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Lecture 8

Lecture 8. Goals: Differentiate between Newton’s 1 st , 2 nd and 3 rd Laws Use Newton’s 3 rd Law in problem solving. Assignment: HW4, (Chapters 6 & 7, due 2/18, Wednesday) Finish Chapter 7 1 st Exam Wed., Feb. 18 th from 7:15-8:45 PM Chapters 1-7 in room 2103 Chamberlin Hall.

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Lecture 8

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  1. Lecture 8 • Goals: • Differentiate between Newton’s 1st, 2nd and 3rd Laws • Use Newton’s 3rd Law in problem solving Assignment: HW4, (Chapters 6 & 7, due 2/18, Wednesday) Finish Chapter 7 1st Exam Wed., Feb. 18th from 7:15-8:45 PM Chapters 1-7 in room 2103 Chamberlin Hall

  2. 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

  3. 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

  4. 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

  5. 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

  6. “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

  7. 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.)

  8. 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) !

  9. 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:

  10. 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!!!

  11. 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)

  12. Example Consider the following two cases (a falling ball and ball on table), Compare and contrast Free Body Diagram and Action-Reaction Force Pair sketch

  13. Example mg FB,T= N mg The Free Body Diagram Ball Falls For Static Situation N = mg

  14. Normal Forces FB,T FT,B Certain forces act to keep an object in place. These have what ever force needed to balance all others (until a breaking point). Main goal at this point : Identify force pairs and apply Newton’s third law

  15. Example FB,E = -mg FB,T= N FT,B= -N FB,E = -mg FE,B = mg FE,B = mg First: Free-body diagram Second: Action/reaction pair forces

  16. 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? • So, what is holding the airplane up in the sky? 

  17. 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.

  18. 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.

  19. 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)

  20. 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?

  21. 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

  22. Example: Friction and Motion v • A box of mass m1 = 1 kg is being pulled by a horizontal string having tension T = 40 N. It slides with friction (mk= 0.5) on top of a second box having mass m2 = 2 kg, which in turn slides on a smooth (frictionless) surface. • What is the acceleration of the secondbox ? Central question: What is force on mass 2? (A) a = 0 N (B) a = 5 N (C) a = 20 N(D) can’t tell slides with friction (mk=0.5) T m1 a = ? m2 slides without friction

  23. ExampleSolution v • First draw FBD of the top box: N1 m1 fk = mKN1 = mKm1g T m1g

  24. ExampleSolution • As we just saw, this force is due to friction: • Newtons 3rd law says the force box 2 exerts on box 1 is equal and opposite to the force box 1 exerts on box 2. Action Reaction f2,1= -f1,2 f1,2 = mKm1g = 5 N m1 m2 (A) a = 0 N(B) a = 5 N (C) a = 20 N(D) can’t tell

  25. ExampleSolution • Now consider the FBD of box 2: N2 f2,1 = mkm1g m2 m1g m2g

  26. ExampleSolution • Finally, solve Fx = ma in the horizontal direction: mK m1g = m2a = 2.5 m/s2 f2,1 = mKm1g m2

  27. Home Exercise Friction and Motion, Replay • A box of mass m1 = 1 kg, initially at rest, is now pulled by a horizontal string having tension T = 10 N. This box (1) is on top of a second box of mass m2 = 2 kg. The static and kinetic coefficients of friction between the 2 boxes are s=1.5 and mk= 0.5.The second box can slide freely (frictionless) on an smooth surface. Compare the acceleration of box 1 to the acceleration of box 2 ? a1 friction coefficients ms=1.5andmk=0.5 T m1 a2 slides without friction m2

  28. Home Exercise Friction and Motion, Replay in the static case a1 friction coefficients ms=1.5andmk=0.5 T m1 a2 slides without friction m2 • A box of mass m1 = 1 kg, initially at rest, is now pulled by a horizontal string having tension T = 10 N. This box (1) is on top of a second box of mass m2 = 2 kg. The static and kinetic coefficients of friction between the 2 boxes are s=1.5 and mk= 0.5.The second box can slide freely on an smooth surface (frictionless). If no slippage then the maximum frictional force between 1 & 2 is (A) 20 N (B) 15 N (C) 5 N (D) depends on T

  29. Home ExerciseFriction and Motion fs = 10 N and the acceleration of box 1 is Acceleration of box 2 equals that of box 1, with |a| = |T| / (m1+m2) and the frictional force f ism2a (Notice that if T were in excess of 15 N then it would break free) N fSS N = S m1 g = 1.5 x 1 kg x 10 m/s2 which is 15 N (so m2 can’t break free) fS T m1 g a1 friction coefficients ms=1.5andmk=0.5 T m1 a2 slides without friction m2

  30. Lecture 8 Recap Assignment: HW4, (Chapters 6 & 7, due 2/18, Wednesday) Finish Chapter 7 1st Exam Wed., Feb. 18th from 7:15-8:45 PM Chapters 1-7 in room 2103 Chamberlin Hall

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