2. Physics 151: Lecture 8, Pg 1 Physics 151: Lecture 8 l Reaminder: çHomework #3 : (Problems from Chapter 5) due Fri. (Sept. 22) by 5.00 PM l Today’s Topics : çReview of Newton’s Laws 1 and 2 - Ch. 5.1-4 çNewton’s third law: action and reaction - Ch. 5.6

3. Physics 151: Lecture 8, Pg 2 Review Newton’s Laws 1 and 2 l Isaac Newton (1643 - 1727) published Principia Mathematica in 1687. In this work, he proposed three “laws” of motion: Law 1: An object subject to no external forces is at rest or moves with a constant velocity if viewed from an inertial reference frame. FFa Law 2: For any object, F NET =  F = ma See text: 5.1-4

4. Physics 151: Lecture 8, Pg 3 Newton’s Third Law: If object 1 exerts a force on object 2 (F 2,1 ) then object 2 exerts an equal and opposite force on object 1 (F 1,2 ) F 1,2 = -F 2,1 See text: 5-6 This is among the most abused concepts in physics. REMEMBER: Newton’s 3rd law concerns force pairs which act on two different objects (not on the same object) ! For every “action” there is an equal and opposite “reaction”

5. Physics 151: Lecture 8, Pg 4 An Example F B,E = - m B g EARTH F E,B = m B g Consider the forces on an object undergoing projectile motion F B,E = - m B g F E,B = m B g

6. Physics 151: Lecture 8, Pg 5 Normal Forces Certain forces act to keep an object in place. These have what ever force needed to balance all others (until a breaking point). F T,B F B,T

7. Physics 151: Lecture 8, Pg 6 Force Pairs Newton’s 3rd law concerns force pairs. Two members of a force pair cannot act on the same object. Don’t confuse gravity (the force of the earth on an object) and normal forces. It’s an extra part of the problem. F T,B F B,T F B,E = -mg F E,B = mg

8. Physics 151: Lecture 8, Pg 7 An Example Consider the following two cases

9. Physics 151: Lecture 8, Pg 8 An Example The Free Body Diagrams mgmgmgmg F B,T = N Ball Falls For Static Situation N = mg

10. Physics 151: Lecture 8, Pg 9 An Example The action/reaction pair forces F B,E = -mg F B,T = N F E,B = mg F B,E = -mg F E,B = mg F T,B = -N

11. Physics 151: Lecture 8, Pg 10 Lecture 8, Act 1 Newton’s 3rd Law l Two blocks are being pushed by a finger on a horizontal frictionless floor. How many action-reaction pairs of forces are present in this system? (a) 2 (b) 4(c) 6 a b

12. Physics 151: Lecture 8, Pg 11 2 4 6 Lecture 8, Act 1 Solution: a b F F a,f F F f,a F F b,a F F a,b F g,a F a,g F g,b F b,g F E,a F a,E F E,b F b,E Is F a,f = F b,a ? (A) YES (B) NO F b,a = F a,f [m b /(m b +m a )]

13. Physics 151: Lecture 8, Pg 12 l You are going to pull two blocks (m A =4 kg and m B =6 kg) at constant acceleration (a= 2.5 m/s 2 ) on a horizontal frictionless floor, as shown below. The rope connecting the two blocks can stand tension of only 9.0 N. Would the rope break ? l (A) YES(B) CAN’T TELL(C) NO Lecture 8, Act 2 A B a= 2.5 m/s 2 rope

14. Physics 151: Lecture 8, Pg 13 l What are the relevant forces ? Lecture 8, Act 2 Solution: mAmA mBmB F app a= 2.5 m/s 2 mAmA rope mBmB T -T T F app = a (m A + m B ) F app = 2.5m/s 2 ( 4kg+6kg) = 25 N total mass ! a A = a = 2.5 m/s 2 T = a m A T = 2.5m/s 2 4kg = 10 N T > 9 N, rope will brake ANSWER (A) F app - T = a m B T = 25N - 2.5m/s 2 6kg=10N T > 9 N, rope will brake a = 2.5 m/s 2 F app a = 2.5 m/s 2 THE SAME ANSWER -> (A)

15. Physics 151: Lecture 8, Pg 14 Exercise: Inclined plane A block of mass m slides down a frictionless ramp that makes angle  with respect to horizontal. What is its acceleration a ?  m a See text: Example 5.7

16. Physics 151: Lecture 8, Pg 15 Inclined plane... l Define convenient axes parallel and perpendicular to plane: ç Acceleration a is in x direction only.  m a i j See text: Example 5.7

17. Physics 151: Lecture 8, Pg 16 Inclined plane... l Consider x and y components separately: i i: mg sin  = ma a = g sin  j j: N - mg cos . N = mg cos  gmggmg N mg sin  mg cos   amaama i j See text: Example 5.7 m 

18. Physics 151: Lecture 8, Pg 17 Angles of an Inclined plane  a = g sin   mg N See text: Example 5.7 m

19. Physics 151: Lecture 8, Pg 18 Free Body Diagram A heavy sign is hung between two poles by a rope at each corner extending to the poles. Eat at Bob’s What are the forces on the sign ?

20. Physics 151: Lecture 8, Pg 19 Free Body Diagram Eat at Bob’s   T1T1 mgmg T2T2 Add vectors : x y T1T1  T2T2  mgmg Vertical (y): mg = T 1 sin  1 + T 2 sin  2 Horizontal (x) : T 1 cos  1 = T 2 cos  2

21. Physics 151: Lecture 8, Pg 20 Example-1 with pulley l Two masses M 1 and M 2 are connected by a rope over the pulley as shown. çAssume the pulley is massless and frictionless. çAssume the rope massless. l If M 1 > M 2 find : çAcceleration of M 1 ? çAcceleration of M 2 ? çTension on the rope ? Free-body diagram for each object M1M1 T2T2 T1T1 M2M2 a Animation Video

22. Physics 151: Lecture 8, Pg 21 Example-2 with pulley l A mass M is held in place by a force F. Find the tension in each segment of the rope and the magnitude of F. çAssume the pulleys massless and frictionless. çAssume the rope massless.  M T5T5 T4T4 T3T3 T2T2 T1T1 F l We use the 5 step method. çDraw a picture: what are we looking for ? çWhat physics idea are applicable ? Draw a diagram and list known and unknown variables. Newton’s 2 nd law : F= m a Free-body diagram for each object

23. Physics 151: Lecture 8, Pg 22 Pulleys: continued l FBD for all objects  M T5T5 T4T4 T3T3 T2T2 T1T1 F T4T4 F=T 1 T2T2 T3T3 T2T2 T3T3 T5T5 M T5T5 MgMg

24. Physics 151: Lecture 8, Pg 23 Pulleys: finally l Step 3: Plan the solution (what are the relevant equations)  F= m a, static (no acceleration: mass is held in place) M T5T5 MgMg T 5 =Mg T2T2 T3T3 T5T5 T 2 +T 3 =T 5 T4T4 F=T 1 T2T2 T3T3 T 1 +T 2 +T 3 =T 4

25. Physics 151: Lecture 8, Pg 24 Pulleys: really finally! l Step 4: execute the plan (solve in terms of variables) çWe have (from FBD): T 5 =MgF=T 1 T 2 +T 3 =T 5 T 1 +T 2 +T 3 =T 4  M T5T5 T4T4 T3T3 T2T2 T1T1 F T 2 =T 3 T 1 =T 3 T 2 =Mg/2 T 2 +T 3 =T 5 gives T 5 =2T 2 =Mg F=T 1 =Mg/2 T 1 =T 2 =T 3 =Mg/2 and T 4 =3Mg/2 T 5 =Mgand çPulleys are massless and frictionless l Step 5: evaluate the answer (here, dimensions are OK and no numerical values)

26. Physics 151: Lecture 8, Pg 25 Lecture 9, ACT 1 Gravity and Normal Forces A woman in an elevator is accelerating upwards The normal force exerted by the elevator on the woman is, A) greater than B) the same as C) less than the force due to gravity acting on the woman

27. Physics 151: Lecture 8, Pg 26 Lecture 9, ACT 1 Gravity and Normal Forces The free body diagram is, For the woman to accelerate upwards, the normal force on the woman must be A) greater than the force due to gravity acting on the woman Note, both of these forces act on the woman, they cannot be an action/reaction pair N mgmg

28. Physics 151: Lecture 8, Pg 27 Lecture 9, ACT 1b Gravity and Normal Forces A woman in an elevator is accelerating upwards The normal force exerted by the elevator on the woman is, A) greater than B) the same as C) less than the force the woman exerts on the elevator.

29. Physics 151: Lecture 8, Pg 28 Lecture 9, ACT 1b Gravity and Normal Forces The action/reaction force diagram for the woman and elevator is, By Newton’s third law these must be (B) equal. N = F W,E F E,W

30. Physics 151: Lecture 8, Pg 29 Recap of today’s lecture l Newton’s 3 Laws l Free Body Diagrams l Action/Reaction Force pairs l Reading for Friday, Ch 5.7-8, pp. 123-139 Applications of Newton’s Laws and Friction çHomework #3 (due next Wed. / Sept. 21 by 11:59 pm)