7.1 Interacting Systems p Action/reaction pairs p Newton s Third Law p Ropes and Pulleys p.

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1 7.1 Interacting Systems p Action/reaction pairs p Newton s Third Law p Ropes and Pulleys p Interacting-system Problems p

2 7.1 Interacting Systems p Action/reaction pairs p

3 One of the most important examples of (almost) circular motion is the motion of planets around the sun None of these orbits is really circular (especially Pluto's), but in most cases it is a very good approximation Many celestial bodies are in a kind of circular motion: moons around planets planets around stars stars around the centre of the galaxy To understand this motion we first have to introduce a new concept: action/reaction pairs of forces 3

4 As we said to understand planets motion we first have to introduce a new concept: action/reaction pairs of forces Any object A that exerts a force on another object B experiences itself a force that is caused by object B Example: if you are sitting on a chair, you (A) feel the chair pressing against you, and the A chair (B) may break if you are too heavy B F B on A is here the normal force, F A on B is related to the weight of A F B on A F A on B 4

5 Two objects with an action/reaction pairs of forces have a mutual influence on each other, We say that the objects interact with each other. Sometimes object B is much larger (usually heavier) than object A. Then A's influence on B can be neglected since the state of B is hardly changed by F A on B Example: a train (B) moving at 5 km/h can easily push you (A) aside without noticeable change in its velocity. But if you walk into the train, it will (usually) hardly move. 5

6 In such a situation we simply can consider object A alone. Object B is then part of the environment which exerts forces on A This is what we implicitly did when drawing free-body diagrams for single-particle dynamics In general we both have forces from the environment (like weight) and action/reaction pairs 6

7 We will also encounter situations where more than two systems are interacting with each other. We then will consider action/reaction pairs for different pairs of systems (see figure) Note that Newton's 2 nd law applies separately to each system: 7

8 Identifying action/reaction pairs (Tactic box 7.1): Draw each object separately Identify all forces and draw them on the object on which they act Identify action/reaction pairs: F A on B goes with F B on A Identify objects that are systems of interest (i.e., which are not part of the environment) Draw a free-body diagram for each object of interest 8

9 Chapter 7: Newton's 3 rd law Example: a person pushes a crate with (action) force F P on C. The reaction force is F C on P The person uses surface friction f S on P to be able to push. There is a reaction force f P on S, but surface and Earth can be considered as part of the environment Other forces: weight, normal f. 9

10 Action/reaction pairs are also behind any mechanism for propulsion (friction, thrust) Ropes can also be considered as a system. The forces acting on a rope are usually tension forces 10

11 Chapter 7: teasers What, if anything, is wrong with the force diagram for a bicycle that is accelerating toward the right? 1. It does not draw each object separately. 2. It draws interactions between one body and itself. 3. It fails to identify all relevant forces. 4. Nothing is wrong. 11

12 Chapter 7: teasers What, if anything, is wrong with the force diagram for a bicycle that is accelerating toward the right? 1. It does not draw each object separately. 2. It draws interactions between one body and itself. 3. It fails to identify all relevant forces. 4. Nothing is wrong. 12

13 Chapter 7: teasers What, if anything, is wrong with this force diagram for a bicycle that is accelerating to the right? A. Nothing is wrong. B. One or more forces have the wrong length. C. One of more forces have the wrong direction. D. One or more action/reaction pairs are wrong. E. Both C and D. 13

14 Chapter 7: teasers What, if anything, is wrong with this force diagram for a bicycle that is accelerating to the right? A. Nothing is wrong. B. One or more forces have the wrong length. C. One or more forces have the wrong direction. D. One or more action/reaction pairs are wrong. E. Both C and D. X X 14

15 End of Week 7 15

16 7.3 Newton s Third Law p Newton s third law (or action/reaction) recognizes force as an INTERACTION BETWEEN OBJECTS rather than as some THING with an independent existence of its own! 16

17 We have seen that the two forces of the gravitational action/reaction pair between two objects are equal in magnitude but have opposite direction This is true for any force and constitutes an example of Newton's 3 rd law: Every force occurs as one member of an action/reaction pair of forces The two forces in an action/reaction pair act on different objects The two forces are equal in magnitude but opposite in direction: F A on B = - F B on A 17

18 It is important to note that Newton's 3 rd law states the equality of the forces, not the acceleration of the two objects The example of the satellite and the planet shows that even if F S on P = F P on S the associated accelerations can be very different in magnitude: Hence a S = (m P /m S ) a P >> a P for m P >> m S. Note that this is true independent of the type of force 18

19 Cliction 7.1 Car B is stopped for a red light. Car A, which has the same mass as car B, doesn t see the red light and runs into the back of B. Which of the following statements is true? 1. B exerts a force on A but A doesn t exert a force on B. 2. B exerts a larger force on A than A exerts on B. 3. B exerts the same amount of force on A as A exerts on B. 4. A exerts a larger force on B than B exerts on A. 5. A exerts a force on B but B doesn t exert a force on A. 19

20 Acceleration Constraint Sometimes two objects A and B are connected and move together In this case their accelerations are the same, but not their net forces: 20

21 Action/reaction pairs Here is another example of an acceleration constraint. In general, the accelerations need not have equal direction. In this example, a Ax = - a By y x 21

22 Cliction 7.2 Boxes A and B are sliding to the right across a frictionless table. The hand H is slowing them down. The mass of A is larger than the mass of B. Rank in order, from largest to smallest, the horizontal forces on A, B, and H. 1. F B on H = F H on B = F A on B = F B on A 2. F B on H = F H on B < F A on B = F B on A 3. F B on H = F H on B < F A on B = F B on A 4. F H on B = F H on A > F A on B 22

23 7.4 Ropes and Pulleys p

24 Let's also take another look at strings with tension. In equilibrium we would have T A on S = T B on S. However, if the string is accelerated this would not be the case: However, the string is often so light that this difference does not matter. In the future we will make the massless string approximation m S =0 so that T A on S = T B on S 24

25 In the example below: the climber is not in contact with the knot. So, strictly speaking forces T and T are not an action/reaction C on K K on C pair! BUT if the ropes are massless then T C on K and T K on Cact as if they are an action/reaction pair. 25

26 In Phys 221 we will also make the approximation that pulleys are massless and have no friction Important principle: the magnitude of the tension in a massless string remains constant as it passes over a massless, frictionless pulley 26

27 Cliction 7.3 All three 50 kg blocks are at rest. Is the tension in rope 2 greater than, less than or equal to the tension in rope 1? 1. greater than 2. less than 3. equal to 27

28 NOTE 1: the force on the axle of the pulley is shared equally by the two lines looping through the pulley. NOTE 2: a single movable pulley allows a unit weight to be lifted with only half the force needed to lift the weight without assistance. The weight lifted divided by the lifting force is defined as the advantage of the pulley system. 28

29 In diagram 3, the addition of a fixed pulley yields a lifting advantage of 3. The tension in each line is ⅓ the unit weight, and the force on the axles of each pulley is ⅔ of a unit weight. As in the case of diagram 2a, another pulley may be added to reverse the direction of the lifting force, but with no increase in advantage. This situation is shown in diagram 3a. 29

30 7.5 Interacting-system Problems p

31 Problem 1 : Pushing a Package The boy needs to push the package with an acceleration of at least 1 m/s in order for the package to make it to the top. Can he give the package a big enough shove/push to reach the top of the ramp? i) The coefficient of static friction of the ground is 0.25 ii) The coefficient of kinetic friction of the ramp is iii) The boy is in static equilibrium. This is the ONLY action/reaction pair in this problem. 31

32 Problem 1 : Pushing a Package x-axis f - F cos(30) = 0 S PonB y-axis n - m g - F sin(30) = 0 B BOY B PonB x-axis -f + F - m gsin(30) = m a k y-axis n - m g cos(30) = 0 P BonP P P x P PACKAGE NEWTON s third law F = F = F BonP PonB STEP 1 From the package equations one can show that F = 139 N STEP 2 STEP 3 From the boy s x-equation we can calculate how much static friction he needs to push the box f = F cos(30) = 120 N S The maximum static friction is f = 0.25n = 115 N CONCLUSION: He CANNOT shove hard enough without slipping! S,max B 32

33 Chapter 7: Newton's 3 rd law Cliction 7.4 A small car is pushing a larger truck that has a dead battery. The mass of the truck is larger than the mass of the car. Which of the following statements is true? 1. The car exerts a force on the truck but the truck doesn t exert a force on the car. 2. The car exerts a larger force on the truck than the truck exerts on the car. 3. The car exerts the same amount of force on the truck as the truck exerts on the car. 4. The truck exerts a larger force on the car than the car exerts on the truck. 5. The truck exerts a force on the car but the car doesn t exert a force on the truck. 33

34 Problem 2 What is the acceleration of the man M hanging at the rope? Here we have an acceleration constraint a Sy = - a My F ceil 34

35 Draw the free-body diagrams of the three systems (man, pulley, set): F net,m M T 1 w M -T 1 P F ceil -T 2 S T 2 w S F net,s F ceil Because the tension is the same on both sides of the pulley we have T 1 = T 2 In principle we also need to consider the rope segments as objects, but since they are massless this is trivial 35

36 Newton's 2 nd law for M, P, S: we need only to consider the y-component of each equation: F ceil Tension remains the same over pulley T 1y = T 2y (F ceil ) y = 2 T 1y F net,m M T 1 w M -T 1 P F ceil -T 2 S T 2 w S 36 F net,s

37 Insert acceleration constraint: F ceil 37

38 Problem 3 Let's a bit elaborate on pulleys. In combination they can create an ancient but still very useful machine: a tackle The principle of a tackle is based on multiple application of our previous example

39 Let's analyze a compound pulley. Pictorial representation: F ceil,2 Chapter 7: Newton's 3 rd law F ceil,1 P 1 F pull P 2 W w F pull 39

40 Free-body diagrams: we don't care about the ceiling forces. In equilibrium we then have Weight W T 3 w Chapter 7: Newton's 3 rd law Pulley P 1 F pull F ceil,1 -T 2 Pulley P 2 T 1 T 2 -T 3 All tensions with a minus sign are part of a respective action/reaction pair: T 2 between P 1 and P 2, and T 3 between P 2 and W F ceil,2 T 1 F pull P 1 P 2 W F ceil,1 T 3 T 2 w 40

41 Pulley principle: same tension on both sides Newton's 2 nd law: F net = 0 for all three systems Weight W T 3 w Eq.(6) is only useful for finding F ceil,1 Pulley P 2 T 1 T 2 -T 3 Pulley P 1 F pull F ceil,1 -T 2 41

42 (5) follows from inserting (3) and (1) into (4) (5) allows to calculate T 2y = m W g/2 from (2) we then find the pulling force (F pull ) y = - m W g/2 Chapter 7: Newton's 3 rd law We therefore need only half the weight force to keep m W in the air 42

43 Chapter 7 Reading Quiz 43

44 The propulsion force on a car is due to 1. static friction. 2. kinetic friction. 3. the car engine. 4. elastic energy. 44

45 Is the tension in rope 2 greater than, less than, or equal to the tension in rope 1? 1. greater than rope 2 2. less than than rope 2 3. equal to rope 2 45

46 An acceleration constraint says that in some circumstances 1. the acceleration of an object has to be positive. 2. two objects have to accelerate in the same direction. 3. the magnitude of the accelerations of two objects have to be equal. 4. an object is prevented from accelerating. 5. Acceleration constraints were not discussed in this chapter. 46

47 Selected Problems 47

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51 End of Chapter 7 IMPORTANT: Print a copy of the SUMMARY page (p. 203) and add it here to your lecture notes. It will save you crucial time when trying to recall: Concepts, Symbols, and Strategies 51