nov 12 '20 at 14:11
-
@VincentThacker Then perhaps it would be best to first ask the OP if this system is massless. – BioPhysicist Nov 12 '20 at 14:17
-
Would it matter if the spring is massless or not? – Hayden Soares Nov 12 '20 at 14:31
-
@HaydenSoares Really what matters is the system. In my answer I do put in mass (to simplify matters, I put it at the ends). That way you can still think of a massless spring. If there is no mass at all then as Vincent points out things get weird. – BioPhysicist Nov 12 '20 at 14:58
-
The straightforward answer to the massless case is that it is just not possible for there to be different forces on each end. It is well-explained for massless strings, which are simply massless springs with a very large spring constant. That's all there is to the answer. However when the spring has mass things get a bit more interesting, but in general the spring still moves to a state where the tension is as uniform as possible. See for example this PDF – Vincent Thacker Nov 13 '20 at 03:46
-
Does this answer your question? Spring force on both sides of spring – Jon Custer Nov 13 '20 at 15:59
1 Answers
1
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 $F_L$.
BioPhysicist
- 56,248
-
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.. – Hayden Soares Nov 12 '20 at 13:28
-
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 – Hayden Soares Nov 12 '20 at 13:32
-
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 – Hayden Soares Nov 12 '20 at 13:34
-
@HaydenSoares Yes, that is correct for the equilibrium length $x=x_R-x_L-\ell$ – BioPhysicist Nov 12 '20 at 13:44
-
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? – Hayden Soares Nov 12 '20 at 13:48
-
@HaydenSoares If the spring isn't connected to anything then how is it being pulled on? Some things to think about 1) In the case of equal masses the mass on either end doesn't matter, as it cancels out. 2) There has to be at least some mass somewhere. If you want, work out the general case with unequal masses, and then do some exploration. Why does the mass on either end matter if $m_R\neq m_L$? What if $m_R\ll m_L$? What if $F_L$ is always restricted to be exactly equal to the spring force? The exploration of this system is a good learning opportunity :) – BioPhysicist Nov 12 '20 at 13:57
-
with unequal masses, I'm getting kx = (m_lF_r + m_rF_l )/m_r + m_l, which suggests that its extension is different than if there were equal masses, so I guess the answer to my question is that it depends on the masses at each ends as well. – Hayden Soares Nov 12 '20 at 14:16
-
also, (this may be more suitable as a standalone question), regarding the 'if the spring isn't connected then how is it pulled' point, what is pulling the masses that are pulling the spring? We could keep adding masses but never reach a 'first cause'. – Hayden Soares Nov 12 '20 at 14:21