Peter's Physics Pages

Lecture 12 (Friction)

In this lecture the following are introduced:
• The force of static friction
• The force of kinetic friction

Friction is a force that opposes motion between two surfaces because of the roughness between the surfaces.

Static friction

When a force is applied to a mass in contact with a rough surface and the mass does not move. It means an opposition force appears which equals the applied force in size.

This opposition force is called the force of static friction.

It arises because the surfaces in contact are not microscopically smooth, as shown in this diagram.

Diagram by Tahir Cagin, Jianwei Che, Michael N. Gardos, Amir Fijany, and William A. Goddard,III at

SEM Tribometer

At spots where the atoms meet, cold-welding occurs which joins the materials together.
Static frictional force is equal to the applied force, up to a maximum force, where the welds break.

Maximum static friction depends on the roughness of the surfaces and how hard the two surfaces are pushed together.

Maximum static friction (Fs) is given by: Fs = μsN

where
• μs is the coefficient of static friction and measures the roughness of the surfaces.
• and N is the Normal reaction force, which pushes the surfaces together.

Note that friction does not normally depend on the area in contact because not all of it is really in contact. The actual contact area may be one thousandth of what it seems to the naked eye. This picture shows that there are peaks on the surface which touch and keep most of the two surfaces apart.

In car racing, tyres are called "wets" and "slicks" and made with rubber that becomes sticky to give greater contact between the two surfaces.

Kinetic friction

When maximum static friction is exceeded the surfaces start to slide past each other and kinetic friction takes over. Kinetic frictional force is reasonably constant but smaller than maximum static friction.

 Fk = μkN where • μk is the coefficient of kinetic friction. • N is the Normal reaction force.

Some coefficients

 Surfaces Static Friction Kinetic Friction Steel on steel (dry) 0.6 0.4 Steel on steel (greasy) 0.1 0.05 Teflon on steel 0.04 0.04 Brake lining on cast iron 0.4 0.3 Rubber tires on dry pavement 0.9 0.8 Metal on ice 0.02 Rubber crutch-tip on rough wood 0.7

Anti-lock brakes on a car maintain static friction instead of kinetic friction. At 25 m.s-1 (90 km/h) the difference between friction coefficients of 0.9 and 0.8 means a 4.4 m difference in stopping (about a small car length).

schematic diagram of ABS

Experimental measurement of static friction

In the diagram below, a block rests on a rough inclined plane. There are three forces acting on the block, the weight (W), the normal reaction of the plane (N), and frictional force opposing the motion of the block down the plane. The weight can be resolved into a force down the plane and a force into the plane. The force into the plane equals the normal reaction.

 At the point of slipping, the force down the plane equals maximum static friction. This gives: W sinθ = Fs= μsN = μs W cosθ Dividing both side by W: sinθ = μscosθ μs = tanθ

Example F1

A 3 kg block lies at rest on a rough horizontal table. The table is raised slowly on one side till it makes an angle of 37° with the horizontal. At this angle the block is at the point of moving. Find the coefficient of friction for the block on the plane.

μs = tan37° = 0.75

Tutorial Questions

61. A 7 kg block lies at rest on a rough horizontal table. The table is raised slowly on one side till it makes an angle of 36°52' with the horizontal. At this angle the block starts to move with constant speed down the table. Find the coefficient of friction for the block on the plane.

62. A 9 kg block lies on a rough plane (coefficient of friction 0.083) which is inclined upwards at an angle of 30° to the horizontal. A light inextensible string attached to the block passes up over a light pulley at the top of the plane and then down to a 0.8 kg mass at the other end. Assume that the system is at rest at the moment it is released. Find
(a) the acceleration of the system after release.
(b) the tension in the string after release.

63. A mass of 6 kg is on a rough horizontal table (coefficient of friction 0.5). A light inextensible string is attached to the mass and passes over a light pulley to a second mass of 3.8 kg. The system moves from rest under the influence of gravity and travels 2.5 m. Use the principle of conservation of energy to find the speed of the system after travelling this distance.

E1. Two people push a freezer weighing 2000 N up a ramp inclined at 36.87° to the horizontal. The coefficient of sliding friction between the object and the ramp is 0.5. Find
(a) the minimum force the people apply to keep the freezer moving up the ramp.
(b) the acceleration that the freezer would have if it were released and started to slide down the ramp.

E2. The planet Mars has an acceleration due to gravity at the surface of 3.75 m.s-2. A rough plane (with coefficient of kinetic friction = 0.2) on Mars is inclined at 330 to the horizontal. A mass of 5 kg lies on the plane and a light inextensible wire is attached to the mass and it passes up the plane and over a pulley at the top and down to a mass of 10 kg hung on the other end as shown in the diagram. Find the acceleration of the 5 kg mass up the plane.

E3. A 60 kg skier, starting from rest, skis 61 m down a rough slope and loses 35 m in altitude. The skier is then travelling at 25 m.s-1. State (in words) the energy conservation law that applies in this case. Find
(a) the work done against friction.
(b) the average frictional force.

E4. An 8 kg block lies on a rough plane (coefficient of friction 0.1) which is inclined upwards at an angle of 30° to the horizontal. A light inextensible string attached to the block passes up over a light pulley at the top of the plane and then down to a 1.8 kg mass at the other end. Assume that the system is at rest at the moment it is released and that the 8 kg block will move down the plane. Find
(a) the acceleration of the system after release.
(b) the tension in the string after release.

E5. A 3 kg block rests on a rough horizontal plane (coefficient of friction 0.33). A light inextensible wire attached to the block passes over a pulley fixed at the end of the plane then down to a 5 kg block suspended at the other end. If the system is released from rest, find
(a) the acceleration of the system, and
(b) the tension in the wire

E6. A 2kg block rests on a rough horizontal plane where the coefficient of static friction is 0.3. Find the angle of the plane where the block just starts to move.

E7. A car is travelling along a straight horizontal plane at 72 km/h. If the minimum distance that the car can stop (without sliding) from this speed is 34 m, find the coefficient of static friction.

E8. A mass of 5 kg lies on a rough plane where the coefficient of kinetic friction is 0.2. A string attached to the mass passes over a light (ie massless) pulley as shown and a 4.8 kg mass is hung on the end of the string. Find
(a) the force, F, that will move the 5 kg mass away from the pulley with an acceleration of 1.2 ms-2, and
(b) the tension in the string when this happens.

E9. A 6 kg block lies on a rough horizontal plane with the coefficient of kinetic friction between it and the plane being 0.36. It is attached with a string via a pulley to a 14 kg block on a smooth (frictionless) plane inclined at 210 to the horizontal as shown in the diagram. Assume the string and the pulley are weightless and frictionless. Find
(a) the acceleration of the blocks, and
(b) the tension in the connecting string.

E10. The coefficient of static friction between the tyres and a dry road is 0.62. The mass of a car is 1500 kg. Find the maximum braking force obtainable on a dry road which slopes down at 90.

Summarising:

 Maximum static friction is given by:  Fs = μsN Kinetic frictional force is reasonably constant but smaller than maximum static friction.

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