Page 1Physics 207 – Lecture 9Physics 207: Lecture 9, Pg 1Lecture 9llGoalsGoalsv Describe Friction in Air (Ch. 6), (not on 1stExam)v Differentiate between Newton’s 1st, 2ndand 3rd Lawsv Use Newton’s 3rdLaw in problem solving1stExam Thurs., Oct. 6thfrom 7:15-8:45 PM Chapters 1-6 & 7(“light”, direct applications of the third law)Rooms: 2103 (302, 303, 306, 309, 310, 313) 2141 (304, 307, 308, 312) , 2223 (311) Chamberlin Hall (plus quiet room)Physics 207: Lecture 9, Pg 2Friction in a viscous mediumDrag Force Quantifiedl With a cross sectional area, A (in m2), coefficient of drag of 1.0 (most objects), ρsea-level density of air, and velocity, v (m/s), the drag force is:D = ½C ρA v2 ≅ c A v2in Newtonsc = ¼ kg/m3In falling, when D = mg, then at terminal velocityl Example: Bicycling at 10 m/s (22 m.p.h.), with projected area of 0.5 m2exerts a force of ~30 Newtonsv At low speeds air drag is proportional to v but at high speeds it is v2v Minimizing drag is often importantPage 2Physics 207 – Lecture 9Physics 207: Lecture 9, Pg 3Newton’s Third Law:If object 1 exerts a force on object 2 (F2,1) then object 2exerts an equal and opposite force on object 1 (F1,2)F1,2= -F2,1IMPORTANT: Newton’s 3rdlaw 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 9, Pg 4Force Pairs vs. Free Body DiagramsConsider the following two cases (a falling ball and ball on table),Compare and contrast Free Body Diagram andAction-Reaction Force Pair sketchPage 3Physics 207 – Lecture 9Physics 207: Lecture 9, Pg 5Forces just on a single body (1stand 2ndLaws only)mgmgFB,T= NBall FallsFor Static SituationN = mgPhysics 207: Lecture 9, Pg 6Force Pairs (3rdLaw)1stand 2ndLaws Free-body diagramRelates force to acceleration3rdLaw Action/reaction pairsShows how forces act between objectsFB,E= -mgFB,T= NFE,B= mgFB,E= -mgFE,B= mgFT,B= -NPage 4Physics 207 – Lecture 9Physics 207: Lecture 9, Pg 7Example (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 9, Pg 8Note on Gravitational ForcesNewton also recognized that gravity is an attractive, long-range force between any two objects.When two objects with masses m1and m2are separated by distance r, each object “pulls” on the other with a force given by Newton’s law of gravity, as follows:Page 5Physics 207 – Lecture 9Physics 207: Lecture 9, Pg 9Example (non-contact)Consider the force on a satellite undergoing projectile motion 40 km above the surface of the earth:FB,E= - mB gEARTHFE,B= mB gFB,E= - mB gFE,B= mB gCompare: g = G m2/ 40002 g’= G m2/ (4000+40)2 g’/ g = 40002 / (4000+40)2 = 0.98Physics 207: Lecture 9, Pg 10A conceptual question: A flying bird in a cagel You 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?l Follow up question:So, what is holding the airplane up in the sky?Page 6Physics 207 – Lecture 9Physics 207: Lecture 9, Pg 113rdLaw : Static Friction with a bicycle wheell You are pedaling hard and the bicycle is speeding up.What is the direction of the frictional force?l You are breaking and the bicycle is slowing downWhat is the direction of the frictional force?Physics 207: Lecture 9, Pg 12Static Friction with a bicycle wheell You are pedaling hard and the bicycle is speeding up.What is the direction of the frictional force?Hint…you are accelerating to the righta = F / mFfriction, on B from Eis to the rightFfriction, on E from,Bis to the leftPage 7Physics 207 – Lecture 9Physics 207: Lecture 9, Pg 13ExerciseNewton’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 9, Pg 14ExerciseNewton’s 3rdLawA. 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 accelerationsPage 8Physics 207 – Lecture 9Physics 207: Lecture 9, Pg 15Exercise Newton’s 3rdLawSolutionBy 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| << | afly| = | F0/mfly |Thus the forces are the sameAnswer for acceleration is (C)Physics 207: Lecture 9, Pg 16Exercise Newton’s 3rd LawA. 2B. 4C. 6D. Something elseabl Two blocks are being pushed by a finger on a horizontal frictionless floor. l How many action-reaction force pairs are present in this exercise?Page 9Physics 207 – Lecture 9Physics 207: Lecture 9, Pg 17Force pairs on an Inclined planeForces on the block (static case)NormalForceFriction Forceθf= µNxyForces on the plane by blockPhysics 207: Lecture 9, Pg 18Force pairs on an Inclined planeForces on the block (sliding case, no friction)NormalForceθxyJust one force on the plane by blockso if plane is to remain stationary these two components must be offset by other force pairs(N cos θ and N sin θ along vertical and horizontal)Page 10Physics 207 – Lecture 9Physics 207: Lecture 9, Pg 19Example: Friction and Motionl A box of mass m1= 1 kg is being pulled by a horizontal string having tension T = 40 N. It slides with friction (µk= 0.5) on top of a second box having mass m2= 2 kg, which in turn slides on a smooth (frictionless) surface.(g is said to be 10 m/s2)v What is the acceleration of the second box ?1stQuestion: What is the force on mass 2 from mass 1?mm22Tmm11slides with friction (µk=0.5 )slides without frictiona = ?vPhysics 207: Lecture 9, Pg 20ExampleSolutionl First draw FBD of the top box:m1N1m1gTfk= µKN1= µKm1gvPage 11Physics 207 – Lecture 9Physics 207: Lecture 9, Pg 21l Newtons 3rdlaw says the force box 2 exerts on box 1 is equal and opposite to
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