Dry Friction - Theory & Concepts

Learning Objectives

  • Understand the characteristics and phases of dry friction.
  • Define and apply equations for static and kinetic friction.
  • Comprehend the concepts of angle of static friction and angle of repose.
  • Analyze engineering applications of friction, including wedges, belts, screws, and bearings.
  • Determine tipping versus slipping failure modes for rigid bodies.

This topic provides a comprehensive overview of dry friction, explaining the principles of static and kinetic friction, the angles of friction and repose, and engineering applications such as wedges, belts, screws, and bearings. It also covers the critical analysis of tipping versus slipping for rigid bodies.

Introduction to Dry Friction

In the previous topics, surfaces were often idealized as "smooth" or frictionless. In reality, whenever two surfaces are in contact and tend to move relative to one another, a resisting force develops at the contact surface. This force is called friction.

Dry friction (or Coulomb friction) occurs between the unlubricated surfaces of two contacting solid bodies. When a horizontal force PP is applied to a block resting on a rough horizontal surface, the block will not move immediately. The surface exerts an equal and opposite static friction force FsF_s that balances PP.

Phases of Friction

  • Static Friction (FsF_s): As the applied force PP increases, the friction force FsF_s increases proportionally to maintain equilibrium (Fs=PF_s = P). The block remains at rest.
  • Impending Motion: There is a limit to how large the static friction force can be. When PP reaches this maximum value, the block is on the verge of slipping. This maximum static friction force is proportional to the normal force NN.
  • Kinetic Friction (FkF_k): If PP slightly exceeds the maximum static friction, the block begins to move. The friction force drops slightly to a constant value called kinetic friction.

Maximum Static Friction

The friction force when the block is on the verge of slipping (impending motion).

Fs,max=μsNF_{s, \text{max}} = \mu_s N

Variables

SymbolDescriptionUnit
Fs,maxF_{s, \text{max}}Maximum static friction forceN
μs\mu_sCoefficient of static frictionunitless
NNNormal forceN

Kinetic Friction

The constant friction force acting on the block when it is sliding.

Fk=μkNF_k = \mu_k N

Variables

SymbolDescriptionUnit
FkF_kKinetic friction forceN
μk\mu_kCoefficient of kinetic friction (typically μk<μs\mu_k < \mu_s)unitless
NNNormal forceN

Actual Friction Force Consideration

If the block is NOT moving and NOT on the verge of moving, the actual friction force FF is determined strictly by the equations of equilibrium (Ī£F=0\Sigma F = 0), and F≤μsNF \le \mu_s N.

Impending Motion Assumption

A common student mistake is to immediately apply F=μsNF = \mu_s N to every friction problem. This equation only applies when the problem explicitly states that motion is "impending," or when you are testing the assumption that slipping occurs first. If you do not know the state of motion, you must solve for the required friction force FF using equilibrium equations first, and then compare it to Fs,maxF_{s, \text{max}}.

Angles of Friction

Instead of components NN and FF, the contact forces can be represented by a single resultant force R\mathbf{R}.

Angle of Static Friction (ϕs\phi_s)

The angle that the resultant force R\mathbf{R} makes with the normal force N\mathbf{N} when motion is impending. It is defined as Ļ•s=tanā”āˆ’1(μs)\phi_s = \tan^{-1}(\mu_s).

Angle of Static Friction

The relationship between the angle of static friction and the coefficient of static friction.

tan⁔ϕs=Fs,maxN=μsNN=μs\tan \phi_s = \frac{F_{s, \text{max}}}{N} = \frac{\mu_s N}{N} = \mu_s

Variables

SymbolDescriptionUnit
ϕs\phi_sAngle of static frictiondeg/rad
Fs,maxF_{s, \text{max}}Maximum static friction forceN
NNNormal forceN
μs\mu_sCoefficient of static frictionunitless

Angle of Repose (Īø\theta)

If a block is placed on an inclined plane and the angle of the plane (Īø\theta) is slowly increased, the angle at which the block is just about to slide is the angle of repose. At this point, the component of gravity parallel to the plane equals the maximum static friction force.

Angle of Repose Equation

The relationship showing that the angle of repose is equal to the angle of static friction.

Īø=Ļ•s=tanā”āˆ’1(μs)\theta = \phi_s = \tan^{-1}(\mu_s)

Variables

SymbolDescriptionUnit
Īø\thetaAngle of reposedeg/rad
ϕs\phi_sAngle of static frictiondeg/rad
μs\mu_sCoefficient of static frictionunitless

Applications of Friction

Friction is not always a hindrance; it is frequently utilized in engineering mechanisms to transmit power, hold objects in place, or decelerate motion. Common applications include wedges, belts, screws, and bearings.

Wedges

A wedge is a simple machine used to transform an applied force into much larger forces, directed at approximately right angles to the applied force. They are used to lift heavy blocks, adjust elevations, or split materials.

Analysis of Wedges

  1. Draw Free-Body Diagrams of the block being lifted and the wedge separately.
  2. Since wedges are usually used to initiate movement, assume impending motion at all contact surfaces. The friction force FF will be at its maximum value: F=μsNF = \mu_s N.
  3. Crucially, ensure the friction force vectors oppose the impending motion on both FBDs. (e.g., if a wedge is pushed in, friction acts outward on the wedge, and inward on the block).
  4. Apply equations of equilibrium (ΣFx=0\Sigma F_x=0, ΣFy=0\Sigma F_y=0) starting from the body with the known forces.

Belt Friction

The transmission of power through belts and pulleys, or the braking of a rotating drum using a band brake, relies entirely on friction over a curved surface. When a flexible belt is wrapped around a rough cylinder and motion is impending (the belt is about to slip), the tension on the pulling side (T2T_2) is greater than the tension on the yielding side (T1T_1).

Belt Friction Equation

The tension relationship across a flexible belt subject to friction.

T2=T1eμβT_2 = T_1 e^{\mu \beta}

Variables

SymbolDescriptionUnit
T2T_2Tension on the pulling side (larger tension)N
T1T_1Tension on the yielding side (smaller tension)N
μ\muCoefficient of static friction between the belt and the surfaceunitless
β\betaAngle of contact between the belt and the cylinderrad
eeBase of the natural logarithm (ā‰ˆ2.718\approx 2.718)unitless

V-Belts

Unlike flat belts that sit on a cylindrical drum, a V-belt sits in a wedged groove with an angle 2α2\alpha. The normal force is magnified by the wedge effect, which significantly increases the friction force without requiring more tension on the belt.

V-Belt Friction Equation

The modified belt friction equation utilizing an effective coefficient of friction for V-belts.

T2=T1eμ′βT_2 = T_1 e^{\mu' \beta}

Variables

SymbolDescriptionUnit
T2T_2Tension on the pulling sideN
T1T_1Tension on the yielding sideN
μ′\mu'Effective coefficient of friction (μ′=μ/sin⁔α\mu' = \mu / \sin \alpha)unitless
β\betaAngle of contactrad

Screws

A square-threaded screw is essentially a wedge wrapped around a cylinder. When a moment MM is applied to tighten the screw against an axial load WW, impending motion is upward along the thread. To loosen the screw, impending motion is downward. If θ≤ϕs\theta \le \phi_s, the screw is self-locking, meaning it will not unscrew under the axial load WW without an applied moment.

Screw Tightening Moment

The moment required to tighten a square-threaded screw against an axial load.

M=Wrtan⁔(Īø+Ļ•s)M = W r \tan(\theta + \phi_s)

Variables

SymbolDescriptionUnit
MMMoment required to tighten the screwNĀ·m
WWAxial loadN
rrMean radius of the threadm
Īø\thetaLead angle of the thread (tan⁔θ=L/(2Ļ€r)\tan \theta = L / (2\pi r) where LL is lead)deg/rad
ϕs\phi_sAngle of static frictiondeg/rad

Screw Loosening Moment

The moment required to loosen a square-threaded screw under an axial load.

M′=Wrtan⁔(Īøāˆ’Ļ•s)M' = W r \tan(\theta - \phi_s)

Variables

SymbolDescriptionUnit
M′M'Moment required to loosen the screwNĀ·m
WWAxial loadN
rrMean radius of the threadm
Īø\thetaLead angle of the threaddeg/rad
ϕs\phi_sAngle of static frictiondeg/rad

Journal Bearings

Friction resists the rotation of a shaft within its bearing. When motion is impending, the reactive force shifts slightly to create a resisting couple moment.

Journal Bearing Resisting Moment

The couple moment resisting the rotation of a shaft in a journal bearing.

M=RfPM = R_f P

Variables

SymbolDescriptionUnit
MMResisting couple momentNĀ·m
RfR_fRadius of the friction circle (Rf=rsin⁔ϕkā‰ˆĪ¼krR_f = r \sin \phi_k \approx \mu_k r)m
PPLoad on the bearingN

Thrust Bearings (Collar Bearings)

Thrust bearings provide axial support to a rotating shaft. The resisting frictional moment acts on the flat circular area.

Thrust Bearing Resisting Moment

The resisting moment on a flat circular thrust bearing.

M=μkP23R23āˆ’R13R22āˆ’R12M = \mu_k P \frac{2}{3} \frac{R_2^3 - R_1^3}{R_2^2 - R_1^2}

Variables

SymbolDescriptionUnit
MMResisting frictional momentNĀ·m
μk\mu_kCoefficient of kinetic frictionunitless
PPAxial loadN
R2R_2Outer radius of the bearingm
R1R_1Inner radius of the bearingm

Rolling Resistance

When a cylinder or wheel rolls on a surface, deformation of both bodies occurs, shifting the normal reaction force slightly forward in the direction of motion by a distance aa. This creates a resisting moment that must be overcome to maintain constant speed.

Rolling Resistance Force

The horizontal pulling force required to overcome rolling resistance.

Pā‰ˆWarP \approx \frac{W a}{r}

Variables

SymbolDescriptionUnit
PPHorizontal pulling force requiredN
WWWeight of the cylinderN
aaCoefficient of rolling resistancem
rrRadius of the cylinderm

Interactive Friction Simulation

Explore how the angle of inclination and the weight of a block affect the normal force, the parallel force component, and whether the block will slide based on the coefficients of friction.

Interactive Physics Simulation

Friction on an Inclined Plane Simulator

Study Coulomb's friction laws. Increase the angle of inclination to find the critical slip angle where static friction yields to sliding kinetic motion.

āœ“ Static Equilibrium
15 deg
120 N
0.50
0.40
Inclined Friction Calculus
Critical Slip Angle:Īøcrit=tanā”āˆ’1(μs)\theta_{crit} = \tan^{-1}(\mu_s)
Active Slip Angle:15° ≤ 26.6°

If the incline angle $\theta$ is smaller than $\theta_{\text{crit}}$, the parallel gravity pull ($W_x$) matches static friction ($f_s$). If it exceeds it, kinetic friction ($f_k$) governs.

FNWxWyfk / fsW = 120 Nθ = 15°
Clamping Normal Force (N)
115.9 N
Parallel Gravity Pull (Wx)
31.1 N
Max Static Limit (μs·N)
58.0 N
Active Friction Force (f)
31.1 N

Tipping vs. Slipping Conditions

When a force PP is applied to a rigid body resting on a rough surface, two failure modes are possible: the body might slide (slip), or it might rotate (tip over). An engineer must always determine which will occur first.

Consider a block of weight WW, height hh, and base width bb. A horizontal force PP is applied at a height yy from the base. The actual force PP required to cause motion is the smaller of PslipP_{\text{slip}} and PtipP_{\text{tip}}. The smaller value determines the dominant failure mode.

Checking for Slipping First

  1. Set the friction force FF to its maximum static value: F=μsNF = \mu_s N.
  2. Use equilibrium (ΣFx=0\Sigma F_x=0, ΣFy=0\Sigma F_y=0) to find the required force PslipP_{\text{slip}}.
  3. Check the moment equilibrium (Ī£M=0\Sigma M = 0) to find the location xx of the normal force NN relative to the center of the block.
  4. Condition: If xx falls within the base of the block (x≤b/2x \le b/2), then the assumption is correct: the block will slip before it tips.

Checking for Tipping First

  1. When tipping is imminent, the entire normal force NN shifts to the extreme front edge (the corner) of the block's base (x=b/2x = b/2).
  2. Take the sum of moments about this corner (Ī£Mcorner=0\Sigma M_{\text{corner}} = 0) to solve for the required force PtipP_{\text{tip}}. The friction and normal forces pass through this point and create no moment.
  3. Check the horizontal equilibrium (Ī£Fx=0\Sigma F_x = 0) to find the required friction force FF.
  4. Condition: If this required FF is less than or equal to μsN\mu_s N, then the assumption is correct: the block will tip before it slips.
Key Takeaways
  • Static Friction (FsF_s) balances applied forces to maintain equilibrium up to a maximum limit: Fs,max=μsNF_{s,\text{max}} = \mu_s N.
  • If the applied force exceeds this limit, motion occurs and Kinetic Friction (FkF_k) takes over: Fk=μkNF_k = \mu_k N.
  • If motion is NOT impending, the friction force must be found using equilibrium equations (Ī£F=0\Sigma F=0), not the friction formula.
  • The Angle of Repose is the maximum incline angle before a block starts sliding, and is equal to the inverse tangent of the static friction coefficient (Īø=tanā”āˆ’1(μs)\theta = \tan^{-1}(\mu_s)).
  • In analysis of blocks or retaining walls, one must always check whether the body will slip (translate) or tip (rotate) first by evaluating the minimum force required for each scenario.