## Static And Kinetic Friction

Coulomb, a famous scientist in the eighteenth century was the first to mention static and kinetic (or sliding) friction, separately. Let a wooden block of weight W rest on a horizontal table.

- Weight W of the block acts on the table vertically downwards; by Newton’s third law of motion, the table also exerts an equal and opposite reaction force R. Force R is called normal force or normal reaction.
- If no other force acts on the system, weight and normal force balance each other and the block rests on the table.
- Let a horizontal force F be applied on the block. If F is below a certain magnitude, the block remains at rest. At the surface of contact, and opposite to the direction of F, a frictional force f develops, which opposes the motion of the block.

**Read and Learn More: Class 11 Physics Notes**

- In the table below, the variation of the frictional force f with the change in the magnitude of the applied force F, and the corresponding state of motion of the block, are illustrated.

**A Study Of The Table Above Brings Out The Following Important Properties Related To Friction:**

- When no external force acts on the block, the block remains stationary and there is no existence of friction [observation number 1].
- Observations 2 to 4 point out an important fact as the applied force is increased, the frictional force also increases automatically to balance it. The block, therefore, remains at rest.
- Frictional force by itself, increases by the exact amount required to balance the applied force, before the motion starts. Hence, before any relative motion between the surfaces begins, the frictional force is a self-adjusting force. The frictional force acting under this condition is called static friction.
**Static Friction Definition:**When two surfaces are relatively at rest but one is trying to slide over the other, then the force that comes into existence between the two surfaces and tries to oppose the motion is called static friction.

- As the applied horizontal force is increased, the frictional force also increases gradually to its maximum limit. When the applied force exceeds this limit, the block can no longer be at rest [observation number 4]. This maximum value of static friction is called limiting friction or the limiting value of static friction.
**Limiting Friction Definition:**The maximum possible magnitude of static friction is called limiting friction.

- As soon as the magnitude of the applied horizontal force exceeds the limiting friction, the horizontal resultant force is no longer zero. Because of this resultant force, the block starts moving over the table with acceleration. This motion is termed as sliding.
- During sliding, the frictional force that comes into play to oppose the motion is called sliding or kinetic friction. An important property of sliding friction is that the magnitude of this frictional force falls a little as soon as motion sets in. Hence, the magnitude of sliding friction is less than that of limiting friction for a given pair of surfaces [observation number 5].
**Sliding Or Kinetic Friction Definition:**When two surfaces slide against each other, the force developed at their surfaces of contact, opposing the relative motion, is called kinetic or sliding friction.

- To keep the block moving with uniform velocity, the resultant of the horizontal forces must be zero (observation 6). To achieve this, the applied force is reduced so that it becomes equal to the kinetic friction.

It is a common experience that, to move a heavy stone over the ground, a large force has to be applied. However, once the stone is moving, a comparatively smaller force (push) is required to maintain its motion.

- Kinetic friction between two surfaces is independent of the speed of their relative motion, and therefore it is nearly a constant for a specific pair of surfaces.
- The graph represents the force of friction acting between two surfaces in contact with each other, against the applied force that acts parallel to the surface of contact.
- The static friction is initially equal to the applied force. This is represented by the line OA. As the magnitude of the applied force becomes equal to OD, the block is on the verge of moving.
- So AD is the magnitude of limiting friction. On increasing the applied force further, the block begins to move and kinetic friction comes into play. Its value is slightly less than that of the limiting friction.

In the graph, EG represents the magnitude of this kinetic friction which is approximately a constant.

## Laws Of Motion – Friction

## Laws Of Friction

Friction, acting between two surfaces in contact, follows certain laws. The eminent fifteenth-century artist and scientist, Leonardo Da Vinci was the first to establish these laws experimentally.

**Laws Of Static Friction**

- The force of static friction between two surfaces in contact always acts against the force that attempts to cause relative motion.
- The force of limiting friction between two surfaces in contact is directly proportional to the normal reaction.
- When normal force remains constant, the force of friction between the surfaces is independent of the area of contact.

**Laws Of Kinetic Friction**

- The force of kinetic friction between two surfaces in contact always acts against their relative motion.
- The force of kinetic friction between two surfaces in contact is directly proportional to the normal force.
- When normal force remains constant, the force of friction between the surfaces is independent of the relative velocity of the surfaces.

**Coefficient Of Static Friction:** On application of a horizontal force on a wooden block kept on a table, the following forces act on the block:

- Normal force (R) acting vertically upward
- Weight (W) of the block acting vertically downward
- Static friction opposite to the direction of the applied force

Let the maximum value of static friction or limiting friction be denoted by f_{s}. When the block is just about to move, let the force applied be \(-\vec{F}\) .Then, \(\overrightarrow{f_s}=-\vec{F}\)

According to the laws of static friction, if a body is in contact with a plane, limiting friction (f_{s}) is directly proportional to the normal reaction (R). With the increase of normal reaction, the number of contact points also increases. So, this increases the adhesion within the area covered by the block. This in turn increases the force of friction.

Hence, f_{s }∝R or, f_{s} = μR

Where μ is a constant for that pair of surfaces. This constant is termed as the coefficient of static friction or simply, the coefficient of friction. The coefficient of friction depends on the materials of the two surfaces in contact and their smoothness.

The coefficient of friction does not depend on the area of the surfaces of contact.

∴ \(\mu=\frac{f_s}{R}=\frac{\text { limiting frictional force }}{\text { normal force }}\)

**Definition Of Coefficient Of Static Friction** is the ratio between the limiting friction and the normal force between the two surfaces in contact.

- The coefficient of friction (μ) being a ratio between two forces, is a dimensionless quantity and has no unit. The value of μ is generally less than 1. In special cases, μ can be equal to or even greater than 1.
- μ may rise up to 10, for two scientifically cleaned metal surfaces, kept in a vacuum. When highly polished and clear metal surfaces are brought together in a vacuum, they instantly form a single piece of metal and cannot slide over each other due to a sudden increase in frictional force.
- This happens because a large number of atoms (lying on both surfaces) come in contact in this case. This results in a stronger adhesive force which increases friction. For example, μ ≈ 1.6 for two copper surfaces, but μ ≈ 1 if the surfaces are made of glass.

**Coefficient Of Kinetic Friction:** As a wooden block slides over a table, two reaction forces act on the block

- Normal force (R) and
- Force of kinetic friction (f
_{k}).

According to the law of kinetic friction, the forces stated above are directly proportional to each other, that is, \(f_k \propto R \quad \text { or, } f_k=\dot{\mu}^{\prime} R \quad \text { and } \mu^{\prime}=\frac{f_k}{R}\)

The constant of proportionality μ’ is called the coefficient of kinetic friction.

This constant μ’ is a ratio between two forces. So, it is dimensionless and has no unit.

As kinetic friction is less than limiting friction, the coefficient of kinetic friction is less than that of static friction μ, i.e., μ’ < μ for a given pair of surfaces. It decreases further if the relative velocity between the surfaces becomes very high. However, for rough estimates, μ’ is often taken to be equal to μ.

**Coefficient Of Kinetic Friction Definition:** For two surfaces in contact, the ratio of the kinetic friction to the normal force, is called the coefficient of kinetic friction.

**Motion Over A Rough Surface:** To move a body over a rough surface, the applied force needs to be greater than the force of friction between the surfaces in contact.

Let the acceleration produced on application of a force F on mass m, be a. As the surface is not smooth, the motion of the body is opposed by kinetic friction. Let the normal force of the plane be R, and the coefficient of kinetic friction be μ’. Hence, the force of kinetic friction =μ’R.

∴ \(F-\mu^{\prime} R=m a \quad \text { or, } a=\frac{F-\mu^{\prime} R}{m}\)

## Class 11 Physics Laws Of Motion Friction

## Laws Of Friction Numerical Examples

**Example 1. To set a body of mass 5 kg in motion over a horizontal surface, a minimum force of 30 N has to be applied. What is the value of the coefficient of friction?**

**Solution:**

Force of limiting friction, f = 30 N; weight of the body, mg =5 x 9.8 = 49 N

Normal force R by the plane is equal to the weight of the body as it is on a horizontal plane.

∴ R = mg = 49 N

Hence, μ = \(\mu=\frac{f}{R}=\frac{30}{49}\) = 0.612.

**Example 2. An iron block of mass 10 kg is kept on a horizontal floor. The block is pulled by a rope at an angle 30° with the floor. What should be the minimum force necessary to set the block in motion? Given μ = 0.5**

**Solution:**

Let the minimum force applied by the rope be T which makes an angle θ with the horizontal.

The horizontal and vertical components of T are T cosθ and T sinθ respectively. T sinθ acts in the direction of the normal force (R) as shown in the diagram. In this problem, θ = 30°.

As the block is about to move, the frictional force becomes equal to the force of limiting friction. The resultant of forces acting on the body is zero, as the block is still stationary.

According to,

net horizontal force = Tcosθ -μR

net vertical force = R + Tsinθ- W

As the block is stationary,

∴ Tcosθ-μR = 0 or, μR = Tcosθ…(1)

and R+ Tsinθ – W = 0 or, R = W- Tsinθ…(2)

Dividing (1) by (2), we get, \(\mu=\frac{T \cos \theta}{W-T \sin \theta}\)

or, \(T=\frac{\mu W}{\mu \sin \theta+\cos \theta}=\frac{0.5 \times 10 \times 9.8}{0.5 \times \frac{1}{2}+\frac{\sqrt{3}}{2}}\)

= \(\frac{49}{0.25+0.866}=43.9 \mathrm{~N}\)

**Example 3. A body moving on the surface of the earth at 14 m • s ^{-1}, comes to rest due to friction after covering 50 m. Find the coefficient of friction between the body and the earth’s surface. Given, acceleration due to gravity = 9.8 m • s^{-2}.**

**Solution:**

Initial velocity, u = 14m · s^{-1}; final velocity, v = 0; displacement, s = 50 m and mass of the body = m.

If retardation is a, then using the formula v² = u²-2as,

a = \(\frac{u^2}{2 s}=\frac{14 \times 14}{2 \times 50}=1.96 \mathrm{~m} \cdot \mathrm{s}^{-2}\)

∴ Kinetic friction, f_{k} = ma = 1.96 m N and normal force, R = mg = 9.8m N

∴ Coefficient of friction, \(\mu^{\prime}=\frac{f_k}{R}=\frac{1.96 m}{9.8 m}=0.2\)

**Example 4. A man holds a book vertically between his two palms so that it does not fall. The mass of the book is 1 kg and j the force exerted by each palm is 2.5 kg x g. Find the coefficient of friction between the book and the palm.**

**Solution:**

The weight of the book is 1 kg x g, which acts vertically downwards.

As two surfaces of the book are passed by two hands, the downward force on each surface of the book = \(\frac{1 \mathrm{~kg} \times g}{2}=0.5 \mathrm{~kg} \times g.\)

As the book is at rest, the frictional force acts upwards to balance this downward force of 0.5 kg x g on each surface.

So the limiting force of friction, f = 0.5 kg x g.

Force exerted by each hand = normal force (R) = 2.5 kg x g

∴ Coefficient of friction, \(\mu=\frac{f}{R}=\frac{0.5 \mathrm{~kg} \times g}{2.5 \mathrm{~kg} \times g}=0.2\)

**Example 5. A block of mass 0.1 kg is kept pressed onto a wall by applying a horizontal force of 5 N. If the coefficient of friction between the block and the wall is 0.5, find the force of friction on the block.**

**Solution:**

Downward force on the block = weight of the block = mg = 0.1 x 9.8 = 0.98 N

Since the block is at rest, the downward force must be balanced by the upward frictional force.

∴ The frictional force on the block = 0.98 N.

**Example 6. Part of a uniform chain of length L is hanging out of a table. If the coefficient of friction between the chain and the table is μ, estimate the maximum length of the chain that can hang without slipping.**

**Solution:**

Let the maximum length of the chain that can hang out without slipping be l.

Length of the chain on the table = L-l.

If m is mass per unit length of the chain, the weight of the chain resting on the table =(L- l)mg

∴ The normal force of the table, R =(L-l)mg

The downward force on the hanging part = weight of the hanging part = lmg

To avoid slipping this downward force has to be balanced by the limiting friction f. i.e., f = lmg

Hence, coefficient of friction, \(\mu=\frac{f}{R}=\frac{l m g}{(L-l) m g}=\frac{l}{L-l}\)

or, \(l=\mu L-\mu l \quad \text { or, } l(1+\mu)=\mu L\)

∴ \(l=\frac{\mu L}{1+\mu} \text {. }\)

**Example 7. A tram is moving with an acceleration of 49 cm · s ^{-2} using 50 % of its motor power; the remaining 50 % is used up to overcome friction. Find the coefficient of friction between the wheel and the tram line.**

**Solution:**

Let the power of the motor be P, and the mass of the tram be m. Let the distance covered by the tram in time t be s.

As per given conditions, work against friction per second, \(\frac{\mu m g \times s}{t}=P \times 50 \%=\frac{P}{2}\)…(1)

and rate of work done for the accelerated motion, \(\frac{m a \times s}{t}=P \times 50 \%=\frac{P}{2}\) [where a = acceleration]…(2)

Now dividing (1) by (2) we get,

∴ \(\frac{\mu g}{a}=1 \text { or, } \mu=\frac{a}{g}=\frac{49}{980}=0.05 \text {. }\)

## Class 11 Physics Laws Of Motion – Angle Of Friction Cone Of Friction

**Angle Of Friction:** Let us consider a body placed on a rough surface. Now if a horizontal force is applied on the body, a frictional force comes into play. If the force applied on the body is increased, the friction also increases until it becomes equal to the limiting friction.

Let the limiting friction be denoted by \(\vec{f}\), the normal force acting on the body be denoted by \(\vec{R}\) and the resultant of the two forces \(\vec{R}\) and \(\vec{f}\) be denoted by \(\vec{Q}\). Now if \(\vec{Q}\) makes an angle A with \(\vec{R}\), then A is called the angle of friction.

**Angle Of Friction Definition:** The resultant of the limiting friction and the normal force between two surfaces in contact, makes an angle with the normal force. This angle is called the angle of friction.

From, R = Q cosλ, f = Q sinλ.

Again, \(\mu=\frac{f}{R}=\frac{Q \sin \lambda}{Q \cos \lambda}=\tan \lambda\)

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

Also, \(R^2+f^2=Q^2\left(\cos ^2 \lambda+\sin ^2 \lambda\right)=Q^2\)

or, \(R^2+\mu^2 R^2=Q^2 \quad \text { or, } Q=R \sqrt{1+\mu^2}\)

**Cone Of Friction:** Let a force be applied on a body at rest on a plane. The force of friction acts on the body exactly in the opposite direction. The applied force acts at point O of the body.

- If the direction of the applied force along the plane changes, the direction of the force of limiting friction f will accordingly change, thereby changing the direction of the resultant Q. But the direction of the normal force and the angle λ between Q and R remains the same, whatever be the direction of Q.
- Hence, taking into consideration the different directions of the applied force in the same plane, it is seen that the resultant Q makes an imaginary cone, with the normal force R as its axis. This imaginary cone is called the cone of friction.
- The vertex of the cone is the contact point O of the body and the plane, its axis is along the normal force R and the semi-vertical angle is the angle of friction λ.