CELE Engineering Mechanics — Force Systems and ResultantsExam Answer Templates

Marks

1

Topic

Resultant of Force Systems

Difficulty

easy

Template Id

T1

Examiner Tip

The word 'equivalent' and the phrase 'same external effect' are the two triggers examiners look for in a 1-mark definition. One of these must appear.

Model Answer

The resultant of a force system is a single equivalent force (and, if necessary, a couple) that produces the same external effect — the same net translation and net rotation — on a body as the entire original system of forces.

Question Type

very_short_answer

Answer Structure

• One concise sentence: identify what the resultant is (single equivalent force) and state the equivalence condition (same net effect on the body) [1 mark]

Scoring Breakdown

Common Mark Deductions

• Saying only 'sum of all forces' without mentioning equivalence of effect
• Omitting that direction/line of action must also be equivalent

Key Phrases To Include

• single equivalent force
• same external effect
• net translation and rotation

Marks

1

Topic

Varignon's Theorem

Difficulty

easy

Template Id

T2

Examiner Tip

The examiner is specifically checking that you distinguish between the force itself and its components. Mention 'components' explicitly.

Model Answer

Varignon's Theorem states that the moment of a force about any point is equal to the algebraic sum of the moments of its rectangular components about that same point.

Question Type

very_short_answer

Answer Structure

• One sentence: name the theorem and state the equality between moment of a force and the sum of moments of its components [1 mark]

Scoring Breakdown

Common Mark Deductions

• Stating it as 'sum of forces' instead of 'sum of moments of components'
• Missing the phrase 'about the same point'

Key Phrases To Include

• moment of a force
• algebraic sum
• moments of its components
• same point

Marks

2

Topic

Couples

Difficulty

easy

Template Id

T3

Examiner Tip

Two marks require two distinct ideas. Examiners split marks: one for definition, one for the free-vector / pure-rotation property. Write them in separate sentences.

Model Answer

A couple consists of two forces that are equal in magnitude, opposite in direction, and parallel, separated by a perpendicular distance d. The moment of a couple M = Fd is a free vector — it has the same magnitude about every point in the plane, regardless of the chosen moment center. A couple produces pure rotation with no net translational force.

Question Type

short_answer

Answer Structure

• Line 1: Define a couple — two equal, opposite, parallel forces separated by distance d [1 mark]
• Line 2: State the key characteristic — couple moment M = Fd is the same about every point (free vector); produces pure rotation [1 mark]

Scoring Breakdown

Common Mark Deductions

• Defining a couple as simply 'two opposite forces' without specifying they must be parallel and equal in magnitude
• Omitting the free-vector characteristic (constant moment about any point)
• Missing the formula M = Fd

Key Phrases To Include

• equal in magnitude, opposite in direction, parallel
• perpendicular distance d
• M = Fd
• free vector
• same about every point
• pure rotation

Marks

2

Topic

Force Components

Difficulty

easy

Template Id

T4

Examiner Tip

For a 2-mark numerical, one mark per component. Direction arrows or words (rightward, upward) are required — a component is a vector, not just a number.

Model Answer

Given: F = 260 N, θ = 67.38° from +x-axis. Horizontal component: Fx = F cos θ = 260 cos 67.38° = 260 × 0.3846 = 100.0 N (→) Vertical component: Fy = F sin θ = 260 sin 67.38° = 260 × 0.9231 = 240.0 N (↑) ∴ Fx = 100 N (rightward), Fy = 240 N (upward).

Question Type

numerical

Answer Structure

• State given data and angle [½ mark implicit setup]
• Apply Fx = F cos θ with substitution and result including unit and direction [1 mark]
• Apply Fy = F sin θ with substitution and result including unit and direction [1 mark]

Scoring Breakdown

Common Mark Deductions

• Swapping sin and cos (common when angle is measured from y-axis instead of x-axis)
• Omitting direction indicators (→ or ↑)
• Rounding error leading to wrong decimal values — keep at least 4 significant figures in intermediate steps

Key Phrases To Include

• Fx = F cos θ
• Fy = F sin θ
• N (unit)
• rightward / upward direction indicators

Marks

3

Topic

Moment of a Force / Varignon's Theorem

Difficulty

medium

Template Id

T5

Examiner Tip

Explicitly label the step 'By Varignon's Theorem'. Examiners are specifically checking that you know the theorem by name and application, not just that you can multiply numbers.

Model Answer

Given: F = 500 N at point A(4, 0) m, θ = 60° from horizontal. Step 1 — Resolve into components: Fx = 500 cos 60° = 500 × 0.5 = 250 N (→) Fy = 500 sin 60° = 500 × 0.8660 = 433.0 N (↑) Step 2 — Apply Varignon's Theorem (CCW positive): Mo = x·Fy − y·Fx Mo = (4)(433.0) − (0)(250) Mo = 1732 − 0 Mo = 1732 N·m (counterclockwise) ∴ The moment of the force about the origin is 1732 N·m counterclockwise.

Question Type

numerical

Answer Structure

• Step 1: Resolve force into Fx and Fy — correct formulas with values and units [1 mark]
• Step 2: Write Varignon's formula Mo = xFy − yFx explicitly, then substitute values [1 mark]
• Step 3: Correct final answer with value, unit (N·m), and rotational direction [1 mark]

Scoring Breakdown

Common Mark Deductions

• Using Mo = yFx − xFy (wrong sign convention without declaring it)
• Forgetting to state rotational direction (CCW or CW)
• Not showing the component resolution step — examiners want to see the work, not just the answer
• Incorrect unit — writing 'N' instead of 'N·m' for moment

Key Phrases To Include

• Varignon's Theorem
• Mo = xFy − yFx
• CCW positive
• 1732 N·m
• counterclockwise

Marks

3

Topic

Resultant of Concurrent Forces

Difficulty

medium

Template Id

T6

Examiner Tip

Show the component table (each force broken into Fx and Fy) even for 3-mark items. This organisation earns partial credit if your arithmetic has a minor error.

Model Answer

Given: F1 = 100 N at 0°, F2 = 150 N at 90°, F3 = 80 N at 200°. Step 1 — Resolve and sum x-components: Rx = ΣFx = 100 cos 0° + 150 cos 90° + 80 cos 200° = 100(1) + 150(0) + 80(−0.9397) = 100 + 0 − 75.18 = 24.82 N Step 2 — Resolve and sum y-components: Ry = ΣFy = 100 sin 0° + 150 sin 90° + 80 sin 200° = 0 + 150 + 80(−0.3420) = 150 − 27.36 = 122.64 N Step 3 — Magnitude of resultant: R = √(Rx² + Ry²) = √(24.82² + 122.64²) R = √(616.0 + 15040.6) = √15656.6 R ≈ 125.1 N Step 4 — Direction (both Rx and Ry are positive → 1st quadrant): θR = arctan(Ry/Rx) = arctan(122.64/24.82) = arctan(4.942) ≈ 78.6° ∴ R = 125.1 N at θ = 78.6° from the positive x-axis (1st quadrant).

Question Type

numerical

Answer Structure

• Step 1: Compute Rx = ΣFx with all terms shown [1 mark]
• Step 2: Compute Ry = ΣFy with all terms shown [1 mark — partial for setup, partial for Ry value]
• Steps 3–4: Correct R = 125.1 N and θ = 78.6° with quadrant check [1 mark]

Scoring Breakdown

Common Mark Deductions

• Using 200° as if it were in the 2nd quadrant — must use 200° directly with cos and sin, which yield negative values
• Omitting quadrant check after arctan — arctan only gives −90° to +90°; quadrant must be confirmed from signs of Rx and Ry
• Not showing individual component terms for each force — partial credit requires visible work

Key Phrases To Include

• Rx = ΣFx
• Ry = ΣFy
• R = √(Rx² + Ry²)
• θR = arctan(Ry/Rx)
• 1st quadrant
• 78.6°

Marks

2

Topic

Couples

Difficulty

easy

Template Id

T7

Examiner Tip

Determination of CW vs CCW: the 40 N upward force at x = 0 and 40 N downward at x = 1.5 m — imagine the rotation they impose; the pair tends to rotate the body clockwise. Always state this.

Model Answer

Given: F = 40 N, separation distance d = 1.5 m. Couple moment: M = F × d = 40 × 1.5 = 60 N·m (clockwise) Significance: Since a couple is a free vector, this moment of 60 N·m clockwise is the same about any point in the plane. The net force of the couple is zero, so it produces pure rotation without translation.

Question Type

short_answer

Answer Structure

• Line 1: Apply M = Fd with substitution and identify rotational direction [1 mark]
• Line 2: State the significance — free vector property (same about every point) and pure rotation [1 mark]

Scoring Breakdown

Common Mark Deductions

• Not stating the rotational direction (CW or CCW)
• Omitting the significance / free vector explanation — this is the second mark
• Incorrect d: using x-coordinates of individual forces rather than the separation distance

Key Phrases To Include

• M = Fd
• 60 N·m
• clockwise
• free vector
• pure rotation
• zero net force

Marks

3

Topic

Resultant of Parallel Force Systems

Difficulty

medium

Template Id

T8

Examiner Tip

Set up a neat three-column table (Force | Position | Moment) even for 3-mark items. This organisation is what separates a 3/3 from a 2/3 answer.

Model Answer

Given: W1 = 30 kN at x1 = 1 m, W2 = 50 kN at x2 = 3 m, W3 = 20 kN at x3 = 6 m. (Taking downward as positive for parallel force sum.) Step 1 — Magnitude of resultant: R = ΣF = 30 + 50 + 20 = 100 kN (downward) Step 2 — Moment sum about left end O (x = 0): ΣMo = 30(1) + 50(3) + 20(6) = 30 + 150 + 120 = 300 kN·m Step 3 — Location of resultant: x̄ = ΣMo / R = 300 / 100 = 3.0 m from O ∴ The resultant is R = 100 kN downward, acting at x̄ = 3.0 m from the left end.

Question Type

numerical

Answer Structure

• Step 1: R = ΣF = 100 kN (downward) — correct magnitude and direction [1 mark]
• Step 2: Moment table ΣMo = 30(1) + 50(3) + 20(6) = 300 kN·m — all terms shown [1 mark]
• Step 3: x̄ = ΣMo/R = 3.0 m — correct location from reference point [1 mark]

Scoring Breakdown

Common Mark Deductions

• Omitting direction of resultant (downward)
• Not referencing the moment sum to a specific point O — must state which end is the reference
• Arithmetic error in moment sum — show all three products to allow partial credit
• Confusing x̄ formula with centroid formula — they are the same principle but context must be stated

Key Phrases To Include

• R = ΣF
• ΣMo = Σ(Wi × xi)
• x̄ = ΣMo / R
• 3.0 m from O
• 100 kN downward

Marks

5

Topic

Resultant of Concurrent Forces

Difficulty

hard

Template Id

T9

Examiner Tip

For 5-mark numerical problems, the marking key awards marks at each step. Even if you get the final R wrong, you earn 4/5 for correct components, correct sums, and correct quadrant. Never skip steps.

Model Answer

Given: Concurrent force system at origin. F1 = 120 N, θ1 = 30° F2 = 90 N, θ2 = 135° F3 = 60 N, θ3 = 240° F4 = 100 N, θ4 = 300° Step 1 — Resolve each force into x- and y-components: F1x = 120 cos 30° = 120(0.8660) = +103.92 N F1y = 120 sin 30° = 120(0.5000) = +60.00 N F2x = 90 cos 135° = 90(−0.7071) = −63.64 N F2y = 90 sin 135° = 90(+0.7071) = +63.64 N F3x = 60 cos 240° = 60(−0.5000) = −30.00 N F3y = 60 sin 240° = 60(−0.8660) = −51.96 N F4x = 100 cos 300° = 100(+0.5000) = +50.00 N F4y = 100 sin 300° = 100(−0.8660) = −86.60 N Step 2 — Sum components: Rx = ΣFx = 103.92 − 63.64 − 30.00 + 50.00 = +60.28 N Ry = ΣFy = 60.00 + 63.64 − 51.96 − 86.60 = −14.92 N Step 3 — Magnitude of resultant: R = √(Rx² + Ry²) = √(60.28² + 14.92²) R = √(3633.7 + 222.6) = √3856.3 R ≈ 62.1 N Step 4 — Quadrant check: Rx = +60.28 N (positive → rightward) Ry = −14.92 N (negative → downward) ∴ Resultant lies in the 4th quadrant. Step 5 — Direction angle: α = arctan(|Ry|/|Rx|) = arctan(14.92/60.28) = arctan(0.2475) = 13.9° θR = 360° − 13.9° = 346.1° (measured from +x-axis, CCW) Or equivalently: 13.9° below the positive x-axis. ∴ R = 62.1 N at θR = 346.1° from the positive x-axis (or 13.9° below +x, 4th quadrant).

Question Type

numerical

Answer Structure

• Step 1: Component table — all four forces resolved into Fx and Fy with correct signs [2 marks]
• Step 2: Correct ΣFx and ΣFy summed with values [1 mark]
• Step 3: Correct magnitude R = 62.1 N using Pythagorean theorem [1 mark]
• Steps 4–5: Correct quadrant identification and direction angle θR = 346.1° (or 13.9° below +x) [1 mark]

Scoring Breakdown

Common Mark Deductions

• Sign errors in components for 2nd, 3rd, or 4th quadrant forces — most common source of error
• Skipping the quadrant check and reporting angle from arctan directly (gives wrong quadrant for 3rd and 4th quadrant results)
• Not showing the component resolution table — bulk of marks (2/5) are here
• Rounding cos/sin values too early (use 4 decimal places minimum in intermediate steps)
• Stating angle as 13.9° without clarifying it is below the +x-axis or giving the standard 346.1° — ambiguous and may lose the direction mark

Key Phrases To Include

• Rx = ΣFx
• Ry = ΣFy
• R = √(Rx² + Ry²)
• 4th quadrant
• 346.1°
• 13.9° below +x-axis

Marks

5

Topic

Moment of a Force / Varignon's Theorem

Difficulty

hard

Template Id

T10

Examiner Tip

The key insight examiners test here is that you CANNOT use |AB| as a moment arm unless verified perpendicular. The unit-vector method with Varignon is the fail-safe approach — show it.

Model Answer

Given: Force F = 300 N directed from A(1,1) to B(4,5). Step 1 — Find the direction vector AB: AB = (4−1, 5−1) = (3, 4) |AB| = √(3² + 4²) = √25 = 5 m Step 2 — Unit vector along AB: û = (3/5, 4/5) = (0.6, 0.8) Step 3 — Force components: Fx = 300 × 0.6 = 180 N (→) Fy = 300 × 0.8 = 240 N (↑) Step 4 — Apply Varignon's Theorem about O, using point A(1,1) as the point of application: (Taking CCW as positive) Mo = x·Fy − y·Fx Mo = (1)(240) − (1)(180) Mo = 240 − 180 Mo = 60 N·m (counterclockwise) Verification using point B(4,5): Mo = (4)(240) − (5)(180) = 960 − 900 = 60 N·m ✓ ∴ The moment of the 300 N force about the origin is 60 N·m counterclockwise.

Question Type

numerical

Answer Structure

• Step 1: Find direction vector AB = (3, 4) and its magnitude |AB| = 5 m [1 mark]
• Step 2: Compute unit vector û = (0.6, 0.8) [1 mark]
• Step 3: Force components Fx = 180 N, Fy = 240 N [1 mark]
• Step 4: Apply Mo = xFy − yFx at point A (or B) = 60 N·m CCW [1 mark]
• Verification or clear final statement with unit and direction [1 mark]

Scoring Breakdown

Common Mark Deductions

• Using the distance |AB| = 5 m as the moment arm directly (wrong — that is not perpendicular to the line of action)
• Not computing unit vector — applying F = 300 N directly without resolving into components
• Using origin O as the point of application instead of a point on the line of action (A or B)
• Forgetting units (N·m) or direction (CCW)

Key Phrases To Include

• direction vector AB
• unit vector
• Fx = F × cosine
• Mo = xFy − yFx
• Varignon's Theorem
• 60 N·m counterclockwise

Marks

2

Topic

Resultant of Non-Concurrent Force Systems

Difficulty

medium

Template Id

T11

Examiner Tip

Examiners want to see that you understand 'concurrent' refers to lines of action, not just points of application — a common exam distinction.

Model Answer

A concurrent force system has all forces whose lines of action pass through a common point. Its resultant is a single force R = √(Rx² + Ry²) passing through that common point, found by vector addition of components. A non-concurrent force system has forces whose lines of action do NOT all intersect at one point. It reduces to a single resultant force R (from ΣFx and ΣFy) whose line of action is located at a perpendicular distance d = ΣMo / R from a chosen reference point, ensuring moment equivalence. If R = 0 but ΣM ≠ 0, the system reduces to a couple.

Question Type

short_answer

Answer Structure

• Sentence 1–2: Define concurrent system and state its resultant — passes through common point via vector addition [1 mark]
• Sentence 3–4: Define non-concurrent system and state resultant location via d = ΣMo/R; mention couple case [1 mark]

Scoring Breakdown

Common Mark Deductions

• Defining non-concurrent simply as 'forces not at one point' without mentioning line of action
• Omitting the couple special case (R = 0, ΣM ≠ 0)

Key Phrases To Include

• common point
• lines of action
• R = √(Rx² + Ry²)
• d = ΣMo / R
• moment equivalence
• couple when R = 0

Marks

2

Topic

Force Components in 3D

Difficulty

medium

Template Id

T12

Examiner Tip

This is a recall-and-explain item. Earn both marks by pairing each formula with its constraint. Do not just list formulas — the examiner asks you to explain what the constraint means.

Model Answer

For a force F in 3D space, the direction cosines are: cos θx = Fx/F, cos θy = Fy/F, cos θz = Fz/F where θx, θy, θz are the angles F makes with the positive x-, y-, z-axes respectively. Constraint: The sum of the squares of the direction cosines must equal unity: cos²θx + cos²θy + cos²θz = 1 This confirms that the unit vector along F has magnitude exactly 1.

Question Type

short_answer

Answer Structure

• State the three formulas cos θx = Fx/F etc. [1 mark]
• State the constraint equation cos²θx + cos²θy + cos²θz = 1 and its meaning [1 mark]

Scoring Breakdown

Common Mark Deductions

• Writing cos θx = F/Fx (inverted ratio)
• Omitting the constraint equation — this is specifically asked in the question
• Not explaining why the constraint must equal 1

Key Phrases To Include

• cos θx = Fx/F
• cos θy = Fy/F
• cos θz = Fz/F
• cos²θx + cos²θy + cos²θz = 1
• unit vector
• magnitude 1

Marks

3

Topic

Moment of a Force

Difficulty

easy

Template Id

T13

Examiner Tip

A 3-mark question expects three distinct, scoreable ideas: definition, scalar formula with variables, component formula with variables. Structure your answer in three clearly separated parts.

Model Answer

The moment of a force about a point O (also called torque) is a measure of the tendency of the force to rotate a body about that point. Scalar formula: Mo = F × d where F = magnitude of the force (N) d = perpendicular distance from O to the force's line of action (m) Mo = moment in N·m Component formula (2D, by Varignon's Theorem): Mo = x·Fy − y·Fx where (x, y) = coordinates of the point of application of F Fx, Fy = rectangular components of the force Sign convention: Counterclockwise (CCW) moments are taken as positive.

Question Type

short_answer

Answer Structure

• Line 1: Definition — tendency to rotate about point O [1 mark]
• Lines 2–4: Scalar formula Mo = Fd with all variables defined [1 mark]
• Lines 5–7: Component formula Mo = xFy − yFx with variables defined and sign convention stated [1 mark]

Scoring Breakdown

Common Mark Deductions

• Defining moment as 'force times distance' without specifying perpendicular distance
• Omitting sign convention — this loses the third mark in most marking schemes
• Not defining variables (F, d, x, y) — define-all is part of the 3-mark expectation

Key Phrases To Include

• tendency to rotate
• Mo = F × d
• perpendicular distance
• Mo = xFy − yFx
• CCW positive
• point of application

Marks

2

Topic

Resultant of Non-Concurrent Force Systems / Couples

Difficulty

medium

Template Id

T14

Examiner Tip

This is a conceptual test — examiners want to see you apply the special case rule: R = 0 + ΣM ≠ 0 → couple. Use those exact terms.

Model Answer

Since ΣFx = 0 and ΣFy = 0, the resultant force R = √(0² + 0²) = 0. The net translational effect is zero — the system has no tendency to translate the body. However, ΣMo = 80 N·m ≠ 0, so the system still has a net rotational tendency. A system with R = 0 but ΣM ≠ 0 reduces to a couple. Therefore, this force system is equivalent to a pure couple of 80 N·m counterclockwise. This couple moment is a free vector — it is the same about every point in the plane.

Question Type

short_answer

Answer Structure

• Line 1: Compute R = 0 from ΣFx = ΣFy = 0; no translational effect [1 mark]
• Line 2: Since ΣMo ≠ 0 with R = 0, system reduces to a couple of 80 N·m CCW; state free-vector property [1 mark]

Scoring Breakdown

Common Mark Deductions

• Saying the system is 'in equilibrium' — incorrect, it is not; the couple produces rotation
• Not identifying it specifically as a couple
• Omitting the free-vector property of the resulting couple

Key Phrases To Include

• R = 0
• no translational effect
• reduces to a couple
• 80 N·m CCW
• free vector

Marks

3

Topic

Resultant of Parallel Force Systems

Difficulty

medium

Template Id

T15

Examiner Tip

Always declare your reference point before summing moments. Examiners check that you know the location result is relative to a stated point — '2.5 m' without a reference is incomplete.

Model Answer

Given: W1 = 10 kN at x1 = 0 m, W2 = 25 kN at x2 = 2 m, W3 = 15 kN at x3 = 5 m. (Sign: downward forces are positive.) Step 1 — Resultant magnitude: R = ΣF = 10 + 25 + 15 = 50 kN (downward) Step 2 — Moment sum about A (x = 0): ΣMA = 10(0) + 25(2) + 15(5) = 0 + 50 + 75 = 125 kN·m Step 3 — Location of resultant from A: x̄ = ΣMA / R = 125 / 50 = 2.5 m from A ∴ The resultant is 50 kN downward, acting at 2.5 m from the left end A.

Question Type

numerical

Answer Structure

• Step 1: R = ΣF = 50 kN downward [1 mark]
• Step 2: ΣMA = 125 kN·m — all three moment terms shown [1 mark]
• Step 3: x̄ = 125/50 = 2.5 m from A [1 mark]

Scoring Breakdown

Common Mark Deductions

• Including the moment of the 10 kN force at x = 0 as non-zero — it is zero (moment arm = 0)
• Using a different reference point without restating it — always declare 'Taking moments about A'
• Omitting 'downward' for direction of R

Key Phrases To Include

• R = ΣF = 50 kN
• ΣMA
• x̄ = ΣMA / R
• 2.5 m from A