CELE Engineering Mechanics — FrictionExam Answer Templates

Marks

1

Topic

Coulomb Dry Friction

Difficulty

easy

Template Id

T1

Examiner Tip

For a 1-mark definition, one precise sentence with the correct formula is enough — lengthy prose wastes time.

Model Answer

The coefficient of static friction μₛ is the ratio of the maximum static friction force Fₘₐₓ to the normal force N at impending motion: μₛ = Fₘₐₓ/N. It is related to the angle of friction φ by tan φ = μₛ.

Question Type

very_short_answer

Answer Structure

• Line 1: State the definition and formula μₛ = Fₘₐₓ/N with the condition 'at impending motion' [1 mark]

Scoring Breakdown

Common Mark Deductions

• Writing μ = F/N without specifying it is the maximum or that motion is impending
• Confusing μₛ with μₖ (kinetic friction coefficient)
• Omitting the condition of impending motion

Key Phrases To Include

• impending motion
• maximum static friction
• μₛ = Fₘₐₓ/N
• tan φ = μₛ

Marks

1

Topic

Block on Inclined Plane

Difficulty

easy

Template Id

T2

Examiner Tip

Always connect the physical meaning ('no holding force needed') with the mathematical condition — examiners reward conceptual clarity even in a 1-mark item.

Model Answer

A block on an inclined plane is self-locking (requires no applied force to remain stationary) when the angle of inclination θ does not exceed the angle of friction φ, i.e., when tan θ ≤ μₛ.

Question Type

very_short_answer

Answer Structure

• Line 1: State the self-locking inequality tan θ ≤ μₛ (or equivalently θ ≤ φ) [1 mark]

Scoring Breakdown

Common Mark Deductions

• Stating the condition as tan θ > μₛ (which is the sliding condition, not the self-locking condition)
• Giving the condition without any mathematical expression

Key Phrases To Include

• tan θ ≤ μₛ
• angle of repose
• self-locking
• no external force required

Marks

2

Topic

Angle of Friction and Repose

Difficulty

easy

Template Id

T3

Examiner Tip

For 2-mark conceptual answers, one mark is typically for the definition and one for the relationship — structure your answer in two distinct statements.

Model Answer

The angle of repose is the maximum angle of inclination at which a block will remain stationary on a rough inclined surface without any applied force. At this limiting angle, the component of gravity along the incline exactly equals the maximum friction force. Therefore: tan φ_repose = μₛ, which shows that the angle of repose is numerically equal to the angle of friction φ.

Question Type

short_answer

Answer Structure

• Line 1: Define angle of repose — maximum inclination angle at which a block just rests [1 mark]
• Line 2: State the relationship tan φ = μₛ and conclude that angle of repose = angle of friction [1 mark]

Scoring Breakdown

Common Mark Deductions

• Defining angle of repose without stating it is a maximum (i.e., limiting) angle
• Stating the relationship without deriving or explaining it
• Confusing angle of repose with the angle of friction angle of the resultant reaction

Key Phrases To Include

• maximum angle of inclination
• remains stationary
• tan φ = μₛ
• angle of repose equals angle of friction

Marks

2

Topic

Coulomb Dry Friction

Difficulty

easy

Template Id

T4

Examiner Tip

Even for a simple 2-mark numerical, always state the formula before substituting — this earns the method mark if your calculator answer is wrong.

Model Answer

Given: W = 500 N, μₛ = 0.35, horizontal surface. FBD: N = W = 500 N (no incline, no vertical applied force). (a) Maximum friction force: Fₘₐₓ = μₛ × N = 0.35 × 500 = 175 N (b) Angle of friction: tan φ = μₛ = 0.35 φ = arctan(0.35) = 19.3°

Question Type

numerical

Answer Structure

• Step 1: Identify N from vertical equilibrium N = W = 500 N [½ mark]
• Step 2: Apply Fₘₐₓ = μₛN = 175 N [½ mark]
• Step 3: State tan φ = μₛ = 0.35 [½ mark]
• Step 4: Compute φ = 19.3° [½ mark]

Scoring Breakdown

Common Mark Deductions

• Using N = W without checking for incline or applied vertical force component
• Reporting the angle in radians instead of degrees
• Omitting units (N) from the friction force

Key Phrases To Include

• Fₘₐₓ = μₛN
• N = 500 N
• tan φ = μₛ
• φ = 19.3°

Marks

3

Topic

Block on Inclined Plane

Difficulty

medium

Template Id

T5

Examiner Tip

The self-locking check is a free mark — always do it first with a single inequality. Then write the force equation with the explicit sign logic (+ for pushing up).

Model Answer

Given: W = 300 N, θ = 15°, μₛ = 0.25. Step 1 — Check for sliding (self-locking condition): tan 15° = 0.268 > μₛ = 0.25 → tan θ > μₛ, so the block WILL slide down without support. Step 2 — Normal force: N = W cos θ = 300 cos 15° = 300 × 0.9659 = 289.8 N Step 3 — Force P for impending motion UP the incline (friction acts DOWN the slope): P = W(sin θ + μₛ cos θ) P = 300(sin 15° + 0.25 × cos 15°) P = 300(0.2588 + 0.25 × 0.9659) P = 300(0.2588 + 0.2415) P = 300 × 0.5003 P ≈ 150.1 N

Question Type

numerical

Answer Structure

• Step 1: Self-locking check — compare tan θ with μₛ; state conclusion [1 mark]
• Step 2: Compute N = W cos θ = 289.8 N [½ mark]
• Step 3: Apply P = W(sin θ + μₛ cos θ) with correct sign (+ for impending up) [1 mark]
• Step 4: Correct numerical answer P ≈ 150 N with unit [½ mark]

Scoring Breakdown

Common Mark Deductions

• Skipping the self-locking check entirely
• Using P = W(sin θ − μₛ cos θ) for the upward motion case (wrong sign — friction acts down, not up, when pushing up)
• Forgetting to note that N = W cos θ on an incline (not N = W)
• Not stating the direction of the friction force on the FBD

Key Phrases To Include

• tan θ > μₛ therefore slides
• N = W cos θ
• friction opposes impending motion
• P = W(sin θ + μₛ cos θ)
• 150 N

Marks

2

Topic

Block on Inclined Plane

Difficulty

medium

Template Id

T6

Examiner Tip

The key distinction between T5 and T6 is the sign: push-up uses +μₛ cos θ, hold-from-sliding uses −μₛ cos θ. Examiners check this sign explicitly.

Model Answer

Since tan 15° = 0.268 > μₛ = 0.25, the block would slide down — a holding force P is required. For impending motion DOWN (friction acts UP the slope, P acts up): P = W(sin θ − μₛ cos θ) P = 300(sin 15° − 0.25 cos 15°) P = 300(0.2588 − 0.2415) P = 300 × 0.0173 P ≈ 5.2 N (directed up the incline)

Question Type

numerical

Answer Structure

• Step 1: Identify friction direction as up the slope (opposing downward impending motion) [½ mark]
• Step 2: Apply P = W(sin θ − μₛ cos θ) with correct minus sign [1 mark]
• Step 3: Correct numerical answer P ≈ 5.2 N [½ mark]

Scoring Breakdown

Common Mark Deductions

• Using a positive sign (+ μₛ cos θ) in the holding-force formula — this is the impending-up formula
• Not specifying the direction of P (must state 'up the incline')
• Ignoring that P would be negative if θ < φ (self-locking case)

Key Phrases To Include

• friction acts up the slope
• P = W(sin θ − μₛ cos θ)
• 5.2 N
• directed up the incline

Marks

3

Topic

Belt Friction

Difficulty

medium

Template Id

T7

Examiner Tip

Always write 'β = 180° = π rad' explicitly on the first line — this shows the examiner you know the radian requirement and earns the conversion mark.

Model Answer

Given: β = 180° = π rad, μ = 0.25, T_tight = 1000 N. Step 1 — State the belt-friction formula: T_tight / T_slack = e^(μβ) Step 2 — Compute the exponent (β MUST be in radians): μβ = 0.25 × π = 0.25 × 3.1416 = 0.7854 Step 3 — Evaluate: e^0.7854 = 2.193 Step 4 — Solve for T_slack: T_slack = T_tight / e^(μβ) = 1000 / 2.193 ≈ 456 N

Question Type

numerical

Answer Structure

• Step 1: State T_tight/T_slack = e^(μβ) and explicitly convert 180° to π rad [1 mark]
• Step 2: Compute μβ = 0.7854 correctly [½ mark]
• Step 3: Evaluate e^0.7854 = 2.193 [½ mark]
• Step 4: Compute T_slack = 456 N with unit [1 mark]

Scoring Breakdown

Common Mark Deductions

• Using β = 180 (degrees) instead of π rad in the exponent — gives a completely wrong answer
• Inverting the formula: writing T_slack/T_tight = e^(μβ) leading to T_slack > T_tight
• Arithmetic error in evaluating the exponential
• Not stating or identifying which side is tight and which is slack

Key Phrases To Include

• T_tight/T_slack = e^(μβ)
• β = π rad
• e^(0.25π) = 2.193
• T_slack ≈ 456 N

Marks

3

Topic

Ladder Problem

Difficulty

medium

Template Id

T8

Examiner Tip

Take moments about the base — this eliminates N_F and F from the moment equation, leaving only N_W, which connects back to μₛ through horizontal equilibrium. This is the most efficient solution path.

Model Answer

Given: uniform ladder, smooth wall (no friction), μₛ = 0.30 at floor. FBD of ladder: • Reaction at wall: N_W (horizontal, perpendicular to wall) • Normal at floor: N_F (vertical) • Friction at floor: F = μₛ N_F (horizontal, toward wall) at impending slip • Weight W at midpoint (L/2 from base) Step 1 — Horizontal equilibrium: N_W = F = μₛ N_F … (i) Step 2 — Vertical equilibrium: N_F = W … (ii) Step 3 — Moment equilibrium about base (taking moments, L = ladder length): N_W × L sin θ − W × (L/2) cos θ = 0 N_W sin θ = (W/2) cos θ Step 4 — Substitute N_W = μₛ N_F = μₛ W: μₛ W sin θ = (W/2) cos θ tan θ = 1 / (2μₛ) = 1 / (2 × 0.30) = 1.667 θ_min = arctan(1.667) ≈ 59.0°

Question Type

numerical

Answer Structure

• Step 1: Draw FBD with N_W, N_F, W at midpoint, F = μₛN_F at floor [½ mark]
• Step 2: Write ΣFx = 0 and ΣFy = 0 to relate N_W and N_F to W [½ mark]
• Step 3: Write ΣM_base = 0 and set up moment equation correctly [1 mark]
• Step 4: Derive tan θ = 1/(2μₛ) and compute θ_min = 59.0° [1 mark]

Scoring Breakdown

Common Mark Deductions

• Applying friction at the wall as well as the floor (wall is stated smooth — zero friction)
• Taking moments about the wrong point (not eliminating two unknowns)
• Using W at the end of the ladder instead of the midpoint for a uniform ladder
• Not setting F = μₛ N_F (forgetting to apply the impending-slip condition)

Key Phrases To Include

• smooth wall — no friction at wall
• F = μₛ N_F at impending slip
• ΣM about base = 0
• tan θ = 1/(2μₛ)
• θ_min = 59.0°

Marks

2

Topic

Belt Friction

Difficulty

medium

Template Id

T9

Examiner Tip

For non-standard wrap angles (not 180°), always write the degree-to-radian conversion explicitly. Examiners check this step first in belt-friction problems.

Model Answer

Given: β = 270° = (3π/2) rad = 4.7124 rad, μ = 0.30, T_tight = 2000 N. Belt-friction formula: T_tight / T_slack = e^(μβ) e^(0.30 × 4.7124) = e^(1.4137) = 4.111 T_slack = 2000 / 4.111 ≈ 486.5 N

Question Type

numerical

Answer Structure

• Step 1: Convert β = 270° = 3π/2 rad = 4.7124 rad [½ mark]
• Step 2: Compute e^(0.30 × 4.7124) = e^1.4137 = 4.111 [½ mark]
• Step 3: Solve T_slack = 2000/4.111 ≈ 487 N [1 mark]

Scoring Breakdown

Common Mark Deductions

• Using β = 270 in degrees directly in the exponential
• Computing 270 × π/180 incorrectly (e.g., forgetting to reduce: 3π/2 ≈ 4.712 rad)
• Swapping tight and slack sides

Key Phrases To Include

• β = 270° = 3π/2 rad
• e^(μβ) = 4.111
• T_slack ≈ 487 N

Marks

3

Topic

Wedges

Difficulty

medium

Template Id

T10

Examiner Tip

In wedge problems, the most common error is drawing one FBD for the whole system. Examiners always award marks for separate FBDs — draw them even if you run out of time for the calculation.

Model Answer

Procedure for Wedge Analysis: 1. Identify all contact surfaces: A wedge typically has two friction surfaces (wedge-block interface and wedge-floor interface) and the block may have a third contact with a wall. 2. Draw separate Free-Body Diagrams (FBDs) for each body (wedge and block), showing: • Normal forces perpendicular to each surface • Friction forces F = μN on every contact surface, directed opposite to impending motion 3. Establish the direction of impending motion for each surface before assigning friction directions — this is critical and changes for raising vs. lowering the load. 4. Write equilibrium equations (ΣFx = 0, ΣFy = 0) for each body and solve simultaneously for the driving force P. Self-locking condition for a wedge: The wedge is self-locking (load does not slide back down when driving force is removed) if the wedge angle α satisfies: α < 2φ (approximately, when both faces have the same μ) where φ = arctan(μ) is the friction angle. If α ≥ 2φ, the wedge is not self-locking.

Question Type

short_answer

Answer Structure

• Point 1: Identify all contact surfaces and the number of friction faces [½ mark]
• Point 2: Draw separate FBDs for wedge and block with friction on ALL surfaces opposing impending motion [1 mark]
• Point 3: Write ΣFx = 0 and ΣFy = 0 for each body [½ mark]
• Point 4: State self-locking condition α < 2φ with definition of φ [1 mark]

Scoring Breakdown

Common Mark Deductions

• Drawing only one combined FBD instead of separate FBDs for wedge and block
• Forgetting friction on one of the contact surfaces (e.g., only applying friction on the wedge face but not the floor)
• Applying the block self-locking condition (tan θ < μ) instead of the wedge self-locking condition (α < 2φ)
• Not specifying the direction of friction on each surface

Key Phrases To Include

• separate FBDs for each body
• friction on every contact surface
• opposes impending motion
• ΣFx = 0, ΣFy = 0
• α < 2φ for self-locking

Marks

5

Topic

Wedges

Difficulty

hard

Template Id

T11

Examiner Tip

For full marks on wedge problems, always show two separate FBDs clearly labeled — one for the block and one for the wedge. The FBDs alone are worth marks even before the calculation. Always close with the self-locking check.

Model Answer

Given: Load W = 10 kN, wedge angle α = 8°, μ = 0.20 on all surfaces. Friction angle: φ = arctan(0.20) = 11.31° --- FBD of the Block (being lifted) --- Contact surfaces on block: (1) vertical wall — friction force opposing downward motion (acts upward), (2) inclined wedge face — friction force opposing upward motion of block (acts downward along the wedge face). Resultant reaction on wedge face with block: R₁ at angle φ = 11.31° from normal to wedge face. Reaction at wall: R₂ at angle φ = 11.31° from horizontal. For the block, the three forces (W, R₁, R₂) must be concurrent and in equilibrium. Using Lami's theorem or resolving forces: Angle analysis (measuring from vertical): • W acts downward (270°) • R₁ acts at (90° + α + φ) from positive x-axis = 90° + 8° + 11.31° = 109.31° from horizontal • R₂ acts at (φ) from horizontal = 11.31° above horizontal Equilibrium of block (ΣFy = 0, ΣFx = 0): R₂ = R₁ × sin(α + 2φ) / cos φ [derived from force triangle] Using the force triangle approach: R₁ sin(α + 2φ) = W cos(α + φ) [not the most direct path] Direct equilibrium of block: ΣFy = 0: R₁ cos(α + φ) + R₂ sin φ − W = 0 … (i) ΣFx = 0: R₁ sin(α + φ) − R₂ cos φ = 0 … (ii) From (ii): R₂ = R₁ sin(α + φ) / cos φ Substitute into (i): R₁ cos(α + φ) + R₁ sin(α + φ) tan φ = W R₁ [cos(α + φ) + sin(α + φ) tan φ] = W R₁ × sin(α + φ + φ) / cos φ = W [using identity] R₁ = W cos φ / sin(α + 2φ) Numerical: α + 2φ = 8° + 22.62° = 30.62° R₁ = 10 × cos(11.31°) / sin(30.62°) = 10 × 0.9806 / 0.5093 = 19.25 kN --- FBD of the Wedge --- Three forces: P (horizontal), R₁' (reaction from block = R₁, acting opposite direction), R₃ (reaction from floor at angle φ from vertical). ΣFx = 0: P = R₁ sin(α + φ) + R₃ sin φ ΣFy = 0: R₃ cos φ = R₁ cos(α + φ) R₃ = R₁ cos(α + φ) / cos φ = 19.25 × cos(19.31°) / cos(11.31°) = 19.25 × 0.9437 / 0.9806 = 18.53 kN P = R₁ sin(α + φ) + R₃ sin φ P = 19.25 × sin(19.31°) + 18.53 × sin(11.31°) P = 19.25 × 0.3305 + 18.53 × 0.1961 P = 6.36 + 3.63 P ≈ 9.99 kN ≈ 10.0 kN (b) Self-locking check: Self-locking requires α < 2φ 2φ = 2 × 11.31° = 22.62° α = 8° < 22.62° ✓ → The wedge IS self-locking.

Question Type

numerical

Answer Structure

• Step 1: Compute friction angle φ = arctan(0.20) = 11.31° [½ mark]
• Step 2: Draw and label FBD of the block with reactions R₁ and R₂, each inclined at φ from their respective surfaces [1 mark]
• Step 3: Apply equilibrium to the block and solve for R₁ [1 mark]
• Step 4: Draw and label FBD of the wedge with P, R₁', and R₃; apply equilibrium [1 mark]
• Step 5: Compute P ≈ 10.0 kN [½ mark]
• Step 6: Self-locking check: α = 8° < 2φ = 22.62°, therefore self-locking [1 mark]

Scoring Breakdown

Common Mark Deductions

• Drawing a single FBD for the combined wedge-block system
• Applying friction only on the wedge-block interface and ignoring the wedge-floor friction
• Using block self-locking criterion (tan θ < μ) instead of wedge criterion (α < 2φ)
• Omitting the friction angle φ and working with μ directly without converting to angle
• Arithmetic errors in angle addition (α + φ, α + 2φ)

Key Phrases To Include

• φ = arctan(μ) = 11.31°
• separate FBDs for block and wedge
• friction opposes impending motion on all surfaces
• P ≈ 10.0 kN
• α < 2φ therefore self-locking

Marks

5

Topic

Block on Inclined Plane

Difficulty

hard

Template Id

T12

Examiner Tip

When P is not parallel to the incline, its normal component changes N — this is the key complication. Always resolve P first, then write the normal equilibrium before writing the friction force.

Model Answer

Given: W = 500 N, θ = 30° (incline), μₛ = 0.40, P acts at α = 20° above the incline surface. Step 1 — FBD: Forces on block: • W = 500 N vertically downward • N = normal force, perpendicular to incline (upward from surface) • F = μₛN = 0.40N, along the incline directed DOWN (opposing upward impending motion) • P at 20° above the incline surface (has components P cos 20° along incline UP, and P sin 20° normal to incline) Step 2 — Normal equilibrium (perpendicular to incline, positive away from surface): N + P sin 20° − W cos 30° = 0 N = W cos 30° − P sin 20° N = 500 × 0.8660 − P × 0.3420 N = 433.0 − 0.3420P … (i) Step 3 — Friction force (impending motion up, F acts down the slope): F = μₛN = 0.40(433.0 − 0.3420P) = 173.2 − 0.1368P … (ii) Step 4 — Tangential equilibrium (along incline, positive up): P cos 20° − W sin 30° − F = 0 P × 0.9397 − 500 × 0.5000 − (173.2 − 0.1368P) = 0 0.9397P − 250.0 − 173.2 + 0.1368P = 0 1.0765P = 423.2 P = 423.2 / 1.0765 P ≈ 393.1 N

Question Type

numerical

Answer Structure

• Step 1: FBD with W, N, F (down the slope), P resolved into components along and perpendicular to incline [1 mark]
• Step 2: Normal equilibrium — N = W cos θ − P sin α (note P sin α reduces N) [1 mark]
• Step 3: Set F = μₛN = 0.40(433 − 0.342P) [1 mark]
• Step 4: Tangential equilibrium — P cos α − W sin θ − F = 0 [1 mark]
• Step 5: Solve for P ≈ 393 N [1 mark]

Scoring Breakdown

Common Mark Deductions

• Treating N as simply W cos θ without accounting for the normal component of P (P sin 20° reduces N)
• Treating the problem as if P is parallel to the incline
• Using F = μₛ × W cos θ (constant N) instead of F = μₛ(W cos θ − P sin α)
• Sign error: adding P sin α to N instead of subtracting (an oblique upward P reduces the normal force)

Key Phrases To Include

• resolve P into components along and normal to incline
• N = W cos θ − P sin α
• F = μₛN depends on P
• P cos α − W sin θ − F = 0
• P ≈ 393 N

Marks

2

Topic

Coulomb Dry Friction

Difficulty

easy

Template Id

T13

Examiner Tip

For comparison questions, use a two-column or two-paragraph structure: one paragraph for static, one for kinetic. End with the physical explanation.

Model Answer

Static friction (Fₛ) is the friction force that prevents relative motion between surfaces in contact. It acts when there is no sliding and can take any value from zero up to its maximum Fₛ,max = μₛN at impending motion. Kinetic friction (Fₖ) is the constant friction force that acts between surfaces in relative (sliding) motion: Fₖ = μₖN, where μₖ < μₛ. Reason μₛ > μₖ: The micro-asperities (surface irregularities) interlock more fully under static contact, requiring a higher force to initiate motion. Once sliding begins, the asperities no longer have time to fully interlock, so the resistance drops — hence kinetic friction is lower than static friction.

Question Type

short_answer

Answer Structure

• Line 1-2: Define static friction — variable up to μₛN, acts before motion [1 mark]
• Line 3-4: Define kinetic friction — constant = μₖN, acts during sliding, plus explain why μₛ > μₖ [1 mark]

Scoring Breakdown

Common Mark Deductions

• Stating μₛ > μₖ without explaining why
• Saying kinetic friction 'varies' (it is treated as constant in Coulomb friction model)
• Not mentioning that static friction is variable (up to a maximum)

Key Phrases To Include

• Fₛ ≤ μₛN
• Fₖ = μₖN (constant)
• impending motion
• μₛ > μₖ
• surface asperities

Marks

3

Topic

Belt Friction

Difficulty

hard

Template Id

T14

Examiner Tip

For capstan problems, always convert turns to radians first: n turns = n × 2π rad. Write this conversion prominently — it is always checked by the examiner.

Model Answer

Given: 3 full turns → β = 3 × 2π = 6π rad = 18.850 rad, μ = 0.15, T_tight = 50 kN. Belt-friction (capstan) formula: T_tight / T_slack = e^(μβ) e^(0.15 × 6π) = e^(0.15 × 18.850) = e^(2.8274) = 16.92 T_slack = T_tight / e^(μβ) = 50,000 / 16.92 ≈ 2,956 N ≈ 2.96 kN A person needs to exert only about 2.96 kN on the slack end to hold a 50 kN ship load — demonstrating the enormous mechanical advantage of the capstan (the exponential capstan effect).

Question Type

numerical

Answer Structure

• Step 1: Convert 3 turns to β = 6π rad = 18.85 rad [1 mark]
• Step 2: Compute e^(μβ) = e^2.827 = 16.92 [½ mark]
• Step 3: Compute T_slack = 50/16.92 ≈ 2.96 kN [1 mark]
• Step 4 (bonus): Comment on the capstan mechanical advantage [½ mark]

Scoring Breakdown

Common Mark Deductions

• Using β = 3 × 360° = 1080° in the exponential without converting to radians
• Using β = 2π for 3 turns (forgetting to multiply by 3)
• Confusing tight and slack sides — T_slack must be the SMALLER tension

Key Phrases To Include

• β = 3 × 2π = 6π rad
• e^(μβ) = 16.92
• T_slack ≈ 2.96 kN
• capstan effect

Marks

3

Topic

Coulomb Dry Friction

Difficulty

easy

Template Id

T15

Examiner Tip

This type of problem tests whether you understand the physical difference between static and kinetic friction. Always label which force corresponds to which condition, then apply the formula — examiners penalize if you mix them up.

Model Answer

Given: W = 200 N (horizontal surface), N = W = 200 N. Force at impending motion = 70 N → Fₛ,max = 70 N. Force to maintain sliding = 60 N → Fₖ = 60 N. (a) Coefficient of static friction: μₛ = Fₛ,max / N = 70 / 200 = 0.35 (b) Coefficient of kinetic friction: μₖ = Fₖ / N = 60 / 200 = 0.30 Note: μₖ = 0.30 < μₛ = 0.35 ✓ (consistent with Coulomb friction model) (c) Angle of (static) friction: tan φₛ = μₛ = 0.35 φₛ = arctan(0.35) = 19.3°

Question Type

numerical

Answer Structure

• Step 1: Identify N = W = 200 N (horizontal surface) [½ mark]
• Step 2: μₛ = 70/200 = 0.35 [½ mark]
• Step 3: μₖ = 60/200 = 0.30 [½ mark]
• Step 4: Check μₖ < μₛ [½ mark]
• Step 5: φₛ = arctan(0.35) = 19.3° [1 mark]

Scoring Breakdown

Common Mark Deductions

• Computing the average of 70 N and 60 N instead of treating them separately
• Not checking or noting that μₖ < μₛ
• Reporting friction angle in radians
• Confusing the forces: 70 N is for static (impending), 60 N is for kinetic (moving)

Key Phrases To Include

• N = W = 200 N (horizontal surface)
• μₛ = Fₛ,max/N = 0.35
• μₖ = Fₖ/N = 0.30
• μₖ < μₛ
• φₛ = arctan(0.35) = 19.3°