Kinetic and Static Friction: Study Guide (Physics C: Mechanics 2.7)

How this differs from the Physics 1 version. Physics C: Mechanics and Physics 1 are separate courses that both have a Topic 2.7 on kinetic and static friction. The friction model is the same. This guide is the calculus-based one: it leans on symbolic derivations, on objects that interact through friction, and on forces that change with time, so you integrate the motion once slipping starts. The algebra-based treatment is in the Physics 1 study guide; do not mix the two when you revise. This topic uses the free-body diagrams and Newton’s second law from earlier in Unit 2, and follows Topic 2.6, Gravitational Force.

Two parts of one contact force

When two surfaces touch, each pushes on the other. It helps to split that contact force into two components:

• The normal force F_N is the component perpendicular to the surface. It points away from the surface that exerts it.
• The friction force F_f is the component parallel to the surface.

The two are linked, but F_N is not a fixed number. You find it from Newton’s second law along the axis perpendicular to the surface. On a level floor with no other vertical forces, F_N = mg. Push down on the object, pull up on it, tilt the surface or accelerate the floor, and F_N changes. Friction changes with it.

Background (not assessed). Real surfaces are rough on a tiny scale. They touch only at many small high points. Pressing harder squashes more high points into contact, so the true contact area grows in proportion to the load. That is one reason friction depends on F_N but not on the apparent area of contact.

Kinetic friction: surfaces sliding

Kinetic friction acts when two surfaces in contact move relative to each other. Its direction on each surface is opposite to that surface’s motion relative to the other surface. Its size is

F_f,k = μ_k F_N

• μ_k is the coefficient of kinetic friction. It has no unit, and it depends on the materials of both surfaces (and their condition, such as wet or dry).
• In this model, F_f,k does not depend on the area of contact or on the sliding speed.

Read the direction rule carefully. It says “relative to the other surface”, not “opposite to the motion”. If a box slides backwards on a truck bed that is speeding up, the box may be moving forwards relative to the road while kinetic friction on it also points forwards. What matters is the box’s motion relative to the bed.

Friction forces come in Newton’s third-law pairs. If the floor exerts friction on a block towards −x, the block exerts friction on the floor towards +x, with the same size.

Static friction: no sliding

Static friction may act between surfaces that are not moving relative to each other. It is a responsive force. It takes whatever size and direction is needed to stop the surfaces from slipping, up to a limit:

|F_f,s| ≤ μ_s F_N, so the maximum is F_f,s,max = μ_s F_N

• “Slipping” or “sliding” means the two surfaces move relative to each other.
• μ_s is the coefficient of static friction. For a given pair of surfaces, μ_s is usually larger than μ_k. That is why it takes a bigger push to start an object moving than to keep it moving.
• Static friction can be zero, for example on a block resting on a level floor with no horizontal push.

How to handle static friction in a problem

1. Assume no slipping. Treat the surfaces as moving together (or both at rest).
2. Use Newton’s second law to find the friction F_f,s needed for that.
3. Find F_N and compute the maximum μ_s F_N.
4. If |F_f,s needed| ≤ μ_s F_N, the assumption holds and the answer from step 2 stands.
5. If not, the surfaces slip. Start again with kinetic friction μ_k F_N, pointing opposite the relative motion.

Never write F_f,s = μ_s F_N unless the problem tells you slipping is just about to start.

Friction on a slope, symbolically

Take a block of mass m on a slope at angle θ. Use +x down the slope and +y perpendicular to the slope, away from it. Perpendicular to the slope there is no acceleration, so

F_N − mg cos θ = 0, which gives F_N = mg cos θ.

Block at rest. Along the slope, mg sin θ − F_f,s = 0, so static friction points up the slope with size mg sin θ. No slipping requires mg sin θ ≤ μ_s mg cos θ, which simplifies to

tan θ ≤ μ_s

The mass cancels, so a heavy block and a light block of the same material start to slip at the same angle.

Block sliding down. Now kinetic friction points up the slope: ma_x = mg sin θ − μ_k mg cos θ, so

a_x = g(sin θ − μ_k cos θ)

Check the limits. With μ_k = 0 you get g sin θ, the frictionless result. With θ = 90° you get g, free fall, because the normal force (and so friction) vanishes on a vertical wall with nothing pressing the block against it.

Worked example 1: a push that grows with time

Question. Take +x in the direction of the push. A 4.0 kg block rests on a level floor. For the block and floor, μ_s = 0.45 and μ_k = 0.30. From t = 0 a horizontal push P = (6.0 N/s)t is applied. (a) Find the friction force at t = 1.0 s. (b) Find when the block starts to slide. (c) Find a_x(t) once it slides. (d) Find the block’s velocity at t = 5.0 s.

1. Normal force. Vertically, F_N − mg = 0, so F_N = 4.0 × 9.8 = 39.2 N.
2. Limits. Maximum static friction: μ_s F_N = 0.45 × 39.2 = 17.64 N. Kinetic friction: μ_k F_N = 0.30 × 39.2 = 11.76 N.
3. (a) At t = 1.0 s, P = 6.0 N. Assume no slipping: friction must balance the push, so F_f,s = 6.0 N towards −x. This is below 17.64 N, so the assumption holds.
4. (b) Slipping starts when the needed friction reaches the maximum: 6.0t = 17.64, so t = 2.94 s (2.9 s to 2 s.f.).
5. (c) After that, kinetic friction 11.76 N acts towards −x. Newton’s second law: 4.0a_x = 6.0t − 11.76, so a_x = 1.5t − 2.94 (m/s², t in s). Just after slipping starts, a_x = 1.5 × 2.94 − 2.94 = 1.47 m/s². The acceleration jumps from 0 to about 1.5 m/s², because friction drops from 17.64 N to 11.76 N.
6. (d) The block starts from rest at t = 2.94 s. Integrate: v_x = ∫ from 2.94 to t of (1.5t − 2.94) dt = 0.75(t² − 2.94²) − 2.94(t − 2.94). At t = 5.0 s: v_x = 0.75(25 − 8.644) − 2.94 × 2.06 = 12.27 − 6.06 = 6.2 m/s.

Check. Do not use v_x = a_x t: the acceleration is not constant. The lower limit of the integral is the moment slipping begins, not t = 0, because before that the velocity stays zero.

Worked example 2: stacked blocks and relative motion

Question. Take +x to the right. A 1.5 kg block A sits on top of a 3.5 kg block B, which rests on a frictionless floor. Between A and B, μ_s = 0.40 and μ_k = 0.25. A horizontal force P to the right acts on B only. (a) Find the largest P for which the blocks move together. (b) Find the acceleration of each block when P = 30 N.

The only horizontal force on A is friction from B. So friction from B on A must point +x: it is what drags A forwards. By Newton’s third law, friction from A on B points −x.

1. (a) For A: F_N = m_A g = 14.7 N, so the largest static friction is 0.40 × 14.7 = 5.88 N. The largest acceleration A can have is a_max = μ_s F_N / m_A = μ_s g = 3.92 m/s².
2. Moving together, the whole system (A + B) has only P as a horizontal external force. So P_max = (m_A + m_B) μ_s g = 5.0 × 3.92 = 19.6 N (20 N to 2 s.f.).
3. (b) Test the no-slip assumption at P = 30 N. Together, a = 30 ÷ 5.0 = 6.0 m/s². A would need 1.5 × 6.0 = 9.0 N of friction, but only 5.88 N is available. So A slips on B.
4. B moves to the right faster than A, so A moves left relative to B. Kinetic friction on A therefore points +x, with size μ_k m_A g = 0.25 × 14.7 = 3.675 N.
5. Block A: a_A = 3.675 ÷ 1.5 = 2.45 m/s² (= μ_k g), to the right.
6. Block B: 3.5a_B = 30 − 3.675, so a_B = 7.5 m/s² to the right.

Check. a_B > a_A, which agrees with B sliding forwards under A. Both blocks move to the right relative to the floor, yet friction on B points left. Friction opposes relative motion, not motion.

Common misconceptions

• “F_N = mg always.” Find F_N from the forces perpendicular to the surface. It changes on slopes, with angled pulls, and in accelerating lifts.
• “Static friction = μ_s F_N.” That is only the maximum. Static friction equals what is needed (Worked example 1, part (a)).
• “Friction always opposes motion.” It opposes the motion of one surface relative to the other. Friction is what makes block A in Worked example 2 speed up.
• “Bigger contact area, more friction.” In this model, friction does not depend on the area of contact.
• “μ_k is larger, because moving objects rub more.” For a given pair of surfaces μ_s is usually the larger one.
• “Use a = constant equations after slipping.” If the applied force changes with time, the acceleration does too. Integrate from the moment slipping starts.
• Forgetting the third-law partner. Friction on A from B has an equal and opposite partner on B from A.

Where this leads

Next, Topic 2.8, Spring Forces adds a force that depends on position. Topic 2.9, Resistive Forces, then treats drag that depends on velocity, which is different from the constant kinetic friction here. In Unit 3 you will find the work done by friction, and in rolling (Unit 6) static friction acts at the contact point. Try the practice questions now, then use the revision notes and the checklist. You can also return to the course roadmap.