Dry Friction, Sliding Loads and Friction Power

SOPEEC 4th Class Power Engineer Part 4A — Free Study Guide 2026 Exam

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Dry Friction, Sliding Loads and Friction Power

Key Takeaways Explain This Topic AI

Static friction adjusts up to its limiting value; it is not always μsN.

For a horizontal pull with no other vertical loads, N = mg.

Sliding-friction power loss equals friction force times speed.

Friction coefficients and lubrication conditions must be supplied or measured.

Last updated: October 2026

Friction in Industrial Machinery

Friction is the tangential resistive force that opposes the relative sliding or rolling motion between two solid surfaces in physical contact.

Microscopic Nature of Friction

Even highly polished metal surfaces (such as honed pump cylinder liners or babbit journal bearings) exhibit microscopic peaks and valleys called asperities . When two surfaces are pressed together under a normal load ( N N N ):

Actual contact occurs only at mating asperity tips, creating an actual contact area that is a small fraction of the apparent geometric surface area.

Extreme localized contact pressures cause asperity tips to plastically deform and form microscopic cold-welds.

Relative motion requires shearing these cold-welds and mechanically plowing harder asperities through the softer opposing metal, manifesting as the macroscopic friction force ( F f F_f F f ​ ).

Classification of Mechanical Friction

Static Friction ( F s F_s F s ​ ): The resistive force opposing the initiation of relative motion between stationary contacting surfaces. As an applied lateral force increases, static friction increases proportionally to maintain equilibrium up to a critical threshold termed limiting friction ( F limiting F_{\text{limiting}} F limiting ​ or F s , max ⁡ F_{s,\max} F s , m a x ​ ).

Kinetic (Sliding) Friction ( F k F_k F k ​ ): The resistive force opposing ongoing relative sliding motion. Once motion begins, microscopic cold-welds have less time to form, so kinetic friction is commonly less than limiting static friction in the elementary dry-friction model.

Rolling Friction ( F r F_r F r ​ ): The resistive force encountered when a cylindrical or spherical element (such as a ball or roller bearing) rolls over a surface. Arising primarily from elastic hysteresis and deformation of contact surfaces, rolling friction is typically 10 to 100 times smaller than sliding friction ( μ r ≪ μ k \mu_r \ll \mu_k μ r ​ ≪ μ k ​ ).

Fluid Friction: Shear resistance generated within viscous fluid lubricant films separating moving solid boundaries. Friction Force (F_f)

• ^
• | Limiting Static Friction (Peak)
• | /\
• | / \................ Kinetic Friction (Sliding)
• | / -------------------------------------
• | / (F_k is relatively constant with speed)
• | /
• | / <-- Static Region (F_s = F_applied)
• | /
• +------+---------------------------------------------> Applied Lateral Force (F)
• |<- Motion Impends (Onset of Sliding)
• The Classical Laws of Dry Friction (Amontons-Coulomb Laws)

The friction force is directly proportional to the perpendicular normal force pressing the contacting surfaces together: F f ∝ N F_f \propto N F f ​ ∝ N .

The friction force is independent of the apparent geometric contact area between the surfaces.

The kinetic friction force is virtually independent of the relative sliding velocity across moderate operating speed ranges.

• Coefficient of Friction and Angles
• Coefficient of Friction ( μ \mu μ )

The coefficient of friction is the dimensionless ratio of the frictional resistance force to the perpendicular normal force acting across the contact plane: μ = F f N \mu = \frac{F_f}{N} μ = N F f ​ ​

Coefficient of Static Friction ( μ s \mu_s μ s ​ ): μ s = F s , max ⁡ N \mu_s = \frac{F_{s,\max}}{N} μ s ​ = N F s , m a x ​ ​

• Coefficient of Kinetic Friction ( μ k \mu_k μ k ​ ): μ k = F k N \mu_k = \frac{F_k}{N} μ k ​ = N F k ​ ​
• Angle of Friction ( ϕ \phi ϕ )

When limiting friction is attained, the resultant contact force ( R R R ) is the vector sum of the normal reaction force ( N N N ) and the limiting friction force ( F f F_f F f ​ ). The angle between R R R and N N N is termed the angle of friction ( ϕ \phi ϕ ) : tan ⁡ ϕ = F f N = μ ⟹ ϕ = arctan ⁡ ( μ ) \tan\phi = \frac{F_f}{N} = \mu \implies \phi = \arctan(\mu) tan ϕ = N F f ​ ​ = μ ⟹ ϕ = arctan ( μ )

Angle of Repose ( θ \theta θ )

The angle of repose is the maximum inclination angle of an inclined plane at which a stationary granular material or solid block will remain resting on the incline without sliding down under its own weight:

• Force pulling body down incline: F down = m g sin ⁡ θ F_{\text{down}} = mg \sin\theta F down ​ = m g sin θ
• Normal force perpendicular to incline: N = m g cos ⁡ θ N = mg \cos\theta N = m g cos θ

Limiting friction resisting slide: F s , max ⁡ = μ s N = μ s m g cos ⁡ θ F_{s,\max} = \mu_s N = \mu_s mg \cos\theta F s , m a x ​ = μ s ​ N = μ s ​ m g cos θ

At the threshold of slipping: m g sin ⁡ θ = μ s m g cos ⁡ θ ⟹ sin ⁡ θ cos ⁡ θ = tan ⁡ θ = μ s mg \sin\theta = \mu_s mg \cos\theta \implies \frac{\sin\theta}{\cos\theta} = \tan\theta = \mu_s m g sin θ = μ s ​ m g cos θ ⟹ c o s θ s i n θ ​ = tan θ = μ s ​ θ repose = ϕ = arctan ⁡ ( μ s ) \theta_{\text{repose}} = \phi = \arctan(\mu_s) θ repose ​ = ϕ = arctan ( μ s ​ )

• Friction Power Loss and Lubrication Regimes
• Parasitic Power Dissipated in Bearings

In rotating plant machinery (such as turbine-generator sets, induction motors, and centrifugal pumps), shaft journals rotate inside stationary sleeve bearings. The normal force on the bearing is the shaft weight plus dynamic loads ( N = W s N = W_s N = W s ​ ).

Friction force: F f = μ N F_f = \mu N F f ​ = μ N

Linear surface velocity of journal of diameter D D D rotating at rotational speed n n n (revolutions per second, rev/s \text{rev/s} rev/s ): v = π D n = π D ( RPM 60 ) v = \pi D n = \pi D \left(\frac{\text{RPM}}{60}\right) v = π D n = π D ( 60 RPM ​ )

Friction power loss dissipated as thermal heat: P loss = F f × v = μ N π D n P_{\text{loss}} = F_f \times v = \mu N \pi D n P loss ​ = F f ​ × v = μ N π D n

Because all dissipated friction work converts into heat, high-speed journal bearings require continuous forced-circulation lube oil cooling systems to prevent bearing babbit melting (the allowable temperature is bearing-material and design dependent).

Lubrication Regimes

To minimize friction and metal wear, plant machinery employs lubrication to establish distinct operational regimes:

Boundary Lubrication: Fluid film thickness is thinner than surface asperity height. Metal-to-metal contact occurs; friction coefficient is high ( μ ≈ 0.08 − 0.15 \mu \approx 0.08 - 0.15 μ ≈ 0.08 − 0.15 ). Experienced during slow-speed machinery startup and coastdown.

Mixed Lubrication: Transitional zone where hydrodynamic oil film carries part of the load while asperity peaks intermittently contact.

Full Fluid (Hydrodynamic) Lubrication: A continuous, pressurized oil wedge completely separates the journal surface from the bearing liner. Zero metal-to-metal contact occurs; the coefficient of friction drops to viscous shear levels ( μ ≈ 0.001 − 0.005 \mu \approx 0.001 - 0.005 μ ≈ 0.001 − 0.005 ), eliminating mechanical wear during normal steady-state operation.

Example 2: Inclined Coal Conveyor Belt Pull and Drive Power

Problem: An industrial coal conveyor moves crushed bituminous coal up an incline of α = 14 ∘ \alpha = 14^\circ α = 1 4 ∘ at a constant belt speed of v = 2.2 m/s v = 2.2\text{ m/s} v = 2.2 m/s . The total loaded belt mass actively supported on the idler roll structure along its length is M = 8 , 500 kg M = 8,500\text{ kg} M = 8 , 500 kg . The conveyor idlers introduce an effective rolling friction coefficient of μ = 0.040 \mu = 0.040 μ = 0.040 . Determine:

(a) The normal reaction force perpendicular to the conveyor structure.

(b) The total pull force required along the belt centerline to transport the coal.

(c) The motor power required if conveyor drive transmission efficiency is η = 88 % \eta = 88\% η = 88% .

Solution:

Step 1: Calculate the total gravitational weight force. W g = M g = 8 , 500 kg × 9.81 m/s 2 = 83 , 385 N W_g = M g = 8,500\text{ kg} \times 9.81\text{ m/s}^2 = 83,385\text{ N} W g ​ = M g = 8 , 500 kg × 9.81 m/s 2 = 83 , 385 N

Step 2: Resolve weight into perpendicular and parallel components.

Normal force perpendicular to belt ( N N N ): N = W g cos ⁡ 14 ∘ = 83 , 385 × 0.970296 ≈ 80 , 908 N N = W_g \cos 14^\circ = 83,385 \times 0.970296 \approx 80,908\text{ N} N = W g ​ cos 1 4 ∘ = 83 , 385 × 0.970296 ≈ 80 , 908 N

Gravitational component acting down the incline ( F gravity , ∥ F_{\text{gravity},\parallel} F gravity , ∥ ​ ): F gravity , ∥ = W g sin ⁡ 14 ∘ = 83 , 385 × 0.241922 ≈ 20 , 173 N F_{\text{gravity},\parallel} = W_g \sin 14^\circ = 83,385 \times 0.241922 \approx 20,173\text{ N} F gravity , ∥ ​ = W g ​ sin 1 4 ∘ = 83 , 385 × 0.241922 ≈ 20 , 173 N

Step 3: Calculate the total frictional resistance along the incline. F f = μ N = 0.040 × 80 , 908 N ≈ 3 , 236 N F_f = \mu N = 0.040 \times 80,908\text{ N} \approx 3,236\text{ N} F f ​ = μ N = 0.040 × 80 , 908 N ≈ 3 , 236 N

Step 4: Calculate the total driving pull force ( F pull F_{\text{pull}} F pull ​ ). Since the conveyor travels at constant velocity ( a = 0 a = 0 a = 0 ), the drive pulley must overcome both the downward gravitational weight component and the opposing friction force: F pull = F gravity , ∥ + F f = 20 , 173 N + 3 , 236 N = 23 , 409 N = 23.41 kN F_{\text{pull}} = F_{\text{gravity},\parallel} + F_f = 20,173\text{ N} + 3,236\text{ N} = 23,409\text{ N} = 23.41\text{ kN} F pull ​ = F gravity , ∥ ​ + F f ​ = 20 , 173 N + 3 , 236 N = 23 , 409 N = 23.41 kN

Step 5: Calculate conveyor drive power output and motor power input. P output = F pull × v = 23 , 409 N × 2.2 m/s ≈ 51 , 500 W = 51.50 kW P_{\text{output}} = F_{\text{pull}} \times v = 23,409\text{ N} \times 2.2\text{ m/s} \approx 51,500\text{ W} = 51.50\text{ kW} P output ​ = F pull ​ × v = 23 , 409 N × 2.2 m/s ≈ 51 , 500 W = 51.50 kW P motor = P output η = 51.50 kW 0.88 = 58.52 kW P_{\text{motor}} = \frac{P_{\text{output}}}{\eta} = \frac{51.50\text{ kW}}{0.88} = 58.52\text{ kW} P motor ​ = η P output ​ ​ = 0.88 51.50 kW ​ = 58.52 kW

Example 3: Parasitic Friction Loss in a Turbine Sleeve Bearing

Problem: A high-pressure steam turbine rotor has a journal diameter of D = 200 mm D = 200\text{ mm} D = 200 mm ( 0.20 m 0.20\text{ m} 0.20 m ) rotating at an operating speed of 3 , 600 RPM 3,600\text{ RPM} 3 , 600 RPM in a sleeve bearing. The static downward weight supported by this bearing is 32.0 kN 32.0\text{ kN} 32.0 kN ( 32 , 000 N 32,000\text{ N} 32 , 000 N ). Under normal operating oil film conditions, the effective hydrodynamic friction coefficient is μ = 0.0045 \mu = 0.0045 μ = 0.0045 . Calculate:

(a) The tangential friction force at the journal surface.

(b) The friction torque resisting shaft rotation.

(c) The rate of mechanical heat generation (friction power loss in kilowatts).

Solution:

Step 1: Calculate the tangential friction force ( F f F_f F f ​ ). F f = μ N = 0.0045 × 32 , 000 N = 144 N F_f = \mu N = 0.0045 \times 32,000\text{ N} = 144\text{ N} F f ​ = μ N = 0.0045 × 32 , 000 N = 144 N

Step 2: Calculate friction resisting torque ( T f T_f T f ​ ). Journal radius r = D / 2 = 0.20 / 2 = 0.10 m r = D / 2 = 0.20 / 2 = 0.10\text{ m} r = D /2 = 0.20/2 = 0.10 m . T f = F f × r = 144 N × 0.10 m = 14.4 N ⋅ m T_f = F_f \times r = 144\text{ N} \times 0.10\text{ m} = 14.4\text{ N}\cdot\text{m} T f ​ = F f ​ × r = 144 N × 0.10 m = 14.4 N ⋅ m

Step 3: Calculate shaft surface linear velocity ( v v v ) and angular velocity ( ω \omega ω ). Rotational speed in revolutions per second: n = 3 , 600 RPM 60 s/min = 60 rev/s n = \frac{3,600\text{ RPM}}{60\text{ s/min}} = 60\text{ rev/s} n = 60 s/min 3 , 600 RPM ​ = 60 rev/s v = π D n = π × 0.20 m × 60 rev/s = 37.699 m/s v = \pi D n = \pi \times 0.20\text{ m} \times 60\text{ rev/s} = 37.699\text{ m/s} v = π D n = π × 0.20 m × 60 rev/s = 37.699 m/s ω = 2 π n = 2 π × 60 = 376.99 rad/s \omega = 2 \pi n = 2 \pi \times 60 = 376.99\text{ rad/s} ω = 2 πn = 2 π × 60 = 376.99 rad/s

Step 4: Calculate friction power loss ( P loss P_{\text{loss}} P loss ​ ). P loss = F f × v = 144 N × 37.699 m/s = 5 , 428.7 W ≈ 5.43 kW P_{\text{loss}} = F_f \times v = 144\text{ N} \times 37.699\text{ m/s} = 5,428.7\text{ W} \approx 5.43\text{ kW} P loss ​ = F f ​ × v = 144 N × 37.699 m/s = 5 , 428.7 W ≈ 5.43 kW

Alternatively using rotational torque: P loss = T f × ω = 14.4 N ⋅ m × 376.99 rad/s = 5 , 428.7 W = 5.43 kW P_{\text{loss}} = T_f \times \omega = 14.4\text{ N}\cdot\text{m} \times 376.99\text{ rad/s} = 5,428.7\text{ W} = 5.43\text{ kW} P loss ​ = T f ​ × ω = 14.4 N ⋅ m × 376.99 rad/s = 5 , 428.7 W = 5.43 kW .

The lube oil system must dissipate 5.43 kW 5.43\text{ kW} 5.43 kW of continuous heat from this single bearing to maintain stable operating temperatures.

Horizontal sliding example

A 500 kg skid rests on a horizontal floor with no other vertical forces. Using g = 9.81 m/s², the normal reaction is 4,905 N. Suppose the supplied static coefficient is 0.40 and the kinetic coefficient is 0.30. The limiting static friction is 1,962 N. A horizontal 1,000 N pull leaves the skid stationary: static friction is 1,000 N, not its maximum 1,962 N. Once it slides, the modelled resistance is 1,471.5 N. A sustained 2,000 N horizontal pull then produces net force 528.5 N and acceleration 1.057 m/s². These values assume the pull remains horizontal; an upward angled pull reduces normal reaction and changes the calculation.

Source notes

Official scope and technical references checked 2026-10-17: Reference 1 . Archived DOE handbooks are used for physical principles, not current Canadian legal requirements.