Wednesday, September 28, 11. Applying Newton s Laws

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1 Applying Newton s Laws

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5 Question 1a Tension I You tie a rope to a tree and you pull on the rope with a force of 100 N. What is the tension in the rope? a) 0 N b) 50 N c) 100 N d) 150 N e) 200 N

6 Question 1a Tension I You tie a rope to a tree and you pull on the rope with a force of 100 N. What is the tension in the rope? a) 0 N b) 50 N c) 100 N d) 150 N e) 200 N The tension in the rope is the force that the rope feels across any section of it (or that you would feel if you replaced a piece of the rope). Because you are pulling with a force of 100 N, that is the tension in the rope.

7 Question 1b Tension II Two tug-of-war opponents each pull with a force of 100 N on opposite ends of a rope. What is the tension in the rope? a) 0 N b) 50 N c) 100 N d) 150 N e) 200 N

8 Question 1b Tension II Two tug-of-war opponents each pull with a force of 100 N on opposite ends of a rope. What is the tension in the rope? a) 0 N b) 50 N c) 100 N d) 150 N e) 200 N This is literally the identical situation to the previous question. The tension is not 200 N!! Whether the other end of the rope is pulled by a person, or pulled by a tree, the tension in the rope is still 100 N!!

9 You and a friend can each pull with a force of 20 N. If you want to rip a rope in half, what is the best way? Question 1c Tension III a) you and your friend each pull on opposite ends of the rope b) tie the rope to a tree, and you both pull from the same end c) it doesn t matter both of the above are equivalent d) get a large dog to bite the rope

10 You and a friend can each pull with a force of 20 N. If you want to rip a rope in half, what is the best way? Question 1c Tension III a) you and your friend each pull on opposite ends of the rope b) tie the rope to a tree, and you both pull from the same end c) it doesn t matter both of the above are equivalent d) get a large dog to bite the rope Take advantage of the fact that the tree can pull with almost any force (until it falls down, that is!). You and your friend should team up on one end, and let the tree make the effort on the other end.

11 Question 2 Three Blocks Three blocks of mass 3m, 2m, and a) T 1 > T 2 > T 3 m are connected by strings and b) T 1 < T 2 < T 3 pulled with constant acceleration a. What is the relationship between the c) T 1 = T 2 = T 3 tension in each of the strings? d) all tensions are zero e) tensions are random a 3m T 3 T 2 T 2m 1 m

12 Question 2 Three Blocks Three blocks of mass 3m, 2m, and a) T 1 > T 2 > T 3 m are connected by strings and b) T 1 < T 2 < T 3 pulled with constant acceleration a. What is the relationship between the c) T 1 = T 2 = T 3 tension in each of the strings? d) all tensions are zero e) tensions are random T 1 pulls the whole set of blocks along, so it must be the largest. T 2 pulls the last two masses, but T 3 only 3m a T 3 T 2 T 2m 1 m pulls the last mass. Follow-up: What is T 1 in terms of m and a?

13 Question 3 Over the Edge In which case does block m experience a larger acceleration? In case (1) there is a 10 kg mass hanging from a rope and falling. In case (2) a hand is providing a constant downward force of 98 N. Assume massless ropes. a) case (1) b) acceleration is zero c) both cases are the same d) depends on value of m e) case (2) m m 10 kg a a F = 98 N Case (1) Case (2)

14 Question 3 Over the Edge In which case does block m experience a larger acceleration? In case (1) there is a 10 kg mass hanging from a rope and falling. In case (2) a hand is providing a constant downward force of 98 N. Assume massless ropes. a) case (1) b) acceleration is zero c) both cases are the same d) depends on value of m e) case (2) In case (2) the tension is 98 N due to the hand. In case (1) the tension is less than 98 N because the block is accelerating down. Only if the block were at rest would the tension be equal to 98 N. 10 kg a m m a F = 98 N Case (1) Case (2)

15 What causes friction?

16 What causes friction?

17 Frictional forces, kinetic and static Friction can keep an object from moving or slow its motion from what we last calculated on an ideal, frictionless surface. Microscopic imperfections cause nonideal motion.

18 Applied force is proportional until the object moves Notice the transition between static and kinetic friction.

19 Notice the effect of friction in horizontal motion Moving a 500 N crate has two parts: getting the motion to begin and then the effect of friction on constant velocity motion.

20 The angle at which tension is applied matters As one varies the angle at which tension is applied, force spent overcoming friction and force lifting the object are interplayed. These components affect each other.

21 The angle at which tension is applied matters As one varies the angle at which tension is applied, force spent overcoming friction and force lifting the object are interplayed. These components affect each other.

22 Consider the toboggan ride two more times Consider the toboggan from earlier with the ski wax worn off (nonzero friction). Finally, consider the toboggan on a much steeper hill, accelerating.

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24 Give some examples of when is friction a nuisance

25 Give some examples of when is friction a nuisance

26 Give some examples of when is friction a nuisance

27 Give some examples of when is friction a nuisance Give some examples of when friction is indispensable

28 Give some examples of when is friction a nuisance Give some examples of when friction is indispensable

29 Want to reduce friction We need friction to run

30 Kinetic Friction

31 Kinetic Friction What do you think happens if we double the weight?

32 Kinetic Friction What do you think happens if we double the weight?

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36 is a vector?

37 is a vector?

38 Question 4 Friction A box sits in a pickup truck on a frictionless truck bed. When the truck accelerates forward, the box slides off the back of the truck because: a) the force from the rushing air pushed it off b) the force of friction pushed it off c) no net force acted on the box d) truck went into reverse by accident e) none of the above

39 Question 4 Friction A box sits in a pickup truck on a frictionless truck bed. When the truck accelerates forward, the box slides off the back of the truck because: a) the force from the rushing air pushed it off b) the force of friction pushed it off c) no net force acted on the box d) truck went into reverse by accident e) none of the above Generally, the reason that the box in the truck bed would move with the truck is due to friction between the box and the bed. If there is no friction, there is no force to push the box along, and it remains at rest. The truck accelerated away, essentially leaving the box behind!!

40 Antilock brakes keep the car wheels from locking and skidding during a sudden stop. Why does this help slow the car down? Question 5 Antilock Brakes a) µ k > µ s so sliding friction is better b) µ k > µ s so static friction is better c) µ s > µ k so sliding friction is better d) µ s > µ k so static friction is better e) none of the above

41 Antilock brakes keep the car wheels from locking and skidding during a sudden stop. Why does this help slow the car down? Question 5 Antilock Brakes a) µ k > µ s so sliding friction is better b) µ k > µ s so static friction is better c) µ s > µ k so sliding friction is better d) µ s > µ k so static friction is better e) none of the above Static friction is greater than sliding friction, so by keeping the wheels from skidding, the static friction force will help slow the car down more efficiently than the sliding friction that occurs during a skid.

42 Question 6 Going Sledding Your little sister wants you to give her a ride on her sled. On level ground, what is the easiest way to accomplish this? a) pushing her from behind b) pulling her from the front c) both are equivalent d) it is impossible to move the sled e) tell her to get out and walk 1 2

43 Question 6 Going Sledding Your little sister wants you to give her a ride on her sled. On level ground, what is the easiest way to accomplish this? a) pushing her from behind b) pulling her from the front c) both are equivalent d) it is impossible to move the sled e) tell her to get out and walk In case 1, the force F is pushing down (in addition to mg), so the normal force is larger. In case 2, the force F is pulling up, against gravity, so the normal force is lessened. Recall that the frictional force is proportional to the normal force. 2 1

44 Question 7 Will It Budge? A box of weight 100 N is at rest on a floor where µ s = 0.4. A rope is attached to the box and pulled horizontally with tension T = 30 N. Which way does the box move? a) moves to the left b) moves to the right c) moves up d) moves down e) the box does not move Static friction (µ s = 0.4 ) m T

45 Question 7 Will It Budge? A box of weight 100 N is at rest on a floor where µ s = 0.4. A rope is attached to the box and pulled horizontally with tension T = 30 N. Which way does the box move? a) moves to the left b) moves to the right c) moves up d) moves down e) the box does not move The static friction force has a maximum of µ s N = 40 N. The tension in the rope is only 30 N. So Static friction (µ s = 0.4 ) m T the pulling force is not big enough to overcome friction. Follow-up: What happens if the tension is 35 N? What about 45 N?

46 A box sits on a flat board. You lift one end of the board, making an angle with the floor. As you increase the angle, the box will eventually begin to slide down. Why? Question 8a Sliding Down I a) component of the gravity force parallel to the plane increased b) coefficient of static friction decreased c) normal force exerted by the board decreased d) both #1 and #3 e) all of #1, #2, and #3 Normal Weight Net Force

47 A box sits on a flat board. You lift one end of the board, making an angle with the floor. As you increase the angle, the box will eventually begin to slide down. Why? Question 8a Sliding Down I a) component of the gravity force parallel to the plane increased b) coefficient of static friction decreased c) normal force exerted by the board decreased d) both #1 and #3 e) all of #1, #2, and #3 As the angle increases, the component of weight parallel to the plane increases and the component perpendicular to the plane decreases (and so does the normal force). Because friction depends on normal force, we see that the friction force gets smaller and the force pulling the box down the plane gets bigger. Weight Normal Net Force

48 A mass m is placed on an inclined plane (µ > 0) and slides down the plane with constant speed. If a similar block (same µ) of mass 2m were placed on the same incline, it would: Question 8b Sliding Down II a) not move at all b) slide a bit, slow down, then stop c) accelerate down the incline d) slide down at constant speed e) slide up at constant speed m

49 A mass m is placed on an inclined plane (µ > 0) and slides down the plane with constant speed. If a similar block (same µ) of mass 2m were placed on the same incline, it would: Question 8b Sliding Down II a) not move at all b) slide a bit, slow down, then stop c) accelerate down the incline d) slide down at constant speed e) slide up at constant speed The component of gravity acting down the plane is double for 2m. However, the f N normal force (and hence the friction force) is also double (the same factor!). This means the two forces still cancel to W y θ W give a net force of zero. W x θ

50 Using Newton s First Law when forces are in equilibrium

51 Using Newton s First Law when forces are in equilibrium

52 Using Newton s First Law when forces are in equilibrium

53 Using Newton s First Law when forces are in equilibrium

54 Using Newton s First Law when forces are in equilibrium

55 Using Newton s First Law when forces are in equilibrium

56 Using Newton s First Law when forces are in equilibrium

57 Using Newton s First Law when forces are in equilibrium

58 Using Newton s First Law when forces are in equilibrium

59 Using Newton s First Law when forces are in equilibrium

60 Using Newton s First Law when forces are in equilibrium

61 Using Newton s First Law when forces are in equilibrium

62 Using Newton s First Law when forces are in equilibrium

63 Using Newton s First Law when forces are in equilibrium

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71 1-D equilibrium considering the rope Figure 5.2 If we re-did the previous example in a less ideal world?

72 A crate on an inclined plane An object on an inclined plane will have components of force in x and y space.

73 Consider a mass on a plane tied to a cord over a pulley You must draw two-free body diagrams. The tension from the mass hanging is the same force that draws the cart up the ramp.

74 Beware incorrect free-body diagrams Only gravity acts on the apple.

75 Straight-line motion with constant force Consider the sailboat mounted on an ice skate.

76 Straight-line motion with constant force Consider the sailboat mounted on an ice skate.

77 Consider motion against friction Earlier, we considered the effect of using real rope in a tension calculation. How about friction at work in a problem that slides an object?

78 Consider motion against friction Earlier, we considered the effect of using real rope in a tension calculation. How about friction at work in a problem that slides an object?

79 Calculation of tension in an elevator cable The elevator mass is balanced by tension in the cable (that s why they have that maximum capacity sign on the wall).

80 Calculation of tension in an elevator cable The elevator mass is balanced by tension in the cable (that s why they have that maximum capacity sign on the wall).

81 More adventures in the elevator The elevator allows riders to experience different weights (without dieting).

82 Acceleration down a hill If you grew up in the northeast, you could often ride on snow and ice in the winter using sleds, toboggans, even inner tubes and plastic sheets.

83 Watch for this common error A good road sign is to be sure that the normal force comes out perpendicular to the surface.

84 Multiple items in motion The same acceleration acts on every surface of your lunch tray (be careful to keep your food in place).

85 Two bodies with the same magnitude of acceleration A glider on the air track connected over a pulley to a falling mass.

86 For the sport of skydiving, terminal speed is vital

87 For the sport of skydiving, terminal speed is vital

88 For the sport of skydiving, terminal speed is vital

89 For the sport of skydiving, terminal speed is vital

90 Many have asked how lethal is a coin dropped from atop a tall building? Urban legends have said that a penny dropped from the top of the Empire State Building can kill. Copyright 2008 Pearson Education Inc., publishing as Pearson Addison-Wesley

91 The dynamics of uniform circular motion In uniform circular motion, both the acceleration and force are centripetal. Cut the cord restraining an object in such motion and observe the object s behavior without the inward force.

92 Examining a misnomer People have adopted the pop-culture use of centrifugal force but it really results from reference frames. It is fictional and results from a car turning while a person continues in straight-line motion (for example).

93 In the game of tetherball, the struck ball whirls around a pole. In what direction does the net force on the ball point? Question 9 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 T W

94 In the game of tetherball, the struck ball whirls around a pole. In what direction does the net force on the ball point? Question 9 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 The vertical component of the tension balances the weight. The horizontal component of tension provides the centripetal force that W T T points toward the center of the circle. W

95 In the game of tetherball, the struck ball whirls around a pole. In what direction does the net force on the ball point? Question 9 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 The vertical component of the tension balances the weight. The horizontal component of tension provides the centripetal force that W T T points toward the center of the circle. W

96 Question 10a Around the Curve I You are a passenger in a car, not wearing a seat belt. The car makes a sharp left turn. From your perspective in the car, what do you feel is happening to you? a) you are thrown to the right b) you feel no particular change c) you are thrown to the left d) you are thrown to the ceiling e) you are thrown to the floor

97 Question 10a Around the Curve I You are a passenger in a car, not wearing a seat belt. The car makes a sharp left turn. From your perspective in the car, what do you feel is happening to you? a) you are thrown to the right b) you feel no particular change c) you are thrown to the left d) you are thrown to the ceiling e) you are thrown to the floor The passenger has the tendency to continue moving in a straight line. From your perspective in the car, it feels like you are being thrown to the right, hitting the passenger door.

98 Question 10a Around the Curve I You are a passenger in a car, not wearing a seat belt. The car makes a sharp left turn. From your perspective in the car, what do you feel is happening to you? a) you are thrown to the right b) you feel no particular change c) you are thrown to the left d) you are thrown to the ceiling e) you are thrown to the floor The passenger has the tendency to continue moving in a straight line. From your perspective in the car, it feels like you are being thrown to the right, hitting the passenger door.

99 Question 10b Around the Curve II During that sharp left turn, you found yourself hitting the passenger door. What is the correct description of what is actually happening? a) centrifugal force is pushing you into the door b) the door is exerting a leftward force on you c) both of the above d) neither of the above

100 Question 10b Around the Curve II During that sharp left turn, you found yourself hitting the passenger door. What is the correct description of what is actually happening? a) centrifugal force is pushing you into the door b) the door is exerting a leftward force on you c) both of the above d) neither of the above The passenger has the tendency to continue moving in a straight line. There is a centripetal force, provided by the door, that forces the passenger into a circular path.

101 Question 10b Around the Curve II During that sharp left turn, you found yourself hitting the passenger door. What is the correct description of what is actually happening? a) centrifugal force is pushing you into the door b) the door is exerting a leftward force on you c) both of the above d) neither of the above The passenger has the tendency to continue moving in a straight line. There is a centripetal force, provided by the door, that forces the passenger into a circular path.

102 Question 10c Around the Curve III You drive your dad s car too fast around a curve and the car starts to skid. What is the correct description of this situation? a) car s engine is not strong enough to keep the car from being pushed out b) friction between tires and road is not strong enough to keep car in a circle c) car is too heavy to make the turn d) a deer caused you to skid e) none of the above

103 The friction force between tires and road provides the centripetal force that keeps the car moving in a circle. If this force is too small, the car continues in a straight line! Question 10c Around the Curve III You drive your dad s car too fast around a curve and the car starts to skid. What is the correct description of this situation? a) car s engine is not strong enough to keep the car from being pushed out b) friction between tires and road is not strong enough to keep car in a circle c) car is too heavy to make the turn d) a deer caused you to skid e) none of the above Follow-up: What could be done to the road or car to prevent skidding?

104 Question 11 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? d a b c e

105 Question 11 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? d a b c e 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?

106 Question 12 Ball and String Two equal-mass rocks tied to strings are whirled in horizontal circles. The radius of circle 2 is twice that of circle 1. If the period of motion is the same for both rocks, what is the tension in cord 2 compared to cord 1? a) T 2 = ¼T 1 b) T 2 = ½T 1 c) T 2 = T 1 d) T 2 = 2T 1 e) T 2 = 4T 1 1 2

107 Question 12 Ball and String Two equal-mass rocks tied to strings are whirled in horizontal circles. The radius of circle 2 is twice that of circle 1. If the period of motion is the same for both rocks, what is the tension in cord 2 compared to cord 1? a) T 2 = ¼T 1 b) T 2 = ½T 1 c) T 2 = T 1 d) T 2 = 2T 1 e) T 2 = 4T 1 The centripetal force in this case is given by the 1 tension, so T = mv 2 /r. For the same period, we find that v 2 = 2v 1 (and this term is squared). However, for the denominator, we see that r 2 = 2r 1 which gives us 2 the relation T 2 = 2T 1.

108 Question 13 Barrel of Fun A rider in a barrel of fun finds herself stuck with her back to the wall. Which diagram correctly shows the forces acting on her? a b c d e

109 Question 13 Barrel of Fun A rider in a barrel of fun finds herself stuck with her back to the wall. Which diagram correctly shows the forces acting on her? a b c d e The normal force of the wall on the rider provides the centripetal force needed to keep her going around in a circle. The downward force of gravity is balanced by the upward frictional force on her, so she does not slip vertically. Follow-up: What happens if the rotation of the ride slows down?

110 Question 14a Going in Circles I 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? a) N remains equal to mg b) N is smaller than mg c) N is larger than mg d) none of the above

111 Question 14a Going in Circles I 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? a) N remains equal to mg b) N is smaller than mg c) N is larger than mg d) none of the above 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?

112 Question 14b Going in Circles II A skier goes over a small round hill with radius R. Because she is in circular motion, there has to be a centripetal force. At the top of the hill, what is F c of the skier equal to? a) F c = N + mg b) F c = mg N c) F c = T + N mg d) F c = N e) F c = mg v R

113 Question 14b Going in Circles II A skier goes over a small round hill with radius R. Because she is in circular motion, there has to be a centripetal force. At the top of the hill, what is F c of the skier equal to? a) F c = N + mg b) F c = mg N c) F c = T + N mg d) F c = N e) F c = mg F c points toward the center of the circle (i.e., downward in this case). The weight vector points down and the normal force (exerted by the hill) points up. The magnitude of the net force, therefore, is F c = mg N. v mg N R Follow-up: What happens when the skier goes into a small dip?

114 Question 14c Going in Circles III You swing a ball at the end of string a) F c = T mg in a vertical circle. Because the ball b) F c = T + N mg is in circular motion there has to be a c) F c = T + mg centripetal force. At the top of the ball s path, what is F d) F c = T c equal to? e) F c = mg v top R

115 Question 14c Going in Circles III You swing a ball at the end of string a) F c = T mg in a vertical circle. Because the ball b) F c = T + N mg is in circular motion there has to be a c) F c = T + mg centripetal force. At the top of the ball s path, what is F d) F c = T c equal to? e) F c = mg F c points toward the center of the circle (i.e., downward in this case). The weight vector points down and the tension (exerted by the string) also points down. The magnitude of the net force, therefore, is F c = T+ mg. v mg T R Follow-up: What is F c at the bottom of the ball s path?