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I am preparing for an exam in classical mechanics and today I have encountered a slight problem when solving a more complex spring-mass system. Consider the following setup:

enter image description here

Then the system of equations should be

\begin{eqnarray} m_1g & = & m_1g-k(z_1-L)-k(z_1-z_2+L) \\ m_2g & = & m_2g-k(z_2-z_1-L) \end{eqnarray}

where $L$ is the natural length of the two springs. My solution was identical with the solution given, however, there was one small difference - I wrote the gravitational force in the first equation as $(m_1+m_2)g$ rather than simply $m_1g$.

Is there an error in my reasoning? The gravitational force on the first mass should involve the sum of the two masses...

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closed as unclear what you're asking by sammy gerbil, Gert, user36790, knzhou, garyp Jul 21 '16 at 3:08

Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.

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    $\begingroup$ Do you mean $m_1a = \ldots$ ?? $\endgroup$ – garyp Jul 20 '16 at 19:11
  • $\begingroup$ Check the sign/direction of the force from the lower spring on $m_1$ in your 1st eqn. $\endgroup$ – sammy gerbil Jul 20 '16 at 19:15
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The systematic way to set up the equations is to draw a free body diagram for each mass. The FBD shows all the forces acting on that particular mass. You can then use Newton's second law to get the acceleration from the resultant force.

The weight of mass 2 acts on mass 2, not on mass 1. Of course the weight of mass 2 will affect the motion of the system, and some of it will get applied to mass 1 via the spring, but applying it as an external force to mass 1 is wrong. (The tension in the lower spring will vary with time as the system moves, of course.)

To see why it is wrong in a simpler example, think about the static equilibrium of the system, if the springs are replaced by strings which don't stretch. Clearly the weight of mass 2 is balanced the tension in the lower string. So that force is applied to mass 1 by the string. If you also apply it to mass 1 as an external force, you are including it twice in your equations.

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Yes, there is an error in your reasoning. The gravitational force on an object is always proportional to the mass of that object. So if you want the gravitational force on $m_1$, you use $m_1 g$. If you want the gravitational force on $m_2$, you use $m_2 g$. If you want the gravitational force on the combined system of $m_1$ and $m_2$, you use $(m_1 + m_2)g$, but that wouldn't make much sense unless they're stuck together.

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  • $\begingroup$ Well I was rather thinking of a tension force which would be equal in magnitude to the gravitational force. Thus, one could add those two and identify the resulting force as the gravitational force acting on the total mass. $\endgroup$ – whypi314 Jul 20 '16 at 19:40
  • $\begingroup$ The tension force in the spring is only equal to the weight of mass 2 when mass 2 is not accelerating. But the only way mass 2 can never accelerate is if it moves with constant velocity, and that doesn't make any physical sense unless its velocity was 0, otherwise the springs would be stretched longer and longer "for ever" during the motion of the system. Common sense says that mass 2 will move during the dynamics solution, but not with constant velocity. $\endgroup$ – alephzero Jul 20 '16 at 19:51

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