--- title: "Newton's Second Law in Two Dimensions" description: "Resolve the forces along the direction of motion and perpendicular to it, then apply F = ma along that direction while the perpendicular components balance. This is what makes a slope problem solvable" canonical: https://lightmysky.com/learn/mathematics/newtons-second-law-in-two-dimensions-mt_yNUFF3lxlj source: https://lightmysky.com/learn/mathematics/newtons-second-law-in-two-dimensions-mt_yNUFF3lxlj.md retrieved: 2026-09-12 --- > **Agent view.** This is the Markdown twin of the page, for tools and assistants. > When to use this site, and the call that answers each job: https://lightmysky.com/agent-instructions.md > API description (OpenAPI 3.1): https://lightmysky.com/openapi.json · Authentication: https://lightmysky.com/auth.md > Pricing: https://lightmysky.com/pricing.md · Catalog: https://lightmysky.com/llms.txt · Full catalog: https://lightmysky.com/llms-full.txt > Every machine-readable file on this domain: https://lightmysky.com/.well-known/ai-catalog.json > Ask for Markdown with `Accept: text/markdown`, a `.md` address, or `?mode=agent`. # Newton's Second Law in Two Dimensions Resolve the forces along the direction of motion and perpendicular to it, then apply F = ma along that direction while the perpendicular components balance. This is what makes a slope problem solvable. Subject: Mathematics · Area: Mechanics · Ages 17 to 18 Page: https://lightmysky.com/learn/mathematics/newtons-second-law-in-two-dimensions-mt_yNUFF3lxlj ## Ready when they can - Find the acceleration of a block sliding down a smooth slope - Resolve perpendicular to the slope to find the normal reaction - Handle a pulling force applied at an angle to the direction of travel ## Lesson: Splitting weight on a slope Tilt your axes along and across the slope: motion happens along the surface. Weight mg splits into mg sin(theta) pulling down the slope and mg cos(theta) pressing into it. For 30 degrees, sin30 = 0.5, so the downslope pull is half the weight. **Example.** On a smooth slope nothing else pushes, so F = ma along the surface gives a = g sin(theta), the same for heavy and light blocks. At 30 degrees with g = 9.8, a = 9.8 times 0.5 = 4.9 m/s squared. Across the slope there is no motion, so the normal reaction balances mg cos(theta) and stays below mg. A rope at an angle resolves the same way: a 20 N pull at 60 degrees to travel gives 20 cos60 = 10 N forward, since cos60 = 0.5. Its upward piece carries some weight and lightens the normal reaction. Forward drives, upward unloads: two effects from one force. **Tip.** Steeper means faster: g sin(theta) grows toward g as the slope steepens, but stays below g for any genuine slope. Against a 30 N pull uphill, a 4 kg block feels 4 times 9.8 times 0.5 = 19.6 N downhill, leaving 30 - 19.6 = 10.4 N up the slope. **Recap.** Resolve weight along and across the slope, then run F = ma down the hill and balance across it. ## Practice 18 questions on this page, each with its working shown. ## Needs first - [Forces as Vectors: Resolving and Resultants](https://lightmysky.com/learn/mathematics/forces-as-vectors-resolving-and-resultants-mt_-qDRgd2g8v) - [Newton's Second Law: Force, Mass and Acceleration](https://lightmysky.com/learn/mathematics/newtons-second-law-force-mass-and-acceleration-mt_SKO85AcKdb) ## Opens up - [Motion on a Rough Inclined Plane](https://lightmysky.com/learn/mathematics/motion-on-a-rough-inclined-plane-mt_0rBy66HFNO) - [Connected Particles and Pulleys](https://lightmysky.com/learn/mathematics/connected-particles-and-pulleys-mt_pZ6zV0RFdC)