Let's say I have a mass spring model like the one in the picture below:

enter image description here

So, there are 3 parts of the spring joined together in an equilateral triangular manner. Each of the joints has a mass of $m$. The resting length of each of the springs is $l$. The top joint point of the spring is fixed to the ceiling.

Now, if I were to pull both the lower points of the spring model such that each of the springs extend proportionally by $\Delta l$ change of length, and release. Now, I want to find an equation for Point A (indicated in the picture above) under gravity and sprint forces when springing back to position. The assumption is all the angles remain at $60$ degrees in every iteration when it springs back.

What I did is:

Let $k$ be the elasticity of the spring.

Then, for the X-axis component of the equation,

$k \cdot \Delta l \cdot cos(60) + k \cdot \Delta l = m\cdot a_x$

The acceleration of the spring going back to original x position would then be dividing both sides by the mass $m$.

For the Y-axis component of the equation with consideration of gravity as $g$,

$k \cdot \Delta l \cdot sin(60) + k \cdot \Delta l - mg = m\cdot a_y$

Similar to the X-axis, I thought I would consider the total of the extended length plus the projection from the x-axis onto the y-axis, and then minus the gravity resistance.

However, it turns out that I am wrong for the Y-axis component. The given answer is:

X: $k \cdot \Delta l \cdot cos(60) + k \cdot \Delta l = m\cdot a_x$

Y: $mg - k \cdot \Delta l \cdot sin(60) = m\cdot a_y$

I don't understand why is it so for the Y-axis, especially when the the gravity turns out to minus the projection and the extended $\Delta l$ is not added as part of the force.


1 Answer 1


you have taken the right approach by summing the forces and setting that equal to mass times acceleration.

For the Y-component you have included a $k$ $⋅$ $\Delta{l}$ which should not be there. In the X-direction, this term represents the force contribution from the bottom spring (the horizontal one), however this spring does not contribute at all to forces in the Y-direction (because it lies completely in the X-direction) thus has no effect on the Y-axis equation.

Also, in the diagram you provided, the Y-axis points down. You wrote your equations as though the Y-axis points up, and this has caused a sign error. Gravity points down along the positive Y-direction, thus the $mg$ term should be positive. An extension of the spring by $\Delta$$l$ leads to a restoring force that will pull point A in the negative Y-direction, thus the $k$ $⋅$ $\Delta{l}$ $⋅$ $sin(60)$ term must have a minus sign.

  • $\begingroup$ Thanks! For the part in which $k \cdot \Delta l$ should not be in the Y-component, it is because it is assumed that the spring is being pulled in the x-direction along the x-axis and therefore has no effect on the Y-axis equation? In other words, if I were to drag diagonally, $k \cdot \Delta l$ will have to be included in the Y-component of the equation? $\endgroup$
    – xenon
    Nov 24, 2012 at 22:37
  • $\begingroup$ We aren't assuming that the springs get pulled in the X-direction. You mentioned in your introduction that we "...pull both the lower points of the spring model such that each of the springs extend proportionally by $Δl$ change of length, and release." You also assumed "all the angles remain at 60 degrees". This second assumption ensures that the lower spring will always remain horizontal and will never contribute to forces in the Y-direction. To visualize this, just imagine the equilateral triangle oscillating in size while maintaining its shape. It will always be equilateral. $\endgroup$ Nov 24, 2012 at 22:46
  • $\begingroup$ Ohh...in other words, if it was pulled in such a way that the lower spring no longer remains horizontal, ie, slanted, the Y-component will have to include the $k \cdot \Delta l$, am I right? $\endgroup$
    – xenon
    Nov 24, 2012 at 22:56
  • $\begingroup$ Almost...you would have to include the sine of the angle of the slant. $\endgroup$ Nov 24, 2012 at 23:01
  • $\begingroup$ Thank you so much for the explanation. I still have one last doubt. Sorry. Earlier, you said $k \cdot \Delta l$ is a term representing the force contribution from the bottom spring. Wouldn't the $k \cdot \Delta l \cdot sin(60)$ term more like the contribution from the bottom spring since it is a projection from the bottom spring? As you said, if there is a slant, then the sine of the angle would give the length to extend in the Y-direction. Did I just confuse myself? $\endgroup$
    – xenon
    Nov 24, 2012 at 23:22

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