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.
What a learner can do afterwards
- 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
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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.
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.
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.
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.
Readings, questions and flashcards written for this dataset. Linked videos: Khan Academy, Static and kinetic friction example, and Force of friction keeping the block stationary.
More to read
Separate the friction demand from its limit
A static coefficient tells you the largest friction force available at a contact. It does not tell you the actual friction force in every rest state.
Consider a block initially at rest on a fixed floor. Suppose its static limit is 26 N. If a horizontal cord pulls with 9 N and friction is the only other horizontal force, balance needs 9 N of friction in the opposite direction. A 20 N pull needs 20 N. With no pull and no other horizontal force, the friction is zero.
At a 26 N pull, rest is still possible in the ideal model. This boundary is limiting equilibrium. A larger pull cannot be balanced by static friction.
With several applied forces, combine their components parallel to the surface first. Compare the friction needed to cancel that resultant with mu_s R. All these decisions concern relative slipping at the contact.
Find the normal reaction before the limit
The normal reaction is a contact force perpendicular to the surface. On a horizontal floor, it points vertically upward while weight points vertically downward. Both forces are vertical, but their directions are opposite and their magnitudes need not be equal.
List every force on the body before calculating the reaction. When there is no acceleration perpendicular to the surface, the force components in that direction sum to zero.
For example, a body rests on a fixed horizontal floor and weighs 70 N. A hand applies another 15 N vertically downward. If the floor reaction is the only upward force, it must be 85 N. With static coefficient 0.2, the friction limit is 0.2 times 85 = 17 N.
Using weight alone would give a different answer because it would omit the hand's force. The reaction must come from the full force balance, not from the surface being horizontal.
Check whether the contact is sliding
A block can slide across a floor while its forces balance. Zero resultant means constant velocity, so it does not necessarily mean rest.
For an already sliding contact, calculate kinetic friction with f_k = mu_k R. Suppose R = 100 N and mu_k = 0.15. Kinetic friction has magnitude 15 N. A 15 N push along the sliding direction balances that friction and keeps the velocity constant. A larger push in that direction increases the speed.
Static friction has a different rule: its magnitude can range from zero up to mu_s R. The static and kinetic coefficients describe the same pair of surfaces in different contact states. Their values need not match.
Each coefficient has no units and depends on the surfaces and their condition. A constant-coefficient model does not track wear, deformation or changes with sliding speed. It also does not explain the microscopic origin of the contact forces.
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