physics-class-11-unit-05-chapter-03-forces-on-bodies-procedure-to-solve-problems-lecture-3-8-chhlqgfhbfy

### <Success style="font-size:25px"> Forces On Bodies Procedure To Solve Problems L-3 </Success>
### <primary style="font-size:24px"> Recap </primary>
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- Case where there is more than one body is involved.

- Draw FBD of each body separately apply Newton's second law to each body.

- Acceleration of body 1 is equal to the acceleration of body 2 because the length of the string does not change.

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<footer style = "font-size: 18px"> $\rightarrow$ Forces on Bodies Procedure to Solve Problems $\rightarrow$ <span style = "color:blue"> Recap </span> $\rightarrow$ Newton's Third Law $\rightarrow$ General Procedure </footer>

### <Success style="font-size:25px"> Forces On Bodies Procedure To Solve Problems L-3 </Success>
### <primary style="font-size:24px"> Block on a Table </primary>
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- For example: Block on a Table, we have the tension T.

- Draw the free body diagram.

- Due to a constraint of the problem that the body moves on a plane, acceleration is in the x direction.

- <span style="color:green">Kinematics</span> $\Rightarrow\vec{a}=a \hat{i}$

- $\sum \vec{F}=m \vec{a}$

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<footer style = "font-size: 18px">General Procedure $\rightarrow$ Forces on Particle $\rightarrow$ <span style = "color:blue"> Block on a Table </span> $\rightarrow$ Block on a Table $\rightarrow$ Block on a Table </footer>

### <Success style="font-size:25px"> Forces On Bodies Procedure To Solve Problems L-3 </Success>
### <primary style="font-size:24px"> Block on a Table </primary>
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- The force T which is being applied, either  be known or will be an unknown force.

- Similarly, the acceleration would either be given or would be an unknown.

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<footer style = "font-size: 18px">Forces on Particle $\rightarrow$ Block on a Table $\rightarrow$ <span style = "color:blue"> Block on a Table </span> $\rightarrow$ Block on a Table $\rightarrow$ Block on a Table </footer>

### <Success style="font-size:25px"> Forces On Bodies Procedure To Solve Problems L-3 </Success>
### <primary style="font-size:24px"> Block on a Table </primary>
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- Number of equations from Newton's law is 2, one in x direction and other in the y direction.

- N and f are two unknowns.

- Either T  or a, one of these will also be unknown.

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<footer style = "font-size: 18px">Block on a Table $\rightarrow$ Block on a Table $\rightarrow$ <span style = "color:blue"> Block on a Table </span> $\rightarrow$ Block on a Table $\rightarrow$ Block on a Table </footer>

### <Success style="font-size:25px"> Forces On Bodies Procedure To Solve Problems L-3 </Success>
### <primary style="font-size:24px"> Block on a Table </primary>
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- We have only two equations and three unknowns.

- Block could be at rest or moving.

- Rest condition: a = 0, So then we have two unknowns N and f.

- If the block is moving, acceleration is not known.

- In this case: f is related to N, $f=\mu_k N$.

- In case for impending slip, $f=\mu_s N$.

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### <Success style="font-size:25px"> Forces On Bodies Procedure To Solve Problems L-3 </Success>
### <primary style="font-size:24px"> Problem </primary>
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- $\vec{a}: \sum \vec{F}_{\text {ext }}=m \vec{a}$

- $\vec{a}$ : If particle moves in a straight line,

- If choose x along motion of particle,

- $\vec{a}=a\hat{i}$ or $-a\hat{i}$

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<footer style = "font-size: 18px">Revisit the problem $\rightarrow$ Revisit the problem $\rightarrow$ <span style = "color:blue"> Problem </span> $\rightarrow$ Problem $\rightarrow$ Particle Moves on a Curved Path </footer>

### <Success style="font-size:25px"> Forces On Bodies Procedure To Solve Problems L-3 </Success>
### <primary style="font-size:24px"> Particle Moves on a Curved Path </primary>
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- When particle moves on a curved path

- $\vec{a}=  \frac{d v}{d t} \hat{e}_t  +\frac{v^2}{R} \hat{e} _n$
 
- $\hat{e}_t$: Tangent to the path.
 
- $\hat{e}_n$: Normal to the path pointing to words the center of curved path.
 
- R: Radius of curvature of path.

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<footer style = "font-size: 18px">Problem $\rightarrow$ Problem $\rightarrow$ <span style = "color:blue"> Particle Moves on a Curved Path </span> $\rightarrow$ Circular Motion of Bodies $\rightarrow$ Centripetal Acceleration </footer>

### <Success style="font-size:25px"> Forces On Bodies Procedure To Solve Problems L-3 </Success>
### <primary style="font-size:24px"> Circular Motion of Bodies </primary>
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- Uniform circular motion means speed is constant.

- Acceleration, $a=\frac{v^2}{R} $

- Particle moves in the circle, the direction of acceleration keeps on changing and because the speed is constant we do not have that tangential component.

- The component of the acceleration towards the center of the circle is called centripetal acceleration.

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<footer style = "font-size: 18px">Problem $\rightarrow$ Particle Moves on a Curved Path $\rightarrow$ <span style = "color:blue"> Circular Motion of Bodies </span> $\rightarrow$ Centripetal Acceleration $\rightarrow$ Angular Velocity </footer>

### <Success style="font-size:25px"> Forces On Bodies Procedure To Solve Problems L-3 </Success>
### <primary style="font-size:24px"> Centripetal Acceleration </primary>
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- $a=\frac{v^2}{R}$, towards the center of the circle.

- If circular motion is not uniform then acceleration has 2 components.

- <span style="color:green">a) Centripetal acceleration towards center</span> = $v^2/R$.

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<footer style = "font-size: 18px">Particle Moves on a Curved Path $\rightarrow$ Circular Motion of Bodies $\rightarrow$ <span style = "color:blue"> Centripetal Acceleration </span> $\rightarrow$ Angular Velocity $\rightarrow$ Angular Acceleration </footer>

### <Success style="font-size:25px"> Forces On Bodies Procedure To Solve Problems L-3 </Success>
### <primary style="font-size:24px"> Angular Velocity </primary>
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- <span style="color:green">b) Tangential component</span> = $\frac{dv}{dt}$ (tangent to the circle at that instant)

- Angular velocity, v = $\omega R$.

- $a_c= \frac{v^2}{R}=\frac{\omega^2 R^2}{R}=\omega^2 R$

- $a_t=\frac{d v}{d t}=\frac{d}{d t}(\omega R)=\frac{d \omega R}{d t} R$

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<footer style = "font-size: 18px">Circular Motion of Bodies $\rightarrow$ Centripetal Acceleration $\rightarrow$ <span style = "color:blue"> Angular Velocity </span> $\rightarrow$ Angular Acceleration $\rightarrow$ Example </footer>

### <Success style="font-size:25px"> Forces On Bodies Procedure To Solve Problems L-3 </Success>
### <primary style="font-size:24px"> Example </primary>
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- Passenger sitting on rear seat of a car, where the car is turning left.

- We assume it is moving in a circular arc.

- We are trying to analyze the situation of the passenger.

- There is a forc,e $a=\frac{v^2}{R}$, acting in the car.

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<footer style = "font-size: 18px">Angular Velocity $\rightarrow$ Angular Acceleration $\rightarrow$ <span style = "color:blue"> Example </span> $\rightarrow$ Example $\rightarrow$ Example </footer>

### <Success style="font-size:25px"> Forces On Bodies Procedure To Solve Problems L-3 </Success>
### <primary style="font-size:24px"> Example </primary>
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- Friction on tyres which produce the centripetal acceleration.

- $f=\frac{m v^2}{R}$.

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<footer style = "font-size: 18px">Angular Acceleration $\rightarrow$ Example $\rightarrow$ <span style = "color:blue"> Example </span> $\rightarrow$ Example $\rightarrow$ Example </footer>

### <Success style="font-size:25px"> Forces On Bodies Procedure To Solve Problems L-3 </Success>
### <primary style="font-size:24px"> Example </primary>
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- We draw the free body diagram of the passenger.

- Normal reaction is coming out of the plane of the paper.

- The weight of the passenger is in the direction perpendicular to the paper.

- Normal reaction and weight will balance each other.

- There should be a friction force between the seat and the passenger.

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<footer style = "font-size: 18px">Example $\rightarrow$ Example $\rightarrow$ <span style = "color:blue"> Example </span> $\rightarrow$ Example $\rightarrow$ Example </footer>

### <Success style="font-size:25px"> Forces On Bodies Procedure To Solve Problems L-3 </Success>
### <primary style="font-size:24px"> Example </primary>
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- N and W are perpendicular to the plane of the paper.

- Friction Force: $f=\frac{m v^2}{R}$, where m is the mass of the passenger.

- We are assuming any frame connected to the surface of the earth is inertial.

- If the velocity of the car is high then $f=\frac{m v^2}{R}$ increases, may exceed $\mu_s N$

- Then friction force will not be able to stop  the relative motion of the body.

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<footer style = "font-size: 18px">Example $\rightarrow$ Example $\rightarrow$ <span style = "color:blue"> Example </span> $\rightarrow$ Example $\rightarrow$ Example </footer>

### <Success style="font-size:25px"> Forces On Bodies Procedure To Solve Problems L-3 </Success>
### <primary style="font-size:24px"> Example </primary>
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- If$\frac{m v^2}{R}>\mu_s N$, then the passenger will start to slip and friction force is acting in a direction.

- That will be the direction of the relative slip.

- So the unbalanced external force acting on the passenger is the force of friction between the seat and the passenger.

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<footer style = "font-size: 18px">Example $\rightarrow$ Example $\rightarrow$ <span style = "color:blue"> Example </span> $\rightarrow$ Example $\rightarrow$ Example </footer>

### <Success style="font-size:25px"> Forces On Bodies Procedure To Solve Problems L-3 </Success>
### <primary style="font-size:24px"> Example </primary>
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- F = ma  =$\frac{m v^2}{R}$

- So friction provides the acceleration to the passenger and the direction of friction is towards the center of the circle.

- If the velocity is such that, $\frac{m v^2}{R} < \mu_s N$,  then we have a case of no slip, $N=mg$.

- $\frac{m v^2}{R} < \mu_smg \Rightarrow \frac{v^2}{R g}<\mu_s $, No slip.

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<footer style = "font-size: 18px">Example $\rightarrow$ Example $\rightarrow$ <span style = "color:blue"> Example </span> $\rightarrow$ Example $\rightarrow$ Example </footer>

### <Success style="font-size:25px"> Forces On Bodies Procedure To Solve Problems L-3 </Success>
### <primary style="font-size:24px"> Problem </primary>
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- Consider is there is a pulley on which we have a string connecting two masses $m_1$ and $m_2$, string is mounted on a pulley. String is light inextansible, frictionless contact with pulley.

- $m_1$ = 5kg

- $m_2$ = 4kg

- Find the tension in the string and the magnitude of acceleration a of the blocks.

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<footer style = "font-size: 18px">Example $\rightarrow$ Example $\rightarrow$ <span style = "color:blue"> Problem </span> $\rightarrow$ Solution $\rightarrow$ Solution </footer>

### <Success style="font-size:25px"> Forces On Bodies Procedure To Solve Problems L-3 </Success>
### <primary style="font-size:24px"> Solution </primary>
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- <span style="color:blue">FBD of mass 2</span>

- Tension is contact.

- Acceleration of 1 and 2 are equal in magnitude.

- $T-m_2 g=m_2 a$

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<footer style = "font-size: 18px">Example $\rightarrow$ Problem $\rightarrow$ <span style = "color:blue"> Solution </span> $\rightarrow$ Solution $\rightarrow$ Solution </footer>

### <Success style="font-size:25px"> Forces On Bodies Procedure To Solve Problems L-3 </Success>
### <primary style="font-size:24px"> Solution </primary>
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- <span style="color:blue">FBD of mass 1</span>

- $T-m_2 g=m_2 a$ -----(1)

- $m_1 g-T=m_1 a$ -----(2)

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<footer style = "font-size: 18px">Problem $\rightarrow$ Solution $\rightarrow$ <span style = "color:blue"> Solution </span> $\rightarrow$ Solution $\rightarrow$ Problem </footer>

### <Success style="font-size:25px"> Forces On Bodies Procedure To Solve Problems L-3 </Success>
### <primary style="font-size:24px"> Problem </primary>
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- Passenger of mass m, stands on an elevator on a weighing scale.

- Acceleration of elevator $W'$ may or may not be equel to mg, where m = mass of passenger.

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<footer style = "font-size: 18px">Solution $\rightarrow$ Solution $\rightarrow$ <span style = "color:blue"> Problem </span> $\rightarrow$ Solution $\rightarrow$ Solution </footer>

### <Success style="font-size:25px"> Forces On Bodies Procedure To Solve Problems L-3 </Success>
### <primary style="font-size:24px"> Solution </primary>
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- <span style="color:blue">Free body diagram of the person</span>

- Acceleration of the elevator as seen from a ground frame = a

- a = 0, the elevator is at rest.

- N is the normal reaction from the weighing scale.

- N is the force being applied by the person

- $N=W^\prime$

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<footer style = "font-size: 18px">Solution $\rightarrow$ Problem $\rightarrow$ <span style = "color:blue"> Solution </span> $\rightarrow$ Solution $\rightarrow$ Solution </footer>

### <Success style="font-size:25px"> Forces On Bodies Procedure To Solve Problems L-3 </Success>
### <primary style="font-size:24px"> Free fall </primary>
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- If the elevator is in a free fall then,

- $mg-W^\prime = mg$

- $W^\prime = 0$

- <span style="color:green">Elevator moves down with acceleration a</span>

- $mg-W^\prime = ma_d$

- $W_1 = (mg - ma_d)$

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<footer style = "font-size: 18px">Solution $\rightarrow$ Solution $\rightarrow$ <span style = "color:blue"> Free fall </span> $\rightarrow$ Thank You $\rightarrow$  </footer>