The forces acting on rigid bodies can be divided into two groups:

Internal: $\vec{F}_{int}$ : (either for translation or rotation)

External: $\vec{F}_{ext}$

Concept of system is in mechanical equilibrium.

$\vec{p}$ is a constant of motion

$\sum_{i=1}^n \vec{F}_i=0 \text { Translational Equilibium }$

$\sum_y F_{i x}=0, \sum_y F_{i y}=0 , \sum_i F_{iz}=0$

$\sum_i \vec{\tau_i}=0$ Rotational Equilibrium

$\sum_i \tau_{i x}=0, \sum_i \tau_{i y}=0, \sum_i \tau_{i z}=0$

Let us say that the system is under equilibrium.

$\sum \vec{F_i}$ = 0 : 2 Conditions

$\vec \tau$ about an axis perpendicular to $\vec{F_1}$ and $\vec{F_2}$ vanishes.

$\sum_i \vec \tau_i$ = 0 (Rotational equilibrium) does it remain valid, if the origin is changed.

Moment of the coplanar =$\vec{r_ 1} \times (-\vec{F}) + (\vec{r_ 2} \times \vec {F})$ =$(\vec{r_ 2} - \vec {r_1}) \times \vec {F}$ = $\vec{AB} \times \vec{F}$

"Rotational equilibrium" is independent to the location of the origin.

Case: 1 $\vec \tau$ =0, $\sum \vec {F}_ i$ = $\vec{F}_{total} \neq 0$

$\sum \vec {F}_ i$ =2 $\vec{F}$

Under rotational equilibrium, but not translational equilibrium.

Case: 2 $\sum \vec {F_i}$ = $\vec{F_{total}}$ = 0

$\sum \vec \tau_i = \vec \tau_{total} \neq 0$

It is under translational equilibrium but not in the rotational equilibrium.

For translation equilibrium : $\vec{R} = \vec {F}_ 1 + \vec{F}_2$

Moment about 0 : $\vec{F}_ 1 \vec{d}_ 2=\vec{F}_ 2 \vec{d}_2$ rotational equilibrium.

$\frac {F_1}{F_2} = \frac {d_2} {l_1}$ known as mechanical advantage.

The center of gravity of a body is the point around which the resultant torque due to gravity forces vanishes.

$\vec{R} = M\vec{g}$ of the book.

Centre of gravity(CG)

$\vec\tau = \sum_i \vec{r_i} \times {m_i} \vec{g}=0$

$\sum_i {m_i} \vec {r_i} =0$

CM = CG in a uniform gravitational field.

$A B=70 \mathrm{cm}$, $A G=35 \mathrm{cm}$, $A P=30 \mathrm{cm}$, and $P G=5 \mathrm{cm}$

$A K_1=3, K_2=10 \mathrm{~cm}$

$K_1 G=K_2 G=25 \mathrm{~cm}$

$T E \Rightarrow R_ 1+R_ 2=W_1+W$

$R_ 1+R_2=10 g$

$R E \Rightarrow -R_ 1(K_ 1 G)+R_ 2(K_ 2 G)+W_1(P G)=0$

$R_ 1 - R_ 2 = 1.2 g$

$R_ 1$ = 54.88N,and $R_ 2$ = 43.12N

$T E \Rightarrow \sum \vec{F}=0$

$\sum F_ x=0 \Rightarrow f = N_2$

$\sum F_ y=0 \Rightarrow N_ 1 = M_g$

$R E \Rightarrow \sum \vec{\tau}_i= 0$

$M g (\frac{L}{2} \cos \theta)-N_2 L \sin \theta=0$

$N_2= f = \frac{Mg \cot \theta }{2}$

$|\vec{F}|=\sqrt{N_1^2+f^2}=\sqrt{N_1^2+N_2^2}$

=$\sqrt{(Mg)^2 (1+ \frac{\cot^2 \theta }{4})}$

$|\vec{F}|= \frac {(Mg) \sqrt {4 + \cot^2 \theta }}{2}$

The direction of the reaction $\vec{R}$ is $\angle OBE.$

TE: $\sum \vec{F} \Rightarrow$

f=Mg $\sin \theta$

$N=Mg \cos \theta$

RE: $|\sum \tau|_c$=0 $\Rightarrow Nx = f \times \frac{h}{2}$