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 Δl 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⋅Δl⋅cos(60)+k⋅Δl=m⋅ax
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⋅Δl⋅sin(60)+k⋅Δl−mg=m⋅ay
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⋅Δl⋅cos(60)+k⋅Δl=m⋅ax
Y: mg−k⋅Δl⋅sin(60)=m⋅ay
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 Δl is not added as part of the force.
Answer
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 ⋅ Δl 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 Δl leads to a restoring force that will pull point A in the negative Y-direction, thus the k ⋅ Δl ⋅ sin(60) term must have a minus sign.
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