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