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A stone of mass $m$ is attached at one end of a vertical spring whose another (upper) end is fixed at horizontal surface. During the motion, a spring takes vertical direction. Force in spring is proportional to extension.

How can I determine equilibrium position?

Will it be in equilibrium when forces that act on it are equal by intensity?

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    $\begingroup$ The equilibrium will be achieved when forces on the stone(Spring force and gravity) exactly cancel out, i.e. net force is zero. $\endgroup$ – udiboy1209 Aug 7 '13 at 16:47
  • $\begingroup$ So, if I take $y$-axis directed vertically downwards, forces that are acting on the stone are $F_g=mg\vec j$ and $F=-k(y-a)\vec j$, so the it will achieve equilibrium position when $mg=k(y-a)$, i.e. $y=\frac{mg}{k}+a$, where $a$ is natural length of spring and $k$ is constant of extension? $\endgroup$ – gov Aug 7 '13 at 16:54
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    $\begingroup$ that is perfectly right! $\endgroup$ – udiboy1209 Aug 7 '13 at 16:55
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Determine all the forces in your problem. Calculate their magnitudes and directions.

Equilibrium = no net force (vector sum of all forces is zero).

NOTE: You can define a "global" equilibrium (no net force at all), or a "partial" equilibrium (no net force along subset of x,y,z axes). In your case (1D problem), the above two are the same.

Good luck

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