class 11 physics notes chapter 5 laws of motion 

Chapter 5 Laws of Motion :  Intuitive concept of force, Inertia, Newton’s first law of motion; momentum
and Newton’s second law of motion; impulse; Newton’s third law of motion. Law of conservation of linear momentum and its applications. Equilibrium of concurrent forces, Static and kinetic friction, laws of friction, rolling friction, lubrication. Dynamics of uniform circular motion: Centripetal force, examples of circular motion (vehicle on a level circular road, vehicle on a banked road).

Hence equal massive bodies having larger velocity will have larger momentum.

•      When two objects of masses  such that  moves with same velocity then

    

As    

Hence the two bodies having same velocities will have larger momentum which have larger mass.

3. When two bodies have equal linear momentum then 

  

Or   

Or  

If   

Thus velocities of the bodies are inversely proportional to masses of the bodies i.e. heavier body will have smaller velocity and smaller body will have larger velocity.

5. Newton’s Laws of Motion: ^M.Imp.

1. Newton’s First Law of Motion:

A body in rest will remain in rest or a body is in motion will remain in motion till any external force is applied on the body.

     How Newton’s first law defines force?

According to Newton’s first law, force is external agency which changes the state of rest or uniform motion of the body. Hence this law gives the qualitative definition of force.

     Newton’s first law sometimes called law of inertia.

2.               Newton’s Second Law of Motion:

According to Newton’s second law of motion, the rate of change of linear momentum is directly proportional to applied force.

Mathematically

Or      ()

Or  

Or  

Or  

Where 

  

Force is a vector quantity. 

Its dimensional formula is 

Unit of Force: There are two types of unit of Force

• Absolute Unit of Force:

 An absolute unit of force is defined as the force which produces unit acceleration in unit mass of a body. In S.I. absolute unit of force is Newton ()

 In absolute unit of force is dyne (1 dyne)

• Gravitational Units of Force:

A gravitational unit of force is defined as the force which produces   (acceleration on due to gravity) acceleration in a body of unit mass.

In S.I. the gravitational unit of force is Kilogram weight  or Kilogram force 

In  the gravitational unit of force is gram weight  or gram force 

Relation between Newton and dyne

Or  

Difference between Gravitational unit and absolute unit of force:

     The absolute units of force remains the same throughout the universe but the Gravitational unit depends upon  so changes from place to place.

     A gravitational unit is  times the corresponding absolute unit. Gravitational unit is also called weight or practical unit of force.

Question: 3 A stone of mass 0.05 kg is thrown vertically upwards. Give the direction and magnitude of the net force on the stone,

(a) during its upward motion, .

(b) During its downward motion,
(c) at the highest point where it is momentarily at rest. 

Do your answers change if the stone was thrown at an angle of 45° with the horizontal direction? 

Answer: (a) When the stone is moving upward, the acceleration g is acting downward, so the force is acting downward is equal to F = mg = 0.05 kg x 10 ms^-2 = 0.5 N.
(b) In this case also F = mg = 0.05 x 10 = 0.5 N. (downwards).
(c) The stone is not at rest at highest point but has horizontal component of velocity. The direction and magnitude of the net force on the stone will not alter even if it is thrown at 45° because no other acceleration except ‘g’ is acting on stone.

Question: 4 give the magnitude and direction of the net force acting on a stone of mass 0.1 kg,

(a) just after it is dropped from the window of a stationary train,
(b) just after it is dropped from the window of a train running at a constant velocity of 36km/h,
(c) just after it is dropped from the window of a train accelerating with 1 ms^-2,
(d) lying on the floor of a train which is accelerating with 1 m s~2, the stone being at rest relative to the train. 
Answer: (a) Mass of stone = 0.1 kg
Net force,  = 0.1 x 10 = 1.0 N. (vertically downwards).
(b) When the train is running at a constant velocity, its acceleration is zero. No force acts on the stone due to this motion. Therefore, the force on the stone is the same (1.0 N.).
(c) The stone will experience an additional force F’ (along horizontal) 

i.e   = 0.1 x l = 0.1 N
As the stone is dropped, the force F’ no longer acts and the net force acting on the stone  = 0.1 x 10 = 1.0 N. (vertically downwards).
(d) As the stone is lying on the floor of the train, its acceleration is same as that of the train. Force acting on the stone, = 0.1 x 1 = 0.1 N. It acts along the direction of motion of the train.

Question: 5 One end of a string of length l is connected to a particle of mass m and the other to a small peg on a smooth horizontal table. If the particle moves in a circle with speed v the net force on the particle (directed towards the centre) is:
(i) T,   (ii) T – mv²/l,   (iii) T +mv²/l,     (iv) 0
T is the tension in the string. [Choose the correct alternative].
Answer: (i) T
The net force T on the particle is directed towards the centre. It provides the centripetal force required by the particle to move along a circle.

Question: 6 A constant retarding force of 50 N is applied to a body of mass 20 kg moving initially with a speed of 15 ms^-3. How long does the body take to stop?
Answer:  Here m = 20 kg, F = – 50 N (retardation force) 

As F = ma     

So 

As 

Using equation         

We get   

Question: 7 constant forces acting on a body of mass 3.0 kg changes its speed from 2.0 ms^-1 to 3.5 ms^-1 in 25 s. The direction of the motion of the body remains unchanged. What is the magnitude and direction of the force?
Answer:  given   ,  ,  t=25s  

As we know  

 

here the force is in direction of motion.

Question: 8 A bodies of mass 5 kg is acted upon by two perpendicular forces 8 N and 6 N. Give the magnitude and direction of the acceleration of the body.
Answer: here 

The resultant force on the body

 =10N

So    in direction of force

The direction of acceleration may be given by

     

So    

Question: 9 the driver of a three-wheeler moving with a speed of 36 km/h sees a child standing in the middle of the road and brings his vehicle to rest in 4.0 s just in time to save the child. What is the average retarding force on the vehicle? The mass of the three-wheeler is 400 kg and the mass of the driver is 65 kg.
Answer: Here mass of three-wheeler m_i = 400 kg, 

Mass of driver = m₂ = 65 kg, 

Initial speed of auto, u = 36 km/h = 36 x = 10 ms^-1, 

Final speed, v = 0 and t = 4s.

As acceleration 

  

Now 

Here –ve sign shows that force is of retarding nature.

3.  Newton’s Third Law of Motion:

It states that to every action, there is always an equal and opposite reaction.

Suppose  is the amount of force exerted by body   on body

  Is the amount of force on   due to  

Then from Newton’s Third law of motion   

I.e. Force on 

e.g.: During swimming a person pushes water back word then the water pushes person in forward direction.

     Here Newton’s third law shows that a single force cannot exist in nature, the force exists in pairs.

     Newton’s third law of motion is applicable irrespective of the nature of the forces.

     Action and reaction always acts on the different bodies.

     The force of action and reaction cannot cancel each other because act on different bodies.

     No action act occurs in the absence of reaction.

Illustrations of Newton’s Third Law of Motion:

     A book kept on table exert a force  due to weight on the table in down ward direction due to which a equal and opposite force  is Act on book due to table in upward direction. Hence both remain in equilibrium.

     While walking, we press the ground in backward direction, them the ground exerts an equal and opposite force on us due to which we walks.

     It is difficult to walk on slippery ground or sand because we are unable to exert force due to which reaction is not sufficient to walk.

6. Horse and Cart Problems:

Suppose a cart connected to a horse by a string. Suppose horse while pull the cart produce T tension in the string, same tension is produce by cart in opposite direction.

Now again suppose that the horse pushes the ground with a force F and R is normal reaction. Resolving R into rectangular components we get:-

Hence if is greater then  then horse will pull cart & if  is greater then  than cart will pull horse.

     On an inclined plane cart may pull horse.

Question :10  Explain why
(a) a horse cannot pull a cart and run in empty space,
(b) passengers are thrown forward from their seats when a speeding bus stops suddenly,
(c) it is easier to pull a lawn mower than to push it,
(d) a cricketer moves his hands backwards while holding a catch.
Answer: (a) A horse by itself cannot move in space due to law of inertia and so cannot pull a cart in space.
(b) The passengers in a speeding bus have inertia of motion. When the bus is suddenly stopped the passengers are thrown forward due to this inertia of motion.
(c) In the case of pull, the effective weight is reduced due to the vertical component of the pull. In the case of push, the vertical component increases the effective weight.
(d) The ball comes with large momentum after being hit by the batsman. When the player takes catch it causes large impulse on his palms which may hurt the cricketer. When he moves his hands backward the time of contact of ball and hand is increased so the force is reduced.

7. Show that Newton’s second law is real law of motion? ^M.Imp

1.                 First law is contained in second law:-

According to Newton’s second law   

 

In the absence of external force     

  Here  

Or  means 

Hence if no external force on the body then if body is rest will remain is rest or if body is moving with uniformly then will remain in motion which is first law. 

2.              Third law is contained in second law:-

Consider an isolated system of two bodies. Let  is force on  and  is force on.

Then from second law   

    

     

    

    

This is Newton’s third law of motion.

Hence both Newton’s first law and third law can be explained by second law, so we can say Newton’s second law is real of motion or fundamental law of motion.

Question: 11 a body of mass 0.40 kg moving initially with a constant speed of 10 ms^-1 to the north is subject to a constant force of 8.0 N directed towards the south for 30 s. Take the instant the force is applied to be t = 0, the position of the body at that time to be x = 0, and predict its position at t = -5 s, 25 s, 100 s. 
Answer: given that m= 0.40 kg,

 

 As  

Also      

     Position at t=-5s  is

 

     Position at t=-25s  is

 

     Position at t=-100s  is

At t=30s,

so   v= 10+ (-20X30) =-590m/s

Now motion from 30s to 100s

 

So total distance

Question: 12   A truck starts from rest and accelerates uniformly at 2.0 ms^-2. At t = 10 s, a stone is dropped by a person standing on the top of the truck (6 m high from the ground). What are the (a) velocity, and (b) acceleration of the stone at t = 11s? (Neglect air resistance.)
Answer: here   u = 0, a = 2 ms^-2, t= 10 s
Using equation, ,

we get 

(a) Initial velocity along x axis   ms^-1Let us now consider vertical motion which is controlled by force of gravity.
=0, a = g = 10 ms^-2, t = (11 — 10) s = 1 s

Using  

we get

Resultant velocity

The direction of resultant velocity can be given by

 

 along horizontal direction.

(b) The moment the stone is dropped from the car, horizontal force on the stone is zero. The only acceleration of the stone is that due to gravity.

This gives a vertically downward acceleration of 10 ms^-2. This is also the net acceleration of the stone.

8. Impulse: ^imp

It is the total effect of a large force which act on a body for a very short time and is equal to the product of force and time. Some time it is called action of force.

Mathematically impulse may be defined as the product of force and time.

From Newton’s second law        

            

 

Integrating both side          

 

 

 

Here   Ft is called impulse, so impulse may be defined as the change in momentum of a body.

Unit:- in S.I. system the unit of impulse is NS or Kgm/s & in c,g,s system dyne/sec or gcm/s.

Usefulness of concept of Impulse:

By the knowledge of impulse we can find out the time for which a force is acted on the body.

9. Application of concept of Impulse: ^imp

1.                 A cricket player lower his hand while catching a ball because by doing this he increase the time of catch due to which he have to apply smaller force to catch and hence in impulse  felled by person decrease.
2.              A person falling from a height on a cemented floor get more injuries as compared when he fall on a help of sand because the time of action of force on sandy surface is large as compared to cement surface.
3. automobiles are provided with shockers because shockers increase the time of action of force due to which we feel small Jerk.
4.            Buffers are provided between the bogies of a train because they increase the time of Jerk due to which action of force decreases and hence we feel small Jerk.
5.              Chinaware’s are packed in straw paper because they increase the time of experience the Jerk during transportation. Hence they strike each other with a small force and get loss damaged.

Question: 13 two billiard balls, each of mass 0.05 kg, moving in opposite directions with speed 6 ms^-1collide and rebound with the same speed. What is the impulse imparted to each ball due to the other?
Answer: Initial momentum of each ball before collision

= 0.05 x 6 kg ms^-1 = 0.3 kg ms^-1
Final momentum of each ball after collision

= – 0.05 x 6 kg ms^-1 = – 0.3 kg ms^-1 

Impulse imparted to each ball due to the other

= final momentum – initial momentum

= 0.3 kg m s-1 – 0.3 kg ms^-1

= – 0.6 kg ms^-1 = 0.6 kg ms^-1 (in magnitude)
The two impulses are opposite in direction.

Question: 14 Figure shows the position-time graph of a particle of mass 0.04 kg. Suggest a suitable physical context for this motion. What is the time between two consecutive impulses received by the particle? What is the magnitude of each impulse?
Answer: This graph can be of a ball rebounding between two walls situated at position 0 cm and 2 cm.

The ball is rebounding from one wall to another, time and again every 2 s with uniform velocity.
Here

So   

And 

The time between two impulse is 2 sec . the ball recieves an impulse every 2 second

Question: 15 A batsmen deflect a ball by an angle of 45° without changing its initial speed which is equal to 54 km/h. What is the impulse imparted to the ball? (Mass of the ball is 0.15 kg.)
Answer:  suppose o is the position of bat. AO is the line shows the path along which the ball strike the bat with velocity u and OB is the path showing deflection such that angle AOB =45 

So initial momentum of the ball

 

Final momentum of the ball 

= along ON 

So Impulse = change in momentum

10. Law of Conservation of Linear Momentum: ^imp

According to law of conservation of linear momentum if there is no any external force is acting on an isolated system then the linear momentum of the system remains constant.

•                 Derivation of Law of Conservation of Linear Momentum form Second law:

Suppose a system of n particles having masses  and velocity  then total linear momentum of n particles will be 

 

If  is the external force acting on the system then    =  

For isolated system F = 0   

Thus in absence of external force the linear momentum of isolated system remains constant.

•              Derivation of Law of Conservation of Linear Momentum form Third law:

Suppose a isolated system of two bodies such that  is the force on body A due to B and  is the force on body B due to A. Then from third law    

Now impulse of 

And impulse of 

As impulse of  = impulse of 

 

 

 

11. Application of Law of Conservation of Linear Momentum: ^imp

                 While firing a bullet, the gun should be hold tight to the shoulder because as we know the recoiling gun can’t hear the shoulder. 

If the gun is hold tight to the body then gun and body becomes a isolated system of large mass and recoil velocity becomes small.

                 When a men jumps out of a boat to the shore, the boat moves slightly away from the shore becomes initially total momentum of both is zero, but during jumping man accrue a momentum in forward direction due to which boat accrue momentum  in backward direction so it happens.

                 Rocket and Jet planes work on the principle of conservation of linear momentum:

In rockets and Jet planes the fuels impart momentum in downward direction due to which rocket get a momentum in upward direction.

                 Recoil of gun: suppose a bullet of mass m is fired from a gun of mass M. initially both are in rest but after firing suppose v is the velocity of bullet and V is the velocity of gun. Then from conservation of linear momentum

Or 

Or 

Here   sign show that both bullet and gun moves in opposite direction and clearly. 

So bullet moves faster from heavy gun.

                 Explosion of a bomb is also based on conservation of linear momentum.

Question: 16 a nucleus is at rest in the laboratory frame of reference. Show that if it disintegrates into two smaller nuclei the products must move in opposite directions.
Answer:  Let be the masses of products and  be their respective velocities.

Therefore, total linear momentum after disintegration = m₁v₁ +m₂ v₂.

Before disintegration, the nucleus is at rest.
Therefore, its linear momentum before disintegration is zero.

 According to the principle of conservation of linear momentum,

 

Negative sign shows that v₁ and v₂ are in opposite directions.

Question: 17 A shell of mass 0.020 kg is fired by a gun of mass 100 kg. If the muzzle speed of the shell is 80 ms^-1 what is the recoil speed of the gun?
Answer:  m = 0.02 kg, M = 100 kg, v = 80 ms^-1, V =?

So  

here negative sign indicates that the gun moves in a direction opposite to the direction of motion of bullet.

12.  Apparent Weight of a Men in An Elevator/Lift: ^M.Imp

Suppose a man of mass  standing on a weighting machine placed in a lift. Then various cases associated with the weight of the person are shown below:

(1) When the lift is in rest or moving with uniform velocity  downward or upward. In this case     in this case the equation becomes  

Or 

In this case the apparent weight of the person is equal to his real weight.

(2) When the lift moves upwards with acceleration. In this case the net force act in upward direction

So      

Or 

Or   

Hence apparent weight of a man increase when lift moves in upward direction.

(3)  When the lift moves down downward with acceleration.

In this case the net force act in downward direction

So      

Or    

Or 

Hence apparent weight of man decreases when moves in downward direction.

(4)  When lift falls freely. In this case the acceleration of the body becomes equal to acceleration due to gravity

i.e.     

Question: 18 A man of mass 70 kg, stands on a weighing machine in a lift, which is moving 
(a) upwards with a uniform speed of 10 ms^-1.
(b) Downwards with a uniform acceleration of 5 ms^-2.
(c) Upwards with a uniform acceleration of 5 ms^-2.
What would be the readings on the scale in each case?
(d) What would be the reading if the lift mechanism failed and it hurtled down freely under gravity?
Answer: Here, m = 70 kg, g = 10 m/s²
(a) When the lift moves upwards with a uniform speed, its acceleration is zero.

 

(b) When the lift moves downwards with a = 5 ms^-2

Than

(c) When the lift moves upwards with a = 5 ms^-2

Than

(d) If the lift were to come down freely under gravity, downward acc. A = g

13. Motion of the connected bodies: ^imp

Suppose two masses  such that  move on a pully. Suppose  is the tension in the string. The mass  moves downward and  upward, the resultant downward force on mass  is

And resultant upward force on mass  is 

Adding eq. (i) and (ii) we get     

 

Or                

Clearly 

Hence acceleration of the two connected bodies is smaller than the acceleration due to gravity.

Also dividing eq. (i) by (ii)              

  

Or 

 

  

 

     A diagram for each body of the system showing all the forces exerted on the body by the remaining part of the body is called free body diagram.

Question: 19 two masses 8 kg and 12 kg are connected at the two ends of a light in extensible string that goes over a friction less pulley. Find the acceleration of the masses, and the tension in the string when the masses are released.
Answer:  For block 

……(1)

And For block 

….(2)

Adding equation

We get

 Or   g

=10

=2

Substituting value of a into equation (ii) we get

 

Question: 20 Two bodies of masses 10 kg and 20 kg respectively kept on a smooth, horizontal surface are tied to the ends of a tight string. A horizontal force F = 600 N is applied to (i) A, (ii) B along the direction of string. What is the tension in the string in each case?
Answer:  

1. When force is applied on 10 kg mass

Than

2. When force is applied on 20 kg mass

than

Question: 21 Ten one rupee coins are put on top of one another on a table. Each coin has a mass m kg. Give the magnitude and direction of
      (a)    the force on the 7th coin (counted from the bottom) due to all coins above it.
      (b)    The force on the 7th coins by the eighth coin and
      (c)     the reaction of the sixth coin on the seventh coin.
Answer:    (a) The force on 7th coin is due to weight of the three coins lying above it

Therefore,     

This force acts vertically downwards.
 (b) The eighth coin is already under the weight of two coins above it and it has its own weight too. Hence force on 7th coin due to 8th coin is sum of the two forces 

i.e.  

  

the force acts vertically downwards.
 (c) The sixth coin is under the weight of four coins above it.
 Reaction, 

 

Minus sign indicates that the reaction acts vertically upwards, opposite to the weight.

Question: 22 A monkeys of mass 40 kg climb on a rope (Fig.) which can stand a maximum tension of 600 N. 

In which of the following cases will the rope break: the monkey
(a) climbs up with an acceleration of 6 ms^-2
(b) climbs down with an acceleration of 4 ms^-2
(c) climbs up with a uniform speed of 5 ms^-1
(d) falls down the rope nearly freely under gravity?
Answer: (a) When the monkey climbs up with acceleration a,

then  so 

But rope can withstand with a force of 600N so will break.

(b) When monkey is climbing down with acceleration, than 

 The rope will not break.

c. When monkey climb up with uniform speed 

than. 

The rope will not break.

D, When the monkey is falling freely, it would be a state of weightlessness. 

So, tension will be zero and the rope will not break.

14. Example of Variable mass situation: Rocket Propulsion: ^imp

 

 

 

 

 

 

 

Suppose at  is mass of rocket including fuel and  is initial velocity of the rocket and at  is mass of rocket left and  is velocity accursed by the rocket.

 As

Now again suppose that in a small interval of time is small decrease in mass of rocket and  is corresponding increase in small increase in velocity and  is the velocity of exhaust gasses w.r.t. the earth.

Now from conservation of linear momentum

Here  sign indicate that gasses escape downward.

  

As  is small so can be neglect

  

Now suppose  is the relative velocity of burnt gas w.r.t. rocket then

i.e. 

Using in (i)  

Or  

Integrating both sides we get      

 

Or    

This is the velocity of rocket at any time 

Now at     

   

     Thus in the absence of any external force, the instantaneous velocity of the rocket depends on exhaust speed of burnt gas and mass of the rocket and burnt fuel.

Burnt out speed of the rocket:-

The speed acquired by the rocket when whole of its fuel gets burnt is called brunt of speed of the rocket.

Let  is the residual mass of the rocket then burnt out speed  can be given as

Burnt out speed of rocket is its maximum speed.

Thrust on the rocket:-

The force with which a rocket moves upward is called thrust of the rocket.

As we know    

Dividing both sides by  we get 

Or  

Or 

Thus upward thrust of the rocket is equal to the product of the exhaust speed of the burnt gas and the rate of consumption of fuel at that instant.

Question: 23 A rockets with a lift-off mass 20,000 kg are blasted upwards with an initial acceleration of 5.0 ms^-2. Calculate the initial thrust (force) of the blast.
Answer: Here,

Initial acceleration = 5 ms^-2
clearly, the thrust should be such that it overcomes the force of gravity besides giving it an upward acceleration of 5 ms^-2.

Thus the force should produce a net acceleration of

.
Since, thrust = force = mass x acceleration

so

15. Equilibrium of Concurrent forces:-

If a number of forces act on a body at a same point and the net unbalanced force is zero, the body will continue in its state of rest or of uniform motion along straight line then concurrent forces (forces acting on single point) are said to be in equilibrium.

Suppose three concurrent forces  are acting on a body shown in fig.

Here resultant of  and  is. 

If the third force having equal magnitude as that of  

And act in opposite to   

Then 

Or 

This is the required condition for equilibrium of the concurrent forces. If there are  forces then

16. Lami’s Theorem:-

Suppose there concurrent forces  are acting on the body. Such that α is angle between, β is angle between γ is angle between  as shown in fig.

Then according to  Lami’s theorem.

At equilibrium   

  

Thus according to Lami’s theorem if three forces are acting on a particle in equilibrium then each force is proportional to the sine of angle between the other two  forces. 

Unit-3(B) Friction:

17. Friction:

An opposing force which comes into play when a body moves or tends to moves on the surface of another body is called frictional force.

Explanation of Origin of Frictional Force:

The force of friction is due to the force of attraction between atoms and molecules of two surfaces at the point of actual contact.

1. Types of Friction:

There are three types of friction

• Static Friction:-

The force of friction between two bodies which comes into play when a body rest on the another body 

• Limiting Friction:-

The maximum value of static friction is called limiting friction, which comes into play when a body is in the position of just motion over the second body 

• Kinetic Friction:-

The force of friction which comes into play when a body moves on the surface of another body

18. Laws of Friction and Coefficient of Friction: ^imp

(i) Law of Limiting Friction:-

•   Limiting friction depends upon the nature of the surface in contact.
•  The direction of limiting friction is opposite to the applied force.
•   The value of limiting friction does not depend upon the area of the contact.
•  The limiting friction is directly proportional to the normal reaction  between the two surfaces.

I.e. 

Or     

 The proportionality constant  is called the coefficient of limiting Friction.

(ii) Law of Kinetic Friction:

•   The kinetic friction opposes the relative motion between two surfaces in contact.
•  The value of kinetic friction is independence on the area of contact.
•   The kinetic friction does not depend upon the velocities provided velocities are not too large or nor too small.
•  The value of kinetic friction is directly proportional  to the normal reaction

i.e.               

Or     

Here proportionality constant  is called coefficient of kinetic friction.

     As 

Hence value of limiting friction is greater than kinetic friction.

19.  ANGLE OF FRICTION AND ANGLE OF REPOSE: ^M.Imp

Angle of Friction: 

The angle of friction may be defined as the angle between normal reaction and resultant of frictional force & normal reaction.

From fig 

Or 

Also we know that  

So 

Thus, the coefficient of friction is equal to tangent of angle of friction.

Angle of Repose:

It is the minimum angle that an inclined plane makes with the horizontal when a body placed on it is just begins to slide down. 

From fig.

Dividing e.q. by (iii)

Or   

Relation between angle of friction and angle of repose:

Clearly from e.q. (ii) and (v)

  And 

So  

   

Hence angle of friction is equal to angle of repose.

20.  Sliding and Rolling Friction:

• Sliding Friction:–

The force of friction which comes into play when a body slide on the surface of another body is called sliding friction.

 e.g.:- When a wooden block is pulled or pushed over a horizontal surface, sliding friction comes into play.

• Rolling Friction:-

The force of friction which comes into play when a body rolls over the surface of another body is called rolling friction.

 e.g.:- When a wheel rolls over a road, rolling friction comes into play.

     For the same magnitude of normal reaction, rolling friction is always much smaller than sliding friction.

Rolling Friction is smaller than Sliding Friction:

When a wheel rolls without sliding over a horizontal plane, the surfaces at contact do not rub each other. 

The relative velocity of the point of contact of the when w.r.t. plane is zero, if there is no slipping. So rolling friction is much smaller than sliding friction.

Cause of Rolling Friction:

When a wheel rolls over the road, then it exerts large pressure over the road due to its small area. 

This produces a slight depression of the rode below a small elevation or mount in front of it. 

In addition to this, the rolling wheel has to continuous detached itself from the surface on which it rolls. 

Due to these two factors a force comes into play which opposes the rolling motion.

21.  Laws of Rolling Friction:

1.                 Rolling friction is directly proportional to the normal reaction

i.e.    

2.              Rolling friction is inversely proportional to the radius of the rolling cylinder or wheel.

From eq. (i) and (ii)

Or 

The proportionality constant  is called coefficient of rolling friction.

22. WORK DONE AGAINST FRICTION:

1.                 Work done sliding a body over a horizontal surface:

Consider a body of weight  resting on a rough horizontal surface. 

The weight  is balanced by the normal reaction  of the horizontal surface. 

Suppose a force  is applied horizontally so that the body just begins to slide. Let  be the kinetic friction. 

Work done against friction in moving the body through distance  will be 

But 

Or  

Question:24 Two bodies A and B of masses 5 kg and 10 kg in contact with each other rest on a table against a rigid wall (Fig.). The coefficient of friction between the bodies and the table is 0.15. A force of 200 N is applied horizontally to A.

What are

 (a) the reaction of the partition 

(b) the action reaction forces between A and B? 

(c) What happens when the wall is removed? Does the answer to (b) change, when the bodies are in motion?

Ignore the difference between  and.
Answer: (i) When the wall exists and blocks A and B are pushing the wall, there can’t be any motion i.e., blocks are at rest. Hence,
(a)  reaction of the partition = – (force applied on A) = 200 N towards left.
(b)  Action reaction forces between A and B are 200 N each. 

A presses B towards right with an action force 200 N and B exerts a reaction force on A towards left having magnitude 200 N.
(ii) When the wall is removed, motion can take place such that net pushing force provides the acceleration to the block system. Hence, taking kinetic friction into account, 

We have   

­

If force exerted by A on B is 

then considering the free body diagram of only A, 

we have              

 

So   

 Toward right

So force exerted on A by B

 toward left 

Question: 24 A blocks of mass 15 kg is placed on a long trolley. The coefficient of static friction between the block and the trolley is 0.18. The trolley accelerates from rest with 0.5 ms^-1 for 20 s and then moves with uniform velocity. Discuss the motion of the block as viewed by 

(a) a stationary observer on the ground, 

(b) an observer moving with the trolley.
Answer:  (a) Force experienced by block,

 

 

Force of friction  

 

. 

i.e., force experienced by block will be less than the friction. 

So the block will not move. It will remain stationary w.r.t. trolley for a stationary observer on ground.
(b) The observer moving with trolley has an accelerated motion 

i.e., he forms non-inertial frame in which Newton’s laws of motion are not applicable. 

The box will be at rest relative to the observer.

Question: 25 A discs revolve with a speed of 33 1/3 rpm and have a radius of 15 cm. Two coins are placed at 4 cm and 14 cm away from the centre of the record. 

If the coefficient of friction between the coins and record is 0.15, which of the coins will revolve with the record?
Answer: If the coin is to revolve with the record, then the force of friction must be enough to provide the necessary centripetal force.       

 

  

Also given that frequency 

So angular frequency 

  

Or     

The condition is satisfied  is satisfied by the coin placed at 4cm from the centre of the record. 

So the coin at 4cm will revolve with the record.         

•              Work done in moving a body up an inclined plane: ^M.Imp

Suppose a body of weight  is placed on an inclined plane. Let  the angle of inclination. A force  applied on the body so that it just begins to slide up the inclined plane. 

The various forces acting on the body are

And  

But        

So  

From eq. (i) we have   

             

Work done in pulling the body through distance  up the inclined plane is                                   

•           Work done in moving a body down an inclined plane: ^M.Imp

Suppose a body of weight  is placed on an inclined plane. Suppose the angle of inclined  be less than angle of repose. A force P is applied to just slide the body down the inclined plane. 

The various forces acting on the body are            

But 

  

The work done in sliding the body through distance S down the inclined plane is 

23. Acceleration of a body sliding down an inclined plane:

Consider a body of weight  placed on an inclined plane. Suppose the angle of inclination  be greater than angle of repose. Let be the acceleration with which the body slide down the inclined plane. 

The various forces acting on the body are   

 

As 

So 

Or 

  

24. Friction is a Necessary Evil: ^M.Imp

Advantage of Friction (Friction is a Necessity):

• It is due to friction between the ground and the feet that we are able to work.
• The breaks of a vehicle cannot work without friction.
• Various parts of machine are able to rotate because of friction between belt and pully.
• The tyres of vehicle are made rough to increase friction.
• Nails and screws join various wooden parts due to friction.
• It will not be possible to write on a proper with pen pencil without friction.

 

Friction is an Evil:

Disadvantage of Friction:

                 A large amount of power is wasted in overcoming friction and the efficiency of the machines decreases considerably.

                 Excessive friction between rotating parts of a machine produces enough heat and causes damage to the machinery.

     As friction has both advantages and disadvantages, we can say that friction is a necessary evil.

25.  Method of Changing Friction: ^imp.

Method of reducing friction:

• By Polishing:-

By making the surface in contact highly polished and smooth, the bumps and depressions get minimum and so friction is reduce considerably.

• Lubrication: 

Lubrication is a substance (solid, liquid or gas) which forms a thin layer between two surfaces in contact. It fills depressions present in the surfaces of contact and hence reduces friction.

• Streamlining: 

Friction due to air is considerably reduced by streamling the shape of the body moving through air. For example jets, aero planes, fast moving cars etc. have stream line shape.

• By using ball bearing.
• By using antifriction alloys.
• By using air cushion (air tube in tyre).

 Some methods of increasing Friction:

                    Threading of tyers: 

Threading of  tyers is done to increase friction between roads and tyres. Moreover synthetic rubber is prefaced over the natural rubber in the manufacture of tyres because of its larger coefficient of friction with the roads.

                    Sand is thrown on track covered with snow: 

This increase the force of friction between the wheels and the track and the driving becomes safer. 

                    On a rainy day, we throw some sand on the slippery ground: 

This increase the friction between over feet and the ground, this reduces the chance of slipping.

26. Why it is easier to pull a lawn roller to push it. ^imp 

Suppose a force  is applied to pull a lawn roller of weight.The force  have two rectangular components.

• Horizontal component  helps to move the lawn roller forward. 
• Vertical component  acts in the upward direction.            

If  is the normal reaction than equating the vertical components have

…………… (1)

So kinetic friction      

Similarly if same force  is applied to push a lawn roller then we have vertical component as        

So the value of kinetic friction is 

 

Comparing eq. (ii) and (iii) we have    

  

  Hence it is easier to pull a lawn roller to push it. 

class 11 physics notes chapter 5 laws of motion