Laws of Motion · Lesson 9 of 14

Sliding Friction, Inclines and Rolling Resistance

“Apply friction models to sliding bodies and inclined planes, and distinguish rolling resistance from static friction.”

Learning Objectives

• Use the kinetic-friction model after sliding begins. • Resolve weight parallel and perpendicular to an incline. • Derive the angle at which static slipping begins. • Explain how pulling, pushing, lubrication and rolling affect resistance.

Once surfaces slide

Kinetic frictionLaTeX

f_{k} = μ_{k} N

The force on one surface opposes its sliding velocity relative to the other surface. μ_k is dimensionless and often smaller than μ_s for the same contact pair.

In the elementary dry-friction model, kinetic friction is approximately independent of apparent contact area and moderately independent of sliding speed. Real contacts can depart from this model, so treat it as an experimentally useful approximation. If a block slides right across a stationary table, friction on the block is left. If a belt moves faster right than a block resting on it, friction on the slipping block can point right, because it opposes their relative slip.

Horizontal sliding under a horizontal pullLaTeX

a = \frac{F - μ _{k} m g}{m}

Here positive is the sliding direction, the pull is horizontal, and there is no vertical acceleration. Check that the assumed sliding direction still holds during the interval.

A sliding block

Problem
A 4 kg block slides right on a horizontal surface with μ_k = 0.20. A 14 N horizontal pull acts right. Use g = 10 m/s².

1.N = mg = 40 N, so kinetic friction is 8 N left.
2.Net force is 14 − 8 = 6 N right, giving acceleration 6/4 = 1.5 m/s².
3.If the pull is removed while the block still slides right, acceleration becomes −8/4 = −2 m/s². Once it stops, reassess static friction rather than continuing the same kinetic force into spontaneous backward motion.

Resolve the weight on an incline

Choose one axis down a plane inclined at θ above horizontal and the other perpendicular outward from the plane. The weight mg has a component mg sinθ down the plane and mg cosθ into it. With no perpendicular acceleration or other perpendicular force, N = mg cosθ. Friction lies along the plane. It is often upward for a released block tending to slide down, but an added pull can change that tendency.

Rest on an inclineLaTeX

f_{s} = m g sin θ, N = m g cos θ

These follow from equilibrium when friction prevents a downward slide and no additional force is present.

Limiting angle of restLaTeX

m g sin θ_{m a x} = μ_{s} m g cos θ_{m a x} ⟹ tan θ_{m a x} = μ_{s}

For smaller angles, friction is mg sinθ rather than automatically μ_sN. The mass cancels in this ideal model.

An incline slowly raised to 15°

Problem
A 4 kg block begins to slip when its plane reaches 15°. Find μ_s.

1.At impending downward slip, f_s = μ_sN.
2.Dividing the parallel and perpendicular equations gives μ_s = tan15° = 0.2679, about 0.27.
3.The stated mass is unnecessary for this coefficient calculation, but would matter if individual force magnitudes were requested.

Forces on a block on a 30-degree incline— The incline rises to the right at 30°. Weight is vertical, normal is tilted up and left, and friction for an impending down-left slide points up and right. Do not rotate the weight arrow to make it perpendicular to the plane. Arrow lengths are schematic unless stated otherwise.

Sliding down an inclineLaTeX

a = g (sin θ - μ_{k} cos θ)

Positive is down the plane. A negative answer means a block already sliding down is slowing; it does not invalidate the sign convention.

Downhill sliding

Problem
A block slides down a 30° incline with μ_k = 0.20. Use g = 10 m/s². Find acceleration.

1.The downhill weight per unit mass is g sin30° = 5 m/s².
2.The uphill friction per unit mass is μ_kg cos30° = 1.732 m/s².
3.Net downhill acceleration is 5 − 1.732 = 3.268 m/s². The mass cancels under these assumptions.

Pulling a handle and pushing a handle

A pull on a lawn mower inclined upward at angle α supplies an upward component F sinα. With no vertical acceleration, N = mg − F sinα. A push through an oppositely inclined handle supplies a downward component, making N = mg + F sinα. For the same force magnitude and angle, pulling reduces normal load and frictional resistance, whereas pushing increases it. Compare the same surface and force conditions before drawing this conclusion.

Normal load for an angled forceLaTeX

N_{pull} = m g - F sin α, N_{push} = m g + F sin α

These expressions require maintained contact and zero vertical acceleration. The forward driving component is F cosα in both cases.

Rolling is not sliding

For ideal rolling without slipping, the point of contact is momentarily stationary relative to the surface, so there is no kinetic friction at that point. On a level ideal rigid surface with no drive or resistance, steady rolling needs no horizontal force and can have zero static friction. Other rolling situations can require static friction, such as an accelerating driven wheel or a rolling object on an incline. No slip does not imply no friction in every situation.

Real wheels and surfaces deform. Deformation and material losses create rolling resistance, which can slow a freely rolling body. For many practical contacts it is much smaller than sliding resistance, which makes wheels and ball bearings useful. Rolling resistance should not simply be identified with kinetic friction μ_kN. Its detailed model depends on material and geometry.

Reducing and using friction

Lubricants reduce direct sliding resistance in machines. Ball bearings replace much sliding with rolling, and an air cushion can reduce solid contact. At the same time, friction is needed for traction and braking. A tyre rolling without slip usually transmits static contact friction, including during ordinary braking; a skidding tyre has kinetic friction. The correct force model follows the contact behaviour, not just whether the car’s centre is moving.

Quick check

For a stationary block on an incline below its slipping angle, which relation gives actual static friction?

Key Takeaways

• Use kinetic friction only while relative sliding occurs. • On an incline, resolve weight into mg sinθ and mg cosθ. • Impending slip gives tanθ_max = μ_s in the simple model. • An upward pull can reduce the normal load and friction. • Rolling without slipping can involve static friction; rolling resistance is a separate real-world effect.

Previous · Lesson 8

Static Friction and the Onset of Slipping

Next · Lesson 10

Connected Bodies, Strings and Pulleys