Friction

Why it matters

Friction determines whether contacts stick or slide. It affects brakes, clutches, belts, wedges, and the stability of objects on slopes.

Dry-friction model

Friction acts tangentially to a contact and opposes relative sliding, or the tendency to slide, between the two surfaces. It need not oppose the motion of the body's centre of mass: static friction can propel a rolling wheel.

For the ideal Coulomb model:

• While a contact sticks, the static-friction magnitude adjusts to the required value: 0 ≤ F_s ≤ μ_s N.
• At impending slip, F_s,max = μ_s N.
• During sliding, F_k = μ_k N, directed opposite the relative sliding velocity.

Here N is the compressive normal reaction in newtons and μ_s and μ_k are dimensionless coefficients. Typically μ_k < μ_s, but actual static friction may be zero or less than kinetic friction because it is not always at its limiting value.

The approximate independence from apparent contact area applies to ordinary dry sliding under the Coulomb model. It is not a universal law for lubricated contacts, rubber, adhesion, or deformable surfaces. Coefficients depend on the materials, surface condition, and operating conditions; use the values given in a problem.

Solving a problem

1. Draw a free-body diagram and find the normal reaction from the equations of motion or equilibrium. N is not always equal to weight.
2. Assume sticking and calculate the tangential force needed for equilibrium or no slip.
3. Check whether its magnitude is at most μ_s N.
4. If the sticking assumption fails, use the sliding model and determine acceleration with Newton's second law.

Worked example: horizontal force

A 100 N block rests on a horizontal floor, with μ_s = 0.4. A horizontal force P is increased slowly. Assume no other vertical forces and no tipping.

Vertical equilibrium gives N = 100 N. The maximum static friction is 0.4 × 100 = 40 N. For P = 25 N, friction is 25 N, not 40 N. The block reaches impending slip at P = 40 N; a force exceeding this limit cannot be balanced by static friction.

If μ_k = 0.3 and P = 45 N after sliding starts, F_k = 30 N. Taking g = 9.81 m/s², m = 100/9.81 kg and a = (45 − 30)/m = 1.4715 m/s².

Inclined plane

For a block acted on only by gravity and the plane's reactions, N = mg cos θ. Rest requires mg sin θ ≤ μ_s mg cos θ, or tan θ ≤ μ_s. At impending downward slip, tan θ = μ_s. If the block slides down, its acceleration is g(sin θ − μ_k cos θ).

Common mistakes

• Setting static friction equal to μ_s N before checking impending motion.
• Using weight instead of the actual normal reaction.
• Choosing the friction direction from the body's absolute velocity instead of relative slip at the contact.
• Ignoring a competing failure mode such as tipping.

Quick check

1. A 100 N block with μ_s = 0.4 experiences a 10 N horizontal push. What is its static friction? 10 N, assuming no other horizontal forces.
2. What is the angle of repose in this model? arctan(μ_s).
3. Does static friction always do zero work on a body? No. Its power depends on the velocity of the point of application in the chosen frame; zero relative sliding alone does not imply zero work on each body.