1. Three blocks of masses M 1 =2 kg, M 2 =4 kg, and M 3 =6 kg are connected by strings on a frictionless inclined plane of 60 o, as shown in the figure below. A force of F=120N is applied upward along the incline, causing the system of three masses to accelerate up the incline. Consider the strings to be massless and taut. What is the acceleration of M 2 ? Find the tensions T 1 and T 2 in the two sections of string.

2. Question 1 Question 2 Question 3 Question 4 Question 5 Question 6 Question 7 Question 8 Question 9 Question 10 Question 11 Question 12 Question 13 Question 14 Question 15 Paired Problem 1

3. Notes *For simplicity when analyzing this system, we choose the the x-axis to be parallel to the surface of the incline and the y-axis to be perpendicular to the surface. Let’s define the positive direction as “upward” for both x and y. *Vectors quantities will be in bold type

4. A : the same as the acceleration of the system as a whole.A B : less than the acceleration of the system as a whole.B C : greater than the acceleration of the system as a whole.C 1.Which one of these statements finishes the following sentence correctly? (there is no acceleration in the y-direction, so “a” refers to the acceleration in the x-direction throughout this tutorial) The acceleration of M 2 is expected to be…

5. Newton’s Second Law tells us: F=Ma where: M = M 1 +M 2 +M 3 (total mass of the system) F is the net force on the system a is the acceleration of the whole system F = ( M 1 +M 2 +M 3 )a = M 1 a+M 2 a+M 3 a Thus, each block accelerates with the same “a,” which is also the acceleration of the system as a whole. Choice: A Correct

6. Remember that the strings are all taut, which allows the masses to move together. Choice: B Incorrect

7. Remember that the strings are all taut, which allows the masses to move together. Choice: C Incorrect

8. 2. Which one of the following free body diagrams is correct for the system of three blocks? Weight (W=Mg) Force of Friction (F f ) Normal Force (F N ) (the length of the arrows does not represent the magnitudes of the forces here) A:A: C:C: D:D: B:B: F MgMg F F F MgMg MgMg MgMg FNFN FNFN FNFN FfFf

9. It is necessary to include the normal force (F N ). Choice: A Incorrect

10. This diagram shows all of the forces acting on the system. Choice: B Correct

11. The gravitational force always acts perpendicular to the ground, not perpendicular to the surface of the plane. Also, there is no friction between the blocks and the inclined plane. Choice: C Incorrect

12. The gravitational force always acts perpendicular to the ground, not perpendicular to the plane of contact. Choice: D Incorrect

13. 3. A good choice of coordinate axes may be to consider the x-axis parallel to the surface of the inclined plane and the y-axis perpendicular to it. (The positive x and y directions should be upwards.) Which forces are aligned with these axes? A: F N and WA B: F and WB C: F N and FC

14. F N is in the positive y-direction (always perpendicular to the surface), but W is not along either coordinate axis. Remember that W=Mg, where g is gravitational acceleration, which always points downward (towards Earth). Choice: A Incorrect

15. F is in the positive x-direction, but W is not along either coordinate axis. Remember that W=Mg, where g is the acceleration due to gravity, which always points downward (towards Earth). Choice: B Incorrect

16. Choice: C Correct F N is directed in the positive y- direction and F is directed in the positive x-direction.

17. 4. When we resolve the gravitational force into its x and y components, we get which one of the following? (the length of the arrows does not represent the magnitudes of the forces here) A:A: C:C: D:D: B:B: Mgcos(  )Mgsin(  ) Mgcos  )Mgsin(  ) Mgcos(  ) Mgsin(  )Mgcos(  )

18. Mgcos(  ) should point in the negative y-direction. Choice: A Incorrect

19. Mgsin(  ) should point in the negative x-direction. Choice: B Incorrect

20. Here, the components are correctly resolved. Choice: C Correct

21. Choice: D Incorrect Check the directions of both components.

22. 5. Which one of the following is the correct equation of motion (x-component) for the whole system (using Newton’s Second Law)? Remember M=M 1 +M 2 +M 3 A: F=MaA B: F-Mgsin(  )=0B C: F-Mgsin(  )=MaC

23. Choice: A Incorrect Don’t forget the effect of gravity

24. Choice: B Incorrect The original problem statement tells us that the applied force is causing upward acceleration. Therefore, there is a net force; it is not equal to zero.

25. Choice: C Correct This equation is correct according to Newton’s Second Law. The applied force on the system of blocks is in the opposite direction to the x-component of the weight of the blocks. We know from the original problem statement that the net force is not zero.

26. 6. The acceleration of the system of blocks can be expressed by which of the following equations? A:A B:B C:C

27. This equation is incorrect because it doesn’t take gravity into consideration. Choice: A Incorrect

28. This is what we get when we rearrange the equation that we found for force in question 6. Choice: B Correct

29. This is just the net force on the system. We are looking for the acceleration of the system. Choice: C Incorrect

30. 7. Determine the value of the acceleration a of the blocks up the incline.

31. Reasoning: Where: F=120N M= M 1 +M 2 +M 3 = 2 kg +4 kg+6 kg= 12 kg g= 9.8 m/s 2

32. As discussed earlier, the acceleration of M 2 is the same as that of the system as a whole because the blocks move together. Now let’s focus on finding the tensions T 1 and T 2.

33. 8. Which one of these is the free body diagram of M 2 ? (the length of the arrows does not represent the magnitudes of the forces here) A:A: C:C: D:D: B:B: M2gM2g T2T2 T1T1 FNFN M2gM2gT1T1 FNFN F+T 2 FNFN T2T2 M1gM1g M2gM2g T2T2 T1T1 M2gM2g FNFN

34. F only acts on M 3. Choice: A Incorrect

35. This diagram shows all of the forces acting upon M 2. Choice: B Correct

36. The gravitational force on M 1 does not act on M 2. Therefore, it does not belong in the free body diagram for M 2. Include T 1. Choice: C Incorrect

37. Choice: D Incorrect The force on an object from contact with a string in tension always acts in the direction away from the object. The directions of T 1 and T 2 should be switched.

38. 9. Which of the following is the correct representation of the x-component of Newton’s second law? You must resolve the gravitational force into components. A :A B :B C :C

39. This expression does not include the gravitational force. Choice: A Incorrect

40. This expression is consistent with the free body diagram. Choice: B Correct

41. Gravitational force and tension T 1 should also be considered because they are acting on M 2. Choice: C Incorrect

42. 10. We have two unknowns, T 1 and T 2. We need two equations. We have one equation from the free body diagram of M 2. Get the other one from the free body diagram of M 1. The free body diagram for M 1 is: (the length of the arrows does not represent the magnitudes of the forces here) A:A: C:C: D:D: B:B: T1T1 T1T1 F FNFN FNFN FNFN M1gM1g M1gM1g M1gM1g T1+T2T1+T2 M1gM1g FNFN

43. All of the forces are shown correctly. Choice: A Correct

44. T 2 doesn’t act on M 1. Choice: B Incorrect

45. F only acts on M 3. Choice: C Incorrect

46. The normal force acts perpendicular to the surface of the plane of contact. Choice: D Incorrect

47. 11. Resolving the gravitational force into its components and applying Newton’s Second Law gives us which of the following as the x- component of the equation of motion for M 1 ? A :A B :B C :C

48. T 1 is not the net force. Remember the force in Newton’s Second Law is the net force. Gravity acts upon M 1. Choice: A Incorrect

49. This equation is consistent with the free body diagram for M 1. Choice: B Correct

50. This would be correct if the angle of inclination were 90 degrees. Choice: C Incorrect

51. 12. Which equation should be solved first? A :A B :B

52. We start with the other equation because this one contains two unknowns. It is not possible to solve for two unknowns using a single equation. Choice: A Incorrect

53. This equation should be solved first since only T 1 is unknown. Choice: B Correct

54. 13. After solving for T 1 you will get : If you plug this back into the equation of motion for M 2 ( ) and simplify, you get which of the following expressions for T 2 ? A :A B :B C :C

55. This expression omits several terms. Check your algebra. Choice: A Incorrect

56. This expression does not include the contribution of gravity. Choice: B Incorrect

57. We get this result using simple algebra and factoring out common variables. Choice: C Correct

58. Now all you have to do is insert the given values and calculate the numerical values of T 1 and T 2.

59. 14) Find the value of T 1 Answer

60. The correct answer is:

61. 15) Now find T 2. Answer

62. The correct answer is :

63. Paired Problem 1) Three blocks of masses M 1 =2 kg, M 2 =4 kg, and M 3 =6 kg are connected by strings on a frictionless inclined plane of 60 o, as shown in the figure below. A force of F=120N is applied upward along the incline, causing the system of three masses to accelerate up the incline. Consider the strings to be massless and taut. What is the acceleration of M 2 ? Find the tensions T 1 and T 2 in the two sections of string.