Physics 1501: Lecture 7 Today ’ s Agenda

1. Physics 1501: Lecture 7Today’s Agenda • Announcements: • Physics Learning Center P201: • Schedule posted on the door • Homeworks • Lowest HW will be dropped • HW 02: due next Monday • Today’s topics: • Newton’s laws • Applications of Newton’s laws

2. Dynamics

3. ReviewNewton’s Laws 1 and 2 • 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. Law 2: For any object, FNET = F = ma Force and mass

4. Lecture 7,ACT 1Newton’s Second Law A constant force is exerted on a cart that is initially at rest on an air table. The force acts for a short period of time and gives the cart a certain final speed. Force Cart AirTrack For a second shot, I can apply a force only half as large (I’m getting tired). To reach the same final speed, for how long must I apply the force ? A) 4 x as long B) 2 x as long C) Same time D) 1/2 as long E) 1/4 x as long

5. Lecture 7,ACT 2Newton’s Second Law I push with a force of 2 Newtons on a cart that is initially at rest on an air table with no air. I push for a second. Because there is no air, the cart stops after I finish pushing. It has traveled a certain distance. Force Cart AirTrack For a second shot, I push just as hard but keep pushing for 2 seconds. The distance the cart moves the second time versus the first is, A) 4 x as long B) 2 x as long C) Same D) 1/2 as long E) 1/4 x as long

6. Lecture 7, ACT 3Newton’s Second Law A constant force is exerted on a cart that is initially at rest on an air table. This force is applied for a short period of time and the cart acquires a certain final speed, which I call vf1. Force Cart AirTrack I repeat the experiment, but this time the cart is already moving with constant speed when I start applying the force. After exerting the same constant force for the same time interval, the cart’s final speed is, A) vf1 B) 2vf1C) vf12 D) cannot be determined from the information given.

7. Newton’s Third Law: If object 1 exerts a force on object 2 (F2,1 ) then object 2 exerts an equal and opposite force on object 1 (F1,2) F1,2 = -F2,1 For every “action” there is an equal and opposite “reaction” 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) !

8. FB,E = - mB g FB,E = - mB g FE,B = mB g FE,B = mB g An Example Consider the forces on an object undergoing projectile motion EARTH

9.  Lecture 7, ACT 4Newton’s Third Law A fly gets smushed onto the windshield of a speeding bus. The force exerted by the bus on the fly is, A) greater than B) the same as C) less than that exerted by the fly on the bus.

10.  Lecture 7, ACT 5Newton’s Third Law A fly gets smushed onto the windshield of a speeding bus. The acceleration due to this collision of the bus is, A) greater than B) the same as C) less than that of the fly.

11. Newton's Third Law... • FA ,B = - FB ,A an example, Fm,w Fw,m Ff,m Fm,f

12. Example of Bad Thinking • Since Fm,b= -Fb,m why isn’t Fnet= 0, and a = 0 ? Fm,b Fb,m a ?? ice

13. Fm,b Example of Good Thinking • Consider only the box as the system! • Free Body Diagram Fb,m ice

14. abox Fg FN Example of Good Thinking • Consider only the box as the system! • Free Body Diagram abox = Fb,m/mbox Fb,m

15. Eat at Bob’s Free Body Diagram A heavy sign is hung between two poles by a rope at each corner extending to the poles. What are the forces on the sign ?

16. T2 T1 T1 Eat at Bob’s mg mg T2 q2 q1 Free Body Diagram q2 q1 Add vectors

17. Eat at Bob’s Free Body Diagram T1 T2 q2 q1 mg Vertical : mg = T1sinq1 + T2sinq2 Horizontal : T1cosq1 = T2cosq2

18. FB,T FT,B Normal Forces and String Tension Certain forces act to keep an object in place. These have what ever force needed to balance all others (until a breaking point).

19. FB,E = -mg FE,B = mg 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. FB,T FT,B

20. a b Lecture 7, Act 6Newton’s 3rd Law • 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

21. An Example Consider the following two cases

22. mg FB,T= N mg An Example The Free Body Diagrams Ball Falls For Static Situation N = mg

23. FB,E = -mg FB,T= N FT,B= -N FB,E = -mg FE,B = mg FE,B = mg An Example The action/reaction pair forces

24. Lecture 7, ACT 7Gravity 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

25. Lecture 7, ACT 7bGravity 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.

26. 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 

27. j i Inclined plane... • Define convenient axes parallel and perpendicular to plane: • Acceleration ais in x direction only. m a 

28. j ma mg sin  i N mg cos  mg Inclined plane... • Consider x and y components separately: • i: mg sin = maa = g sin  • j: N - mg cos= 0. N = mg cos 

29. N mg  Angles of an Inclined plane ma = mg sin    + f = 90 f 