--- title: "Newton's Second Law in Two Dimensions" description: "Apply F = ma along two perpendicular axes. For motion on a fixed straight slope, resolve forces along and perpendicular to the surface. While contact remains, perpendicular acceleration is zero." 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: 2026-11-12 --- > **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 Apply F = ma along two perpendicular axes. For motion on a fixed straight slope, resolve forces along and perpendicular to the surface. While contact remains, perpendicular acceleration is zero. 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 You have used F = ma for resultant force and resolved forces into components. Apply F = ma along each perpendicular axis. For a fixed straight slope, use axes along and perpendicular to it. Let theta be its angle above horizontal, with 0 < theta < 90 degrees. Weight mg acts vertically down. Its components are mg sin(theta) downhill and mg cos(theta) into the slope. The normal reaction R points outward, perpendicular to the slope. Smooth means no friction. If only weight and R act, mg sin(theta) = ma downhill, so a = g sin(theta). Mass cancels. With uphill positive, write -mg sin(theta) = ma. Acceleration is downhill even when velocity is uphill; the block then slows down. For a given g, increasing theta from 0 toward 90 degrees increases sin(theta) and downhill acceleration. Without extra forces, this acceleration is less than g. *(drawing: Only weight and R act, with g = 10 m/s squared. Bars show downhill acceleration in m/s squared, rounded to two decimal places.)* While a block follows a fixed straight slope, perpendicular acceleration is zero. If only weight and R act perpendicular to it, R = mg cos(theta). Thus R < mg for 0 < theta < 90 degrees, but R = mg on level ground. To find mass, rearrange: m = R / (g cos(theta)). With a parallel uphill pull P, choose uphill positive: P - mg sin(theta) = ma. For m = 6 kg, theta = 30 degrees, g = 10 m/s squared and P = 42 N, the resultant is 42 - 30 = 12 N uphill. With an extra parallel downhill force D, use downhill positive: D + mg sin(theta) = ma. Extra forces can make acceleration exceed g. On a smooth level floor, a rope force F at angle alpha above horizontal has components F cos(alpha) forward and F sin(alpha) upward. If only rope, weight and reaction act, contact gives a = F cos(alpha)/m and R = mg - F sin(alpha). A 16 N pull at 60 degrees gives 8 N forward. At fixed F and m, increasing alpha from 0 toward 90 degrees reduces forward acceleration. *(drawing: A 12 N rope pulls a 3 kg body on a smooth level floor. With g = 10, contact remains. Bars show forward acceleration in m/s squared.)* For a rope above a slope, measure alpha above the uphill direction. Then a = (F cos(alpha) - mg sin(theta))/m and R = mg cos(theta) - F sin(alpha), with uphill positive. An extra force Q perpendicular into a slope instead gives R = mg cos(theta) + Q. On level ground, the lifting threshold is F sin(alpha) = mg, where R = 0. A negative required R means contact cannot persist: the surface can push, but cannot pull the body toward itself. **Recap.** Apply F = ma to the resultant force along each axis. On a fixed straight surface, balance perpendicular forces while contact remains, and include every applied force. *Lesson, questions, worked solutions and flashcards written for this dataset.* ## 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)