Lesson 6, Topic 2
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# Friction

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Friction is the opposing force between two surfaces in contact when they move over one another.

There are two types of friction namely:

Static or Limiting Friction: This is the maximum force that must be overcome before a body can just start to move over another.

Kinetic Friction: This is the force needed to keep an object moving at uniform speed.

A larger force is always required to move a static object than to keep an object moving, hence static friction is greater than kinetic friction.

### Laws of Solid Friction:

1. Friction opposes motion.

2. Friction is directly proportional to the normal frictional force.

3. Frictional force is directly proportional and perpendicular to the normal reaction.

4. Friction is independent of the area of surfaces in contact.

5. Friction depends on the nature of surfaces in contact.

### Coefficient of Friction:

The coefficient of friction is the ratio of the frictional force resisting the motion of two surfaces in contact, to the normal reaction pressing the two surfaces together. It is usually symbolized by the Greek letter mu (μ).

### Relationship Between Friction Force, Normal Reaction and Coefficient of Friction:

F $$\scriptsize \propto R$$

F = Frictional Force, R = Normal reaction

F = $$\scriptsize \mu R$$

μ = $$\frac{F}{R}$$

μ = Coefficient of friction

Because both F and R are measured in units of force (such as newtons or pounds), the coefficient of friction is dimensionless.

Worked Examples:

a. A box is placed on an inclined plane such that the frictional force opposing its motion is 60N. If the normal reaction of the plane on the box is 80N, Calculate the coefficient of friction.

Solution

Frictional force = 60N

Normal reaction force = 80N

Coefficient of friction = μ

F = $$\scriptsize \mu R$$

Substituting the values

60N = $$\scriptsize \mu \; \times \; 80N$$

Divide both sides by 80

:- $$\frac{\mu \; \times \; 80N}{80N} = \frac{60N}{80N}$$

μ = $$\frac{60N}{80N}$$

μ = 0.75

b. A wooden block of mass 10kg rests on a rough horizontal surface. If the limiting frictional force between the block and the surface is 12N, calculate the coefficient of friction (g = 10m/s2)

Frictional force, F = 12N

Normal reaction, R = W = mg = 10kg x 10 m/s2 = 100N

Where W = Weight of the body, m = Mass of the body, g = Acceleration due to Gravity, R = Normal Reaction Force

F = $$\scriptsize \mu R$$

Coefficient of friction, µ = $$\frac{F}{R} \\ = \frac{12N}{100N} \\ \scriptsize = 0.12$$

c. A cement block of mass 30kg rests on a horizontal floor. If the coefficient of friction, µ, is 0.15, determine the minimum force required to move the cement block when pulled horizontally.

Normal reaction, R = W = mg = 30kg x 10 m/s2 = 300N

Where W = Weight of the body, m = Mass of the body, g = Acceleration due to Gravity, R = Normal Reaction Force

E is the effort required to move the cement block. The question is asking for the minimum Force therefore F = E

F = E = $$\scriptsize \mu R$$

Coefficient of friction, µ, is 0.15, R = 300N

F = E = $$\scriptsize 0.15 \; \times \; 300N \\ \scriptsize = 45N$$

The minimum force required to move the cement block = 45N error: