Newton's Laws and Friction: 1D Motion Lecture

1. Lecture 8 • Goals: • Solve 1D motion with friction • Differentiate between Newton’s 1st, 2nd and 3rd Laws • Begin to use Newton’s 3rd Law in problem solving Assignment: HW4, (Chapter 6, due 2/17, Wednesday) Finish Chapter 7 1st Exam Wed., Feb. 17th from 7:15-8:45 PM Chapters 1-6 in room 2103 Chamberlin Hall

2. Static and Kinetic Friction • Friction exists between objects and its behavior has been modeled. At Static Equilibrium: A block, mass m, with a horizontal force F applied, Direction: A force vector  to the normal force vector N and the vector is opposite to the direction of acceleration if m were 0. Magnitude:fis proportional to the applied forces such that fs≤ms N ms called the “coefficient of static friction”

3. Friction: Static friction FBD N F m1 fs mg Static equilibrium: A block with a horizontal force F applied, As F increases so does fs S Fx = 0 = -F + fs fs = F S Fy = 0 = - N + mg  N = mg

4. Static 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 and the vector is opposite to the direction of acceleration if m were 0. Magnitude:fSis proportional to the magnitude of N fs = ms N N F fs m mg

5. Kinetic or Sliding friction (fk < fs) Dynamic equilibrium, moving but acceleration is still zero As F increases fk remains nearly constant (but now there acceleration is acceleration) FBD S Fx = 0 = -F + fk fk = F S Fy = 0 = - N + mg  N = mg v N F m1 fk mg fk = mk N

6. Sliding Friction: Modeling • Direction: A force vector  to the normal force vector N and the vector is opposite to the velocity. • Magnitude: fkis proportional to the magnitude of N • fk= kN ( = Kmg in the previous example) • The constant k is called the “coefficient of kinetic friction” • Logic dictates that S > Kfor any system

7. Coefficients of Friction

8. An experiment (with a ≠ 0) N m2 m2g Two blocks are connected on the table as shown. The table has unknown static and kinetic friction coefficients. Design an experiment to find mK. T Non-equilibrium: Set m2 and adjust m1 to find regime where a ≠ 0 Requires two FBDs fk T m1 m1g Mass 1 S Fy = m1a = T – m1g T = m1g + m1a = mkm2g – m2a mk= (m1(g+a)+m2a)/m2g Mass 2 S Fx = m2a = -T + fk= -T + mkN S Fy = 0 = N – m2g

9. Inclined plane with “Normal” and Frictional Forces Static Equilibrium Case Dynamic Equilibrium (see 1) Dynamic case with non-zero acceleration “Normal” means perpendicular Normal Force Friction Force f S F = 0 Fx= 0 = mg sin q – f Fy= 0 = –mg cos q + N with mg sin q = f ≤ mS N if mg sin q> mS N, must slide Critical angle ms = tan qc mg sin q q y mg cos q q q x Block weight is mg

10. Inclined plane with “Normal” and Frictional Forces Static Equilibrium Case Dynamic Equilibrium Friction opposite velocity (down the incline) “Normal” means perpendicular Normal Force v Friction Force fK S F = 0 Fx= 0 = mg sin q – fk Fy= 0 = –mg cos q + N fk= mk N = mk mg cos q Fx= 0 = mg sin q – mk mg cos q mk = tan q(only one angle) mg sin q q y mg cos q q q x mg

11. Inclined plane with “Normal” and Frictional Forces Normal Force Friction Force Sliding Down v mg sin q fk Sliding Up q q Weight of block ismg 3. Dynamic case with non-zero acceleration Result depends on direction of velocity Fx= max = mg sin q± fk Fy= 0 = –mg cos q + N fk= mk N = mk mg cos q Fx= max = mg sin q±mk mg cos q ax = g sin q±mk g cos q

12. Velocity and acceleration plots: Notice that the acceleration is always down the slide and that, even at the turnaround point, the block is always motion although there is an infinitesimal point at which the velocity of the block passes through zero. At this moment, depending on the static friction the block may become stuck.

13. Friction in a viscous mediumDrag Force Quantified • With 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  Av2 c A v2in Newtons c = ¼ kg/m3 In falling, when D = mg, then at terminal velocity • Example: Bicycling at 10 m/s (22 m.p.h.), with projected area of 0.5 m2 exerts ~30 Newtons • Minimizing drag is often important

14. Fish Schools

15. By swimming in synchrony in the correct formation, each fish can take advantage of moving water created by the fish in front to reduce drag. Fish swimming in schools can swim 2 to 6 times as long as individual fish.

16. “Free” Fall • Terminal velocity reached when Fdrag = Fgrav (= mg) • For 75 kg person with a frontal area of 0.5 m2, vterm 50 m/s, or 110 mph which is reached in about 5 seconds, over 125 m of fall

17. Newton’s Laws Read: Force of B on A Law 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, FNET = F = ma Law 3: Forces occur in pairs: FA , B = - FB , A (For every action there is an equal and opposite reaction.)

18. Newton’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 = -F2,1 For every “action” there is an equal and opposite “reaction” IMPORTANT: Newton’s 3rd law concerns force pairs which act ontwo different objects(not on the same object) !

19. Gravity Newton also recognized that gravity is an attractive, long-range force between any two objects. When two objects with masses m1and m2 are separated by distance r, each object “pulls” on the other with a force given by Newton’s law of gravity, as follows:

20. Cavendish’s Experiment F = m1 g = G m1 m2 / r2 g = G m2 / r2 If we know big G, little g and r then will can find m2 the mass of the Earth!!!

21. Example (non-contact) FB,E = - mB g FB,E = - mB g FE,B = mB g FE,B = mB g EARTH Consider the forces on an object undergoing projectile motion Question: By how much does g change at an altitude of 40 miles? (Radius of the Earth ~4000 mi)

22. Example (non-contact) FB,E = - mB g FB,E = - mB g FE,B = mB g FE,B = mB g EARTH Consider the forces on an object undergoing projectile motion Compare: g= G m2 / 40002 g’ = G m2 / (4000+40)2 g / g’ = / (4000+40)2 / 40002 = 0.98

23. The flying bird in the cage • 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? • Follow up question: So, what is holding the airplane up in the sky? 

24. ExerciseNewton’s Third Law  • greater than • equal to • less than A fly is deformed by hitting the windshield of a speeding bus. v The force exerted by the bus on the fly is, that exerted by the fly on the bus.

25. ExerciseNewton’s Third Law  B. equal to A fly is deformed by hitting the windshield of a speeding bus. v The force exerted by the bus on the fly is, that exerted by the fly on the bus.

26. Exercise 2Newton’s Third Law  • greater than • equal to • less than Same scenario but now we examine the accelerations A fly is deformed by hitting the windshield of a speeding bus. v The magnitude of the acceleration, due to this collision, of the bus is that of the fly.

27. Exercise 2Newton’s Third LawSolution By Newton’s third law these two forces form an interaction pair which are equal (but in opposing directions).  Thus the forces are the same However, by Newton’s second law Fnet = ma or a = Fnet/m. So Fb, f = -Ff, b = F0 but |abus | = |F0 / mbus | << | afly | = | F0/mfly | Answer for acceleration is (C)

28. Exercise 3Newton’s 3rd Law a b • 2 • 4 • 6 • Something else • Two blocks are being pushed by a finger on a horizontal frictionless floor. • How many action-reaction force pairs are present in this exercise?

29. Exercise 3Solution: Fa,f Fb,a Ff,a Fa,b FE,a FE,b Fg,b Fg,a Fb,g Fa,g Fa,E Fb,E a b 6

30. Lecture 8 Recap Assignment: HW4, (Chapter 6, due 2/17, Wednesday) Finish Chapter 7 1st Exam Wed., Feb. 17th from 7:15-8:45 PM Chapters 1-6 in room 2103 Chamberlin Hall