1. Two-Body Pulley Problems

2. Consider the following system: M A = 3.0 kg M B = 1.5 kg Pulley a) frictionless b) μ K = 0.200 table A B

3. Two-Body Pulley Problems Consider the following system: What is a pulley? M A = 3.0 kg M B = 1.5 kg Pulley a) frictionless b) μ K = 0.200 table A B

4. Two-Body Pulley Problems Consider the following system: M A = 3.0 kg M B = 1.5 kg Pulley: a machine that changes the direction of the tension force but not its size. a) frictionless b) μ K = 0.200 table A B

5. Two-Body Pulley Problems Consider the following system: Assuming the system accelerates from rest, find the acceleration of each mass and the size of the tension force in the string if the coefficient of friction between table and block A is... a) 0 (table is frictionless) b) μ K = 0.200 M A = 3.0 kg M B = 1.5 kg Pulley: a machine that changes the direction of the tension force but not its size. a) frictionless b) μ K = 0.200 table A B

6. Two-Body Pulley Problems Consider the following system: Assuming the system accelerates from rest, find the acceleration of each mass and the size of the tension force in the string if the coefficient of friction between table and block A is... a) 0 (table is frictionless) b) μ K = 0.200 Step One : ??? M A = 3.0 kg M B = 1.5 kg Pulley: a machine that changes the direction of the tension force but not its size. a) frictionless b) μ K = 0.200 table A B

7. Two-Body Pulley Problems Consider the following system: Assuming the system accelerates from rest, find the acceleration of each mass and the size of the tension force in the string if the coefficient of friction between table and block A is... a) 0 (table is frictionless) b) μ K = 0.200 Step One : determine the direction of the acceleration of each body and label on the diagram. M A = 3.0 kg M B = 1.5 kg Pulley: a machine that changes the direction of the tension force but not its size. a) frictionless b) μ K = 0.200 table A B

8. Two-Body Pulley Problems Consider the following system: Assuming the system accelerates from rest, find the acceleration of each mass and the size of the tension force in the string if the coefficient of friction between table and block A is... a) 0 (table is frictionless) b) μ K = 0.200 Step One : determine the direction of the acceleration of each body and label on the diagram. M A = 3.0 kg M B = 1.5 kg Pulley: a machine that changes the direction of the tension force but not its size. a) frictionless b) μ K = 0.200 table A B aAaA aBaB

9. Two-Body Pulley Problems Consider the following system: Assuming the system accelerates from rest, find the acceleration of each mass and the size of the tension force in the string if the coefficient of friction between table and block A is... a) 0 (table is frictionless) b) μ K = 0.200 Step One : determine the direction of the acceleration of each body and label on the diagram. Step Two: ??? M A = 3.0 kg M B = 1.5 kg Pulley: a machine that changes the direction of the tension force but not its size. a) frictionless b) μ K = 0.200 table A B aAaA aBaB

10. Two-Body Pulley Problems Consider the following system: Assuming the system accelerates from rest, find the acceleration of each mass and the size of the tension force in the string if the coefficient of friction between table and block A is... a) 0 (table is frictionless) b) μ K = 0.200 Step One : determine the direction of the acceleration of each body and label on the diagram. Step Two: compare the magnitude of the acceleration of each block M A = 3.0 kg M B = 1.5 kg Pulley: a machine that changes the direction of the tension force but not its size. a) frictionless b) μ K = 0.200 table A B aAaA aBaB

11. Two-Body Pulley Problems Consider the following system: Assuming the system accelerates from rest, find the acceleration of each mass and the size of the tension force in the string if the coefficient of friction between table and block A is... a) 0 (table is frictionless) b) μ K = 0.200 Step One : determine the direction of the acceleration of each body and label on the diagram. Step Two: compare the magnitude of the acceleration of each block If block B descends by 1 cm, it is clear that block A will move to the right by 1 cm. (conservation of rope) Therefore | a A | = | a B | = a Green a will represent the positive size of the acceleration. M A = 3.0 kg M B = 1.5 kg Pulley: a machine that changes the direction of the tension force but not its size. a) frictionless b) μ K = 0.200 table A B aAaA aBaB

12. Two-Body Pulley Problems Step 3: ??? M A = 3.0 kg M B = 1.5 kg Pulley: a machine that changes the direction of the tension force but not its size. a) frictionless b) μ K = 0.200 table A B aAaA aBaB

13. Two-Body Pulley Problems Step 3: Draw an FBD of block A M A = 3.0 kg M B = 1.5 kg Pulley: a machine that changes the direction of the tension force but not its size. a) frictionless b) μ K = 0.200 table B aAaA aBaB A

14. Two-Body Pulley Problems Step 3: Draw an FBD of block A M A = 3.0 kg M B = 1.5 kg Pulley: a machine that changes the direction of the tension force but not its size. a) frictionless b) μ K = 0.200 table A B aAaA aBaB aAaA a) M A = 3.0 kg T F g = ? FNFN A

15. Two-Body Pulley Problems Step 3: Draw an FBD of block A M A = 3.0 kg M B = 1.5 kg Pulley: a machine that changes the direction of the tension force but not its size. a) frictionless b) μ K = 0.200 table A B aAaA aBaB aAaA a) M A = 3.0 kg T F g = 30 N FNFN A

16. Two-Body Pulley Problems Step 3: Draw an FBD of block A Step 4: ??? M A = 3.0 kg M B = 1.5 kg Pulley: a machine that changes the direction of the tension force but not its size. a) frictionless b) μ K = 0.200 table A B aAaA aBaB aAaA a) M A = 3.0 kg T F g = 30 N FNFN A

17. Two-Body Pulley Problems Step 3: Draw an FBD of block A Step 4: Label an XY plane with the positive x axis in the direction of acceleration M A = 3.0 kg M B = 1.5 kg Pulley: a machine that changes the direction of the tension force but not its size. a) frictionless b) μ K = 0.200 table A B aAaA aBaB aAaA a) M A = 3.0 kg T F g = 30 N FNFN A

18. Two-Body Pulley Problems Step 3: Draw an FBD of block A Step 4: Label an XY plane with the positive x axis in the direction of acceleration M A = 3.0 kg M B = 1.5 kg Pulley: a machine that changes the direction of the tension force but not its size. a) frictionless b) μ K = 0.200 table A B aAaA aBaB aAaA a) M A = 3.0 kg T F g = 30 N FNFN X Y A

19. Two-Body Pulley Problems Step 3: Draw an FBD of block A Step 4: Label an XY plane with the positive x axis in the direction of acceleration Step 5: ??? M A = 3.0 kg M B = 1.5 kg Pulley: a machine that changes the direction of the tension force but not its size. a) frictionless b) μ K = 0.200 table A B aAaA aBaB aAaA a) M A = 3.0 kg T F g = 30 N FNFN X Y A

20. Two-Body Pulley Problems Step 3: Draw an FBD of block A Step 4: Label an XY plane with the positive x axis in the direction of acceleration Step 5: Express the unknown vectors in terms of positive scalar magnitudes. M A = 3.0 kg M B = 1.5 kg Pulley: a machine that changes the direction of the tension force but not its size. a) frictionless b) μ K = 0.200 table A B aAaA aBaB aAaA a) M A = 3.0 kg T F g = 30 N FNFN X Y A

21. Two-Body Pulley Problems Step 3: Draw an FBD of block A Step 4: Label an XY plane with the positive x axis in the direction of acceleration Step 5: Express the unknown vectors in terms of positive scalar magnitudes. M A = 3.0 kg M B = 1.5 kg Pulley: a machine that changes the direction of the tension force but not its size. a) frictionless b) μ K = 0.200 table A B aAaA aBaB a A = + a a) M A = 3.0 kg T = +T F g = 30 N FNFN X Y A

22. Two-Body Pulley Problems Step 3: Draw an FBD of block A Step 4: Label an XY plane with the positive x axis in the direction of acceleration Step 5: Express the unknown vectors in terms of positive scalar magnitudes. Step 6: Find F N in case we need to find friction, so deal with y-forces M A = 3.0 kg M B = 1.5 kg Pulley: a machine that changes the direction of the tension force but not its size. a) frictionless b) μ K = 0.200 table A B aAaA aBaB a A = + a a) M A = 3.0 kg T = +T F g = 30 N FNFN X Y A

23. Two-Body Pulley Problems Step 3: Draw an FBD of block A Step 4: Label an XY plane with the positive x axis in the direction of acceleration Step 5: Express the unknown vectors in terms of positive scalar magnitudes. Step 6: Find F N in case we need to find friction, so deal with y-forces What is a y = ? M A = 3.0 kg M B = 1.5 kg Pulley: a machine that changes the direction of the tension force but not its size. a) frictionless b) μ K = 0.200 table A B aAaA aBaB a A = + a a) M A = 3.0 kg T = +T F g = 30 N FNFN X Y A

24. Two-Body Pulley Problems Step 3: Draw an FBD of block A Step 4: Label an XY plane with the positive x axis in the direction of acceleration Step 5: Express the unknown vectors in terms of positive scalar magnitudes. Step 6: Find F N in case we need to find friction, so deal with y-forces a y = 0 M A = 3.0 kg M B = 1.5 kg Pulley: a machine that changes the direction of the tension force but not its size. a) frictionless b) μ K = 0.200 table A B aAaA aBaB a A = + a a) M A = 3.0 kg T = +T F g = 30 N FNFN X Y A

25. Two-Body Pulley Problems Step 3: Draw an FBD of block A Step 4: Label an XY plane with the positive x axis in the direction of acceleration Step 5: Express the unknown vectors in terms of positive scalar magnitudes. Step 6: Find F N in case we need to find friction, so deal with y-forces a y = 0 According to Newton's first law, what is F net y = ? M A = 3.0 kg M B = 1.5 kg Pulley: a machine that changes the direction of the tension force but not its size. a) frictionless b) μ K = 0.200 table A B aAaA aBaB a A = + a a) M A = 3.0 kg T = +T F g = 30 N FNFN X Y A

26. Two-Body Pulley Problems Step 3: Draw an FBD of block A Step 4: Label an XY plane with the positive x axis in the direction of acceleration Step 5: Express the unknown vectors in terms of positive scalar magnitudes. Step 6: Find F N in case we need to find friction, so deal with y-forces If a y = 0 then F net y = 0 M A = 3.0 kg M B = 1.5 kg Pulley: a machine that changes the direction of the tension force but not its size. a) frictionless b) μ K = 0.200 table A B aAaA aBaB a A = + a a) M A = 3.0 kg T = +T F g = 30 N FNFN X Y A

27. Two-Body Pulley Problems Step 3: Draw an FBD of block A Step 4: Label an XY plane with the positive x axis in the direction of acceleration Step 5: Express the unknown vectors in terms of positive scalar magnitudes. Step 6: Find F N in case we need to find friction, so deal with y-forces If a y = 0 then F net y = 0 Vector Statement ??? M A = 3.0 kg M B = 1.5 kg Pulley: a machine that changes the direction of the tension force but not its size. a) frictionless b) μ K = 0.200 table A B aAaA aBaB a A = + a a) M A = 3.0 kg T = +T F g = 30 N FNFN X Y A

28. Two-Body Pulley Problems Step 3: Draw an FBD of block A Step 4: Label an XY plane with the positive x axis in the direction of acceleration Step 5: Express the unknown vectors in terms of positive scalar magnitudes. Step 6: Find F N in case we need to find friction, so deal with y-forces If a y = 0 then F net y = 0 Vector Statement F N + F g = 0 M A = 3.0 kg M B = 1.5 kg Pulley: a machine that changes the direction of the tension force but not its size. a) frictionless b) μ K = 0.200 table A B aAaA aBaB a A = + a a) M A = 3.0 kg T = +T F g = 30 N FNFN X Y A

29. Two-Body Pulley Problems Step 3: Draw an FBD of block A Step 4: Label an XY plane with the positive x axis in the direction of acceleration Step 5: Express the unknown vectors in terms of positive scalar magnitudes. Step 6: Find F N in case we need to find friction, so deal with y-forces If a y = 0 then F net y = 0 Vector Statement F N + F g = 0 Scalar Statement ??? M A = 3.0 kg M B = 1.5 kg Pulley: a machine that changes the direction of the tension force but not its size. a) frictionless b) μ K = 0.200 table A B aAaA aBaB a A = + a a) M A = 3.0 kg T = +T F g = 30 N FNFN X Y A

30. Two-Body Pulley Problems Step 3: Draw an FBD of block A Step 4: Label an XY plane with the positive x axis in the direction of acceleration Step 5: Express the unknown vectors in terms of positive scalar magnitudes. Step 6: Find F N in case we need to find friction, so deal with y-forces If a y = 0 then F net y = 0 Vector Statement F N + F g = 0 Scalar Statement F N - 30 = 0 M A = 3.0 kg M B = 1.5 kg Pulley: a machine that changes the direction of the tension force but not its size. a) frictionless b) μ K = 0.200 table A B aAaA aBaB a A = + a a) M A = 3.0 kg T = +T F g = 30 N FNFN X Y A

31. Two-Body Pulley Problems Step 3: Draw an FBD of block A Step 4: Label an XY plane with the positive x axis in the direction of acceleration Step 5: Express the unknown vectors in terms of positive scalar magnitudes. Step 6: Find F N in case we need to find friction, so deal with y-forces If a y = 0 then F net y = 0 Vector Statement F N + F g = 0 Scalar Statement F N - 30 = 0 F N = 30 N [up ] M A = 3.0 kg M B = 1.5 kg Pulley: a machine that changes the direction of the tension force but not its size. a) frictionless b) μ K = 0.200 table A B aAaA aBaB a A = + a a) M A = 3.0 kg T = +T F g = 30 N FNFN X Y A

32. Two-Body Pulley Problems Step 3: Draw an FBD of block A Step 4: Label an XY plane with the positive x axis in the direction of acceleration Step 5: Express the unknown vectors in terms of positive scalar magnitudes. Step 6: Find F N in case we need to find friction, so deal with y-forces If a y = 0 then F net y = 0 Vector Statement F N + F g = 0 Scalar Statement F N - 30 = 0 F N = 30 N [up ] Label on FBD M A = 3.0 kg M B = 1.5 kg Pulley: a machine that changes the direction of the tension force but not its size. a) frictionless b) μ K = 0.200 table A B aAaA aBaB a A = + a a) M A = 3.0 kg T = +T F g = 30 N FNFN X Y A

33. Two-Body Pulley Problems Step 3: Draw an FBD of block A Step 4: Label an XY plane with the positive x axis in the direction of acceleration Step 5: Express the unknown vectors in terms of positive scalar magnitudes. Step 6: Find F N in case we need to find friction, so deal with y-forces If a y = 0 then F net y = 0 Vector Statement F N + F g = 0 Scalar Statement F N - 30 = 0 F N = 30 N [up ] Label on FBD M A = 3.0 kg M B = 1.5 kg Pulley: a machine that changes the direction of the tension force but not its size. a) frictionless b) μ K = 0.200 table A B aAaA aBaB a A = + a a) M A = 3.0 kg T = +T F g = 30 N F N = 30 N X Y A

34. Two-Body Pulley Problems Step 3: Draw an FBD of block A Step 7: ??? M A = 3.0 kg M B = 1.5 kg Pulley: a machine that changes the direction of the tension force but not its size. a) frictionless b) μ K = 0.200 table A B aAaA aBaB a A = + a a) M A = 3.0 kg T = +T F g = 30 N F N = 30 N X Y A

35. Two-Body Pulley Problems Step 3: Draw an FBD of block A Step 7: Look at x-forces. Start by stating Newton's second law equation or Equation of motion. M A = 3.0 kg M B = 1.5 kg Pulley: a machine that changes the direction of the tension force but not its size. a) frictionless b) μ K = 0.200 table A B aAaA aBaB a A = + a a) M A = 3.0 kg T = +T F g = 30 N F N = 30 N X Y A

36. Two-Body Pulley Problems Step 3: Draw an FBD of block A Step 7: Look at x-forces. Start by stating Newton's second law equation or Equation of motion. F net x = m a x M A = 3.0 kg M B = 1.5 kg Pulley: a machine that changes the direction of the tension force but not its size. a) frictionless b) μ K = 0.200 table A B aAaA aBaB a A = + a a) M A = 3.0 kg T = +T F g = 30 N F N = 30 N X Y A

37. Two-Body Pulley Problems Step 3: Draw an FBD of block A Step 7: Look at x-forces. Start by stating Newton's second law equation or Equation of motion. F net x = m a x Vector Statement ???? M A = 3.0 kg M B = 1.5 kg Pulley: a machine that changes the direction of the tension force but not its size. a) frictionless b) μ K = 0.200 table A B aAaA aBaB a A = + a a) M A = 3.0 kg T = +T F g = 30 N F N = 30 N X Y A

38. Two-Body Pulley Problems Step 3: Draw an FBD of block A Step 7: Look at x-forces. Start by stating Newton's second law equation or Equation of motion. F net x = m a x Vector Statement T = 3 a M A = 3.0 kg M B = 1.5 kg Pulley: a machine that changes the direction of the tension force but not its size. a) frictionless b) μ K = 0.200 table A B aAaA aBaB a A = + a a) M A = 3.0 kg T = +T F g = 30 N F N = 30 N X Y A

39. Two-Body Pulley Problems Step 3: Draw an FBD of block A Step 7: Look at x-forces. Start by stating Newton's second law equation or Equation of motion. F net x = m a x Vector Statement T = 3 a Scalar Statement ???? M A = 3.0 kg M B = 1.5 kg Pulley: a machine that changes the direction of the tension force but not its size. a) frictionless b) μ K = 0.200 table A B aAaA aBaB a A = + a a) M A = 3.0 kg T = +T F g = 30 N F N = 30 N X Y A

40. Two-Body Pulley Problems Step 3: Draw an FBD of block A Step 7: Look at x-forces. Start by stating Newton's second law equation or Equation of motion. F net x = m a x Vector Statement T = 3 a Scalar Statement T = 3 a M A = 3.0 kg M B = 1.5 kg Pulley: a machine that changes the direction of the tension force but not its size. a) frictionless b) μ K = 0.200 table A B aAaA aBaB a A = + a a) M A = 3.0 kg T = +T F g = 30 N F N = 30 N X Y A

41. Two-Body Pulley Problems Step 3: Draw an FBD of block A Step 7: Look at x-forces. Start by stating Newton's second law equation or Equation of motion. F net x = m a x Vector Statement T = 3 a Scalar Statement T = 3 aequation #1 M A = 3.0 kg M B = 1.5 kg Pulley: a machine that changes the direction of the tension force but not its size. a) frictionless b) μ K = 0.200 table A B aAaA aBaB a A = + a a) M A = 3.0 kg T = +T F g = 30 N F N = 30 N X Y A

42. Two-Body Pulley Problems Step 7b: ??? M A = 3.0 kg M B = 1.5 kg Pulley: a machine that changes the direction of the tension force but not its size. a) frictionless b) μ K = 0.200 table B aAaA aBaB A

43. Two-Body Pulley Problems Step 7b: Draw an FBD of block B M A = 3.0 kg M B = 1.5 kg Pulley: a machine that changes the direction of the tension force but not its size. a) frictionless b) μ K = 0.200 table B aAaA aBaB A

44. Two-Body Pulley Problems Step 7b: Draw an FBD of block B M A = 3.0 kg M B = 1.5 kg Pulley: a machine that changes the direction of the tension force but not its size. a) frictionless b) μ K = 0.200 table B aAaA aBaB M B = 1.5 kg B aBaB T FgFg A

45. Two-Body Pulley Problems Step 7b: Draw an FBD of block B M A = 3.0 kg M B = 1.5 kg Pulley: a machine that changes the direction of the tension force but not its size. a) frictionless b) μ K = 0.200 table B aAaA aBaB M B = 1.5 kg B aBaB T F g = ? A

46. Two-Body Pulley Problems Step 7b: Draw an FBD of block B M A = 3.0 kg M B = 1.5 kg Pulley: a machine that changes the direction of the tension force but not its size. a) frictionless b) μ K = 0.200 table B aAaA aBaB M B = 1.5 kg B aBaB T F g = 15 N A

47. Two-Body Pulley Problems Step 7b: Draw an FBD of block B Step 8: Draw a y-axis and indicate the + and – directions M A = 3.0 kg M B = 1.5 kg Pulley: a machine that changes the direction of the tension force but not its size. a) frictionless b) μ K = 0.200 table B aAaA aBaB M B = 1.5 kg B aBaB T F g = 15 N A

48. Two-Body Pulley Problems Step 7b: Draw an FBD of block B Step 8: Draw a y-axis and indicate the + and – directions M A = 3.0 kg M B = 1.5 kg Pulley: a machine that changes the direction of the tension force but not its size. a) frictionless b) μ K = 0.200 table B aAaA aBaB M B = 1.5 kg B aBaB T F g = 15 N +Y -Y A

49. Two-Body Pulley Problems Step 7b: Draw an FBD of block B Step 8: Draw a y-axis and indicate the + and – directions Step 9: Express the unknown vectors in terms of positive scalar magnitudes. M A = 3.0 kg M B = 1.5 kg Pulley: a machine that changes the direction of the tension force but not its size. a) frictionless b) μ K = 0.200 table B aAaA aBaB M B = 1.5 kg B aBaB T F g = 15 N +Y -Y A

50. Two-Body Pulley Problems Step 7b: Draw an FBD of block B Step 8: Draw a y-axis and indicate the + and – directions Step 9: Express the unknown vectors in terms of positive scalar magnitudes. M A = 3.0 kg M B = 1.5 kg Pulley: a machine that changes the direction of the tension force but not its size. a) frictionless b) μ K = 0.200 table B aAaA aBaB M B = 1.5 kg B a B = -a T = + T F g = 15 N +Y -Y A

51. Two-Body Pulley Problems Step 7b: Draw an FBD of block B Step 8: Draw a y-axis and indicate the + and – directions Step 9: Express the unknown vectors in terms of positive scalar magnitudes. Step 10: ??? M A = 3.0 kg M B = 1.5 kg Pulley: a machine that changes the direction of the tension force but not its size. a) frictionless b) μ K = 0.200 table B aAaA aBaB M B = 1.5 kg B a B = -a T = + T F g = 15 N +Y -Y A

52. Two-Body Pulley Problems Step 7b: Draw an FBD of block B Step 8: Draw a y-axis and indicate the + and – directions Step 9: Express the unknown vectors in terms of positive scalar magnitudes. Step 10: Look at y-forces. Start by stating Newton's second law equation or Equation of motion. M A = 3.0 kg M B = 1.5 kg Pulley: a machine that changes the direction of the tension force but not its size. a) frictionless b) μ K = 0.200 table B aAaA aBaB M B = 1.5 kg B a B = -a T = + T F g = 15 N +Y -Y A

53. Two-Body Pulley Problems Step 7b: Draw an FBD of block B Step 8: Draw a y-axis and indicate the + and – directions Step 9: Express the unknown vectors in terms of positive scalar magnitudes. Step 10: Look at y-forces. Start by stating Newton's second law equation or Equation of motion.F net y = m a y M A = 3.0 kg M B = 1.5 kg Pulley: a machine that changes the direction of the tension force but not its size. a) frictionless b) μ K = 0.200 table B aAaA aBaB M B = 1.5 kg B a B = -a T = + T F g = 15 N +Y -Y A

54. Two-Body Pulley Problems Step 7b: Draw an FBD of block B Step 8: Draw a y-axis and indicate the + and – directions Step 9: Express the unknown vectors in terms of positive scalar magnitudes. Step 10: Look at y-forces. Start by stating Newton's second law equation or Equation of motion.F net y = m a y Vector statement ??? M A = 3.0 kg M B = 1.5 kg Pulley: a machine that changes the direction of the tension force but not its size. a) frictionless b) μ K = 0.200 table B aAaA aBaB M B = 1.5 kg B a B = -a T = + T F g = 15 N +Y -Y A

55. Two-Body Pulley Problems Step 7b: Draw an FBD of block B Step 8: Draw a y-axis and indicate the + and – directions Step 9: Express the unknown vectors in terms of positive scalar magnitudes. Step 10: Look at y-forces. Start by stating Newton's second law equation or Equation of motion.F net y = m a y Vector statement T + F g = 1.5 a M A = 3.0 kg M B = 1.5 kg Pulley: a machine that changes the direction of the tension force but not its size. a) frictionless b) μ K = 0.200 table B aAaA aBaB M B = 1.5 kg B a B = -a T = + T F g = 15 N +Y -Y A

56. Two-Body Pulley Problems Step 7b: Draw an FBD of block B Step 8: Draw a y-axis and indicate the + and – directions Step 9: Express the unknown vectors in terms of positive scalar magnitudes. Step 10: Look at y-forces. Start by stating Newton's second law equation or Equation of motion.F net y = m a y Vector statement T + F g = 1.5 a M A = 3.0 kg M B = 1.5 kg Pulley: a machine that changes the direction of the tension force but not its size. a) frictionless b) μ K = 0.200 table B aAaA aBaB M B = 1.5 kg B a B = -a T = + T F g = 15 N +Y -Y A

57. Two-Body Pulley Problems Step 7b: Draw an FBD of block B Step 8: Draw a y-axis and indicate the + and – directions Step 9: Express the unknown vectors in terms of positive scalar magnitudes. Step 10: Look at y-forces. Start by stating Newton's second law equation or Equation of motion.F net y = m a y Scalar statement ??? Vector statement T + F g = 1.5 a M A = 3.0 kg M B = 1.5 kg Pulley: a machine that changes the direction of the tension force but not its size. a) frictionless b) μ K = 0.200 table B aAaA aBaB M B = 1.5 kg B a B = -a T = + T F g = 15 N +Y -Y A

58. Two-Body Pulley Problems Step 7b: Draw an FBD of block B Step 8: Draw a y-axis and indicate the + and – directions Step 9: Express the unknown vectors in terms of positive scalar magnitudes. Step 10: Look at y-forces. Start by stating Newton's second law equation or Equation of motion.F net y = m a y Scalar statement T - 15 = -1.5 a Vector statement T + F g = 1.5 a M A = 3.0 kg M B = 1.5 kg Pulley: a machine that changes the direction of the tension force but not its size. a) frictionless b) μ K = 0.200 table B aAaA aBaB M B = 1.5 kg B a B = -a T = + T F g = 15 N +Y -Y A

59. Two-Body Pulley Problems Step 7b: Draw an FBD of block B Step 8: Draw a y-axis and indicate the + and – directions Step 9: Express the unknown vectors in terms of positive scalar magnitudes. Step 10: Look at y-forces. Start by stating Newton's second law equation or Equation of motion.F net y = m a y Scalar statement T - 15 = -1.5 a Vector statement T + F g = 1.5 aT = ? M A = 3.0 kg M B = 1.5 kg Pulley: a machine that changes the direction of the tension force but not its size. a) frictionless b) μ K = 0.200 table B aAaA aBaB M B = 1.5 kg B a B = -a T = + T F g = 15 N +Y -Y A

60. Two-Body Pulley Problems Step 7b: Draw an FBD of block B Step 8: Draw a y-axis and indicate the + and – directions Step 9: Express the unknown vectors in terms of positive scalar magnitudes. Step 10: Look at y-forces. Start by stating Newton's second law equation or Equation of motion.F net y = m a y Scalar statement T - 15 = -1.5 a Vector statement T + F g = 1.5 aT = 15 -1.5 a M A = 3.0 kg M B = 1.5 kg Pulley: a machine that changes the direction of the tension force but not its size. a) frictionless b) μ K = 0.200 table B aAaA aBaB M B = 1.5 kg B a B = -a T = + T F g = 15 N +Y -Y A

61. Two-Body Pulley Problems Step 7b: Draw an FBD of block B Step 8: Draw a y-axis and indicate the + and – directions Step 9: Express the unknown vectors in terms of positive scalar magnitudes. Step 10: Look at y-forces. Start by stating Newton's second law equation or Equation of motion.F net y = m a y Scalar statement T - 15 = -1.5 a Vector statement T + F g = 1.5 aequation #2 T = 15 -1.5 a M A = 3.0 kg M B = 1.5 kg Pulley: a machine that changes the direction of the tension force but not its size. a) frictionless b) μ K = 0.200 table B aAaA aBaB M B = 1.5 kg B a B = -a T = + T F g = 15 N +Y -Y A

62. Two-Body Pulley Problems Now we have two scalar equations: T = 3 a equation #1 T = 15 -1.5 a equation #2 M A = 3.0 kg M B = 1.5 kg Pulley: a machine that changes the direction of the tension force but not its size. a) frictionless b) μ K = 0.200 table B aAaA aBaB A

63. Two-Body Pulley Problems Now we have two scalar equations: T = 3 a equation #1 T = 15 -1.5 a equation #2 Step 11: Solve by sub #1 into #2 M A = 3.0 kg M B = 1.5 kg Pulley: a machine that changes the direction of the tension force but not its size. a) frictionless b) μ K = 0.200 table B aAaA aBaB A

64. Two-Body Pulley Problems Now we have two scalar equations: T = 3 a equation #1 T = 15 -1.5 a equation #2 Step 11: Solve by sub #1 into #2 3 a = 15 -1.5 a M A = 3.0 kg M B = 1.5 kg Pulley: a machine that changes the direction of the tension force but not its size. a) frictionless b) μ K = 0.200 table B aAaA aBaB A

65. Two-Body Pulley Problems Now we have two scalar equations: T = 3 a equation #1 T = 15 -1.5 a equation #2 Step 11: Solve by sub #1 into #2 3 a = 15 -1.5 a 4.5 a = 15 M A = 3.0 kg M B = 1.5 kg Pulley: a machine that changes the direction of the tension force but not its size. a) frictionless b) μ K = 0.200 table B aAaA aBaB A

66. Two-Body Pulley Problems Now we have two scalar equations: T = 3 a equation #1 T = 15 -1.5 a equation #2 Step 11: Solve by sub #1 into #2 3 a = 15 -1.5 a 4.5 a = 15 a = 3.3 m/s 2 M A = 3.0 kg M B = 1.5 kg Pulley: a machine that changes the direction of the tension force but not its size. a) frictionless b) μ K = 0.200 table B aAaA aBaB A

67. Two-Body Pulley Problems Now we have two scalar equations: T = 3 a equation #1 T = 15 -1.5 a equation #2 Step 11: Solve by sub #1 into #2 3 a = 15 -1.5 a 4.5 a = 15 a = 3.3 m/s 2 a A = ? M A = 3.0 kg M B = 1.5 kg Pulley: a machine that changes the direction of the tension force but not its size. a) frictionless b) μ K = 0.200 table B aAaA aBaB A

68. Two-Body Pulley Problems Now we have two scalar equations: T = 3 a equation #1 T = 15 -1.5 a equation #2 Step 11: Solve by sub #1 into #2 3 a = 15 -1.5 a 4.5 a = 15 a = 3.3 m/s 2 a A = 3.3 m/s 2 [ right ] M A = 3.0 kg M B = 1.5 kg Pulley: a machine that changes the direction of the tension force but not its size. a) frictionless b) μ K = 0.200 table B aAaA aBaB A

69. Two-Body Pulley Problems Now we have two scalar equations: T = 3 a equation #1 T = 15 -1.5 a equation #2 Step 11: Solve by sub #1 into #2 3 a = 15 -1.5 a 4.5 a = 15 a = 3.3 m/s 2 a A = 3.3 m/s 2 [ right ] a B = ? M A = 3.0 kg M B = 1.5 kg Pulley: a machine that changes the direction of the tension force but not its size. a) frictionless b) μ K = 0.200 table B aAaA aBaB A

70. Two-Body Pulley Problems Now we have two scalar equations: T = 3 a equation #1 T = 15 -1.5 a equation #2 Step 11: Solve by sub #1 into #2 3 a = 15 -1.5 a 4.5 a = 15 a = 3.3 m/s 2 a A = 3.3 m/s 2 [ right ] a B = 3.3 m/s 2 [ down ] M A = 3.0 kg M B = 1.5 kg Pulley: a machine that changes the direction of the tension force but not its size. a) frictionless b) μ K = 0.200 table B aAaA aBaB A

71. Two-Body Pulley Problems Now we have two scalar equations: T = 3 a equation #1 T = 15 -1.5 a equation #2 Step 11: Solve by sub #1 into #2 3 a = 15 -1.5 a 4.5 a = 15 a = 3.3 m/s 2 a A = 3.3 m/s 2 [ right ] a B = 3.3 m/s 2 [ down ] How can we find the magnitude of the tension force in the string? M A = 3.0 kg M B = 1.5 kg Pulley: a machine that changes the direction of the tension force but not its size. a) frictionless b) μ K = 0.200 table B aAaA aBaB A

72. Two-Body Pulley Problems Now we have two scalar equations: T = 3 a equation #1 T = 15 -1.5 a equation #2 Step 11: Solve by sub #1 into #2 3 a = 15 -1.5 a 4.5 a = 15 a = 3.3 m/s 2 a A = 3.3 m/s 2 [ right ] a B = 3.3 m/s 2 [ down ] Back -sub into #1 T = 3 a M A = 3.0 kg M B = 1.5 kg Pulley: a machine that changes the direction of the tension force but not its size. a) frictionless b) μ K = 0.200 table B aAaA aBaB A

73. Two-Body Pulley Problems Now we have two scalar equations: T = 3 a equation #1 T = 15 -1.5 a equation #2 Step 11: Solve by sub #1 into #2 3 a = 15 -1.5 a 4.5 a = 15 a = 3.3 m/s 2 a A = 3.3 m/s 2 [ right ] a B = 3.3 m/s 2 [ down ] Back -sub into #1 T = 3 a tension in string T = 3 ( 3.3) = 9.9 N M A = 3.0 kg M B = 1.5 kg Pulley: a machine that changes the direction of the tension force but not its size. a) frictionless b) μ K = 0.200 table B aAaA aBaB A

74. Two-Body Pulley Problems Now we have two scalar equations: T = 3 a equation #1 T = 15 -1.5 a equation #2 Step 11: Solve by sub #1 into #2 3 a = 15 -1.5 a 4.5 a = 15 a = 3.3 m/s 2 a A = 3.3 m/s 2 [ right ] a B = 3.3 m/s 2 [ down ] Back -sub into #1 T = 3 a tension in string T = 3 ( 3.3) = 9.9 N M A = 3.0 kg M B = 1.5 kg Pulley: a machine that changes the direction of the tension force but not its size. a) frictionless b) μ K = 0.200 table B aAaA aBaB Step 12 Conclusion: For part a) where there is no friction between block A and the table, a A = 3.3 m/s 2 [ right ] and a B = 3.3 m/s 2 [ down ] and the tension in the string is 9.9 N A

75. Two-Body Pulley Problems Now let's look at part b): Assuming the system accelerates from rest, find the acceleration of each mass and the size of the tension force in the string if the coefficient of friction between table and block A is... a) 0 (table is frictionless) b) μ K = 0.200 M A = 3.0 kg M B = 1.5 kg Pulley: a machine that changes the direction of the tension force but not its size. a) frictionless b) μ K = 0.200 table A B aAaA aBaB

76. Two-Body Pulley Problems Now let's look at part b): Assuming the system accelerates from rest, find the acceleration of each mass and the size of the tension force in the string if the coefficient of friction between table and block A is... a) 0 (table is frictionless) b) μ K = 0.200 Note | a A | = | a B | = a as before. Which free-body diagram will need to be modified in part b) ? M A = 3.0 kg M B = 1.5 kg Pulley: a machine that changes the direction of the tension force but not its size. a) frictionless b) μ K = 0.200 table A B aAaA aBaB

77. Two-Body Pulley Problems Now let's look at part b): Assuming the system accelerates from rest, find the acceleration of each mass and the size of the tension force in the string if the coefficient of friction between table and block A is... a) 0 (table is frictionless) b) μ K = 0.200 Note | a A | = | a B | = a as before. The FBD for block A will need to be modified because kinetic friction now acts. The equation of motion developed for block A will also change. The FBD for block B will not change and the equation of motion developed for block B will still be valid. M A = 3.0 kg M B = 1.5 kg Pulley: a machine that changes the direction of the tension force but not its size. a) frictionless b) μ K = 0.200 table A B aAaA aBaB

78. Two-Body Pulley Problems What addition must be made to our block A FBD ? M A = 3.0 kg M B = 1.5 kg Pulley: a machine that changes the direction of the tension force but not its size. a) frictionless b) μ K = 0.200 table A B aAaA aBaB a A = + a a) M A = 3.0 kg T = +T F g = 30 N F N = 30 N X Y A

79. Two-Body Pulley Problems What addition must be made to our block A FBD ? M A = 3.0 kg M B = 1.5 kg Pulley: a machine that changes the direction of the tension force but not its size. a) frictionless b) μ K = 0.200 table A B aAaA aBaB a A = + a a) M A = 3.0 kg T = +T F g = 30 N F N = 30 N X Y fkfk A

80. Two-Body Pulley Problems What addition must be made to our block A FBD ? Does F N change as a result of adding f k to our FBD? M A = 3.0 kg M B = 1.5 kg Pulley: a machine that changes the direction of the tension force but not its size. a) frictionless b) μ K = 0.200 table A B aAaA aBaB a A = + a a) M A = 3.0 kg T = +T F g = 30 N F N = 30 N X Y fkfk A

81. Two-Body Pulley Problems What addition must be made to our block A FBD ? F N does not change as a result of adding f k to our FBD b/c there are no changes with the y-forces. F N stays at 30 N [up]. M A = 3.0 kg M B = 1.5 kg Pulley: a machine that changes the direction of the tension force but not its size. a) frictionless b) μ K = 0.200 table A B aAaA aBaB a A = + a a) M A = 3.0 kg T = +T F g = 30 N F N = 30 N X Y fkfk A

82. Two-Body Pulley Problems What addition must be made to our block A FBD ? F N does not change as a result of adding f k to our FBD b/c there are no changes with the y-forces. F N stays at 30 N [up]. M A = 3.0 kg M B = 1.5 kg Pulley: a machine that changes the direction of the tension force but not its size. a) frictionless b) μ K = 0.200 table A B aAaA aBaB a A = + a a) M A = 3.0 kg T = +T F g = 30 N F N = 30 N X Y f k = ? A

83. Two-Body Pulley Problems What addition must be made to our block A FBD ? F N does not change as a result of adding f k to our FBD b/c there are no changes with the y-forces. F N stays at 30 N [up]. M A = 3.0 kg M B = 1.5 kg Pulley: a machine that changes the direction of the tension force but not its size. a) frictionless b) μ K = 0.200 table A B aAaA aBaB a A = + a a) M A = 3.0 kg T = +T F g = 30 N F N = 30 N X Y f k = μ k F N A

84. Two-Body Pulley Problems What addition must be made to our block A FBD ? F N does not change as a result of adding f k to our FBD b/c there are no changes with the y-forces. F N stays at 30 N [up]. M A = 3.0 kg M B = 1.5 kg Pulley: a machine that changes the direction of the tension force but not its size. a) frictionless b) μ K = 0.200 table A B aAaA aBaB a A = + a a) M A = 3.0 kg T = +T F g = 30 N F N = 30 N X Y f k = μ k F N =.2X30 A

85. Two-Body Pulley Problems What addition must be made to our block A FBD ? F N does not change as a result of adding f k to our FBD b/c there are no changes with the y-forces. F N stays at 30 N [up]. M A = 3.0 kg M B = 1.5 kg Pulley: a machine that changes the direction of the tension force but not its size. a) frictionless b) μ K = 0.200 table A B aAaA aBaB a A = + a a) M A = 3.0 kg T = +T F g = 30 N F N = 30 N X Y f k = μ k F N =.2X30 = 6.0 N A

86. Two-Body Pulley Problems What addition must be made to our block A FBD ? F N does not change as a result of adding f k to our FBD b/c there are no changes with the y-forces. F N stays at 30 N [up]. Look at x-forces. Start by stating Newton's second law equation or Equation of motion. F net x = m a x Vector Statement ??? M A = 3.0 kg M B = 1.5 kg Pulley: a machine that changes the direction of the tension force but not its size. a) frictionless b) μ K = 0.200 table A B aAaA aBaB a A = + a a) M A = 3.0 kg T = +T F g = 30 N F N = 30 N X Y f k = μ k F N =.2X30 = 6.0 N A

87. Two-Body Pulley Problems What addition must be made to our block A FBD ? F N does not change as a result of adding f k to our FBD b/c there are no changes with the y-forces. F N stays at 30 N [up]. Look at x-forces. Start by stating Newton's second law equation or Equation of motion. F net x = m a x Vector Statement T + f K = 3 a M A = 3.0 kg M B = 1.5 kg Pulley: a machine that changes the direction of the tension force but not its size. a) frictionless b) μ K = 0.200 table A B aAaA aBaB a A = + a a) M A = 3.0 kg T = +T F g = 30 N F N = 30 N X Y f k = μ k F N =.2X30 = 6.0 N A

88. Two-Body Pulley Problems What addition must be made to our block A FBD ? F N does not change as a result of adding f k to our FBD b/c there are no changes with the y-forces. F N stays at 30 N [up]. Look at x-forces. Start by stating Newton's second law equation or Equation of motion. F net x = m a x Vector Statement T + f K = 3 a Scalar Statement ??? M A = 3.0 kg M B = 1.5 kg Pulley: a machine that changes the direction of the tension force but not its size. a) frictionless b) μ K = 0.200 table A B aAaA aBaB a A = + a a) M A = 3.0 kg T = +T F g = 30 N F N = 30 N X Y f k = μ k F N =.2X30 = 6.0 N A

89. Two-Body Pulley Problems What addition must be made to our block A FBD ? F N does not change as a result of adding f k to our FBD b/c there are no changes with the y-forces. F N stays at 30 N [up]. Look at x-forces. Start by stating Newton's second law equation or Equation of motion. F net x = m a x Vector Statement T + f K = 3 a Scalar Statement T – 6 = 3 a M A = 3.0 kg M B = 1.5 kg Pulley: a machine that changes the direction of the tension force but not its size. a) frictionless b) μ K = 0.200 table A B aAaA aBaB a A = + a a) M A = 3.0 kg T = +T F g = 30 N F N = 30 N X Y f k = μ k F N =.2X30 = 6.0 N A

90. Two-Body Pulley Problems What addition must be made to our block A FBD ? F N does not change as a result of adding f k to our FBD b/c there are no changes with the y-forces. F N stays at 30 N [up]. Look at x-forces. Start by stating Newton's second law equation or Equation of motion. F net x = m a x Vector Statement T + f K = 3 a Scalar Statement T – 6 = 3 a M A = 3.0 kg M B = 1.5 kg Pulley: a machine that changes the direction of the tension force but not its size. a) frictionless b) μ K = 0.200 table A B aAaA aBaB a A = + a a) M A = 3.0 kg T = +T F g = 30 N F N = 30 N X Y f k = μ k F N =.2X30 = 6.0 N A

91. Two-Body Pulley Problems What addition must be made to our block A FBD ? F N does not change as a result of adding f k to our FBD b/c there are no changes with the y-forces. F N stays at 30 N [up]. Look at x-forces. Start by stating Newton's second law equation or Equation of motion. F net x = m a x Vector Statement T + f K = 3 a Scalar Statement T – 6 = 3 aT = ??? M A = 3.0 kg M B = 1.5 kg Pulley: a machine that changes the direction of the tension force but not its size. a) frictionless b) μ K = 0.200 table A B aAaA aBaB a A = + a a) M A = 3.0 kg T = +T F g = 30 N F N = 30 N X Y f k = μ k F N =.2X30 = 6.0 N A

92. Two-Body Pulley Problems What addition must be made to our block A FBD ? F N does not change as a result of adding f k to our FBD b/c there are no changes with the y-forces. F N stays at 30 N [up]. Look at x-forces. Start by stating Newton's second law equation or Equation of motion. F net x = m a x Vector Statement T + f K = 3 a Scalar Statement T – 6 = 3 aT = 3 a + 6 M A = 3.0 kg M B = 1.5 kg Pulley: a machine that changes the direction of the tension force but not its size. a) frictionless b) μ K = 0.200 table A B aAaA aBaB a A = + a a) M A = 3.0 kg T = +T F g = 30 N F N = 30 N X Y f k = μ k F N =.2X30 = 6.0 N A

93. Two-Body Pulley Problems What addition must be made to our block A FBD ? F N does not change as a result of adding f k to our FBD b/c there are no changes with the y-forces. F N stays at 30 N [up]. Look at x-forces. Start by stating Newton's second law equation or Equation of motion. F net x = m a x Vector Statement T + f K = 3 a Scalar Statement T – 6 = 3 aT = 3 a + 6 new equation #1 M A = 3.0 kg M B = 1.5 kg Pulley: a machine that changes the direction of the tension force but not its size. a) frictionless b) μ K = 0.200 table A B aAaA aBaB a A = + a a) M A = 3.0 kg T = +T F g = 30 N F N = 30 N X Y f k = μ k F N =.2X30 = 6.0 N A

94. Two-Body Pulley Problems Now we have two equations of motion: T = 3 a + 6 new equation #1 T = 15 -1.5 a equation #2 (no change) M A = 3.0 kg M B = 1.5 kg Pulley: a machine that changes the direction of the tension force but not its size. a) frictionless b) μ K = 0.200 table B aAaA aBaB A

95. Two-Body Pulley Problems Now we have two equations of motion: T = 3 a + 6 new equation #1 T = 15 -1.5 a equation #2 (no change) Solve by sub #1 into #2 M A = 3.0 kg M B = 1.5 kg Pulley: a machine that changes the direction of the tension force but not its size. a) frictionless b) μ K = 0.200 table B aAaA aBaB A

96. Two-Body Pulley Problems Now we have two equations of motion: T = 3 a + 6 new equation #1 T = 15 -1.5 a equation #2 (no change) Solve by sub #1 into #2 3 a + 6 = 15 -1.5 a M A = 3.0 kg M B = 1.5 kg Pulley: a machine that changes the direction of the tension force but not its size. a) frictionless b) μ K = 0.200 table B aAaA aBaB A

97. Two-Body Pulley Problems Now we have two equations of motion: T = 3 a + 6 new equation #1 T = 15 -1.5 a equation #2 (no change) Solve by sub #1 into #2 3 a + 6 = 15 -1.5 a 4.5 a = 9 M A = 3.0 kg M B = 1.5 kg Pulley: a machine that changes the direction of the tension force but not its size. a) frictionless b) μ K = 0.200 table B aAaA aBaB A

98. Two-Body Pulley Problems Now we have two equations of motion: T = 3 a + 6 new equation #1 T = 15 -1.5 a equation #2 (no change) Solve by sub #1 into #2 3 a + 6 = 15 -1.5 a 4.5 a = 9 a = 2.0 m/s 2 M A = 3.0 kg M B = 1.5 kg Pulley: a machine that changes the direction of the tension force but not its size. a) frictionless b) μ K = 0.200 table B aAaA aBaB A

99. Two-Body Pulley Problems Now we have two equations of motion: T = 3 a + 6 new equation #1 T = 15 -1.5 a equation #2 (no change) Solve by sub #1 into #2 3 a + 6 = 15 -1.5 a 4.5 a = 9 a = 2.0 m/s 2 a A = ____ m/s 2 [ ____ ] a B = ____ m/s 2 [ ____ ] M A = 3.0 kg M B = 1.5 kg Pulley: a machine that changes the direction of the tension force but not its size. a) frictionless b) μ K = 0.200 table B aAaA aBaB A

100. Two-Body Pulley Problems Now we have two equations of motion: T = 3 a + 6 new equation #1 T = 15 -1.5 a equation #2 (no change) Solve by sub #1 into #2 3 a + 6 = 15 -1.5 a 4.5 a = 9 a = 2.0 m/s 2 a A = 2.0 m/s 2 [ right ] a B = ____ m/s 2 [ ____ ] M A = 3.0 kg M B = 1.5 kg Pulley: a machine that changes the direction of the tension force but not its size. a) frictionless b) μ K = 0.200 table B aAaA aBaB A

101. Two-Body Pulley Problems Now we have two equations of motion: T = 3 a + 6 new equation #1 T = 15 -1.5 a equation #2 (no change) Solve by sub #1 into #2 3 a + 6 = 15 -1.5 a 4.5 a = 9 a = 2.0 m/s 2 a A = 2.0 m/s 2 [ right ] a B = 2.0 m/s 2 [ down ] M A = 3.0 kg M B = 1.5 kg Pulley: a machine that changes the direction of the tension force but not its size. a) frictionless b) μ K = 0.200 table B aAaA aBaB A

102. Two-Body Pulley Problems Now we have two equations of motion: T = 3 a + 6 new equation #1 T = 15 -1.5 a equation #2 (no change) Solve by sub #1 into #2 3 a + 6 = 15 -1.5 a 4.5 a = 9 a = 2.0 m/s 2 a A = 2.0 m/s 2 [ right ] a B = 2.0 m/s 2 [ down ] Back-sub in equation #1 to get T M A = 3.0 kg M B = 1.5 kg Pulley: a machine that changes the direction of the tension force but not its size. a) frictionless b) μ K = 0.200 table B aAaA aBaB A

103. Two-Body Pulley Problems Now we have two equations of motion: T = 3 a + 6 new equation #1 T = 15 -1.5 a equation #2 (no change) Solve by sub #1 into #2 3 a + 6 = 15 -1.5 a 4.5 a = 9 a = 2.0 m/s 2 a A = 2.0 m/s 2 [ right ] a B = 2.0 m/s 2 [ down ] Back-sub in equation #1 to get T T = 3 ( 2.0) + 6 = ? M A = 3.0 kg M B = 1.5 kg Pulley: a machine that changes the direction of the tension force but not its size. a) frictionless b) μ K = 0.200 table B aAaA aBaB A

104. Two-Body Pulley Problems Now we have two equations of motion: T = 3 a + 6 new equation #1 T = 15 -1.5 a equation #2 (no change) Solve by sub #1 into #2 3 a + 6 = 15 -1.5 a 4.5 a = 9 a = 2.0 m/s 2 a A = 2.0 m/s 2 [ right ] a B = 2.0 m/s 2 [ down ] Back-sub in equation #1 to get T T = 3 ( 2.0) + 6 = 12 N M A = 3.0 kg M B = 1.5 kg Pulley: a machine that changes the direction of the tension force but not its size. a) frictionless b) μ K = 0.200 table B aAaA aBaB A

105. Two-Body Pulley Problems Now we have two equations of motion: T = 3 a + 6 new equation #1 T = 15 -1.5 a equation #2 (no change) Solve by sub #1 into #2 3 a + 6 = 15 -1.5 a 4.5 a = 9 a = 2.0 m/s 2 a A = 2.0 m/s 2 [ right ] a B = 2.0 m/s 2 [ down ] Back-sub in equation #1 to get T T = 3 ( 2.0) + 6 = 12 N M A = 3.0 kg M B = 1.5 kg Pulley: a machine that changes the direction of the tension force but not its size. a) frictionless b) μ K = 0.200 table B aAaA aBaB Conclusion: For part b) where there is friction between block A and the table, a A = 2.0 m/s 2 [ right ] and A B = 2.0 m/s 2 [ down ] and the tension in the string is 12 N A

106. A Related Statics Problem What minimum coefficient of static friction is needed between block A and the table to keep this system at rest? M A = 3.0 kg M B = 1.5 kg Pulley table A B

107. A Related Statics Problem What minimum coefficient of static friction is needed between block A and the table to keep this system at rest? Use a “systems” approach to solve: To balance the 15 N pull of gravity on block B, f s = 15 N on block A M A = 3.0 kg M B = 1.5 kg Pulley table A B fsfs F g = 15 N

108. A Related Statics Problem What minimum coefficient of static friction is needed between block A and the table to keep this system at rest? Use a “systems” approach to solve: To balance the 15 N pull of gravity on block B, f s = 15 N on block A f s ≤ μ s F N M A = 3.0 kg M B = 1.5 kg Pulley table A B fsfs F g = 15 N

109. A Related Statics Problem What minimum coefficient of static friction is needed between block A and the table to keep this system at rest? Use a “systems” approach to solve: To balance the 15 N pull of gravity on block B, f s = 15 N on block A f s ≤ μ s F N 15 ≤ μ s 30 M A = 3.0 kg M B = 1.5 kg Pulley table A B fsfs F g = 15 N

110. A Related Statics Problem What minimum coefficient of static friction is needed between block A and the table to keep this system at rest? Use a “systems” approach to solve: To balance the 15 N pull of gravity on block B, f s = 15 N on block A f s ≤ μ s F N 15 ≤ μ s 30 μ s 30 ≥ 15 M A = 3.0 kg M B = 1.5 kg Pulley table A B fsfs F g = 15 N

111. A Related Statics Problem What minimum coefficient of static friction is needed between block A and the table to keep this system at rest? Use a “systems” approach to solve: To balance the 15 N pull of gravity on block B, f s = 15 N on block A f s ≤ μ s F N 15 ≤ μ s 30 μ s 30 ≥ 15 M A = 3.0 kg M B = 1.5 kg Pulley table A B fsfs F g = 15 N

112. A Related Statics Problem What minimum coefficient of static friction is needed between block A and the table to keep this system at rest? Use a “systems” approach to solve: To balance the 15 N pull of gravity on block B, f s = 15 N on block A f s ≤ μ s F N μ s ≥ 15/30 15 ≤ μ s 30 μ s 30 ≥ 15 M A = 3.0 kg M B = 1.5 kg Pulley table A B fsfs F g = 15 N

113. A Related Statics Problem What minimum coefficient of static friction is needed between block A and the table to keep this system at rest? Use a “systems” approach to solve: To balance the 15 N pull of gravity on block B, f s = 15 N on block A f s ≤ μ s F N μ s ≥ 15/30 15 ≤ μ s 30μ s ≥ 0.50 μ s 30 ≥ 15 M A = 3.0 kg M B = 1.5 kg Pulley table A B fsfs F g = 15 N

114. A Related Statics Problem What minimum coefficient of static friction is needed between block A and the table to keep this system at rest? Use a “systems” approach to solve: To balance the 15 N pull of gravity on block B, f s = 15 N on block A f s ≤ μ s F N μ s ≥ 15/30 15 ≤ μ s 30μ s ≥ 0.50 μ s 30 ≥ 15The minimum coefficient of static friction is 0.50 M A = 3.0 kg M B = 1.5 kg Pulley table A B fsfs F g = 15 N