Dynamics · Chapter 1 of 10 · Beginner
Introduction to Dynamics
Statics asks what keeps a body still. Dynamics keeps the same free-body habit but adds one term, mass times acceleration, and with it the entire study of motion.
Readiness check
Dynamics builds straight on statics and calculus. Tick only what you can do closed-notes.
- Draw a free-body diagram and resolve forces into components.
- State that in statics the forces sum to zero.
- Differentiate and integrate basic functions of time.
- Distinguish a vector from a scalar.
- Work consistently in SI units.
The core idea
A body accelerates in proportion to the net force on it and in inverse proportion to its mass; statics is just the special case where that acceleration is zero.
ΣF = maW = mgF = Gm₁m₂/r²Newton's second law is the engine of the whole course. Mass m is the fixed measure of how much a body resists acceleration; weight W = mg is the force gravity exerts on that mass and changes with location. Keeping the two apart, and the units consistent, prevents most beginner errors.
The skills, taught in order
The first chapter sets the vocabulary and habits the rest of the course assumes. Five short skills cover the laws, the mass-weight distinction, units, gravitation, and the solution method.
1.1 Newton's three laws
First law: a body keeps its velocity unless a net force acts. Second law: ΣF = ma, the working equation. Third law: forces come in equal and opposite pairs. The second law is what you compute with; the first and third tell you when it applies and how bodies interact.
1.2 Mass versus weight
Mass (kilograms) measures inertia and never changes. Weight (newtons) is the gravitational force W = mg and depends on g. A 1 kg object has the same mass on the Moon but about one sixth the weight. In ΣF = ma the m is always mass, so convert any weight to mass first.
1.3 Units and dimensional homogeneity
Work in SI: metre, kilogram, second, with force in newtons (1 N = 1 kg·m/s²). Every equation must be dimensionally homogeneous, so checking units is a free error-detector. Standard gravity is g = 9.81 m/s².
| Quantity | SI unit | Note |
|---|---|---|
| Mass m | kilogram (kg) | invariant measure of inertia |
| Force, weight | newton (N = kg·m/s²) | W = mg, varies with g |
| Acceleration a | m/s² | g = 9.81 m/s² at sea level |
1.4 The law of gravitation
Any two masses attract with F = Gm₁m₂/r², where G = 6.673×10⁻¹¹ m³/(kg·s²). Applied to a body of mass m near Earth, this gives g = GMe/r², so g falls off with the square of the distance from Earth's centre. At everyday altitudes the change is small, which is why g is treated as constant.
1.5 The solution method
Every problem in this course follows four steps: model (draw the body, choose coordinates and a positive sense), relate (write the governing equation), solve (substitute numbers), and check (units, sign, and magnitude). Resisting the urge to compute first is the single most useful habit.
Engineering connection: every later chapter, and downstream controls, vehicle, and robotics work, rests on ΣF = ma and a clean separation of mass from weight.
Worked example 1: mass, weight, and acceleration
A 1200 kg car experiences a net forward force of 3000 N. Find its acceleration and its weight on Earth, then state what changes if the same car sits on the Moon (g = 1.62 m/s²).
- ProblemFind the acceleration and weight of the car in Figure 1, and state what the Moon changes.
- Given / findm = 1200 kg, ΣF = 3000 N, g = 9.81 m/s². Find a, W on Earth, and the Moon comparison.
- AssumptionsThe 3000 N is the net horizontal force; the car is treated as a particle; g constant.
- ModelApply ΣF = ma horizontally for acceleration and W = mg vertically for weight.
- Equationsa = ΣF/m W = mg
- Solvea = 3000/1200 = 2.5 m/s². W = 1200 × 9.81 = 11 772 N ≈ 11.8 kN. On the Moon the mass is still 1200 kg, but W = 1200 × 1.62 = 1944 N, about one sixth of the Earth weight.
- CheckUnits: N/kg = m/s², correct. The acceleration depends only on force and mass, so it would be the same on the Moon for the same 3000 N; only the weight changed. That is the mass-weight distinction in action.
- ConclusionAcceleration is governed by mass, not weight. Confusing the two, for instance dividing the 3000 N by the weight, would give an answer wrong by a factor of g.
Worked example 2: how gravity varies with altitude
Using the law of gravitation (G = 6.673×10⁻¹¹, Earth mass Me = 5.976×10²⁴ kg, mean radius R = 6371 km), find g at the surface and at the 400 km altitude of the Space Station. What does the result say about "weightless" astronauts?
- ProblemFind g at the surface and at 400 km altitude, and interpret the "weightlessness" of orbit.
- Given / findG = 6.673×10⁻¹¹, Me = 5.976×10²⁴ kg, R = 6.371×10⁶ m, h = 4.0×10⁵ m. Find gsurface and g at R+h.
- AssumptionsEarth is a uniform sphere; only Earth's gravity matters.
- ModelApply g = GMe/r², evaluating r at the surface and at the orbital radius.
- Equationsg = GMe/r²
- SolveAt the surface, g = (6.673×10⁻¹¹ × 5.976×10²⁴)/(6.371×10⁶)² = 9.83 m/s², matching the standard 9.81. At r = 6.771×10⁶ m, g = 8.70 m/s², about 88.5% of the surface value.
- Checkg decreased as 1/r², as it must, and only slightly because 400 km is small next to the 6371 km radius. The model recovers the textbook surface value, a good sign.
- ConclusionAstronauts in orbit still feel about 89% of surface gravity; they float because the station and everything in it are in continuous free fall together, not because gravity has vanished. The word "weightless" describes the sensation, not the physics.
Misconceptions and diagnostics
| Mistake | Symptom | Diagnostic question | Correction |
|---|---|---|---|
| Using weight as mass | Answer off by a factor of g | "Is my m in kilograms?" | In ΣF = ma the m is mass; divide any weight by g first. |
| Thinking mass changes with location | Mass recomputed on the Moon | "Did the inertia change, or the gravity?" | Mass is invariant; only weight W = mg changes with g. |
| Skipping the free body | Forces missed or double-counted | "Have I drawn every force on the body?" | Always draw the free-body diagram before writing ΣF = ma. |
| Mixing units | Dimensionally wrong result | "Do the units reduce correctly?" | Work in SI throughout; check that both sides reduce to the same units. |
Practice ladder
A net force of 50 N acts on a 4 kg block on a frictionless surface. Find its acceleration.
Show answer
a = ΣF/m = 50/4 = 12.5 m/s². The force and mass alone set the acceleration; weight does not enter a horizontal frictionless problem.
An object weighs 196 N on Earth. What is its mass, and what net horizontal force gives it an acceleration of 3 m/s²?
Show answer
m = W/g = 196/9.81 = 20 kg. Then ΣF = ma = 20 × 3 = 60 N. The weight is only used to recover the mass; the horizontal law uses mass directly.
At what altitude does g fall to 9.0 m/s²? Use g = GMe/r² with the surface value 9.83 m/s² at R = 6371 km.
Show answer
r = R√(gs/g) = 6371√(9.83/9.0) = 6371 × 1.045 = 6660 km, so the altitude is about 290 km. Even a few hundred kilometres barely dents g, confirming why it is treated as constant near the surface.
Pick a moving object around you (a lift, a bike, a drone). Estimate its mass, the net force acting, and the resulting acceleration, then state one assumption that most affects your answer.
What good work looks like
A free-body sketch, mass in kilograms, ΣF = ma applied with a positive direction stated, and a named assumption (such as neglecting drag or friction) with its likely effect.
Working with AI, and proving it yourself
Use AI as an examiner, not a solver
Portfolio task
Write a half-page note that takes one real object through model-relate-solve-check, computing both an acceleration and a weight, and explicitly separating mass from weight.
Retrieval and spaced review
Closed notes. Answer out loud, then reveal.
1. State Newton's three laws in one line each.
Inertia: velocity is constant without a net force. Second law: ΣF = ma. Action-reaction: forces occur in equal, opposite pairs.
2. What is the difference between mass and weight?
Mass (kg) is invariant inertia; weight (N) is the gravitational force W = mg and varies with g.
3. Give the SI unit of force in base units.
1 N = 1 kg·m/s².
4. How does g vary with distance from Earth's centre?
As 1/r², from g = GMe/r².
5. Name the four steps of the solution method.
Model, relate, solve, check.
Textbook mapping
| Item | Mapping |
|---|---|
| Primary source | Meriam and Kraige, Engineering Mechanics: Dynamics (7th ed), Chapter 1 (Introduction to Dynamics) |
| Cross-reference | Hibbeler, Dynamics, Ch. 12 intro · Beer and Johnston, Ch. 11 intro |
| Core topics | 1.1 Newton's laws · 1.2 Mass vs weight · 1.3 Units · 1.4 Gravitation · 1.5 Solution method |
| Engineering connection | The foundation for every later chapter and for controls, vehicle, and robotics modelling. |
| Read next | Chapter 2: Kinematics of Particles. |