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Physics 207, Lecture 6, Sept. 24

Physics 207, Lecture 6, Sept. 24. Chapter 5 Friction (a external force that opposes motion) Chapter 6 (Dynamics II) Motion in two (or three dimensions) Frames of reference. Agenda:. Assignment: For Wednesday read Chapter 7 MP Problem Set 3 due Wednesday

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Physics 207, Lecture 6, Sept. 24

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  1. Physics 207, Lecture 6, Sept. 24 • Chapter 5 • Friction (a external force that opposes motion) • Chapter 6 (Dynamics II) • Motion in two (or three dimensions) • Frames of reference Agenda: Assignment: For Wednesday read Chapter 7 • MP Problem Set 3 due Wednesday • MP Problem Set 4 available soon • MidTerm Thursday, Oct. 4, Chapters 1-7, 90 minutes, 7:15-8:45 PM

  2. Friction... • Friction is caused by the “microscopic” interactions between the two surfaces:

  3. A special contact force, friction • What does it do? • It opposes motion ! • Parallel to a surface • Perpendicular to a surface Normal force • How do we characterize this in terms we have learned? • A resulting force in a direction opposite to the direction of motion (actual or implied)! j N FAPPLIED i ma fFRICTION mg

  4. Sliding Friction (i.e., with motion) • Direction: A force vector  to the normal force vector N • Magnitude: |fK|is proportional to the magnitude of |N | • |fK|= K| N |( = K|mg | in the previous example) • The constant K is called the “coefficient of kinetic friction” • As the normal force varies so does the frictional force

  5. Case study ... big F • Dynamics: x-axis i :max= F KN y-axis j :may = 0 = N–mgor N = mg soFKmg=m ax fk v j N F i max fk Kmg mg

  6. Case study ... little F • Dynamics: x-axis i :max= F KN y-axis j :may = 0 = N–mgor N = mg soFKmg=m ax fk v j N F i max fk Kmg mg

  7. Static Friction... • So far we have considered friction acting when something has a non-zero velocity • We also know that it acts in fixed or “static” systems: • In these cases, the force provided by friction depends on the forces applied on the system (magnitude: fs≤msN) • Opposes motion that would occur if ms were zero j N Fapplied i fS mg

  8. Static Friction... • Just like in the sliding case except a = 0. i :FappliedfS = 0 j :N = mg • While the block is static:fS Fapplied (unlike kinetic friction) j N Fapplied i fS mg

  9. Static Friction... • The maximum possible force that the friction between two objects can provide is fMAX = SN, where sis the “coefficient of static friction”. • SofSS N. • As one increases F, fS gets bigger until fS=SNand the object “breaks loose” and starts to move. j N F i fS mg

  10. j N FMAX i Smg mg Static Friction... • S is discovered by increasing F until the block starts to slide: i :FMAXSN = 0 j :N = mg SFMAX / mg ActiveFigure http://romano.physics.wisc.edu/winokur/fall2006/AF_0516.html

  11. Additional comments on Friction: • The force of friction does not depend on the area of the surfaces in contact (a relatively good approximation if there is little surface deformation) • Logic dictates that S > Kfor any system

  12. Coefficients of Friction

  13. Forces at different angles Case1: Downward angled force with friction Case 2: Upwards angled force with friction Cases 3,4: Up against the wall Questions: Does it slide? What happens to the normal force? What happens to the frictional force? Cases 3, 4 Case 2 Case 1 ff F F N N N F ff ff mg mg mg

  14. Forces at different angles • Draw a Force Body Diagram • Choose directions for x, y and z axes • Write down Newton’s 2nd Law for each axis • If no acceleration sum of the forces is zero, ma otherwise Cases 3, 4 Case 2 Case 1 ff F F N N N q F q ff ff mg mg mg

  15. Lecture 6, Exercise 1Dragging a block A person pulls a block across a rough horizontal surface at constant velocity by applying a non-zero force F. The arrows in the diagram accurately indicate the direction of the forces but not their magnitude. Which of the relationship between the magnitudes ofN and W holds true ? • N > W • N = W • N < W • dependent on f • dependent on F constant

  16. Lecture 6, Exercise 2Dragging a block A person pulls a block across a rough horizontal surface atconstant velocity by applying a non-zero force F. The arrows in the diagram accurately indicate the direction of the forces but not their magnitude. Which of the relationship between the magnitudes of F and f holds true ? • f > F • f = F • f < F • dependent on N • dependent on W constant

  17. Frictionless inclined plane • A block of mass m slides down a frictionless ramp that makes angle  with respect to horizontal. What is its acceleration a ? m a 

  18. N mg  Angles of the inclined plane max = mg sin    + f = 90 f 

  19. j i mg Frictionless inclined plane... • Use a FBD and consider x and y components separately: • Fx i: max =mg sin ax = g sin  • Fy j: may =0 = N –mg cos N =mg cos max mg sin  N  mg cos  q

  20. j i mg Inclined plane...static friction • Use a FBD and consider x and y components separately:\ • Fx i: max = 0 =mg sin  - f0 = g sin  - f • Fy j: may =0 = N –mg cos N =mg cos Friction Force Special case: At the breaking point f = ms N = msmg cos  g sin q = f = msmg cos  max= 0 mg sin  N  mg cos  q

  21. “Reaction” force From Ramp Normal Force Friction Force Decompose Vector Weight of block Weight of block “Normal” Forces and Frictional Forces:Sitting still “Normal” means perpendicular q q Friction Force ≤ Normal Force  coeff. of static friction Ffriction = mg sinq≤ N mk

  22. “Reaction” Force From Ramp Normal Force Friction Force ax mg sin  Decompose Vector q q Weight of block Weight of block ax “Normal” Forces and Frictional Forces:Sliding down with acceleration Sliding Friction Force < Normal Force  coeff. of static friction Ffriction = Nmk<mgsinq S Fx = m ax = mg sinq - Nmk

  23. Weight of block Weight of block vx “Normal” Forces and Frictional Forces:Sliding down with constant speed(ax=0) “Reaction” Force From Ramp Normal Force Friction Force vx mg sin  Decompose Vector q q Sliding Friction Force = weight vector component along the block Ffriction = Nmk=mgsinq SFx = 0 = mg sinq - Nmk (with respect to previous slide we must increase mk and/or reduce q)

  24. m Lecture 6, Exercise 3Test your intuition • A block of mass m, is placed on a rough inclined plane (m > 0) and given a brief push. It motion thereafter is down the plane with a constant speed. • If a similar block (samem) of mass 2m were placed on the same incline and given a brief push with v0 down the block, it will (A)decrease its speed (B) increase its speed (C) move with constant speed

  25. m Lecture 6, Exercise 3Test your intuition • A block of mass m, is placed on a rough inclined plane (m > 0) and given a brief push. It motion thereafter is down the plane with a constant speed. • If a similar block (samem) of mass 2m were placed on the same incline and given a brief push with v0 down the block, it will • decrease its speed • increase its speed • move with constant speed

  26. Lecture 6, Exercise 3Solution • Draw FBD and find the total force in the x-direction FTOT,x= 2mg sin q - mK 2mgcos q = 2 ma mKN = ma = 0 (case when just m) Doubling the mass will simplydouble both terms…net forcewill still be zero ! Speed will still be constant ! (C) j N q 2 mg q i

  27. “Normal” Forces and Frictional Forces At first the velocity is v up along the slide Can you draw a velocity time plot? (Will be done in the review session) What is the acceleration versus time? Normal Force Friction Force Sliding Down fk Sliding Up v q q mg sin q Weight of block ismg Friction Force = Normal Force  (coefficient of friction) Ffriction = kFnormal = mk mg sin q

  28. Air Resistance and Drag • So far we’ve “neglected air resistance” in physics • Can be difficult to deal with • Affects projectile motion • Friction force opposes velocity through medium • Imposes horizontal force, additional vertical forces • Terminal velocity for falling objects • Dominant energy drain on cars, bicyclists, planes • This issue has been with a very long time.

  29. Drag 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 • Requires (F v) of power 300 Watts to maintain speed • Minimizing drag is often important

  30. Fish Schools

  31. 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.

  32. “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

  33. 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.

  34. Lecture 6 Recap Assignment: • Wednesday class: Read Chapter 6, Start Chapter 7 • MP Problem Set 3 due Wednesday • MP Problem Set 4 available soon • MidTerm Thursday, Oct. 4, Chapters 1-7, 90 minutes, 7:15-8:45 PM

  35. Physics 207, Lecture 7, Sept. 26 • Chapter 6 (Dynamics II) • Motion in two (or three dimensions) • Frames of reference • Start Chapter 7 Agenda: Assignment: For Wednesday read Chapter 7 • MP Problem Set 3 due tonight • MP Problem Set 4 available now • MidTerm Thursday, Oct. 4, Chapters 1-7, 90 minutes, 7:15-8:45 PM Rooms: B102 & B130 in Van Vleck.

  36. Chapter 6: Motion in 2 (and 3) dimensions, Dynamics II • Recall instantaneous velocity and acceleration • These are vector expressions reflecting x, y and z motion r = r(t) v = dr / dt a = d2r / dt2

  37. Kinematics • The position, velocity, and acceleration of a particle in 3-dimensions can be expressed as: r = x i + y j + z k v = vx i + vy j + vz k(i,j,kunit vectors ) a = ax i + ay j + az k • All this complexity is hidden away in r = r(t) v = dr / dt a = d2r / dt2

  38. xvst x 4 0 t Special Case Throwing an object with x along the horizontal and y along the vertical. x and y motion both coexist and t is common to both Let g act in the –y direction, v0x= v0 and v0y= 0 xvsy yvst t = 0 y 4 y x 4 t 0

  39. Another trajectory Can you identify the dynamics in this picture? How many distinct regimes are there? xvsy t = 0 y t =10 x

  40. xvsy xvst t = 0 x 4 y x 4 0 t Trajectory with constant acceleration along the vertical How does the trajectory appear to different observers ? What if the observe is moving with the same x velocity?

  41. yvst t = 0 xvsy t = 0 4 4 y y x Trajectory with constant acceleration along the vertical This observer will only see the y motion x In an inertial reference frames everyone sees the acceleration

  42. xvsy t = 0 4 x y Trajectory with constant acceleration along the vertical What do the velocity and acceleration vectors look like? Velocity vector is always tangent to the curve! Acceleration may or may not be!

  43. v Instantaneous Velocity • But how we think about requires knowledge of the path. • The direction of the instantaneous velocity is alonga line that is tangent to the path of the particle’s direction of motion. • The magnitude of the instantaneous velocity vector is the speed, s. (Knight uses v) s = (vx2 + vy2 + vz )1/2

  44. Average Acceleration: Review • The average acceleration of particle motion reflects changes in the instantaneous velocity vector (divided by the time interval during which that change occurs). • The average acceleration is a vector quantity directed along ∆v

  45. Instantaneous Acceleration • The instantaneous acceleration is the limit of the average acceleration as ∆v/∆t approaches zero • The instantaneous acceleration is a vector with components parallel (tangential) and/or perpendicular (radial) to the tangent of the path • Changes in a particle’s path may produce an acceleration • The magnitude of the velocity vector may change • The direction of the velocity vector may change (Even if the magnitude remains constant) • Both may change simultaneously (depends: path vs time)

  46. v2 v1 Motion along a path ( displacement, velocity, acceleration ) • 3-D Kinematics : vectorequations: r = r(t) v = dr / dt a = d2r / dt2 Velocity : y path vav = r / t v2 v = dr / dt -v1 v Acceleration : aav = v / t x a = dv / dt

  47. = + = 0 = 0 = 0 a a a a a a path andtime t v v v a a a General 3-D motion with non-zero acceleration: • Uniform Circular Motion (Ch. 7) is one specific case: Two possible options: Change in the magnitude of Change in the direction of Animation

  48. Lecture 6 Recap Assignment: • Wednesday class: Read Chapter 7 • MP Problem Set 3 due Wednesday • MP Problem Set 4 available soon • MidTerm Thursday, Oct. 4, Chapters 1-7, 90 minutes, 7:15-8:45 PM

  49. Relative Velocity, equations • The positions as seen from the two reference frames are related through the velocity • r’ = r – vot • The derivative of the position equation will give the velocity equation • v’ = v – vo • These are called the Galilean transformation equations

  50. Central concept for problem solving: “x” and “y” components of motion treated independently. • Again: man on the cart tosses a ball straight up in the air. • You can view the trajectory from two reference frames: y(t) motion governed by 1) a = -gy 2) vy = v0y – g t 3) y = y0 + v0y – g t2/2 Reference frame on the moving train. x motion: x = vxt Reference frame on the ground. Net motion: R = x(t) i + y(t) j(vector)

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