Friction is a fundamental concept in mechanics that affects how objects move and interact in all professional settings. Understanding friction allows professionals across fields such as healthcare, agriculture, business, and hospitality to optimize equipment use, improve safety, and enhance operational efficiency. For instance, friction influences how hospital beds are moved, how vehicles grip roads in county services, and even how agricultural machinery operates on different terrains. This chapter explores the nature of friction and its practical applications in diverse Kenyan workplaces.

3.1 Friction

Friction plays a critical role in everyday activities and industrial operations by resisting motion between contacting surfaces. It is a force that can both aid and hinder processes, making its understanding essential for professionals who seek to manage energy use, reduce wear and tear, or improve safety. Friction's impact is evident in many Kenyan workplaces, such as retail businesses where conveyor belts rely on friction to transport goods, or in county government offices where friction in door hinges affects maintenance schedules.

3.1.1 Definition

Friction is the force that opposes the relative motion or tendency of such motion between two surfaces in contact. It arises because of the microscopic roughness and interactions at the surfaces, which create resistance when one surface tries to slide over another. Friction is not a fundamental force like gravity but results from electromagnetic forces between molecules in contact areas.

Nature of Friction

The microscopic roughness on surfaces means that even seemingly smooth objects have peaks and valleys, called asperities, that catch on each other. Where the peaks touch, the material is pressed together hard enough to stick, and sliding requires these tiny welds to be sheared apart continuously.

Friction depends on the nature of the surfaces involved and the force pressing them together. Rougher surfaces generally produce more friction, while smoother surfaces reduce it. For example, in a county hospital, the rubber soles of healthcare workers' shoes increase friction with the floor, preventing slips and falls during busy shifts. In a county government office, the friction between a chair's legs and the floor prevents the chair from sliding unintentionally.

Types of Friction

There are several types of friction that affect motion differently:

• Static friction: The force preventing motion when two surfaces are at rest relative to each other. It must be overcome to start movement, such as pushing a heavy hospital trolley or a safe resting on a bank floor.
• Kinetic (sliding) friction: Acts when surfaces slide past each other, like a cleaning staff sliding a mop bucket across a tiled floor.
• Rolling friction: Occurs when an object rolls over a surface, such as wheels of delivery carts in a supermarket or luggage wheels in an airport hotel. It is far smaller than sliding friction, which is why wheels are used at all.
• Fluid friction: Resistance when objects move through fluids like air or water, relevant in irrigation pumps on farms.
• Internal friction: Resistance within materials when they deform, important in structural integrity assessments and in quality control of building materials.

Static friction is always the larger of the first two. This is why a trolley takes a hard initial shove and then moves easily, and it is the single most commonly tested point in this topic.

Factors Affecting Friction

Several factors influence frictional force magnitude:

• Surface roughness or texture: Rougher surfaces increase friction due to greater interlocking of asperities.
• Normal force: The perpendicular force pressing surfaces together. Heavier objects experience more friction.
• Material properties: Different materials interact differently. Rubber on concrete has far higher friction than steel on ice, and softer materials may deform and increase the real contact area.
• Lubrication: Presence of lubricants like oils or greases reduces friction by creating a separating film.
• Temperature and environment: Moisture, dust, contaminants and heat alter material properties and the effectiveness of lubrication.

Note what is not on this list. Within the normal range, friction does not depend on the apparent area of contact, and kinetic friction does not depend much on the sliding speed. A brick sliding on its wide face and on its narrow edge experiences the same friction, which surprises most students and is examinable.

The Laws of Friction

Three experimental laws summarise the behaviour of dry sliding friction:

1. Friction is proportional to the normal force. Double the normal force and the frictional force doubles.
2. Friction is independent of the apparent area of contact.
3. Kinetic friction is largely independent of the speed of sliding, and is less than the maximum static friction.

These are experimental laws rather than exact physical truths. They hold well for ordinary dry surfaces at moderate loads, and they break down at very high loads, at very high speeds and with very clean or very soft surfaces.

Friction as a Contact Force

Friction acts parallel to the contact surface and opposite to the direction of motion or intended motion. It is a reactive force, existing only when surfaces try to move relative to each other. For instance, at a retail shop, friction between the cashier's counter and the customer's purse prevents the purse from sliding off, and friction prevents documents from sliding off a desk in a busy county office.

Measurement and Calculation of Friction

Friction is quantified by the coefficient of friction, written μ (the Greek letter mu), a dimensionless ratio of the frictional force to the normal force pressing the surfaces together.

The Basic Equations

Kinetic (sliding) friction

$$F_k = \mu_k N$$

Static friction, which varies from zero up to a maximum:

$$F_s \leq \mu_s N$$

where:

• F is the frictional force, in newtons (N)
• N is the normal reaction force, in newtons (N)
• μ is the coefficient of friction, which has no units because it is a ratio of two forces

Rearranged to find the coefficient:

$$\mu = \frac{F}{N}$$

Finding the Normal Force

The normal force is the force pressing the two surfaces together, acting perpendicular to them. It is not always equal to the weight.

Throughout this topic, take the acceleration due to gravity as g = 9.8 m/s².

Typical Coefficients of Friction

These are typical values for guidance. Measured values vary with cleanliness, moisture and surface finish, which is why a laboratory determines the coefficient for the actual surfaces in use rather than reading it from a table.

Worked Examples

Worked Example 1: Finding the frictional force

A crate of mass 25 kg rests on a concrete floor. The coefficient of kinetic friction between the crate and the floor is 0.40. Calculate the force needed to keep the crate sliding at a steady speed.

Step 1. Find the normal force. The surface is horizontal, so N = mg.

N = 25 × 9.8 = 245 N

Step 2. Apply F = μN.

F = 0.40 × 245 = 98 N

At steady speed the applied force exactly balances friction, so the force required is 98 N.

Worked Example 2: Finding the coefficient of friction

A loaded trolley weighs 400 N. A spring balance shows that a steady horizontal pull of 120 N keeps it moving at constant speed. Calculate the coefficient of kinetic friction.

Given: F = 120 N, N = 400 N (the weight is given directly, so no need to multiply by g)

$$\mu_k = \frac{F}{N} = \frac{120}{400} = \textbf{0.30}$$

The coefficient has no units.

Worked Example 3: Static friction, and whether the object moves at all

A box of mass 40 kg rests on a floor. The coefficient of static friction is 0.55 and the coefficient of kinetic friction is 0.35. A worker pushes horizontally with a force of 180 N. Determine whether the box moves, and state the frictional force acting.

Step 1. Find the normal force.

N = 40 × 9.8 = 392 N

Step 2. Find the maximum static friction.

F_s(max) = 0.55 × 392 = 215.6 N

Step 3. Compare with the applied force.

Applied force 180 N is less than 215.6 N, so the box does not move.

Step 4. State the actual frictional force.

Since the box is in equilibrium, friction must exactly balance the push. The frictional force is 180 N, not 215.6 N.

This is the point most often missed. Static friction is not a fixed value. It adjusts itself to match whatever force is applied, up to its maximum, and only then does the object break away.

Worked Example 4: Net force and acceleration

The same box is now pushed with a force of 250 N. Calculate its acceleration.

Step 1. Check that it moves. 250 N is greater than the maximum static friction of 215.6 N, so the box breaks away and begins to slide.

Step 2. Find the kinetic friction, which now applies.

F_k = 0.35 × 392 = 137.2 N

Step 3. Find the net force.

Net force = applied force − friction = 250 − 137.2 = 112.8 N

Step 4. Apply Newton's second law, F = ma.

$$a = \frac{F_{net}}{m} = \frac{112.8}{40} = \textbf{2.82 m/s}^2$$

Worked Example 5: Coefficient from the angle of repose

A flat specimen is placed on a horizontal plank. The plank is slowly tilted and the specimen begins to slide when the plank reaches 22° to the horizontal. Calculate the coefficient of static friction.

At the point of sliding, the component of weight down the slope exactly equals the maximum friction:

$$mg\sin\theta = \mu_s \, mg\cos\theta$$

The mass and g cancel from both sides, leaving:

$$\mu_s = \tan\theta$$

$$\mu_s = \tan 22° = \textbf{0.40}$$

The angle at which sliding just begins is called the angle of repose or the angle of friction. Because the mass cancels, the result does not depend on how heavy the specimen is, which makes this the simplest laboratory method of finding μ.

Worked Example 6: An object on an inclined plane

A block of mass 15 kg is released on a slope inclined at 20° to the horizontal. The coefficient of kinetic friction is 0.25. Calculate the acceleration of the block down the slope.

Step 1. Find the weight.

W = mg = 15 × 9.8 = 147 N

Step 2. Resolve the weight into components.

Perpendicular to the slope: N = mg cos 20° = 147 × 0.9397 = 138.1 N

Down the slope: mg sin 20° = 147 × 0.3420 = 50.3 N

Step 3. Find the frictional force, which acts up the slope, opposing motion.

F_k = μN = 0.25 × 138.1 = 34.5 N

Step 4. Find the net force down the slope.

Net force = 50.3 − 34.5 = 15.8 N

Step 5. Find the acceleration.

$$a = \frac{15.8}{15} = \textbf{1.05 m/s}^2$$

Note that the normal force here is 138.1 N, not the full weight of 147 N. Using the weight instead of its perpendicular component is the commonest error in inclined plane problems.

Worked Example 7: Work done against friction

The crate from Worked Example 1, experiencing 98 N of friction, is dragged 12 m across the floor. Calculate the work done against friction, and state what becomes of that energy.

$$W = F \times d = 98 \times 12 = \textbf{1176 J}$$

All 1176 J is converted into heat at the sliding surfaces, with a small part going to sound and to wear of the materials. This energy is not recoverable, which is why friction in machinery is both an energy cost and a cause of overheating.

Worked Example 8: Determining μ from laboratory readings

In a laboratory determination, a block is pulled at constant speed across a bench by a spring balance while different weights are placed on it. The following readings are obtained:

Calculate the mean coefficient of kinetic friction.

$$\mu_{mean} = \frac{0.360 + 0.355 + 0.360 + 0.363}{4} = \frac{1.438}{4} = \textbf{0.36}$$

The four values agree closely, which confirms the first law of friction: the ratio F ÷ N stays constant as the load changes. If the readings had drifted steadily as the load increased, this would point to an experimental fault such as the spring balance not being pulled horizontally.

Plotted as a graph of F against N, the points lie on a straight line through the origin, and the gradient of that line is μ. Drawing the graph is more reliable than averaging, because it reveals any faulty reading at a glance.

Worked Example 9: Power needed to overcome friction

A conveyor belt in a warehouse must overcome a total frictional resistance of 250 N while running at 1.5 m/s. Calculate the power the motor must supply to overcome friction.

$$P = F \times v = 250 \times 1.5 = \textbf{375 W}$$

Worked Example 10: Braking distance

A vehicle travelling at 20 m/s brakes hard on a dry road where the coefficient of friction between the tyres and the road is 0.70. Calculate the shortest distance in which it can stop.

Step 1. Find the deceleration. The only horizontal force is friction, so:

$$ma = \mu mg \quad \text{which gives} \quad a = \mu g$$

a = 0.70 × 9.8 = 6.86 m/s²

Step 2. Use the equation of motion v² = u² + 2as, with final velocity v = 0.

$$d = \frac{u^2}{2a} = \frac{20^2}{2 \times 6.86} = \frac{400}{13.72} = \textbf{29.2 m}$$

Notice that the mass cancelled in step 1. A loaded lorry and a small car stop in the same distance on the same surface, provided the tyres and brakes are equally good, because the extra weight increases both the friction and the inertia in the same proportion.

On a wet road where μ falls to 0.50, the same calculation gives a stopping distance of 40.8 m, an increase of about 40 per cent. This is the calculation behind road safety advice to increase the following distance in rain.

Determining the Coefficient of Friction in the Laboratory

Two standard methods are used, and a Science Laboratory Technology student is expected to perform both.

Method A: Horizontal Pull with a Spring Balance

Apparatus: wooden block, spring balance or force meter, set of slotted masses, plane horizontal bench surface, string.

Procedure

1. Weigh the block and record its weight in newtons.
2. Attach the spring balance to the block with a short string, keeping the string horizontal.
3. Pull gently and increase the pull slowly. Record the highest reading reached just before the block starts to move. This is the maximum static friction.
4. Once moving, continue pulling so the block travels at a steady speed, and record the steady reading. This is the kinetic friction.
5. Add a known mass to the block, so increasing the normal force, and repeat.
6. Repeat for at least five different normal forces.
7. Tabulate N against F, calculate μ for each, and plot F against N. The gradient gives μ.

Precautions

• Keep the string horizontal, since a sloping pull changes the normal force and corrupts every reading.
• Move at genuinely constant speed when reading kinetic friction, since any acceleration means the reading is not equal to friction.
• Clean the surfaces and keep them free of oil, dust and water between runs.
• Use the same face of the block each time, so that any surface variation does not enter the results.
• Take several readings at each load and use the mean.

Method B: The Tilting Plane

Apparatus: plane board hinged so it can be tilted, specimen block, protractor or angle scale.

Procedure

1. Place the block on the horizontal board.
2. Raise one end slowly and steadily.
3. Note the angle at which the block just begins to slide.
4. Repeat several times and take the mean angle.
5. Calculate μ_s = tan θ.

Advantages and limitations. The tilting plane needs no force meter and the mass of the block does not matter, so it is quick and free of calibration error. Against that, the angle at which sliding begins is a matter of judgement, it gives only the static coefficient, and a board raised jerkily will shake the block loose early and give a reading that is too low.

3.1.2 Applications

Friction's dual role as both a helpful and adverse force makes it crucial in designing and operating equipment and processes across Kenyan industries. Proper management of friction can lead to improved safety, energy savings, and equipment longevity.

Friction in Transportation and Mobility

Friction enables vehicles to move safely on roads by providing the necessary grip between tires and the surface, for acceleration, steering and braking alike. County government vehicles depend on adequate friction, especially on wet or dusty rural roads. As Worked Example 10 shows, a fall in the coefficient from 0.70 to 0.50 lengthens the stopping distance by about 40 per cent, which is why tyre tread depth is a legal requirement rather than a recommendation.

Friction in Healthcare Equipment

Hospital beds and wheelchairs rely on controlled friction to ensure ease of movement without uncontrolled slipping. At a county referral hospital, wheelchairs use tires that balance friction for smooth rolling and stability on tiled floors, while the brakes work by applying a large normal force to create friction on demand.

Friction in Agricultural Machinery

Farmers use friction to their advantage in machinery like threshers and ploughs, where frictional forces drive belts and components. A belt drive transmits power only because of friction between belt and pulley, which is why a slack or oily belt slips and the machine loses power. Excessive friction from poor lubrication causes wear and energy loss, so cooperative farms schedule regular lubrication.

Friction in Industrial and Office Equipment

Office printers and conveyor belts in warehouses function because of friction. At a large retail business, conveyor belts rely on friction between the belt surface and rollers to move goods, and the motor must supply power to overcome it, as calculated in Worked Example 9. Friction in printer rollers ensures paper feeds correctly without jams, and worn rollers with reduced friction are the usual cause of misfeeds.

Friction in Everyday Workplace Safety

Friction between footwear and floor surfaces prevents slips and falls, a common workplace hazard. Hotels install non-slip mats or textured flooring in kitchens and bathrooms to raise the coefficient of friction and reduce accidents among staff and guests. Water, oil and dust all act as unintended lubricants and lower it sharply.

Friction in Energy and Resource Management

In environments regulated by NEMA, friction reduction in mechanical systems such as pumps and fans contributes to energy efficiency and lower emissions. Every joule of work done against friction becomes waste heat, as shown in Worked Example 7, so lubrication, bearings and surface treatments pay for themselves in reduced running cost and in lower emissions.

Methods of Reducing Unwanted Friction

• Lubrication with oil, grease or graphite, which separates the surfaces with a fluid film.
• Ball and roller bearings, which replace sliding contact with rolling contact.
• Polishing and smoothing the contacting surfaces.
• Low-friction materials such as Teflon, with a coefficient around 0.04.
• Streamlining, where the resistance is fluid friction rather than solid contact.
• Air cushions, as in hovercraft and air bearings, which remove solid contact altogether.

Methods of Increasing Useful Friction

• Treading and texturing, as on tyres, shoe soles and non-slip floor tiles.
• Increasing the normal force, as a brake does by pressing the pad harder against the disc.
• Using high-friction materials such as rubber and brake lining compounds.
• Keeping surfaces clean and dry, since water, oil and dust all reduce grip.

Chapter Summary

This chapter explored the concept of friction, defining it as the resistance that occurs when two surfaces move against each other and highlighting its practical applications in everyday life and machinery. Newton's laws of motion were examined with a focus on circular motion, detailing how angular displacement, angular velocity, and angular acceleration describe the movement of objects along curved paths. The chapter also covered the concept of tension, explaining its role in forces transmitted through strings or cables. Further, the definition of Newton's laws was reinforced along with their various applications in analyzing forces and motion. The discussion then shifted to shear, describing the force that causes layers of material to slide past one another. Finally, the chapter addressed the bulk modulus, which measures a material's resistance to uniform compression, providing insight into material properties under pressure. Together, these topics form a comprehensive foundation for understanding mechanics in both theoretical and practical contexts.