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Physics 151: Lecture 10

Physics 151: Lecture 10. Homework #3 (9/22/06, 5 PM) from Chapter 5 Today’s Topics: Example with a pulley and kinetic Static Friction Circular motion and Newton’s Laws - Ch 6. m 2. T 1. m 1. m 3. Example with pulley and kinetic friction Problem from the textbook.

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Physics 151: Lecture 10

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

  2. m2 T1 m1 m3 Example with pulley and kinetic frictionProblem 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 m1 = 4.00 kg, m2 = 1.00kg and m3 = 2.00kg. What is the magnitude and direction of acceleration on the three blocks ? What is the tension on the two cords ? a = 2.31 m/s2 T12 = = 30.0 N , T23 = 24.2 N

  3. See text: Ch 5.8 Static Friction... • The maximum possible force that the friction between two objects can provide is fMAX = SN, where s is the “coefficient of static friction”. • So fSS N. • As one increases F, fS gets bigger until fS=SN and the object “breaks loose” and starts to move. j N F i fS mg

  4. See text: Ch 5.8 Static Friction... • S is discovered by increasing F until the block starts to slide: i :FMAXSN = 0 j :N = mg S FMAX / mg j N FMAX i Smg mg

  5. See text: 6-1 Additional comments on Friction: • 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 > Kfor any system (think about it...).

  6. v aC R Newton’s Laws and Circular Motion(Chapter 6) • Centripetal Acceleration • aC = v2/R • What is Centripetal Force ? • FC = maC = mv2/R

  7. See Example 6.2 Example Problem 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 FBS is large versus the weight of the soap)

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

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

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

  11. An example before we considered a race car going around a curve on a flat track. N Ff mg What’s differs on a banked curve ?

  12. N ma Ff CAR mg Banked Corners Free Body Diagram for a banked curve. For small banking angles, you can assume that Ff is parallel to ma. This is equivalent to the small angle approximation sinq = tanq. Can you show that ?

  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. y x mg sin q N q mg Lecture 10, ACT 3Solution B) For icy road => no friction ! • Draw FBD and find the total force in the x-direction v = 18.5 m/s ~ 40 mi / hr r =500m Q = 4o mk =0.5 ms =0.7 M =1,000 kg q

  15. Static Friction Kinetic Friction N N F F fk fS mg mg Lecture 10, ACT 3Remember • While the block is static: a = 0, fS F fS,max ms N • The object is moving : a = 0 , fk <F fk  mkN dry road => static friction keeps objects from moving

  16. y x N q mg sin q mg Lecture 10, ACT 3Solution A) dry road => static friction keeps objects from moving • Draw FBD and find the total force in the x-direction v = 61.3 m/s ~ 120 mi / hr ! msN r =500m Q = 4o mk =0.5 ms =0.7 M =1,000 kg

  17. Lecture 10, ACT 4 • 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. NII mg Fc Fc 2.7 km Fc NI mg Lecture 10, ACT 4Solution II) ANSWER (C) A) Larger B) Same C) Smaller ANSWER (A) I)

  19. ExampleGravity, Normal Forces etc. Consider a women on a swing: Animation When is the tension on the rope largest ? 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)

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

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