Frames of reference and key terms
A frame of reference is the point of view from which we measure motion.
- Inertial frame: not accelerating (the ground, roughly). Newton's laws work directly.
- Non-inertial frame: accelerating (a braking bus, a turning car). Objects seem pushed by fictitious forces that have no source, such as being thrown forward when a bus brakes.
Key words: net force ΣF (the sum of all forces), normal force N (surface push, at right angles), tension T (pull in a rope), weight Fg = mg, coefficient of friction μ.
Newton's second law: ΣF = ma. Force in newtons (N), mass in kg, acceleration in m/s².
Free-body diagrams and Newton's second law
A free-body diagram (FBD) shows one object as a dot or box with every force on it as an arrow.
- Choose the object.
- Draw weight (down), normal force (away from surface), applied forces, tension, friction (against motion).
- Choose axes; for slopes use along and across the slope.
- Write ΣFx = max and ΣFy = may.
On a flat floor with no vertical motion: N = mg. If you push down at an angle, N becomes bigger than mg; if you pull up at an angle, N becomes smaller.
Static and kinetic friction
- Static friction stops a still object from starting to slide. It grows to match the push, up to a maximum: fs ≤ μsN.
- Kinetic friction acts while sliding: fk = μkN. Usually μk < μs, so it is easier to keep something sliding than to start it.
Advantages: walking, tyre grip, brakes, holding nails and knots. Disadvantages: wear, heat and wasted energy in engines and machines; we reduce it with oil, ball bearings and smooth surfaces.
Inclines and connected objects
Inclined plane
For angle θ: parallel part of weight = mg sinθ (down the slope), perpendicular part = mg cosθ. So N = mg cosθ and friction = μN. Without friction a = g sinθ.
Connected objects
Objects joined by a light rope move with the same acceleration and the rope has one tension T. Method: treat the whole system to find a, then one object to find T.
Block m₁ on a frictionless table pulled by hanging m₂: a = m₂g ÷ (m₁ + m₂), T = m₁a.
Example (step 5): m₁ = 3 kg, m₂ = 2 kg → a = 2 × 9.8 ÷ 5 = 3.9 m/s², T = 3 × 3.9 ≈ 11.8 N.
Uniform circular motion
An object moving at constant speed in a circle is still accelerating, because its direction keeps changing.
Derivation (short)
In a small time Δt the velocity turns by angle Δθ. The change in velocity has size Δv = vΔθ and points to the centre. Since Δθ = vΔt ÷ r, a = Δv ÷ Δt = v²/r.
- Centripetal acceleration: ac = v²/r = 4π²r/T²
- Speed: v = 2πr/T (T = period)
- Centripetal force: Fc = mv²/r, always toward the centre. It is not a new force; it is the net force supplied by tension, friction, gravity or a normal force.
Orbits: gravity supplies Fc, so GMm/r² = mv²/r gives v = √(GM/r).
Devices, society and the environment
Many devices apply linear and circular dynamics: seat belts and airbags (lengthen stopping time to cut force), elevators, cranes and pulleys, centrifuges in hospitals and dairies, washing-machine spin cycles, banked race tracks and satellites.
Impact: seat belts and ABS save many lives; satellites give GPS and weather forecasts but leave space debris; faster vehicles need more fuel and safer roads.
Try it
Put a coin on a book and tilt it slowly. Note the angle where it starts to slide: tanθ = μs. Then swing a bottle cap on a string in a circle and feel the pull grow when you spin faster. Compare with the sliders in step 6.
Key formulas and definitions
- ΣF = m a
- F_g = m g ; on a flat surface N = m g
- f_s ≤ μ_s N ; f_k = μ_k N
- Incline: parallel = m g sinθ, perpendicular = m g cosθ
- Pulley system: a = m₂ g ÷ (m₁ + m₂)
- a_c = v²/r = 4π² r / T² ; F_c = m v²/r ; v = 2π r / T
Worked examples
1. A 10 kg box is pushed with 50 N on a floor with μₖ = 0.30. Find its acceleration.
N = mg = 98 N. fₖ = 0.30 × 98 = 29.4 N. ΣF = 50 − 29.4 = 20.6 N. a = 20.6 ÷ 10 = 2.06 m/s².
2. A 20 kg crate sits on a floor with μₛ = 0.5. What is the smallest horizontal push that starts it moving?
fₛ max = μₛ mg = 0.5 × 20 × 9.8 = 98 N. A push just over 98 N starts it.
3. A 4 kg block slides down a frictionless 30° slope. Find its acceleration and the normal force.
a = g sin30° = 9.8 × 0.5 = 4.9 m/s². N = mg cos30° = 4 × 9.8 × 0.866 ≈ 33.9 N.
4. Same 30° slope but μₖ = 0.2. Find the acceleration.
a = g(sinθ − μ cosθ) = 9.8 × (0.5 − 0.2 × 0.866) = 9.8 × 0.327 ≈ 3.2 m/s².
5. A 3 kg block on a frictionless table is tied over a pulley to a hanging 2 kg block. Find a and T.
a = 2 × 9.8 ÷ (3 + 2) = 3.92 m/s². T = 3 × 3.92 ≈ 11.8 N. Check with the hanging block: 2 × 9.8 − 11.8 = 7.8 = 2 × 3.92 ✓.
6. A 1200 kg car turns on a flat curve of radius 50 m at 15 m/s. What friction force is needed, and what is the minimum μₛ?
F_c = mv²/r = 1200 × 225 ÷ 50 = 5400 N. μₛ ≥ F_c ÷ mg = 5400 ÷ 11 760 ≈ 0.46.
7. A satellite orbits at radius 7.0 × 10⁶ m from Earth's centre with period 5.8 × 10³ s. Find its speed and centripetal acceleration.
v = 2πr/T = 2π × 7.0×10⁶ ÷ 5.8×10³ ≈ 7.6 × 10³ m/s. a = v²/r ≈ (7.6×10³)² ÷ 7.0×10⁶ ≈ 8.2 m/s².
Common mistakes
- Adding "centripetal force" as an extra arrow on a free-body diagram. It is the net inward force from real forces like tension or friction.
- Always writing N = mg. On a slope N = mg cosθ, and angled pushes or pulls change N.
- Using μₖ before the object moves. While it is still, use static friction (up to μₛN).
- Thinking a body in uniform circular motion has zero acceleration because its speed is constant. Its direction changes, so it accelerates toward the centre.