* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
Download Lecture 8 Static and Kinetic Friction fs ≤ µs N
Survey
Document related concepts
Inertial frame of reference wikipedia , lookup
Jerk (physics) wikipedia , lookup
Fictitious force wikipedia , lookup
Modified Newtonian dynamics wikipedia , lookup
Equations of motion wikipedia , lookup
Newton's theorem of revolving orbits wikipedia , lookup
Centrifugal force wikipedia , lookup
Mass versus weight wikipedia , lookup
Fundamental interaction wikipedia , lookup
Classical mechanics wikipedia , lookup
Renormalization group wikipedia , lookup
Rigid body dynamics wikipedia , lookup
Classical central-force problem wikipedia , lookup
Work (physics) wikipedia , lookup
Centripetal force wikipedia , lookup
Transcript
Physics 207 – Lecture 8 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 Physics 207: Lecture 8, Pg 1 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 µ were 0. Magnitude: f is proportional to the applied forces such that fs ≤ µs N µs called the “coefficient of static friction” Physics 207: Lecture 8, Pg 2 Page 1 Physics 207 – Lecture 8 Case study ... big F Dynamics: x-axis i : y-axis j : max = F − µKN may = 0 = N – mg or N = mg F − µKmg = m ax so fk v j N F i ma ax fk µK mg mg g Physics 207: Lecture 8, Pg 3 Case study ... little F Dynamics: x-axis i : y-axis j : so max = F − µKN may = 0 = N – mg or N = mg F − µKmg = m ax fk v j N F i ma ax fk µK mg mg g Physics 207: Lecture 8, Pg 4 Page 2 Physics 207 – Lecture 8 Friction: Static friction Static equilibrium: A block with a horizontal force F applied, Σ Fx = 0 = -F + fs fs = F FBD Σ Fy = 0 = - N + mg N = mg N As F increases so does fs F m fs 1 mg Physics 207: Lecture 8, Pg 5 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 µ were 0. Magnitude: fS is proportional to the magnitude of N fs = µs N N F m mg Physics 207: Lecture 8, Pg 6 Page 3 fs Physics 207 – Lecture 8 Kinetic or Sliding friction (fk < fs) Dynamic equilibrium, moving but acceleration is still zero Σ Fx = 0 = -F + fk FBD fk = F v Σ Fy = 0 = - N + mg N = mg N F As F increases fk remains nearly constant (but now there acceleration is acceleration) m fk 1 mg fk = µk N Physics 207: Lecture 8, Pg 7 Sliding Friction: Modeling Direction: A force vector ⊥ to the normal force vector N and the vector is opposite to the velocity. Magnitude: fk is proportional to the magnitude of N fk = µk N ( = µK mg g in the previous example) The constant µk is called the “coefficient of kinetic friction” Logic dictates that µS > µK for any system Physics 207: Lecture 8, Pg 8 Page 4 Physics 207 – Lecture 8 Coefficients of Friction Material on Material µs = static friction µk = kinetic friction steel / steel 0.6 0.4 add grease to steel 0.1 0.05 metal / ice 0.022 0.02 brake lining / iron 0.4 0.3 tire / dry pavement 0.9 0.8 tire / wet pavement 0.8 0.7 Physics 207: Lecture 8, Pg 9 An experiment Two blocks are connected on the table as shown. The table has unknown static and kinetic friction coefficients. Design an experiment to find µS N T Static equilibrium: Set m2 m2 and add mass to m1 to reach the breaking point. Requires two FBDs Mass 1 Σ Fy = 0 = T – m1g fS T m1 m2g m1g Mass 2 Σ Fx = 0 = -T + fs = -T + µS N Σ Fy = 0 = N – m2g T = m1g = µS m2g µS = m1/m2 Page 5 Physics 207: Lecture 8, Pg 10 Physics 207 – Lecture 8 A 2nd experiment Two blocks are connected on the table as shown. The table has unknown static and kinetic friction coefficients. Design an experiment to find µK. T Dynamic equilibrium: Set m2 and adjust m1 to find place when T a = 0 and v ≠ 0 fk m1 Requires two FBDs Mass 1 Σ Fy = 0 = T – m1g N m2 m2g m1g Mass 2 Σ Fx = 0 = -T + ff = -T + µk N Σ Fy = 0 = N – m2g T = m1g = µk m2g µk = m1/m2 Physics 207: Lecture 8, Pg 11 An experiment (with a ≠ 0) Two blocks are connected on the table as shown. The table has unknown static and kinetic friction coefficients. N Design an experiment to find µK. T Non-equilibrium: Set m2 T and adjust m1 to find regime where a ≠ 0 m1 Requires two FBDs m1g Mass 1 Σ Fy = m1a = fk m2 m2g Mass 2 T – m1g Σ Fx = m2a = -T + fk Σ Fy = 0 = N – m2g = -T + µk N T = m1g + m1a = µk m2g – m2a µk = (m1(g+a)+m2a)/m2g Physics 207: Lecture 8, Pg 12 Page 6 Physics 207 – Lecture 8 Sample 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 µk for jelloium on steel? Physics 207: Lecture 8, Pg 13 Sample Problem Σ Fx =ma = F - ff = F - µk N = F - µk mg Σ Fy = 0 = N – mg µk = (F - ma) / mg & x = ½ a t 2 0.80 m = ½ a 4 s 2 a = 0.40 m/s2 µk = (1.00 - 0.20 · 0.40 ) / (0.20 ·10.) = 0.46 Physics 207: Lecture 8, Pg 14 Page 7 Physics 207 – Lecture 8 Inclined plane with “Normal” and Frictional Forces 1. Static Equilibrium Case “Normal” means perpendicular 2. Dynamic Equilibrium (see 1) Normal Force 3. Dynamic case with non-zero acceleration Friction ΣF=0 f Force Fx= 0 = mg sin θ – f mg sin θ Fy= 0 = –mg cos θ + N θ y mg cos θ θ with mg sin θ = f ≤ µS N if mg sin θ > µS N, must slide θ x Block weight is mg Critical angle µs = tan θc Physics 207: Lecture 8, Pg 15 Inclined plane with “Normal” and Frictional Forces 1. Static Equilibrium Case “Normal” means perpendicular Normal Force 2. Dynamic Equilibrium Friction opposite velocity v (down the incline) ΣF=0 fK Friction Force mg sin θ Fx= 0 = mg sin θ – fk Fy= 0 = –mg cos θ + N θ mg cos θ θ fk = µk N = µk mg cos θ Fx= 0 = mg sin θ – µk mg cos θ µk = tan θ (only one angle) Page 8 θ mg Physics 207: Lecture 8, Pg 16 y x Physics 207 – Lecture 8 Inclined plane with “Normal” and Frictional Forces 3. Dynamic case with non-zero acceleration Normal Result depends on direction of velocity Force Friction Force Sliding Down v mg sin θ θ Fx= max = mg sin θ ± fk θ fk Sliding Up Weight of block is mg Fy= 0 = –mg cos θ + N fk = µk N = µk mg cos θ Fx= max = mg sin θ ± µk mg cos θ ax = g sin θ ± µk g cos θ Physics 207: Lecture 8, Pg 17 The inclined plane coming and going (not static): the component of mg along the surface > kinetic friction Σ Fx = max = mg sin θ ± uk N Σ Fy = may = 0 = -mg cos θ + N Putting it all together gives two different accelerations, ax = g sin θ ± uk g cos θ. A tidy result but ultimately it is the process of applying Newton’s Laws that is key. Physics 207: Lecture 8, Pg 18 Page 9 Physics 207 – Lecture 8 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. Physics 207: Lecture 8, Pg 19 Friction in a viscous medium Drag Force Quantified With a cross sectional area, A (in m2), coefficient of drag of 1.0 (most objects), ρ sea-level density of air, and velocity, v (m/s), the drag force is: D = ½ C ρ A v2 ≅ c A v2 in Newtons 3 c = ¼ kg/m 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 Physics 207: Lecture 8, Pg 20 Page 10 Physics 207 – Lecture 8 Fish Schools Physics 207: Lecture 8, Pg 21 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. Physics 207: Lecture 8, Pg 22 Page 11 Physics 207 – Lecture 8 “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 Physics 207: Lecture 8, Pg 23 Trajectories with Air Resistance Baseball launched at 45° with v = 50 m/s: Without air resistance, reaches about 63 m high, 254 m range With air resistance, about 31 m high, 122 m range Vacuum trajectory vs. air trajectory for 45° launch angle. Physics 207: Lecture 8, Pg 24 Page 12 Physics 207 – Lecture 8 Newton’s Laws 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 = Σ F = ma a Law 3: Forces occur in pairs: FA , B = - FB , A (For every action there is an equal and opposite reaction.) Read: Force of B on A Physics 207: Lecture 8, Pg 25 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 on two different objects (not on the same object) ! Physics 207: Lecture 8, Pg 26 Page 13 Physics 207 – Lecture 8 Gravity Newton 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 27 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!!! Physics 207: Lecture 8, Pg 28 Page 14 Physics 207 – Lecture 8 Example (non-contact) Consider the forces on an object undergoing projectile motion FB,E = - mB g FB,E = - mB g FE,B = mB g FE,B = mB g EARTH Question: By how much does g change at an altitude of 40 miles? (Radius of the Earth ~4000 mi) Physics 207: Lecture 8, Pg 29 Example (non-contact) Consider the forces on an object undergoing projectile motion FB,E = - mB g FB,E = - mB g FE,B = mB g FE,B = mB g EARTH Compare: g = G m2 / 40002 g’ = G m2 / (4000+40)2 g / g’ = / (4000+40)2 / 40002 = 0.98 Physics 207: Lecture 8, Pg 30 Page 15 Physics 207 – Lecture 8 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? Physics 207: Lecture 8, Pg 31 Exercise Newton’s Third Law A fly is deformed by hitting the windshield of a speeding bus. ✿ v The force exerted by the bus on the fly is, A. B. C. greater than equal to less than that exerted by the fly on the bus. Physics 207: Lecture 8, Pg 32 Page 16 Physics 207 – Lecture 8 Exercise Newton’s Third Law A fly is deformed by hitting the windshield of a speeding bus. ✿ v The force exerted by the bus on the fly is, B. equal to that exerted by the fly on the bus. Physics 207: Lecture 8, Pg 33 Exercise 2 Newton’s Third Law 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 A. greater than B. equal to C. less than that of the fly. Physics 207: Lecture 8, Pg 34 Page 17 Physics 207 – Lecture 8 Exercise 2 Newton’s Third Law Solution 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) Physics 207: Lecture 8, Pg 35 Exercise 3 Newton’s 3rd Law Two blocks are being pushed by a finger on a horizontal frictionless floor. How many action-reaction force pairs are present in this exercise? a A. B. C. D. b 2 4 6 Something else Physics 207: Lecture 8, Pg 36 Page 18 Physics 207 – Lecture 8 Exercise 3 Solution: Fa,f Fb,a Ff,a a FE,a Fa,b bF Fg,a Fg,b Fa,g Fb,g Fa,E E,b Fb,E 6 Physics 207: Lecture 8, Pg 37 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 Physics 207: Lecture 8, Pg 38 Page 19