1. PHYS 20 LESSONS Unit 3: Dynamics Lesson 3: Newton’s 2 nd law

2. Reading Segment: Newton’s 2 nd Law To prepare for this section, please read: Unit 3: p.8

3. B2. Newton's 2nd Law From Newton's 1st law, we discovered that if the forces are balanced, then the object moves at a constant velocity. Clearly, then, if the forces are unbalanced, the velocity cannot remain constant. i.e. The object must accelerate in some way. This is summarized in Newton's 2nd law.

4. Newton's 2nd Law: If an object experiences unbalanced forces, the object accelerates in the same direction as the net external force. The acceleration is calculated using the formula: a = F net or F net = m a m

5. Units for force: Based on the formula F net = m a, 1 N = 1 kg  m s 2

6. Summary: If an object experiences acceleration (unbalanced forces), then: F net = m a F net = F 1 + F 2 +... These are vector equations. Be certain to use a reference system for 1-D problems.

7. Ex. 1A 12 kg cart experiences a forward force of 48 N. If it accelerates backward at 1.8 m/s 2, then find the backward force.

8. a = 1.8 m/s 2 F 1 = 48 N F 2 12 kg

9. Ref: a = 1.8 m/s 2 Forward +F 1 = 48 N F 2 Backward -12 kg 2 nd law: F net = m aF net = F 1 + F 2 Show both equations. Then, choose which equation you can use first.

10. Ref: a = 1.8 m/s 2 Forward +F 1 = 48 N F 2 Backward -12 kg 2 nd law: F net = m a = (12 kg) (-1.8 m/s 2 ) = -21.6 N

11. F net = m a = (12 kg) (-1.8 m/s 2 ) = -21.6 N Then, F net = F 1 + F 2 F 2 = F net - F 1 = (-21.6 N) - (+48 N) = -69.6 N So,F 2 = 70 N backwards

12. Practice Problems Try these problems in the Physics 20 Workbook: Unit 3 p. 9 #1 - 5

13. Ex. 2A 120 kg cart experiences a forward force of 550 N and a backward force of 370 N. If its initial velocity is 8.00 m/s backwards, then find the time required for it to come to rest.

14. This problem has two major stages: 1. Dynamics - using forces and Newton's Laws 2. Kinematics - making a list of kinematic variables - choosing the proper acceleration equation and solving

15. Ref: a Forward +F 1 = 550 N F 2 = 370 N Backward - 120 kg 2 nd law: F net = m aF net = F 1 + F 2

16. Ref: a Forward +F 1 = 550 N F 2 = 370 N Backward - 120 kg F net = F 1 + F 2 = (+550 N) + (-370 N) = 180 N

17. F net = F 1 + F 2 = (+550 N) + (-370 N) = 180 N F net = m a a = F net = 180 N = 1.50 m/s 2  m120 kg

18. Kinematics: v i = -8.00 m/s v f = 0 (comes to rest) a = +1.50 m/s 2 d t ? Equation: a = v f - v i t

19. a = v f - v i t at = v f - v i t = v f - v i = 0 - (-8.00 m/s) a 1.50 m/s 2 = 5.33 s

20. Practice Problems Try these problems in the Physics 20 Workbook: Unit 3 p. 9 #6 - 9

21. Understanding Newton's 2nd law According to the equation a = F net, m the acceleration of an object depends on two factors: 1. F net 2. Mass

22. 1. Net External force (F net ) What is the relationship between a and F net ? If F net  5, what happens to the acceleration? What assumption is made here? What does the graph of this relationship look like?

23. Based on the equation a = F net, m acceleration has a direct relationship with F net. i.e. a  F net This is assuming that the mass of the object remains constant (controlled).

24. a  F net (direct relationship) So, if the magnitude of F net increases, then the acceleration increases i.e. the bigger the force, the bigger the acceleration In fact, if F net  5, then a  5 (if m stays constant)

25. a  F net (direct relationship) Graph: The graph of a direct relationship is a straight line through the origin. (F net is the manipulated variable) a F net

26. 2. Mass (m) What is the relationship between a and m ? If m  5, what happens to the acceleration? What assumption is made here? What does the graph of this relationship look like?

27. Based on the equation a = F net, m acceleration has a inverse relationship with mass. i.e. a  1 m This is assuming that the net force on the object remains constant (controlled).

28. a  1(inverse relationship) m So, if the mass of an object increases, then the acceleration decreases i.e. the bigger the mass, the smaller the acceleration In fact, if m  5, then a  5 (if F net stays constant) or a  1 5

29. a  1(inverse relationship) m Graph: The smaller the mass, the larger the acceleration. The bigger the mass, the smaller the acceleration. a m

30. Ex. 3The acceleration of an object is 0.60 m/s 2. If the net force on the object is tripled (i.e. multiplied by 3) and the mass is fifthed (divided by 5), find the new acceleration.

31. Based on a = F net, m a  F net : If F net  3, a  3 a  1 : If m  5, a  5 m So, new acceleration = 0.60 m/s 2  3  5 = 9.0 m/s 2

32. 2-D Acceleration Questions Recall, Newton's 2nd Law tells us that an object will always accelerate in the same direction as F net. So, when we find F net using 2-D vector sums, its direction will be the same as its acceleration

33. Ex. 1Two forces act on a 4.6 kg object: 18 N directed West 13 N directed South What is the acceleration of the object (Mag and Dir) ?

34. Force Diagram: F 1 = 18 N 4.6 kg F 2 = 13 N Since they are at right angles, find F net using tail-to-tip.

35. 1. Find F net :F 1 = 18 N  F 2 = 13 N R (= F net ) Using Pythag:Using SohCahToa: c 2 = a 2 + b 2 tan  = 13 R 2 = 18 2 + 13 2 18 R = 22.2036 N  = tan -1 (0.722) = 35.8  So, F net = 22.2 N at 35.8  S of W (54.2  W of S, 216  )

36. 2. Find acceleration: Magnitude:Direction: F net = m aBased on Newton's 2nd Law, it accelerates in the a = F net = 22.2036 Nsame direction as F net m 4.6 kg = 4.8 m/s 2 So, the acceleration is 4.8 m/s 2 at 36  S of W

37. Practice Problems Try these problems in the Physics 20 Workbook: Unit 3 p. 9 #10

38. Diagonal Force Questions Any time you have diagonal forces, you must find their x- and y-components. Then, analyze the horizontal and the vertical separately.

39. Example A 2.0 kg object experiences 2 forces If the acceleration is 40 m/s 2 West, then determine F 2 and . F 1 = 60 N 25  F2F2  2.0 kg a = 40 m/s 2

40. F 1 = 60 N 25  2.0 kg Ref: Right + Left  F 1x F 1y = 54.378 N = 25.375 N

41. F 1 = 60 N 25  2.0 kg F 1x =  54.378 N F 1y = 25.375 N F2F2  Vertical: Newton’s 1 st law (balanced forces) There is no vertical acceleration. = 25.375 N F 2y =  25.375 N The vertical forces must cancel out.

42. F 1 = 60 N 25  2.0 kg F 1x =  54.378 N F 1y = 25.375 N F2F2  Horizontal: Newton’s 2 nd law (unbalanced forces) F 2y =  25.375 N a =  40 m/s 2 F 2x

43. 2.0 kg F 1x =  54.378 N a =  40 m/s 2 F 2x = (2.0 kg) (  40 m/s 2 )  (  54.378 N) =  25.62 N

44. F 1 = 60 N 25  2.0 kg F 1x =  54.378 N F 1y = 25.375 N F2F2  F 2y =  25.375 N F 2x =  25.62 N We can now find F 2.

45. F2F2  F 2y = 25.375 N F 2x = 25.62 N Magnitude of F 2 : = 36 N Direction of F 3 : = 45 

46. Practice Problems Try these problems in the Physics 20 Workbook: Unit 3 p. 9 #11 - 12