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The Coefficient of Friction The Coefficient of Friction (1)(1)The Coefficient of Friction The Coefficient of Friction (1)(1)

Volume 4: Mechanics 1Volume 4: Mechanics 1The Coefficient of Friction The Coefficient of Friction

(1)(1)

Volume 4: Mechanics 1Volume 4: Mechanics 1The Coefficient of Friction The Coefficient of Friction

(1)(1)

table

Parcel at rest

Suppose we have a parcel resting on a table . . .

W

R

and we now attach a light inextensible string to the parcel and start to pull.

table

W

RT

If our model assumes the contact between the parcel and the table is smooth there will be no frictional force and the tension makes the parcel accelerate.

Parcel moving a

v

table

W

R

Parcel still at rest

TFr

When we introduce a force trying to move the parcel, we say there is a tendency to move.

However, if the contact between the surfaces is rough, the attempt to move the parcel produces a frictional force.

table

W

Parcel still at rest

T1Fr1

R1

The harder we pull, the larger the frictional force becomes until it cannot get any larger and the parcel moves.

table

W

Parcel accelerating

T2Fr2

avR2

T2 T1

When the parcel is still at rest but about to move, we say it is in limiting

equilibrium.

Why does R change when T does. Does R get larger or smaller as T increases ?

Ans: The vertically upward forces are R and the component of T. They are balanced by W, so if T increases, R decreases.

table

W

R

Parcel about to move

TFr

In limiting equilibrium, Fr R

Limiting Equilibrium

is a Greek letter pronounced “mew” and its value depends on the slipperiness of the two surfaces.

In limiting equilibrium, the frictional force, Fr

is proportional to the normal reaction, R.

is called the coefficient of friction

mass of block M

wt = reaction Mg = R

pulling force = friction Fr

1.49 14.602 3.91.69 16.562 4.41.89 18.522 4.92.09 20.482 5.52.29 22.442 6

Spreadsheet

If you ride a bike or drive a car you will know that the coefficient of friction is less when the road and tyres are wet.

Here are some approximate values:

The values are for dry conditions before movement starts.

Rubber on Asphalt: 0·7 ( e.g. a tyre on a road )

Metal on snow : 0·1 ( e.g. a ski on snow )Leather on Wood: 0·35 ( e.g. a shoe on a floor )

Values of can only be found by experiment.

Values of also usually decrease when movement starts but we will ignore this in our models.

If we just know that a body is in equilibrium, Fr may not have reached its largest value, so

Fr R

Fr R

So, in limiting equilibrium and during movement, Fr is at its greatest value and is given by

in equilibrium,

e.g.1 A particle of mass 2 kg rests on a rough plane which is inclined at 25 to the horizontal.

Find the coefficient of friction between the particle and plane if the particle is about to slip, giving the answer correct to 2 decimal places.

25

25

Solution:

2g

FrR

Weight: 2g newtons

The particle is about to slip down the plane so the frictional force is up the plane.

Limiting equilibrium

Fr RTo find from this equation we need Fr and R,

so we need 2 more equations.

Resolving:

Fr 2gsin 25 0

------ (1)

Fr =

R 2gcos 25 0 R 17·7…Substitute in

(1): Fr R

8·28…

RFr

0·47 ( 2 d.p. )

“About to slip” limiting equilibrium

Find the coefficient of friction between the particle and plane if the particle is about to slip.

25

e.g.2 A force of magnitude 2 N, at an angle of 50 to the horizontal, acts on a particle as shown in the diagram

What is the range of values of the coefficient of friction between the particle and table ?

table

2

50

The particle is of mass 300 g andstays at rest on a rough horizontal table.

table

2

50

mass: 300 g

0·3g

Fr

R

grams

. . . . . . (1)

“The particle stays at rest” equilibrium

Fr R

( but not necessarily limiting equilibrium )

Fr

RFr

R

Solution:

What is the range of values of the coefficient of friction between the particle and table ?

table

2

50

0·3g

Fr

R

2 sin 50

Resolving:

R 0 0·3g

2 cos 50Fr 0Fr 1·28…

R 0·3g 2 sin 50 R ·40…Substitute in

(1):

------ (1) Fr

R

Fr

R 0·91

2 cos 50Fr

( 2 d.p. )

What is the range of values of the coefficient of friction between the particle and table ?

e.g.3 A trunk of mass 20 kg rests on a rough horizontal floor. It is pushed with a force of magnitude P newtons from an angle of 30 above the horizontal. (a) What is the maximum value of the

magnitude of the frictional force if P = 300 and the coefficient of friction is 0·8 ?

(b)Will the trunk move ?

( Model the trunk as a particle )

table

P

30

(c) What is the actual size of the frictional force ?

R P sin 30 20g

table

= 300

30

mass 20

kg

20g

R

Fr

P sin 30Fr RResolvi

ng:

R 0

R 300 sin 30 20

9·8Substitute in (1)

277 N ( 3 s.f. )

Solution: The maximum value of Fr is

given by 20g

R 346

------ (1)

0·8

(a) Find the maximum value of the magnitude of the frictional force.

P

Fr 0·8 346

P 300

table

= 300

30

20g

R

Fr

P

Fr (max) 277

Resolving P: 300cos 30

(b)Will the trunk move ?

The trunk moves if the horizontal component of P is greater than the maximum value of friction.

Since the component of P is less than the maximum value of Fr, the trunk will not move.

= 260

Solution:

Fr (max) 277

The frictional force cannot be greater than the component of the pushing force, so, Fr 260 newtons.

Solution:

table

= 300

30

20g

R

Fr

P

Horizontal component of P 260(c) What is the actual size of the frictional

force ?

SUMMARY In equilibrium ( that is the body is either at

rest or moving with constant velocity ) we have the inequality

Fr R

Fr R

If the body is on the point of moving we have limiting equilibrium and Fr is at its largest value, so

When movement starts, Fr stays at its

maximum value, Fr R.

where Fr is the force due to friction, is the

coefficient of friction and R is the normal reaction. We assume that is constant for any pair of surfaces.

EXERCISE

1. A skier of mass 80 kg is in limiting equilibrium on a slope inclined at 7 to the horizontal.

(a) the magnitudes of the frictional force and the normal reaction.

(b)the coefficient of friction between the ski and the snow.

Modelling the skier as a particle, find

7

7

EXERCISE

Solution:

mass: 80 kg

80g

R Fr

(b) Limiting equilibrium

Fr 80g sin 7 0 Fr 95·5 ( 3 s.f. )

R 80g cos 7 0 R 778 ( 3 s.f. )

RFr 0·12 ( 2 d.p. )

(a) Resolving:

Fr R

EXERCISE2. A box is being dragged at constant speed

in a straight line along a rough horizontal surface by a horizontal cord tied to the top corner. The coefficient of

friction between the box and the ground is 0·4 and the weight of the box is 200 newtons.

By modelling the box as a particle, find the tension in the cord.

EXERCISE

200

R

Fr

Solution:

cord

0·4

W 200 T

Resolving:

R 0 200 Fr 0

R 200T

The particle is moving, so,

Fr R ------ (1)

T FrSubstitute for R in

(1) :Fr R

------ (2)

Fr 0·4 200 Fr 80

Substitute in (2) :

T Fr T 80 newtons

The following page contains the summary in a form suitable for photocopying.

Summary

THE COEFFICIENT OF FRICTION (1)

TEACH A LEVEL MATHS – MECHANICS 1

In equilibrium ( that is the body is either at rest or moving with constant velocity ) we have the inequality

Fr R

Fr R

If the body is on the point of moving we have limiting equilibrium and Fr is at its largest value, so

When movement starts, Fr stays at its maximum value, Fr R .

where Fr is the force due to friction, is the coefficient of

friction and R is the normal reaction. We assume that is constant for any pair of surfaces.