PHY 101 1nd Edition Lecture 12 Outline of Last Lecture I. Newton’s Third LawII. Application of Newton’s LawsIII. Objects in EquilibriumOutline of Current Lecture VI. 4.6 Forces of Frictiona. What causes friction?VII. Static FrictionVIII. Kinetic FrictionCurrent Lecture4.6 Forces of Friction- Friction: the resistance an object experiences when moving on a surface or through a viscous medium- Friction is not necessarily a bad thingo It is important because without friction, we would not be able to walk, drive, slow down a car, etc.- What causes friction?o Roughness of contact surfaceo Less friction on a smooth contact surfaceo Normal force: easier to push forward lighter objects Ex. Bicycle brakes vs. car brakes- Ex. Pulling a crateo Before pulling a crate, no friction forceo Start pulling with small force crate may not moveThese notes represent a detailed interpretation of the professor’s lecture. GradeBuddy is best used as a supplement to your own notes, not as a substitute.o Magnitude of friction force = tension (opposite direction)o Pull harder, the friction force on the crate grows until it reaches the maxo Then it will start movingStatic Friction- The max friction force: fs, max = μsFno μs = coefficient of static frictiono Fn = normal force- Therefore, before an object moves: o F = Fso Fs < Fs, max- ** Fs and Fs, max are always opposite in direction**Kinetic Friction- Once an object starts moving, friction force is constant called kinetic friction forceo Fk = μkFno μk = coefficient of kinetic friction, *General μk <μs- need greater force to push an object to start moving than to keep it movingExample:- a block of 2.5kg is at rest on an incline with angle 30°o a) what is the magnitude of the friction force Fs acting on the block?o Analysis: object at rest = equilibrium = net force zeroo Draw coordinate system, label forces, indicate important angleso How many forces acting on the block? Free body diagramo Forces all along the x-direction 1st component of gravitational force- Fgx = mgsin30 = mgcos60 Equilibrium condition: Fgx - Fs = 0- Mgsin30 - Fs = 0- Fs = mgsin30 = (2.5)(9.8)(0.5) = 12.3N NOTE** Fgx is not a separate force, it is the x-component of gravitational forceo B) what is the magnitude of the normal force, Fn? Use equilibrium condition in the y-direction- Fn = Fgy = 0- Fgy = -mgcos30 = -mgsin60- Fn = mgcos30 = (2.5)(9.8)(0.866) =
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