CELE Engineering Mechanics — FrictionRevision Notes

Formulas

Exam Tips

• Always start with the FBD — correctly drawn FBDs solve 80% of friction problems automatically.
• State explicitly 'at impending motion, F = μN' before substituting — examiners award process marks.
• When μ is not given but the angle of repose is given, use μ = tan(φ_repose) directly.
• Check self-locking: if tan θ < μ on an incline, the block will NOT slide — no holding force is needed.
• For board exam multiple-choice: compute both the 'push-up' force and the 'hold-from-sliding-down' force — different scenarios are often offered as distractors.

Key Points

• Friction force F acts tangent to the contact surface and always opposes the direction of impending or actual sliding motion.
• Below impending slip, F is statically indeterminate — it equals whatever value equilibrium demands, up to a maximum of μ_s × N.
• At impending slip (just about to move): F_max = μ_s × N (static friction law).
• During sliding: F_k = μ_k × N where μ_k < μ_s (kinetic friction law).
• The coefficient of friction μ is dimensionless and depends on the material pair and surface condition, NOT on the contact area.
• Typical values for board exam: μ_s ≈ 0.2–0.5 for common structural interfaces (wood on concrete, steel on steel, etc.).
• The total reaction R at a surface is the resultant of N and F: R = √(N² + F²), inclined at angle φ = arctan(F/N) from the normal.
• Three friction states exist: (a) No motion — F < μ_s N; (b) Impending motion — F = μ_s N; (c) Sliding motion — F = μ_k N.

Definitions

Section Title

1. Coulomb (Dry) Friction — Fundamental Theory

Common Mistakes

• Assuming N = W regardless of inclination or applied force direction — always resolve forces normal to the surface first.
• Assigning friction in the wrong direction — friction always opposes the IMPENDING motion, not the applied force.
• Using μ_k when μ_s is required (impending motion problems use μ_s).
• Forgetting that below impending motion, F is not equal to μN — it is whatever equilibrium requires.
• Using contact area in the friction formula — Coulomb friction is independent of contact area.

Formulas

Exam Tips

• If a problem states 'the block is on the verge of sliding down the incline under its own weight,' the incline angle equals the angle of repose, and you can immediately extract μ = tan(θ_given).
• The phrase 'rough surface' means friction is present; 'smooth surface' means frictionless (μ = 0, no friction force).
• Angle of friction problems often appear in context with the graphical (force triangle) method — practice both analytical and graphical approaches.

Key Points

• The angle of friction φ is defined as the angle that the total surface reaction R makes with the normal to the surface at impending slip: tan φ = μ_s.
• Physically, φ represents the maximum angle from the normal at which the resultant reaction can act without slip occurring.
• The angle of repose is the steepest inclination angle θ_r at which a block will rest on a surface under its own weight alone — it equals the angle of friction: θ_r = φ.
• Derivation of angle of repose: at impending slide down the incline, ΣF along slope = 0 gives W sin θ = μ W cos θ, so tan θ = μ = tan φ, hence θ_r = φ.
• Practical significance: soil mechanics uses the angle of repose extensively (angle of internal friction in Mohr-Coulomb failure criterion for soils).
• If the incline angle θ > φ (angle of friction), the block slides. If θ ≤ φ, the block remains stationary without any applied force.

Definitions

Section Title

2. Angle of Friction and Angle of Repose

Common Mistakes

• Confusing tan φ = μ with sin φ = μ or cos φ = μ — it is specifically the tangent.
• Forgetting that angle of repose equals angle of friction — these are the same angle, not two separate values.
• Mixing up degrees and radians when computing arctan(μ) on a calculator.

Formulas

Exam Tips

• Always check: is tan θ > μ? If no → self-locking. If yes → holding force is needed. This one check eliminates wrong answer choices.
• Memorize the pattern: UP case adds μ cos θ; HOLD (down) case subtracts μ cos θ. The sin θ term is always present.
• For incline problems with horizontal force P_h: resolve P_h along and normal to the slope (P_h cos θ along slope, P_h sin θ into slope), then update N = W cos θ + P_h sin θ.
• Board exam frequently presents both (a) force to push up and (b) minimum force to hold as two parts — solve both systematically.

Key Points

• Standard setup: block of weight W on an incline at angle θ to the horizontal. Applied force P acts parallel to the incline (most common board exam scenario).
• Normal force on the incline: N = W cos θ (for P parallel to slope).
• Two critical cases exist: (a) impending motion UP the slope, and (b) impending motion DOWN the slope (minimum holding force).
• For impending motion UP: friction acts DOWN the slope (opposing upward tendency). ΣF along slope = P - W sin θ - μN = 0.
• For impending motion DOWN (holding case): friction acts UP the slope. ΣF along slope = P + μN - W sin θ = 0.
• Self-locking condition: if tan θ < μ_s, the block will not slide down even with P = 0. No holding force is needed.
• If P is applied at an angle α to the incline, N changes: N = W cos θ - P sin α (for P directed away from surface). This modifies both F_max and the equilibrium equation.
• The minimum force to move a block up an incline is achieved when P is inclined at angle φ (angle of friction) above the incline surface — useful for optimization problems.

Definitions

Section Title

3. Block on an Inclined Plane

Common Mistakes

• Using N = W instead of N = W cos θ on inclined planes.
• Applying the wrong sign in the friction force — friction direction flips between the 'up' and 'hold' cases.
• Failing to check self-locking — computing a negative P_hold and reporting it without realizing it means no holding force is needed.
• When P is at an angle to the incline, not updating the normal force N to include the perpendicular component of P.
• Forgetting that P_up and P_hold bracket the range of P values for which the block remains stationary: P_hold ≤ P_static ≤ P_up.

Formulas

Exam Tips

• Label every friction force with an arrow and a direction on your FBD — examiners and you yourself need to track these carefully.
• For wedge problems with a smooth wall: the wall provides only a horizontal normal reaction (no friction). This simplifies the block FBD significantly.
• The graphical force polygon method can be faster for simple wedges — draw the known forces and close the polygon to find unknowns.
• Self-locking check is a quick 2-second calculation: α vs. 2φ. Do this before anything else in a wedge problem.

Key Points

• A wedge converts a small driving force P into a large lifting force on a load — it is essentially an inclined plane mechanism.
• Analysis requires separate FBDs for each body: (a) the wedge, and (b) the block/load being lifted.
• Friction must be assigned on ALL contact surfaces, opposing the impending motion of each respective body at each surface.
• Typical wedge problem: flat wedge of angle α under a block. Three surfaces may have friction: (1) wedge bottom on floor, (2) wedge top on block, (3) block side on wall (often smooth wall).
• Procedure: (1) Identify impending motion direction for each body. (2) Draw FBD with all normals and friction forces. (3) Write equilibrium equations for each body. (4) Set F = μN on each surface. (5) Solve simultaneously.
• Self-locking for a wedge: the wedge will NOT slide back out (is self-locking) when the wedge angle α ≤ 2φ for a double-friction-surface wedge. For a single friction surface: α ≤ φ.
• Once self-locking is established, no force is needed to hold the load — the wedge maintains position.
• Wedge problems often involve summing forces graphically using the force polygon — master both analytical and graphical methods.

Definitions

Section Title

4. Wedge Friction

Common Mistakes

• Drawing only one FBD for the entire wedge-block system instead of separate FBDs for the wedge and the block.
• Assigning friction forces in the same direction on both bodies at a shared surface — action-reaction: friction on the wedge top points in the opposite direction to friction on the block bottom.
• Forgetting friction on the floor surface of the wedge.
• Using the self-locking criterion α < φ (single-surface rule) for a double-surface wedge — it should be α < 2φ.
• Not checking self-locking before computing a withdrawal force.

Formulas

Exam Tips

• Memorize key values: e^(0.25π) ≈ 2.19, e^(0.3π) ≈ 2.57, e^(0.5π) ≈ 4.81 — these are common exam configurations.
• The belt friction equation can also be written: T_tight = T_slack × e^(μβ) — choose the form that has the unknown on the left.
• To find μ from given tensions: μ = ln(T_tight/T_slack) / β — inverse of the standard form.
• For problems asking 'minimum number of turns to hold a load,' set up T_tight/T_slack = e^(μ × n × 2π) and solve for n.
• Always state: 'At impending slip, T_tight/T_slack = e^(μβ)' before substituting — good exam practice.

Key Points

• A belt or rope wrapped around a curved surface (drum, pulley, bollard) develops a tension difference due to friction between the tight side and the slack side.
• The contact angle β MUST be in radians in all calculations. Convert degrees to radians: β (rad) = β (deg) × π/180.
• At impending slip: T_tight / T_slack = e^(μβ) where e is Euler's number (≈ 2.718).
• The tight side always has the higher tension; the slack side has the lower tension.
• The exponential relationship (capstan effect) means small increases in wrap angle dramatically increase the tension ratio — practical in rigging, mooring, and braking applications.
• Common contact angles in board exams: 90° = π/2 rad, 180° = π rad, 270° = 3π/2 rad, 360° = 2π rad.
• For a flat belt drive: the effective (net) force transmitted = T_tight - T_slack; this drives the friction load.
• For a V-belt: the effective friction coefficient is μ_eff = μ / sin(β_groove/2), which amplifies the tension ratio. (Less common on board exams but appears occasionally.)

Definitions

Section Title

5. Belt and Rope Friction (Capstan Equation)

Common Mistakes

• Using β in degrees instead of radians in the exponential — the most common belt friction error on exams.
• Confusing tight and slack sides — T_tight is always the LARGER tension (ratio > 1).
• Using μ_k for the coefficient when the problem specifies impending slip (use μ_s).
• Forgetting to convert partial turns: 1.5 turns = 1.5 × 2π = 3π rad (not 1.5 × 360° in the formula).
• In V-belt problems, using μ directly instead of the effective friction coefficient μ/sin(groove half-angle).

Formulas

Exam Tips

• For a uniform ladder, smooth wall: memorize tan θ_min = 1/(2μ) — this formula saves significant time.
• The moment center (point A at the base) eliminates both N_f and F_f = μN_f if you substitute the slip condition before taking moments — alternatively, take moments, then substitute — both work.
• When a rider is placed on the ladder, θ_min INCREASES (the ladder must be more upright) — use this as a check on your answer.
• A common variant: given the angle and μ, find the maximum safe height a person can climb. Set up the moment equation with person at unknown position d and solve.
• Ladder problems with friction on BOTH surfaces appear rarely but require two simultaneous slip conditions — read the problem carefully.

Key Points

• Classic three-force equilibrium problem: a uniform ladder leaning against a smooth (frictionless) wall with a rough floor.
• Forces acting: (1) Weight W at midpoint (uniform ladder) or at center of gravity (non-uniform load); (2) Normal reaction N_w from wall (horizontal, since wall is smooth); (3) Normal reaction N_f and friction F_f from floor.
• At impending slip at the floor: F_f = μ_s × N_f (friction acts horizontally toward the wall, opposing the tendency of the base to slide outward).
• Equilibrium equations: ΣF_x = 0 → N_w = F_f = μ N_f; ΣF_y = 0 → N_f = W; ΣM_A = 0 (moments about base).
• Moment equation about the base: N_w × L sin θ = W × (L/2) cos θ (uniform ladder, no rider). Where L is ladder length and θ is angle with horizontal.
• Solving for minimum angle: tan θ_min = W / (2 N_w) = W / (2μ N_f) = W / (2μW) = 1/(2μ).
• Formula: tan θ_min = 1/(2μ) for a uniform ladder against a smooth wall with friction only at the floor.
• If a person stands on the ladder: the weight distribution changes. The moment equation must include the person's weight at their position — this modifies the minimum angle formula.
• If both wall and floor are rough: two friction conditions, more complex analysis required.

Definitions

Section Title

6. The Ladder Problem

Common Mistakes

• Taking moments about the wrong point — always take moments about the base of the ladder to eliminate both floor reactions (N_f and F_f) from the moment equation.
• Using the wrong lever arms — the horizontal distance from base to W is (L/2) cos θ (not L/2), and the vertical height of the wall reaction is L sin θ.
• Forgetting that for a smooth wall, N_w is horizontal (not perpendicular to the ladder).
• For a person on the ladder: forgetting to include the person's weight in ΣF_y when computing N_f.
• Reporting the angle from the vertical instead of from the horizontal.