Dynamics — Topic Summary

1 Forces Overview

A force is a push or pull that acts on an object. Forces are measured in Newtons (N) and are vector quantities — they have both magnitude and direction.

Contact vs Field Forces

Type	Description	Examples
Contact forces	Require physical touching between objects	Friction, normal force, tension, applied force
Field forces	Act at a distance — no touching required	Gravity, magnetic force, electric force

Net Force & Vector Addition

The net force (ΣF) is the vector sum of all forces acting on an object. Forces in the same direction add; forces in opposite directions subtract.

Net force (x-axis)

ΣFₓ = F₁ₓ + F₂ₓ + …

Net force (y-axis)

ΣFₔ = F₁ₔ + F₂ₔ + …

💡Forces at an angle must be resolved into x and y components before adding. Use sin and cos relative to the angle from the positive x-axis.

2 Newton's First Law — Inertia

📝Law of Inertia: An object at rest stays at rest, and an object in motion stays in motion at constant velocity, unless acted on by a net external force.

Inertia is the tendency of an object to resist changes in its state of motion. A more massive object has more inertia and is harder to start or stop moving.

If ΣF = 0 → acceleration = 0 (object moves at constant velocity or stays at rest)
A book sitting on a table is in equilibrium: gravity pulls down, normal force pushes up, ΣF = 0
A hockey puck sliding on ice with no friction would slide forever — no net force, constant velocity
Seatbelts exist because of inertia: your body continues forward when a car stops suddenly

Equilibrium condition

ΣF = 0 → a = 0

3 Newton's Second Law

Second Law

ΣF = ma (units: N = kg · m/s²)

x-component

ΣFₓ = maₓ

y-component

ΣFₔ = maₔ

A larger net force produces greater acceleration. A larger mass produces smaller acceleration for the same force. Acceleration is always in the same direction as the net force.

Free-Body Diagram (FBD) — Step by Step

Isolate the object — draw it as a simple dot or box.
Draw all forces as arrows from the centre — label each one (F_g, N, T, f_k, F_app).
Choose a positive direction — typically right = +x, up = +y. On an incline, choose along and perpendicular to the surface.
Write ΣF = ma for each axis separately, substituting values.
Solve for the unknown (usually acceleration or a force).

Worked Example

✏️

5 kg box, applied force F_app = 30 N, friction f_k = 10 N
FBD: F_app = 30 N right, f_k = 10 N left, F_g = 49 N down, N = 49 N up
x-axis: ΣFₓ = 30 − 10 = 20 N
ΣFₓ = maₓ → 20 = 5a → a = 4 m/s² to the right

4 Newton's Third Law

📝Action-Reaction: For every force that object A exerts on object B, object B exerts an equal and opposite force back on object A.

Third Law

F_AB = −F_BA (equal magnitude, opposite direction)

Action-reaction pairs always act on different objects — they never cancel each other
You push the floor down → the floor pushes you up (this is the normal force)
A rocket pushes exhaust gases backward → exhaust gases push rocket forward
You push a wall with 50 N → the wall pushes back on you with 50 N

⚠️Common confusion: Action-reaction pairs are NOT the same as equilibrium. A book on a table has F_g (gravity down) balanced by N (normal force up) — these are NOT a Newton's 3rd law pair. The pair to gravity on the book is the book pulling Earth upward.

5 Weight, Normal Force & Applied Force

Weight (gravity)

F_g = mg (g = 9.8 m/s² downward)

Normal force (flat)

N = mg (only on a horizontal surface with no vertical applied force)

Normal force (incline)

N = mg cosθ (perpendicular to the surface)

The normal force is always perpendicular to the surface — it is the surface pushing back on the object. It is NOT always equal to mg.

Apparent Weight in an Elevator

Situation	Net force equation	Apparent weight
Accelerating upward	N − mg = ma	N = m(g + a) > mg
Accelerating downward	mg − N = ma	N = m(g − a) < mg
Constant velocity	N − mg = 0	N = mg (feels normal)
Free fall	mg − N = mg (a = g)	N = 0 (weightless)

6 Tension

Tension is the pulling force transmitted through a rope, string, or cable. It always pulls — never pushes. For a massless rope, the tension is identical at every point along the rope.

Atwood Machine

Two masses m₁ and m₂ connected over a frictionless, massless pulley. Assume m₂ > m₁, so m₂ falls and m₁ rises.

Acceleration

a = (m₂ − m₁)g / (m₁ + m₂)

Tension (rising mass)

T = m₁(g + a)

Tension (falling mass)

T = m₂(g − a)

✏️

Example: m₁ = 3 kg, m₂ = 5 kg
a = (5 − 3)(9.8) / (3 + 5) = 19.6 / 8 = 2.45 m/s²
T = 3(9.8 + 2.45) = 3 × 12.25 = 36.75 N

7 Friction

Friction is caused by microscopic surface irregularities. It always opposes relative motion (or attempted motion) between surfaces.

Static friction (max)

f_s ≤ μ_sN (object not moving — friction matches applied force up to this limit)

Kinetic friction

f_k = μ_kN (object sliding — constant value)

Key Properties

μ_s > μ_k always — it takes more force to start sliding than to keep sliding
Kinetic friction direction always opposes the direction of motion
Friction coefficients (μ) are dimensionless — they have no units
Friction does NOT depend on speed or contact area (for simple models)

Surface pair	μ_s (approx.)	μ_k (approx.)
Rubber on dry concrete	0.70	0.60
Wood on wood	0.40	0.30
Steel on steel	0.15	0.10
Ice on ice	0.10	0.03

8 Inclined Planes

On a ramp at angle θ, gravity pulls straight down (mg) but the surface is tilted. Resolve mg into two components:

Perpendicular to surface

F_⊥ = mg cosθ (balanced by N)

Parallel to surface

F_∥ = mg sinθ (pulls object down the slope)

Normal force (incline)

N = mg cosθ

Kinetic friction on incline

f_k = μ_k mg cosθ

Acceleration (sliding down)

a = g(sinθ − μ_k cosθ)

✏️

Example: 10 kg box on 30° ramp, μ_k = 0.2
N = 10 × 9.8 × cos 30° = 84.9 N
F_∥ = 10 × 9.8 × sin 30° = 49 N
f_k = 0.2 × 84.9 = 17 N
Net F = 49 − 17 = 32 N → a = 3.2 m/s² down the slope

💡When θ = 0 (flat surface): N = mg cos 0° = mg cos 0 = mg. This confirms the flat surface formula is just a special case of the incline formula.

9 Pendulum Forces

A pendulum bob moves in a circular arc. At every point two forces act: tension T along the string toward the pivot, and gravity mg downward. The net centripetal force points toward the centre of the circular path.

At angle θ (general)

T − mg cosθ = mv²/r → T = mg cosθ + mv²/r

At rest / slow swing (v ≈ 0)

T ≈ mg cosθ

At the bottom (θ = 0)

T − mg = mv²/r → T = mg + mv²/r

💡At the bottom of the swing, tension must support the weight AND provide the centripetal force — so T > mg always at the bottom. At the top of a full circular loop, tension and gravity both point inward.

✏️

Example: 0.5 kg bob at bottom of swing, v = 2 m/s, r = 0.8 m
T = mg + mv²/r = (0.5)(9.8) + (0.5)(4)/(0.8)
T = 4.9 + 2.5 = 7.4 N

10 Common Mistakes to Avoid

Mistake	What to do instead
N = mg always	N = mg only on a horizontal surface with no vertical applied force. On an incline: N = mg cosθ.
Friction always points left	Friction opposes the direction of motion (or attempted motion). If an object moves left, friction points right.
Action-reaction pairs cancel	They act on different objects, so they never cancel. Only forces on the same object cancel if ΣF = 0.
Tension can push	Tension only pulls. A rope goes slack if you try to push with it.
Forgetting to resolve forces on inclines	Always resolve gravity into F_∥ = mg sinθ and F_⊥ = mg cosθ before writing ΣF = ma.
μ has units	μ is dimensionless — no units. It is just a ratio.
Using μ_s when object is moving	Once the object is sliding, use μ_k (kinetic). μ_s only applies when the object has not yet started moving.