# Solve this exercise in Mechanics?

Dec 14, 2016

See below.

#### Explanation:

Recalling $\theta$ as the angle between the $x$ axis and the rod, (this new definition is more according to the positive angle orientation) , and considering $L$ as the rod length, the rod's center of mass is given by

$\left(X , Y\right) = \left({x}_{A} + \frac{L}{2} \cos \left(\theta\right) , \frac{L}{2} \sin \left(\theta\right)\right)$

the horizontal sum of intervening forces is given by

$\mu N \text{sign} \left({\dot{x}}_{A}\right) = m \ddot{X}$

the vertical sum gives

$N - m g = m \ddot{Y}$

Considering the origin as the moment reference point we have

$- \left(Y m \ddot{X} + X m \ddot{Y}\right) + {x}_{A} N - X m g = J \ddot{\theta}$

Here $J = m {L}^{2} / 3$ is the inertia moment.

Now solving

$\left\{\begin{matrix}\mu N \text{sign} \left({\dot{x}}_{A}\right) = m \ddot{X} \\ N - m g = m \ddot{Y} \\ - \left(Y m \ddot{X} + X m \ddot{Y}\right) + {x}_{A} N - X m g = J \ddot{\theta}\end{matrix}\right.$

for $\ddot{\theta} , {\ddot{x}}_{a} , N$ we obtain

$\ddot{\theta} = \frac{L m \left(\cos \left(\theta\right) + \mu \text{sign} \left({\dot{x}}_{A}\right) \sin \left(\theta\right)\right) {f}_{1} \left(\theta , \dot{\theta}\right)}{f} _ 2 \left(\theta , {\dot{x}}_{A}\right)$
$N = - \frac{2 J m {f}_{1} \left(\theta , \dot{\theta}\right)}{f} _ 2 \left(\theta , {\dot{x}}_{A}\right)$

${\ddot{x}}_{A} = {f}_{3} \frac{\theta , \dot{\theta} , {\dot{x}}_{A}}{2 {f}_{2} \left(\theta , {\dot{x}}_{A}\right)}$

with

${f}_{1} \left(\theta , \dot{\theta}\right) = L \sin \left(\theta\right) {\dot{\theta}}^{2} - 2 g$
f_2(theta,dot x_A)=mL^2(cos^2(theta)+mu cos(theta)sin(theta)"sign"(dot x_A)+4J
f_3(theta,dot theta,dot x_A)=(g mu (8 J - L^2 m + L^2 m Cos(2theta) "sign"(dot x_A)- g L^2 m Sin(2theta)+ L ((4 J + L^2 m) Cos(theta) + (L^2 m-4J)mu "sign"(dot x_A) Sin(theta)) dot theta^2)