Copyright R. Janow – Spring 2012Physics 111 Lecture 05Force and Motion II – Friction, Circular MotionSJ 8th Ed.: Ch. 5.8, 6.1 – 6.3• Dynamics Summary• Friction Basics– Static Friction– Kinetic Friction• Properties of Friction• Sample Problems• Uniform Circular Motion -Centripetal Force– Free Body Diagrams– Sample Problems• Non-Uniform Circular Motion• Accelerated Frames of Reference • Drag Forces and Terminal Speed5.8 Friction Forces6.1 Extending the Uniform CircularMotion Model6.2 Non-uniform Circular Motion6.3 Motion in Accelerated FramesCopyright R. Janow – Spring 2012Summary – Dynamics so far Now: Add friction….Mass: inertia, resistance to acceleration, inherent propertyForce: pushes or pulls on bodies, contact or through field, causes a body to accelerateFBD: Free Body Diagram, vital to problem analysis. Include ALL forces acting on a bodyNet Force: vector sum of all forces acting ON a body (superposition). ∑∑∑∑≡≡≡≡iinetF FrvFirst Law: A body’s velocity is constant if the net external force acting on it is zero (equilibrium)Second Law: am F netrr====Acceleration resulting from FnetThird Law: If body A exerts a force on body B, then body B exerts a force equal in magnitude and opposite in direction on body A.For each Cartesian componentNewton’s 3 Lawshold in inertial (un-accelerated) reference framesUnits:SYSTEM FORCE MASS ACCELERATIONSI Newton (N) Kg m/s2CGS Dyne gm cm/s2British Pound (lb) slug ft/s2ConceptsCopyright R. Janow – Spring 2012Friction BasicsA contact force between surfacesdue to their roughness Friction force f always opposes the motionMotion may be:• Actual motion  kinetic (sliding) friction fk• Impending motion  static friction fsv’fvfModel for solid surfaces in contact:•Surface roughness measured by “friction coefficients µµµµkand µµµµs• Friction force proportional to pressure between surface and µµµµ• Friction force independent of speed v• Other dissipative forces (e.g. in liquids) depend on speed and viscosity N fssµµµµ≤≤≤≤WNfsFSTATIC FRICTIONfsself-varies tocancel F, up to a “breakaway”limitfk< fs,maxµµµµk< µµµµs N f kkµµµµ====KINETIC FRICTIONWNfk0≠≠≠≠vFCopyright R. Janow – Spring 2012Friction force as a function of applied force FWNfF0 W N F iy====−−−−====∑∑∑∑ ma f F F ix====−−−−====∑∑∑∑Slowly increase F and observe friction force 0 a ==== N f F ssµµµµ≤≤≤≤====Static friction: N f F 0 a smax,sµµµµ============Impending motion:µµµµkand µµµµsdepend on surface conditions and materialsµµµµk< µµµµs otherwise breakaway could not happenKinetic friction after “breakaway”: f F max,s≥≥≥≥fk< fs,maxand0 mf F a 0 a k>>>>−−−−====≠≠≠≠ N f F skµµµµ====≥≥≥≥Copyright R. Janow – Spring 2012Friction force as a function of applied force FWNfFSlowly increase F and observe friction forceay= 0 so N = WBREAKAWAYfs,maxfs= F fk= µµµµkNfs,max= µµµµsNF N f ssµµµµ≤≤≤≤fk< fs,max skµµµµ<<<<µµµµCopyright R. Janow – Spring 2012Some Friction Coefficients skµµµµ<<<<µµµµCopyright R. Janow – Spring 2012Example: Kinetic FrictionFg= msledgfk= µµµµkNThe sled and load are pulled at constant velocityFind the tension T in the cordSubstitute (1) into (2))sin(T F )cos(Tkgkφφφφµµµµ−−−−µµµµ====φφφφ F ] )sin( )cos( [ Tgkkµµµµ====φφφφµµµµ++++φφφφ )sin( )cos( gm T kkφφφφµµµµ++++φφφφµµµµ====Evaluate:T = 91 N.Msled= 75 kgµµµµk= 0.10φφφφ = 42oFBDNFriction force depends on N – the normal force – which often does NOT = the weightma )sin(TFNFgy========φφφφ++++−−−−====∑∑∑∑0 )sin(TFN gφφφφ−−−−====(1)Apply Second Law0========−−−−φφφφ====∑∑∑∑xkxmaf)cos(TFvxis constant N)cos(T kµµµµ====φφφφ(2)Copyright R. Janow – Spring 2012Do you push or pull (for smallest F)?Assume child slides at constant velocity v, and ay= 0WNfkF30o030 ========−−−−−−−−====∑∑∑∑yoyma)sin(FWNF W )sin(FWN o>>>>++++==== 30Larger N, Larger frictionWN’fk’F30o030 ========++++−−−−====∑∑∑∑yoyma)sin(FW'NF W )sin(FW'N o<<<<−−−−==== 30Smaller N, smaller friction030 ====µµµµ−−−−====∑∑∑∑N)cos(FFkox )cos(N F ok30µµµµ====How much force F is required to overcome frictionCopyright R. Janow – Spring 2012Example: Static Friction on a RampA coin is just about to slide down the book when the angle θθθθ = 15o. Find the static friction coefficient µµµµs.Model as a block on a rampChoose x-axis along the ramp’s surfaceN = the normal force is along y-axisWeight Fgacts straight downN f is fsmax,ssµµµµ====gm F coin of weightg========The y and x accelerations = 0motion) (opposespositive issnegative isgx f F⇒⇒⇒⇒FBDN0====θθθθ−−−−====∑∑∑∑)cos(FNFgymotion) (impendingxgsx ma )sin(F N F 0========θθθθ−−−−µµµµ====∑∑∑∑ )cos(F N gθθθθ====(1) )sin(F N gsθθθθ====µµµµ(2)Divide (2) by (1) )cos(F)sin(F NN ggsθθθθθθθθ====µµµµ .23 )(nta s≈≈≈≈θθθθ====µµµµCopyright R. Janow – Spring 2012Acceleration of Connected Objects with FrictionA block of mass m2on a rough horizontal surfaceis connected to hanging ball m1by a cord passing over a massless pulley. Force F is applied to block m2at an angle θθθθ.Determine the magnitude of the acceleration a.Assume a is positive (rightward) and the same for bothm1m2m1Tm1gaFBDamgmTFy111====−−−−====∑∑∑∑ )ga(mT ++++====1(1) mm]mm[ g)]sin()[cos( F a 21kk++++µµµµ++++−−−−θθθθµµµµ++++θθθθ====21For F = 0, a is negativeFor µµµµk= 0 mmgm)cos(F a 21++++−−−−θθθθ====1If F also = 0 mmgm a 21++++−−−−====10 )Fsin( gmNFy====θθθθ++++−−−−====∑∑∑∑22m2gm2TNθθθθfkaFBDF )Fsin( gmN θθθθ−−−−====2(2)am fT)Fcos( F2kx====−−−−−−−−θθθθ====∑∑∑∑2 N f kkµµµµ==== fT )Fcos( am k2−−−−−−−−θθθθ====(3)Copyright R. Janow – Spring 2012Friction force of a wall5 – 1: A student presses her physics book against a rough vertical wall, with her hand exerting a force normal to the wall. What is the