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Resolving Forces — Essential Mechanics
Resolving Forces — Essential Mechanics
Resolving forces sounds fancy, but it just means splitting one slanted force into two easy directions so your equations stay simple.
Do this well and ramps, rough surfaces, strings, and even projectiles become routine.
Right — let’s break it down the way an examiner would expect you to think it through.
🔙 Previous topic:
“Review Newton’s Laws before resolving individual forces.”
🧭 What “Resolving” Really Means
A force has both size and direction — it’s a vector. Resolving means replacing one angled force by two perpendicular components along chosen axes.
📏 Typical axes choices:
- Horizontal and vertical (flat ground, hanging objects)
- Along a plane and perpendicular to it (inclined surfaces)
⚙️ If a force F makes an angle θ to your x-axis:
- Component along x-axis: F cos θ
- Component along y-axis: F sin θ
🧠 Only use sine/cosine relative to the angle you’ve actually marked on your diagram — that’s where most sign mistakes come from.
❗ Exam insight: Mark schemes reward (1) a clear diagram, (2) correct components, (3) one equation per axis.
Be consistent and you’ll collect method marks even if the arithmetic slips.
⚙️ The Three Rules You’ll Reuse
1️⃣ Newton’s 2nd Law (per component):
Sum of forces along an axis = mass × acceleration along that axis.
2️⃣ Equilibrium:
If it’s at rest or constant speed, the sum of forces on each axis = 0.
3️⃣ Friction (rough surfaces):
At the limit, friction = μ × reaction (R). Direction: opposite intended or actual motion.
✅ If you write those three lines clearly, you’ll have 70 % of Mechanics 1 covered already.
📘 Worked Example 1 — Pulling a Crate on Smooth Ground
Mass 25 kg. Tension 120 N at 30° above horizontal. Find acceleration.
Horizontal: 120 cos 30° ≈ 103.9 N = m a → a ≈ 103.9 ÷ 25 = 4.16 m/s².
Vertical balance (if asked for normal reaction):
R + 120 sin 30° = mg → R ≈ (25 × 9.81) − 60 ≈ 185 N.
🧠 Tip: On flat ground, a vertical pull reduces the normal reaction; a downward push increases it.
📘 Worked Example 2 — Rough Incline with an Extra Pull
Mass 10 kg on a plane at 25°. Coefficient of friction μ = 0.3.
A force P acts up the plane at 15° above it. Find the least P to hold it (limiting equilibrium).
Perpendicular to plane:
R + P sin 15° = mg cos 25° → R = mg cos 25° − P sin 15°.
Along plane (taking “up” as positive):
P cos 15° − mg sin 25° − friction = 0.
Friction at limit: μR = 0.3 [mg cos 25° − P sin 15°].
Substitute and solve:
P (cos 15° + μ sin 15°) = mg (sin 25° + μ cos 25°).
With g = 9.81:
Right side ≈ 98.1 × (0.423 + 0.3×0.906) ≈ 68.2.
Left bracket ≈ 0.966 + 0.3×0.259 ≈ 1.044.
So P ≈ 68.2 ÷ 1.044 ≈ 65.4 N.
✅ Tip: On slopes, always choose axes along and perpendicular to the plane — it cuts sign errors in half.
📘 Worked Example 3 — Two Strings Meeting at a Ring (Equilibrium)
Weight 80 N supported by two light strings.
The left string makes 40° to horizontal, right string 65°.
Find tensions T₁ (left) and T₂ (right).
Horizontal: T₂ cos 65° = T₁ cos 40° → T₂ = T₁ (cos 40° / cos 65°).
Vertical: T₁ sin 40° + T₂ sin 65° = 80.
Using sin 40° = 0.643, cos 40° = 0.766, sin 65° = 0.906, cos 65° = 0.423:
T₂ ≈ 1.81 T₁.
Then 0.643T₁ + 0.906(1.81T₁) = 80 → 2.286T₁ = 80 → T₁ ≈ 35 N, T₂ ≈ 63 N.
🧠 Tip: Resolve horizontally and vertically rather than jumping to vector triangles — it’s clearer for marks and easier to correct mid-way.
🪜 Choosing Axes — Quick Rules
📏 Flat ground or hanging objects → horizontal/vertical.
📏 Any slope → along/perpendicular to the plane.
📏 Motion constrained by a string or track → take an axis along that direction so “ma” lives on just one line.
✅ Choosing sensible axes keeps signs consistent and your algebra half as long.
⚙️ Friction Done Cleanly
Direction → opposite motion (or intended motion).
Size at limit → μR.
R comes from the perpendicular equation.
❗ If R turns negative, your setup or direction guess is wrong — fix the sketch first, not the algebra.
🧠 I tell my classes every year: friction questions aren’t nasty; they just punish sloppy diagrams.
❗ Common Traps (and the Fixes)
- Using the wrong angle with sine/cosine → sketch a tiny right triangle on the force arrow.
- Assuming R = mg when there’s a vertical pull → check the vertical balance.
- Degrees vs radians → Mechanics defaults to degrees.
- Putting “ma” on the wrong axis → acceleration belongs along the motion.
- Friction the wrong way → if your answer comes out negative, the direction just flips.
✅ Good habit: Write “Check sign → OK if consistent” beside your final line. Examiners love that clarity.
📘 From Forces to Motion — The Pipeline
Resolve → write component equations → find net force along motion →
a = F_net ÷ m → use SUVAT for time, distance, and speed.
🧠 That chain is how multi-part exam questions hang together. Once you see it, Mechanics starts feeling predictable.
🪜 Quick Practice (Answers in Brackets)
- 60 N at 35° above horizontal. Components?
Fx = 60 cos 35° ≈ 49.1 N, Fy = 60 sin 35° ≈ 34.4 N. - 8 kg on a smooth 20° slope. Acceleration down the plane?
a = g sin 20° ≈ 3.35 m/s². - 5 kg on 30° rough slope, μ = 0.25, held by P.
P ≈ mg (sin 30° + μ cos 30°) ≈ 35.6 N. - 50 N on two strings at 30° and 50° to horizontal.
T₁ ≈ 32 N, T₂ ≈ 48 N. - 12 kg on rough horizontal ground, pulled by 90 N at 40° above horizontal, no motion.
R = mg − 90 sin 40° ≈ 59.8 N.
✅ Teacher tip: Check each answer’s direction and size — that’s often where careless slips live.
🧠 Exam Board Tips (AQA | Edexcel | OCR)
- Start with a neat, labelled diagram.
- Show the resolution step explicitly — it earns method marks.
- State modelling assumptions (smooth plane, light string, constant g).
- Keep three significant figures until the end.
- Always check whether equilibrium or motion is implied.
🧠 Examiners often use phrases like “unreliable if extrapolated” or “limiting equilibrium” — use those same words in your reasoning.
🚀 Next Steps
If this topic still feels patchy, our A Level Maths Revision Course walks through resolving forces one diagram at a time.
🚀 You’ll get guided examples, board-style problems, and live explanations of tricky slope questions for AQA, Edexcel, and OCR.
Right — practise one “crate on a slope” problem tonight. Once you can set it up confidently, Mechanics suddenly feels easy.
✅ Quick Recap Table
Concept | Formula / Idea | Key Reminder |
Component along axis | F cos θ or F sin θ | Use correct angle from sketch |
Newton’s 2nd Law | ΣF = m a | One equation per axis |
Equilibrium | ΣF = 0 | Object at rest or constant speed |
Friction | F = μR | Opposes motion; limit when about to move |
Axes choice | Along/perpendicular | Reduces sign errors |
Net force link | F_net = m a | Leads straight to SUVAT |
🧭 Next topic:
“Next, explore forces acting in two dimensions.”
About the Author
S. Mahandru is Head of Maths at Exam.tips and has more than 15 years of experience in simplifying difficult subjects such as pure maths, mechanics and statistics. He gives worked examples, clear explanations and strategies to make students succeed.
