Frictional Forces

Chapter 4, Section 4

Pg. 141-149

Friction

There are two types of frictional forces in the physical world that effect us.

- Static/Kinetic Friction

- Air Resistance

Air ResistanceAir ResistanceIs a resistance force that acts in the

opposite direction of gravitational forces.

Force of Gravity

Air Resistance

Help!!

SPLAT!!

Force of Gravity

Air Resistance

v1 > v2

Static and Kinetic FrictionStatic and Kinetic Friction

Forces that oppose motion between two surfaces that are touching each other.

Fx

Fr

Fr (non-moving) = Static Friction (Fs)Fr (moving) = Kinetic

Friction (Fk)

Frictional force depends on the size and mass of the object moving across a surface.

Fk

Fx

Fx

Fk

The amount of friction occurring between an object and the surface depends on the surface type.

Tile Floor

Carpet

Fk

Fk

Fx

Fx

The value that expresses the dependence of frictional forces on the surface type the object is in contact with is called the coefficient of friction (µ).

It is the ratio between the force of friction and the normal force.

Coefficient of Kinetic FrictionCoefficient of Kinetic Friction

μk = Fk/Fn coefficient of kinetic friction

μs = Fs,max/Fn coefficient of static friction

Ff = μFn frictional force

Sample problem

While redecorating her apartment, Suzy slowly pushes an 82 kg cabinet across a wooden dining room floor, which resists the motion with a force of friction of 320 N. What is the coefficient of kinetic friction between the cabinet and the floor?

FappliedFk = 320 N

m = 82 kgg = 9.81 m/s²

w = mg

w = (82 kg) (9.81 m/s²)w = 804 N w = Fn = 804 N

μk = Fk/Fn

μk = 320 N/804 N

μk = 0.40

Fn

mg

Sample Problem 2 (Pg. 146)

A student moves a box of books by attaching a rope to the box by pulling with a force of 90 N at an angle 30° to the horizontal. The box of books has a mass of 20.0 kg and the coefficient of kinetic friction between the bottom of the box and the sidewalk is 0.50. What is the acceleration of the box?

Fapplied = 90 N

Fk

mg

Fn

30°

g = 9.81 m/s²m = 20.0 kg

μk = 0.50

w = mg = 196 N

Fapp,y30°

90 N

Fapp, x

Step 1: Solve for Weight and X & Y components of applied force

Fapp, y = (90N) (sin 30°) = 45.0 N

Fapp, x = (90N) (cos 30°) = 77.9 N

Step 2: Find the Kinetic Friction

∑Fy = Fn + w + Fapp, y = 0

0 = Fn – 196 N + 45.0 N

Fn = 196 N – 45 N = 151 N

μk = Fk/Fn

Fk = (μk)(Fn) = (0.50)(151 N) = 75.5 N

Fapp, y = 45.0 N Fapp, x = 77.9 Nμk = 0.50

w = 196 NFapp,y

30°

90 N

Fapp, x

Step 3: Solve for acceleration

∑F = ma

a = (Fapp, x – Fk) / m

a = (77.9 N – 75.5 N) / 20.0 kg

a = 0.12 m/s² to the right

a = ∑F/m

Fapp,y30°

90 N

Fapp, x

m = 20.0 kg

Fk = 75.5 N

Fapp, x = 77.9 N