Normal forces in friction oscillator I am looking at the demonstration on this website：
https://demonstrations.wolfram.com/TheFrictionOscillator/
I don’t understand how the formulas for the normal forces are derived. The formulas are on the website, can someone help me?
 A: The formulas for the normal reactions are derived by using $torque=0$ and $F_{vertical}=0$ on the rod.
Consider that x is a small displacement.

Here, we know that
$$
troque_{COM}=0\\
N_1(L-x)-N_2(L+x)=0\rightarrow equation\ 1
$$
Also,
$$
F_{vertical}=0\\
N_1+N_2-Mg=0\\
\implies N_2 = Mg-N_1 \rightarrow equation\ 2
$$
Now, if we substitute value of $N_2$ in equation 1 using equation 2, we get,
$$
N_1(L-x)-(Mg-N_1)(L+x)=0\\
N_1(L-x+L+x)=Mg(L+x)\\
N_1(2L)=Mg(L+x)\\
\implies N_1=\frac{Mg(L+x)}{2L}\\
using\ equation\ 2,\\
\implies N_2=\frac{Mg(L-x)}{2L}
$$
Since the rotors are constantly spinning, the friction must be kinetic at both the point of contacts, so,($\mu$ is coefficient of friction)
$$
f_1 = \mu N_1\\
f_2 = \mu N_2
$$
Now, we can get the expression for net force in the horizontal direction,
$$
F_{horizontal}= f_1-f_2\\
F_{horizontal}= \mu N_1-\mu N_2\\
F_{horizontal}= \mu (N_1-N_2)\\
F_{horizontal}= \mu \left(\frac{Mg(L+x)}{2L} - \frac{Mg(L-x)}{2L} \right)\\
F_{horizontal}= \mu \left(\frac{Mg(2x)}{2L}\right)\\
F_{horizontal}= \left(\frac{\mu Mg}{L}\right)x\\
So,\ k=\frac{\mu Mg}{L}
$$
Since we have now obtained k, we can solve for the time period with the formula below and the question is done.
$$
T = 2 \pi \sqrt{\frac{M}{k}}
$$
Feel free to ask any questions regarding this.
Hope this helps!
