$F_{B E}$ and $F_{B T}$ cancel each other $\rightarrow$ Book is at rest.

$F_{B E}$ and $F_{B T}$ do NOT constitute an action reaction pair.

Both $F_{B E}$ and $F_{B T}$ act on the same body $\rightarrow$ book.

Action and Reaction on pair are on 2 different bodies.

$F_{B E} \rightarrow$ Force on book by earth.

Reaction to $F_{B E}$ is $F_{E B}$ the force on earth by book.

$F_{BT} \rightarrow$ Force On book by table.

1. Forces which act from a distance

2. Arise out of intraction with a body 2 where body 1 and body 2 are not in contact

Any body which is on surface of earth or close to the surface will experience a gravitational force due to earth

$m$ = mass of earth

$R_e$ = radius of earth

$F_{b e}=\frac{G m_b M}{r^2}$

If body on surface of earth, r=$R_e$

$R_e =6370 \mathrm{km} =6.37 \times 10^6 \mathrm{m}$

$F_{b_e} =\frac{G M}{R_e^2} \cdot{m_b}$

$\frac{G M}{R_e^2}$ = g

g is called the gravitational constant.

$\frac{G M}{R_e^2} \rightarrow g$

g is called the gravitational constant.

$g=9.81 \frac {m}{s^2}$

$F_{b e}=m_b g \rightarrow$ weight.

$F=\frac{1}{4 \pi \epsilon_0} \frac{q_1 q_2}{r^2}$

$q_1, q_2 \rightarrow$ charges

(Force due to $q_1$ on $q_2$ or $q_2$ on $q_1$ )

$F=\frac{q}{C} (\vec{v} \times \vec{B})$

Where:

$\vec{v} \rightarrow$ Velocity

$\vec{B} \rightarrow$ Magnetic field intensity

Three basic forces which act from a distance gravity, electrostatic forces and electromagnetic forces.

We have a body A and body B.

Let the velocity of body $A$ is $v_A$ and the velocity of body $B$ is $v_B$.

Both are the tangential components of the velocities.

"t" shows that it is a tangential component.

The two bodies are not slipping with respect to each other.

$v_{A,t}$= $v_{B,t}$ at time t, at rest time $t^+$

This condition is said to be no slip.

Slip at time t, $v_{A, t} \neq v_{B, t}$

Impending slip at time t: $v_{A, t} = v_{B, t}$

$v_{A, t}=v_{B, t}$ at time t

But at time $t^{+}, v_{A t} \neq v_{B, t}$

Example: Duster is kept on the block

Tangential force it will try to stop the relative motion

When no other force is being applied on this duster, it has two forces, weight of the duster and the contact force which will have two components a normal reaction and a friction

When we apply a small force the body does not move that means the friction force is automatically acting on the tangential surface which opposes the motion.

Frictional force is a self adjusting force and it tries to oppose the motion.

As F increases reach a stage where duster is about to move, at this stage we have the state of impending slip.

Coulomb's laws say that, when we have impending slip that means the body is just about to move then the force of friction is equal to a constant = $\mu_s \times N$

Where N is the normal force or normal reaction on the body

Body is just about to move

$f=\mu_s N$

where:

$\mu_s \rightarrow$ coefficient static friction

N $\rightarrow$ Normal force (reaction body)

$v_A$=5 $\hat{i}$

$v_B$=0 $\hat{i}$

Velocity of A with respect to B = $v_A-v_B$=5$\hat{i}$

$a_A=2 {m}/{s^2} \hat{i}$

$a_B=2 {m}/{s^2} \hat{i}$

$\vec{v}_{A / B}=5-5=0$

$\vec{a}_{A B}=2-2=0$

No slip

Slip: $\vec{v}_A=\vec{V}_B$, not slipping.

$\overrightarrow{a_A}-\overrightarrow{a_B}=a_{A / B}=(2-3) \hat{i} = -\hat{i}$

Friction force on body $\rightarrow \hat{i}$.