Page 1Physics 207 – Lecture 8Physics 207: Lecture 8, Pg 1Lecture 8Goals:Goals: Solve 1D motion with friction Differentiate between Newton’s 1st, 2ndand 3rd Laws Begin to use Newton’s 3rdLaw in problem solvingAssignment: HW4, (Chapter 6, due 2/17, Wednesday)Finish Chapter 7 1stExam Wed., Feb. 17thfrom 7:15-8:45 PM Chapters 1-6in room 2103 Chamberlin HallPhysics 207: Lecture 8, Pg 2Static and Kinetic Friction Friction exists between objects and its behavior has been modeled.At Static Equilibrium: A block, mass m, with a horizontal force Fapplied, Direction: A force vector ⊥⊥⊥⊥ to the normal force vector N N and the vector is opposite to the direction of acceleration if µ were 0. Magnitude: f is proportional to the applied forces such thatfs≤ µsN µscalled the “coefficient of static friction”Page 2Physics 207 – Lecture 8Physics 207: Lecture 8, Pg 3Case study ... big F Dynamics:x-axis i : max= F − µKNy-axis j : may= 0 = N – mg or N = mgso F − µKmg = m axmaaxxFFmggNNiij j µKmgvvfkfkPhysics 207: Lecture 8, Pg 4Case study ... little F Dynamics:x-axis i : max= F − µKNy-axis j : may= 0 = N – mg or N = mgso F − µKmg = m axmaaxxFFmggNNiij j µKmgvvfkfkPage 3Physics 207 – Lecture 8Physics 207: Lecture 8, Pg 5Friction: Static frictionStatic equilibrium: A block with a horizontal force F applied, As F increases so does fsFm1FBDfsNmgΣ Fx= 0 = -F + fs fs= F Σ Fy= 0 = - N + mg N = mgPhysics 207: Lecture 8, Pg 6Static friction, at maximum (just before slipping)Equilibrium: A block, mass m, with a horizontal force F applied, Direction: A force vector ⊥⊥⊥⊥ to the normal force vector N N and the vector is opposite to the direction of acceleration if µ were 0. Magnitude: fSis proportional to the magnitude of N fs= µsN FmfsNmgPage 4Physics 207 – Lecture 8Physics 207: Lecture 8, Pg 7Kinetic or Sliding friction (fk< fs)Dynamic equilibrium, moving but acceleration is still zeroAs F increases fkremains nearly constant (but now there acceleration is acceleration)Fm1FBDfkNmgΣ Fx= 0 = -F + fk fk= F Σ Fy= 0 = - N + mg N = mgvfk= µkN Physics 207: Lecture 8, Pg 8Sliding Friction: Modeling Direction: A force vector ⊥ to the normal force vector N N and the vector is opposite to the velocity. Magnitude: ffkis proportional to the magnitude of N N ffk= µkN N ( = µK mgg in the previous example) The constant µkis called the “coefficient of kinetic friction” Logic dictates that µS > µKfor any systemPage 5Physics 207 – Lecture 8Physics 207: Lecture 8, Pg 9Coefficients of Friction0.70.8tire / wet pavement0.80.9tire / dry pavement0.30.4brake lining / iron0.020.022metal / ice0.050.1add grease to steel0.40.6steel / steelµk= kinetic frictionµs= static frictionMaterial on MaterialPhysics 207: Lecture 8, Pg 10An experimentTwo blocks are connected on the table as shown. Thetable has unknown static and kinetic friction coefficients.Design an experiment to find µµµµSStatic equilibrium: Set m2and add mass to m1to reach the breaking point. Requires two FBDsm1m2m2gNm1gTTMass 2Σ Fx= 0 = -T + fs= -T + µSNΣ Fy= 0 = N – m2gfSMass 1Σ Fy= 0 = T – m1gT = m1g = µSm2g µS= m1/m2Page 6Physics 207 – Lecture 8Physics 207: Lecture 8, Pg 11A 2ndexperimentTwo blocks are connected on the table as shown. Thetable has unknown static and kinetic friction coefficients.Design an experiment to find µK.Dynamic equilibrium: Set m2and adjust m1to find place whena = 0 and v ≠ 0Requires two FBDsm1m2m2gNm1gTTMass 2Σ Fx= 0 = -T + ff= -T + µkNΣ Fy= 0 = N – m2gfkMass 1Σ Fy= 0 = T – m1gT = m1g = µkm2g µk= m1/m2Physics 207: Lecture 8, Pg 12An experiment (with a ≠ 0)Two blocks are connected on the table as shown. Thetable has unknown static and kinetic friction coefficients.Design an experiment to find µK.Non-equilibrium: Set m2and adjust m1to find regime where a ≠ 0Requires two FBDsTMass 2Σ Fx= m2a = -T + fk= -T + µkNΣ Fy= 0 = N – m2gm1m2m2gNm1gTfkMass 1Σ Fy= m1a = T – m1gT = m1g + m1a = µkm2g – m2a µk= (m1(g+a)+m2a)/m2gPage 7Physics 207 – Lecture 8Physics 207: Lecture 8, Pg 13Sample Problem You have been hired to measure the coefficients of friction for the newly discovered substance jelloium. Today you will measure the coefficient of kinetic friction for jelloium sliding on steel. To do so, you pull a 200 g chunk of jelloium across a horizontal steel table with a constant string tension of 1.00 N. A motion detector records the motion and displays the graph shown. What is the value of µkfor jelloium on steel?Physics 207: Lecture 8, Pg 14Sample ProblemΣ Fx=ma = F - ff= F - µkN = F - µkmgΣ Fy= 0 = N – mgµk= (F - ma) / mg & x = ½ a t2 0.80 m = ½ a 4 s2a = 0.40 m/s2 µk= (1.00 - 0.20 · 0.40 ) / (0.20 ·10.) = 0.46Page 8Physics 207 – Lecture 8Physics 207: Lecture 8, Pg 15Inclined plane with “Normal” and Frictional Forces1. Static Equilibrium Case2. Dynamic Equilibrium (see 1)3. Dynamic case with non-zero accelerationBlock weight is mgNormalForceFriction Force“Normal” means perpendicularθθmg cos θfθxymg sin θΣ F = 0Fx= 0 = mg sin θ – fFy= 0 = –mg cos θ + Nwith mg sin θ = f ≤ µSNif mg sin θ > µSN, must slideCritical angle µs= tan θcPhysics 207: Lecture 8, Pg 16Inclined plane with “Normal” and Frictional Forces1. Static Equilibrium Case2. Dynamic EquilibriumFriction opposite velocity (down the incline)mgNormalForceFriction Force“Normal” means perpendicularθθmg cos θfKθxymg sin θΣ F = 0Fx= 0 = mg sin θ – fkFy= 0 = –mg cos θ + Nfk= µkN = µkmg cos θFx= 0 = mg sin θ – µkmg cos θµk= tan θ (only one angle)vPage 9Physics 207 – Lecture 8Physics 207: Lecture 8, Pg 17Inclined plane with “Normal” and Frictional Forces3. Dynamic case with non-zero accelerationResult depends on direction of velocityWeight of block is mgNormalForceFriction ForceSliding Down θθvmg sin θfkSliding Up Fx= max= mg sin θ ± fkFy= 0 = –mg cos θ + Nfk= µkN = µkmg cos θFx= max= mg sin θ ± µkmg cos θax= g sin θ ± µkg cos θPhysics 207: Lecture 8, Pg 18The inclined plane coming and going (not static):the component of mg along the surface > kinetic frictionPutting it all together gives two different accelerations, ax= g sin θ ± ukg cos θ. A tidy result but ultimately it is the process of applying Newton’s Laws that is key.Σ Fx= max= mg sin θ ± ukN Σ Fy= may= 0 = -mg
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