🎯 Hook – why a plane doesn't fly where it points
A pilot points the nose of a small aircraft due north and holds 250 km h⁻¹. If a crosswind blows from the west, the plane's actual path over the ground is north-east of where it's pointing, and faster or slower than 250 km h⁻¹ depending on the wind. What actually happens is the vector sum of the plane's velocity through the air and the wind's velocity over the ground — neither vector alone tells the story.
➕ Adding vectors: triangle and parallelogram rules
Along one line: add with signs — 5 N right + 3 N left = \(+5+(-3)=2\) N right. Triangle (tip-to-tail) rule: draw the first vector, start the second at the tip of the first; the resultant runs from the tail of the first to the tip of the last — works for any number of vectors. Parallelogram rule: draw both vectors from the same point, complete the parallelogram; the diagonal from the common point is the resultant. The resultant R of two vectors a and b always satisfies \(|a-b| \le R \le a+b\).
📐 Resolving a vector into components
A vector F at angle θ to the x-axis has components — the one next to the angle uses cos, the one opposite the angle uses sin:
Adding several vectors by components: resolve every vector into x and y (with signs), sum each set separately (\(R_x=\Sigma F_x\), \(R_y=\Sigma F_y\)), then recombine: \(R=\sqrt{R_x^2+R_y^2}\), \(\theta=\tan^{-1}(R_y/R_x)\) — check the quadrant from the signs (the quadrant trap: calculator \(\tan^{-1}\) only returns −90° to +90°; add 180° if \(R_x\) is negative).
⚖️ Equilibrium and inclined planes
An object is in translational equilibrium when the resultant force is zero: \(\Sigma F_x=0\) and \(\Sigma F_y=0\). Graphically, the force vectors drawn tip-to-tail form a closed triangle or polygon.
For a mass m on a slope at angle θ, resolve weight mg parallel and perpendicular to the slope — down the slope: \(mg\sin\theta\); into the slope: \(mg\cos\theta\) (balanced by the normal reaction \(N=mg\cos\theta\) with no other perpendicular force).
Projectiles (preview of A.1): a launch velocity u at angle θ above horizontal has \(u_x=u\cos\theta\) (constant, ignoring air resistance) and \(u_y=u\sin\theta\) (changes at \(-g\)) — the two directions are independent because a horizontal force cannot change vertical motion.
🌊 Deeper notes — river crossing and sagging ropes
Heading straight across a river gives the shortest time (full boat speed contributes to the crossing), but the current carries the boat downstream. To reach the point directly opposite (shortest path), head upstream at angle \(\sin^{-1}(v_{\text{current}}/v_{\text{boat}})\) — only possible if the boat is faster than the current.
Tension in a supported load: a load hung from the middle of a rope sagging at a small angle θ below horizontal needs \(T=W/(2\sin\theta)\), which grows without limit as \(\theta\to0\) — you can never pull a loaded rope perfectly straight.
📊 Graph mastery — resultants and components vs angle
| Graph type | Axes (X, Y) | Shape | What to read |
|---|---|---|---|
| Resultant of two equal forces F vs angle between them | angle · R | \(R=2F\cos(\phi/2)\): falls from 2F at 0° to 0 at 180° | R at φ=90° gives \(F\sqrt2\) |
| Components vs angle | θ · component | \(F\cos\theta\) falling, \(F\sin\theta\) rising, crossing at 45° | Crossing point: components equal |
| Weight components vs slope angle | θ · component | \(mg\sin\theta\) rising, \(mg\cos\theta\) falling | Equal at θ = 45° |
✍️ Worked example 1 — perpendicular forces and three-vector sum
| Force | x / N | y / N |
|---|---|---|
| 10.0 N at 0° | +10.00 | 0.00 |
| 8.0 N at 90° | 0.00 | +8.00 |
| 6.0 N at 225° | −4.24 | −4.24 |
| Sum | +5.76 | +3.76 |
✍️ Worked example 2 — inclined plane and river crossing
Resultant speed \(=\sqrt{3.0^2+1.2^2}=\mathbf{3.2\ m\,s^{-1}}\) at \(\tan^{-1}(1.2/3.0)=22^\circ\) downstream of straight across.
🌍 Real-world: a hanging sign
A 40 N sign hangs from two cables, each at 30° above horizontal. Vertically: \(2T\sin30^\circ=40 \Rightarrow T=40\) N; the horizontal components cancel automatically by symmetry.
🚀 Going further — non-perpendicular pairs
For two vectors with angle φ between them (tail to tail), the cosine rule gives \(R^2=a^2+b^2+2ab\cos\phi\). At \(\phi=0\), \(R=a+b\); at 90°, Pythagoras; at 180°, \(R=|a-b|\). The component method avoids ever needing this rule.
Challenge: two 10 N forces act with 120° between them. What is R? \(R=\sqrt{10^2+10^2+2(10)(10)\cos120^\circ}=\sqrt{200-100}=10\ \text{N}\).
📋 Quick practice — answers in italics
(a) Find the resultant of 5.0 N north and 12 N east. 13 N at 23° east of north (tan⁻¹(5/12)).
(b) Resolve a 200 N force at 40° below horizontal into components. Horizontal = 200cos40° ≈ 153 N; vertical = 200sin40° ≈ 129 N.
(c) A ball is kicked at 18 m s⁻¹ at 35° above horizontal — find its velocity components. \(u_x=18\cos35°≈14.7\ \text{m s}^{-1}\); \(u_y=18\sin35°≈10.3\ \text{m s}^{-1}\).
(d) Three 4.0 N forces act at 0°, 120°, 240° — show the point is in equilibrium. Each pair of components cancels by symmetry: Σ cos(0°,120°,240°) = 0 and Σ sin(0°,120°,240°) = 0.
⚠️ Trap corner – common mistakes
| Misconception | Correct view |
|---|---|
| sin and cos can be swapped freely | If the angle is measured from the vertical, the vertical component uses cos — check which side of the angle each component is on |
| Adding magnitudes of non-parallel vectors (6 N + 8 N = 14 N always) | Only true when the vectors are parallel; otherwise resolve into components first |
| Normal reaction always equals mg | On a slope with no other perpendicular force, \(N=mg\cos\theta\), not mg |
| Calculator angle mode doesn't matter | Always set degrees/radians to match the angle given, or every component is wrong |
| tan⁻¹ on a calculator always gives the right quadrant | It only returns −90° to +90°; add 180° when \(R_x<0\), and always sketch to check |
| Δv = v_initial − v_final | Change in velocity is always final minus initial: \(\Delta v=v_{\text{final}}-v_{\text{initial}}\) |
Complete the statements:
✓ (b) Resultant of AB and AC: \(\sqrt{9.0^2+12.0^2}=\mathbf{15.0\ N}\) — the crane cable tension equals this, 15.0 N.
✓ (c) \(\tan^{-1}(12/9)=\mathbf{53^\circ}\) between the crane cable's line and cable AB.
✓ (b) \(N=mg\cos20^\circ \approx \mathbf{111\ N}\).
✓ (c) Along the slope: \(60 = 40.3+f \Rightarrow f \approx \mathbf{19.7\ N}\), acting down the slope (opposing the upward motion).
✓ (d) Constant speed means zero acceleration, so by Newton's first law the resultant force must be zero.
✓ (b) \(\sin\theta=0.90/1.5=0.60 \Rightarrow \theta=\mathbf{36.9^\circ}\) upstream of straight across.
✓ Resultant speed across \(=1.5\cos36.9^\circ=1.2\ \text{m s}^{-1}\); time \(=45/1.2=\mathbf{37.5\ s}\).
| String | Force/N | Angle/° |
|---|---|---|
| 1 | 3.0 | 0 |
| 2 | 4.0 | 90 |
| 3 | 4.9 | 235 |
✓ (b) \(\Sigma F_x=0.19\) N, \(\Sigma F_y=-0.01\) N; resultant \(\approx\mathbf{0.19\ N}\).
✓ (c)(i) \(\Delta F_x=\cos235°\times0.1\approx0.057\) N. (ii) \(\Delta F_x\approx4.9\times\sin235°\times(2\pi/180)\approx0.14\) N.
✓ (d) The 0.19 N resultant is small next to the ~3–5 N string forces and within the combined uncertainty (~0.2 N) — consistent with equilibrium.
✓ (e) Friction in the pulleys (reduces transmitted tension); a miscalibrated angle scale on the force table.
✓ For a projectile this splits one curved 2D problem into a simple constant-velocity horizontal motion and a simple constant-acceleration vertical motion, each solved with ordinary SUVAT equations.
✓ Only the weight needs resolving (into \(mg\sin\theta\) and \(mg\cos\theta\)), so there are fewer components to calculate and fewer chances for sign errors than resolving every force horizontally/vertically.
✓ Any angle between these extremes (0° and 180°) gives an intermediate resultant, so R is always bounded between these two values — a triangle-inequality argument.
✓ As θ → 0° (line pulled straight), \(\sin\theta\to0\) and T would need to become infinite — impossible — so the line must always sag at some small but non-zero angle.
✓ Shortest path: head upstream at \(\theta=\sin^{-1}(v_{\text{current}}/v_{\text{boat}})\) so the current's effect is exactly cancelled, landing directly opposite the start, but taking longer.
✓ This requires \(v_{\text{current}}/v_{\text{boat}}\le1\); if the current is faster than the boat, \(\sin\theta>1\) has no solution, so the boat can never fully cancel the drift — the shortest-path strategy is impossible.