Lecture 8Inclined plane with “Normal” and Frictional ForcesSlide 3Slide 4Friction in a viscous medium Drag Force Quantified“Free” FallNewton’s LawsNewton’s Third Law:GravityCavendish’s ExperimentExample (non-contact)ExampleSlide 16Normal ForcesSlide 18The flying bird in the cageExercise Newton’s Third LawExercise 2 Newton’s Third LawExercise 2 Newton’s Third Law SolutionExercise 3 Newton’s 3rd LawExercise 3 Solution:Example: Friction and MotionExample SolutionSlide 27Slide 28Slide 29Home Exercise Friction and Motion, ReplayHome Exercise Friction and Motion, Replay in the static caseHome Exercise Friction and MotionLecture 8 RecapPhysics 207: Lecture 8, Pg 1Lecture 8Goals:Goals: Differentiate between Newton’s 1st, 2nd and 3rd Laws Use Newton’s 3rd Law in problem solvingAssignment: HW4, (Chapters 6 & 7, due 2/18, Wednesday)Finish Chapter 7 1st Exam Wed., Feb. 18th from 7:15-8:45 PM Chapters 1-7in room 2103 Chamberlin HallPhysics 207: Lecture 8, Pg 2Inclined plane with “Normal” and Frictional Forces1. Static Equilibrium Case2. Dynamic Equilibrium (see 1)3. Dynamic case with non-zero acceleration Block weight is mgNormalForceFriction Force“Normal” means perpendicularmg cos f xymg sin F = 0Fx= 0 = mg sin – f Fy= 0 = mg cos – Nwith mg sin = f ≤ S Nif mg sin > S N, must slideCritical angle s = tan cPhysics 207: Lecture 8, Pg 3Inclined plane with “Normal” and Frictional Forces1. Static Equilibrium Case2. Dynamic Equilibrium Friction opposite velocity (down the incline)mgNormalForceFriction Force“Normal” means perpendicularmg cos fK xymg sin F = 0Fx= 0 = mg sin – fk Fy= 0 = mg cos – Nfk = k N = k mg cos Fx= 0 = mg sin – k mg cos k = tan (only one angle)vPhysics 207: Lecture 8, Pg 4Inclined plane with “Normal” and Frictional Forces3. Dynamic case with non-zero accelerationResult depends on direction of velocityWeight of block is mgNormalForceFriction ForceSliding Down vmg sin fk Sliding Up Fx= max = mg sin ± fk Fy= 0 = mg cos – Nfk = k N = k mg cos Fx= max = mg sin ± k mg cos ax = g sin ± k g cos Physics 207: Lecture 8, Pg 7Friction in a viscous mediumDrag Force QuantifiedWith a cross sectional area, A (in m2), coefficient of drag of 1.0 (most objects), sea-level density of air, and velocity, v (m/s), the drag force is:D = ½ C A v2 c A v2 in Newtons c = ¼ kg/m3 In falling, when D = mg, then at terminal velocityExample: Bicycling at 10 m/s (22 m.p.h.), with projected area of 0.5 m2 exerts ~30 Newtons Minimizing drag is often importantPhysics 207: Lecture 8, Pg 8“Free” FallTerminal velocity reached when Fdrag = Fgrav (= mg)For 75 kg person with a frontal area of 0.5 m2,vterm 50 m/s, or 110 mphwhich is reached in about 5 seconds, over 125 m of fallPhysics 207: Lecture 8, Pg 10Newton’s LawsLaw 1: An object subject to no external forces is at rest or moves with a constant velocity if viewed from an inertial reference frame.Law 2: For any object, FFNET = FF = ma a Law 3: Forces occur in pairs: FFA , B = - FFB , A (For every action there is an equal and opposite reaction.)Read: Force of B on APhysics 207: Lecture 8, Pg 11Newton’s Third Law:If object 1 exerts a force on object 2 (F2,1 ) then object 2 exerts an equal and opposite force on object 1 (F1,2)F1,2 = -F2,1IMPORTANT: Newton’s 3rd law concerns force pairs whichact on two different objects (not on the same object) ! For every “action” there is an equal and opposite “reaction”Physics 207: Lecture 8, Pg 12GravityNewton 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 13Cavendish’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 14Example (non-contact)Consider the forces on an object undergoing projectile motionFB,E = - mB gEARTHFE,B = mB gFB,E = - mB gFE,B = mB gQuestion: By how much does g change at an altitude of 40 miles? (Radius of the Earth ~4000 mi)Physics 207: Lecture 8, Pg 15ExampleConsider the following two cases (a falling ball and ball on table),Compare and contrast Free Body Diagram andAction-Reaction Force Pair sketchPhysics 207: Lecture 8, Pg 16ExampleThe Free Body DiagrammgmgFB,T= NBall FallsFor Static SituationN = mgPhysics 207: Lecture 8, Pg 17Normal ForcesCertain forces act to keep an object in place. These have what ever force needed to balance all others (until a breaking point).FT,BFB,TMain goal at this point : Identify force pairs and apply Newton’s third lawPhysics 207: Lecture 8, Pg 18ExampleFirst: Free-body diagramSecond: Action/reaction pair forcesFB,E = -mgFB,T= NFE,B = mgFB,E = -mgFE,B = mgFT,B= -NPhysics 207: Lecture 8, Pg 19The flying bird in the cageYou have a bird in a cage that is resting on your upward turned palm. The cage is completely sealed to the outside (at least while we run the experiment!). The bird is initially sitting at rest on the perch. It decides it needs a bit of exercise and starts to fly. Question: How does the weight of the cage plus bird vary when the bird is flying up, when the bird is flying sideways, when the bird is flying down?So, what is holding the airplane up in the sky?Physics 207: Lecture 8, Pg 20Exercise Newton’s Third LawA. greater thanB. equal to C. less thanA fly is deformed by hitting the windshield of a speeding bus. vThe force exerted by the bus on the fly is,that exerted by the fly on the bus.Physics 207: Lecture 8, Pg 21Exercise 2Newton’s Third LawA. greater thanB. equal to C. less thanA fly is deformed by hitting the windshield of a speeding bus. vThe magnitude of the acceleration, due to this collision, of the bus isthat of the fly.Same scenario but now we examine the accelerationsPhysics 207: Lecture 8, Pg 22Exercise 2Newton’s Third LawSolutionBy Newton’s third law these two forces form an interaction pair which are equal (but in opposing directions). However, by Newton’s second law Fnet = ma or a = Fnet/m.So Fb, f = -Ff, b = F0but |abus | = |F0 / mbus |
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