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Short Version : 5. Newton's Laws Applications

Short Version : 5. Newton's Laws Applications . 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. F h.

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Short Version : 5. Newton's Laws Applications

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  1. Short Version : 5. Newton's Laws Applications

  2. 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

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

  4. 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. 

  5.  Tension T = 1N throughout

  6. 5.3. Circular Motion Uniform circular motion centripetal 2nd law:

  7. 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

  8. 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

  9. 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 ( a  g ) Consider hair as mass point connected to head by massless string. Then T  mg =  m a where T is tension on string. Thus, T = m ( g  a )  0 . ( downward ) This means hair points upward ( opposite to that shown in cartoon).

  10. 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

  11. 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

  12. Rolling wheel:

  13. Skidding wheel 滑動的輪子 : kinetic friction 動摩擦 k 0.8 Rolling wheel 滾動的輪子 : static friction 靜摩擦 s 1 Rolling friction滾動摩擦 r 0.01

  14. Dynamics of Wheels F fs fr

  15. 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 :

  16. Steering Car turning to the left. Bicycle turning to the left. More details

  17. 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)

  18. 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

  19. Simple Machines

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