STATICS

$T_{tight}=\ T_{loose}\;e^{\mu\beta_{(rads)}}$

MODE-2 (CMPLX)

Put all the Forces in a CERTAIN origin.

The MOMENT is EXCLUDED when
finding the RESULTANT

$\int_{0}^{x_{lft\ or\ ryt}}{\sqrt{1+\left(\frac{wx}{T_0}\right)^2}dx}$

$s\ =c\ sinh\ \left(\frac{x}{c}\right)$

$y\ =c\ cosh\ \left(\frac{x}{c}\right)$

$T=wy$

$y^2=\ s^2+c^2$

$I_{max\ }\;\;\;\&\;\;\; I_{min\ }=\ \frac{{\bar{I}}_x\ +\ {\bar{I}}_y}{2}\ \pm\sqrt{\left(\frac{{\bar{I}}_x\ -\ {\bar{I}}_y}{2}\right)^2+\left({\bar{I}}_{xy}\right)^2}$

$tan\ \left(2\theta\right)=\frac{-2{\bar{\ I}}_{xy}}{{\bar{I}}_x\ -\ {\bar{I}}_y}$

$tan\ \phi\ =\ \mu $
$ f_{\ s/k}\ =\ N\left(\mu_{s/k}\right)\ $

Static Friction
is always LARGER than
Kinetic Friction
$\mu_s>\ \mu_k$

The FRICTION FORCE ONLY RESIST the APPLIED FORCE

the APPLIED FORCE REACHES the
MAXIMUM STATIC FRICTIONAL FORCE
that will result to IMPENDING MOTION

the FRICTIONAL FORCE only PROVIDES KINETIC FRICTION,
because it is now DYNAMIC (moving).

ALL the DOWNWARD FORCE is
RESISTED by the NORMAL FORCE