Lesson 2 of 5Dynamics

Unit 1 · Lesson 2 of 5

Forces on Inclines and Friction

Drawing free-body diagrams on a slope, splitting gravity into parts along and across the surface, the difference between static and kinetic friction, and the angle at which a resting block finally slides.

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Key things to remember

1On an incline: $F_{g ∥} = m g sin θ$ along the slope and $F_{g ⊥} = m g cos θ$ into it.
2With no other perpendicular forces, $F_{N} = m g cos θ$ .
3Static friction: $F_{s} \leq μ_{s} F_{N}$ . Kinetic friction: $F_{k} = μ_{k} F_{N}$ , and usually $μ_{k} < μ_{s}$ .
4Sliding down with friction: $a = g (sin θ - μ_{k} cos θ)$ .
5A resting block starts to slide when $tan θ = μ_{s}$ .

A car parked on a steep driveway, a skier on a hill, a box sliding down a loading ramp: each is a force problem where the surface is tilted. Gravity still pulls straight down, but the object can only move along the slope. The trick that makes every incline problem manageable is to stop using horizontal and vertical axes and tilt your axes to match the slope.

Resolving gravity on an incline

Fig 1Gravity is split into a part along the slope and a part into the slope. The angle between gravity and the perpendicular equals the incline angle.

Choose one axis parallel to the slope and one perpendicular to it. The normal force and friction already lie along these axes, so only gravity needs to be resolved. Because the perpendicular to the slope is tilted by the same angle $θ$ as the slope itself, the angle between $F_{g}$ and the perpendicular axis is also $θ$ . That gives the two components below.

F_{g ∥} = m g sin θ (down the slope) F_{g ⊥} = m g cos θ (into the slope)

Check the extremes: at 0 degrees nothing pulls along the slope, and at 90 degrees nothing presses into it.

The normal force on a slope

Nothing accelerates perpendicular to the slope, so the forces across it balance:

F_{N} = m g cos θ

(when no other force pushes into or pulls away from the surface). This is smaller than

m g

, which is why friction on a slope is smaller than friction on level ground for the same object.

Static and kinetic friction

Fig 2Static friction grows to match the push, then kinetic friction takes over at a lower, roughly steady value.

F_{s} \leq μ_{s} F_{N} F_{k} = μ_{k} F_{N}

Static friction can take any value up to its maximum. Kinetic friction has a single, roughly constant value. Both equations are good approximations, not exact laws.

AStatic friction

Acts when the surfaces are not sliding relative to each other
Adjusts its size to match whatever is trying to cause sliding
Has a maximum value of $μ_{s} F_{N}$ , reached just before slipping
Useful: tires gripping a road, shoes on a floor, a car parked on a hill

BKinetic friction

Acts while the surfaces are sliding
Has a steady size of $μ_{k} F_{N}$ , roughly independent of speed
Usually smaller than the maximum static friction, since for most surfaces $μ_{k} < μ_{s}$
A drawback in machines (heating, wear), useful in brakes

Static friction is not automatically mu s times FN

A box resting on a ramp has static friction equal to exactly

m g sin θ

, whatever that is, because the box is not accelerating. The expression

μ_{s} F_{N}

is only the largest static friction can get. Use it only for the moment an object is on the verge of slipping.

The angle at which a block starts to slide

Tilt a board slowly with a block resting on it. As $θ$ increases, the pull down the slope $m g sin θ$ grows while the maximum static friction $μ_{s} m g cos θ$ shrinks. The block slips when they become equal: $m g sin θ = μ_{s} m g cos θ$ . The mass and $g$ cancel, leaving $tan θ = μ_{s}$ . The same reasoning with kinetic friction shows that a block already sliding moves at constant velocity when $tan θ = μ_{k}$ . Measuring these angles is a simple lab method for finding both coefficients.

Situation	Direction of friction	Net force along the slope
Sliding down	Up the slope	$m g sin θ - μ_{k} m g cos θ$ , down the slope
Sliding up (after a push)	Down the slope	$m g sin θ + μ_{k} m g cos θ$ , down the slope, so it slows
At rest	Up the slope, static	Zero, so $F_{s} = m g sin θ$
Pushed up at constant velocity	Down the slope	Zero, so $F_{p u s h} = m g sin θ + μ_{k} m g cos θ$

A pull at an angle changes the normal force

On level ground, pulling a sled with a rope angled upward lifts part of its weight, so

F_{N} = m g - F sin ϕ

and friction drops. Pushing down at an angle does the opposite. Whenever an applied force has a component perpendicular to the surface, find

F_{N}

from the perpendicular forces before you calculate friction.

Worked examples

Reveal one step at a time and try to beat it to the next line.

Example 1

5.0 kg

block slides down a frictionless ramp inclined at 25°. Find the normal force and the block's acceleration.

Example 2

8.0 kg

crate slides down a ramp inclined at 30° with a coefficient of kinetic friction of

0.20

. Find its acceleration.

Example 3

A block rests on a board. The coefficient of static friction between them is

0.45

. The board is tilted slowly. At what angle does the block begin to slide?

Example 4

A worker pushes a

12 kg

box up a ramp inclined at 20° at constant velocity, pushing parallel to the ramp. The coefficient of kinetic friction is

0.30

. How hard must the worker push?

Where marks get lost

Swapping sine and cosine: using $m g cos θ$ as the pull down the slope. Check with 0°: nothing should pull along a flat surface, and $sin 0^{\circ} = 0$ .
Taking $F_{N} = m g$ on a slope. On an incline the normal force is $m g cos θ$ , which is smaller.
Using $μ_{s} F_{N}$ as the static friction on an object that is not about to slip. Static friction is only as large as it needs to be.
Pointing friction down the slope for an object sliding down. Friction always opposes the relative sliding.
Forgetting that an angled pull or push changes the normal force, and so changes friction.

Flashcards 7

Component of gravity along an incline

Component of gravity perpendicular to an incline

Static friction

Kinetic friction

+ 3 more

Practice

15 kg

block slides down a frictionless incline at 35°. What is its acceleration, in

m/s^{2}

? Give three significant figures.

m/s²

10 kg

box rests on a ramp inclined at 40°. What is the normal force on it, in newtons? Give three significant figures.

A coin on a book starts to slide when the book is tilted to 31°. What is the coefficient of static friction? Give three significant figures.

6.0 kg

block slides down a 28° incline with

μ_{k} = 0.15

. What is its acceleration, in

m/s^{2}

? Give three significant figures.

m/s²

20 kg

sled is pulled across level snow by a rope at 30° above the horizontal with a tension of

80 N

. The coefficient of kinetic friction is

0.10

. What is the sled's acceleration, in

m/s^{2}

? Give three significant figures.

m/s²

4.0 kg

box sits at rest on a ramp. The coefficient of static friction is

0.60

. Which expression gives the static friction acting on it?

Why is it harder to start a heavy crate sliding than to keep it sliding?

A block is given a push so that it slides up a rough incline. While it slides up, which way does friction act?

Which type of friction lets car tires grip the road without skidding when a car turns or brakes smoothly: static or kinetic?

Match each quantity for a block on an incline of angle theta.

Pull of gravity along the slope

Normal force (no other forces)

Kinetic friction

Condition to start sliding

Order the steps for finding the acceleration of a block sliding down a rough incline.

1Apply Newton's second law along the slope
2Find the normal force from the perpendicular balance, then friction
3Resolve gravity into mg sin θ and mg cos θ
4Draw a free-body diagram showing gravity, normal force and friction
5Tilt the axes so one lies along the slope

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