# How to set up equations of forces for 3 springs connected to each other in an elevator? In the above diagram the elevator is moving downward with acceleration $a_E$ and velocity $v_E$. All 3 masses are given as $m1,m2,m3$. The question is how much has the middle spring stretched from equilibrium.

I immediately attempt to draw a free body diagram for $m2$ and hoping somebody could correct me if something looks wrong: My net force equations is like so:

$F_{net} = m2*a_E = -k(x1+x2+x3)-(m2+m3)g$

Here is my thought process: Since the elevator is moving downward, all the springs are compressed and their restoring forces want to push downward in hopes to reach their respective equilibriums. Then you see the $-k(x1+x2+x3)$ term, because these all 3 spring forces acting on mass $m2$. Then there is the force of gravity on $m2$, but you have to realize $m3$ is right below so we need to factor that into the weight. Hence the $-(m2+m3)g$ term. Finally, we know the elevator itself is moving downward, and so the net force has to be the $m2*a_E$.

So am I thinking about this correctly? Also, I'm not sure how to solve for $x1,x2,x3$ with only 1 force equation.

Would appreciate all / any advise from the community.

• Are the springs stable at a new position, or is this system oscillating as the elevator descends? – David White Sep 24 '15 at 1:14
• Is $a_E=g$? Or is $a_E<g$? – Gert Sep 24 '15 at 1:24

## 1 Answer

This is a simple system. You did not say about any further extention of the springs. And also the elevator moving with a constant acceleration. So in this case, the system is at equilibrium and weight act on the system is cancelled by the restoring force on spring.
Now, as $m_1$ is in equilibrium, extention of second spring is due to massess $m_2$ and $m_3$ only. Thus the total weight act on the spring is $(m_2+m_3)(g-a_E)$ . This is because elevator is moving down with an acceleration $a_E$ so there is a change in weight. If extension on the spring 2 is $x_2$ and its force constant is $k$, then the restoring force will be $F=kx_2$ . Then using our earlier assumptions, we can write
$$kx_2=(m_2+m_3)(g-a_E)\\ \Rightarrow{x_2}=\frac{(m_2+m_3)(g-a_E)}{k}$$
it is interested to not that if $a_E=g$, then there will be no extension, as the weight is zero. And also If $a_E\lt{g}$ , then the spring expand and if$a_E\gt{g}$ , spring compressess