--- title: "Friction and the Coefficient of Friction" description: "Use the static friction limit to decide whether a body can remain at rest, and use a separate kinetic coefficient when surfaces slide." canonical: https://lightmysky.com/learn/mathematics/friction-and-the-coefficient-of-friction-mt_EwFUd9N24W source: https://lightmysky.com/learn/mathematics/friction-and-the-coefficient-of-friction-mt_EwFUd9N24W.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`. # Friction and the Coefficient of Friction Use the static friction limit to decide whether a body can remain at rest, and use a separate kinetic coefficient when surfaces slide. Subject: Mathematics · Area: Mechanics · Ages 17 to 18 Page: https://lightmysky.com/learn/mathematics/friction-and-the-coefficient-of-friction-mt_EwFUd9N24W ## Ready when they can - State the static friction inequality and explain limiting equilibrium - Use force balance and the static or kinetic friction model to decide whether a body stays at rest or changes velocity on a horizontal surface - Explain what the coefficient of friction is a property of, and what the model leaves out ## Lesson: How hard surfaces can grip Static friction acts along a contact surface when the surfaces are not sliding relative to each other. It opposes the tendency to slip. Let f_s be its magnitude, R the normal reaction and mu_s the static coefficient. The rule is 0 <= f_s <= mu_s R. The upper value is the static limit. Equality means limiting equilibrium: the body can still be at rest, but no larger static friction is available. **Example.** An initially resting 4 kg block is on a fixed horizontal floor, with g = 10 N/kg. Only weight and reaction act vertically, so R = mg = 4 times 10 = 40 N. With static coefficient mu_s = 0.3, multiply by R: the static limit is mu_s R = 0.3 times 40 = 12 N. With just a 5 N horizontal push, friction is 5 N against it and the block stays at rest. A push above 12 N cannot be balanced by static friction. *(drawing: The bars compare magnitudes, not directions. Friction opposes the push and remains below its limit.)* To test an initially resting body, add all applied forces parallel to the surface, subtracting opposite directions. The friction needed for balance is equal and opposite to their resultant. Rest is possible if that magnitude is at or below mu_s R. With no tangential force to balance, f_s = 0. A pull of 18 N right and 7 N left needs 11 N friction left. At the limit, rest is still possible. The normal reaction is perpendicular to the surface. Resolve forces in that direction. If acceleration is zero, they balance. On a fixed horizontal floor, R = mg when weight and reaction are the only vertical forces. An extra downward force D gives R = mg + D if these are the only vertical forces. A 5 kg block with g = 10 N/kg and an extra 10 N downward push has R = 60 N. The coefficient has no units. It describes the pair of surfaces and their condition; changing either surface or adding lubricant can change it. To compare two contacts at the same R, calculate each static limit, then subtract. For R = 90 N, coefficients 0.2 and 0.6 give limits of 18 N and 54 N. Their difference is 36 N. An observed rest state can give a lower bound for mu_s. First find the friction needed for balance. Since f_s <= mu_s R, divide by R to get mu_s >= f_s/R when R > 0. If balance needs 21 N and R = 70 N, mu_s must be at least 21/70 = 0.30. A larger coefficient also allows rest, so the observation does not give its exact value. During relative sliding, use the kinetic coefficient mu_k: f_k = mu_k R. It acts against relative sliding and is often below the static limit for the same R. For a block already sliding right on a fixed floor, a rightward push P gives resultant P - f_k if friction is the only other horizontal force. If P = f_k, acceleration is zero and velocity stays constant. For R = 70 N and mu_k = 0.2, a 14 N push right balances 14 N friction left. This model uses constant coefficients for the stated surface pair and condition. It does not track wear, surface deformation or changes with sliding speed. It does not explain the microscopic contact forces. A changed contact may need a different coefficient or a more detailed model. **Recap.** Static friction supplies the force needed for balance up to mu_s R. At the limit, rest is still possible. During sliding, use f_k = mu_k R. *Lesson, questions and flashcards 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) - [Equilibrium and Newton's First Law](https://lightmysky.com/learn/mathematics/equilibrium-and-newtons-first-law-mt_e6RFRQkRPo) - [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)