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5.7 Some Applications of Newton’s Law, cont

5.7 Some Applications of Newton’s Law, cont. Multiple Objects. When two or more objects are connected or in contact, Newton’s laws may be applied to the system as a whole and/or to each individual object Whichever you use to solve the problem, the other approach can be used as a check.

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5.7 Some Applications of Newton’s Law, cont

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  1. 5.7 Some Applications of Newton’s Law, cont

  2. Multiple Objects • When two or more objects are connected or in contact, Newton’s laws may be applied to the system as a whole and/or to each individual object • Whichever you use to solve the problem, the other approach can be used as a check

  3. Example 5.16Multiple Objects • First treat the system as a whole: • Apply Newton’s Laws to the individual blocks • Solve for unknown(s) • Check: |P21| = |P12|

  4. Example 5.17 Two Boxes Connected by a Cord • Boxes A & B are connected by a cord (mass neglected). Boxes are resting on a frictionless table. • FP = 40.0 N • Find: • Acceleration (a) of each box • Tension(FT) in the cord connecting the boxes

  5. Example 5.17 Two Boxes Connected by a Cord, final • There is only horizontal motion With: aA = aB = a • Apply Newton’s Laws for box A: ΣFx = FP –FT = mAa(1) • Apply Newton’s Laws for box B: ΣFx = FT = mBa(2) Substituting (2) into (1): FP –mBa = mAa FP = (mA + mB)a a = FP/(mA + mB) = 1.82m/s2 Substituting a into (2)  FT = mBa = (12.0kg)(1.82m/s2)= 21.8N

  6. Example 5.18 The Atwood’s Machine • Forces acting on the objects: • Tension (same for both objects, one string) • Gravitational force • Each object has the same acceleration since they are connected • Draw the free-body diagrams • Apply Newton’s Laws • Solve for the unknown(s)

  7. Example 5.18 The Atwood’s Machine, 2 • The Atwood’s Machine: • Find:aandT • Apply Newton’s 2nd Law to each Mass. ΣFy =T– m1g = m1a(1) ΣFy =T– m2g = – m2a(2) • Then: T= m1g + m1a(3) T= m2g– m2a(4)

  8. Example 5.18 The Atwood’s Machine, 3 • The Atwood’s Machine: Equating: (3) = (4) and Solving for a m1g + m1 a = m2g– m2 a  m1 a + m2 a = m2g – m1g a (m1 + m2) = (m2 – m1)g  (5)

  9. Example 5.18 The Atwood’s Machine, final • The Atwood’s Machine: Substituting (5) into (3) or (4): T= m1g + m1a(3)

  10. Active Figure 5.14

  11. Example 5.19 Two Objects and Incline Plane • Find: a and T • One cord: so tension is the same for both objects • Connected: so acceleration is the same for both objects • Apply Newton’s Laws • Solve for the unknown(s)

  12. Example 5.19 Two Objects and Incline Plane, 2 • xy plane: ΣFx = 0&ΣFy = m1 a T– m1g = m1 a T= m1g + m1 a(1) x’y’ plane: ΣFx = m2 a&ΣFy = 0 m2gsinθ –T = m2 a(2) n – m2gcosθ = 0(3)

  13. Example 5.19 Two Objects and Incline Plane, Final • Substituting (1)in (2) gives: m2gsinθ– (m1g + m1 a) = m2a m2gsinθ– m1g – m1 a = m2a  a (m1 + m2) = m2gsinθ – m1g • Substituting ain (1) we’ll get: T= m1g + m1 a(1)

  14. Problem-Solving Hints Newton’s Laws • Conceptualize the problem – draw a diagram • Categorize the problem • Equilibrium (SF = 0) or • Newton’s Second Law (SF = m a) • Analyze • Draw free-body diagrams for each object • Include only forces acting on the object

  15. Problem-Solving Hints Newton’s Laws, cont • Analyze, cont. • Establish coordinate system • Be sure units are consistent • Apply the appropriate equation(s) in component form • Solve for the unknowns. • This always requires Kinder Garden Algebra (KGA). Like solving two linear equations with two unknowns • Finalize • Check your results for consistency with your free- body diagram • Check extreme values

  16. 5.8 Forces of Friction • When an object is in motion on a surface or through a viscous medium, there will be a resistance to the motion • This is due to the interactions between the object and its environment • This resistance is called the Force of Friction

  17. Forces of Friction, 2 • Frictionexists between any 2 sliding surfaces. • Two types of friction: • Static(no motion) friction • Kinetic(motion) friction • The size of the friction force depends on: • The microscopic details of 2 sliding surfaces. • The materials they are made of • Are the surfaces smooth or rough? • Are they wet or dry?

  18. Forces of Friction, 3 • Friction is proportional to the normal force • ƒs£µsn(5.8)andƒk = µk n(5.9) • These equations relate the magnitudes of the forces, THEY ARE NOT vector equations • The force of static friction(maximum) is generally greater than the force of kinetic frictionƒs>ƒk • The coefficients of friction (µk,s) depends on the surfaces in contact

  19. Forces of Friction, final • The direction of the frictional force is opposite the direction of motion and parallel to the surfaces in contact • The coefficients of friction (µk,s)are nearly independent of the area of contact

  20. Static Friction • Static friction acts to keep the object from moving: ƒs = F • If F increases, so doesƒs • If F decreases, so does ƒs • ƒs µs n where the equality holds when the surfaces are on the verge of slipping • Called impending motion

  21. Static Friction, cont • Experiments determine the relation used to compute friction forces. • The friction force ƒsexists ║ to the surfaces, even if there is no motion. Consider the applied force F  ∑F = ma = 0& also v = 0 • There must be a friction forceƒs to oppose F  F –ƒs= 0ƒs=F

  22. Kinetic Friction • The force of kinetic friction (ƒk ) acts when the object is in motion • Friction force ƒk is proportional to the magnitude of the normal forcen between 2 sliding surfaces. • ƒknƒk kn(magnitudes) • k Coefficient of kinetic friction • k : depends on the surfaces & their conditions • k : is dimensionless & < 1

  23. Static & Kinetic Friction • Experiments find that the Maximum Static Friction Forceƒs,maxis proportional to the magnitude (size) of the normal forcenbetween the 2 surfaces. • DIRECTIONS:ƒs,max n • Then: ƒs,max= sn (magnitudes)

  24. Static & Kinetic Friction • s Coefficient of static friction • s : depends on the surfaces & their conditions • s : is dimensionless & < 1 • Always: • ƒs,max>ƒk s n>kn (Cancel n)  s>k • ƒs ƒs,max=µsnƒs µs n

  25. Some Coefficients of Friction

  26. Active Figure 5.16

  27. Friction in Newton’s Laws Problems • Friction is a force, so it simply is included in the Net Force (SF ) in Newton’s Laws • The rules of friction allow you to determine the direction and magnitudeof the force offriction

  28. Assume:mg = 98.0N  n = 98.0 N, s= 0.40, k = 0.30  ƒs,max= sn= 0.40(98N)= 39N Find Force of Friction if the force applied FAis: FA= 0ƒs=FA= 0ƒs=0 Box does not move!! FA = 10N FA < ƒs,max or (10N< 39N) ƒs–FA= 0ƒs=FA=10N The box still does not move!! Example 5.20 Pulling Against Friction n ƒs,k

  29. FA = 38N < ƒs,maxƒs–FA= 0 ƒs=FA=38N This force is still not quite large enough to move the box!!! FA = 40N> ƒs,maxkinetic friction. This one will start moving the box!!! ƒk kn= 0.30(98N) =29N. The net force on the box is: ∑F = max40N – 29N = max11N = maxax =11 kg.m/s2/10kg =1.10 m/s2 Example 5.20 Pulling Against Friction, 2 n ƒs,k

  30. Example 5.20 Pulling Against Friction, final ƒs,max= 39N ƒs,k ƒk=29N ƒs µs n

  31. Example 5.21To Push or Pull a Sled • Similar to Quiz 5.14 • Will you exerts less force if you push or pull the girl? • θ is the same in both cases • Newton’s 2nd Law: ∑F = ma Pushing Pulling

  32. Example 5.21To Push or Pull a Sled, 2 • x direction:∑Fx= max • Fx– ƒs,max = max • Pushing • y direction:∑Fy = 0 n – mg – Fy= 0  n = mg + Fy ƒs,max =μsn ƒs,max =μs (mg + Fy ) n Fx ƒs,max Fy Pushing

  33. Example 5.21To Push or Pull a Sled, final • Pulling • y direction:∑Fy = 0 n +Fy – mg= 0  n = mg – Fy ƒs,max =μsn  ƒs,max =μs (mg–Fy ) NOTE: ƒs,max (Pushing) > ƒs,max (Pulling) Friction Force would be less if you pull than push!!! Fy n ƒs,max Fx Pulling

  34. Conceptual Example 5.22 Why Does the Sled Move? (Example 5.11 Text Book) • To determine ifthehorse (sled)moves:consider onlythe horizontal forces exerted ONthehorse (sled), then apply 2nd Newton’s Law:ΣF = m a. • Horse:T : tension exerted by the sled. fhorse : reaction exerted by the Earth. • Sled:T : tension exerted by the horse. fsled : friction between sled and snow.

  35. Conceptual Example 5.22 Why Does the Sled Move? final • Horse: If fhorse > T , the horse accelerates to the right. • Sled: If T> fsled , the sled accelerates to the right. • The forces that accelerates the system (horse-sled) is the net force fhorse  fsled • If fhorse = fsled the system will move with constant velocity.

  36. Example 5.23 Sliding Hockey Puck • Example 5.13 (Text Book) • Draw the free-body diagram, including the force of kinetic friction • Opposes the motion • Is parallel to the surfaces in contact • Continue with the solution as with any Newton’s Law problem

  37. Example 5.23 Sliding Hockey Puck, 2 • Given:vxi = 20.0 m/s vxf= 0, xi= 0, xf = 115 m • Find μk? • y direction:(ay = 0) ∑Fy = 0  n – mg = 0 n =mg(1) • x direction:∑Fx = max– μkn = max(2) • Substituting(1)in(2) : – μk(mg) = max ax = –μk g

  38. Example 5.23 Sliding Hockey Puck, final • ax = –μk g • To the left (slowing down) & independent of the mass!! • Replacing ax in the Equation: vf2 = vi2 + 2ax(xf – xi)  0 = (20.0m/s)2 + 2(–μk g)(115m) μk 2(9.80m/s2)(115m) = 400(m2/s2)  μk =400(m2/s2) / (2254m2/s2)  • μk = 0.177

  39. Example 5.24 Two Objects Connected with Friction • Example 5.13 (Text Book) • Known: ƒk kn • Find: a Mass 1:(Block) • y direction:∑Fy = 0,ay= 0 n + Fsinθ– m1g = 0 n = m1g – Fsinθ(1) • x direction:∑Fx = m1a Fcosθ– T–ƒk= m1a Fcosθ– T–kn= m1a T = Fcosθ–kn–m1a(2)

  40. Example 5.24 Two Objects Connected with Friction, 2 Mass 2:(Ball) • y direction:∑Fy = m2a T– m2g = m2a T= m2g + m2a(3) • x direction:∑Fx = 0,ax= 0 n = m1g – Fsinθ(1) T = Fcosθ–kn–m1a(2) • Substitute (1) into (2): T = Fcosθ– k(m1g – Fsinθ) – m1a(4) • Equate:(3) = (4)and solve for a:

  41. Example 5.24 Two Objects Connected with Friction, final

  42. Inclined Plane Problems • Tilted coordinate system: Convenient, but not necessary. • K-Trigonometry: Fgx= Fgsinθ = mgsinθ Fgy= Fgcosθ = – mgcosθ • Understand: • ∑F = m a , ƒk kn ax ≠ 0ay = 0 • y direction:∑Fy = 0 n – mgcosθ= 0 n = mgcosθ(1) • x direction:∑Fx = max mgsinθ –ƒ= max(2) ax Is the normal force n equal & opposite to the weight Fg ? NO!!!!

  43. Experimental Determination of µs and µk • The block is sliding down the plane, so friction acts up the plane • This setup can be used to experimentally determine the coefficient of friction µs,k • µs,k = tan qs,k • For µs use the angle where the block just slips • For µk use the angle where the block slides down at a constant speed

  44. Active Figure 5.19

  45. Example 5.25 The Skier • Assuming: FG= mg , ay = 0 ƒk kn,k= 0.10 • Find: ax • Components: FGx= FGsin30o = mgsin30o FGy= FGcos30o = – mgcos30o • Newton’s 2nd Law • y direction:∑Fy = 0 n – mgcos30o= 0 n = mgcos30o(1) • x direction:∑Fx = max mgsin30o –ƒk= max(2) ax ax n ƒk kn

  46. Example 5.25 The Skier n = mgcos30o(1) mgsin30o –ƒk= max(2) • Replacingƒk kn in (2) mgsin30o –kn = max (3) • Substituting (1) into (3) mgsin30o– kmgcos30o= max ax=gsin30o–μkgcos30o ax=g(0.5)–0.10g(0.87) ax=0.41g ax =4.00m/s2 ax ax n ƒk kn

  47. Material for the Midterm • Examples to Read!!! • Example 5.2(Page 120) • Example 5.3(Page 122) • Example 5.12(Page 134) • Homework to be solved in Class!!! • NONE

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