**CBSE NOTES CLASS 11 PHYSICS CHAPTER 5**

**NEWTONS LAWS OF MOTION**

**Newton’s First Law or Law of Inertia**

An object continues to be in its state of rest or of uniform motion in a straight line unless compelled by some external force to act otherwise.

Or

If the net external force on a body is zero, its acceleration is zero. Acceleration can be non zero only if there is a net external force on the body.

**Inertia **

The property by virtue of which a body opposes any change in its state of rest or of uniform motion is known as inertia.

Greater the mass of the body, greater is its inertia. That is mass is the measure of the inertia of the body.

If F = 0; u = constant (In the absence of external applied force velocity of body remains unchanged.)

**Examples of Inertia**

- When a moving vehicle suddenly stops, passenger’s head gets jerked in the forward direction.
- When a stationery vehicle suddenly starts moving passenger’s head gets jerked in the backward direction.
- On hitting used mattress by a stick, dust particles come out of it.
- In order to catch a moving bus safely we must run forward in the direction of motion of bus.
- Whenever it is required to jump off a moving bus, we must always run for a short distance after jumping on road to prevent us from falling in the forward direction.
- A spaceship out in interstellar space, far from all other objects and with all its rockets turned off, has no net external force acting on it. If it is in motion, it must continue to move with a uniform velocity.

**Linear Momentum**

Momentum of a body is defined to be the product of its mass m and velocity $\overrightarrow{\mathrm{v}}$, and is denoted by $\overrightarrow{\mathrm{p}}$,

**$$\overrightarrow{\mathrm{p}}\mathrm{}=\mathrm{}\mathrm{m}\mathrm{}\overrightarrow{\mathrm{v}}\mathrm{}\Rightarrow \mathrm{}\mathrm{\Delta}\overrightarrow{\mathrm{p}}\mathrm{}=\mathrm{}\mathrm{m}\mathrm{}\mathrm{\Delta}\overrightarrow{\mathrm{v}}$$**

**Momentum is a vector quantity. SI unit is kg ms ^{-1}**

**Physical Significance of Momentum**

- Much greater force is needed to push a truck than a car to bring them to the same speed in same time. Similarly, a greater opposing force is needed to stop a heavy body than a lighter body in the same time, if they are moving with the same speed.
- If two stones, one light and the other heavy, are dropped from the top of a building, a person on the ground will find it easier to catch the lighter stone than the heavier stone. The mass of a body is thus an important parameter that determines the effect of force on its motion.
- A bullet fired by a gun can easily pierce human tissue before it stops, whereas the same bullet fired with moderate speed will not cause much damage. Thus for a given mass, the greater the speed, the greater is the opposing force needed to stop the body in a certain time.
- Taken together, the product of mass and velocity, that is momentum, is a relevant variable of motion. The greater the change in the momentum in a given time, the greater is the force that needs to be applied to bring in that change.
- While catching a ball,

**Newton’s Second Law of motion**

The rate of change of momentum of a body is proportional to the applied force and takes place in the direction in which force acts.

$$\overrightarrow{\mathrm{F}}\mathrm{}\propto \frac{\mathrm{\Delta}\overrightarrow{\mathrm{p}}}{\mathrm{\Delta}\mathrm{t}}\mathrm{}\mathrm{}\mathrm{}\Rightarrow \mathrm{}\overrightarrow{\mathrm{F}}\mathrm{}=\mathrm{}\mathrm{k}\frac{\mathrm{\Delta}\overrightarrow{\mathrm{p}}}{\mathrm{\Delta}\mathrm{t}}$$

As Δt → 0, we have,

$$\overrightarrow{\mathrm{F}}\mathrm{}=\underset{\mathrm{\Delta}\mathrm{t}\to \mathrm{}0}{\mathrm{lim}}\frac{\mathrm{\Delta}\overrightarrow{\mathrm{p}}}{\mathrm{\Delta}\mathrm{t}}=\mathrm{}\frac{\mathrm{d}\overrightarrow{\mathrm{p}}}{\mathrm{d}\mathrm{t}}\mathrm{}$$

For a body of fixed mass,

$$\overrightarrow{\mathrm{F}}\mathrm{}=\mathrm{}\mathrm{k}\mathrm{}\mathrm{m}\frac{\mathrm{d}\overrightarrow{\mathrm{v}}}{\mathrm{d}\mathrm{t}}\mathrm{}\mathrm{}\Rightarrow \mathrm{}\overrightarrow{\mathrm{F}}\mathrm{}=\mathrm{}\mathrm{k}\mathrm{}\mathrm{m}\mathrm{}\overrightarrow{\mathrm{a}}$$

Here k is a constant of proportionality.

The unit of force has been defined in such a way that k = 1.

$$\Rightarrow \mathrm{}\overrightarrow{\mathrm{F}}\mathrm{}=\mathrm{}\mathrm{m}\mathrm{}\overrightarrow{\mathrm{a}}$$

The SI unit of force, Newton (N), is one that causes an acceleration of 1 m s^{-2} to a mass of 1 kg,

i.e. 1 N = 1 kg m s^{-2}**.**

- The second law is consistent with the First law, i.e. F = 0 implies a = 0.
- The second law of motion is a vector law. It is equivalent to three equations, one for each component of the vectors, i.e.,
$${\mathrm{F}}_{\mathrm{x}}=\frac{\mathrm{d}{\mathrm{p}}_{\mathrm{x}}}{\mathrm{d}\mathrm{t}}=\mathrm{m}\mathrm{}{\mathrm{a}}_{\mathrm{x}}$$

$${\mathrm{F}}_{\mathrm{y}}=\frac{\mathrm{d}{\mathrm{p}}_{\mathrm{y}}}{\mathrm{d}\mathrm{t}}=\mathrm{m}\mathrm{}{\mathrm{a}}_{\mathrm{y}}$$

$${\mathrm{F}}_{\mathrm{z}}=\frac{\mathrm{d}{\mathrm{p}}_{\mathrm{z}}}{\mathrm{d}\mathrm{t}}=\mathrm{m}\mathrm{}{\mathrm{a}}_{\mathrm{z}}$$

- The force changes only the component of velocity along the direction of force. The component of velocity normal to the force remains unchanged. For example, in the motion of a projectile under the gravitational force, the horizontal component of velocity remains unchanged.
- The second law of motion is applicable to a single point particle as well as to a system of particles. The force
**F**in the law stands for the net external force on the system of particles and**a**stands for acceleration of the system as a whole. **F**at a point at a certain instant determines acceleration at that point and at that instant. Acceleration at an instant does not depend on the history of motion.- Force is not always in the direction of motion. Depending on the situation
**F**may be along**v**, opposite to**v**, normal to**v**, or may make some other angle with**v**. In every case it is parallel to acceleration. - If v = 0 at an instant, i.e., if a body is momentarily at rest, it does not mean that force or acceleration are necessarily zero at that instant. For example, when a ball thrown upward reaches its maximum height, velocity is zero, but the force continues to be its weight mg and the acceleration is ‘g’ the acceleration due to gravity.

**Examples**

(i) Body kept on horizontal plane is at rest, *F _{x}* = 0 and

*F*.

_{y}= mg(ii) Body kept on horizontal plane is accelerating horizontally under single horizontal force, *F*_{x}* = ma*_{x }and* F*_{y}* = mg.*

(iii) Body kept on horizontal plane is accelerating horizontally towards right under two horizontal forces. *(F*_{1}* > F*_{2}*), F*_{x}* = F*_{1}* - F*_{2}* = ma*_{x}* and F*_{y}* = mg.*

**Impulse**

The change in momentum of an object, when a force acts on it for a short duration, is called impulse.

*J = p*_{2}* - p*_{1}* = mv – mu*

$$\mathrm{F}\mathrm{}=\frac{\mathrm{m}\mathrm{v}\mathrm{}\u2013\mathrm{}\mathrm{m}\mathrm{u}}{\mathrm{\Delta}\mathrm{t}}$$

*⇒ F *Δ*t = mv – mu*

$$\mathrm{J}=\mathrm{}{\int}_{\mathrm{t}1}^{\mathrm{t}2}\mathrm{F}\mathrm{}\mathrm{d}\mathrm{t}$$

A large force, acting for a short time to produce a finite change in momentum is called an **impulsive force**.

**Examples of Impulsive Forces **

- Force applied by foot on hitting a football.
- Force applied by boxer on a punching bag.
- Force applied by bat on a ball in hitting it to the boundary.
- While catching a ball a player lowers his hand to save himself from getting hurt.
- A person falling on a cemented floor receives more jerk as compared to one falling on a sandy floor.

**Newton’s Third Law of Motion**

To every action, there is always an equal and opposite reaction.

Here action refers to the force applied by first body on the second body and reaction refers to the force applied by second body on the first one.

Or

Forces always occur in pairs. Force on a body A by B is equal and opposite to the force on the body B by A.

- There is no cause effect relation implied in the third law. The force on A by B and the force on B by A, act at the same instant.
- Action and reaction forces act on different bodies, not on the same body.
- Consider a pair of bodies A and B. According to the third law,

$\overrightarrow{\mathrm{F}}$_{AB} (force on A by B) = – $\overrightarrow{\mathrm{F}}$_{BA} (force on B by A)

**Examples**

- When a bullet is fired from the gun, the gun pushes the bullet forward and the bullet pushes the gun in backward direction (recoil).
- When we push any block in the forward direction then block pushes us in the backward direction with an equal and opposite force.
- Horse pulls the rod attached to the cart in the forward direction and the tension of the rod pulls the horse in the backward direction.
- Earth pulls a body on its surface in vertically downward direction and the body pulls the earth with the same force in vertically upward direction.
- While walking, we push the ground in the backward direction using static frictional force and the ground pushes us in the forward direction using static frictional force.
- When a person sitting on a horse whips the horse and horse suddenly accelerates, the saddle on the back of the horse pushes the person in the forward direction using static frictional force and the person pushes the saddle in the backward direction using static frictional force.

**Principle of Conservation of Momentum**

The total momentum of an isolated system of interacting particles is conserved.

Or

When two bodies collide, the total momentum of the system before the collision is equal to the total momentum of system after the collision,

**Proof**

Consider two bodies A and B, with initial momenta $\overrightarrow{{\mathrm{p}}_{\mathrm{A}}}$ and $\overrightarrow{{\mathrm{p}}_{\mathrm{B}}}$. The bodies collide for a duration of time Δt and get apart, with final momenta $\overrightarrow{{\mathrm{p\text{'}}}_{\mathrm{A}}}$ and $\overrightarrow{{\mathrm{p\text{'}}}_{\mathrm{B}}}$respectively.

By the second Law

$\overrightarrow{{\mathrm{F}}_{\mathrm{A}\mathrm{B}}}$_{ }Δt = $\overrightarrow{{{\mathrm{p}}^{\mathrm{\text{'}}}}_{\mathrm{A}}}$ − $\overrightarrow{{\mathrm{p}}_{\mathrm{A}}}$

and

$\overrightarrow{{\mathrm{F}}_{\mathrm{B}\mathrm{A}}}$, _{ }Δt = $\overrightarrow{{{\mathrm{p}}^{\mathrm{\text{'}}}}_{\mathrm{B}}}$− $\overrightarrow{{\mathrm{p}}_{\mathrm{B}}}$

By third law, $\overrightarrow{{\mathrm{F}}_{\mathrm{A}\mathrm{B}}}$ = − $\overrightarrow{{\mathrm{F}}_{\mathrm{B}\mathrm{A}}}$, we have

$\overrightarrow{{{\mathrm{p}}^{\mathrm{\text{'}}}}_{\mathrm{A}}}$ − $\overrightarrow{{\mathrm{p}}_{\mathrm{A}}}$ = − ($\overrightarrow{{{\mathrm{p}}^{\mathrm{\text{'}}}}_{\mathrm{B}}}$− $\overrightarrow{{\mathrm{p}}_{\mathrm{B}}}$ )

i.e. $\overrightarrow{{{\mathrm{p}}^{\mathrm{\text{'}}}}_{\mathrm{A}}}$ + ${\mathrm{p}}^{\mathrm{\text{'}}}}_{\mathrm{B}$ = $\overrightarrow{{\mathrm{p}}_{\mathrm{A}}}$ + $\overrightarrow{{\mathrm{p}}_{\mathrm{B}}}$_{,}

This could also be written as,

$${\mathrm{m}}_{1}\overrightarrow{{\mathrm{u}}_{1}}\mathrm{}+\mathrm{}{\mathrm{m}}_{2}\overrightarrow{{\mathrm{u}}_{2}}\mathrm{}=\mathrm{}{\mathrm{m}}_{1}\overrightarrow{{\mathrm{v}}_{1}}\mathrm{}+\mathrm{}{\mathrm{m}}_{2}\overrightarrow{{\mathrm{v}}_{2}}$$

**Equilibrium of a Particle**

Equilibrium (translational) of a particle refers to the situation when the net external force on the particle is zero.

[Actually equilibrium of a body requires not only translational equilibrium (zero net external force) but also rotational equilibrium (zero net external torque)].

According to the first law, this means that, the particle is either at rest or in uniform motion.

If only two forces are acting on a particle then the two forces must be equal and opposite.

If n forces $\overrightarrow{{\mathrm{F}}_{1}}$, $\overrightarrow{{\mathrm{F}}_{2}}$, …. $\overrightarrow{{\mathrm{F}}_{\mathrm{n}}}$, act on a particle, then

$\overrightarrow{{\mathrm{F}}_{1}}+\mathrm{}\mathrm{}\mathrm{}\overrightarrow{{\mathrm{F}}_{2}}+\dots \mathrm{}+\overrightarrow{{\mathrm{F}}_{\mathrm{n}}}=0$

This implies that

F_{1x} + F_{2x} + … + F_{nx} = 0

F_{1y} + F_{2y} + … + F_{ny} = 0

F_{1z} + F_{2z} + … + F_{nz} = 0

where F_{1x}, F_{1y} and F_{1z} are the components of F_{1} along *x*, *y* and *z* directions respectively and so on.

- If the net force on an object is $\overrightarrow{{\mathrm{F}}_{\mathrm{n}\mathrm{e}}}$
_{t}then$\overrightarrow{{\mathrm{F}}_{\mathrm{n}\mathrm{e}}}$

_{t}=*m*$\overrightarrow{\mathrm{a}}$

**Examples of Conservation of Momentum**

(i) Recoil of gun – when bullet is fired in the forward direction gun recoils in the backward direction.

(ii) When a person jumps on the boat from the shore of river, boat along with the person on it moves in the forward direction.

When a person on the boat jumps forward on the shore of river, boat starts moving in the backward direction.

(iii) In rocket propulsion fuel is ejected out in the downward direction due to which rocket is propelled up in vertically upward direction.

**Second Law of Motion is the Real Law of Motion**

Newton's second law is the real law of motion in the sense that the first and third law of motion can be derived from the 2nd law of motion**.**

**First Law from Second Law**

From Newton’s 2^{nd} law*, *F* = m a*

If F = 0, then *a* = 0

This means that if no force is applied on the body, its acceleration will be zero. If the body is at rest, it will remain in the state of rest and if it is moving, it will remain moving in the same direction with the same speed. Thus, a body not acted upon by an external force, does not change its state of rest or motion. This is the statement of Newton’s first law of motion.

**Third Law of Motion from Second Law of Motion**

Consider an isolated system of two bodies A and B. An isolated system is such that no external force acts on the system.

Let us consider that there is an interaction between object A and B.

Let F_{AB} bet the force applied by B on A and F_{BA} be the force applied by A on B.

Now,

$${\mathrm{F}}_{\mathrm{B}\mathrm{A}}\mathrm{}=\frac{\mathrm{d}{\mathrm{p}}_{\mathrm{B}}}{\mathrm{d}\mathrm{t}}$$

And

$${\mathrm{F}}_{\mathrm{A}\mathrm{B}}\mathrm{}=\frac{\mathrm{d}{\mathrm{p}}_{\mathrm{A}}}{\mathrm{d}\mathrm{t}}$$

Adding the two, we get,

$${\mathrm{F}}_{\mathrm{B}\mathrm{A}}+{\mathrm{F}}_{\mathrm{A}\mathrm{B}}\mathrm{}=\frac{\mathrm{d}{\mathrm{p}}_{\mathrm{B}}}{\mathrm{d}\mathrm{t}}+\frac{\mathrm{d}{\mathrm{p}}_{\mathrm{A}}}{\mathrm{d}\mathrm{t}}$$

$$=\mathrm{}\frac{\mathrm{d}{(\mathrm{p}}_{\mathrm{B}}+{\mathrm{p}}_{\mathrm{A}})}{\mathrm{d}\mathrm{t}}$$

But total momentum of the system is constant, i.e.,

$$\frac{\mathrm{d}{(\mathrm{p}}_{\mathrm{B}}+{\mathrm{p}}_{\mathrm{A}})}{\mathrm{d}\mathrm{t}}=0\mathrm{}\mathrm{}$$

$$\Rightarrow \mathrm{}{\mathrm{F}}_{\mathrm{B}\mathrm{A}}+{\mathrm{F}}_{\mathrm{A}\mathrm{B}}=0$$

$$\Rightarrow \mathrm{}{\mathrm{F}}_{\mathrm{B}\mathrm{A}}=-{\mathrm{F}}_{\mathrm{A}\mathrm{B}}$$

Free Body Diagrams

Free-body diagrams are diagrams used to show the relative magnitudes and directions of all forces acting upon an object in a given situation.

**Common Forces in Mechanics**

**Non Contact Forces**

Gravitational force, electrical force and magnetic force.

**Contact Forces**

**Normal Reaction Forces (N or R)**

When bodies are in contact (e.g. a book resting on a table, a system of rigid bodies connected by rods, hinges etc.), there are mutual contact forces, for each pair of bodies, satisfying the third law. The component of contact force normal to the surfaces in contact is called normal reaction.

- Upward bouyant force equal to the weight of the fluid displaced, the viscous force, air resistance, etc are also examples of contact forces

**Tension Force in String (T)**

Force applied by string, rope, chain, rod etc. on an object is known as tension. Tension of the string always acts away from the body to which it is attached irrespective of the direction.

**Examples of Tension **

- Flexible wire holding an object pulls the object in upward direction and pulls the point of suspension in the downward direction.
- Rope holding the bucket in the well pulls the bucket in the upward direction and the pulley in the downward direction.
- Rope attached between the cattle and the peg pulls the cattle towards the peg and peg towards the cattle.
- When a block is pulled by the chain, the chain pulls the block in forward direction and the person holding the chain in reverse direction.
- Pillars supporting the house push the house in the upward direction and push the ground in the downward direction.
- Wooden bars used in the chair push the ground in the downward direction and pushes the seating top in the upward direction.
- Parallel bars, attached to the ice-cream trolley, push the trolley in the forward direction and pushes the ice-cream vendor in the backward direction, when the trolley is being pushed by the vendor.
- Rod holding the ceiling fan pulls the fan in the upward direction and pulls the hook attached to the ceiling in the downward direction.
- Parallel rods attached between the cart and the bull pulls the cart in the forward direction and pulls the bull in the backward.
- In case of light string, rope, chain, rod etc. tension is same all along their lengths.

**Fixed Pulley**

It is a simple machine in the form of a circular disc or rim supported by spokes having groove at its periphery. It is free to rotate about an axis passing through its center and perpendicular to its plane.

In case of light pulley, tension in the rope on both the sides of the pulley is same.

T_{1} = T_{2}

**Spring Force**

Restoring force F** = - ***k x*, where *k*** **is the spring constant and ** x** is the extension or compression (displacement) in the spring.

**Frictional Forces**

Friction is property by virtue of which relative motion between two objects is resisted.

Frictional Forces are tangential forces developed between the two surfaces in contact, so as to oppose their relative motion.

**Types of Frictional Forces **

- Static frictional force
- Kinetic frictional force
- Rolling frictional force

**Static Frictional Force **

Frictional force acting between the two surfaces in contact, which are relatively at rest, so as to oppose their impending relative motion under the effect of an external force is known as static frictional force.

If there is no external force, there is no static friction.

*f*_{s}* ≤ μ*_{s}* *N* *

where *f*_{s} is the static friction,** N **is the normal reaction and μ_{s }is** **called coefficient of static friction. This is constant for two given surfaces.

The limiting value of static friction (*f*_{s})_{max}, when the object is just about to move, is independent of the area of contact and is given by

(*f*_{s})_{max} = μ_{s}N

Or 0 ≤ *f*_{s} ≤ µ_{s}N

**Kinetic Frictional Force**

Frictional force that opposes relative motion between surfaces in contact and moving relative to each other, is called kinetic or sliding friction and is denoted by *f _{k}*

*f*_{k} = µ_{k}N

*f*_{k} < (*f*_{s})_{max} ⇒ µ_{k}N < µ_{s}N

Hence **(f**_{s}**)**_{max} is also known as limiting frictional force.

**Rolling Frictional Force**

Frictional force which opposes the rolling of bodies (like cylinder, sphere, ring etc.) over any surface is called rolling frictional force and is given by µ_{r}N.

µ_{s }≥ µ_{k }≥_{ }µ_{r}

**Laws of limiting friction**

- The direction of limiting frictional force is always opposite the direction of motion.
- Limiting friction acts tangential to the two surfaces in contact.
- The magnitude of limiting friction is directly proportional to the normal reaction between the two surfaces.
- The limiting friction depends upon the material, the nature of the surfaces in contact and their smoothness.
- For any two given surfaces, the magnitude of limiting friction is independent of the shape or the area of the surfaces in contact so long as the normal reaction remains the same.

**Solving Problems Related to Forces**

- Draw a diagram showing schematically the various parts of the assembly of bodies, the links, supports, etc.
- Choose a convenient part of the assembly as one system.
- Draw a separate diagram which shows this system and all the forces on the system by the remaining part of the assembly. Include also the forces on the system by other agencies.
**Do not include the forces on the environment by the system**. - In a free-body diagram, include information about forces (their magnitudes and directions. The rest should be treated as unknowns to be determined using laws of motion.
- If necessary, follow the same procedure for another choice of the system. In doing so, employ Newton’s third law. That is, if in the free-body diagram of A, the force on A due to B is shown as
**F,**then, in the free-body diagram of B, the force on B due to A should be shown as –**F.** - While drawing the diagram choose the coordinate system in such a way so that one or more of the axes are along directions in which no motion of the object is taking place.

**Force in Circular Motion**

Acceleration of a body moving in a circle of radius R with uniform speed v is v^{2}/R directed towards the centre.

According to the second law, the force *f*, providing this acceleration is

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

This force directed forwards the centre is called the centripetal force.

For a stone rotated in a circle by a string, the centripetal force is provided by the tension in the string.

**Motion of a car on a level road**

Three forces act on the car.

(i) The weight of the car, mg

(ii) Normal reaction, N

(iii) Frictional force, *f*

As there is no acceleration in the vertical direction

N – mg = 0

⇒ N = mg

The centripetal force required for circular motion is along the surface of the road, and is provided by the component of the contact force between road and the car tyres along the surface. The static friction (since the car is not going out of the circular path) provides this force.

$$\mathrm{f}\mathrm{}\mathrm{}\le \mathrm{}{\mathrm{\mu}}_{\mathrm{s}}\mathrm{N}\mathrm{}=\frac{\mathrm{m}{\mathrm{v}}^{2}}{\mathrm{R}}\mathrm{}$$

$$\Rightarrow \mathrm{}{\mathrm{\mu}}_{\mathrm{s}}\mathrm{}\mathrm{m}\mathrm{}\mathrm{g}\mathrm{}=\frac{\mathrm{m}\mathrm{}{{\mathrm{v}}_{\mathrm{m}\mathrm{a}\mathrm{x}}}^{2}}{\mathrm{R}}$$

$$\Rightarrow \mathrm{}\mathrm{}{\mathrm{v}}_{\mathrm{m}\mathrm{a}\mathrm{x}}=\sqrt{{\mathrm{\mu}}_{\mathrm{s}}\mathrm{R}\mathrm{}\mathrm{g}}\mathbf{}$$

**Motion of a Car on Banked Road**

In case of horizontal road necessary centripetal force mv^{2}/r is provided by static frictional force. When heavy vehicles move with high speed on a sharp turn (small radius) then all the factors contribute to huge centripetal force which if provided by the static frictional force may result in the fatal accident. To prevent this roads are banked by lifting their outer edge.

Due to this, normal reaction of road on the vehicle gets tilted inwards such that it’s vertical component balances the weight of the body and the horizontal component provides the necessary centripetal force.

Since there is no acceleration along the vertical direction, the net force along this direction must be zero. Hence,

N cos θ = mg + *f*_{s} sin θ …(1)

The centripetal force is provided by the horizontal components of N and f.

N sin θ + *f*_{s} cos θ = mv^{2}/R … (2)

But *f*_{s }≤ μ_{s} N

Thus to obtain v_{max} we put *f*_{s }= μ_{s} N, in (1) and (2)

⇒ N cos θ = mg + μ_{s} N sin θ … (3) and

N sin θ + μ_{s} N cos θ = mv^{2}/R … (4)

So we get,

$$\mathrm{N}=\frac{\mathrm{m}\mathrm{g}}{\mathrm{cos}-\mathrm{}{\mathrm{\mu}}_{\mathrm{s}}\mathrm{sin}}$$

Substituting value of N in (4)

$$\frac{\mathrm{m}\mathrm{g}}{\mathrm{cos}-\mathrm{}{\mathrm{\mu}}_{\mathrm{s}}\mathrm{sin}}\times \left(\mathrm{sin}+\mathrm{}{\mathrm{\mu}}_{\mathrm{s}}\mathrm{cos}\right)=\mathrm{}\frac{\mathrm{m}{\mathrm{v}}^{2}}{\mathrm{R}}$$

$$\Rightarrow \mathrm{}{\mathrm{v}}_{\mathrm{m}\mathrm{a}\mathrm{x}}=\sqrt{\mathrm{R}\mathrm{}\mathrm{g}\left(\frac{\left(\mathrm{sin}+\mathrm{}{\mathrm{\mu}}_{\mathrm{s}}\mathrm{cos}\right)}{\mathrm{cos}-\mathrm{}{\mathrm{\mu}}_{\mathrm{s}}\mathrm{sin}}\right)}$$

$$\Rightarrow \mathrm{}{\mathrm{v}}_{\mathrm{m}\mathrm{a}\mathrm{x}}=\sqrt{\mathrm{R}\mathrm{}\mathrm{g}\left(\frac{\left({\mathrm{\mu}}_{\mathrm{s}}+\mathrm{}\mathrm{tan}\right)}{1-\mathrm{}{\mathrm{\mu}}_{\mathrm{s}}\mathrm{tan}}\right)}$$

**Take examples**

What is the acceleration of the block and trolley system, if the coefficient of kinetic friction between the trolley and the surface is 0.04? What is the tension in the string? (Take g = 10 m s^{-2}). Neglect the mass of the string.

**[NCERT]**

**Motion of a Block Sliding on an Inclined Plane**

Choose *x*-axis along OB and *y*-axis perpendicular to it.

Now balance the forces along both axes.

N = mg cos θ

*f*_{s }**=**_{ }μ_{s}** N = **μ_{s }mg cos θ

ma = mg sin θ - *f*_{s }**= **mg sin θ - μ_{s }mg cos θ

$\mathrm{a}=\mathrm{g}\mathrm{}(\mathrm{sin}-\mathrm{}{\mathrm{\mu}}_{\mathbf{s}}\mathbf{}\mathrm{}\mathrm{c}\mathrm{o}\mathrm{s}\mathrm{}$)

**Angle of Repose**

The maximum angle, at which an object can rest on an inclined plane, without sliding down, is called **angle of repose**.

For *a* = 0,

$$\mathrm{sin}-\mathrm{}{\mathrm{\mu}}_{\mathrm{s}}\mathrm{}\mathrm{}\mathrm{c}\mathrm{o}\mathrm{s}\mathrm{}=0$$

μ_{s} = tan θ or θ = tan^{-1} μ_{s}