Friction

definition

Static Friction

Static friction is the force which acts between two bodies when a force is applied on a body but there is no relative motion between the surfaces. Static friction opposes an impending motion. It is denoted by

f_{s}

. Law of static friction is given by:

f_{s} \leq μ_{s} N

where

μ_{s}

is the coefficient of static friction and depends only on the nature of the surfaces in contact.

N

is the normal force.

example

Rolling Friction

A solid sphere of mass

m

is placed on a rough inclined plane as shown in the figure. The coefficient

μ

is insufficient to start pure rolling. The sphere slides a length

ℓ

on the incline from rest and its kinetic energy becomes

K

. Then, the work done by friction will be given by:
Work done by friction + Work done by gravity + work done by normal force = Kinetic energy

K

(Work energy theorem)
work done by normal force=0
Work done by gravity is mglsin

θ

Work done by friction =

- m g l s i n θ + K

example

Methods to increase or reduce friction

To reduce friction we can make the surfaces in contact smooth. Also adding a lubricant to two surfaces in contact reduces friction between them. Grease is added between the moving parts in machines, to reduce friction. Another method of reducing friction is by using bearings. When ball bearings are introduced between two surfaces friction is reduced because of the freely rotating metal balls.

To increase friction, the surfaces in contact have to be made rough. The two surfaces in contact can also be pressed harder to increase force between them. Use of adhesive materials that stick surfaces , also increases friction between them.

definition

Self-adjusting nature of friction

When a moving body is suddenly stopped, then frictional force arises when there is a relative motion between two surfaces. When body is stopped means there is no relative motion and no resultant force is there hence frictional force acting becomes zero.

example

Laws of friction and example

Laws of Friction are:

When an object is moving, the friction is proportional and perpendicular to the normal force (N)
Friction is independent of the area of contact as long as there is an area of contact.
The coefficient of static friction is slightly greater than the coefficient of kinetic friction.
Within rather large limits, kinetic friction is independent of velocity.
Friction depends upon the nature of the surfaces in contact.

Example: A body of mass

M

is placed on a rough inclined plane of inclination

θ

and coefficient friction

μ_{k}

. A force of (

m g s i n θ + μ_{k} m g c o s θ

) is applied in the upward direction, the acceleration of the body is:
Equating forces along X and Y axes,

N = m g c o s θ

Also,

- m g s i n θ - μ_{k} m g c o s θ - m a + m g s i n θ + μ_{k} m g c o s θ = 0

∴ m a = 0

example

Total contact force between surfaces

Example: For the arrangements shown in figure, acceleration of block B is

\sqrt{3} m / s^{2}

upwards. Find the normal reaction (in

k N

) between the
surfaces of contact of the two blocks.

Solution: Let horizontal direction is

x

and vertical direction is

y

and

N

is normal force between inclined surfaces, perpendicular to surface
direction,then constrained motion is defined as

x = y c o t θ

then

a_{x} = a_{y} c o t θ

a_{x} = c o t 30 \times \sqrt{3}

a_{x} = 3 m s^{-} 2

F - N s i n θ = m a_{x}

5000 - N s i n 30 = 1000 \times 3

N = 4000 N = 4 k N

definition

Angle of Friction

It is the angle (

α

), measured between the normal force (N) and resultant force (R).

A body rests on a rough horizontal plane. A force is applied to the body directed towards the plane at an angle

α

with the vertical.
The body can be moved along the plane:
Consider a block placed on a rough horizontal plane. Now, the reaction force

\vec{R}

is because it is equal and opposite to the
weight

\vec{W}

. If the force

\vec{F}

is applied to the block towards the plane at an angle

α

, the resolved forces will act along vertical and horizontal direction.
The horizontal component of force

F \sin α

have to overcome the frictional force so that the block just begins to slide. Frictional force is equal to limiting friction

F_{l i m i t i n g}

, when this condition is satisfied the angle of applied force

α

will be greater than angle of friction

α

and block can move along the plane.

t a n α = μ

diagram

Graph of total external force Vs frictional force

definition

Static Friction

Static friction is the force which acts between two bodies when a force is applied on a body but there is no relative motion between the surfaces. Static friction opposes impending motion. It is denoted by

f_{s}

. law of static friction is given by:

f_{s} \leq μ_{s} N

where

μ_{s}

is the coefficient of static friction and depends only on the nature of the surfaces in contact.

N

is the normal force.

definition

Limiting Friction

The maximum static friction that a body can exert on the other body in contact with it is called limiting friction. This limiting friction is proportional to the normal contact force between the two bodies, It is given by:

(f_{s})_{m a x} = μ_{s} N

definition

Coefficient of static friction

Coefficient of static friction is a property of pair of surfaces. The coefficient of friction depends on the materials used; for example, ice on steel has a low coefficient of friction, while rubber on pavement has a high coefficient of friction. Coefficients of friction range from near zero to greater than one.

example

Maximum Possible Static Friction

A horizontal force of

20

N

is applied to a block of mass

4

k g

resting on a rough horizontal table. The value of coefficient of static friction between the block and the table for the block remains at rest [

g = {10 m s}^{- 2}

]

Mass,

m = 4 k g

. Friction is a self-adjusting force.So, normal reaction =

4 g

.Since the block is at rest, so limiting friction force

\geq

applied force.i.e.,

μ_{S} m g \geq 20

\Rightarrow μ_{S} \geq 0.5

. But value of coefficient of static friction is fixed. We can definitely say that

μ_{S} \geq 0.5

example

Example of maximum static friction in non-inertial frame

Example:
A block of metal is lying on the floor of a bus. Find the maximum acceleration which can be given to the bus so that the block may remain at rest.
Solution:Let

a

be the acceleration of the bus. The contact between the metal block and the bus floor is due to the frictional force. The forces acting on the block are the inertia force and the frictional force. For equilibrium we have

m a = μ m g \Rightarrow a = μ g

where

μ

is the friction coefficient in between block and the floor of bus.

definition

Angle of Repose

The angle of repose is the steepest angle of descent or dip relative to the horizontal plane to which a body doesn't slide on a slope.It is found that a body on an inclined plane just starts sliding down if inclination is

s i n^{- 1} (\frac{3}{5})

, the angle of repose (friction) is: We know,
Angle of repose = angle of friction
Also,

μ = t a n θ

θ = s i n^{- 1} (\frac{3}{5})

Also,
By trigonometry,

θ = t a n^{- 1} (\frac{3}{4})

definition

Friction that involve multiple forces

A block of mass

m

is pressed against a vertical wall by applying a force

P

at an angle

θ

to the horizontal as shown in the figure. As a result, if that block is prevented from falling down and

μ

is the coefficient of static friction between the block and the wall, the value of

P

will be given as:
From the FBD of block, we have

P c o s θ = N

where

N

is normal reaction.

μ N - P s i n θ = m g

Or,

μ P c o s θ - P s i n θ = m g

P = \frac{m g}{μ c o s θ - s i n θ}

definition

Kinetic Friction

Frictional force that opposes relative motion between surfaces in contact is called kinetic or sliding friction and is denoted by

f_{k}

. It is given by:

f_{k} = μ_{k} N

where

μ_{k}

is the coefficient of kinetic friction, which depends only on the surfaces in contact and

N

is the normal force.
Example: When a wooden block is sliding on the floor the friction acting between the wood surface and the floor will be kinetic friction.

example

Kinetic Friction: Explained

Frictional force that opposes relative motion between surfaces in contact is called kinetic or sliding friction and is denoted by

f_{k}

. It is given by:

f_{k} = μ_{k} N

where

μ_{k}

is the coefficient of kinetic friction, which depends only on the surfaces in contact and

N

is the normal force.
Example: When a wooden block is sliding on the floor the friction acting between the wood surface and the floor will be kinetic friction.

result

Qualitative Difference Between Static and Kinetic Friction

SI.	Static Friction	Kinetic Friction
Nature of Force	It is the frictional force acting between two surfaces which are attempting to move, but are not moving.	It is the frictional force acting between two surfaces which are in motion against each other.
Action	It acts when the surfaces are not in relative motion against each other.	It acts when the surfaces are in relative motion against each other.
Range	It increases linearly with the force applied until it reaches a maximum value.	It remains constant regardless of the force applied.

definition

Coefficient of kinetic friction

Coefficient of kinetic friction (as given in fig(1)) is defined by

μ_{k} = \frac{F_{k}}{N}

example

Retardation due to friction

A car of mass

1000

k g

moving with a velocity of

10 m s^{- 1}

is acted upon by a forward force of

1000

N

due to engine and retarding force of

500

N

due to friction. It's velocity after

10

s

is:

For a body in equilibrium,
Equating forces along Y-axis,
Normal reaction

N = m g

Equating forces along X-axis,

m a + μ N = F

m a + μ m g = F

m a = F - μ m g

a = \frac{F - μ m g}{m}

a = \frac{1000 - 500}{1000}

∴ a = 0.5 m / s^{2}

Now,

v = u + a t

∴ v = u + a \times t

∴ v = 10 + 0.5 \times 10

∴ v = 15 m / s

example

Finding coefficient of friction

Example. A car begins to move at time

t = 0

and then accelerates along a straight track with a speed given by

V (t) = 2 t^{2} m s^{-} 1

for

0 \leq t \leq 2

. After the end of acceleration, the car continues to move at constant speed. A small block initially at rest on the floor of the car begins to slip at

t = 1 s e c

. and stops slipping at

t = 3 s e c

. Find the coefficient of static and kinetic friction between the block and the floor.
Solution:

a (t) = \frac{d v (t)}{d t} = 4 t

t = 1

a = 4

When the block is just starting to slip we have,

μ_{s} N = m a

\Rightarrow μ_{s} m g = 4 m

\Rightarrow μ_{s} = 0.4

Now, it goes on slipping till 3 seconds, therefore in 2 seconds the block's
velocity changes from

2 t o 2 \times (2)^{2} = 8 m / s e c

Hence, acceleration from

t = 1

t = 3

secs is:

a_{a v g} = \frac{8 - 2}{2} = 3 m / s^{2}

Applying equation of motion

μ_{k} \times m \times g = m \times a_{a v g} = 3 m

\Rightarrow μ_{k} = 0.3

example

Friction acting at one fixed horizontal surface

The maximum value of the force

F

such that the block shown in the arrangement, does not move is given by:
Block does not move till the horizontal force on it becomes more than the maximum static frictional force.

F_{m a x} \cos \frac{π}{3} = μ m g N μ (m g + F_{m a x} \sin \frac{π}{3}) = \frac{1 \times (\sqrt{3} \times 10 + \frac{{\sqrt{3} F}_{m a x}}{2})}{2 \sqrt{3}} \Rightarrow F_{m a x} = 20 N

example

Friction on Multiple horizontal surfaces

A block A of mass

2

kg rests on another block B of mass

8

kg which rests on a horizontal floor. The coefficient of friction between A and B is

0.2

, while that between B and the floor is

0.5

. When a horizontal force of

25

N is applied on B, the force of friction between A and B is:
Here

F = μ m g = 0.5 \times 10 \times 10

50 N

To move block of

8

kg along

2

kg,

50 N

force is required
But only

25 N

is applied.

∴

Block 'B' will not move.
Hence relative motion between A and B is not there hence friction

= 0

definition

Friction at fixed inclined surface

A plane surface is inclined at an angle of

60^{0}

. A body of mass

10

kg is placed on it. If the value of coefficient of friction

μ_{K}

, between the body and the inclined surface is

0.2

, calculate the downward acceleration of the body, along the inclined plane surface. (Take

g = 10 m s^{- 2}

) At angles greater than the the critical angle of inclination, the block slides down the incline with uniform acceleration

a

.The frictional force is

μ_{K} N

. Here

N

is the normal reaction.
The net force acting on the body in a direction along the plane is:

F_{x} = m g s i n θ - μ_{K} m g c o s θ = m a

Hence, the acceleration

a

of the body is related to

θ

μ_{K}

by the equation:

a = g (s i n θ - μ_{k} c o s θ)

On substituting the respective values:

a = 10 (\frac{\sqrt{3}}{2} - 0.2 \times \frac{1}{2})

a = 7.66 m s^{- 2}

definition

Kinetic friction in accelerated frame

Block A of mass

35

kg is resting on a frictionless floor. Another block B of mass

7

kg is resting on it as shown in the figure. The coefficient of friction between the blocks is

0.5

while kinetic friction is

0.4

. If a horizontal force of

100

N. is applied to block B, the acceleration of the block A will be

(g = 10 m s^{- 2})

f = 0.4 \times 7 \times 10 = 28

a = \frac{f}{35} = \frac{28}{35} = 0.8 m / s^{2}

example

Problems on friction where magnitude of force is variable

Example: A variable force F

= 10 t

is applied to block B placed on a smooth surface. The coefficient of friction between A and B is

0.5

. (

t

is time in seconds. Initial velocities are zero). At what instant block A starts sliding on B?

Solution:

f_{m a x} = μ \times 3 g = 0.5 \times 3 \times 10 N = 15 N

where

f_{m a x}

is max force of static friction
Block A starts sliding when friction becomes

f_{m a x}

\Rightarrow 15 = 3 a

\Rightarrow a = 5 m / s^{2}

Both the blocks will move with same acceleration at this instant

F - 15 = 7 a

10 t - 15 = 7 a = 7 \times 5

\Rightarrow 10 t = 50

\Rightarrow t = 5 s

Work done

= \int F . d s = \int 10 t . d s (a = \frac{F}{m} = \frac{10 t}{10} = t)

= \int_{0}^{5} 10 t . V d t (d s = v d t)

= \int_{0}^{5} 10 t . \frac{t^{2}}{2} d t (v = \int a d t = \int t d t = \frac{t^{2}}{2})

= \int_{0}^{5} 5 t^{3} d t = 5 {[\frac{t^{4}}{4}]}_{0}^{5} = \frac{5}{4} [625 - 0] = \frac{625 \times 5}{4} = 781.25 J

.
This is the amount of heat produced due to friction.

TB - UG

Monday, July 31, 2017

Friction

Static Friction

Rolling Friction

Methods to increase or reduce friction

Self-adjusting nature of friction

Laws of friction and example

Total contact force between surfaces

Angle of Friction

Graph of total external force Vs frictional force

Static Friction

Limiting Friction

Coefficient of static friction

Maximum Possible Static Friction

Example of maximum static friction in non-inertial frame

Angle of Repose

Friction that involve multiple forces

Kinetic Friction

Kinetic Friction: Explained

Qualitative Difference Between Static and Kinetic Friction

Coefficient of kinetic friction

Retardation due to friction

Finding coefficient of friction

Friction acting at one fixed horizontal surface

Friction on Multiple horizontal surfaces

Friction at fixed inclined surface

Kinetic friction in accelerated frame

Problems on friction where magnitude of force is variable

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