Lecture 8Friction with no accelerationFriction...Friction: Static frictionSlide 5Static friction, at maximum (just before slipping)Model of Static FrictionKinetic (fk < fs)Model of Sliding FrictionCoefficients of FrictionSliding friction (fk < fs) but now |a| > 0Acceleration, Inertia and MassExercise Newton’s 2nd Law1st: Frictionless experiment (with a ≠ 0)Experiment with friction (with a ≠ 0)To repeat, net force accelerationHome Exercise Newton’s 2nd LawRemember: Forces are ConditionalForces at different anglesExample (non-contact)GravityCavendish’s ExperimentSlide 31RecapPhysics 207: Lecture 8, Pg 1Lecture 8Goals:Goals: Solve 1D & 2D motion with friction Utilize Newton’s 2nd Law Differentiate between Newton’s 1st, 2nd and 3rd Laws Begin to use Newton’s 3rd Law in problem solvingPhysics 207: Lecture 8, Pg 2Friction with no accelerationNo net forceSo frictional force just cancels applied forceFFAPPLIEDffFRICTIONmggNNii j jPhysics 207: Lecture 8, Pg 3Friction...Friction is caused by the “microscopic” interactions between the two surfaces:Physics 207: Lecture 8, Pg 4Friction: Static frictionStatic equilibrium: A block with a horizontal force F applied, As F increases so does fs FmFBDfsNmgPhysics 207: Lecture 8, Pg 5Friction: Static frictionStatic equilibrium: A block with a horizontal force F applied, FmFBDfsNmg Fx = 0 = -F + fs fs = F Fy = 0 = - N + mgN = mg Force appliedFrictionForcePhysics 207: Lecture 8, Pg 6Static friction, at maximum (just before slipping)fS is proportional to the magnitude of N fs = s N FmfsNmgForce appliedFrictionForcePhysics 207: Lecture 8, Pg 7Model of Static Friction Magnitude: f is proportional to the applied forces such that fs ≤ s N s called the “coefficient of static friction” Direction: If just a single “applied” force, friction is in opposite directionPhysics 207: Lecture 8, Pg 8Kinetic (fk < fs)Dynamic equilibrium, moving but acceleration is still zero FmFBDfkNmg Fx = 0 = -F + fk fk = F Fy = 0 = - N + mg N = mg vfk = k NPhysics 207: Lecture 8, Pg 9Model of Sliding FrictionDirection: to the normal force vector N N and opposite to the velocity. Magnitude: ffk is proportional to the magnitude of N N ffk = k NNThe constant k is called the “coefficient of kinetic friction” Logic dictates that S > K for any systemPhysics 207: Lecture 8, Pg 10Coefficients of FrictionMaterial on Materials = static friction k = kinetic frictionsteel / steel 0.6 0.4add grease to steel 0.1 0.05metal / ice 0.022 0.02brake lining / iron 0.4 0.3tire / dry pavement 0.9 0.8tire / wet pavement 0.8 0.7Physics 207: Lecture 8, Pg 11Sliding friction (fk < fs) but now |a| > 0A change in velocityAs F increases fk remains nearly constant (but now there is acceleration) FmFBDfkNmg Fx = -F + fk = net Force Fy = 0 = - N + mg N = mg vfk = k NPhysics 207: Lecture 8, Pg 12Acceleration, Inertia and MassThe tendency of an object to resist any attempt to change its velocity is called InertiaMass is that property of an object that specifies how much resistance an object exhibits to changes in its velocity (acceleration)If mass is constant thenIf force constant Mass is an inherent property of an objectMass is independent of the method used to measure itMass is a scalar quantityThe SI unit of mass is kgnetFama1|| |a|mPhysics 207: Lecture 8, Pg 14ExerciseNewton’s 2nd LawA. increasingB. decreasingC. constant in timeD. Not enough information to decideAn object is moving to the right, and experiencing a net force that is directed to the right. The magnitude of the force is decreasing with time (read this text carefully). The speed of the object isPhysics 207: Lecture 8, Pg 151st: Frictionless experiment (with a ≠ 0)Two blocks are connected on the table as shown. Thetable is frictionless. Find the acceleration of mass 2.Requires two FBDsTMass 2 Fx = m2ax = -T Fy = 0 = N – m2gm1m2m2gNm1gTMass 1 Fy = m1ay = T – m1gNotice ay = ax = aEliminate Tm1a + m2a = m1a = m1 / (m2+m1)gPhysics 207: Lecture 8, Pg 16Experiment with friction (with a ≠ 0)Two blocks, of m1 & m2 , are connected on the table as shown. The table has unknown static and kinetic friction coefficients.Given an a, find K.Similar but now with friction.TMass 2 Fx = m2a = -T + fk = -T + k N Fy = 0 = N – m2gm1m2m2gNm1gTfkMass 1 Fy = m1a = T – m1gT = m1g + m1a = k m2g – m2a k = (m1(g+a)+m2a)/m2gPhysics 207: Lecture 8, Pg 18To repeat, net force accelerationIn physics: A force is an action which causes an object to accelerate (translational & rotational)This is Newton’s Second LawzzyyxxmaFmaFmaFamFF 0netPhysics 207: Lecture 8, Pg 19Home Exercise Newton’s 2nd LawA. AB. BC. DD. FE. GA mass undergoes motion along a line with velocities as given in the figure below. In regards to the stated letters for each region, in which is the magnitude of the force on the mass at its greatest?Physics 207: Lecture 8, Pg 20Remember: Forces are ConditionalNotice what happens if we change the direction of the applied forceThe normal force can increase or decreaseHere the normal force exceeds mgLet a=0FfFmgN F sin +mgii j j F sin Physics 207: Lecture 8, Pg 27Forces at different anglesCase 1Case 2FmgNCase1: Downward angled force with frictionCase 2: Upwards angled force with frictionCases 3,4: Up against the wallQuestions: Does it slide? What happens to the normal force?What happens to the frictional force?mgCases 3, 4mgNNFFffffffPhysics 207: Lecture 8, Pg 28Example (non-contact)Consider the forces on an object undergoing projectile motionFB,E = - mB gEARTHFE,B = mB gFB,E = - mB gFE,B = mB gPhysics 207: Lecture 8, Pg 29GravityNewton also recognized that gravity is an attractive, long-range force between any two objects. When two objects with masses m1 and m2 are separated by distance r, each object “pulls” on the other with a force given by Newton’s law of gravity, as follows:Physics 207: Lecture 8, Pg 30Cavendish’s ExperimentF = m1 g = G m1 m2 / r2g = G m2 / r2If we know big G, little g and r then will can find m2 the mass of the Earth!!!Physics 207: Lecture 8, Pg 31Example (non-contact)FB,E = - mB gEARTHFE,B = mB gFB,E = - mB gFE,B = mB gCompare: g = G m2 /
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