physics-class-11-unit-07-chapter-05-equilibrium-ofa-rigid-body-moments-and-center-of-gravity-l5-10-vu2yw9f27ig

### <Success style="font-size:25px"> Equilibrium Ofa Rigid Body Moments And Center Of Gravity L-5 </Success>
### <primary style="font-size:24px"> Equilibrium of a rigid body </primary>
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- The forces acting on rigid bodies can be divided into two groups:

- <span style="color:green">Internal</span>: $\vec{F}_{int} $ : (either for translation or rotation)

- <span style="color:green">External</span>: $\vec{F}_{ext}$

- <span style="color:blue">Concept of system is in mechanical equilibrium.</span>

- $\vec{p}$ 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

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

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<footer style = "font-size: 18px"> $\rightarrow$ Equilibrium of a Rigid Body, Moments and Center of Gravity $\rightarrow$ <span style = "color:blue"> Equilibrium of a rigid body </span> $\rightarrow$ Coplanar Forces $\rightarrow$ Illustration </footer>

### <Success style="font-size:25px"> Equilibrium Ofa Rigid Body Moments And Center Of Gravity L-5 </Success>
### <primary style="font-size:24px"> Coplanar Forces </primary>
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- 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.

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<footer style = "font-size: 18px">Equilibrium of a Rigid Body, Moments and Center of Gravity $\rightarrow$ Equilibrium of a rigid body $\rightarrow$ <span style = "color:blue"> Coplanar Forces </span> $\rightarrow$ Illustration $\rightarrow$ Principle of Moment </footer>

### <Success style="font-size:25px"> Equilibrium Ofa Rigid Body Moments And Center Of Gravity L-5 </Success>
### <primary style="font-size:24px"> Illustration </primary>
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- **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.

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<footer style = "font-size: 18px">Equilibrium of a rigid body $\rightarrow$ Coplanar Forces $\rightarrow$ <span style = "color:blue"> Illustration </span> $\rightarrow$ Principle of Moment $\rightarrow$ Centre of Gravity </footer>

### <Success style="font-size:25px"> Equilibrium Ofa Rigid Body Moments And Center Of Gravity L-5 </Success>
### <primary style="font-size:24px"> Principle of Moment </primary>
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- 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.

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<footer style = "font-size: 18px">Coplanar Forces $\rightarrow$ Illustration $\rightarrow$ <span style = "color:blue"> Principle of Moment </span> $\rightarrow$ Centre of Gravity $\rightarrow$ Illustration </footer>

### <Success style="font-size:25px"> Equilibrium Ofa Rigid Body Moments And Center Of Gravity L-5 </Success>
### <primary style="font-size:24px"> Centre of Gravity </primary>
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- 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.

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<footer style = "font-size: 18px">Illustration $\rightarrow$ Principle of Moment $\rightarrow$ <span style = "color:blue"> Centre of Gravity </span> $\rightarrow$ Illustration $\rightarrow$ Illustration </footer>

### <Success style="font-size:25px"> Equilibrium Ofa Rigid Body Moments And Center Of Gravity L-5 </Success>
### <primary style="font-size:24px"> Illustration </primary>
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- $ 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

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<footer style = "font-size: 18px">Principle of Moment $\rightarrow$ Centre of Gravity $\rightarrow$ <span style = "color:blue"> Illustration </span> $\rightarrow$ Illustration $\rightarrow$ Illustration </footer>

### <Success style="font-size:25px"> Equilibrium Ofa Rigid Body Moments And Center Of Gravity L-5 </Success>
### <primary style="font-size:24px"> Illustration </primary>
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- $ 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$

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<footer style = "font-size: 18px">Centre of Gravity $\rightarrow$ Illustration $\rightarrow$ <span style = "color:blue"> Illustration </span> $\rightarrow$ Illustration $\rightarrow$ Block placed on an inlcined plane </footer>

### <Success style="font-size:25px"> Equilibrium Ofa Rigid Body Moments And Center Of Gravity L-5 </Success>
### <primary style="font-size:24px"> Illustration </primary>
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- $ 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 $\angle OBE.$

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<footer style = "font-size: 18px">Illustration $\rightarrow$ Illustration $\rightarrow$ <span style = "color:blue"> Illustration </span> $\rightarrow$ Block placed on an inlcined plane $\rightarrow$ Thank you </footer>

### <Success style="font-size:25px"> Equilibrium Ofa Rigid Body Moments And Center Of Gravity L-5 </Success>
### <primary style="font-size:24px"> Block placed on an inlcined plane </primary>
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- <span style="color:green">TE:</span> $\sum \vec{F} \Rightarrow$

- f=Mg $\sin \theta $

- $N=Mg \cos \theta$

- <span style="color:green">RE:</span> $|\sum \tau|_c $=0 $\Rightarrow Nx = f \times \frac{h}{2}$

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<footer style = "font-size: 18px">Illustration $\rightarrow$ Illustration $\rightarrow$ <span style = "color:blue"> Block placed on an inlcined plane </span> $\rightarrow$ Thank you $\rightarrow$  </footer>