Static Friction & Dynamic Friction:

Limiting static friction: the maximal frictional force experienced by a stationary object

Dynamic friction: the friction force experienced by a moving object.

• Limiting static frictional force: $F_s=\mu_s N$
• Dynamic frictional force: $F_k=\mu_k N$

Graph:

$F_f=F_s=F_a \; \text{if} \; F_a \le \mu_s N$

Then:

$F_f=F_k \; \text{if} \; F_a\ge\mu_s N$

Where:

$F_a:\text{applied force} \\ F_f:\text{frictional force} \\ F_s:\text{static friction} \\ F_k:\text{dynamic friction}$

Finding $\mu_s$ :

If we take a block that is at the point where it starts falling due to the slant $\theta$ then we can deduce that:

$$ \mu_s =\frac{mg \sin\theta}{mg\cos\theta}=\tan\theta $$

Friction in a non-uniform circular motion:

Particle on a String

Finally by looking at the highest point we deduce:

$$ T=0=m\left(\frac{v^2}{R}-g\right) $$

Which allows us to find:

$$ v_c=\sqrt{Rg} $$

Forces on the system:

$$ \begin{align*} F_\theta &=mg\sin\theta \\ F_r&= mg\cos\theta-T \end{align*} $$

Using $F=ma$:

$$ \begin{aligned} a_\theta&=\frac{F_\theta}{m}=g\sin\theta=R\ddot{\theta} \\ a_r&=\frac{F_r}{m}=\frac{mg \cos\theta-T}{m} \end{aligned} $$

We find tension using $a_r=-\frac{v^2}{R}$:

$$ T=m\left(\frac{v^2}{R}+g\cos\theta\right) $$