Chapter 6 Applications of Newton’s Laws

1. Chapter 6 Applications of Newton’s Laws

2. Midterm # 1 Average = 16.35 Standard Deviation = 3.10 Well done!

3. Who Wins? Kevin Parks (5’6”, 90kg) takes a handoff from Michael Rocco, but the hole closes as he hits the line. He is pinned between to O-linemen behind him, and 3 D-linemen ahead... and everyone is pushing, lifting, grabbing, and kicking. His acceleration is 1 m/s2 upfield. What is the net force on KP? a) I need to know a lot more about angles, masses of the line men, etc. b) I could answer this if only I had the initial velocity c) I would also need to know what planet this happened on. d) 90 N, upfield. e) This isn’t the Tech game, is it?

4. Who Wins? Kevin Parks (5’6”, 90kg) takes a handoff from Michael Rocco, but the hole closes as he hits the line. He is pinned between to O-linemen behind him, and 3 D-linemen ahead... and everyone is pushing, lifting, grabbing, and kicking. His acceleration is 1 m/s2 upfield. What is the net force on KP? a) I need to know a lot more about angles, masses of the line men, etc. b) I could answer this if only I had the initial velocity c) I would also need to know what planet this happened on. d) 90 N, upfield. e) This isn’t the Tech game, is it? If you know the acceleration of a body, you know the direction of the net force. If you also know the mass, then you know the magnitude.

5. Static friction(s= 0.4) T m Will It Budge? a) moves to the left, because the force of static friction is larger than the applied force b) moves to the right, because the applied force is larger than the static friction force c) the box does not move, because the static friction force is larger than the applied force d) the box does not move, because the static friction force is exactly equal the applied force e) The answer depends on the value for μk. A box of weight 100 N is at rest on a floor where μs = 0.4. A rope is attached to the box and pulled horizontally with tension T = 30 N. Which way does the box move?

6. Static friction(s= 0.4) T m Will It Budge? a) moves to the left, because the force of static friction is larger than the applied force b) moves to the right, because the applied force is larger than the static friction force c) the box does not move, because the static friction force is larger than the applied force d) the box does not move, because the static friction force is exactly equal the applied force e) The answer depends on the value for μk. A box of weight 100 N is at rest on a floor where μs = 0.4. A rope is attached to the box and pulled horizontally with tension T = 30 N. Which way does the box move? The static friction force has a maximum of sN = 40 N. The tension in the rope is only30 N. So the pulling force is not big enough to overcome friction. Follow-up: What happens if the tension is 35 N? What about 45 N?

7. Springs Hooke’s law for springs states that the force increases with the amount the spring is stretched or compressed: The constant k is called the spring constant.

8. Springs Note: we are discussing the force of the spring on the mass. The force of the spring on the wall are equal, and opposite.

9. A mass M hangs on spring 1, stretching it length L1 Springs and Tension Mass M hangs on spring 2, stretching it length L2 S1 Now spring1 and spring2 are connected end-to-end, and M1 is hung below. How far does the combined spring stretch? S2 a) (L1 + L2) / 2 b) L1 or L2, whichever is smaller c) L1 or L2, whichever is bigger d) depends on which order the springs are attached e) L1 + L2

10. A mass M hangs on spring 1, stretching it length L1 Fs=T Springs and Tension W Mass M hangs on spring 2, stretching it length L2 S1 Now spring1 and spring2 are connected end-to-end, and M1 is hung below. How far does the combined spring stretch? S2 a) (L1 + L2) / 2 b) L1 or L2, whichever is smaller c) L1 or L2, whichever is bigger d) depends on which order the springs are attached e) L1 + L2

11. A mass M hangs on spring 1, stretching it length L1 Fs=T Springs and Tension W Mass M hangs on spring 2, stretching it length L2 S1 Now spring1 and spring2 are connected end-to-end, and M1 is hung below. How far does the combined spring stretch? S2 a) (L1 + L2) / 2 b) L1 or L2, whichever is smaller c) L1 or L2, whichever is bigger d) depends on which order the springs are attached e) L1 + L2 Spring 1 supports the weight. Spring 2 supports the weight. Both feel the same force, and stretch the same distance as before.

12. Recall -- Instantaneous acceleration Velocity vector is always in the direction of motion; acceleration vector can points in the direction velocity is changing:

13. Circular Motion and Centripetal Force An object moving in a circle must have a force acting on it; otherwise it would move in a straight line! The net force must have a component pointing to the center of the circle centripetal force This force may be provided by the tension in a string, the normal force, or friction, or....

14. The magnitude of this centripetal component of the force is given by: If the speed is constant, the direction of the acceleration (which is due to the net force) is towards the center of the circle. a a For circular motion problems, it is often convenient to choose coordinate axes with one pointing along the direction of this centripetal force Circular Motion and Centripetal Acceleration An object moving in a circle must have a force acting on it; otherwise it would move in a straight line.

15. Circular Motion An object may be changing its speed as it moves in a circle; in that case, there is a tangential acceleration as well:

16. e d c a b Missing Link A Ping-Pong ball is shot into a circular tube that is lying flat (horizontal) on a tabletop. When the Ping-Pong ball leaves the track, which path will it follow?

17. e d c a b Missing Link A Ping-Pong ball is shot into a circular tube that is lying flat (horizontal) on a tabletop. When the Ping-Pong ball leaves the track, which path will it follow? Once the ball leaves the tube, there is no longer a force to keep it going in a circle. Therefore, it simply continues in a straight line, as Newton’s First Law requires! Follow-up: What physical force provides the centripetal force?

18. T W Tetherball a) toward the top of the pole b) toward the ground c) along the horizontal component of the tension force d) along the vertical component of the tension force e) tangential to the circle In the game of tetherball, the struck ball whirls around a pole. In what direction does the net force on the ball point?

19. T Tetherball T W W In the game of tetherball, the struck ball whirls around a pole. In what direction does the net force on the ball point? a) toward the top of the pole b) toward the ground c) along the horizontal component of the tension force d) along the vertical component of the tension force e) tangential to the circle The vertical component of the tension balances the weight. The horizontal component of tension provides the necessary centripetal force that accelerates the ball to towards the pole and keeps it moving in a circle.

21. Examples of centripetal force no friction is needed to hold the track! when

22. A hockey puck of mass m is attached to a string that passes through a hole in the center of a table, as shown in the figure. The hockey puck moves in a circle of radius r. Tied to the other end of the string, and hanging vertically beneath the table, is a mass M. Assuming the tabletop is perfectly smooth, what speed must the hockey puck have if the mass M is to remain at rest? 22

23. To support mass M, the necessary tension is: necessary centripetal force: Only force on puck is tension in the string! A hockey puck of mass m is attached to a string that passes through a hole in the center of a table, as shown in the figure. The hockey puck moves in a circle of radius r. Tied to the other end of the string, and hanging vertically beneath the table, is a mass M. Assuming the tabletop is perfectly smooth, what speed must the hockey puck have if the mass M is to remain at rest?

24. Circular motion and apparent weight Car at the bottom of a dip Note: at any specific point, any curve can be approximated by a portion of a circle This normal force is the apparent, or perceived, weight

25. Going in Circles I a) N remains equal to mg b) N is smaller than mg c) N is larger than mg d) none of the above You’re on a Ferris wheel moving in a vertical circle. When the Ferris wheel is at rest, the normal force N exerted by your seat is equal to your weight mg. How does N change at the top of the Ferris wheel when you are in motion?

26. Going in Circles I a) N remains equal to mg b) N is smaller than mg c) N is larger than mg d) none of the above You’re on a Ferris wheel moving in a vertical circle. When the Ferris wheel is at rest, the normal force N exerted by your seat is equal to your weight mg. How does N change at the top of the Ferris wheel when you are in motion? You are in circular motion, so there has to be a centripetal force pointing inward. At the top, the only two forces are mg (down) and N (up), so N must be smaller than mg. Follow-up: Where is N larger than mg? 26

27. A So, as long as: vertical (up) vertical (down) horizontal at C: Centripetal acceleration must be at the top, then N>0 and pointing down. (note: before falling, apparent weight is in the opposite direction to true weight!) B C Vertical circular motion Condition for falling: N=0 27

28. Chapter 7 Work and Kinetic Energy http://people.virginia.edu/~kdp2c/downloads/WorkEnergySelections.html

29. Work Done by a Constant Force The definition of work, when the force is parallel to the displacement: SI unit: newton-meter (N·m) = joule, J 29

30. Working Hard... or Hardly Working Sisyphus pushes his rock up a hill. Atlas holds up the world. (b) Who does more work? (a)

31. Working Hard... or Hardly Working Sisyphus pushes his rock up a hill. Atlas holds up the world. (b) With no displacement, Atlas does no work (a)

32. Forces not along displacement If the force is at an angle to the displacement: 32

33. Friction and Work I a) friction does no work at all b) friction does negative work c) friction does positive work A box is being pulled across a rough floor at a constant speed. What can you say about the work done by friction?

34. Displacement N Pull f mg Friction and Work I a) friction does no work at all b) friction does negative work c) friction does positive work A box is being pulled across a rough floor at a constant speed. What can you say about the work done by friction? Friction acts in the opposite direction to the displacement, so the work is negative. Or using the definition of work (W = F (Δr)cos ), because = 180º, thenW <0.

35. Friction and Work II Can friction ever do positive work? a) yes b) no

36. Friction and Work II Can friction ever do positive work? a) yes b) no Consider the case of a box on the back of a pickup truck. If the box moves along with the truck, then it is actually the force of friction that is making the box move. ... but .... the friction isn’t really doing the work, it is only transmitting the forces to the box, while work is done by the truck engine. My view: the language is confusing so I’m not interested in arguing the point.

37. Convenient notation: the dot product The work can also be written as the dot product of the force and the displacement: vector “dot” operation: project one vector onto the other

38. Force and displacement The work done may be positive, zero, or negative, depending on the angle between the force and the displacement: 38

39. Sum of work by forces = work by sum of forces If there is more than one force acting on an object, the work done by each force is the same as the work done by the net force:

40. Units of Work Lifting 0.5 L H2O up 20 cm = 1 J

41. Play Ball! In a baseball game, the catcher stops a 90-mph pitch. What can you say about the work done by the catcher on the ball? a) catcher has done positive work b) catcher has done negative work c) catcher has done zero work

42. Play Ball! In a baseball game, the catcher stops a 90-mph pitch. What can you say about the work done by the catcher on the ball? a) catcher has done positive work b) catcher has done negative work c) catcher has done zero work The force exerted by the catcher is opposite in direction to the displacement of the ball, so the work is negative. Or using the definition of work (W = F (Δr)cos  ), because = 180º, thenW < 0. Note that the work done on the ball is negative, and its speed decreases. Follow-up: What about the work done by the ball on the catcher?

43. Tension and Work a) tension does no work at all b) tension does negative work c) tension does positive work A ball tied to a string is being whirled around in a circle with constant speed. What can you say about the work done by tension?

44. T v Tension and Work a) tension does no work at all b) tension does negative work c) tension does positive work A ball tied to a string is being whirled around in a circle with constant speed. What can you say about the work done by tension? No work is done because the force acts in a perpendicular direction to the displacement. Or using the definition of work (W = F (Δr)cos ), = 90º, thenW = 0. Follow-up: Is there a force in the direction of the velocity?

45. N A ball of mass m rolls down a ramp of height h at an angle of 45o. What is the total work done on the ball by gravity? Fgx = Fg sinθ W = Fd = Fgx L = (Fg sinθ) (h / sinθ) W = Fd = Fgx h a W = mgh h = L sinθ Fg a W = Fg h = mgh Fg h h θ Work by gravity A ball of mass m drops a distance h. What is the total work done on the ball by gravity? Path doesn’t matter when asking “how much work did gravity do?” Only the change in height!

46. Motion and energy When positive work is done on an object, its speed increases; when negative work is done, its speed decreases.

47. Kinetic Energy As a useful word for the quantity of work we have done on an object, thereby giving it motion, we define the kinetic energy:

48. Work-Energy Theorem Work-Energy Theorem: The total work done on an object is equal to its change in kinetic energy. (True for rigid bodies that remain intact)

49. FHAND r v = const a = 0 mg Lifting a Book You lift a book with your hand in such a way that it moves up at constant speed. While it is moving, what is the total work done on the book? a) mg×r b) FHAND×r c) (FHAND + mg) ×r d) zero e) none of the above

50. FHAND r v = const a = 0 mg Lifting a Book You lift a book with your hand in such a way that it moves up at constant speed. While it is moving, what is the total work done on the book? a) mg×r b) FHAND×r c) (FHAND + mg) ×r d) zero e) none of the above The total work is zero because the net force acting on the book is zero. The work done by the hand is positive, and the work done by gravity is negative. The sum of the two is zero. Note that the kinetic energy of the book does not change either! Follow-up: What would happen if FHAND were greater than mg?