--- title: "Equilibrium and Newton's First Law" description: "For a particle, zero resultant force means constant velocity, including rest. Balance force components in two perpendicular directions to find unknown forces." canonical: https://lightmysky.com/learn/mathematics/equilibrium-and-newtons-first-law-mt_e6RFRQkRPo source: https://lightmysky.com/learn/mathematics/equilibrium-and-newtons-first-law-mt_e6RFRQkRPo.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`. # Equilibrium and Newton's First Law For a particle, zero resultant force means constant velocity, including rest. Balance force components in two perpendicular directions to find unknown forces. Subject: Mathematics · Area: Mechanics · Ages 17 to 18 Page: https://lightmysky.com/learn/mathematics/equilibrium-and-newtons-first-law-mt_e6RFRQkRPo ## Ready when they can - Show three given forces are in equilibrium by resolving in two directions - Find the tension in each of two strings holding a hanging mass - Name every force on a book resting on a table and say what exerts it ## Lesson: Balancing force components You have resolved forces into perpendicular components. Now use them to test equilibrium. Work in a frame that is not accelerating or rotating. For a particle, equilibrium means zero resultant external force. Its velocity stays constant: it remains at rest or moves at constant speed in a straight line. Take east and north as positive. Add signed horizontal components and signed vertical components separately. Both totals must be zero. *(drawing: East is positive. Each arrow starts at zero and shows one component: +9 N or -9 N. Their sum is 0 N. Check the vertical total separately.)* A book resting on a level table, with no other applied forces, has weight downward from Earth and a normal force upward from the table. Normal means perpendicular to the surface. If a hand also presses down, the normal force balances the weight plus that push. Weight and normal both act on the book. They are not an action-reaction pair: such a pair acts on two different bodies. The book's push on the table acts on the table. Weight is mg downward, where m is mass and g is gravitational field strength. A taut string pulls along its length toward its support. For a mass at rest on one vertical string, tension T = mg. For two vertical strings, T_L + T_R = mg. Each tension equals mg/2 only if equal sharing is stated. Without that condition, force balance determines their sum but not each tension. *(drawing: An 8 kg mass has weight 80 N when g = 10 N/kg. Two vertical strings sharing equally pull 40 N each: 40 + 40 - 80 = 0.)* If the existing horizontal and vertical totals are R_x and R_y, a balancing force has components -R_x and -R_y. Its magnitude is sqrt(R_x² + R_y²), because the components are perpendicular. These are two components of one extra force. Check both signed totals after adding it. To test three given forces, add all three horizontal components and all three vertical components. For taut strings sloping up left and up right, let a and b be their angles above the horizontal. Let T_L and T_R be their tensions and W the weight. Horizontal balance gives T_L cos(a) = T_R cos(b). Vertical balance gives T_L sin(a) + T_R sin(b) = W. Use the horizontal equation to express one tension in terms of the other, then substitute into the vertical equation. Equal angles give equal tensions. **Example.** An 8 kg particle rests on two strings, each at 30° above the horizontal on opposite sides. Use g = 10 N/kg, so W = 80 N. Horizontal balance gives T_L = T_R = T. Since sin30° = 1/2, vertical balance gives T/2 + T/2 = 80. Each tension is 80 N, but each upward component is only 40 N. **Example.** Keep W = 80 N, but use a left angle of 30° and a right angle of 60°. Use sin30° = cos60° = 1/2 and cos30° = sin60° = √3/2. Horizontally, T_L√3/2 = T_R/2, so T_R = √3 T_L. Substitute vertically: T_L/2 + 3T_L/2 = 80. Thus T_L = 40 N and T_R = 40√3 N. Their upward components are 20 N and 60 N; their horizontal components cancel. Being motionless for one instant does not prove equilibrium. At the highest point of a vertical throw, a ball's velocity is briefly zero. Ignoring air resistance, its weight still acts downward with no balancing force. Its velocity therefore changes. Equilibrium requires zero resultant force, not just a moment with zero velocity. **Recap.** For a particle, zero resultant force means constant velocity, including rest. List forces on the chosen body, then balance signed components in both directions. *The lesson, practice questions, worked solutions and flashcards were written for this dataset.* ## Practice 14 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) ## Optional connections - [Proving Geometry Facts with Vectors](https://lightmysky.com/learn/mathematics/proving-geometry-facts-with-vectors-mt_A00MA7Ecgf) ## Opens up - [Friction and the Coefficient of Friction](https://lightmysky.com/learn/mathematics/friction-and-the-coefficient-of-friction-mt_EwFUd9N24W) - [Moments and the Equilibrium of a Rigid Body](https://lightmysky.com/learn/mathematics/moments-and-the-equilibrium-of-a-rigid-body-mt_FX5egGL4LI)