Stick and slip motion: mass and spring inside a box model I am trying to determine a set of differential equation which can describe the motion of a mechanical system as below. Here, at the bottom we have a plate, and a box on top of it. Inside the box, there is a ball supported by 2 springs. Then we try to oscillate the bottom plate. 
If needed, the assumptions are as below
1. Suppose that only horizontal motion is permitted.
2. The box's mass is negligible.
3. The ball has mass m.
4. There exist static and dynamic friction force between the box and the plate.
   The coefficient of static and dynamic force are $\mu_{s}$ and $\mu_{k}$, respectively.
5. The input oscillation for the bottom plate is $\gamma(t)$
This is not a homework, I am trying to make a mechanical model for droplet oscillation.
I understand that the box will move when the force on it is greater than the static force. But I am still confused on when the box move and and its velocity when it is moving.  

 A: Here are the steps you can take.


*

*Degrees of Freedom. There are 3 degrees of freedom, one for the base plate, one for the box and one for the mass. Hence there are 3 variables that you need to track, as well as their derivatives. I will name them $x_0=\gamma(t)$ for the plate, $x_1$ for the box and $x_2$ for the ball.

*Free Body Diagrams. For the moving parts draw free body diagrams enumerating the forces involved. An FBD for the plate is not required because its motion is completely prescribed. $$ F_1 = m_1 \ddot{x}_1 + F_2 \\ F_2 = m_2 \ddot{x}_2 $$ The forces from the spring are $F_2 = -k (x_2-x_1) - d (\dot{x}_2-\dot{x}_1 )$ with stiffness $k$ and damping $d$ (for realism). Friction force can be split into magnitude and direction as $$F_1 =-F_{f}\; {\rm sign}(\dot{x}_1-\dot{x}_0)$$ using the sign function that returns -1,0 or 1. The negative sign is because friction opposes motion.

*Static Conditions. Assume that no slipping occurs, with $\ddot{x}_1=\ddot{x}_0$ given, to find $$F_{f} ={\rm abs} \left(m_1 \ddot{x}_2+m_1 \ddot{x}_0\right) $$. If $F_f\ge\mu_s (m_1+m_2) g$ the slipping occurs

*Slipping Conditions. With slipping $x_1$ is decoupled with $x_0$, and friction magnitude is known $F_f=\mu_k (m_1+m_2) g$ with motion
$$ \ddot{x}_1 = \frac{1}{m_1} \left(\mu_k (m_1+m_2) g\; {\rm sign}(\dot{x}_1-\dot{x}_0)-m_2\ddot{x}_2\right) $$


Summary
Evaluate sticking conditions with
$$ \boxed{ \begin{aligned} 
\ddot{x}_2 &=\frac{1}{m_2} \left(- k (x_2-x_1) - d (\dot{x}_2-\dot{x}_1) \right) \\
\ddot{x}_1 &=\ddot{\gamma}(t) \\
F_f & =  {\rm abs}\left(m_2 \ddot{x}_2+m_1 \ddot{x}_0\right)   \end{aligned} } $$
and if $F_f \ge \mu_s (m_1+m_2) g$ then evaluate slipping conditions instead, with
$$ \boxed{ \begin{aligned} 
\ddot{x}_2 &=\frac{1}{m_2} \left(- k (x_2-x_1) - d (\dot{x}_2-\dot{x}_1) \right) \\
F_f & =  \mu_k (m_1+m_2) g  \\
\ddot{x}_1 &=\frac{1}{m_1} \left( F_f {\rm sign}(\dot{x}_1-\dot{x}_0) + k (x_2-x_1) + d (\dot{x}_2-\dot{x}_1)  \right) \\
 \end{aligned} } $$
A: The behavior of the system (not surprisingly)depends on the initial conditions. (For the sake of argument, we can assume the box starts stationary with respect to the table and $\gamma(0)=0$) I am assuming the problem is $1d$; this way we will end up with two coupled equations of motion.
Let's show the box's coordinate with $\chi$ and ball's coordinate with $x$. These are the equations of motion we have to deal with:
$$m\ddot x=2k(\chi-x) \\ -2k(\chi-x)=F_f$$
where $F_f$ is the friction force. Now we have some constraints on $F_f$, if the box is stationary with respect to the table, it will remain stationary, unless $F_f$ gets greater than $\mu_s N=\mu_s mg$. And if they are moving with respect to each other then $F_f=\mu_k m g$.
These equations should be enough, together with the initial conditions, to determine the behavior of the system.
