Peter's Physics Pages

An Introductory Physics Course with Peter Eyland
Lecture 14 (Friction and Equilibrium)

In this lecture the following are introduced:
• The force of static friction
• The force of kinetic friction
• The conditions for equilibrium
• Couples

"Friction is the source of most pleasures in life" [E.P.E. 1975]. Without friction we couldn't walk, talk, eat, paint, play a musical instrument, drive or do most physical things. Friction is the 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 of surface atom roughness by Tahir Cagin, Jianwei Che, Michael N. Gardos, Amir Fijany, and William A. Goddard,III.

At spots where the atoms meet, cold-welding occurs which joins the materials together so that they do not slip relative to each other. The static frictional force is equal to the applied force, up to a maximum force, where the welds break.

The maximum static friction force (Fs) is given by: Fs = μs·N, where
• μs is the coefficient of static friction and is a pure number (i.e. no unit) that measures the roughness of the surfaces, and
• N is the Normal reaction force, which pushes the surfaces together.
Friction does not depend strongly 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 why.

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 and smaller than maximum static friction.

Fk = μk·N

where
• μk is the coefficient of kinetic friction.
• N is the Normal reaction force, pushing the surfaces together.

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 Braking Systems

Anti-lock brakes on a car maintain static friction instead of kinetic friction. This means that the wheels do not skid. 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

The driving force down the plane is the component of the weight down the plane, i.e. W·sinθ.
The Normal force pushing the two surfaces together equals the component of the weight into the plane, i.e. W·cosθ
At the point of slipping, the force down the plane (W·sinθ) equals the maximum static friction force.
So               W·sinθ = Fs = μs·N = μs·W·cosθ
Simplifying:  sinθ = μs·cosθ
From this:    μs = tanθ

Example:

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.

From above:

μs = tanθ = tan37° = 0.75

Equilibrium

There are two conditions to be met for a body to be in equilibrium.
• it does not accelerate (force rule)
• it does not rotate (moment rule)

The force rule says that for equilibrium, the sum of the forces on a body is zero, i.e. the force vectors form a closed polygon. The moment rule says that for equilibrium, the sum of the moments about any axis is zero, with anticlockwise moments taken as positive and clockwise moments taken as negative (the maths angle convention). Free-body force diagrams are used which show only the forces acting on the body.

Example
A car has mass 1200kg and the distance between the front and rear wheels is 4m. There is 2.5m from the rear wheels to the centre-of-mass. Find the reaction forces that the front and rear wheels apply to the car.

Using the force law for the vertical forces:

Using the moment law by taking moments about NR:

From the second equation,

From the first equation,

Example
A person's horizontal forearm is 0.35m long with a weight of 12N which acts at 0.15m from the elbow. The biceps muscle provides a vertical force which acts at 0.05m from the elbow. The person holds a mass of 5kg in their hand.
Find
(a) the tension force in the biceps, and
(b) the force exerted by the elbow joint.

Taking moments about an axis at the elbow

There are no horizontal components of force. For the forces at the elbow, take "up" components as "plus" and "down" as "minus".

The force required in the biceps is 318 N, which is 7.7 times the 5 x 9.8 = 49 N weight lifted! This type of lever sacrifices mehanical advantage for speed.

Example
A weightless beam of length 3m is attached to a vertical wall by a hinge. The beam is set horizontal by a cable attached 2m from the hinge and making an angle of 26.5O upwards from the beam. A weight of 500N is hung from the free end of the beam.
Find
(a) the reaction force at the hinge.
(b) the tension in the cable.

Taking moments about the hinge:

Taking horizontal forces:

Taking vertical forces:

Example
A plank of length 10m and mass 10kg has a mass of 5kg placed on it 2m from one end. Find the centre of gravity for the combination.

The centre of gravity is the point where the total weight is acts. Taking moments about the left hand side.

Example
A wooden beam 2m long has a weight of 100N. It is suspended by wires at each end, and a weight of 200N is hung 0.2m from one end. Find the tensions in the wires.

Taking moments about the left hand end:

Taking the forces in equilibrium:

Example
The deltoid muscle raises the upper arm to a horizontal position. It acts at 0.15m from the shoulder joint upwards at an angle of 18°. The weight of the arm is 35N acting at 0.35m from the shoulder.
Find
(a) the tension force in the deltoid muscle.
(b) the force exerted by the shoulder joint.

Taking moments about the shoulder:

Taking horizontal forces:

Taking vertical forces:
In Cartesian coordinates the force is (251,47) N and in polar coordinates (255,10.6O) N.

Couples

A couple has two equal and opposite forces, which have a perpendicular separation, so that they act anti-parallel to each other and cause a rotation.

Taking moments about O:

The size of the couple depends on the size of one of the equal forces and the perpendicular distance between them, hence "the closer the couple, the less the torque".

Summarising:

Maximum static friction is given by:  Fs = μs·N
Kinetic frictional force is reasonably constant and smaller than maximum static friction.
There are two conditions to be met for a body to be in equilibrium.
• it does not accelerate (force rule).
• it does not rotate (moment rule).
The size of a couple (of equal forces) equals the size of one of the forces times the perpendicular distance between them.

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