1EXS 387 - BiomechanicsDr. MoranThursday March 1, 2007Spring 2007Linear KineticsFrictionEXS 387 2Lecture ObjectivesFriction• Static Friction– Limiting Value• Dynamic Friction• Coefficient of Friction• Determining Normal Force• Example Problems• Rolling Friction2EXS 387 3What do we remember?• A man has a mass of 82.5 kg, what would be an expected vertical impact force during normal gait?(a) 690 N(b) 790 N(c) 890 N(d) 990 N(e) None of the above• Which of the following is NOT necessary to fully define a FORCE?(a) point of application(b) direction(c) time(d) magnitude(e) all of the aboveEXS 387 4Friction• Def: acts at the interface of surfaces in contact• FORCE (units – N)• Consider the following:Would just a small horizontally directed force get the book moving to the right? Would a LARGE horizontally directed force get the book moving to the right?3EXS 387 5Friction(con’t)• Frictional force acts in a direction parallel to the area of contact, and opposes the motion or the tendency to move. The friction force depends on two things:•• Nature of the surfaces involved (µ) (rough, smooth, etc.)• Friction = µ · Rn•EXS 387 6Normal force (Rn)WeightDirection of Impending MotionLimiting Value of Static FrictionAs F increases the magnitude of opposing friction also increases until it reaches a MAXIMUM STATIC FRICTION (Fm).Three Scenarios:(1) F < Fm(2) F = Fm (3) F > Fm4EXS 387 7Friction Force Applied External Force STATIC DYNAMICNO MOTIONMOTIONEXS 387 8µCoefficient of Friction• Unitless number•• Influencing Factors:• Surface Roughness•• Molecular Interaction between Surfaces• µK< µSfor two bodies in contact• Human joints have a µ value less than two pieces of ICE in contact because of5EXS 387 9Determining Normal Force• Weight: always directed vertically down• Normal Force = force to surfaceNormal Force = Weight15oWeightNormalEXS 387 1015oWeightNormalIf the mass is 80 Kg, then what is the Normal Force?Nwtwtmgwt8.784)81.9)(80(===15oWtN6EXS 387 11Example Problem #1• The coefficient of friction between a sled and the snow is 0.18, with a coefficient of kinetic friction of 0.15. A 250N boy sits on the 200N sled. How much force directed parallel to the horizontal surface is required to start the sled in motion? How much force is required to keep the sled in motion? EXS 387 12Example Problem #2• If a skier has a weight of 800N and the static coefficient of friction between the skis and the snow is 0.20 will the skier move down the hill?15o7EXS 387 13Rolling Friction• Ex: between bicycle/automobile tire and the ground• Due to deformation of the tire• Rolling Friction Values• 1/100thto 1/1000th versus static/kinetic values• Decrease Rolling Friction By:• Decrease load (normal force)• Increase air pressure• Increase wheel diameter• Increase smoothness of surfaces EXS 387 14Review Questions• Introductory Problems – page 415 #3,4• Additional Problem – page 415 #3• How is a normal force oriented to the surface?• Does friction depend on the amount of surface area (contact area)?• What does influence frictional force?• Why do human joints have such a low frictional