1. Physics 207: Lecture 9, Pg 1 Lecture 10 l Today:  Review session Assignment: For Monday, Read through Chapter 8 There will be a reading quiz posted at Mastering Physics. Exam Thursday, Oct. 6 th from 7:15-8:45 PM Chapters 1-6,7 One 8½ X 11 hand written note sheet and a calculator (for trig.)

2. Physics 207: Lecture 9, Pg 2 Textbook Chapters l Chapter 1 Concept of Motion l Chapter 2 1D Kinematics l Chapter 3 Vector and Coordinate Systems l Chapter 4 Dynamics I, Two-dimensional motion l Chapter 5 Forces and Free Body Diagrams l Chapter 6 Force and Newton’s 1 st and 2 nd Laws l Chapter 7 Newton’s 3 rd Law Exam will reflect most key points (but not all) 25-30% of the exam will be more conceptual 70-75% of the exam is problem solving

3. Physics 207: Lecture 9, Pg 3 Example with pulley l A mass M is held in place by a force F. Find the tension in each segment of the massless ropes and the magnitude of F.  Assume the pulleys are massless and frictionless. The action of a massless frictionless pulley is to change the direction of a tension. This is an example of static equilibrium.  M T5T5 T4T4 T3T3 T2T2 T1T1 F

4. Physics 207: Lecture 9, Pg 4 Example with pulley l A mass M is held in place by a force F. Find the tension in each segment of the rope and the magnitude of F.  Assume the pulleys are massless and frictionless.  Assume the rope is massless. The action of a massless frictionless pulley is to change the direction of a tension. Here F = T 1 = T 2 = T 3 = T Equilibrium means  F = 0 for x, y & z For example: y-dir ma = 0 = T 2 + T 3 – T 5 and ma = 0 = T 5 – Mg So T 5 = Mg = T 2 + T 3 = 2 F  T = Mg/2  M T5T5 T4T4 T3T3 T2T2 T1T1 F

5. Physics 207: Lecture 9, Pg 5 Another example with a pulley Three blocks are connected on the table as shown. The table is frictionless & the masses are m 1 = 4.0 kg, m 2 = 1.0 kg and m 3 = 2.0 kg. (A) 3 Free Body Diagrams m1m1 T1T1 m2m2 m3m3 m2gm2g N m3gm3g m1gm1g T3T3 T1T1

6. Physics 207: Lecture 9, Pg 6 Another example with a pulley Three blocks are connected on the table as shown. The table is frictionless & the masses are m 1 = 4.0 kg, m 2 = 1.0 kg and m 3 = 2.0 kg. (1) m 1 a 1y = -m 1 g + T 1 (2) m 2 a 2yx = -T 1 + T 3 (3) m 3 a 3y = -m 3 g + T 3 Let a = a 1y = a 12y = - a 3y or m 3 a= m 3 g - T 3 Add (1) & (2) (m 1 +m 2 )a = -m 1 g + T 3 Now add (3) (m 1 +m 2 +m 3 )a = -m 1 g + m 3 g a = (-m 1 g + m 3 g)/(m 1 +m 2 +m 3 )= -20 / 7 m/s 2 m1m1 T1T1 m2m2 m3m3 m2gm2g N m3gm3g m1gm1g T3T3 T1T1 T1T1 T3T3

7. Physics 207: Lecture 9, Pg 7 Problem recast as 1D motion Three blocks are connected on the table as shown. The center table has a coefficient of kinetic friction of  K =0.40, the masses are m 1 = 4.0 kg, m 2 = 1.0 kg and m 3 = 2.0 kg. m1m1 m2m2 m3m3 m2gm2g N m3gm3g m1gm1g T3T3 T1T1 frictionless m 1 g > m 3 g and m 1 g > (  k m 2 g + m 3 g) and friction opposes motion (starting with v = 0) so f f is to the right and a is to the left (negative) f

8. Physics 207: Lecture 9, Pg 8 Another example with a pulley Three blocks are connected on the table as shown. The table has a coefficient of kinetic friction of  K =0.40, the masses are m 1 = 4.0 kg, m 2 = 1.0 kg and m 3 = 2.0 kg. (A) FBD (except for friction) (B) So what about friction ? m1m1 T1T1 m2m2 m3m3 m2gm2g N m3gm3g m1gm1g T3T3 T1T1

9. Physics 207: Lecture 9, Pg 9 Problem recast as 1D motion Three blocks are connected on the table as shown. The center table has a coefficient of kinetic friction of  K =0.40, the masses are m 1 = 4.0 kg, m 2 = 1.0 kg and m 3 = 2.0 kg. m1m1 m2m2 m3m3 m2gm2g N m3gm3g m1gm1g T3T3 T1T1 frictionless x-dir: 1.  F x = m 2 a =  k m 2 g - T 1 + T 3 m 3 a = m 3 g - T 3 m 1 a =  m 1 g + T 1 Add all three: ( m 1 + m 2 + m 3 ) a =  k m 2 g  m 3 g – m 1 g f T3T3 T1T1

10. Physics 207: Lecture 9, Pg 10 Analyzing motion plots l The graph is a plot of velocity versus time for an object. Which of the following statements is correct? A The acceleration of the object is zero. B The acceleration of the object is constant. C The acceleration of the object is positive and increasing in magnitude. D The acceleration of the object is negative and decreasing in magnitude. E The acceleration of the object is positive and decreasing in magnitude. Velocity Time

11. Physics 207: Lecture 9, Pg 11 Chapter 2

12. Physics 207: Lecture 9, Pg 12 Chapter 2 Also average speed and average velocity

13. Physics 207: Lecture 9, Pg 13 Chapter 3

14. Physics 207: Lecture 9, Pg 14 Chapter 3

15. Physics 207: Lecture 9, Pg 15 Chapter 4

16. Physics 207: Lecture 9, Pg 16 Chapter 4

17. Physics 207: Lecture 9, Pg 17 Chapter 5

18. Physics 207: Lecture 9, Pg 18 Chapter 5 & 6

19. Physics 207: Lecture 9, Pg 19 Chapter 6 Note: Drag in air is proportional to v 2

20. Physics 207: Lecture 9, Pg 20 Chapter 7

21. Physics 207: Lecture 9, Pg 21 Conceptual Problem The pictures below depict cannonballs of identical mass which are launched upwards and forward. The cannonballs are launched at various angles above the horizontal, and with various velocities, but all have the same vertical component of velocity.

22. Physics 207: Lecture 9, Pg 22 Graphing problem The figure shows a plot of velocity vs. time for an object moving along the x-axis. Which of the following statements is true? (A) The average acceleration over the 11.0 second interval is -0.36 m/s 2 (B) The instantaneous acceleration at t = 5.0 s is -4.0 m/s 2 (C) Both A and B are correct. (D) Neither A nor B are correct. Note:  x ≠ ½ a avg  t 2

23. Physics 207: Lecture 9, Pg 23 Conceptual Problem A block is pushed up a 20º ramp by a 15 N force which may be applied either horizontally (P1) or parallel to the ramp (P2). How does the magnitude of the normal force N depend on the direction of P? (A) N will be smaller if P is horizontal than if it is parallel the ramp. (B) N will be larger if P is horizontal than if it is parallel to the ramp. (C) N will be the same in both cases. (D) The answer will depend on the coefficient of friction. 20 °

24. Physics 207: Lecture 9, Pg 24 Conceptual Problem A cart on a roller-coaster rolls down the track shown below. As the cart rolls beyond the point shown, what happens to its speed and acceleration in the direction of motion? A. Both decrease. B. The speed decreases, but the acceleration increases. C. Both remain constant. D. The speed increases, but acceleration decreases. E. Both increase. F. Other

25. Physics 207: Lecture 9, Pg 25 Sample Problem l A 200 kg wood crate sits in the back of a truck. The coefficients of friction between the crate and the truck are μ s = 0.9 and μ k = 0.5. The truck starts moving up a 20° slope. What is the maximum acceleration the truck can have without the crate slipping out the back? l Solving:  Visualize the problem, Draw a picture if necessary  Identify the system and make a Free Body Diagram  Choose an appropriate coordinate system  Apply Newton’s Laws with conditional constraints (friction)  Solve

26. Physics 207: Lecture 9, Pg 26 Sample Problem l A physics student on Planet Exidor throws a ball that follows the parabolic trajectory shown. The ball’s position is shown at one-second intervals until t = 3 s. At t = 1 s, the ball’s velocity is v = (2 i + 2 j) m/s. a. Determine the ball’s velocity at t = 0 s, 2 s, and 3 s. b. What is the value of g on Planet Exidor?

27. Physics 207: Lecture 9, Pg 27 Sample Problem l You have been hired to measure the coefficients of friction for the newly discovered substance jelloium. Today you will measure the coefficient of kinetic friction for jelloium sliding on steel. To do so, you pull a 200 g chunk of jelloium across a horizontal steel table with a constant string tension of 1.00 N. A motion detector records the motion and displays the graph shown. l What is the value of μ k for jelloium on steel?

28. Physics 207: Lecture 9, Pg 28 Sample Problem  F x =ma = F - f f = F -  k N = F -  k mg  F y = 0 = N – mg  k = (F - ma) / mg & x = ½ a t 2  0.80 m = ½ a 4 s 2 a = 0.40 m/s 2  k = (1.00 - 0.20 · 0.40 ) / (0.20 ·10.) = 0.46

29. Physics 207: Lecture 9, Pg 29 Exercise: Newton’s 2 nd Law A. 8 x as far B. 4 x as far C. 2 x as far D. 1/4 x as far A force of 2 Newtons acts on a cart that is initially at rest on an air track with no air and pushed for 1 second. Because there is friction (no air), the cart stops immediately after I finish pushing. It has traveled a distance, D. Air Track Cart Force Next, the force of 2 Newtons acts again but is applied for 2 seconds. The new distance the cart moves relative to D is:

30. Physics 207: Lecture 9, Pg 30 Exercise: Solution Air Track Cart Force (B) 4 x as long We know that under constant acceleration,  x = a (  t) 2 /2 (when v 0 =0) Here  t 2 =2  t 1, F 2 = F 1  a 2 = a 1

31. Physics 207: Lecture 9, Pg 31 Another question to ponder How high will it go? l One day you are sitting somewhat pensively in an airplane seat and notice, looking out the window, one of the jet engines running at full throttle. From the pitch of the engine you estimate that the turbine is rotating at 3000 rpm and, give or take, the turbine blade has a radius of 1.00 m. If the tip of the blade were to suddenly break off (it occasionally does happen with negative consequences) and fly directly upwards, then how high would it go (assuming no air resistance and ignoring the fact that it would have to penetrate the metal cowling of the engine.)

32. Physics 207: Lecture 9, Pg 32 Another question to ponder How high will it go?   = 3000 rpm = (3000 x 2  / 60) rad/s = 314 rad/s  r = 1.00 m v o =  r = 314 m/s (~650 mph!) l h = h 0 + v 0 t – ½ g t 2 l v h = 0 = v o – g t  t = v o / g So l h = v 0 t – ½ g t 2 = ½ v o 2 / g = 0.5 x 314 2 / 9.8 = 5 km or ~ 3 miles

33. Physics 207: Lecture 9, Pg 33 Sample exam problem An object is at first travelling due north, turns and finally heads due west while increasing its speed. The average acceleration for this maneuver is pointed A directly west. B somewhere between west and northwest. C somewhere between west and southwest. D somewhere between northwest and north. E somewhere between southwest and south. F None of these are correct

34. Physics 207: Lecture 9, Pg 34 Sample exam problem An object is at first travelling due north, turns and finally heads due west while increasing its speed. The average acceleration for this maneuver is pointed a = (v f – v i ) /  t A directly west. B somewhere between west and northwest. C somewhere between west and southwest. D somewhere between northwest and north. E somewhere between southwest and south. F None of these are correct

35. Physics 207: Lecture 9, Pg 35 Sample exam problem A small block moves along a frictionless incline which is 45 ° from horizontal. Gravity acts down at 10 m/s 2. There is a massless cord pulling on the block. The cord runs parallel to the incline over a pulley and then straight down. There is tension, T 1, in the cord which accelerates the block at 2.0 m/s 2 up the incline. The pulley is suspended with a second cord with tension, T 2. A. What is the tension magnitude, T 1, in the 1 st cord? B. What is the tension magnitude,T 2, in the 2 nd cord? (Assume T 1 = 50. N if you don’t have an answer to part A.)

36. Physics 207: Lecture 9, Pg 36 Sample exam problem a = 2.0 m/s 2 up the incline. What is the tension magnitude, T 1, in the 1 st cord? Use a FBD! Along the block surface  F x = m a x = -mg sin  + T T = 5 x 2 N + 5 x 10 x 0.7071 N = (10 + 35) N = 45 N

37. Physics 207: Lecture 9, Pg 37 Sample exam problem a = 0.0 m/s 2 at the pulley. What is the tension magnitude,T 2, in the 2 nd cord? Use a FBD!

38. Physics 207: Lecture 9, Pg 38 Conceptual Problem l A person initially at point P in the illustration stays there a moment and then moves along the axis to Q and stays there a moment. She then runs quickly to R, stays there a moment, and then strolls slowly back to P. Which of the position vs. time graphs below correctly represents this motion?

39. Physics 207: Lecture 9, Pg 39 Sample exam problem l You have a 2.0 kg block that moves on a linear path on a horizontal surface. The coefficient of kinetic friction between the block and the path is μ k. Attached to the block is a horizontally mounted massless string as shown in the figure below. The block includes an accelerometer which records acceleration vs. time. As you increase the tension in the rope the block experiences an increasingly positive acceleration. At some point in time the rope snaps and then the block slides to a stop (at a time of 10 seconds). Gravity, with g = 10 m/s2, acts downward.

40. Physics 207: Lecture 9, Pg 40 Sample exam problem A. At what time does the string break and, in one sentence, explain your reasoning? B. What speed did the block have when the string broke? C. What is the value of μ k ? D. Using μ k above (or a value of 0.25 if you don’t have one), what was the tension in the string at t = 2 seconds?

41. Physics 207: Lecture 9, Pg 41 Sample exam problem B. What speed did the block have when the string broke? Don’t know initial v (t=0) so can’t integrate area at t < 4 sec. v f = 0 m/s and from t = 4 to 10 sec (6 second) a = - 2 m/s 2 0 = v i + a t = v i – 2 x 6 m/s  v i = 12 m/s

42. Physics 207: Lecture 9, Pg 42 Sample exam problem C. What is the value of μ k ?Use a FBD!  F x = m a x = - f k = - μ k N  F y = 0 = mg – N  N = mg So m a x = - f k = - μ k mg  μ k = - a x / g = - (-2)/10 = 0.20

43. Physics 207: Lecture 9, Pg 43 Sample exam problem D. What was the tension in the string at t = 2 seconds? Again a FBD!  F x = m a x = - f k + T  F y = 0 = mg – N  N = mg T = f k + m a x = (0.20 x 2 x 10 + 2 x 3 ) N = 10 N

44. Physics 207: Lecture 9, Pg 44 Recap Exam Thursday, Oct. 6 th from 7:15-8:45 PM Chapters 1-6, 7 One 8½ X 11 hand written note sheet and a calculator (for trig.)