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Image of masses on spring

I'm learning about springs in Physics and have come across something I do not fully understand.

As in the picture, there are three masses (A, B and C), with mass ma, mb and mc, respectively. In the configuration above, masses A and B are connected by a light inelastic string. The masses of the string and the two springs are negligible and the whole system is stationary (in equilibrium).

What I want to work out is the acceleration of the three masses when the string between A and B is suddenly cut.

I currently understand from slinkies that C would stay stationary while S2 contracts. So B is accelerating downwards. Therefore, C would have no acceleration.

Also, S1 is carrying all three masses but after the string is cut, it only carries ma, so it would contract upwards with a force equivalent to g(mb+mc), because they are the forces taken away from the spring. But when I calculate the acceleration of A using F=ma, I end up with a=g and I am not sure if that is correct, because I also get a=g for the acceleration of B.

I am trying to express each acceleration in terms of ma, mb, mc and g.

I am not looking for answers, as I know this is not a homework site. All I ask is for pointers in the right direction and perhaps correct me if my understanding of the spring system is flawed.

Thank you.

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  • $\begingroup$ I think your reasoning is sound so far. I'm a bit confused how you wound up with the acceleration for A as g. The force on A should have been g(mb + mc), which would make A's acceleration g(mb + mc)/ma (so if mb + mc = ma then it will be just g). $\endgroup$
    – JMac
    Jan 24, 2017 at 19:31
  • $\begingroup$ Why is A's acceleration the force divided by ma? Would it not be the force divided by (mb + mc), which is how I got to g? $\endgroup$
    – AkThao
    Jan 24, 2017 at 19:35
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    $\begingroup$ We know the string had a tension of g(mb + mc), so we know that is the force that was removed from ma. We then use the relationship F = maa (confusing wording, ma is mass A while the second a is acceleration of mass A), we know the force is g(mb + mc) so we equate that to get maa = g(mb + mc), when you solve for a you get g(mb + mc)/ma. $\endgroup$
    – JMac
    Jan 24, 2017 at 19:40
  • $\begingroup$ Oh I see now. So A accelerates upwards with a force equal to ma times a. But then why is this force equal to the previous force g(mb + mc)? $\endgroup$
    – AkThao
    Jan 24, 2017 at 19:47
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    $\begingroup$ Yes, because the spring was countering that force but no longer has to, causing the spring to contract and the mass to accelerate. $\endgroup$
    – JMac
    Jan 24, 2017 at 20:31

1 Answer 1

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When the connection between A & B is cut, mass of B & C is removed from the spring. When the masses B & C were attached, they exerted some force on the spring causing extension. They are removed so the spring will relax and extension in it will be reduced. Mass A will be accelerated upwards and will start performing simple harmonic motion. As for the masses B & C, gravitational force will pull them down. They will be in free- fall and therefore mass C will not pull the spring downward as it was doing in initial conditions. So, the spring will relax and come back to it's natural length. Coming back to it's natural length, it will exert an upward force on C and downward force on B.

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  • $\begingroup$ Ok so B and C both have an acceleration downwards of g. But then is C pulled up by the spring with a force equal to its weight? Therefore, the forces would balance so it has no velocity, making its acceleration zero. $\endgroup$
    – AkThao
    Jan 27, 2017 at 7:55
  • $\begingroup$ And is the force from the spring acting on B equal to its weight as well, thus making its acceleration 2g? Or is there something wrong with that? It's difficult to analyse the forces without numbers. $\endgroup$
    – AkThao
    Jan 27, 2017 at 7:55
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    $\begingroup$ you can calculate the forces on B & C by using center of mass. Center of mass of the system would have the acceleration g due to only external force acting on system- gravity. Also, force due to spring on both the blocks would be same. $\endgroup$ Jan 28, 2017 at 9:12
  • $\begingroup$ I guess that is another way of thinking about it. If my understanding is correct, since the overall acceleration of the system containing B and C is g, then any other forces within the system must balance each other out. $\endgroup$
    – AkThao
    Jan 28, 2017 at 17:07
  • $\begingroup$ I think I have figured it out now. Initially, C applies a force on the spring equal to mcg. When the string is cut, C no longer applies that force. So the spring pulls up C with a force F=mcg. Thus, the spring also pulls B down with the same force F=mcg. So if the force is mcg and the mass is mb, then according to F=ma, the acceleration of B due to the spring's force is a=mcg/mb. PLUS the acceleration from gravity means the total acceleration of B equals (mcg/mb)+g. $\endgroup$
    – AkThao
    Jan 28, 2017 at 17:14

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