Chapter 6: Frictional Forces  Two types: - static – applies to stationary objects - kinetic – applies to sliding (moving) objects  Like the normal force, the Frictional Force is a contact force, but acts parallel to the interface of two objectsy x n mg FA f Apply Newton’s 2nd Law € Fy= N − mg = may= 0∑N = mgFx= FA− f = max∑ If applied force is small, book does not move (static), ax=0, then f=fs  Increase applied force, book still does not move  Increase FA more, now book moves, ax≠0 SAfF =SxSAxSAfmafFmafF >+=⇒=− There is some maximum static frictional force, fsmax. Once the applied force exceeds it, the book moves € fSmax=µSN• µs is the coefficient of static friction, it is a dimensionless number, different for each surface-object pair (wood-wood, wood-metal); also depends on surface preparation • µs does not depend on the mass or surface area of the object • Has value: 0 < µs < 1.5 • If no applied vertical force Magnitudes not vectors mgfSSµ=max Push down on book  Apply Newton’s 2nd Law y x N mg FA f € Fy= N − mg − FP= may= 0∑N = mg + FPfSmax=µSN =µS(mg + FP)FP  What is FA needed just to start book moving? ∑+==⇒=−= )(0maxPSSASAxFmgfFfFFµWhen an Object is Moving?  fsmax is exceeded so the object can move, but friction force is still being applied.  However, less force is needed to keep an object moving (against friction) than to get it started  We define kinetic friction  µk is the coefficient of kinetic friction, similar to µS but always less than µS (Table 6.1) • Now, let’s consider incline plane problem, but with friction NkkFfµ=y y FBD θ mg N x N mg θ θx fs fs  Book is at rest  m = 1.00 kg, θ = 10.0°, µs = 0.200 € Fy= N − mgcosθ= 0∑N = mgcosθ= (1.0kg)(9.80ms2)(cos10.0o)= 9.65 NN 70.1)0.10)(sin80.9)(kg00.1(sin0sin2sm====−=∑oSSxmgffmgFθθ Book can move (slide) if mgsinθ ≥ fsmax  What is fsmax? € fSmax=µSN = (0.200)(9.65 N) = 1.93 N > fSBook does not move.  What angle is needed to cause book to slide? € mgsinθ≥ fSmaxmgsinθ≥µSNmgsinθ≥µSmgcosθtanθ≥µSo111.3 )(tan≥≥−Sµθ As θ is increased, N decreases, therefore fsmax decreases Once book is moving, we need to use the kinetic coefficient of friction  Lets take, θ = 15.0° and µk = 0.150 < µs 22smsm12.1 )0.15cos150.00.15)(sin80.9( )cos(sinor cossin sin=−=−==−==−=∑ookxxkxkxgamamgmgmafmgFθµθθµθθExample Problem A skier is pulled up a slope at a constant velocity by a tow bar. The slope is inclined at 25.0° with respect to the horizontal. The force applied to the skier by the tow bar is parallel to the slope. The skier’s mass is 55.0 kg , and the coefficient of kinetic friction between the skis and the snow is 0.120. Find the magnitude of the force that the tow bar exerts on the skier. Given: m = 55.0 kg, µk = 0.120, θ = 25.0°  Infer: since velocity is constant, ax=0; also ay=0 since skier remains on slope ⇒ equilibrium ∑ ∑ ∑==⇒= 0,00yxFFFDraw FBD, apply Newton’s 2nd Law θ mg N x N mg θ θx Fp Fp fk fk€ Fy= 0∑N − mgcosθ= 0N = mgcosθ€ Fx= 0∑fk− FP+ mgsinθ= 0FP= fk+ mgsinθ=µkN + mgsinθ=µkmgcosθ+ mgsinθ= mg(µkcosθ+ sinθ)= 286