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5. Newton's Laws Applications

5. Newton's Laws Applications. Using Newton’s 2 nd Law Multiple Objects Circular Motion Friction Drag Forces. Why doesn’t the roller coaster fall its loop-the loop track?. Ans. The downward net force is just enough to make it move in a circular path. 5.1. Using Newton’s 2 nd Law.

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5. Newton's Laws Applications

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  1. 5. Newton's Laws Applications Using Newton’s 2nd Law Multiple Objects Circular Motion Friction Drag Forces

  2. Why doesn’t the roller coaster fall its loop-the loop track? Ans. The downward net force is just enough to make it move in a circular path.

  3. 5.1. Using Newton’s 2nd Law • Example 5.1. Skiing • A skier of mass m = 65 kg glides down a frictionless slope of angle  = 32. Find • The skier’s acceleration • The force the snow exerts on him.  y x : n a y :  x  Fg

  4. Example 5.2. Bear Precautions Mass of pack in figure is 17 kg. What is the tension on each rope? since x :  y y : T2 T1   x Fg

  5. Example 5.3. Restraining a Ski Racer A starting gate acts horizontally to restrain a 60 kg ski racer on a frictionless 30 slope. What horizontal force does the gate apply to the skier? since y n x :   y :  x Fh Fg

  6. Alternative Approach Net force along slope (x-direction) : y n   Fh x Fg

  7. GOT IT? 5.1. • A roofer’s toolbox rests on a frictionless 45 ° roof, • secured by a horizontal rope. • Is the rope tension • greater than, • less than, or • equal to • the box’s weight? n x :   T Smaller   smaller T Fg x

  8. 5.2. Multiple Objects • Example 5.4. Rescuing a Climber • A 70 kg climber dangles over the edge of a frictionless ice cliff. • He’s roped to a 940 kg rock 51 m from the edge. • What’s his acceleration? • How much time does he have before the rock goes over the edge? • Neglect mass of the rope. 

  9.  Tension T = 1N throughout

  10. GOT IT? 5.1. • What are • the rope tension and • the force exerted by the hook on the rope? 1N 1N

  11. 5.3. Circular Motion Uniform circular motion centripetal 2nd law:

  12. Example 5.5. Whirling a Ball on a String Mass of ball is m. String is massless. Find the ball’s speed & the string tension. x : y : y  T  a x Fg

  13. Example 5.6. Engineering a Road At what angle should a road with 200 m curve radius be banked for travel at 90 km/h (25 m/s)? y x : y : n   x a Fg

  14. Example 5.7. Looping the Loop Radius at top is 6.3 m. What’s the minimum speed for a roller-coaster car to stay on track there? Minimum speed  n = 0

  15. Conceptual Example 5.1. Bad Hair Day What’s wrong with this cartoon showing riders of a loop-the-loop roller coaster? From Eg. 5.7: n + mg = m a = m v2 / r Consider hair as mass point connected to head by massless string. Then T + mg = m a where T is tension on string. Thus, T = n. Since n is downward, so is T. This means hair points upward ( opposite to that shown in cartoon ).

  16. 5.4. Friction Some 20% of fuel is used to overcome friction inside an engine. The Nature of Friction

  17. Frictional Forces • Pushing a trunk: • Nothing happens unless force is great enough. • Force can be reduced once trunk is going. Static friction s= coefficient of static friction Kinetic friction k= coefficient of kinetic friction k: < 0.01 (smooth), > 1.5 (rough) Rubber on dry concrete : k= 0.8, s= 1.0 Waxed ski on dry snow: k= 0.04 Body-joint fluid: k= 0.003

  18. Application of Friction Walking & driving require static friction. No slippage: Contact point is momentarily at rest  static friction at work foot pushes ground ground pushes you

  19. Example 5.8. Stopping a Car • k& sof a tire on dry road are 0.61 & 0.89, respectively. • If the car is travelling at 90 km/h (25 m/s), • determine the minimum stopping distance. • the stopping distance with the wheels fully locked (car skidding). (a)  = s :  (b)  = k :

  20. Application: Antilock Braking Systems (ABS) Skidding wheel: kinetic friction Rolling wheel: static friction

  21. Example 5.9. Steering A level road makes a 90 turn with radius 73 m. What’s the maximum speed for a car to negotiate this turn when the road is (a) dry ( s = 0.88 ). (b) covered with snow ( s = 0.21 ). (a) (b)

  22. Example 5.10. Avalanche! Storm dumps new snow on ski slope. s between new & old snow is 0.46. What’s the maximum slope angle to which the new snow can adhere? y n x : y : fs   x Fg

  23. Example 5.11. Dragging a Trunk Mass of trunk is m. Rope is massless. Kinetic friction coefficient is k. What rope tension is required to move trunk at constant speed? y y : x : n T fs  x Fg

  24. GOT IT? 5.4 • Is the frictional force • less than, (b) equal to , or (c) greater than • the weight multiplied by the coefficient of friction? Reason: Chain is pulling downward, thus increasing n.

  25. 5.5. Drag Forces Drag force: frictional force on moving objects in fluid. Depends on fluid density, object’s cross section area, & speed. Terminal speed: max speed of free falling object in fluid. Parachute: vT ~ 5 m/s. Ping-pong ball: vT ~ 10 m/s. Golf ball: vT ~ 50 m/s. Sky-diver varies falling speed by changing his cross-section. Drag & Projectile Motion

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