1. Physics 151: Lecture 10, Pg 1 Physics 151: Lecture 10 l Homework #3 (9/22/06, 5 PM) from Chapter 5 Today’s Topics: çExample with a pulley and kinetic çStatic Friction çCircular motion and Newton’s Laws - Ch 6

2. Physics 151: Lecture 10, Pg 2 Example with pulley and kinetic friction Problem from the textbook Three blocks are connected on the table as shown. The table has a coefficient of kinetic friction of 0.350, the masses are m 1 = 4.00 kg, m 2 = 1.00kg and m 3 = 2.00kg. a) What is the magnitude and direction of acceleration on the three blocks ? b) What is the tension on the two cords ? m1m1 T1T1 m2m2 m3m3 T 12 = = 30.0 N, T 23 = 24.2 N a = 2.31 m/s 2

3. Physics 151: Lecture 10, Pg 3 Static Friction... F gmggmg N i j fSfS The maximum possible force that the friction between two objects can provide is f MAX =  S N, where  s is the “coefficient of static friction”.  So f S   S N.  As one increases F, f S gets bigger until f S =  S N and the object “breaks loose” and starts to move. See text: Ch 5.8

4. Physics 151: Lecture 10, Pg 4 Static Friction... F  S is discovered by increasing F until the block starts to slide: i :F MAX  S N = 0 j :N = mg  S  F MAX / mg F F MAX gmggmg N i j  S mg See text: Ch 5.8

5. Physics 151: Lecture 10, Pg 5 Additional comments on Friction: See text: 6-1 Since f =  N, the force of friction does not depend on the area of the surfaces in contact. By definition, it must be true that  S >  K for any system (think about it...).

6. Physics 151: Lecture 10, Pg 6 Newton’s Laws and Circular Motion (Chapter 6) Centripetal Acceleration a C = v 2 /R What is Centripetal Force ? F C = ma C = mv 2 /R v R aCaC

7. Physics 151: Lecture 10, Pg 7 Example Problem See Example 6.2 I am feeling very energized while I shower. So I swing a soap on a rope around in a horizontal circle over my head. Eventually the soap on a rope breaks, the soap scatters about the shower and I slip and fall after stepping on the soap. To decide whether to sue ACME SOAP I think about how fast I was swinging the soap (frequency) and if the rope should have survived. From the manufacturers web site I find a few details such as the mass of the soap is 0.1 kg (before use), the length of the rope is 0.1 m and the rope will break with a force of 40 N. (assume F BS is large versus the weight of the soap)

8. Physics 151: Lecture 10, Pg 8 Example Problem See Example 6.2 Step 2 : Diagram. v T Step 1: we need to find the frequency of the soap’s motion that caused the rope to break. Step 3 – Solve Symbolically Step 4 – Numbers Step 5 –It seems that the suit is in trouble. Being able to twirl your soap safely 10 rev/s is pretty good. f~ 10 rev./s

9. Physics 151: Lecture 10, Pg 9 Lecture 10, ACT 2 Circular Motion Forces l How fast can the race car go ? (How fast can it round a corner with this radius of curvature ?) m car = 1500 kg  S = 0.5 for tire/road R = 80 m R A) 10 m/s B) 20 m/s C) 75 m/s D) 750 m/s

10. Physics 151: Lecture 10, Pg 10 Lecture 10, ACT 2 Circular Motion Forces l This is just like the soap on a rope problem but friction replaces the tension in the rope as the centripedal force. N mgmg F?F? Answer is (B) m car = 1500 kg  S = 0.5 for tire/road R = 80 m A) 10 m/s B) 20 m/s C) 75 m/s D) 750 m/s

11. Physics 151: Lecture 10, Pg 11 An example before we considered a race car going around a curve on a flat track. N mgmg FfFf What’s differs on a banked curve ?

12. Physics 151: Lecture 10, Pg 12 Banked Corners Free Body Diagram for a banked curve. N mgmg FfFf For small banking angles, you can assume that F f is parallel to ma. This is equivalent to the small angle approximation sin  = tan . Can you show that ? CAR mama

13. Physics 151: Lecture 10, Pg 13 Lecture 10, ACT 3 Because of your physics background, you have been hired as a member of the team the state highway department has assigned to review the safety of Connecticut roads. This week you are studying US- 6, trying to redesign the section from Willimantic to I-384 so fewer folks die on it. The proposed new road has a curve that is essentially 1/8 of a circle with a radius of 0.3 miles (1/2 km). The road has been designed with a banked curve so that the road makes an angle of 4° to the horizontal throughout the curve. To begin the study, the head of your department asks that you calculate the maximum speed (in km/hr) for a standard passenger car (a little more than 2000 lbs or about 1,000 kg ) to complete the turn while maintaining without sliding off the road. She asks that you first consider the case of a slick, ice and slush covered road. When you have completed that calculation she wants you to do the case of a dry, clear road where the coefficient of kinetic friction is 0.50 and the coefficient of static friction is 0.70 between the tires and the road. This will give her team the two extremes of Connecticut driving conditions on which to base the analysis. What is the maximum speed you can drive through the curve for A) dry road condition, and for B) icy road condition ?

14. Physics 151: Lecture 10, Pg 14 Lecture 10, ACT 3 Solution l Draw FBD and find the total force in the x-direction mg N   v = 18.5 m/s ~ 40 mi / hr xy mg sin  B) For icy road => no friction ! r =500m  = 4 o  k =0.5  s =0.7 M =1,000 kg

15. Physics 151: Lecture 10, Pg 15 Static Friction Kinetic Friction Lecture 10, ACT 3 Remember dry road => static friction keeps objects from moving gmggmg N F fSfS l The object is moving : a = 0, f k  F f k  k  N N F gmggmg fkfk l While the block is static: a = 0, f S  F f S,max   s N

16. Physics 151: Lecture 10, Pg 16 Lecture 10, ACT 3 Solution l Draw FBD and find the total force in the x-direction mg N  sNsN v = 61.3 m/s ~ 120 mi / hr ! A) dry road => static friction keeps objects from moving xy mg sin  r =500m  = 4 o  k =0.5  s =0.7 M =1,000 kg

17. Physics 151: Lecture 10, Pg 17 Lecture 10, ACT 4 l When a pilot executes a loop- the-loop (as in figure on the right) the aircraft moves in a vertical circle of radius R=2.70 km at a constant speed of v=225 m/s. Is the force exerted by the seat on the pilot: A) Larger B) Same C) Smaller then pilot’s weight (mg) at : I) the bottom and II) at the top of the loop. 2.7 km

18. Physics 151: Lecture 10, Pg 18 Lecture 10, ACT 4 Solution 2.7 km FcFc FcFc I) II) mg ANSWER (C) NINI N II ANSWER (A) FcFc A) Larger B) Same C) Smaller

19. Physics 151: Lecture 10, Pg 19 Example Gravity, Normal Forces etc. Consider a women on a swing: 1.When is the tension on the rope largest ? 2. Is it :A) greater than B) the same as C) less than the force due to gravity acting on the woman (neglect the weight of the swing) Animation

20. Physics 151: Lecture 10, Pg 20 Recap of today’s lecture l Example with pulley l Forces and Circular Motion - text Ch 6.1 l Reading for next class: Section 6.2 and 6.3 l Homework #3 (due 9/22/06, 5 PM) from Chapter 5 l Homework #4 (due 10/02/06, 5 PM) from Chapter 6