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© 2013 Pearson Education, Inc. Chapter Goal: To use Newton’s third law to understand interacting objects. Chapter 7 Newton’s Third Law Slide 7-2.

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Presentation on theme: "© 2013 Pearson Education, Inc. Chapter Goal: To use Newton’s third law to understand interacting objects. Chapter 7 Newton’s Third Law Slide 7-2."— Presentation transcript:

1 © 2013 Pearson Education, Inc. Chapter Goal: To use Newton’s third law to understand interacting objects. Chapter 7 Newton’s Third Law Slide 7-2

2 © 2013 Pearson Education, Inc. Chapter 7 Preview Slide 7-3

3 © 2013 Pearson Education, Inc. Chapter 7 Preview Slide 7-4

4 © 2013 Pearson Education, Inc. Chapter 7 Preview Slide 7-6

5 © 2013 Pearson Education, Inc. Chapter 7 Preview Slide 7-6

6 © 2013 Pearson Education, Inc. Chapter 7 Preview Slide 7-7

7 © 2013 Pearson Education, Inc. Chapter 7 Preview Slide 7-8

8 © 2013 Pearson Education, Inc. Interacting Objects  When a hammer hits a nail, it exerts a forward force on the nail.  At the same time, the nail exerts a backward force on the hammer.  If you don’t believe it, imagine hitting the nail with a glass hammer.  It’s the force of the nail on the hammer that would cause the glass to shatter! Slide 7-21

9 © 2013 Pearson Education, Inc. Interacting Objects  When a bat hits a ball, the ball exerts a force on the bat.  When you pull someone with a rope in a tug-of- war, that person pulls back on you.  When your chair pushes up on you (the normal force), you push down on the chair.  All forces come in pairs, called action/reaction pairs.  These forces occur simultaneously, and we cannot say which is the “action” and which is the “reaction.” Slide 7-22 The bat and the ball are interacting with each other.

10 © 2013 Pearson Education, Inc. Interacting Objects  If object A exerts a force on object B, then object B exerts a force on object A.  The pair of forces, as shown, is called an action/reaction pair. Slide 7-23

11 © 2013 Pearson Education, Inc. Interacting Objects  Long-range forces, such as gravity, also come in pairs.  If you release a ball, it falls because the earth’s gravity exerts a downward force.  At the same time, the ball pulls upward on the earth with a force.  The ocean tides are an indication of the long-range gravitational interaction of the earth and moon. Slide 7-24

12 © 2013 Pearson Education, Inc. Example 7.1 Pushing a Crate Slide 7-32

13 © 2013 Pearson Education, Inc. Example 7.1 Pushing a Crate 4. Below are the free-body diagrams of the person and the crate. For each, three forces are external forces. Subscripts label which object each force acts on. There is one internal interaction, labeled as an action/reaction pair. ASSESS The completed free-body diagrams above could now be the basis for a quantitative analysis. Slide 7-34

14 © 2013 Pearson Education, Inc. Propulsion  If you try to walk across a frictionless floor, your foot slips and slides backward.  In order to walk, your foot must stick to the floor as you straighten your leg, moving your body forward.  The force that prevents slipping is static friction.  The static friction force points in the forward direction.  It is static friction that propels you forward! Slide 7-35 What force causes this sprinter to accelerate?

15 © 2013 Pearson Education, Inc. Examples of Propulsion Slide 7-36

16 © 2013 Pearson Education, Inc. Newton’s Third Law  Every force occurs as one member of an action/reaction pair of forces.  The two members of an action/reaction pair act on two different objects.  The two members of an action/reaction pair are equal in magnitude, but opposite in direction:  A catchy phrase, which is less precise, is: “For every action there is an equal but opposite reaction.” Slide 7-37

17 © 2013 Pearson Education, Inc. Reasoning with Newton’s Third Law  When you release a ball, it falls down.  The action/reaction forces of the ball and the earth are equal in magnitude.  The acceleration of the ball is  The acceleration of the earth is  If the ball has a mass of 1 kg, the earth accelerates upward at 2  10 −24 m/s 2. Slide 7-38

18 © 2013 Pearson Education, Inc. A mosquito runs head-on into a truck. Which is true during the collision? A.The magnitude of the mosquito’s acceleration is larger than that of the truck. B.The magnitude of the truck’s acceleration is larger than that of the mosquito. C.The magnitude of the mosquito’s acceleration is the same as that of the truck. D.The truck accelerates but the mosquito does not. E.The mosquito accelerates but the truck does not. QuickCheck 7.2 Slide 7-41

19 © 2013 Pearson Education, Inc. A mosquito runs head-on into a truck. Which is true during the collision? A.The magnitude of the mosquito’s acceleration is larger than that of the truck. B.The magnitude of the truck’s acceleration is larger than that of the mosquito. C.The magnitude of the mosquito’s acceleration is the same as that of the truck. D.The truck accelerates but the mosquito does not. E.The mosquito accelerates but the truck does not. QuickCheck 7.2 Slide 7-42 Newton’s second law: Don’t confuse cause and effect! The same force can have very different effects.

20 © 2013 Pearson Education, Inc. Example 7.3 The Forces on Accelerating Boxes Slide 7-43

21 © 2013 Pearson Education, Inc. Example 7.3 The Forces on Accelerating Boxes Slide 44

22 © 2013 Pearson Education, Inc. Acceleration Constraints  If two objects A and B move together, their accelerations are constrained to be equal: a A = a B.  This equation is called an acceleration constraint.  Consider a car being towed by a truck.  In this case, the acceleration constraint is a Cx = a Tx = a x.  Because the accelerations of both objects are equal, we can drop the subscripts C and T and call both of them a x. Slide 7-46

23 © 2013 Pearson Education, Inc. Acceleration Constraints  Sometimes the acceleration of A and B may have different signs.  Consider the blocks A and B in the figure.  The string constrains the two objects to accelerate together.  But, as A moves to the right in the +x direction, B moves down in the −y direction.  In this case, the acceleration constraint is a Ax = −a By. Slide 7-47

24 © 2013 Pearson Education, Inc. Example 7.5 Pulling a Rope Slide 7-55

25 © 2013 Pearson Education, Inc. Example 7.5 Pulling a Rope Slide 7-56

26 © 2013 Pearson Education, Inc. Example 7.5 Pulling a Rope Slide 7-57

27 © 2013 Pearson Education, Inc. Example 7.5 Pulling a Rope The rope’s tension is the same in both situations. Slide 7-58

28 © 2013 Pearson Education, Inc. The Massless String Approximation  Often in problems the mass of the string or rope is much less than the masses of the objects that it connects.  In such cases, we can adopt the following massless string approximation: Slide 7-59

29 © 2013 Pearson Education, Inc. The Massless String Approximation  Two blocks are connected by a massless string, as block B is pulled to the right.  Forces and act as if they are an action/reaction pair:  All a massless string does is transmit a force from A to B without changing the magnitude of that force.  For problems in this book, you can assume that any strings or ropes are massless unless it explicitly states otherwise. Slide 7-60

30 © 2013 Pearson Education, Inc. Boxes A and B are being pulled to the right on a frictionless surface. Box A has a larger mass than B. How do the two tension forces compare? QuickCheck 7.5 Slide 7-61 A. T 1 > T 2 B. T 1 = T 2 C. T 1 < T 2 D. Not enough information to tell.

31 © 2013 Pearson Education, Inc. Boxes A and B are being pulled to the right on a frictionless surface. Box A has a larger mass than B. How do the two tension forces compare? QuickCheck 7.5 Slide 7-62 A. T 1 > T 2 B. T 1 = T 2 C. T 1 < T 2 D. Not enough information to tell.

32 © 2013 Pearson Education, Inc. Pulleys  Block B drags block A across a frictionless table as it falls.  The string and the pulley are both massless.  There is no friction where the pulley turns on its axle.  Therefore, T A on S = T B on S. Slide 7-69

33 © 2013 Pearson Education, Inc. Pulleys  Since T A on B = T B on A, we can draw the simplified free- body diagram on the right, below.  Forces and act as if they are in an action/reaction pair, even though they are not opposite in direction because the tension force gets “turned” by the pulley. Slide 7-70

34 © 2013 Pearson Education, Inc. All three 50-kg blocks are at rest. The tension in rope 2 is QuickCheck 7.7 Slide 7-71 A.greater than the tension in rope 1. B.equal to the tension in rope 1. C.less than the tension in rope 1.

35 © 2013 Pearson Education, Inc. A.greater than the tension in rope 1. B.equal to the tension in rope 1. C.less than the tension in rope 1. All three 50-kg blocks are at rest. The tension in rope 2 is QuickCheck 7.7 Slide 7-72 Each block is in static equilibrium, with.

36 © 2013 Pearson Education, Inc. The acceleration constraint here is QuickCheck 7.9 Slide 7-75 A. a Ay = a By. B. –a Ay = –a By. C. a Ay = –a By. D. a By = –a Ay. E. Either C or D.

37 © 2013 Pearson Education, Inc. The acceleration constraint here is QuickCheck 7.9 Slide 7-76 A. a Ay = a By. B. –a Ay = –a By. C. a Ay = –a By. D. a By = –a Ay. E. Either C or D. Either says that the acceleration vectors point in opposite directions.


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