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 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 Whichever you use to solve the problem, the other approach can be used as a check

3. Example 5.16 Multiple Objects First treat the system as a whole: Apply Newton’s Laws to the individual blocks Solve for unknown(s) Solve for unknown(s) |P 21 | = |P 12 | Check: |P 21 | = |P 12 |

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. F P = 40.0 N F P = 40.0 N Find: Find: Acceleration (a) Acceleration (a) of each box Tension(F T ) Tension (F T ) in the cord connecting the boxes

5. Example 5.17 Two Boxes Connected by a Cord, final There is only horizontal motion There is only horizontal motion With: a A = a B = a Apply Newton’s Laws for box A: ΣF x = F P – F T = m A a (1) Apply Newton’s Laws for box B: ΣF x = F T = m B a (2) (2)(1): Substituting (2) into (1): F P – m B a = m A a F P = (m A + m B )a F P – m B a = m A a  F P = (m A + m B )a  a = F P /( m A + m B ) = 1.82m/s 2 a(2) Substituting a into (2)  F T = m B a = (12.0kg)(1.82m/s 2 ) = 21.8N

6. Example 5.18 The Atwood’s Machine Forces acting on the objects: Tension Tension (same for both objects, one string) Gravitational force Gravitational force same acceleration 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 Example 5.18 The Atwood’s Machine, 2 The Atwood’s Machine: The Atwood’s Machine: Find: a T Find: a and T Apply Newton’s 2 nd Law to each Mass. ΣF y = T – m 1 g = m 1 a (1) ΣF y = T – m 2 g = – m 2 a (2) ΣF y = T – m 2 g = – m 2 a (2) Then: T = m 1 g + m 1 a (3) T = m 2 g – m 2 a (4) T = m 2 g – m 2 a (4)

8. Example 5.18 The Atwood’s Machine, 3 Example 5.18 The Atwood’s Machine, 3 The Atwood’s Machine: The Atwood’s Machine: Equating: (3) = (4) a Equating: (3) = (4) and Solving for a m 1 g + m 1 a = m 2 g – m 2 a  m 1 a + m 2 a = m 2 g – m 1 g  a (m 1 + m 2 ) = (m 2 – m 1 )g  (5)

9. Example 5.18 The Atwood’s Machine, final Example 5.18 The Atwood’s Machine, final The Atwood’s Machine: The Atwood’s Machine: (5)(3)(4): Substituting (5) into (3) or (4): T = m 1 g + m 1 a (3) 

10. Active Figure 5.14

11. Example 5.19 Two Objects and Incline Plane aT Find: a and T One cord:tension same One cord: so tension is the same for both objects Connected: acceleration same 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 xy plane: ΣF x = 0 ΣF y = m 1 a ΣF x = 0 & ΣF y = m 1 a  T – m 1 g = m 1 a T – m 1 g = m 1 a  T = m 1 g + m 1 a(1) T = m 1 g + m 1 a (1) x’y’ x’y’ plane: ΣF x = m 2 a ΣF y = 0  ΣF x = m 2 a & ΣF y = 0  m 2 gsinθ – T = m 2 a (2) m 2 gsinθ – T = m 2 a (2) n – m 2 gcosθ = 0 (3) n – m 2 gcosθ = 0 (3)

13. Example 5.19 Two Objects and Incline Plane, Final (1)(2) Substituting (1) in (2) gives: m 2 gsinθ – (m 1 g + m 1 a) = m 2 a m 2 gsinθ – (m 1 g + m 1 a) = m 2 a  m 2 gsinθ – m 1 g – m 1 a = m 2 a m 2 gsinθ – m 1 g – m 1 a = m 2 a  a (m 1 + m 2 ) = m 2 gsinθ – m 1 g a (m 1 + m 2 ) = m 2 gsinθ – m 1 g  a (1) Substituting a in (1) we’ll get: T = m 1 g + m 1 a(1) T = m 1 g + m 1 a (1)

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

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

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

17. Forces of Friction, 2 Friction Friction exists between any 2 sliding surfaces. Two types of friction: Two types of friction: Static Static (no motion) friction Kinetic Kinetic (motion) friction The size of the friction force 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 Frictionnormal force Friction is proportional to the normal force ƒ s  µ s n (5.8) ƒ k = µ k n (5.9) ƒ s  µ s n (5.8) and ƒ k = µ k n (5.9) magnitudes THEY ARE NOT These equations relate the magnitudes of the forces, THEY ARE NOT vector equations force of static friction(maximum) force of kinetic friction ƒ s > ƒ k The force of static friction (maximum) is generally greater than the force of kinetic friction ƒ s > ƒ k The coefficients of friction (µ k,s ) The coefficients of friction (µ k,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 direction of the frictional force is opposite the direction of motion and parallel to the surfaces in contact The coefficients of friction (µ k,s ) are nearly independent of the area of contact The coefficients of friction (µ k,s ) are nearly independent of the area of contact

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

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

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

23. Static & Kinetic Friction Maximum Static Friction Force ƒ s,max magnitude (size) nbetween Experiments find that the Maximum Static Friction Force ƒ s,max is proportional to the magnitude (size) of the normal force n between the 2 surfaces. DIRECTIONS: ƒ s,max n DIRECTIONS: ƒ s,max  n ƒ s,max =  s n (magnitudes) Then: ƒ s,max =  s n (magnitudes)

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

25. Some Coefficients of Friction

26. Active Figure 5.16

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

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

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

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

31. Example 5.21 To Push or Pull a Sled Similar to Quiz 5.14 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 2 nd Law: Newton’s 2 nd Law: ∑F = ma Pushing Pulling

32. Example 5.21 To Push or Pull a Sled, 2 x direction:∑F x = ma x  x direction: ∑F x = ma x  F x – ƒ s,max = ma x F x – ƒ s,max = ma x Pushing Pushing y direction:∑F y = 0  y direction: ∑F y = 0  n – mg – F y = 0  n = mg + F y  ƒ s,max = μ s n  ƒ s,max = μ s (mg + F y ) Pushing n ƒ s,max FxFxFxFx FyFyFyFy

33. Example 5.21 To Push or Pull a Sled, final Pulling Pulling y direction:∑F y = 0  y direction: ∑F y = 0  n F y – mg = 0  n + F y – mg = 0  n = mg – F y  ƒ s,max = μ s n  ƒ s,max = μ s (mg – F y ) NOTE: ƒ s,max (Pushing) > ƒ s,max (Pulling) Friction Force would be less if you pull than push!!! Pulling FxFxFxFx FyFyFyFy n ƒ s,max

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

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

36. Example 5.23 Sliding Hockey Puck Example 5.13 Example 5.13 (Text Book) free-body kinetic friction 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 v xi = 20.0 m/s v xf = 0, x i = 0, x f = 115 m Given: v xi = 20.0 m/s v xf = 0, x i = 0, x f = 115 m μ k ? Find μ k ? y direction:(a y = 0) y direction: (a y = 0) ∑F y = 0  ∑F y = 0  n – mg = 0  n = mg (1) n – mg = 0  n = mg (1) x direction:∑F x = ma x  – μ k n = ma x (2) x direction: ∑F x = ma x  – μ k n = ma x (2) Substituting (1) in (2) : – μ k (mg) = ma x  a x = – μ k g

38. Example 5.23 Sliding Hockey Puck, final a x = – μ k g a x = – μ k g To the left (slowing down) & independent of the mass!! To the left (slowing down) & independent of the mass!! a x v f 2 = v i 2 + 2a x (x f – x i )  Replacing a x in the Equation: v f 2 = v i 2 + 2a x (x f – x i )  0 = (20.0m/s) 2 + 2(– μ k g)(115m)  μ k (9.80m/s 2 )(115m)400(m 2 /s 2 )  μ k 2(9.80m/s 2 )(115m) = 400(m 2 /s 2 )  μ k =400(m 2 /s 2 ) / (2254m 2 /s 2 )  μ k = 400(m 2 /s 2 ) / (2254m 2 /s 2 )  μ k = 0.177 μ k = 0.177

39. Example 5.24 Two Objects Connected with Friction Example 5.13 Example 5.13 (Text Book) ƒ k   k n Known: ƒ k   k n a Find: a Mass 1:(Block) Mass 1: (Block) y direction:∑F y = 0a y = 0  y direction: ∑F y = 0, a y = 0  n + Fsinθ – m 1 g = 0  n = m 1 g – Fsinθ(1) n = m 1 g – Fsinθ (1) x direction:∑F x = m 1 a  x direction: ∑F x = m 1 a  Fcosθ– T –ƒ k = m 1 a  Fcosθ – T – ƒ k = m 1 a  Fcosθ – T –  k n = m 1 a  T = Fcosθ –  k n a(2) T = Fcosθ –  k n – m 1 a (2)

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

41. Example 5.24 Two Objects Connected with Friction, final

42. Inclined Plane Problems Tilted coordinate system Tilted coordinate system: Convenient, but not necessary. K-Trigonometry: K-Trigonometry: F gx = F g sinθ = mgsinθ F gy = F g cosθ = – mgcosθ Understand: Understand: ∑F = m a, ƒ k   k n ∑F = m a, ƒ k   k n a x ≠ 0 a y = 0 y direction:∑F y = 0  y direction: ∑F y = 0  – mgcosθ= 0  n – mgcosθ = 0  n = mgcosθ(1) n = mgcosθ (1) x direction:∑F x = ma x  x direction: ∑F x = ma x  mgsinθ –ƒ = ma x (2) mgsinθ – ƒ = ma x (2) axaxaxax Is the normal force n equal & opposite to the weight Fg ? NO!!!! NO!!!!

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

44. Active Figure 5.19

45. Example 5.25 The Skier Assuming: F G = mg, a y = 0 Assuming: F G = mg, a y = 0 ƒ k   k n,  k = 0.10 ƒ k   k n,  k = 0.10 Find: a x Find: a x Components: Components: F Gx = F G sin30 o = mgsin30 o F Gy = F G cos30 o = – mgcos30 o Newton’s 2 nd Law Newton’s 2 nd Law y direction:∑F y = 0  y direction: ∑F y = 0  – mgcos30 o = 0  n – mgcos30 o = 0  n = mgcos30 o (1) x direction:∑F x = ma x  x direction: ∑F x = ma x  mgsin30 o –ƒ k = ma x (2) mgsin30 o – ƒ k = ma x (2) axaxaxax ƒk  k nƒk  k nƒk  k nƒk  k n n nn n axaxaxax

46. Example 5.25 The Skier n = mgcos30 o (1) n = mgcos30 o (1) mgsin30 o –ƒ k = ma x (2) mgsin30 o – ƒ k = ma x (2) ƒ k   k n (2) Replacing ƒ k   k n in (2) mgsin30 o –  k n = ma x (3) Substituting (1) into (3) Substituting (1) into (3) mgsin30 o –  k mgcos30 o = ma x a x =gsin30 o –μ k gcos 30 o a x = gsin30 o – μ k gcos 30 o a x =g(0.5)–0.10g(0.87)  a x = g(0.5) – 0.10g(0.87)  a x =0.41g  a x = 0.41g  a x =4.00m/s 2 a x = 4.00m/s 2 axaxaxax ƒk  k nƒk  k nƒk  k nƒk  k n n nn n axaxaxax

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