Homework P. 146 30, 33, 34, 35, 37

What causes friction? Why is there Friction? Surface roughness Electronic interactions at the atomic level Friction is caused by the temporary electrostatic bonds created between two objects in contact with one another. Examples of Friction - Desirable - Undesirable

Examples of Friction. - Desirable. - Walking. - Driving. - Braking Examples of Friction - Desirable - Walking - Driving - Braking - Undesirable - Engine Efficiency - Coasting - Pushing a heavy object

Why would I want to change friction? - How would I do it?

Friction & Applying Newton’s 2nd Law System Chapter 6.2

Friction How does friction affect the motion of objects? It can slow an object down like the friction between the tires and the road. It is responsible for increasing the speed of an object like a car. It is also responsible for objects being able to change direction.

Static Friction Static Friction: The resistive force that keeps an object from moving. Fforward Ffriction Fnet = FAPPLIED – Ffriction Since the crate is not accelerating, Fnet = 0 FAPPLIED = Ffriction FAPPLIED Ffriction Fground-on-crate Fgravity System Note: As long as the crate does not move, FAPPLIED = Ffriction

Fnet = FAPPLIED – Ffriction Kinetic Friction Kinetic Friction: The resistive force that opposes the relative motion of two contacting surfaces that are moving past one another. Since the crate will initially accelerate, Fnet > 0. Fforward Ffriction Fground-on-crate Fgravity Fforward Ffriction Fnet = FAPPLIED – Ffriction Fnet System Note: If the crate moves at a constant speed, then FAPPLIED = Ffriction and Fnet = 0.

An Important Term APPLIED FORCE Usually whatever is pushing or pulling NOT the same as Net Force

Determining the Frictional Force For people who had a lot of wrong ideas about Physics the Greek alphabet sure gets used a lot! The force of friction is proportional to the normal force and a proportionality constant ( - pronounced mu) called the coefficient of friction. For static friction: 0 < Ff, static < s FN For kinetic friction: Ff, kinetic = k FN Note: FN = the force normal (perpendicular) to the frictional force on the object.  is dimensionless Ff, static > Ff, kinetic Ff FN

Frictional Force For static friction: 0 < Ff, static < s FN For kinetic friction: Ff, kinetic = k FN

Determining the Frictional Force  (the coefficient of friction) is usually in the range of 0<=  <= 1, but this is not always the case Material 1 Material 2  Tire, dry Road, dry 1 Tire, wet Road, wet 0.2 Rubber Steel 1.6 Teflon 0.04 Ice Wood 0.05 Glass Metal 0.5 - 0.7 Chromium 0.41 Aluminum 1.3

Determining the Frictional Force Sketch a graph of Fs vs applied force Sketch a graph of Fk versus applied force Sketch a graph showing the transition from Fs to Fk Do this as a class exercise

Ff versus applied force

Ff versus applied force

The Normal Force The normal force is a force that opposes the Earth’s gravitational attraction and is perpendicular to the surface that an object rests or is moving on. For a horizontal surface, FN = Fg = mg. For a surface that is not perpendicular to gravity, FN = Fgcos FN 

The Normal Force  FN FN Fg Fg FN = Fg = mg FN = Fg cos = mg cos cos = adj/hyp Fg FN = Fg = mg FN = Fg cos = mg cos

Example 2: Determining Friction (Balanced Forces) Assume that the man in the figure is pushing a 25 kg wooden crate across a wooden floor at a constant speed of 1 m/s. How much force is exerted on the crate? FAPPLIED Ff FN Fg System

Diagram the Problem FAPPLIED Ff FN Fg FAPPLIED Ff FN Fg +y +x FAPPLIED Ff FN Fg FAPPLIED Ff FN Fg System y-direction: FN = Fg x-direction: Fnet = FAPPLIED - Ff Since the crate is moving with constant speed, a = 0, Fnet = 0, and FAPPLIED = Ff

State the Known and Unknowns What is known? Mass (m) = 25 kg Speed = 1 m/s Acceleration (a) = 0 m/s2 k = 0.3 (wood on wood) What is not known? FAPPLIED = ?

Perform Calculations y-direction: x-direction: a = 0 Fg = FN = mg Fnet = Fforward – Ff FAPPLIED = Ff FAPPLIED = kFN; FAPPLIED = kmg FAPPLIED = (0.3)(25 kg)(9.8 m/s2) FAPPLIED = 74 N

Example 3: Determining Friction (Unbalanced Forces) Assume that the man in the figure is pushing a 25 kg wooden crate across a wooden floor at a speed of 1 m/s with a force of 74 N. If he doubled the force on the crate, what would the acceleration be? Assume Constant speed FAPPLIED Ff FN Fg System

Diagram the Problem FN FAPPLIED Ff FN Fg Ff FAPPLIED Fg +y +x FN FAPPLIED Ff FN Fg System Ff FAPPLIED Fg y-direction: FN = Fg x-direction: Since a > 0, Fnet = Fforward - Ff

State the Known and Unknowns What is known? Force = 148 N Mass (m) = 25 kg Speed = 1 m/s k = 0.3 (wood on wood) What is not known? a ?

Perform Calculations y-direction: x-direction: a > 0 Fg = FN = mg x-direction: a > 0 Fnet = FAPPLIED – Ff ma = FAPPLIED – Ff ma = FAPPLIED – kmg a = FAPPLIED – kmg m a = (148N)/(25kg) – (0.3)(9.8 m/s2) a = 2.96 m/s2 Fnet = 148N – 74N ma = 74N a = 74N/25kg a = 2.96 m/s2

Key Ideas Friction is an opposing force that exists between two bodies. Friction is proportional to the normal force and the coefficient of friction; static or kinetic. The force required to overcome static friction is greater than that required to overcome kinetic friction.