FRICTION - the force that present whenever two surfaces are in contact and always acts opposite to the direction of motion.

Depends on:• Type of materials in contact • Surfaces of materials

Does NOT depend on:•Surface area in contact•Speed

Ff = μFn

Fn = Normal Force; Force acting perpendicular to the two surfaces in contact. In level surfaces, this is equal in magnitude but acting opposite to the weight

Ff = Force of Friction in N

μ = Coefficient of Friction (depends on material 0 < μ < 1

TYPES OF FRICTIONTYPES OF FRICTIONKinetic Friction - Force needed to keep it going at a constant velocity.

Ff = μ kFn

Static Friction - Force needed to start motion.Ff < μ sFn

Keeps the object at rest.Only relevant when object is stationary.Calculated value is a maximum

Fri

ctio

nal

For

ce

Res

isti

ng

Mot

ion

Force Causing the Object to Move

Kinetic RegionStatic Region

Max

kFF s

Fs ≥ Fk

Fk

Fs

Frictional Forces Occur When Materials are in Contact

Fw

FsFa

Fn

Surfaces in Contact

M1

Spring Scale

Fa = Force Causing Motion (Pull on Scale)

Fs = Force of Static Friction (Resists Motion)Fn = Normal Force (Perpendicular to Surfaces)Fw = Weight of Object ( Mass x Gravity)

Friction is a Force That Always Resists Motion

Surfaces in Contact

The Block M will only move if the Applied Force (Fa) is greater the Force of Static Friction (Fs).

M

Spring Scale

Fn

FsFa

Fw

FRICTION FRICTION converts kinetic converts kinetic energy to heat energy.energy to heat energy.

Banana peel reduces Banana peel reduces friction.friction.

INCREASING FRICTIONINCREASING FRICTION

REDUCING FRICTIONREDUCING FRICTION

The coefficient of friction (μ) is the ratio of the Applied Force (Fa) over the Normal Force (Fn).

Surfaces in Contact

M

Spring Scale

Fn

FsFa

Fwμ = Fa

Fn

Determine the amount of force needed to move 12.0 kg of rubber at a constant velocity across dry concrete?

Solution:F = ma FFr = μkFN

FN = weight = mg = (12 kg)(9.8 N/kg) = 117.6 N

FFr = μkFN = (0.8)(117.6 N) = 94.08 N = 90 N90 N

What is the force needed to slide a stationary 150 kg rubber block across wet concrete?

SOLUTION:F = ma FFr < μsFN

FN = weight = mg = (150 kg)(9.8 N/kg) = 1470 N

FFr = μkFN = (.7)(1470) = 1029 N = 1000 N

1000 N

Given the following :μs = .62, μk = .48Determine the acceleration of a sliding 8.50 kg block if a force of 72 N is applied to it?

FFr = μkFN FN = mg FFr = μkmgFFr = (.48)(8.5 kg)(9.8 N/kg) = 39.984 NF = ma<72 N - 39.984 N> = (8.5 kg)a, a = 3.77 = 3.8 m/s2

3.8 m/s/s

72 N8.5 kg

v

FFr