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.

The connecting element between these two bodies will be;

$F_{A B}=-F_{B A}$

We will learn about the procedure for solving mechanics problems.

Body/particle on which forces are acting, due to which the body accelerates.

a) Weight (gravity)

b) Contact force

Normal reaction

Friction

c) Other contact forces like

String

Spring

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.

Kinematics $\Rightarrow\vec{a}=a \hat{i}$

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

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.

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.

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, $f=\mu_s N$.

You do not know if the body is movig or not?

Applied forces are given.

Find value of f at contact surface.

Assume: No motion. i.e.a=0

FBD and kinematics,

$\sum F_x=0 \quad \sum F y=0$

Find value of f.

Find value of N.

Check if $f \leqslant \mu_s N$.

If this is satisfied assumption ok.

If $f>\mu_s N$, than no slip assumption is incorrect.

$a \neq 0$

$f=\mu_s N$

Direction has to be correct, opposite of relative slip between body and contact surface.

Acceleration will become the unknown for which problem will be solved.

$\vec{a}: \sum \vec{F}_{\text {ext }}=m \vec{a}$

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

If choose x along motion of particle,

• $\vec{a}=a\hat{i}$ or $-a\hat{i}$

When particles moves in straight line,$\vec{a} \neq 0$ if velocity is changing in magnitude with time.

If speed of particle is constant, when it moves in a straight line $\vec{a} = 0.$

$|\vec{a}|=\frac{d v}{d t}$

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.

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.

$a=\frac{v^2}{R}$, towards the center of the circle.

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

a) Centripetal acceleration towards center = $v^2/R$.

b) Tangential component = $\frac{dv}{dt}$ (tangent to the circle at that instant)

Angular velocity, v = $\omega R$.

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

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

Rate of change of angular velocity is called the angular acceleration.

$\frac{d \omega}{d t}=$ Angular acceleration

$a_t=R \alpha, \quad \alpha=\frac{d \omega}{d t}$.

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.

Friction on tyres which produce the centripetal acceleration.

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

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.

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.

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.

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.

Suppose, $\frac{m v^2}{R} < \mu_smg$, normal reaction = mg

Then the passenger will start to slip

Acceleration of the car, $a_{car}=\frac{v^2}{R}$

We have the acceleration of the passenger with respect to the car.

Let $a_p$ be the acceleration of the passenger with respect to the car.

The net acceleration of the passenger with respect to ground this will become $\frac{v^2}{R} - a_p$ in the direction toward the center of the circle.

Net acceleration of passenger w.r.t. ground =$(\frac{v^2}{R}- a_p)$

Friction force, $f=m(\frac{v^2}{R}- a_p)$ = $\mu_k N=\mu_k m g$

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.

FBD of mass 2

Tension is contact.

Acceleration of 1 and 2 are equal in magnitude.

$T-m_2 g=m_2 a$

FBD of mass 1

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

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

$(m_1-m_2) g = (m_1+m_2) a$

$a=\frac{(m_1-m_2)}{(m_1+m_2)} g$

$T=\frac{2 m_1 m_2}{(m_1+m_2)} g$

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.

Free body diagram of the person

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

$N=W^\prime$

When acceleration, a = 0, $W^\prime = mg$

If acceleration is upwards, it is positive.

$W^\prime - mg = ma$

$W^\prime = m(a+g)$

It looks as if the person has got some extra weight.

Elevator moves down with acceleration a.

$mg-W^\prime = ma_d$

$W^\prime = (mg - ma_d)$

It shows as if the person has lost weight.

If the elevator is in a free fall then,

$mg-W^\prime = mg$

$W^\prime = 0$

Elevator moves down with acceleration a

$mg-W^\prime = ma_d$

$W_1 = (mg - ma_d)$