0
$\begingroup$

enter image description here

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:

enter image description here

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.

$\endgroup$
2
  • $\begingroup$ Are the springs stable at a new position, or is this system oscillating as the elevator descends? $\endgroup$ – David White Sep 24 '15 at 1:14
  • $\begingroup$ Is $a_E=g$? Or is $a_E<g$? $\endgroup$ – Gert Sep 24 '15 at 1:24
1
$\begingroup$

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

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.