P. 146 30, 33, 34, 35, 37 Homework. What causes friction? Why is there Friction? Surface roughness Electronic interactions at the atomic level  Friction.

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P , 33, 34, 35, 37 Homework

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 - Undesirable - Engine Efficiency - Coasting - Pushing a heavy object

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

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

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 System F forward F friction F ground-on-crate F gravity F forward F friction F net = F forward – F friction Since the crate is not accelerating, F net = 0 F forward = F friction Note: As long as the crate does not move, F forward = F friction Static Friction: –The resistive force that keeps an object from moving.

Kinetic Friction System F forward F friction F ground-on-crate F gravity 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, F net > 0. F forward F friction F net = F forward – F friction F net Note: If the crate moves at a constant speed, then F forward = F friction and F net = 0.

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

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

Frictional Force For static friction: –0–0 < F f, static <  s F N For kinetic friction: –F–F f, kinetic =  k F N

Determining the Frictional Force  (the coefficient of friction) is usually in the range of 0<=  <= 1, but this is not always the case Material 1Material 2  Tire, dryRoad, dry1 Tire, wetRoad, wet0.2 RubberSteel1.6 Teflon 0.04 IceWood0.05 GlassMetal 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

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, F N = F g = mg. –For a surface that is not perpendicular to gravity, F N = F g cos  FNFN 

The Normal Force FNFN FgFg FNFN  FgFg F N = F g = mg cos  = adj/hyp F N = F g 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? System F forward FfFf FNFN FgFg

Diagram the Problem System F forward FfFf FNFN FgFg FfFf FNFN FgFg y-direction: F N = F g x-direction: F net = F forward - F f Since the crate is moving with constant speed, a = 0, F net = 0, and F forward = F f +y +x

State the Known and Unknowns What is known? oMass (m) = 25 kg oSpeed = 1 m/s oAcceleration (a) = 0 m/s 2 o  k = 0.3 (wood on wood) What is not known? oF forward = ?

Perform Calculations y-direction: oF g = F N = mg x-direction: a = 0 oF net = F forward – F f oF forward = F f oF forward =  k F N ; F forward =  k mg oF forward = (0.3)(25 kg)(9.8 m/s 2 ) oF forward = 74 N 0

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? System F forward FfFf FNFN FgFg Assume Constant speed

Diagram the Problem System F forward FfFf FNFN FgFg FfFf FNFN FgFg y-direction: F N = F g x-direction: Since a > 0, F net = F forward - F f +y +x

State the Known and Unknowns What is known? oForce = 148 N oMass (m) = 25 kg oSpeed = 1 m/s o  k = 0.3 (wood on wood) What is not known? oa ?

Perform Calculations oF net = 148N – 74N oma = 74N oa = 74N/25kg oa = 2.96 m/s 2 y-direction: oF g = F N = mg x-direction: a > 0 oF net = F forward – F f oma = F forward – F f oma = F forward –  k mg oa = F forward –  k mg m oa = (148N)/(25kg) – (0.3)(9.8 m/s 2 ) oa = 2.96 m/s 2

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.