4003-02. Statics & Dynamics

⚙️ Statics & Dynamics — Mechanical Engineering Systems

LO2a covers the mechanical foundations of engineering: units and SI prefixes, scalars and vectors, 2D force systems, free-body diagrams, equilibrium, beam reactions under combined loading, and dynamics including Newton's laws, SUVAT, D'Alembert's principle, and energy conservation.

SI units & prefixesScalars & vectors 2D force systemsFBD & space diagrams Beam reactionsD'AlembertSUVAT

3Topics

8Examples

4Calculators

12Practice Qs

📜 This file covers the full range of LO2a static and dynamic fundamentals — from units and scalar/vector distinction through to beam support reactions, Newton's laws, and D'Alembert's principle applied to free-body diagrams.

⚙️Forces, Vectors & EquilibriumUnits, FBD, beams, centroids

🚀Dynamics & D'AlembertNewton, SUVAT, energy

⚖D'Alembert Deep DiveProof, method, all scenarios

🏢Industrial Case StudiesNewton, D'Alembert, energy in practice

🔌Applications4 mechanics calculators

🎮Game12 randomised questions

📝Practice12 graded questions

⚙️

Forces, Vectors, Free-Body Diagrams & Beam Reactions

Statics analyses structures at rest. Before drawing a free-body diagram or resolving forces, you must understand units, distinguish scalars from vectors, and be able to resolve 2D force systems. Support reactions for beams are then found by equilibrium (ΣF = 0, ΣM = 0).

SI unitsScalar vs vectorResolutionFBDΣF = 0ΣM = 0UDLCentroid

📜 Key rules — statics and force systems

SI prefixes: k (10³), M (10⁶), G (10⁹), m (10⁻³), µ (10⁻⁶), n (10⁻⁹). Always state units. Force: N, moment: N·m, pressure: Pa = N/m².

Scalar: magnitude only (mass, speed, energy). Vector: magnitude + direction (force, velocity, acceleration, displacement). Vectors add by components, not arithmetic.

Resolving 2D forces: F𝑥 = F cosθ, F𝑦 = F sinθ. Resultant: R = √(F𝑥² + F𝑦²), direction: θ = tan⁻¹(F𝑦/F𝑥).

Equilibrium: ΣF𝑥 = 0, ΣF𝑦 = 0, ΣM = 0. Moment M = F × d. Couple = F × d (independent of pivot).

UDL: total force = w × L acting at midpoint. Take moments about a support to eliminate its unknown — this is the key beam reaction strategy.

Centroid of a uniform shape: geometric centre. Composite bodies: ¯x = Σ(A𝑛¯x𝑛) ÷ ΣA𝑛. Centre of mass = centroid when density is uniform.

📚 Key vocabulary

Scalar: a quantity fully described by magnitude alone — examples: mass (kg), speed (m/s), temperature (K), energy (J). No direction.
Vector: a quantity with both magnitude and direction — examples: force (N), velocity (m/s), displacement (m), acceleration (m/s²). Represented by an arrow.
Free-body diagram (FBD): a sketch of a single body isolated from its surroundings, showing every external force and moment acting on it. The starting point for any equilibrium analysis.
Space diagram: a scale drawing showing the geometry of a structure — positions of supports, loads and members — before simplifying to an FBD.
Couple: two equal, opposite, parallel forces separated by distance d. Moment of couple = F × d. Acts as a pure moment — same value about any point.
Centroid: the geometric centre of a shape — the point where the shape would balance. For uniform density, centroid = centre of mass. Composite areas: weighted average of individual centroids.

👁️ Visual aid — space diagram (left) and free-body diagram (right): every force and reaction shown explicitly

📊 Common SI units and prefixes used in statics

Quantity	SI Unit	Symbol	Common prefixes
Force	Newton	N = kg·m/s²	kN (10³), MN (10⁶)
Moment / Torque	Newton-metre	N·m	kN·m
Pressure / Stress	Pascal	Pa = N/m²	kPa, MPa, GPa
Length	Metre	m	mm, cm, km
Mass	Kilogram	kg	g, Mg (tonne = 10³ kg)
Energy / Work	Joule	J = N·m	kJ, MJ
Power	Watt	W = J/s	kW, MW

✍️Worked example 1 — beam reactions: concentrated load + UDL step through it

Free-body diagram — all forces and reactions shown

QuestionBeam AB, span 6 m, simply supported. Point load 24 kN at 2 m from A. UDL 4 kN/m over full span. Draw the FBD and find R𝐴 and R𝐵.

FBD

Show: W=24kN↓ at 2m; UDL=4kN/m over 6m; R𝐴↑ at A; R𝐵↑ at B. Space diagram shows geometry; FBD adds all forces.

UDL totalW𝑢𝑑𝑙 = 4 × 6 = 24 kN acting at 3 m from A (midpoint of span)

Moments ⌀AR𝐵 × 6 = 24 × 2 + 24 × 3 = 48 + 72 = 120 kN·m

R𝐵R𝐵 = 120 ÷ 6 = 20 kN

ΣF𝑦 = 0

Vertical equilibrium: total upward force must equal total downward force, so R𝐴 + R𝐵 = W + W𝑢𝑑𝑙.   R𝐴 + 20 = 24 + 24 = 48 kN   →   R𝐴 = 28 kN

AnswerR𝐴 = 28 kN R𝐵 = 20 kN ✓

Step 1 of 7

✍️Worked example 2 — resolving a 2D force system

Free-body diagram — forces at the bracket, resultant shown dashed (parallelogram method)

QuestionForces on a bracket: F₁ = 40 kN at 0° and F₂ = 30 kN at 90°. Find the resultant magnitude and direction.

ComponentsΣF𝑥 = 40cos0° + 30cos90° = 40 + 0 = 40 kN

ΣF𝑦 = 40sin0° + 30sin90° = 0 + 30 = 30 kN

ResultantR = √(40² + 30²) = √2500 = 50 kN

Directionθ = tan⁻¹(30/40) = tan⁻¹(0.75) = 36.87° above horizontal

AnswerR = 50 kN at 36.87°

✍️Worked example 3 — moment, torque and couple

QuestionA couple: two 500 N forces separated by 0.8 m. A wrench: 60 N at 0.25 m. An angled force: 80 N at 35° acts at 0.6 m from a pivot. Find all three moments.

CoupleM = F × d = 500 × 0.8 = 400 N·m (same about any point)

Wrench torqueT = F × r = 60 × 0.25 = 15 N·m

Angled forceM = F sinθ × d = 80 × sin35° × 0.6 = 80 × 0.5736 × 0.6 = 27.53 N·m

AnswerCouple = 400 N·m, Wrench = 15 N·m, Angled = 27.5 N·m

✍️Worked example 4 — centroid of a composite area

QuestionAn L-section consists of a 100×20 mm horizontal flange and a 20×80 mm vertical web below it. Find the centroid from the bottom.

WebA₂ = 20×80 = 1600 mm²

The web sits on the bottom, so its own centroid is at half its own height: ¯y₂ = 80 ÷ 2 = 40 mm from bottom

FlangeA₁ = 100×20 = 2000 mm²

The flange sits on top of the 80 mm web, so its own centroid is: (height it sits on) + (half its own thickness) = 80 + (20 ÷ 2) = 80 + 10 = 90 mm from bottom

Composite¯y = (2000×90 + 1600×40)/(2000+1600) = (180000+64000)/3600 = 244000/3600 = 67.8 mm

AnswerCentroid = 67.8 mm from the bottom

🔧 Try it yourself — interactive calculators

Now you can try these calculators with your own values — enter numbers and watch the results update live.

⚙️Beam reaction calculator — point load + UDLUpdates as you type

Span L (m)

Point load W (kN)

Distance from A (m)

UDL w (kN/m)

Enter beam parameters — reactions R𝐴 and R𝐵 update automatically.

📊SI prefix quick converterUpdates as you type

Value

From

Enter a value — converts automatically.

📏Force resolver — try your own valuesUpdates as you type

Force F (N)

Angle θ (°)

Enter magnitude and angle — components update automatically.

⌖Composite centroid calculator — two areasUpdates as you type

Area 1 (mm²)

¯y₁ (mm)

Area 2 (mm²)

¯y₂ (mm)

Enter both areas and centroid positions — updates automatically.

🚀

Dynamics, Newton's Laws & D'Alembert's Principle

Dynamics analyses bodies in motion. Newton's Second Law (F = ma), the SUVAT equations (constant acceleration), D'Alembert's principle (treats dynamics as statics by adding the inertia force −ma to the FBD), and energy conservation are the core tools at HN Level 4.

Newton I/II/III1D Kinematics1D KineticsSUVATD'Alembert −maKE & PEConservation

Newton's Three Laws of Motion

One-Dimensional Particle Kinematics — SUVAT

👁️ SUVAT applies only when acceleration a is constant. Establish + direction first.

One-Dimensional Particle Kinetics — Force, Mass, Acceleration

👁️ Kinetics links kinematics to forces. Draw FBD first, resolve along motion axis, apply ΣF = ma (or D'Alembert: ΣF − ma = 0).

📜 Key rules — dynamics and D'Alembert

Newton's Second Law: F𝑟𝑒𝑠 = ma. Net force = mass × acceleration. Direction matters — establish a positive direction first.

SUVAT (constant acceleration only): v = u+at, s = ut+½at², v² = u²+2as, s = (u+v)t/2.

D'Alembert: add fictitious inertia force −ma opposing acceleration to the FBD. Apply ΣF = 0 as if static. Gives identical answer to Newton but uses familiar static method.

Energy: KE = ½mv², PE = mgh. Conservation (no friction): KE₁ + PE₁ = KE₂ + PE₂.

Inclined plane: driving force along slope = W sinθ. Normal reaction = W cosθ. Friction force = µN = µW cosθ.

📚 Key vocabulary

Kinematics: describes motion — position, velocity, acceleration — without reference to forces. SUVAT equations apply when acceleration is constant.
Kinetics: relates forces to motion via Newton's Second Law F = ma. Kinematics + kinetics together allow complete analysis of a dynamic system.
D'Alembert's principle: a dynamics problem is converted to a statics problem by adding an inertia force −ma to the free-body diagram, in the direction opposing acceleration. Then ΣF = 0 is applied.
Inertia force: the fictitious force −ma introduced in D'Alembert's method. It always acts opposite to the acceleration direction. Not a real force — a mathematical device.
Conservation of energy: total mechanical energy KE + PE is constant when no non-conservative forces act (no friction, no drag). A ball dropped from height h arrives with v = √(2gh).
Work-energy theorem: W_net = ΔKE. Net work done on a body equals the change in its kinetic energy. Includes work done against friction (negative work).

👁️ Visual aid — D'Alembert: inertia force −ma added to FBD opposes motion; system treated as static with ΣF = 0

✍️Worked example 5 — D'Alembert's principle applied to a free-body diagram step through it

Full FBD — T, f, W and N are the real forces; −ma (dashed, purple) is the inertia force added by D'Alembert, always opposing the acceleration

QuestionA 200 kg crate is pulled by cable tension T = 1200 N along a floor. Friction = 400 N. Use D'Alembert's principle, draw the FBD with all forces including the inertia force, and find acceleration.

Step 1: FBDDraw body isolated. Forces: T = 1200 N →, f = 400 N ←, W = mg ↓, N ↑, inertia = −ma ← (opposes motion)

Step 2: D'AlembertApply ΣF = 0 (horizontal): T − f − ma = 0

Step 3: Substitute1200 − 400 − 200a = 0 → 200a = 800

Step 4: Solvea = 800 ÷ 200 = 4 m/s²

Check (Newton)ΣF = ma → 1200 − 400 = 200 × 4 = 800 N ✓

Step 1 of 6

✍️Worked example 6 — SUVAT: stopping distance

Motion diagram — constant deceleration from u to rest over distance s

QuestionA vehicle decelerates at 6 m/s² from 30 m/s to rest. Find stopping distance using v² = u² + 2as.

Knownv = 0, u = 30 m/s, a = −6 m/s²

Equation0 = 900 + 2×(−6)×s → 12s = 900 → s = 75 m

AnswerStopping distance = 75 m

💡 Why is it mg sinθ down the slope but mg cosθ into the slope?

Weight always points straight down — gravity doesn't care that the surface is tilted. To analyse motion on the incline, we split that vertical weight into two components measured relative to the slope itself: one along the slope surface, one perpendicular to it.

The geometry trick: the incline surface is tilted by θ from the horizontal ground. That means the direction perpendicular to the slope is also tilted by θ from the true vertical. So the angle between the weight vector (straight down) and the slope's perpendicular is θ — the same θ as the incline angle. That's why it reappears in the force triangle.

Because the perpendicular component sits at angle θ from the weight vector, it equals mg cosθ (adjacent side). The component 90° away from it — along the slope — equals mg sinθ (opposite side).

Physical meaning: mg sinθ is the part of gravity that actually drives sliding down the slope (unopposed, so on a frictionless incline it causes a = g sinθ). mg cosθ is the part pressing the object into the surface — it's exactly cancelled by the normal reaction, so N = mg cosθ, and it plays no role in the motion.

Quick check: at θ=0° (flat ground), sin0°=0 and cos0°=1 — all weight is normal force, nothing slides. At θ=90° (a vertical wall), sin90°=1 and cos90°=0 — it's just free fall, no normal force at all. Both match what we'd expect.

✍️Worked example 7 — D'Alembert on an inclined plane

Full FBD — frictionless incline, inertia force (dashed, purple) always opposes acceleration (down-slope)

QuestionA 50 kg block slides down a frictionless 30° incline. Use D'Alembert's principle with a clear FBD to find acceleration.

FBD (along incline)Forces: W sin30° down-slope; N perpendicular; inertia force ma up-slope (opposes motion).

Why sinθ?

Weight W always points straight down. Splitting it into slope-axes: the angle between W and the slope's perpendicular is also θ (same tilt as the incline itself) → component along the slope = W sinθ, component into the slope = W cosθ = N.

D'Alembert (ΣF𝑠𝑙𝑜𝑝𝑒 = 0)W sin30° − ma = 0 → 50×9.81×0.5 = 50a

Solve245.25 = 50a → a = 4.905 m/s²

Answera = 4.905 m/s² = g sin30° ✓

✍️Worked example 8 — energy conservation: projectile to max height

Vertical launch — kinetic energy at ground converts fully to potential energy at max height

QuestionA 5 kg projectile is launched vertically at 20 m/s. Find (a) max height by energy conservation, (b) time to reach max height by SUVAT.

EnergyKE₁ = PE₂ → ½mv² = mgh → h = v²/(2g) = 400/19.62 = 20.39 m

SUVATv = u + at → 0 = 20 − 9.81t → t = 20/9.81 = 2.04 s

Answerh = 20.39 m, t = 2.04 s

🔧 Try it yourself — interactive calculators

Now you can try these calculators with your own values — enter numbers and watch the results update live.

🚀Dynamics calculator — Newton, SUVAT & energySelect what to find

Find

Net force F (N)

Mass m (kg)

Select what to find, then enter values.

🔌

Engineering Applications

Four interactive mechanics calculators.

⚙️ Beam reactions — combined loading

Find support reactions for a simply supported beam with a point load at any position and a full-span UDL.

R𝐵×L = W×a + wL×(L/2) R𝐴 = W + wL − R𝐵

🧮 Calculator

Span L (m)

Point load W (kN)

Distance from A (m)

UDL w (kN/m)

Enter beam parameters.

📈 2D force resultant

Find the resultant of two forces at given angles to the horizontal.

R = √((ΣF𝑥)²+(ΣF𝑦)²) θ = tan⁻¹(ΣF𝑦/ΣF𝑥)

🧮 Calculator

F₁ (N or kN)

θ₁ (°)

F₂ (same unit)

θ₂ (°)

Enter forces and angles from horizontal.

🚀 D'Alembert — horizontal motion with friction

Find acceleration using D'Alembert's principle. Enter applied force and friction; the inertia force −ma is added automatically.

F𝐟 − f − ma = 0 → a = (F𝐟 − f)/m

🧮 Calculator

Mass m (kg)

Applied F (N)

Friction f (N)

Enter mass, applied force and friction force.

🔋 Energy method — height to velocity

Find velocity from a height drop or launch speed using conservation of mechanical energy (no friction).

v = √(2gΔh + v₀²)

🧮 Calculator

Height Δh (m)

Initial vel v₀ (m/s)

Enter height drop and initial velocity.

🎮

Mechanics Challenge

12 randomised questions — units, vectors, beams, Newton and D'Alembert.

⚙️ Quick-fire round

Question 1 / 12Score: 0Streak: 0 🔥

Press Start to begin!

📝

Practice Questions

13 questions — type your answer, check, then reveal full working.

⚙️ Draw an FBD and a space diagram for every question involving forces. State all equations used. g = 9.81 m/s².

Q1 · Units & SI1 mark

Convert 45 kN·m to N·m. State the SI unit of pressure and its equivalent in base units.

45 kN·m = 45 × 10³ = 45 000 N·m
Pressure unit: Pa = N/m² = kg/(m·s²)

Q2 · Scalar vs vector1 mark

Classify each as scalar or vector: (a) 500 N vertical load, (b) 80 kg mass, (c) 30 m/s velocity, (d) 200 J energy, (e) 15 m displacement north.

Vectors (magnitude + direction): (a) force, (c) velocity, (e) displacement
Scalars (magnitude only): (b) mass, (d) energy

Q3 · Force resolution2 marks

A force of 60 N acts at 40° above the horizontal. Find its horizontal and vertical components.

F𝑥 = 60 cos40° = 60 × 0.766 = 45.96 N
F𝑦 = 60 sin40° = 60 × 0.643 = 38.57 N

Q4 · Resultant force2 marks

Two forces act at a point: F₁ = 60 N at 0°, F₂ = 80 N at 90°. Find the resultant and its direction.

R = √(60²+80²) = √10000 = 100 N
θ = tan⁻¹(80/60) = 53.13° above horizontal

Q5 · Moment & couple2 marks

A couple consists of two 400 N forces separated by 0.5 m. Calculate the moment of the couple. Does the moment change if the pivot point is moved? Justify your answer.

M = F × d = 400 × 0.5 = 200 N·m
No — a couple's moment is independent of the pivot point. It is a pure moment (no resultant force).

Q6 · Beam reactions3 marks

Beam AB, span 8 m, simply supported. Point load 40 kN at 3 m from A. UDL 5 kN/m over full span. Draw the FBD and find R𝐴 and R𝐵.

UDL total = 5×8 = 40 kN at 4 m from A
Moments ⌀A: R𝐵×8 = 40×3 + 40×4 = 120+160 = 280
R𝐵 = 280/8 = 35 kN
R𝐴 = 40+40−35 = 45 kN

Q7 · Beam reactions3 marks

Beam AB, span 5 m. Point load 15 kN at 2 m from A. UDL 8 kN/m over full span. Find R𝐵.

UDL = 8×5 = 40 kN at 2.5 m from A
Moments ⌀A: R𝐵×5 = 15×2 + 40×2.5 = 30+100 = 130
R𝐵 = 26 kN R𝐴 = 15+40−26 = 29 kN

Q8 · Centroid2 marks

A T-section: flange 120×20 mm at top, web 20×100 mm below. Find the centroid from the bottom.

Flange: A₁=120×20=2400mm², ¯y₁=100+10=110mm
Web: A₂=20×100=2000mm², ¯y₂=50mm
¯y = (2400×110+2000×50)/(2400+2000) = (264000+100000)/4400 = 364000/4400 = 82.7 mm
(Accept 79–83 mm depending on exact section dimensions assumed)

Q9 · SUVAT2 marks

A machine part starts from rest and accelerates at 3 m/s². Find (a) velocity after 4 s, (b) distance covered.

v = u+at = 0+3×4 = 12 m/s
s = ut+½at² = 0+½×3×16 = 24 m

Q10 · D'Alembert3 marks

A 150 kg block is pulled by cable T = 900 N. Friction = 300 N. Using D'Alembert's principle, sketch the FBD including the inertia force and find acceleration.

FBD: T=900N→, f=300N←, −ma← (inertia)
D'Alembert: 900−300−150a = 0
150a = 600 → a = 4 m/s²

Q11 · D'Alembert — incline3 marks

An 80 kg crate slides down a frictionless 25° ramp. Using D'Alembert's principle with a clear FBD, find acceleration. (sin 25° = 0.4226)

FBD along slope: W sin25° down, −ma up-slope (inertia)
D'Alembert: 80×9.81×0.4226 − 80a = 0
331.7 = 80a → a = 4.146 m/s²

Q12 · Combined4 marks

Beam AB 10 m: point load 50 kN at 4 m from A, UDL 6 kN/m full span. Find reactions. Also: a 300 kg block is accelerated by 1500 N force with 300 N friction. Find acceleration using D'Alembert. State the inertia force value.

UDL = 6×10 = 60 kN at 5 m
R𝐵×10 = 50×4+60×5 = 200+300 = 500 → R𝐵 = 50 kN
R𝐴 = 50+60−50 = 60 kN
Block: 1500−300−300a=0 → a = 4 m/s²
Inertia force = ma = 300×4 = 1200 N (acting opposite to motion)

Q13 · D'Alembert — horizontal3 marks

A 200 kg crate is pulled by cable tension T = 1200 N along a floor. Friction = 400 N. Using D'Alembert's principle, draw the FBD with all forces including the inertia force, and find acceleration.

FBD: T = 1200 N →, f = 400 N ←, −ma ← (inertia force, opposes direction of acceleration)
D'Alembert (horizontal): T − f − ma = 0
1200 − 400 − 200a = 0 → 800 = 200a
a = 4 m/s²
Inertia force = ma = 200×4 = 800 N (acting opposite to the direction of acceleration)
Newton check: ΣF = ma → 1200−400 = 200×4 = 800 N ✓

⚖

D'Alembert's Principle — Complete Guide

D'Alembert's principle converts a dynamics problem into a statics problem by introducing an inertia force −ma onto the free-body diagram. Every moving-body problem that Newton solves with ΣF = ma can be solved with D'Alembert using ΣF = 0 — the same answer, the same FBD, but using the equilibrium approach engineers are most familiar with.

Inertia force −maFBD methodHorizontalVerticalInclined planeConnected bodiesRotating system

📜 D'Alembert — the essential rules

The core idea: add a fictitious inertia force −ma to the real forces on the FBD. Apply ΣF = 0 (static equilibrium). The answer is identical to Newton's 2nd Law.

Direction of −ma: ALWAYS opposite to the direction of acceleration. If the body accelerates right, −ma points left. If it accelerates up, −ma points down.

Sign convention: establish a positive direction first. Write all forces with correct signs. The inertia force magnitude is ma and it always opposes positive acceleration.

On an incline: resolve along the slope. The inertia force ma acts up the slope (opposing downward acceleration). Normal forces and weight components perpendicular to slope cancel in the usual way.

Connected bodies: apply D'Alembert to EACH body separately. The tension T is the same in both FBDs. Add the equations to eliminate T if needed, or solve simultaneously.

Why use it? Engineers trained in statics find ΣF = 0 more intuitive than ΣF = ma. D'Alembert lets you use the full statics toolkit — moments, equilibrium, FBDs — on dynamic problems.

📚 D'Alembert vocabulary

Inertia force: the fictitious force −ma added to the FBD in D'Alembert's method. It has magnitude ma and always acts opposite to acceleration. Not a real force — it is a mathematical device to convert ΣF = ma into ΣF = 0.
Dynamic equilibrium: the state described by D'Alembert's equation: real forces + inertia force = 0. The body is not truly in equilibrium — it is accelerating — but the equation has the same form as static equilibrium.
Reference frame: D'Alembert's principle applies in an inertial (non-accelerating) reference frame. The inertia force arises because we are analysing a body that is accelerating relative to an inertial frame.
Newton's 2nd Law (comparison): Newton: ΣF = ma (net real force = ma, forces on one side, inertia on other). D'Alembert: ΣF − ma = 0 (all terms on same side, treated as equilibrium). Mathematically identical.
Atwood machine: two unequal masses connected by a string over a frictionless pulley. D'Alembert applied to each mass separately gives two equations. Adding them eliminates tension T and gives acceleration a = (m₁−m₂)g/(m₁+m₂).
Normal force on incline: on a slope at angle θ, the normal force N = mg cosθ. Why cos and not sin? Weight always acts straight down; resolved relative to the tilted slope, the angle between the weight vector and the slope's perpendicular is also θ (same tilt as the incline) — so the into-surface component is mg cosθ and the along-slope component is mg sinθ. This is unchanged whether the body is static or accelerating along the slope (D'Alembert only adds a force along the slope, not perpendicular to it).

👁️ Visual — Newton's 2nd Law (left) vs D'Alembert (right): same FBD, same answer, different organisation

👁️ D'Alembert on an inclined plane — all force arrows start at the block centre, labels placed well clear.

✍️Worked example A — Horizontal motion: applied force + friction step through it

Full FBD — four real forces (F, f, W, N) plus the D'Alembert inertia force −ma opposing acceleration

QuestionA 120 kg pallet is pushed across a warehouse floor by a force of 480 N. Friction force = 120 N. Using D'Alembert's principle with a fully labelled FBD, find the acceleration.

Step 1: FBD

Draw 120 kg body. Label all real forces: F = 480 N →, f = 120 N ←, W = 1177.2 N ↓, N = 1177.2 N ↑. Positive direction: right (direction of F).

Step 2: InertiaAdd inertia force: −ma ← (opposing positive/rightward acceleration). Magnitude = 120a, direction = left.

Step 3: ΣF = 0Apply D'Alembert (horizontal): F − f − ma = 0 → 480 − 120 − 120a = 0

Step 4: Solve360 = 120a → a = 3 m/s²

Step 5: VerifyNewton check: ΣF = ma → 480−120 = 120×3 = 360 N ✓

Answera = 3 m/s² (rightward) ✓

Step 1 of 7

✍️Worked example B — Vertical lift: cable tension during acceleration step through it

FBD — cable tension T, weight W; inertia force −ma flips direction with each acceleration phase

QuestionA 600 kg lift cage is accelerated upward at 2 m/s² by a cable. Using D'Alembert's principle with a clear FBD, find the cable tension T. Then find T at constant speed and when decelerating at 1.5 m/s² while ascending.

Step 1: FBD (phase 1)

Body: 600 kg cage. Real forces: T ↑ (unknown), W = mg = 600×9.81 = 5886 N ↓. Positive direction: upward (direction of acceleration).

Step 2: Inertia forceAcceleration is upward (+), so inertia force = −ma acts downward. Magnitude = 600×2 = 1200 N ↓.

Step 3: D'Alembert (phase 1)ΣF = 0: T − W − ma = 0 → T = W + ma = 5886 + 1200 = 7086 N

Step 4: Constant speeda = 0, inertia force = 0. T − W = 0 → T = W = 5886 N. Cable only supports weight.

Step 5: Decelerating up

Decelerating means acceleration is downward (−1.5 m/s²). Inertia force = −ma acts upward = +900 N ↑. D'Alembert: T − W + ma = 0 → T = 5886 − 900 = 4986 N

AnswerAccelerating up: T = 7086 N Constant: T = 5886 N Decelerating: T = 4986 N ✓

Step 1 of 7

✍️Worked example C — Inclined plane with friction

Full FBD — friction and the inertia force both act up-slope, opposing the crate's downward acceleration

QuestionA 90 kg crate accelerates down a 20° ramp. Coefficient of kinetic friction μ = 0.18. Using D'Alembert's principle, draw the FBD and find acceleration.

FBD (along slope)

Positive direction: down the slope (direction of motion). Forces: mg sin20° (down-slope), friction f = μN (up-slope, opposing motion), inertia −ma (up-slope, opposing acceleration).

Why sinθ/cosθ?

W acts straight down. Split relative to the slope: the angle between W and the slope's perpendicular equals θ (same tilt as the incline) → along-slope component = W sinθ (drives sliding), perpendicular component = W cosθ = N (balanced by the surface, feeds into f = μN).

Normal forceN = mg cos20° = 90×9.81×cos20° = 829.7 N

Frictionf = μN = 0.18×829.7 = 149.4 N (up-slope)

D'Alembert (ΣF=0 along slope)mg sin20° − f − ma = 0 → 90×9.81×sin20° − 149.4 − 90a = 0

Solve301.8 − 149.4 = 90a → 152.4 = 90a → a = 1.693 m/s²

Answera = 1.69 m/s² down the slope

✍️Worked example D — Atwood machine (two connected masses)

Frictionless pulley — same tension T on both sides; heavier mass m₁ accelerates down, m₂ accelerates up

QuestionTwo masses m₁ = 5 kg and m₂ = 3 kg hang on either side of a frictionless pulley. Using D'Alembert's principle applied to each body, find acceleration a and tension T.

FBD for m₁ (heavier, accelerates down)

Real forces: m₁g ↓, T ↑. Positive direction: down. Inertia force = −m₁a ↑ (opposes downward acceleration). D'Alembert: m₁g − T − m₁a = 0

FBD for m₂ (accelerates up)

Real forces: m₂g ↓, T ↑. Positive direction: up. Inertia force = −m₂a ↓ (opposes upward acceleration). D'Alembert: T − m₂g − m₂a = 0

Add equationsm₁g − m₂g = (m₁ + m₂)a → a = (m₁−m₂)g/(m₁+m₂) = (5−3)×9.81/(5+3) = 19.62/8 = 2.45 m/s²

Tension TFrom m₂ equation: T = m₂(g+a) = 3×(9.81+2.45) = 3×12.26 = 36.79 N

CheckFrom m₁ equation: T = m₁(g−a) = 5×(9.81−2.45) = 5×7.36 = 36.80 N ✓

Answera = 2.45 m/s², T = 36.8 N

✍️Worked example E — Pushing a body UP a rough incline

■ mg sin15° = 177.6 N ■ f = μN = 132.6 N ■ −ma = m×a = 175 N

Full FBD — F drives the body up-slope; weight component, friction and inertia all act down-slope, opposing it

QuestionA 70 kg trolley is pushed UP a 15° ramp at 2.5 m/s² by force F. μ = 0.20. Draw the FBD with all D'Alembert forces and find F.

FBD (along slope, +ve up)Up-slope: F. Down-slope: mg sin15°, f = μmg cos15°, inertia −ma (all opposing upward acceleration).

Why sinθ/cosθ?

Weight W acts straight down regardless of the ramp's tilt. Resolved relative to the slope, the angle between W and the slope's perpendicular equals θ (the incline angle itself) → along-slope component = W sinθ, perpendicular (into-surface) component = W cosθ = N. That's why W's "down the slope" share always carries sinθ and its "into the slope" share always carries cosθ.

NormalN = mg cos15° = 70×9.81×0.9659 = 663.0 N f = 0.20×663.0 = 132.6 N

D'Alembert (ΣF=0 up-slope)F − mg sin15° − f − ma = 0

SolveF = 70×9.81×sin15° + 132.6 + 70×2.5 = 177.6 + 132.6 + 175 = 485.2 N

AnswerF = 485 N (up the ramp to produce 2.5 m/s² acceleration against gravity and friction)

📊 D'Alembert FBD — what to add in each scenario

Scenario	Inertia force direction	D'Alembert equation
Horizontal, accelerates right	Left (←)	F − f − ma = 0
Vertical, accelerates up	Down (↓)	T − mg − ma = 0
Vertical, decelerates while ascending	Up (↑)	T − mg + ma = 0
Slides down smooth incline	Up the slope	mg sinθ − ma = 0 → a = g sinθ
Slides down rough incline	Up the slope	mg sinθ − μmg cosθ − ma = 0
Pushed up rough incline	Down the slope	F − mg sinθ − μmg cosθ − ma = 0
Atwood machine (heavier mass)	Up (opposes downward a)	m₁g − T − m₁a = 0
Atwood machine (lighter mass)	Down (opposes upward a)	T − m₂g − m₂a = 0

🔧 Try it yourself — interactive calculators

Now you can try these calculators with your own values — enter numbers and watch the results update live.

⚖D'Alembert calculator — select your scenarioFull working shown

Scenario

Mass m (kg)

Select a scenario and enter values — full D'Alembert working appears here.

🏢

Industrial Case Studies

Real engineering scenarios where Newton's laws, 1D kinematics, 1D kinetics, D'Alembert's principle, and conservation of energy are applied to solve practical problems. Each case study follows the full professional analysis workflow: define system → FBD → equations → solution → interpretation.

Newton's lawsKinematicsKineticsD'AlembertEnergy conservationIndustrial application

🏢 Professional analysis workflow

Step 1 — Define the system: identify the body of interest, establish a positive direction, note all known and unknown quantities with units.

Step 2 — Draw the FBD: isolate the body, show every external force (applied, weight, normal, friction, tension). Never skip this step — it is the foundation of every solution.

Step 3 — Select method: Newton (ΣF = ma) or D'Alembert (ΣF − ma = 0). Use energy method when displacement matters more than time. Use SUVAT for uniform acceleration kinematics.

Step 4 — Solve & verify: substitute values, solve for unknowns, check units, verify equilibrium (forces balance, energy conserved). State answer with correct units and direction.

🏢 Case Study 1 — Conveyor belt system: kinetics + kinematics (manufacturing)

👁️ A product is placed on a moving conveyor — the belt applies friction to accelerate the package until they reach the same speed.

🏢Case Study 1 — Full solution

ScenarioA 4 kg package is placed at rest on a conveyor belt moving at 2.5 m/s. The coefficient of kinetic friction between package and belt is 0.35. Find: (a) acceleration of the package, (b) time to reach belt speed, (c) distance slid on belt, (d) energy dissipated as heat.

FBD

Forces on package: weight W = mg = 39.24 N↓, Normal N = 39.24 N↑, Friction f = µN = 0.35×39.24 = 13.73 N→ (in direction of belt motion)

Newton (kinetics)ΣF = ma → 13.73 = 4×a → a = 3.43 m/s²

Kinematicsv = u + at → 2.5 = 0 + 3.43×t → t = 0.729 s

Distance (package)s𝑝 = ut + ½at² = 0 + ½×3.43×0.729² = 0.911 m

Distance (belt)s𝐵 = v𝐵×t = 2.5×0.729 = 1.823 m. Slip = 1.823−0.911 = 0.912 m

Heat energyQ = f × slip = 13.73 × 0.912 = 12.5 J (energy dissipated by friction = heat generated at belt-package interface)

Check: EnergyKE gained by package = ½×4×2.5² = 12.5 J. Work by friction = 13.73×0.911 = 12.5 J ✓ (energy balance confirmed)

🏢 Case Study 2 — Vehicle emergency braking: D'Alembert + SUVAT (automotive)

👁️ D'Alembert converts the braking dynamics problem into a static equilibrium — the inertia force −ma acts forward, friction force acts rearward.

🏢Case Study 2 — Full solution

ScenarioA 1200 kg car travelling at 90 km/h (25 m/s) brakes to a halt. All four wheels lock (µ = 0.75, wet road). Using D'Alembert's principle, find: (a) deceleration, (b) stopping distance, (c) time to stop, (d) KE dissipated as heat.

FBD + D'AlembertBraking friction f = µmg = 0.75×1200×9.81 = 8829 N. Inertia force = −ma (forward). D'Alembert: f − ma = 0

Decelerationa = f/m = 8829/1200 = 7.36 m/s² (= µg = 0.75×9.81 ✓)

Stopping distancev² = u² + 2as → 0 = 625 − 2×7.36×s → s = 625/14.71 = 42.5 m

Stopping timev = u + at → 0 = 25 − 7.36×t → t = 3.40 s

Energy dissipatedKE = ½mv² = ½×1200×25² = 375 000 J = 375 kJ (all converted to heat in brakes and tyres)

CheckW𝑓 = f×s = 8829×42.5 = 375 000 J ✓ (work done by friction = KE lost)

🏢 Case Study 3 — Mine hoist (cage on incline): Newton II + energy (mining/lifting)

👁️ Mine cage raised vertically by cable — accelerates from rest, then holds constant speed.

FBD — cable tension T and weight W; Newton II (ΣF = ma) applied during the acceleration phase

🏢Case Study 3 — Full solution

ScenarioA 2500 kg mine cage is raised vertically by a cable from rest to 4 m/s over 8 s (uniform acceleration), then travels at constant speed. Find: (a) cable tension during acceleration, (b) power required at constant speed, (c) PE gained in 8 s phase. g = 9.81 m/s².

Kinematicsa = (v−u)/t = (4−0)/8 = 0.5 m/s² | s = ut+½at² = 0+½×0.5×64 = 16 m

FBD (Newton II)Cable tension T (upward), Weight W = mg = 24 525 N (downward). Net force = T − W = ma

Cable tensionT = ma + W = 2500×0.5 + 24525 = 1250 + 24525 = 25 775 N ≈ 25.8 kN

Power (const. speed)At const. speed a = 0, so T = W = 24 525 N. Power P = T×v = 24525×4 = 98 100 W = 98.1 kW

PE gainedPE = mgh = 2500×9.81×16 = 392 400 J = 392.4 kJ

Engineering noteDuring acceleration the motor supplies energy for both PE gain and KE gain. KE = ½×2500×4² = 20 000 J. Total energy = 392 400 + 20 000 = 412 400 J over 8 s.

🏢 Case Study 4 — Drop weight impact testing: energy conservation (materials testing)

👁️ A hammer is released from rest and falls freely before impacting the test specimen.

Free fall (energy conservation) followed by rapid deceleration on impact over 3 mm

🏢Case Study 4 — Full solution

ScenarioA Charpy/Izod-style drop weight: a 5 kg hammer is released from rest at 1.8 m above a test specimen. Find: (a) velocity at impact using energy conservation, (b) verify using SUVAT, (c) impact force if the hammer decelerates over 3 mm.

Energy conservationPE₁ = KE₂ → mgh = ½mv² → v = √(2gh) = √(2×9.81×1.8) = √35.316 = 5.943 m/s

Verify (SUVAT)v² = u²+2as = 0+2×9.81×1.8 = 35.316 → v = 5.943 m/s ✓

Impact decelerationv² = u²+2as → 0 = 35.316−2×a×0.003 → a = 35.316/0.006 = 5886 m/s²

Impact forceF = ma = 5×5886 = 29 430 N = 29.4 kN (+ weight 49N, negligible at this scale)

Engineering noteThis is why impact forces are so much larger than static loads — a 5 kg hammer weighing 49 N produces a 29.4 kN impact force. The energy method (PE = KE) is the most direct approach for this class of problem. The work-energy theorem then gives the impact force from the deceleration distance.

🏢 Case Study 5 — D'Alembert on a crane load (construction)

👁️ Crane cable accelerates a concrete slab upward — D'Alembert gives the tension for each phase.

FBD — cable tension T, weight W, inertia force −ma; direction of −ma flips with each phase (lift / hold / lower)

🏢Case Study 5 — Full solution

ScenarioA crane lifts a 3000 kg concrete slab. The cable accelerates the load upward at 1.2 m/s². Using D'Alembert's principle, find the cable tension. Then find the tension at constant speed and during lowering at 0.8 m/s² deceleration.

FBDCable tension T (up), Weight W = 3000×9.81 = 29 430 N (down). Positive direction: upward.

D'Alembert (accelerating up)

Inertia force = −ma = −3000×1.2 = −3600 N (opposes upward motion, acts downward). ΣF=0: T − W − ma = 0 → T = W + ma = 29430 + 3600 = 33 030 N

Constant speeda = 0, inertia force = 0. T = W = 29 430 N

Decelerating (lowering)

Lowering and decelerating means net acceleration is upward (+). D'Alembert: T − W + ma = 0 → T = W − ma = 29430 − 3000×0.8 = 29430 − 2400 = 27 030 N

Engineering noteCable tension varies with acceleration. Always size cable and attachments for the acceleration phase — the maximum tension (33 kN here) governs the design. D'Alembert gives a clear static FBD for each phase.

📊 Method selection — when to use each approach

Situation	Best method	Key equations
Forces on a moving body, find a	Newton 2nd Law	ΣF = ma
Same, but want familiar static layout	D'Alembert's principle	ΣF − ma = 0 (add −ma to FBD)
Uniform acceleration, find v, s, t	1D Kinematics (SUVAT)	v=u+at, v²=u²+2as, s=ut+½at²
Height drop or launch, find speed	Conservation of energy	mgh = ½mv² → v = √(2gh)
Impact force from known KE	Work-energy theorem	F = ma, a from v²=u²+2as over stop dist.
Cable tension during lift/lower	Newton II or D'Alembert	T = m(g ± a) depending on direction