The Coefficient of Friction (1) Teach A Level Maths The Coefficient of Friction (1)

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

R W Suppose we have a parcel resting on a table . . . Parcel at rest and we now attach a light inextensible string to the parcel and start to pull.

table W R Parcel moving a v T 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.

However, if the contact between the surfaces is rough, the attempt to move the parcel produces a frictional force. table W R Parcel still at rest T When we introduce a force trying to move the parcel, we say there is a tendency to move. Fr

The harder we pull, the larger the frictional force becomes until it cannot get any larger and the parcel moves. table W Parcel still at rest T1 Fr1 R1 table W Parcel accelerating T2 Fr2 a v R2 T2 > T1 Ans: The vertically upward forces are R and the component of T. They are balanced by W, so if T increases, R decreases. Why does R change when T does. Does R get larger or smaller as T increases ? When the parcel is still at rest but about to move, we say it is in limiting equilibrium.

m is called the coefficient of friction Limiting Equilibrium table W R Parcel about to move T Fr In limiting equilibrium, the frictional force, Fr is proportional to the normal reaction, R. In limiting equilibrium, Fr = mR m is a Greek letter pronounced “mew” and its value depends on the slipperiness of the two surfaces. m is called the coefficient of friction

pulling force = friction Fr mass of block M wt = reaction Mg = R pulling force = friction Fr 1.49 14.602 3.9 1.69 16.562 4.4 1.89 18.522 4.9 2.09 20.482 5.5 2.29 22.442 6 Spreadsheet

Values of m can only be found by experiment. Here are some approximate values: Metal on snow : 0·1 ( e.g. a ski on snow ) Leather on Wood: 0·35 ( e.g. a shoe on a floor ) Rubber on Asphalt: 0·7 ( e.g. a tyre on a road ) The values are for dry conditions before movement starts. 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. Values of m also usually decrease when movement starts but we will ignore this in our models.

So, in limiting equilibrium and during movement, Fr is at its greatest value and is given by Fr = mR If we just know that a body is in equilibrium, Fr may not have reached its largest value, so in equilibrium, Fr mR <

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 R Fr 2g Fr = mR Fr - 2gsin 25 = 0  Fr = 8·28… R - 2gcos 25 = 0 Find the coefficient of friction between the particle and plane if the particle is about to slip. 25 Weight: 2g newtons R Fr   Solution:   “About to slip”  limiting equilibrium 25 2g   The particle is about to slip down the plane so the frictional force is up the plane. Limiting equilibrium  Fr = mR ------ (1) To find m from this equation we need Fr and R, so we need 2 more equations. Resolving: Fr - 2gsin 25 = 0  Fr = 8·28… R - 2gcos 25 = 0  R = 17·7… R Fr Substitute in (1): Fr = mR  m =  m = 0·47 ( 2 d.p. )

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 table 2 50 The particle is of mass 300 g and stays at rest on a rough horizontal table. What is the range of values of the coefficient of friction between the particle and table ?

2 R Fr  equilibrium 0·3g Fr mR <  Fr R Fr R  < m  m > What is the range of values of the coefficient of friction between the particle and table ? grams table 2 50 mass: 300 g R Solution: Fr “The particle stays at rest”  equilibrium 0·3g ( but not necessarily limiting equilibrium ) Fr mR <  Fr R Fr R  < m  m > . . . . . . (1)

2 R m > Fr R Fr 0·3g Fr - 2 cos 50 = 0 Fr =  = 2 cos 50 What is the range of values of the coefficient of friction between the particle and table ? table 2 50 R     ------ (1) m > Fr R Fr   0·3g   Resolving: Fr - 2 cos 50 = 0 Fr =  = 2 cos 50 Fr = 1·28…  R + 2 sin 50 - 0·3g = 0  R = 0·3g - 2 sin 50  R = 1·40… m Fr R > Substitute in (1): >  m 0·91 ( 2 d.p. )

( Model the trunk as a particle ) 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. table P 30 (a) What is the maximum value of the magnitude of the frictional force if P = 300 and the coefficient of friction is 0·8 ? ( Model the trunk as a particle ) (b) Will the trunk move ? (c) What is the actual size of the frictional force ?

R P = 300 m = 0·8 Fr P = 300 20g Fr = mR R - P sin 30 - 20g = 0     P = 300 mass = 20 kg 30 m = 0·8 Fr table P = 300 20g   (a) Find the maximum value of the magnitude of the frictional force. Solution: The maximum value of Fr is given by Fr = mR ------ (1) Resolving: R - P sin 30 - 20g = 0  R = P sin 30 + 20g  R = 300 sin 30 + 20  9·8  R = 346 Substitute in (1)  Fr = 0·8  346 = 277 N ( 3 s.f. )

R P = 300 Fr (max) = 277 Fr 20g 300cos 30 = 260 30 table = 300 30 20g R Fr P     Fr (max) = 277   (b) Will the trunk move ? Solution: The trunk moves if the horizontal component of P is greater than the maximum value of friction. Resolving P: 300cos 30 = 260 Since the component of P is less than the maximum value of Fr, the trunk will not move.

R P = 300 Fr (max) = 277 Fr 20g 30 Horizontal component of P = 260 table = 300 30 20g R Fr P Fr (max) = 277 Horizontal component of P = 260 (c) What is the actual size of the frictional force ? Solution: The frictional force cannot be greater than the component of the pushing force, so, Fr = 260 newtons.

Fr mR < Fr = mR SUMMARY In equilibrium ( that is the body is either at rest or moving with constant velocity ) we have the inequality Fr mR < where Fr is the force due to friction, m is the coefficient of friction and R is the normal reaction. We assume that m is constant for any pair of surfaces. If the body is on the point of moving we have limiting equilibrium and Fr is at its largest value, so Fr = mR When movement starts, Fr stays at its maximum value, Fr = mR.

EXERCISE A skier of mass 80 kg is in limiting equilibrium on a slope inclined at 7 to the horizontal. 7 Modelling the skier as a particle, find (a) the magnitudes of the frictional force and the normal reaction. (b) the coefficient of friction between the ski and the snow.

7 R Fr 80g Fr - 80g sin 7 = 0  Fr = 95·5 ( 3 s.f. ) R - 80g cos 7 EXERCISE 7 R Fr mass: 80 kg Solution: 80g (a) Resolving: Fr - 80g sin 7 = 0  Fr = 95·5 ( 3 s.f. ) R - 80g cos 7 = 0  R = 778 ( 3 s.f. ) (b) Limiting equilibrium  Fr = mR R Fr  m =  m = 0·12 ( 2 d.p. )

EXERCISE 2. 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.

R W = 200 T m = 0·4 Fr 200 Fr = mR R - 200 = 0  R = 200 T - Fr = 0 EXERCISE R Solution: W = 200 T cord m = 0·4 Fr 200 The particle is moving, so, Fr = mR ------ (1) Resolving: R - 200 = 0  R = 200 T - Fr = 0  T = Fr ------ (2) Substitute for R in (1) : Fr = mR  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.

THE COEFFICIENT OF FRICTION (1) 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 mR < Fr = mR 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 = mR . where Fr is the force due to friction, m is the coefficient of friction and R is the normal reaction. We assume that m is constant for any pair of surfaces.