NEET Physics Class 11 Chapter 6 Friction Notes

Friction Contact Force

When two bodies are kept in contact, electromagnetic forces act between the charged particles (molecules) at the surfaces of the bodies.

Thus, each body exerts a contact force on the other. The magnitudes of the contact forces acting on the two bodies are equal but their directions are opposite and therefore the contact forces obey Newton’s third law.

• The direction of the contact force acting on a particular body is not necessarily perpendicular to the contact surface.
• We can resolve this contact force into two components, one perpendicular to the contact surface and the other parallel to it.
• In the figure, the perpendicular component to the contact surface is called the normal contact force or normal force (generally written as N) and the parallel component is called friction (generally written as f). Therefore if R is the contact force then

⇒ \(\mathrm{R}=\sqrt{f^2+N^2}\)

Reasons For Friction

1. Ιnter-locking of extended parts of one object into the extended parts of the other object.
2. Bonding between the molecules of the two surfaces or objects in contact.

Friction Force Is Of Two Types.

1. Kinetic
2. Static

1. Kinetic Friction Force

Kinetic friction exists between two contact surfaces only when there is relative motion between the two contact surfaces. It stops acting when relative motion between two surfaces ceases.

Direction Of Kinetic Friction On An Object

• Ιt is opposite to the relative velocity of the object considered with respect to the other object in contact.
• Note that its direction is not opposite to the force applied it is opposite to the relative motion of the body considered which is in contact with the other surface.

Magnitude Of Kinetic Friction

The magnitude of the kinetic friction is proportional to the normal force acting between the two bodies. We can write

f_k= μ_k N

where N is the normal force. The proportionality constant μ_k is called the coefficient of kinetic friction and its value depends on the nature of the two surfaces in contact.

Question 1. Find the direction of the kinetic friction force

1. On the block, exerted by the ground.
2. On the ground, exerted by the block.

Answer:

where f₁ and f₂ are the friction forces on the block and ground respectively.

Question 2. In the above example, the correct relation between the magnitude of f₁ and f₂ is

1. f₁> f₂
2. f₂> f₁
3. f₁= f₂
4. It is not possible to decide due to insufficient data.

Answer: By Newton‘s third law the above friction forces are action-reaction pairs, equal but opposite to each other in direction. Hence (3).

Also, note that the direction of kinetic friction has nothing to do with applied force F.

Question 3. All surfaces as shown in the figure are rough. Draw the friction force on A and B

Answer:

Kinetic friction acts in such a way so as to reduce relative motion.

Question 4. Find out the distance travelled by the blocks shown in the figure before it stops.

Answer:

N – 10 g = 0

N = 100 N

f_k= µ_kN

f_k= 0.5 × 100 = 50 N

f_k= ma

50 = 10 a

⇒ a = 5

∴ v² = u² + 2as

0² = 10² + 2 (–5) (S)

∴ S = 10 m

Question 5. Find out the distance travelled by the block on the incline before it stops. The initial velocity of the block is 10 m/s and the coefficient of friction between the block and incline is μ = 0.5.

Answer:

N = mg cos37°

∴ mg sin 37° + µN = ma

a = 10 m/s² down the incline

Now v² = u² + 2as

0 = 10² + 2(–10) S

∴ S = 5 m

Question 6. Find the time taken in the above example by the block before it stops.

Answer:

a = g sin 37° + µg cos 37°

∴ a = 10 m/s² down the incline

∴ S = \(u t+\frac{1}{2} at^2\)

5 = \(\frac{1}{2} \times 10 \times \mathrm{t}^2\)

∴ t = 1sec.

Question 7. A block is given a velocity of 10 m/s and a force of 100 N in addition to friction force is also acting on the block. Find the retardation of the block?

Answer:

As there is relative motion

∴ kinetic friction will act to reduce this relative motion.

f_k = µN = 0.1 × 10 × 10 = 10 N

100 + 10 = 10a

a = 11 m/s²

Static Friction: Ιt exists between the two surfaces when there is a tendency of relative motion but no relative motion occurs along the two contact surfaces.

• For Question consider a bed inside a room; when we gently push the bed with a finger, the bed does not move.
• This means that the bed has a tendency to move in the direction of the applied force but does not move as there exists static friction force acting in the opposite direction of the applied force.

Question 8. What is the value of static friction force on the block?

Answer:

In the horizontal direction acceleration is zero.

Therefore Σ F = 0.

∴ ƒ = 0

Direction Of Static Friction Force: The static friction force on an object is opposite to its impending motion relative to the surface.

The following steps should be followed in determining the direction of static friction force on an object.

1. Draw the free body diagram with respect to the other object on which it is kept.
2. Include pseudo force also if the contact surface is accelerating.
3. Decide the resultant force and the component parallel to the surface of this resultant force.
4. The direction of static friction is opposite to the above component of the resultant force.

Note: Here once again the static friction is involved when there is no relative motion between two surfaces.

Question 9. In the following figure, an object of mass M is kept on a rough table as seen from above. Forces are applied to it as shown. Find the direction of static friction if the object does not move.

Answer:

In the above problem, we first draw the free-body diagram to find the resultant force.

As the object does not move this is not a case of limiting friction. The direction of static friction is opposite to the direction of the resultant force F_R as shown in the figure by f_s. Its magnitude is equal to 25 N.

Magnitude Of Kinetic And Static Friction

Kinetic Friction:

The magnitude of the kinetic friction is proportional to the normal force acting between the two bodies. We can write

f_k= μ_k N

where N is the normal force. The proportionality constant μ_k is called the coefficient of kinetic friction and its value depends on the nature of the two surfaces in contact.

If the surfaces are smooth μ_k will be small, if the surfaces are rough μ_k will be large. It also depends on the materials of the two bodies in contact.

Static Friction:

The magnitude of static friction is equal and opposite to the external force exerted, till the object at which force is exerted is at rest. This means it is a variable and self-adjusting force. However, it has a maximum value called limiting friction.

f_max = μ_sN

The actual force of static friction may be smaller than μ_sN and its value depends on other forces acting on the body. The magnitude of frictional force is equal to that required to keep the body at relative rest.

0, f_s, f_smax

Here μsand μk are proportionality constants. μ_s is called the coefficient of static friction and μ_k is called the coefficient of kinetic friction. They are dimensionless quantities independent of shape and area of contact. It is a property of the two contact surfaces.

Note: μ_s> μ_k for a given pair of surfaces. If not mentioned then μ_s= μ_k can be taken. The value of μ can be from 0 to ∞.

The following table gives a rough estimate of the values of the coefficient of static friction between certain pairs of materials.

The actual value depends on the degree of smoothness and other environmental factors. For example, wood may be prepared at various degrees of smoothness and the friction coefficient will vary.

table

Rolling Friction: When a body (say wheel) rolls on a surface the resistance offered by the surface is called rolling friction.

• Rolling friction forces arise as, for example, a rubber tyre rolls on pavement, primarily because the tyre deforms as the wheel rolls. The sliding of molecules against each other within the rubber causes energy to be lost.
• The velocity of the point of contact with respect to the surface remains zero.
• The rolling friction is negligible in comparison to static or kinetic friction which may be present simultaneously i.e., µ_R< µ_k< µ_S

Angle Of Friction

The angle of friction is the angle which the resultant of limiting friction F_S and normal reaction N makes with the normal reaction. It is represented by λ, Thus from the figure.

⇒ \(\tan \lambda=\frac{F_S}{N}\) (∵ F_s= µ N) ortan

or \(\frac{F_S}{N}\) = θtan λ = µ

For smooth surfaces, λ = 0 (zero)

Angle Of Repose (θ)

If a body is placed on an inclined plane and if its angle of inclination is gradually increased, then at some angle of inclination θ the body will just begin to slide down this angle is called the angle of repose (θ).

F_S = mg sinθ and N = mg cosθ

So, or µ = tanθ

Relation between angle of friction (λ) and angle of repose (θ)

We know that tan λ = µ and µ = tan θ

hence tan λ = tan θ or θ = λ

Thus, angle of repose = angle of friction

Question 1. Find the acceleration of the block. Initially, the block is at rest.

Answer: Zero

Question 2. Find out the acceleration of the block. Initially, the block is at rest.

Answer:

N + 24 – 100 = 0 for vertical direction

∴ N = 76 N

Now 0 ≤ f_s ≤ µ_s N

0 ≤ f_s ≤ 76 × 0.5

0 ≤ f_s≤ 38 N

∴ 32 < 38 Hence f = 32

∴ acceleration of the block is zero.

Question 3. Find out the acceleration of the block for different ranges of F.

Answer:

0 ≤ f ≤ µ_sN

⇒ 0 ≤ f ≤ µ_smg

a = 0 if F ≤ µ_smg

⇒ a = \(\frac{F-\mu M g}{M}\) if F > µMg M

Question 4. Find out the acceleration of the block. Initially, the block is at rest.

Answer:

0 ≤ f_s ≤ µ_sN

⇒ 0 ≤ f_s≤ 50

Now 51 > 50

∴ The block will move but if the block starts moving then kinetic friction is involved.

f_k= µ_k N = 0.3 × 100 = 30 N

∴ 51 – 30 = 10 a

∴ a = 2.1 m/s²

Question 5. Find out the minimum force that must be applied on the block vertically downwards so that the block doesn’t move.

Answer:

100-f_s = 0

∴ f_s= 100 ……..

F + 10 g = N ⇒ N = 100 + F ……….. (2)

Now 0 ≤ f_s≤ μN

100 ≤ 0.5 N

100 ≤ 0.5 [100 + F]

200 ≤ 100 + F

F ≥ 100 N

∴ Minimum F = 100 N

Question 6. A particle of mass 5 kg is moving on a rough fixed inclined plane with a constant velocity of 5 m/s as shown in the figure. Find the friction force acting on a body by plane.

Answer:

f_k= μ_kN = μ_k mg cos 37° = mg sin30° = 5 (10)\(\left(\frac{1}{2}\right)\)

⇒ f_k= 25 N

Question 7. The angle of inclination is slowly increased. Find out the angle at which the block starts moving.

Answer:

0 ≤ f ≤ µ_s N

mg sinθ > f_smax

mg sin θ > µN

mg sinθ > µ mg cos θ

∴ tan θ > µ

θ = tan^-1 µ

for tan θ ≤ µ no sliding on an inclined plane.

This method is used to find out the value of µ practically.

Question 8. Find out the acceleration of the block. If the block is initially at rest.

Answer: (FBD of the block excluding friction) N = 10 g cos 37° = 80 N

Now 0 ≤ f_s≤ µN

0 ≤ f_s≤ 0.5 × 80

∴ f_s≤ 40 N

We will put a value of f in the last i.e. in the direction opposite to the resultant of other forces. f acts down the incline and its value is of = 75 – 60 = 15 N

So acceleration is zero

Question 9. In the above problem, how much force should be added to 75 N force so that the block starts to move up the incline?

Answer:

∴ 60 + 40 = 75 + f_extra

∴ f_s= 25 N

Question 10. In the above problem, what is the minimum force by which 75 N force should be replaced so that the block does not move?
Answer:

In this case, the block has a tendency to move downwards.

Hence friction acts upwards.

∴ F + 40 = 60

∴ F = 20 N

Question 11. The top view of a block on a table is shown (g = 10 m/s²). Find out the acceleration of the block.

Answer:

Now f_s≤ µN

∴ f_s≤ 50

Hence the block will move.

a = \(\frac{40 \sqrt{2}-50}{10}\)

⇒ \((4 \sqrt{2}-5) \mathrm{m} / \mathrm{s}^2\)

Question 12. Find minimum µ so that the blocks remain stationary.

Answer: T = 100 g = 1000 N

∴ f = 1000 to keep the block stationary

Now f_max= 1000

µN = 1000

µ = 2

Can µ be greater than 1?

Yes 0 < µ ≤ ∝

Question 13. Find out the minimum acceleration of block A so that the 10 kg block doesn’t fall.

Answer:

Applying NL in a horizontal direction

N = 10 a ………(1)

Applying NL in a vertical direction

10 g = µ N …….(2)

10 g = µ 10 a from (1) and (2)

∴ a = 20 m/s²

Question 14. In the following figure force F is gradually increased from zero. Draw the graph between applied force F and tension T in the string. The coefficient of static friction between the block and the ground is μ_s.

Answer:

As the external force F is gradually increased from zero it is compensated by the friction and the string bears no tension. When limiting friction is achieved by increasing force F to a value till μ_smg, the further increase in F is transferred to the string.

Question 15. Find the acceleration of the two blocks. The system is initially at rest and the friction coefficient is as shown in the figure.

Answer:

f_max= 50 N

∴ f ≤ 50 N

If they move together a = \(\frac{101}{20}\)

= 5.05 m/s²

Check friction on B

f = 10 × 5.05 = 50.5 (required)

50.5 > 50 (therefore required > available)

Hence they will not move together.

Hence they move separately so kinetic friction is involved.

∴ for \(\mathrm{a}_{\mathrm{B}}=\frac{50}{10}=5.1 \mathrm{~m} / \mathrm{s}^2\)

⇒ \(a_A=\frac{101-50}{10}=5 \mathrm{~m} / \mathrm{s}^2\)

Also, a_A > a_B as force is applied on A.

Question 16. Find the acceleration of the two blocks. The system is initially at rest and the friction coefficient is as shown in the figure.

Answer:

Move Together Move Separately No need to calculate.

a = 2 m/s²

Check friction on 20 kg.

f = 20 × 2

f = 40 (which is required)

40 < 50 (therefore required < available)

∴ will move together.

Question 17. In the above example, find the maximum F for which two blocks will move together.
Answer:

Observing the critical situation where friction becomes limiting.

∴ F – f_max= 10 a ………(1)

f_max = 20 a ……….(2)

∴ F = 75 N

Question 18. Initially, the system is at rest. find out the minimum value of F for which sliding starts between the two blocks.

Answer:

At just sliding conditions limiting friction is acting.

F – 50 = 20 a ………..(1)

f = 10 a ………………(2)

50 = 10 a

∴ a = 5 m/s²

Hence F = 50 + 20 × 5 = 150 N

∴ F_min = 150 N

It is Easier to Pull Than to Push a Body. Why –

Let a force P be applied to pull a body of weight Mg.

The applied force is resolved into two components : P cosθ and P sinθ The normal reaction, R = (Mg – P sinθ)

Now, the kinetic force of friction is given by

F₁= μ_k R = μ_k(Mg – P sinθ) …….(1)

On the other hand, when the same force is applied to push a body of weight Mg, then a normal reaction,

R = (Mg + P sinθ) …….(2)

∴ Kinetic force of friction is F₂= μ_k R = μ_k(mg + P sinθ)

From eqn (1) and (2), it is clear and F₂> F₁

That is, the force of friction in the case of push is more than that in the case of pull.

Hence, it is easier to pull than to push the body.

Friction is a Necessary Evil :

Friction is a necessary evil. It means it has advantages as well as disadvantages. In other words, friction is not desirable but without friction, we cannot think of survival.

Disadvantages :

1. A significant amount of energy of a moving object is wasted in the form of heat energy to overcome the force of friction.
2. The force of friction restricts the speed of moving vehicles like buses, trains, aeroplanes, rockets etc.
3. The efficiency of machines decreases due to the presence of force of friction.
4. The force of friction causes a lot of wear and tear in the moving parts of a machine.
5. Sometimes, a machine gets burnt due to the friction force between different moving parts.

Advantages :

1. The force of friction helps us to move on the surface of the earth. In the absence of friction, we cannot think of walking on the surface. That is why, we fall down while moving on a smooth surface.
2. The force of friction between the tip of a pen and the surface of paper helps us to write on the paper. It is not possible to write on the glazed paper as there is no force of friction.
3. The force of friction between the tyres of a vehicle and the road helps the vehicle to stop when the brake is applied. In the absence of friction, the vehicle skid off the road when the brake is applied.
4. Moving belts remain on the rim of a wheel because of friction.
5. The force of friction between a chalk and the blackboard helps us to write on the board. Thus, we observe that irrespective of the various disadvantages of friction, it is very difficult to part with it. So, friction is a necessary evil.

Methods Of Reducing Friction

As friction causes the wastage of energy it becomes necessary to reduce the friction. Friction can be reduced by the following methods.

1. Polishing the surface. We know, that friction between rough surfaces is much greater than between the polished surfaces. So we polish the surface to reduce the friction. The irregularities on the surface are filled with polish and hence the friction decreases.
2. Lubrication. To reduce friction, lubricants like oil or grease are used. When the oil or grease is put in between the two surfaces, the irregularities remain apart and do not interlock tightly. Thus, the surfaces can move over each other with less friction between them.
3. By providing a streamlined shape. When a body (e.g. bus, train, aeroplane etc.) moves with high speed, air resistance (friction) opposes its motion. The effect of air resistance on the motion of the objects (stated above) is decreased by providing them with a streamlined shape.
4. Converting sliding friction into rolling friction. Since rolling friction is much less than sliding friction, we convert the sliding friction into rolling friction. This is done by using a ball bearings arrangement. Ball bearings are placed in between the axle (A) and hub (B) of the wheel as shown in the figure. The ball bearing tends to roll around the axle as the wheel turns and as such the frictional force is diminished.

Friction Key Concept

Part of the contact force that is tangential to the surface is called friction force. Microscopically friction force because of attraction between molecules of the two surfaces.

The friction force is of two types:

1. Kinetic friction
2. Static friction

Kinetic Friction Force:

Kinetic friction exists between two surfaces (in the case of a block), two points (in the case of a sphere), or two lines (in the case of a cylinder) when there is relative motion. It stops acting when relative motion ceases to exist.

Direction Of Kinetic Friction:

It is opposite to the relative velocity of contact surfaces.

Note: Its direction is not opposite to the force applied it is opposite to the motion of the body considered which is in contact with the other surface.

Kinetic friction f_k= μ_kN

The proportionality constant μ_k is called the coefficient of kinetic friction and its value depends on the nature of the two surfaces in contact.

Static Friction:

When two surfaces in contact have a relative velocity of zero, but there is a tendency of relative motion then the friction force acting will be static.

Direction Of Static Friction: If there is a tendency to slide between the contact surfaces, it will act in such a direction to prevent sliding.

Static friction is a variable and self-adjusting force. It can adjust its value upto a limit which is called limiting friction force (f_smax).

f_smax= μ_sN

Here μ_s is the coefficient of static friction which depends on the nature of two contact surfaces. The actual force of static friction may be smaller than μ_sN and its value depends on other forces acting on the body. The magnitude of frictional force is equal to that required to keep the body at relative rest.

0 f_s f_smax

The following steps should be followed in determining the direction of static friction force on an object.

1. Draw the free-body diagram to the other object on which it is kept.
2. Include pseudo force also if the contact surface is accelerating.
3. Decide the resultant force and the component parallel to the surface of this resultant force.
4. The direction of static friction is opposite to the above component of the resultant force.

Here μ_s and μ_k are dimensionless quantities independent of shape and area of contact. It is a property of the two contact surfaces. In general μ_s > μ_k for a given pair of surfaces. If it is not mentioned separately the μ_s = μ_k can be taken.

μ_s and μ_k can also be represented as angles. If θ_s and θ_k are angles of static friction and kinetic friction respectively, then

θ_s= tan^-1 μ_s

θ_k= tan^-1 μ_k

θ_s is also called the angle of repose.

Rolling Friction:

When a body rolls on a surface the resistance offered by the surface is called rolling friction.

Rolling Friction Is Less Than Sliding Friction

In sliding motion, elevation collides. This introduces friction. In rolling motion, elevations are crossed over. This avoids friction.

Rolling Friction Zero In Ideal Case:

• In ideal rolling the contact with the cylindrical surface of the body and lower surface must be along a straight line. Elevations must be crossed over and no friction be present.
• But no rolling is ideal. Due to the deformation of the moving cylindrical surface (wheel of a loaded truck), or the deformation of the lower surface (mud street), contact becomes over a flat surface. This introduces sliding, which causes friction.