It means that when the object 1 exerts force on the object 2, the object 2 exerts equal and opposite force on the body 1 at the same instant.
(a) Hammer and the nail
(b) Ball bouncing off the wall
(c) Walking on the floor with friction.

(i) The car is moving with uniform velocity v with respect to a person standing (at rest) on the ground.
(ii) As the car is moving with constant velocity with respect to ground to the person is at rest on the ground, both frames (with respect to the car and to the ground) are inertial frames.

False

It is to avoid slipping. Smaller step causes more normal force and thereby more friction.

The surface area of parachute is very large. And when it descends downwards, the air provides resistance to it and so it descends slowly.

Limiting static/kinetic friction between any pair of dry unlubricated surfaces is independent of apparent area in contact.
Limiting static/kinetic friction between any pair of unlubricated surfaces depends on nature of material of the surfaces in contact.
Limiting static/kinetic friction between any pair of dry unlubricated surfaces is directly proportional to the normal reaction.
Static Friction: If Fs is static friction and R (N) is a normal reaction
Then
F_s ∝ R
∴ F_s = μ_sR
μ_s = coefficient of static friction.
∴ μ_s = \(\frac{F_s}{R}\)
The coefficient of static friction is defined as the ratio of the static friction to normal reaction
Kinetic friction: If Fs is static friction and R (N) is a normal reaction 
Then
F_k ∝ R
∴ F_k = μ_sR
μ_k = coefficient of static friction.
∴ μ_k = \(\frac{F_k}{R}\)
The coefficient of kinetic friction is defined as the ratio of the kinetic friction to normal reaction

The centrifugal force appears to act on the particle, only when we analyses the motion from a rotating frame. With respect to an inertial frame there is only centripetal force which is given by the tension in the string. For this reason centrifugal force is called as a 'pseudo force'. A pseudo force has no origin, It arises due to the non-inertial nature of the frame considered.

When a vehicle travels in a curved path, there must be a centripetal force acting on it. This centripetal force is provided by the frictional force between tyre and surface of the road. Consider a vehicle of mass 'm' moving at a speed 'v' in the circular track of radius 'r'. There are three forces acting on the vehicle when it moves as shown in the figure.
(i) Gravitational force (mg) acting downwards
(ii) Normal force (mg) acting upwards
(iii) Frictional force (F_s) acting horizontally inwards along the road
Suppose the road is horizontal then the normal force and gravitational force are exactly equal and opposite. The centripetal force is provided by the force of static friction Fs between the tyre and surface of the road which acts towards the center of the circular track,
\(\frac { m{ v }^{ 2 } }{ r } ={ F }_{ s }\)
As we have already seen in the previous section, the static friction can increase from zero to a maximum value
F_s ≤ μ_smg
There are two conditions possible namely without skidding and skidding.
For without skidding \(\frac { m{ v }^{ 2 } }{ r } \le { \mu }_{ s }mg\), or \({ \mu }_{ s }\ge \frac { { v }^{ 2 } }{ rg } \) or \(\sqrt { { \mu }_{ s }rg } \ge v\)
The static friction would be able to provide necessary-centripetal force to bend the car on the road.

(i) One of the important applications is suitcases with rolling on coasters. Rolling wheels makes it easier than carrying luggage.
(ii) When an object moves on a surface, essentially it is sliding on it. But wheels move on the -surface through rolling motion.
(iii) In rolling motion when a wheel moves on a surface, the point of contact with surface is always at rest.
(iv) Since the point of contact is at rest, there is no relative motion between the wheel and surface. Hence the frictional force is very less. At the same time if an object moves without a wheel, there is a relative motion between the object and the surface.
(v) As a result frictional force is larger. This makes it difficult to move the object.
(vi) Ideally in pure rolling, motion of the point of contact with the surface should be at rest, but in practice it is not so.
(vii) Due to the elastic nature of the surface at the point of contact there will be some deformation on the object at this point on the wheel or surface.
(viii) Due to this deformation, there will be minimal friction between wheel and surface. It is called 'rolling friction'. In fact, 'rolling friction' is much smaller than kinetic friction.

i) Vertical motion
Consider two blocks of masses m₁ and m₂ (m₁ > m₂) connected by a light and inextensible string that passes over a pulley as shown in fig. A little consideration will show, that the greater mass m₁ will move downwards, whereas the smaller one will move upwards. Since the string is inextensible, the upward acceleration of the mass m₂ will be equal to the downward acceleration of the mass m₁

Condition	Acceleration	Tension
	T - m2g = m2a .......(1) T - m1g = m1a   m1g - T = m1a .......(2) Adding equations (1) and (2), we get m1g - m2g = m1a + m2a (m1 - m2)g = (m1 + m2)a  .......(3) (vi) From equation (3), the acceleration of both the masses is \(a=\left( {{m_1-m_2}\over{m_1+m_2}} \right)g\) .......(4) If both the masses are equal (m1 = m2), from equation then a = 0	T - m2g = m2a .......(1)\(a=\left({m_1-m_2 \over m_1+m_2 } \right)g\)  .......(4) Substituting the equation (4) into(1) \(T-m_2g=m_2\left({m_1-m_2 \over m_1+m_2 } \right)g\) \(T=m_2 g\left(1+{m_1-m_2 \over m_1+m_2 } \right)\) \(T=g\left({2m_1m_2\over m_1+m_2} \right)\)

ii) Horizontal Motion: Mass m, is kept on a horizontal table and mass m, is hanging through a small pulley as shown in the figure. Assume that there is no friction on the surface. As both blocks are connected to the unstretchable string, if m, moves with an accelerating 'a' downward them m₂ move with the same acceleration 'a' horizontally.

Condition	Acceleration	Tension
(i) The forces acting on mass m2 are 1. Downward gravitational force (m2g) 2. Upward normal force (N) exerted by the surface 3. Horizontal tension (T) exerted by the string. (ii) The forces acting on mass m1 are 1. Downward gravitational force (m1g). 2. Tension (T) acting upwards	T-m1g= -m1a ...........(1) T = m2 a  .......(2) There is no acceleration along y-direction for m2 N - m2g = 0 ⇒ N= m2g .... (3) m2a-m1g= -m1a m2a+m1a= m1g \(a={m_1 \over m_1+m_2}g\)    ......(4)	Tension in the string can be obtained by substituting equation (4) in equation (2). \(T=m_2({m_1 \over m_1+m_2})g\) \(T=g({m_1m_2 \over m_1+m_2})\) This result has an important application in industries. The ropes used in conveyor belts (horizontal motion) work for longer duration than those of cranes and lifts (vertical motion).

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

(i) Centrifugal force is a pseudo force, its effects are real.
(ii) When a car takes a turn in a curved road, person inside the car feels an outward force which pushes the person away.
(iii) This outward force is also called centrifugal force.
(iv) If there is sufficient friction between the person and the seat, it will prevent the person from moving-outwards.

(i) If a particle is in uniform circular motion, there must be centripetal acceleration towards the center of the circle.
(ii) If there is acceleration then there must be some force acting on it with respect to an inertial frame. This force is called centripetal force.
(iii) The centripetal acceleration of a particle in the circular motion is given by \(a=\frac { { V }^{ 2 } }{ r } \) and it acts towards center of the circle. According to Newton's second law, the centripetal force is given by
\({ F }_{ Cp }={ ma }_{ cp }=\frac { { MV }^{ 2 } }{ r } \quad a=\frac { { V }^{ 2 } }{ r } \)  
(iv) The word Centripetal force means center seeking force in vector notation \({ F }_{ Cp }=\frac { { MV }^{ 2 } }{ r } \hat { r } \). For Uniform circular Motion  \({ F }_{ Cp }=-m\omega ^{ 2 }r\hat { r } \)