Chapter 8: Newton s Third law

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1 Warning: My approach is a soewhat abbreviated and siplified version of what is in the text, yet just as coplete. Both y treatent and the text s will prepare you to solve the sae probles. Restating Newton s Third law fro Chapter 4: When two bodies interact, the forces exerted by each body on the other are always equal in agnitude and opposite in direction. Less precisely: For every action there is an equal and opposite reaction. SMU PHYS1100.1, fall 2008, Prof. Clarke 1

2 A syste and its environent. Often, there will be ore than one object of interest in a proble, such as these two asses being pushed by a single force. F M Each object of interest (e.g., and M) is referred to as a syste; everything else (e.g., earth, table) is referred to as the environent. Typically, systes will interact (exert forces) with each other and with their environent. SMU PHYS1100.1, fall 2008, Prof. Clarke 2

3 Internal and external forces Forces exerted by one syste of interest on another are internal forces. e.g., ass pushes on ass M: P on M ass M pushes on ass : P M on Forces exerted on systes of interest by agents in the environent are external forces. e.g., the Earth exerts a gravitational force on : w E on the table exerts a noral force on : n T on the table exerts a frictional force on M: f k,t on M the external force F exerted on : F X on etc. The notation: P on M eans the force P exerted by on M. SMU PHYS1100.1, fall 2008, Prof. Clarke 3

4 Free-body diagras for and M: y P M on f k,t on n T on w E on a x F X on Using the F A on B notation, it is easy assign the correct forces to each syste. Only forces with on are put on the FBD for, and only forces with on M are put on the FBD for M. y M f k,t on M n T on M a P on M w E on M x In particular, the force driving the two asses, F, becoes F X on in this notation (X being soe unknown, external agent), and is applied only to the FBD for! SMU PHYS1100.1, fall 2008, Prof. Clarke 4

5 Action/reaction pairs y P M on f k,t on n T on a x F X on Note the syetry between the two internal forces: P on M y M f k,t on M n T on M a P on M x w E on P M on w E on M The two internal forces constitute an action/reaction pair. Newton s 3 rd Law states that every force has a corresponding reaction force, but only internal forces have their reaction forces labeled on the FBDs, and never on the sae FBD. SMU PHYS1100.1, fall 2008, Prof. Clarke 5

6 External forces (e.g., n T on ) are also part of an action-reaction pair, but the reaction force (e.g., n on T, the noral force exerts on the table) is irrelevant to the dynaics of or M. Thus, external forces do not appear in action/reaction pairs on the FBDs, only the internal forces do. Action/reaction: a bit of a isnoer? This ter sees to iply that one force, being the reaction to the action, soehow happens after the action. This is incorrect: one cannot exist without the other. The two forces co-exist siultaneously, no exceptions. SMU PHYS1100.1, fall 2008, Prof. Clarke 6

7 Identifying action/reaction pairs Easy!!! In our notation F A on B, just swap the labels A and B! Thus, if F A on B is the action force, the reaction force is F B on A e.g.: You (Y) are pushing a box (B) across the floor. If the action force is P Y on B, the reaction force is the box pushing back on you, P B on Y. P B on Y Y B P Y on B B Y Note that P Y on B and P B on Y appear on different FBDs!! P Y on B ended here, 16/10/08 SMU PHYS1100.1, fall 2008, Prof. Clarke 7 B P B on Y Y

8 More exaples. Two boxes, A and B, sit on a table, T, which sits on the earth, E. a) If the action force is the noral force of the table, T, on box A, (n T on A ), what is the reaction force? A n T on A n A on T = the noral force of box A on the table, T. T E B w E on B b) If the action force is the gravitational force the earth, E, exerts on box B (w E on B, otherwise known B s weight), what is the reaction force? w B on E = the gravitational force box B exerts on the earth, E. SMU PHYS1100.1, fall 2008, Prof. Clarke 8

9 Alert: A very coon isconception: Regardless of what you ay have learned or thought you learned elsewhere, the reaction force to the weight of is NOT the noral force fro the table! First, gravitational and noral forces are entirely different kinds of forces. Action/reaction pairs are ALWAYS the sae type of force. Second, both the weight and noral force are put on the sae FBD. Action/reaction pairs are NEVER put on the sae FBD. w n T E n SMU PHYS1100.1, fall 2008, Prof. Clarke 9 w

10 Alert: A very coon isconception: Regardless of what you ay have learned or thought you learned elsewhere, the reaction force to the weight of is NOT the noral force fro the table! First, gravitational and noral forces are entirely different kinds of forces. Action/reaction pairs are always the sae type of force. n T on w E on T Second, both the weight and noral force are put on the sae FBD. Action/reaction pairs are NEVER put on the sae FBD. E n T on Using the notation F A on B will help you avoid this trap. w E on SMU PHYS1100.1, fall 2008, Prof. Clarke 10

11 Newton s Third Law does ore than just pair up forces into action/reaction pairs. It also states that the two forces are equal in agnitude and opposite in direction. Thus, if P on M and P M on are an action/reaction pair, the two are related by: P on M = P M on exaple: If your weight is 150 pounds, this eans the earth is pulling you down with a gravitational force of 150 pounds. At the very sae tie, your body is pulling up on the earth with a gravitational force of the sae agnitude: 150 pounds! Who knew! SMU PHYS1100.1, fall 2008, Prof. Clarke 11

12 Clicker question 8.1 Solve a high-schooler s conundru: If I try to push a box across the floor, doesn t its reaction force on e cancel y action force on it? If so, why a I able to push boxes across the floor? a) the box doesn t push back with quite the sae force, and so the difference in forces allows it to ove. b) The friction force on y feet is greater than the friction force on the botto of the box, so it oves. c) The two forces act on different bodies (e and the box), and thus don t even get the chance to cancel out. d) Newton s 3 rd Law doesn t apply in this case. SMU PHYS1100.1, fall 2008, Prof. Clarke 12

13 Clicker question 8.1 Solve a high-schooler s conundru: If I try to push a box across the floor, doesn t its reaction force on e cancel y action force on it? If so, why a I able to push boxes across the floor? Buffalo uffins! a) the box doesn t push back with quite the sae force, and so the difference in forces allows it to ove. b) The friction force on y feet is greater than the friction force on the botto of the box, so it oves. ay be correct, but doesn t answer the question c) The two forces act on different bodies (e and the box), and thus don t even get the chance to cancel out. d) Newton s 3 rd Law doesn t apply in this case. Oh yes it does! SMU PHYS1100.1, fall 2008, Prof. Clarke 13

14 Clicker question 8.2 Car B is stopped for a red light. Car A, whose ass is greater than Car B, doesn t see the red light and runs into the back of Car B. Which of the following stateents is true? v a) B exerts a force on A, but A doesn t exert a force on B. b) B exerts a larger force on A than A exerts on B. c) B exerts the sae aount of force on A as A exerts on B. d) A exerts a larger force on B than B exerts on A. e) A exerts a force on B but B doesn t exert a force on A. SMU PHYS1100.1, fall 2008, Prof. Clarke 14

15 Clicker question 8.2 Car B is stopped for a red light. Car A, whose ass is greater than Car B, doesn t see the red light and runs into the back of Car B. Which of the following stateents is true? v a) B exerts a force on A, but A doesn t exert a force on B. b) B exerts a larger force on A than A exerts on B. c) B exerts the sae aount of force on A as A exerts on B. d) A exerts a larger force on B than B exerts on A. e) A exerts a force on B but B doesn t exert a force on A. SMU PHYS1100.1, fall 2008, Prof. Clarke 15

16 Clicker question 8.3 Consider yourself sitting in your chair. If the action force is your weight (i.e., the Earth s pull on you), what is the reaction force? a) the noral force you exert against the chair (pointing down); b) the noral force the chair exerts against you (pointing up); c) your gravitational pull on the Earth; d) the chair pushing against the floor. SMU PHYS1100.1, fall 2008, Prof. Clarke 16

17 Clicker question 8.3 Consider yourself sitting in your chair. If the action force is your weight (i.e., the Earth s pull on you), what is the reaction force? a) the noral force you exert against the chair (pointing down); b) the noral force the chair exerts against you (pointing up); c) your gravitational pull on the Earth; d) the chair pushing against the floor. SMU PHYS1100.1, fall 2008, Prof. Clarke 17

18 Challenge exaple. a) In the syste shown, ass is pushed to the right with a force F X. Given the asses and M and the coefficient of kinetic friction between the table and M (µ k,tm ), what conditions ust exist on µ k,m (coefficient of kinetic friction between M and ) and F X if is to slide across M and M is to slide across the table? y f k,m on n M on w E on a x F X on y f k,t on M notes 8.1 n T on M M a M x F X on M µ k,m µ k,tm T n on M f k, on M w E on M E SMU PHYS1100.1, fall 2008, Prof. Clarke 18

19 y f k,m on n M on w E on a x F X on FBD for Two forces internal to the syste + M: n M on (noral force exerted by M on ) f k,m on (kinetic friction exerted by M on ) The action/reaction counterparts to these forces ust appear on the FBD for M. They do not appear here!!! All other forces are external to the syste + M their action/ reaction counterparts do not appear on either FBD. x/ F X f k,m = a y/ n M g = 0 n M = g 1 f k,m = µ k,m n M = µ k,m g 2 a = F X µ k,m g F a = X µ k,m g 3 SMU PHYS1100.1, fall 2008, Prof. Clarke 19

20 f k,t on M y n on M n T on M M a M f k, on M w E on M x FBD for M In the equations, I drop the on and on M part of the notation to ake the less awkward to write. They were needed to help us put the forces on the correct FBD but serve no purpose in the equations. Newton s 2 nd Law x/ f k, f k,t = Ma M y/ n T n Mg = 0 n T = n + Mg Newton s 3 rd Law n = n M = g (fro 1 ) f k, = f k,m = µ k,m g (fro 2 ) f k,t = µ k,tm n T = µ k,tm (n + Mg) = µ k,tm ( + M)g f Thus, a M = k, f k,t a M = g µ k,m µ [ k,tm ( + 1)] 4 M M M SMU PHYS1100.1, fall 2008, Prof. Clarke 20

21 We can now answer the question asked: What are the constraints on F X and µ k,m so that slips on M and M slips on the table, i.e., for a > a M > 0? Fro 4 : a M = g [ µ k,m µ ( k,tm + 1 )] > 0 M M µ k,m > µ ( k,tm + 1 ) µ k,m > µ ( M k,tm 1 + ) 5 M M Next, using 3 and 4 : a > a M F X µ k,m g > g [ µ k,m µ ( k,tm + 1 )] M M Solving for F X, we get: F X > g (µ k,m µ k,tm )( + 1 ) 6 M SMU PHYS1100.1, fall 2008, Prof. Clarke 21

22 Challenge exaple. b) Let = 2.0 kg, M = 3.0 kg, F X = 20 N, µ k,tm = 0.2, and µ k,m = 0.8. Verify that both asses will slip over their respective surfaces. fro 5 : µ ( M k,tm 1 + ) 3 = (0.2)(1+ ) = 0.5 < 0.8 = µ 2 k,m fro 6 : g (µ k,m µ k,tm )( + 1 ) 2 = (2.0)(9.8)( )( +1) M 3 = 19.6 N < 20 N = F X c) If M has width W = 0.45 and starts directly over the centre of M, how uch tie passes before is pushed off M? Fro 3, a = 2.16 s 2 and fro 4, a M = 1.96 s 2. Thus, 1 2 s = a t 2 ; s M = a M t 2 s s M = = (a a M ) t 2 t 2 = W/(a a M ) = 0.45/(0.2) = t = 1.5 s SMU PHYS1100.1, fall 2008, Prof. Clarke 22 W 2 1 2

23 Acceleration Constraints If objects are connected together as they ove, their accelerations will be related to each other: perhaps but not necessarily equal. The relationships aong the accelerations are called acceleration constraints. So long as the rope is under tension, the accelerations of the truck and car will be equal. a C = a T So long as the rope doesn t stretch, the agnitudes of the accelerations of blocks A and B are equal, though their directions are different. a A = a B SMU PHYS1100.1, fall 2008, Prof. Clarke 23

24 Note: In the iddle of page 216 where the text discusses this exaple, it gives the acceleration constraint as: a Ax = a By which relates the x-coponent of a A with the y- coponent of a B : The constraint on the previous slide (a A = a B ) relates the agnitudes of the accelerations, and thus no negative sign. SMU PHYS1100.1, fall 2008, Prof. Clarke 24

25 Acceleration constraints can often provide the issing piece of inforation needed to solve a proble, but soeties they can be tricky. Consider the proble below. a 1 s 1 s 2 a 2 SMU PHYS1100.1, fall 2008, Prof. Clarke 25

26 Acceleration constraints can often provide the issing piece of inforation needed to solve a proble, but soeties they can be tricky. Consider the proble below. In tie t, the length of rope that disappears in front of 1, s 1, ust be accounted for by the two extra lengths of rope, s 2, that appear above 2. a 1 s 1 s 1 = 2 s 2 s 2 but s 1 = ½ a 1 t 2 ; s 2 = ½ a 2 t 2 a 1 = 2a 2. and this is the acceleration constraint. a 2 SMU PHYS1100.1, fall 2008, Prof. Clarke 26

27 Clicker question 8.4 Rope 1 is fixed to a wall at one end and pulled at the other end with a force of 100N (top). Rope 2 is pulled at both ends by a force of 100 N each (botto). Which stateent is correct? a) The tension in rope 2 is the sae as the tension in rope 1. b) The tension in rope 2 is twice that of the tension in rope 1. c) The tension in rope 2 is less than the tension in rope 1. SMU PHYS1100.1, fall 2008, Prof. Clarke 27

28 Clicker question 8.4 Rope 1 is fixed to a wall at one end and pulled at the other end with a force of 100N (top). Rope 2 is pulled at both ends by a force of 100 N each (botto). Which stateent is correct? a) The tension in rope 2 is the sae as the tension in rope 1. b) The tension in rope 2 is twice that of the tension in rope 1. c) The tension in rope 2 is less than the tension in rope 1. SMU PHYS1100.1, fall 2008, Prof. Clarke 28

29 Explanation to Clicker question 8.4 : Break up the rope into three bits: left (L), iddle (M), and right (R) (or as any bits as you like). Newton s 3 rd Law identifies two action/reaction pairs: F M on L = F L on M and F R on M = F M on R. Nothing oves. Thus Newton s 2 nd Law requires that F S on L = F M on L, F L on M = F R on M, and F M on R = F W on R. Thus 100 N = F S on L = F M on L = F L on M = F R on M = F M on R = F W on R, and this is equivalent to the situation where two people are pulling. 100 N F S on L L F M on L F L on M M F R on M F M on R R F W on R Sa action/reaction pairs Wall SMU PHYS1100.1, fall 2008, Prof. Clarke 29

30 8.4 Ropes and Pulleys 1. A stationary rope transits an action/reaction pair so that the pull at one end of the rope is equal in agnitude and opposite in direction to the pull at the other end. If a rope of ass is used to accelerate an object of ass M, and is not negligible, then T 1 T 2 = a, and T 1 > T 2. T T Massless string approxiation: If = 0, we recover T 1 = T 2. Thus, the tension along a assless rope (string) is constant even if the rope is accelerating. M a T 2 T 2 a T 1 SMU PHYS1100.1, fall 2008, Prof. Clarke 30

31 If a ass M hangs at the end of a rope of ass, the tension at the top of the rope is greater than at the botto since the top has to support both and M, while the botto only needs to support M. T top = T bot + g; T bot = Mg T top = Mg + g a/r T top T top M If we use the assless string approxiation, = 0 and T top = T bot. Once again the tension in the string is constant. For a vertical syste, we need to assue the string is assless even if the syste is not accelerating. SMU PHYS1100.1, fall 2008, Prof. Clarke 31 Mg g a/r T bot T bot

32 2. An ideal pulley is assless and frictionless, so that its only effect on the proble is to redirect the tension in the string. The approxiations of ideal pulleys and assless strings allow us to write: T S on A = T S on B = T. We ll do this particular proble as our first exaple. SMU PHYS1100.1, fall 2008, Prof. Clarke 32

33 Exaple. Mass A (1.0 kg) is held in place on a frictionless table and is attached via a assless string that passes over an ideal pulley to ass B (0.5 kg) that dangles freely, as shown. a) When A is released, find the acceleration of the asses and the tension in the string. For ass A: For ass B: y x/ T = A a 1 x/ irrelevant y A n T on A w E on A a T S on A x y/ irrelevant y/ T B g = B a 2 Substitute 1 into 2 : A a B g = B a B Solve for a: a = g = 3.27 s -2 A + B Evaluate T fro 1 : T = (1.0)(3.27) = 3.27 N a B T S on B x w E on B SMU PHYS1100.1, fall 2008, Prof. Clarke 33

34 b) What was the tension in the string when ass A was being held in place? B y T S on B x w E on B Look at ass B again, but this tie with a = 0: x/ irrelevant y/ T B g = 0 T = (0.5)(9.8) = 4.9 N The tension in the string is less when the asses are accelerating than when they are being held still. SMU PHYS1100.1, fall 2008, Prof. Clarke 34

35 Clicker question 8.5 Blocks A and B are pulled with a force F across a frictionless table by assless strings. Let the tension in string 1 be T 1 and the tension in string 2 be T 2. Which of the following is true? a) T 2 < T 1 < F b) T 2 = T 1 = F c) T 2 > T 1 > F d) T 2 < T 1 = F e) We need to know about the asses before we can tell. F SMU PHYS1100.1, fall 2008, Prof. Clarke 35

36 Clicker question 8.5 Blocks A and B are pulled with a force F across a frictionless table by assless strings. Let the tension in string 1 be T 1 and the tension in string 2 be T 2. Which of the following is true? a) T 2 < T 1 < F b) T 2 = T 1 = F c) T 2 > T 1 > F d) T 2 < T 1 = F e) We need to know about the asses before we can tell. F SMU PHYS1100.1, fall 2008, Prof. Clarke 36

37 Clicker question 8.6 In the diagra, consider the tension in the (assless) string when A is held in place (T held ) and the tension in the string when A is let go and B falls (T fall ). Which of the following stateents is true? a) T held > T fall b) T held = T fall c) T held < T fall SMU PHYS1100.1, fall 2008, Prof. Clarke 37

38 Clicker question 8.6 In the diagra, consider the tension in the (assless) string when A is held in place (T held ) and the tension in the string when A is let go and B falls (T fall ). Which of the following stateents is true? a) T held > T fall b) T held = T fall c) T held < T fall SMU PHYS1100.1, fall 2008, Prof. Clarke 38

39 Clicker question 8.7 In the syste shown, the string is assless and all pulleys are ideal. T 1 is the tension in the string between 1 and the pulley on the table. T 2 is the tension in the string between the pulley on the table and 2. T 3 is the tension in the string between the ceiling and 2. Which tension is the greatest? a) T 1 b) T 2 c) T 3 d) They are all the sae. T 1 T 2 T 3 SMU PHYS1100.1, fall 2008, Prof. Clarke 39

40 Clicker question 8.7 In the syste shown, the string is assless and all pulleys are ideal. T 1 is the tension in the string between 1 and the pulley on the table. T 2 is the tension in the string between the pulley on the table and 2. T 3 is the tension in the string between the ceiling and 2. Which tension is the greatest? a) T 1 b) T 2 c) T 3 d) They are all the sae. T 1 T 2 T 3 ended here, 21/10/08 SMU PHYS1100.1, fall 2008, Prof. Clarke 40

41 Exaple: Find the tension in the string and accelerations in ters of 1, 2, and g. Assue a assless string, ideal pulleys, and a frictionless table. a 1 y n T on 1 a 1 1 : x/ T = 1 a 1 y/ irrelevant 1 T x a 2 1 g SMU PHYS1100.1, fall 2008, Prof. Clarke 41

42 Exaple: Find the tension in the string and accelerations in ters of 1, 2, and g. Assue a assless string, ideal pulleys, and a frictionless table. a 1 y n T on 1 a 1 1 : x/ T = 1 a 1 y/ irrelevant 2 : x/ no forces 1 T x y/ 2T 2 g = 2 a 2 a 2 1 g y T T 2 x a 2 2 g SMU PHYS1100.1, fall 2008, Prof. Clarke 42

43 Exaple: Find the tension in the string and accelerations in ters of 1, 2, and g. Assue a assless string, ideal pulleys, and a frictionless table. y n T on 1 a 1 1 : x/ T = 1 a 1 y/ irrelevant a 1 2 : x/ no forces 1 T 1 g x 2 1 a 1 2 g = 2 a 2 y/ 2T 2 g = 2 a 2 acceleration constraint fro slide 26: a 1 = 2a 2 y a a 2 2 g = 2 a 2 a 2 ( ) = 2 g T T a 2 = 2 g 2 a 1 = 2 g T = g a 2 2 x 2 g SMU PHYS1100.1, fall 2008, Prof. Clarke 43

44 Exaple: Atwood s achine a) If 1 = 1.0 kg and 2 = 1.5 kg, with what iniu force F ust the ideal pulley be pulled so that both asses lift off the ground? b) What is the acceleration of the pulley? F P T 1 1 g F a 1 T 1 g = 1 a 1 T 2 g = 0 T = 2 g T a 1 = g = 2 1 g 1 ( ) 1 F 2T = 0 (assless pulley) F = 2 2 g = 29.4 N 1 s 1 P a 1 P acceleration constraint: s P = s 1 1 a P = a 1 = 1 = 2.45 s 2 ( ) g 2 2 T T 2 SMU PHYS1100.1, fall 2008, Prof. Clarke 44 T 2 2 g a 2 = 0 n G on 2 = 0 s P 1 2 F P

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