The object is at rest with respect to Earth (inertial coordinate system). There are two forces that act on the object.

(i) Gravity acting downward (negative y-direction)
(ii) Normal force by the surface of the Earth acting upward (positive y-direction) The free body diagram for this object is

\(\vec { F } _{ g }=-mg\hat { j } \)
\(\vec { N } =N\vec { j } \)
Net force \({ \vec { F } }_{ net }=-mg\hat { j } +N\vec { j } \)
But there is no acceleration on the ball
So \(\vec { a }\) = 0. By applying newton's second law \(\left( { \vec { F } }_{ net }=m\vec { a } \right) \)
Since \(\vec { a } =0,{ \vec { F } }_{ net }=-mg\hat { j } +N\vec { j } \)
\((-mg+N)\vec { j } =0\)
(iii) By comparing the components on both sides of the equation we get, 
-mg + N = 0
N = mg
We can conclude that if the object is at rest, the magnitude of the normal force is exactly equal to the magnitude of gravity

In a leveled circular road, skidding mainly depends on the coefficient of static friction \({ \mu }_{ s }\). The coefficient of static friction depends on the nature of the surface which has a maximum limiting value. If the speed of car exceeds this safe speed, then it starts to skid outward but frictional force comes into effect and provides an additional centripetal force to prevent the outward skidding. 
\(tan \ \theta> { \mu }_{ s }\)
When the tangent of the angle of banking is greater than the coefficient of friction, skidding occurs.

(a) A frame of reference which is at rest or which is moving with a uniform velocity along a straight line is called an inertial frame of reference.
(b) In the inertial frame of reference Newton's laws of motion holds good.
Example: The lift at rest, lift moving (up or down) with constant velocity, car moving with constant velocity on a straight road.

Mass of a bullet (m) = \(\frac { 40 }{ 1000 } kg\) = 0.04kg
Initial velocity (u) = 250 ms^-1
Final velocity (v) = 0.
displacement (s) = 20cm = 20\(​​\times\)10^-2m
Acceleration a bullet \(a=\frac { { v }^{ 2 }-{ u }^{ 2 } }{ 2s } \)
\(=\frac { { 0 }^{ 2 }-\left( 250 \right) ^{ 2 } }{ 2\times (20\times 10^{ -2 }m) } \)
= \(\frac { -250\times 250 }{ 2\times 20\times { 10 }^{ -2 }m } \)
= \(\frac { -62,50,000 }{ 40\times { 10 }^{ -2 } } \)
Acceleration 'a'  = - 156.25 x10⁵ ms^-2
Now, Average force F = ma
F = 0.04 \(​​\times\)(-156.25 x 10⁵)
= 6.25\(​​\times\)10⁵N

Mass of a car = 1000 kg
Initial velocity u = 90 kmh^-1 = 25 m/s
Final velocity v = 0
Time t = 40s
Impulse = change in momentum
= m (v - u)
Impulse = 1000 (0 - 25) = - 25\(​​\times\)10³

The common acceleration of the masses is,
 \(a=\frac { \left( { m }_{ 1 }-{ m }_{ 2 } \right) }{ \left( { m }_{ 1 }+{ m }_{ 2 } \right) } \times g\)     [\(\therefore\) g \(\longrightarrow \) acceleration due to gravity]
= \(\frac { \left( 0.5-0.5 \right) }{ \left( 0.5+0.5 \right) } \times 9.8\)
= \(\frac { 0 }{ 1 } \times 9.8\)
The acceleration of the masses = 0

(i) Choose a suitable inertial coordinate system to analyse the problem. For most of the cases we can take Earth as an inertial coordinate system.
(ii) Identify the system to which Newton's laws need to be applied. The system can be a single object or more than one object.
(iii) Draw the free body diagram
(iv) Once the forces acting on the system are identified, and the free body diagram is drawn, apply Newton's second law. In the left hand side of the equation, write the forces acting on the system in vector notation and equate it to the right hand side of equation which is the product of mass and acceleration. Here, acceleration should also be in vector notation.
(v) If acceleration is given, the force can be calculated. If the force is given, acceleration can be calculated.

Mass of a body m = 2 kg
Magnitude of the two force, F₁= 3N and F₂ =5N,
Angle (\(\theta\) ) = 90°
Resultant force on the body = 
F = \(\sqrt { { F }_{ 1 }^{ 2 }={ F }_{ 2 }^{ 2 }+2{ F }_{ 1 }{ F }_{ 2 }cos\theta } \)
= \(\sqrt { { 3 }^{ 2 }+{ 5 }^{ 2 }+2\times 3\times 5\times cos90° } \)
= \(\sqrt { 9+25+2\times 3\times 5\times 0 } \)   \(\left[ \because \cos90°=0 \right] \)
= \(\sqrt { 9+25 } \)
F = \(\sqrt { 34 } \)
Net acceleration experienced by the body
a =\(\frac { F }{ m } \)
= \(\frac { \sqrt { 34 } }{ 3.5 } \) = 1.66 ms^-2

When the lift is moving down with an acceleration
' a ' the weight of the person W is, given by
W = mg = ma = m(g-a)
Fractional change is the weight of the person
= \(\frac { mg-w }{ mg } \)
= \(1-\frac { w }{ mg } -\frac { mg-(mg-ma) }{ mg } \)

 \(=\frac { a }{ g } =\frac { 5.0 }{ 9.8 } =0.51\)
=Percentage change = 0.51\(\times\)100 
= 51 % 

Force F = F₁ = F₂
F₁ = \(\frac { dp }{ { dt }_{ 1 } } ,{ F }_{ 2 }=\frac { dp }{ { dt }_{ 2 } } \)
\({ F }_{ 2 }={ F }_{ 1 }=\frac { { dt }_{ 1 } }{ { dt }_{ 2 } } \)
\(=10\times \frac { 8 }{ 4 } =20N\)